Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers

Big Ideas Math Answers Grade 3 Chapter 8

Are you searching for the solutions of Big Ideas Math Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers questions? If yes, then you are at the correct place. We are giving the solution and explanation for each and every question mentioned at BIM 3rd Grade 8th Chapter Add and Subtract Multi-Digit Numbers Book. Students who want to complete their homework in time can download the Big Ideas Math  Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers Solutions PDF.

Big Ideas Math Book 3rd Grade Answer Key Chapter 8 Add and Subtract Multi-Digit Numbers

BIM 3rd Grade 8th Chapter Add and Subtract Multi-Digit Numbers Answers are not only useful for the children but also for teachers and parents. If parents or teachers find any question difficult to solve, then they can refer to Big Ideas Math Book Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers Solution Key.

The topics included in the answer key are Identify Addition Properties, Use Number Lines to Addition, Use Mental Math to Add, Use Partial Sums to Add, Add Three-Digit Numbers, Add Three or More Numbers, Use Number Lines to Subtract, Use Mental Math to Subtract, Subtract Three-Digit Numbers, Relate Addition and Subtraction. The problem solving or performance task sections mentioned after these lessons are helpful to check skills.

Lesson 1 – Identify Addition Properties

Lesson 2 – Use Number Lines to Addition

Lesson 3 –  Use Mental Math to Add

Lesson 4 – Use Partial Sums to Add

Lesson 5 – Add Three-Digit Numbers

Lesson 6 – Add Three or More Numbers

Lesson 7 – Use Number Lines to Subtract

Lesson 8 – Use Mental Math to Subtract

Lesson 9 – Subtract Three-Digit Numbers

Lesson 10 – Relate Addition and Subtraction

Lesson 11 – Problem Solving: Addition and Subtraction

Performance Task

Lesson 8.1 Identify Addition Properties

Explore and Grow

Use the addition table to write all of the addition equations that have a sum of 13.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 1
What do you notice now?
Answer:
By adding a vertical number and the horizontal number we can find the sum of 13.
0 + 0 = 0
0 + 1 = 0
Add blue line and yellow line.
Add 10 in the blue column and 3 in the yellow column.
10 + 3 = 13
Thus the addition equation has the sum of 13.

Structure
Use the addition table to write all of the equations that have a sum of 12. What do you notice?

Answer:
Add blue line and yellow line.
Add 10 in the blue column and 2 in the yellow column.
10 + 2 = 12
Thus the addition equation has the sum of 12.

Think and Grow: Addition Properties
Changing the order of addends does not change the sum.
3 + 5 = 5 + 3
Associative Property of Addition
Changing the grouping of addends does not change the sum.
(7 + 6) + 4 = 7 + (6 + 4)
Addition Property of Zero
The sum of any number and 0 is that number.
9 + 0 = 9
Example
Identify the property.
56 + 0 = ___
12 + 29 = 29 + 12 _____
(24 + 17) + 23 = 24 + (17 + 23) ____

Answer:
56 + 0 = 56
It shows Addition Property of Zero. The Addition Property of Zero defines the sum of any number and 0 is that number.
12 + 29 = 29 + 12
Changing the order of addends does not change the sum.
(24 + 17) + 23 = 24 + (17 + 23)
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Show and Grow

Identify the property.

Question 1.
16 + (14 + 19) = (16 + 14) + 19

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 2.
11 + 54 = 54 + 11

Answer:
Associative Property of Addition defines the changing the order of addends does not change the sum.

Question 3.
0 + 43 = 43

Answer: Addition Property of Zero
The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 4.
(27 + 18) + 22 = 27 + (18 + 22)

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Apply and Grow: Practice

Identify the property.

Question 5.
(28 + 16) + 14 = 28 + (16 + 14)

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 6.
12 + 35 = 35 + 12

Answer:
Associative Property of Addition defines the changing the order of addends does not change the sum.

Question 7.
36 + 0 = 36

Answer:
The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 8.
11 + (9 + 57) = (11 + 9) + 57

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Find the missing number.

Question 9.
23 + 45 = 45 + ___

Answer: 23

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
23 + 45 = 45 + x
23 + 45 = 45 + 23
Thus the missing number is 23.

Question 10.
(13 + 12) + __ = 13 + (12 + 45)

Answer: 45

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
(13 + 12) + x = 13 + (12 + 45)
Thus the missing number is 45.

Question 11.
4 + (76 + 10) = (___ + 76) + 10

Answer: 4

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
4 + (76 + 10) = (x + 76) + 10
4 + (76 + 10) = (4 + 76) + 10
Thus the missing number is 4.

Question 12.
98 + ___ = 98

Answer: 0

Explanation:
Let the missing number be x.
The Addition Property of Zero defines the sum of any number and 0 is that number.
98 + x = 98
x = 98 – 98
x = 0
Thus the missing number is 0.

Question 13.
(___ + 0) + 32 = 6 + 32

Answer: 6

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
(x + 0) + 32 = 6 + 32
x + 32 = 38
x = 38 – 32
x = 6
Thus the missing number is 6.

Question 14.
64 + (5 + 23) = (23 + ___) + 64

Answer: 5

Explanation:
Let the missing number be x.
By using the Associative Property of Addition property we can find the missing number.
64 + (5 + 23) = (23 + x) + 64
64 + 28 = 23 + x + 64
82 = x + 87
x = 87 – 82
x = 5
Thus the missing number is 5.

Question 15.
DIG DEEPER!
Use the numbers 24, 54, and 11 to write an equation that shows the Associative Property of Addition.

Answer:
We can write the equation by using the Associative Property of Addition.
24 + (54 + 11) = (24 + 54) + 11

Writing
Use a property to find the sum. Which property did you use? Why?

Question 16.
54 + 0 = __

Answer: 54

Explanation:
We can find the sum of the 54 + 0 by using the addition property of zero.
The Addition Property of Zero defines the sum of any number and 0 is that number.
54 + 0 = 54

Question 17.
(46 + 17) + 33 = ___

Answer: 96

Explanation:
We can find the sum of the (46 + 17) + 33 by using the Associative Property of Addition.
(46 + 17) + 33 = 96

Question 8.
20 + 63 = ___

Answer: 83

Explanation:
We can find the sum of the Associative Property of Addition.
20 + 63 = 83

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 2
A tourist visits 13 museums, 19 memorials, and 11 monuments. Explain how to use a property to find the total number of sites the tourist visits.
(13 + 19) + 11 = ?
Explain:
The tourist visits ___ sites.

Answer:
Given,
A tourist visits 13 museums, 19 memorials, and 11 monuments.
We can find the total number of sites the tourist visits.
(13 + 19) + 11 = 13 + (19 + 11) = 43
Thus the tourist visits 43 sites.

Show and Grow

Question 19.
A farmer sells 34 cucumbers, 48 ears of corn, and 26 bell peppers at a farmer’s market. Explain how to use properties to find the total number of vegetables the farmer sells.
(34 + 48) + 26 = ?

Answer:
Given,
A farmer sells 34 cucumbers, 48 ears of corn, and 26 bell peppers at a farmer’s market.
We can find the sum of the Associative Property of Addition.
(34 + 48) + 26 = 34 + (48 + 26)
34 + 48 + 26 = 108
Thus the total number of vegetables the farmer sells is 108.

Question 20.
How many people go on the field trip?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 3

Answer:
The number of adults = 20
The number of grade 2 students = 47
The number of grade 3 students = 53
20 + 47 + 53 = 120
Thus 120 people go on the field trip.

DIG DEEPER!
The Grade 2 and Grade 3 students are divided into 10 equal groups. How many students are in each group? Explain.

Answer:
Given that,
The Grade 2 and Grade 3 students are divided into 10 equal groups.
The number of grade 2 students = 47
The number of grade 3 students = 53
47 + 53 = 100
100/10 = 10
Thus there are 10 students in each group.

Identify Addition Properties Homework & Practice 8.1

Identify the property.

Question 1.
(79 + 12) + 13 = 79 + (12 + 13)

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 2.
24 + 63 = 63 + 24

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 3.
0 + 64 = 64

Answer: Associative property of zero
The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 4.
37 + (43 + 19) = (37 + 43) + 19

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 5.
17 + 38 = 38 + 17

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 6.
18 + 48 = 48 + 18

Answer: Associative property of addition
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Find the missing number.

Question 7.
36 + __ = 36

Answer: 0

Explanation:
Let the missing number be x.
36 + x = 36
x = 36 – 36
x = 0
Thus the missing number is 0.

Question 8.
25 + __ + 11 = 25 + 11

Answer: 0

Explanation:
Let the missing number be y.
25 + y + 11 = 25 + 11
y + 36 = 36
y = 36 – 36
y = 0
Thus the missing number is 0.

Question 9.
0 + 43 = __ + 0

Answer: 43

Explanation:
Let the missing number be t.
0 + 43 = t + 0
43 = t
Thus the missing number is 43.

Question 10.
(22 + 19) + 28 = 19 + (___ + 28)

Answer: 22

Explanation:
Let the missing number be p.
22 + 19 + 28 = 19 + p + 28
69 = 47 + p
p = 69 – 47
p = 22
Thus the missing number is 22.

Question 11.
Number Sense
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 4
Newton uses two properties. Identify the properties he uses.

Answer:
Given the expression (18 + 27) + 12 = 27 + (18 + 12)
By seeing the above expression we can say that Newton used Associative Property of Addition and Commutative Property of Addition.

Question 12.
Open-Ended
Write an equation that shows the Commutative Property of Addition.

Answer: 4 + 2 = 2 + 4
Commutative property of addition: Changing the order of addends does not change the sum.

Question 13.
Structure
Explain how the Associative Property of Addition and the Associative Property of Multiplication are alike and how they are different.

Answer:
Associative property explains that addition and multiplication of numbers are possible regardless of how they are grouped.
Example:
2 × (3 × 5) = (2 × 3) × 5
2 + (3 + 5) = (2 + 3) + 5
The method is same but the solution for both the equations are different.

Question 14.
Modeling Real Life
A florist uses 11 roses, 12 lilies, and 19 daisies to make bouquets. How many flowers does he use?

Answer:
Given that,
A florist uses 11 roses, 12 lilies, and 19 daisies to make bouquets.
Add all the flowers to find the total number of flowers he used.
11 + 12 + 19 = 42
Thus he used 42 flowers.

DIG DEEPER!
The florist uses 6 flowers for each bouquet. How many bouquets does he make? Explain.

Answer:
Given,
The florist uses 6 flowers for each bouquet.
Total flowers = 42
Divide the total number of flowers by the number of flowers in each bouquet
42/6 = 7
Thus he made 7 bouquets.

Review & Refresh

Find the product.

Question 15.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 400

Answer:
Multiply the two numbers 0 and 3.
0 × 3 = 0

Question 16.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 5

Answer:
Multiply the two numbers 7 and 7.
7 × 7 = 49

Question 17.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 6

Answer:
Multiply the two numbers 10 and 4.
10 × 4 = 40

Question 18
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 7

Answer:
Multiply the two numbers 7 and 6.
7 × 6 = 42

Lesson 8.2 Use Number Lines to Addition

Explore and Grow

Color to find 33 + 25. Then model your jumps on the number line.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 8

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-8
First, you have to color the two numbers 33 and 25.
Now add 33 and 25
33 + 25 = 58
Now try to solve the problem by using the number line.
Start at 33. Count by twenties, then by 5s.
33 + 20 + 5 = 58

Reasoning

How can finding 33 + 25 help you find 533 + 25?

Answer:
We can find the sum of 533 and 25 with the help of the above problem.
33 + 25 = 58
533+ 25 = 538
Just add 5 on the left side to get the sum.

Think and Grow: Adding on a Number Line

Example
Find 245 + 38
One Way: Use the count on strategy. Start at 245. Count on by tens, then by ones.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 9
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-9

Another Way: Use the make a ten strategy. Start at 245. Count on to the nearest ten. Then count on by tens and by ones.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 10
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-10

Show and Grow

Question 1.
Use the count on strategy to find 368 + 24.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 11

Answer: 392
Use the count on strategy. Start at 368. Count on by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-11
368 + 10 + 10 + 2 + 2 = 392

Question 2.
Use the count on strategy to find 57 + 179.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 12

Answer: 236
Use the count on strategy. Start at 57. Count on by hundreds, tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-12

Apply and Grow: Practice

Question 3.
Use the count on strategy to find 47 + 216.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 13

Answer: 263
Use the count on strategy. Start at 47. Count on by hundreds, tens, then by ones.
47 + 100 + 100 + 10 + 6 = 263
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-13

Question 4.
Use the make a ten strategy to find 478 + 64.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 14

Answer: 542
Use the count on strategy. Start at 478. Count on by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-14

Find the sum

Question 5.
395 + 85 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 15

Answer: 480
Use the count on strategy. Start at 395. Count on by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-15

Question 6.
653 + 109 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 16

Answer: 762
Use the count on strategy. Start at 353. Count on by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-15

Question 7.
Humans have 24 rib bones and 182 other types of bones. How many bones do humans have in all?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 17

Answer:
Given,
Humans have 24 rib bones and 182 other types of bones.
182 + 24 = 206
Thus Humans have 206 bones.

Question 8.
Structure
Write the equation shown by the number line.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 18

Answer:
436 + 4 + 20 + 4
440 + 20 + 4 = 464

Question 9.
Structure
Show two different ways to find 225 + 39 using a number line.

Answer:
Use the count on strategy. Start at 225. Count on by tens, then by ones.
225 + 30 + 9 = 264
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-15

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 19
The Leaning Tower of Pisa has 294 steps. A visitor climbs 156 steps, takes a break, and then climbs 78 more steps. Does the visitor reach the top of the tower?
Addition equation:
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 20
The visitor ___ reach the top of the tower.

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-16
156 + 78 = 234
294 – 234 = 60
Thus the visitor does not reach the top of the tower.

Show and Grow

Question 10.
A book has 216 pages. You have already read 167 pages. You read 49 more pages. Do you finish reading the book?

Answer:
Given that,
A book has 216 pages. You have already read 167 pages. You read 49 more pages.
167 + 49 = 216 pages
216 – 216 = 0
Thus you finish reading the book.

Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 21
DIG DEEPER!
A puzzle has 350 pieces. You put 95 pieces together. Your friend puts 185 pieces together. Do you and your friend complete the puzzle? If not, how many pieces are left?

Answer:
Given that,
A puzzle has 350 pieces. You put 95 pieces together. Your friend puts 185 pieces together.
95 + 185 = 280 pages
350 – 280 = 70 pages
No you and your friend do not complete the puzzle.
70 pieces are left to complete the puzzle.

Question 12.
A music library has 483 songs. You listen to162 different songs one week and 171 more songs the next week. How many songs are left?

Answer:
Given that,
A music library has 483 songs.
You listen to162 different songs one week and 171 more songs the next week.
162 + 171 = 333 songs
483 – 333 = 150 songs
Thus 150 songs are left.

Use Number Lines to Addition Homework & Practice 8.2

Question 1.
Use the count on strategy to find 402 + 39.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 22

Answer:
Use the count on strategy. Start at 402. Count on by tens, then by ones.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-22

Question 2.
Use the make a ten strategy to find 81 + 647.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 23

Answer:
Use the count on strategy. Start at 81. Count on by hundreds, tens, then by ones.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-23

Question 3.
Find the sum.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 24

Answer:

Question 4.
718 + 226 = __
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 25

Answer:
Use the count on strategy. Start at 718. Count on by hundreds, tens, then by ones.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-25
718 + 226 = 944

Question 5.
YOU BE THE TEACHER
Your friend uses a number line to find 435 + 27. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 26

Answer:
Your friend uses a number line to find 435 + 27.
435 + 27 = 462
Your friend is not correct.

Question 6.
YOU BE THE TEACHER
Your friend says she can find 64 + 691 by starting at 691 on a number line because of the Associative Property of Addition. Is your friend correct? Explain.

Answer:
Your friend says she can find 64 + 691 by starting at 691 on a number line because of the Associative Property of Addition.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-25
Yes, your friend is correct.

Question 7.
Modeling Real Life
Your cousin needs to write a 400-word essay. She writes 318 words during class. She finishes her essay by writing 94 words at home. Does your cousin’s essay have enough words?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 401

Answer:
Given,
Your cousin needs to write a 400-word essay.
She writes 318 words during class. She finishes her essay by writing 94 words at home.
318 + 94 = 412
412 – 400 = 12
Yes your cousin’s essay have enough words.

Question 8.
DIG DEEPER!
There are 275 apartments in an apartment building. There are 203 two-bedroom apartments rented, and 56 one-bedroom not apartments rented. How many apartments are rented?

Answer:
Given that,
There are 275 apartments in an apartment building.
There are 203 two-bedroom apartments rented, and 56 one-bedroom not apartments rented.
203 + 56 = 259
Thus 259 apartments are rented.

Find the quotient

Question 9.
18 ÷ 6 = ___

Answer: 3

Explanation:
Divide the two numbers 18 and 6.
18/6 = 3
Thus the quotient is 3.

Question 10.
32 ÷ 8 = ___

Answer: 4

Explanation:
Divide the two numbers 32 and 8.
32/4 = 4
Thus the quotient is 4.

Question 11.
72 ÷ 9 = __

Answer: 8

Explanation:
Divide the two numbers 72 and 9.
72/9 = 8
Thus the quotient is 8.

Lesson 8.3 Use Mental Math to Add

Explore and Grow

What addition problem is shown? How can you find the sum?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 27

Answer: 366 + 504 = 870

Explanation:
By seeing the above figure we can find the sum.
First count the number of blocks
There are 100 blocks in each figure
There are three 100 blocks, six 10 blocks, and 6 blocks.
300 + 60 + 6 = 366
There are five 100 blocks and 4 blocks.
500 + 4 = 504
Now add both
366 + 504 = 870

Reasoning
Show how to find 402 + 248.

Answer:
You can find the sum of 402 and 248 by using the above arrays.
Take 10 × 10 block
402 – Four 10 × 10 blocks, 2 blocks
248 – Two 10 × 10 blocks, Four 10 blocks, and eight blocks
402 + 248 = 650

Think and Grow: Mental Math Strategies for Addition

Example
Find 247 + 328.
Use compensation to add.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 28

Answer:
You can find the sum of 247 and 328 by using mental math strategies.
247 + 3 = 250
328 – 3 = 325
250
+325
575
Thus the sum of 247 and 328 is 575.

Example
Find 119 + 163.
Make a ten and count on to add.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 29

Answer:
119 + 163 = 119 + (1 + 100 + 60 + 2)
(119 + 1) + 100 + 60 + 2
120 + 100 + 60 + 2
382

Show and Grow

Question 1.
Use compensation to find 322 + 158.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 30

Answer:
322 – 2 = 320
158 + 2 = 160
320
+160
480
So, 322 + 158 = 480

Question 2.
Make a ten and count on to find 695 + 187.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 31

Answer:
695 + 187 = 695 + (5 + 100 + 80 + 2)
(695 + 5) + (100 + 80 + 2)
700 + 100 + 80 + 2
882
So, 695 + 187 = 882

Apply and Grow: Practice

Question 3.
Use compensation to find the sum.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 32

Answer:
604 – 4 = 600
275 + 4  = 279
600
+279
879

Question 4.
Make a ten and count on to find the sum.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 33

Answer:
359 + 1 = 360
360 + 300 = 660
660 + 17 = 677
So, 359 + 318 = 677

Use mental math to find the sum.

Question 5.
436 + 248 = __

Answer:
436 + 248 = 436 + (8 + 40 + 200)
436 + (4 + 4 + 40 + 200)
(436 + 4) + 4 + 40 + 200
440 + 244
684
So, 436 + 248 = 684

Question 6.
795 + 102 = ___

Answer:
102 – 2 = 100
795 + 2 = 797
797
+100
897
So, 795 + 102 = 897

Question 7.
503 + 71 = ___

Answer:
503 – 3 = 500
71 + 3 = 74
500
+74
574
So, 503 + 71 = 574

Question 8.
589 + 407 = ___

Answer:
589 + 1 = 590
407 – 1 = 406
590
+406
996

Question 9.
734 + 97 = ___

Answer:
734 – 3 = 731
97 + 3 = 100
731
+100
831

Question 10.
352 + 164 = ___

Answer:
352 – 2 = 350
164 + 2 = 166
350
+166
516

Question 11.
297 + 211 = ___

Answer:
297 + 3 = 300
211 – 3 = 208
300
+208
508

Question 12.
426 + 364 = ___

Answer:
426 + 4 = 430
364 – 4 = 360
430
+360
790

Question 13.
159 + 104 = ___

Answer:
159 + 1 = 160
104 – 1 = 103
160
+103
263

Question 14.
A community shelter collects 101 shirts and 109 pairs of pants from a clothing drive. How many clothing items does the community shelter collect?

Answer:
Given,
A community shelter collects 101 shirts and 109 pairs of pants from a clothing drive.
101 – 1 = 100
109 + 1 = 110
100
+110
210
Thus the community shelter collect 210 clothing items.

Question 15.
Number Sense
Descartes wants to use compensation to find 238 + 127. Which numbers could he use?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 34

Answer:
Descartes wants to use compensation to find 238 + 127.
The number near 238 is 240, 127 is 130.
Thus he could use 240 and 130

Think and Grow: Modeling Real Life

A company manager has $900. Does he have enough money to buy the laptop and the cell phone?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 35
Addition equation:
Compare:
The manager ___ have enough money.

Answer:
Given,
A company manager has $900.
The cost of the laptop is $595
The cost of the mobile is $249.
595 + 5 = 600
249 – 5 = 244
600
+244
844
900 – 844 = 56
Thus the manager does not have enough money.

Show and Grow

Question 16.
A USB drive holds 600 photos. You have 279 photos on a digital camera and 337 photos on a cell phone. Can the USB drive hold all of your photos?

Answer:
Given,
A USB drive holds 600 photos. You have 279 photos on a digital camera and 337 photos on a cell phone.
279 + 1 = 280
337 – 1 = 336
280
+336
616
Yes the USB drive can hold all of your photos

Question 17.
A school board president has $1,000. Which two items can she buy?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 36

DIG DEEPER!
The president buys the two items. How much money does she have left?

Answer:
A school board president has $1,000.
The cost of a swing set is $648
The cost of the seesaw is $372
The cost of a Dome Climber is $498
498
+372
870
1000 – 870 = 130
Thus she has left $130.

Use Mental Math to Add Homework & Practice 8.3

Use compensation to find the sum.

Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 37

Answer:
248 + 2 = 250
137 – 2 = 135
250
+135
385
So, 248 + 137 = 385

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 38

Answer:
401 – 1 = 400
165 + 1 = 166
400
+166
566

Make a ten and count on to find the sum

Question 3.
506 + 369 = ?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 39

Answer:
506 + 369 = 506 + (4 + 300 + 60 + 5)
(506 + 4) + 300 + 60 + 5
510 + 365
875
So, 506 + 369 = 875

Question 4.
617 + 348 = ?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 40

Answer:
617 + 348 = 617 + (3 + 300 + 40 + 5)
(617 + 3) + 300 + 40 + 5
620 + 345
965
617 + 348 = 965

Use mental math to find the sum.

Question 5.
478 + 219 = ___

Answer:
478 + 2 = 480
219 – 2 = 217
480
+217
697

Question 6.
503 + 64 = ___

Answer:
503 – 3 = 500
64 + 3 = 67
500
+67
567

Question 7.
288 + 242 = ___

Answer:
288 + 2 = 290
242 – 2 = 240
290
+240
530

Question 8.
Structure
Explain how to make a ten to find the sum.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 41

Answer:
738 + 126 = 738 + (6 + 20 + 100)
(738 + 2) + 4 + 20 + 100
740 + 124
864

Question 9.
Writing
How is compensation make a ten similar to the strategy? How is it different?

Answer:
Compensation is a mental math strategy for multi-digit addition that involves adjusting one of the addends to make the equation easier to solve. The methods are different but the solutions are the same.

Question 10.
Modeling Real Life
A binder holds 250 sheets of paper. You have 107 science papers and 142 math papers. Can the binder hold all of your papers?

Answer:
Given,
A binder holds 250 sheets of paper. You have 107 science papers and 142 math papers.
107 – 2 = 105
142 + 2 = 144
105 + 144 = 249
Yes, the binder can hold all of your papers.

Question 11.
Modeling Real Life
A school nurse has $450. Which two items can she buy?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 42

DIG DEEPER!
The school nurse buys the two items. How much money does she have left?

Answer:
A school nurse has $450.
The cost of a Stethoscope = $119
The cost of thermometer = $ 308
119 + 1 = 120
308 – 1 = 307
120
+307
427
450 – 427 = 23
She has left $23

Review & Refresh

Question 12.
It costs $1 to get into each football game. Newton buys a chicken wrap for $2 and a drink for $1 each game. How much money does Newton spend in 4 games?

Answer:
Given,
It costs $1 to get into each football game. Newton buys a chicken wrap for $2 and a drink for $1 each game.
1 + 2 + 1 = 4
4 × $4 = $8
Thus Newton spent $8 for 4 games.

Lesson 8.4 Use Partial Sums to Add

Explore and Grow

Model each number. Draw to show each model.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 43
How can you use the models to find 341 + 227?

Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-43
You can find the sum of 341 and 227 is

300 + 40 + 1
200 + 20 + 7
500 + 60 + 8 = 568

Reasoning
How can breaking apart addends help you add three-digit numbers?

Answer: Breaking apart addends helps you add three-digit numbers easily. Mind math is possible with this breaking apart addends.

Think and Grow: Use Partials Sums to Add

Example
Find 356 + 408.
Step 1: Write each number in expanded form.
Step 2: Find the partial sums.
Step 3: Add the partial sums.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 44

Answer: 764

Show and Grow

Use partial sums to add.

Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 45

Answer:
319 = 300 + 10 + 9
234 = 200 + 30 + 4
553 = 500 + 40 + 13

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 46

Answer:
568 = 500 + 60 + 8
173 = 100 + 70 + 3
741 = 600 + 130 + 11

Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 47

Answer:
424 = 400 + 20 + 4
450 = 400 + 50 + 0
874 = 800 + 70 + 4

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 48

Answer:
281 = 200 + 80 + 1
365 = 300 + 60 + 5
646 = 500 + 140 + 6

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 49

Answer:
127 = 100 + 20 + 7
609 = 600 + 0 + 9
736 = 700 + 20 + 16

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 50

Answer:
276 = 200 + 70 + 6
39 =  000 + 30 + 9
315 = 200 + 100 + 15

Apply and Grow: Practice

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 51

Answer:
759 = 700 + 50 + 9
202 = 200 + 00 + 2
961 = 900 + 50 + 11

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 52

Answer:
864 = 800 + 60 + 4
131 = 100 + 30 + 1
995 = 900 + 90 + 5

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 53

Answer:
387 = 300 + 80 + 7
94 = 000 + 90 + 4
481 = 300 + 170 + 11

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 54

Answer:
560 = 500 + 60 + 0
273 = 200 + 70 + 3
833 = 700 + 130 + 3

Question 11.
498 + 375 = ___

Answer:
498 = 400 + 90 + 8
375 = 300 + 70 + 5
873 = 700 + 160 + 13

Question 12.
209 + 158 = ___

Answer:
209 = 200 + 00 + 9
158 = 100 + 50 + 8
367 = 300 + 50 + 17

Think and Grow: Modeling Real Life

A giant panda weighs 696 pounds less than a polar bear. How much does the polar bear weigh?
Addition equation:
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 55

Answer:
Given,
A giant panda weighs 696 pounds less than a polar bear.
696 – 263 = 433
Thus the polar bear weighs 433 pounds.

Show and Grow

Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 56
Question 15.
A herd of wildebeests has 258 fewer members than a herd of zebras has. There are 335 wildebeests in the herd. How many zebras are in their herd?

Answer:
Given,
A herd of wildebeests has 258 fewer members than a herd of zebras has. There are 335 wildebeests in the herd.
335 – 258 = 77
Therefore there are 77 zebras in their herd.

Question 16.
There are three candidates in an election. Candidate A receives 184 fewer votes than Candidate B. Who wins the election?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 57

Answer:
Given,
There are three candidates in an election. Candidate A receives 184 fewer votes than Candidate B.
Number of votes candidate A receives = 347
347 + 184 = 631
Thus Candidate B wins the election.

Question 17.
DIG DEEPER!
You, your friend, and your cousin play a video game. Your friend scores 161 fewer points than you. Your cousin scores 213 more points than your friend. What is each player’s score? Who wins?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 58

Answer:
Your friend score is 579
Your friend scores 161 fewer points than you.
579 + 161 = 740
Your score is 740
Your cousin scores 213 more points than your friend.
213 + 579 = 592
Your cousin’s score is the highest among all three.
So, your cousin wins the game.

Use Partial Sums to Add Homework & Practice 8.4

Use partials sums to add

Question 1.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 59

Answer:
We can find the sum by using the partial sum model.
479 = 400 + 70 + 9
356 = 300 + 50 + 6
835 = 700 + 120 + 15
So, 479 + 356 = 835

Question 2.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 60

Answer:
We can find the sum by using the partial sum model.
674 = 600 + 70 + 4
321 = 300 + 20 + 1
995 = 900 + 90 + 5

Question 3.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 61

Answer:
We can find the sum by using the partial sum model.
396 = 300 + 90 + 6
278 = 200 + 70 + 8
674 = 500 + 160 + 14

Question 4.
564 + 218 = ___

Answer:
We can find the sum by using the partial sum model.
564 = 500 + 60 + 4
218 = 200 + 10 + 8
782 = 700 + 70 + 12

Question 5.
190 + 123 = ___

Answer:
We can find the sum by using the partial sum model.
190 = 100 + 90 + 0
123 = 100 + 20 + 3 
313 = 200 + 110 + 3

Question 6.
YOU BE THE TEACHER
Your friend uses partial sums to find 205 + 124. Is your friend correct? Explain.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 62

Answer: Your friend is incorrect.
205 = 200 + 00 + 5
124 = 100 + 20 + 4
329 = 300 + 20 + 9
The sum of 205 + 124 = 329

Question 7.
Patterns
Write and solve the next problem in the pattern.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 63

Answer:
The addend is increased by 100.
So, the next pattern is
516
+178
694

Question 8.
Modeling Real Life
There are worker bees and drone bees in a beehive. A hive has 268 fewer drones than workers. There are 351 drone bees. How many worker bees are there?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 64

Answer:
Given,
There are worker bees and drone bees in a beehive. A hive has 268 fewer drones than workers. There are 351 drone bees.
351
-268
83
Therefore there are 83 worker bees.

Question 9.
Modeling Real Life
Three athletes compete in Olympic weight lifting. Weight lifter A lifts 104 fewer pounds than Weight lifter B. Who lifts the most weight?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 65

Answer:
Three athletes compete in Olympic weight lifting. Weight lifter A lifts 104 fewer pounds than Weight lifter B.
Weight lifter A lifts 368 pounds
368 + 104 = 472
Therefore, Weight Lifter B lifts the most weight.

Review & Refresh

Circle the value of the underlined digit.

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 66

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-66
In the given value 2 is in the ones place so the answer is 2.

Question 11.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 67

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-67
In the given value 4 is in the hundreds place so the answer is 400.

Question 12.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 68

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-68
In the given value 0 is in tens place so the answer is 0.

Lesson 8.5 Add Three-Digit Numbers

Explore and Grow

Model the equation. Draw your model. Then find the sum.
195 + 308 = ___

Answer:
195 = 100 + 90 + 5
308 = 300 + 00 + 8
503 = 400 + 90 + 13

Reasoning
How can you use an estimate to check whether your answer reasonable?

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
195 = 200
308 = 300
200
+300
500
Ths sum is about 500.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
195
+308
503
503 is close to 500. So, the answer is reasonable.

Think and Grow: Add Three-Digit Numbers

Example
Find 236 + 378. Check whether your answer is reasonable.
Step 1: Estimate. Round each addend to the nearest hundred.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 69

Show and Grow

Find the sum. Check whether your answer is reasonable.

Question 11.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 70

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
457 – 500
133 – 100
500
+100
600
The sum is about 600.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
457
+133 
590
590 is close to 600. So, the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 71

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
269 – 300
354 – 300
300
+300
300
Step 2: Find the sum. Add the ones, tens, then the hundreds.
269
+354
623
623 is close to 600. So, the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 72

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
560 – 600
274 – 300
600
+300
900
Step 2: Find the sum. Add the ones, tens, then the hundreds.
560
+274 
834
834 is close to 900. So, the answer is reasonable.

Question 4.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 73

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
345 – 300
286 – 300
300
+300
600
Step 2: Find the sum. Add the ones, tens, then the hundreds.
345
+286
631
631 is close to 600. So, the answer is reasonable.

Question 5.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 74

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
129 – 100
668 – 700
100
+700
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
129
+668
797
797 is close to 800. So, the answer is reasonable.

Question 6.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 75

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
383 – 400
539 – 500
400
+500
900
Step 2: Find the sum. Add the ones, tens, then the hundreds.
383
+539
922
922 is close to 900. So, the answer is reasonable.

Apply and Grow: practice

Find the sum. Check whether your answer is reasonable.

Question 7.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 76

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
803 – 800
179 – 200
800
+200
1000
Step 2: Find the sum. Add the ones, tens, then the hundreds.
803
+179
982
982 is close to 1000. So, the answer is reasonable.

Question 8.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 77

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
608 – 600
239 – 200
600
+200
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
608
+239
847
847 is close to 800. So, the answer is reasonable.

Question 9.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 78

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
427 – 400
385 – 400
400
+400
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
427
+385
812
812 is close to 800. So, the answer is reasonable.

Question 10.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 79

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
401 – 400
109 – 100
400
+100
500
Step 2: Find the sum. Add the ones, tens, then the hundreds.
401
+109
510
510 is close to 500. So, the answer is reasonable.

Question 11.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 80

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
265 – 300
157 – 100
300
+100
400
Step 2: Find the sum. Add the ones, tens, then the hundreds.
265
+157
422
422 is close to 400. So, the answer is reasonable.

Question 12.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 81

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
375 – 400
64 – 100
400
+100
500
Step 2: Find the sum. Add the ones, tens, then the hundreds.
375
+64
439
439 is close to 500. So, the answer is reasonable.

Question 13.
Estimate: ___
469 + 284 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
469 – 500
284 – 300
500
+300
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
469
+284 
753
753 is close to 800. So, the answer is reasonable.

Question 14.
Estimate: ___
580 + 246 = __

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
580 – 600
246 – 200
600
+200
800
Step 2: Find the sum. Add the ones, tens, then the hundreds.
580
+246
826
826 is close to 800. So, the answer is reasonable.

Question 15.
Estimate: ___
796 + 135 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
796 – 800
135 – 100
800
+100
900
Step 2: Find the sum. Add the ones, tens, then the hundreds.
796
+135
931
931 is close to 900. So, the answer is reasonable.

Question 16.
A truck driver travels 428 miles on Monday. He travels 473 miles on Tuesday. How many miles does he travel in all on Monday and Tuesday?

Answer:
Given that,
A truck driver travels 428 miles on Monday. He travels 473 miles on Tuesday.
428
+473
901
Thus he travels 901 miles on Monday and Tuesday.

Question 17.
Reasoning
Your friend finds a sum. Is her answer reasonable? If not, describe her mistake.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 85

Answer:
119
+ 187
306
Your answer is not reasonable.

Think and Grow: Modeling Real Life

A construction team builds an 825-meter-long boardwalk on a beach. The team builds 408 meters one week and 377 meters the next week. Is the boardwalk complete?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 86
Addition equation:
The boardwalk ___ complete.

Answer:
Given,
A construction team builds an 825-meter-long boardwalk on a beach. The team builds 408 meters one week and 377 meters the next week.
408
+377
785
The boardwalk did not complete.

Show and Grow

Question 18.
A road crew repaves the road on a 547-meter-long bridge. The crew repaves 318 meters the first day and 229 meters the second day. Is the road on the bridge completely repaved?

Answer:
Given,
A road crew repaves the road on a 547-meter-long bridge.
The crew repaves 318 meters on the first day and 229 meters on the second day.
318
+229
547
Yes, the road on the bridge completely repaved.

Question 19.
A family drives from St. Louis to Orlando for a vacation. The family drives 363 miles the first day and 386 miles the second day. How many miles does the family have left to drive?

Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 87
Answer:
Given,
A family drives from St. Louis to Orlando for a vacation. The family drives 363 miles the first day and 386 miles the second day.
363 + 386 = 749 miles
922 miles – 749 miles = 173 miles
173 miles the family have left to drive.

Question 20.
Which booth had more visitors in all?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 88

Answer:
Day 1 – 468 + 527 = 995
Day 2 – 416 + 374 = 790
995 – 790 = 205
Thus day 1 has more visitors in all.

Add Three-Digit Numbers Homework & Practice 8.5

Find the sum. Check whether your answer is answerable.

Question 1.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 89

Answer: 630

Explanation:
The estimated number of 493 is 490
The estimated number of 142 is 140.
490
+ 140
630

Question 2.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 90

Answer: 820

Explanation:
The estimated number of 763 is 760.
The estimated number of 58 is 60.
760
+ 60
820

Question 3.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 91

Answer: 580

Explanation:
The estimated number of 308 is 310
The estimated number of 273 is 270
310
+270
580

Question 4.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 92

Answer: 420

Explanation:
The estimated number of 276 is 280
The estimated number of 138 is 140
280
+ 140
420

Question 5.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 93

Answer: 700

Explanation:
The estimated number of 532 is 530
The estimated number of 167 is 170.
530
+170
700

Question 6.
Estimate: ___
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 94

Answer: 980

Explanation:
The estimated number of 680
The estimated number of 296 is 300.
680
+300
980

Question 7.
Estimate: ___
595 + 280 = ___

Answer: 880

Explanation:
The estimated number of 595 is 600
The estimated number of 280
600
+ 280
880

Question 8.
Estimate: ___
419 + 295

Answer: 720

Explanation:
The estimated number of 419  is 420
The estimated number of 295 is 300
420
+ 300
720

Question 9.
Estimate: ___
498 + 305 = ___

Answer: 800

Explanation:
The estimated number of 498 is 500
The estimated number of 305 is 300
500
+300
800

Question 10.
Open-Ended
Complete the addends so you need to regroup to add. Then find the sums.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 95

Answer:
Let the addend be 3
479
+283
762
If you take 3 as addend then you need to regroup to find the sum.
Let the addend be 9
697
+135
832
If you take 9 as addend then you need to regroup to find the sum.

Question 11.
DIG DEEPER!
Find the missing digits.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 96

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-96
You will get 466 if you add 107 and 359.
You will get 982 if you add 748 and 234
You will get 962 if you add 670 and 292.
You will get 982 if you add 809 and 173.

Question 12.
Modeling Real Life
Newton wants to complete a 770-mile hike in 2 months. He hikes 423 miles the first month and 347 miles the second month. Does he complete the hike?

Answer:
Given,
Newton wants to complete a 770-mile hike in 2 months.
He hikes 423 miles the first month and 347 miles the second month
423
+347
770
Thus he Newton completed the hike.

Question 13.
Modeling Real Life
You ship a package 750 miles from San Diego to Salt Lake City. The package is now in Las Vegas. How many miles are left until your package is delivered?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 97

Answer:
Given,
You ship a package 750 miles from San Diego to Salt Lake City. The package is now in Las Vegas.
121 + 270 = 391 miles until Las Vegas.
750 – 391 = 359 miles
Therefore 359 miles are left until your package is delivered.

Review & Refresh

Find the quotient.

Question 14.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 98

Answer: 10

Explanation:
Divide the two numbers 100 and 10.
100/10 = 10
Thus the quotient is 10.

Question 15.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 99

Answer: 9

Explanation:
Divide the two numbers 45 and 5.
45/5 = 9
Thus the quotient is 9.

Question 16.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 100

Answer: 7

Explanation:
Divide the two numbers 14 and 7.
14/2 = 7
Thus the quotient is 7.

Question 17.
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 101

Answer: 8

Explanation:
Divide the two numbers 10 and 80.
80/10 = 8
Thus the quotient is 8.

Question 18.
Divide 25 by 5.

Answer: 5

Explanation:
Divide the two numbers 5 and 25.
25/5 = 5
Thus the quotient is 5.

Question 19.
Divide 30 by 10.

Answer: 3

Explanation:
Divide the two numbers 10 and 30.
30/10 = 3
Thus the quotient is 3.

Question 20.
Divide 8 by 2.

Answer: 4

Explanation:
Divide the two numbers 8 and 2.
8/2 = 4
Thus the quotient is 4.

Lesson 8.6 Add Three or More Numbers

Find the sum of the numbers. Which two numbers should you add first?
Big Ideas Math Answers 3rd Grade Chapter 8 Add and Subtract Multi-Digit Numbers 102

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
348 = 350
478 = 480
152 = 150
350 + 480 + 150 = 980
The sum is about 980.
517 = 520
117 = 120
283 = 280
520 + 120 + 280 = 920
The sum is about 920
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
348
+478
+152
978
978 is close to 980. So, the answer is reasonable.
517
+117
+283
917
917 is close to 920, So, the answer is reasonable.

Reasoning
Why did you choose those numbers? Compare your strategy to your partner’s strategy.

Answer:
348 = 350
478 = 480
152 = 150
I choose these numbers because they are nearest to ten. This strategy will help to improve the mental math. After solving the problem you can verify the answer with the actual sum.

Think and Grow: Add Three or More Numbers

Example
Find 138 + 221 + 176 + 92. Check whether your answer is reasonable.
Step 1: Estimate. Round each addend to the nearest ten.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 103

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
140 + 220 + 180 + 90 = 630
The sum is about 630.
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
138
+221
+176
+92
627
627 is close to 630. So, the answer is reasonable.

Show and Grow

Find the sum. Check whether your answer is reasonable.

Question 1.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 104

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
342 = 340
73 = 70
267 = 270
340 + 70 + 270 = 680
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
342
+73
+267
682
682 is close to 680. So, the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 105

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
191 = 190
452 = 450
206 = 210
190 + 450 + 210 = 850
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 105
849
849 is close to 850. So, the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 106

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
65 = 60
98 = 100
637 = 640
640
+100
+60
800
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
637
+98
+65
800
800 is close to 800. So, the answer is reasonable.

Question 4.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 107

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
241 = 240
394 = 390
85 = 80
193 = 190
390
240
190
+80
900
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 107
913
913 is close to 900. So, the answer is reasonable.

Question 5.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 108

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
136 = 140
51 = 50
64 = 60
410 = 410
140
410
+60
+50
660
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 108
661
661 is close to 660. So, the answer is reasonable.

Question 6.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 109

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
105 = 100
113 = 110
222 = 220
307 = 310
100 + 110 + 220 + 310 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 109
747
747 is close to 740. So, the answer is reasonable.

Apply and Grow: Practice

Find the sum. Check whether your answer is answerable.

Question 7.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 110

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
557 = 560
79 = 80
283 = 280
560 + 80 + 280 = 920
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 110
919
919 is close to 920. So, the answer is reasonable.

Question 8.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 111

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
382 = 380
357 = 360
160 + 380 + 360 = 900
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
160
382
357
899
899 is close to 900. So, the answer is reasonable.

Question 9.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 112

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
35 = 30
68 = 70
827 = 830
30 + 70 + 830 = 930
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
35 + 68 + 827 = 930
930 is close to 930. So, the answer is reasonable.

Question 10.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 113

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
153 = 150
235 = 230
458 = 460
67 = 70
150 + 230 + 460 + 70 = 910
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
153 + 235 + 458 + 67 = 913
913 is close to 910. So, the answer is reasonable.

Question 11.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 114

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
108 = 110
172 = 170
200 = 200
263 = 260
110 + 170 + 200 + 260 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
108 + 172 + 200 + 263 = 743
743 is close to 740. So, the answer is reasonable.

Question 12.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 115

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
181 = 180
629 = 630
140 = 140
23 = 20
180 + 630 + 140 + 20 = 970
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
181 + 629 + 140 + 23 = 973
973 is close to 970. So, the answer is reasonable.

Question 13.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 116

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
213 = 210
208 = 210
462 = 460
111 = 110
210 + 210 + 460 + 110 = 990
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 116
994
994 is close to 990. So, the answer is reasonable.

Question 14.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 117

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
108 = 110
172 = 170
200 = 200
263 = 260
110 + 170 + 200 + 260 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 117
743
743 is close to 740. So, the answer is reasonable.

Question 15.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 118

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
259 = 260
233 = 230
223 = 220
147 = 150
260 + 230 + 220 + 150 = 860
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 118
862
862 is close to 860. So, the answer is reasonable.

Question 16.
Number Sense
Use the Associative Property of Addition to find (345 + 234) + 206.

Answer:
We can find the (345 + 234) + 206 by using the Associative Property of Addition.
(a + b) + c = a + (b + c)
(345 + 234) + 206 = 345 + (234 + 206)
345 + 440 = 785

Question 17.
YOU BE THE TEACHER
Your friend finds 364 + 109 + 27. Is your friend correct? Explain.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 119

Answer:
Your friend is incorrect. The order of the sum is wrong. 27 is placed in the wrong pattern. 2 and 7 must be places on tens place and ones place.
364
109
+27
503

Think and Grow: Modeling Real Life

An elevator has a weight limit of 1,000 pounds. A 186-pound man has three 265-pound boxes to deliver. Can he bring all 3 boxes on the elevator at once?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 120

Understand the problem:
Make a plan:
Solve:
He ___ bring all 3 boxes on the elevator at once.

Answer:
Given that,
An elevator has a weight limit of 1,000 pounds.
A 186-pound man has three 265-pound boxes to deliver.
265 + 265 + 265 + 186 = 981
1000 – 981 = 19 pounds
Thus he can bring all 3 boxes on the elevator at once.

Show and Grow

Question 18.
An auditorium has 650 seats. 175 students from each of 3 schools compete in a math competition. 68 teachers assist. Are there enough seats for all of the students and teachers?

Answer:
Given,
An auditorium has 650 seats. 175 students from each of 3 schools compete in a math competition. 68 teachers assist.
175 + 175 + 175 + 68 = 593
Thus the seats are enough for all of the students and teachers.

Question 19.
DIG DEEPER!
Four students at a school organize a petition for more lunch food options. They need 500 signatures. How many more signatures do they need?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 121

Answer:
Add the number of signatures of all the students
A  + B + C + D = 77 + 108 + 112 + 96 = 393
500 – 393 = 107
Thus they need 107 signatures more.

Add Three or More Numbers Homework & Practice 8.6

Find the sum. Check whether your answer is reasonable

Question 1.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 122

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
862 is close to 860. So, the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 123

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
65 = 60
41 = 40
786 = 790
60 + 40 + 790 = 890
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 123
892
892 is close to 890. So, the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 124

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
409 = 410
87 = 90
463 = 460
410 + 90 + 460 = 960
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 124
959
959 is close to 960. So, the answer is reasonable.

Question 4.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 125

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
302 = 300
253 = 250
169 = 170
18 = 20
300 + 250 + 170 + 20 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 125
742
742 is close to 740. So, the answer is reasonable.

Question 5.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 126

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
353 = 350
121 = 120
154 = 150
116 = 120
350 + 120 + 150 + 120 = 740
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 126
744
744 is close to 740. So, the answer is reasonable.

Question 6.
Estimate: ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 127

Answer:
Step 1: Estimate. Round each addend to the nearest ten.
213 = 210
251 = 250
139 = 140
210 + 270 + 250 + 140 = 870
Step 2: Find the sum, add the ones, then the tens, then the hundreds.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 127
873
873 is close to 870. So, the answer is reasonable.

Question 7.
Structure
Which problem can you solve without regrouping?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 128

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-128

Question 8.
Reasoning
You add 602 + 125 + 231. Your friend adds 231 + 602 + 125. Do you both get the same answer? Use an additional property to explain.

Answer:
There are four mathematical properties that involve addition. The properties are the commutative, associative, additive identity and distributive properties. Additive Identity Property: The sum of any number and zero is the original number.
602 + 125 + 231 = 958
231 + 602 + 125 = 958
The sum will be the same irrespective of the change of order.

Question 9.
Modeling Real Life
A firefighter’s ladder has a weight limit of 750 pounds. One firefighter weighs 196 pounds. Another firefighter weighs 243 pounds. They each have67 pounds of gear. If both firefighters wear their gear, can they climb the ladder at the same time?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 129

Answer:
A firefighter’s ladder has a weight limit of 750 pounds.
One firefighter weighs 196 pounds. Another firefighter weighs 243 pounds. They each have 67 pounds of gear.
196 + 243 + 67 + 67 = 573
750 – 573 = 177 pounds
If both firefighters wear their gear, they can climb the ladder at the same time.

Question 10.
DIG DEEPER!
Your principal agrees to make a lip-sync video if the school’s social media page reaches 1,000 likes in 5 days. How many more likes does the school’s page need?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 130

Answer:
Given,
Your principal agrees to make a lip-sync video if the school’s social media page reaches 1,000 likes in 5 days.
Add all the number of likes
573 + 168 + 201 + 47 = 989
1000 – 989 = 19 likes
Thus the school’s page needs 19 likes more.

Review & Refresh

Question 11.
Find the area of the shape
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 131

Answer:
From the figure, we can observe that each block = 1 sq. cm
We have to find the area of the shaded part.
The shaded region is in the form of a rectangle.
So, we have to find the area of the rectangle.
A = l × b
A = 7 × 3
A = 21 sq.cm
Thus the area of the shape is 21 sq. cm

Lesson 8.7 Use Number Lines to Subtract

Explore and Grow

Color to find 79 – 47. Then model your jumps on the number line.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 132

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-132 (1)

Reasoning
How can finding 79 – 47 help you find 379 – 47?

Answer:
You can subtract 47 from 79 by using the number line.
79 – 47 = 32
Just add 300 to 79 or add 3 to the left and subtract 47 from 379.
379 – 47 = 332

Think and Grow: Subtracting on a Number Line

Example
Find 358 – 82.
One Way:
Use the count back strategy. Start at 358. Count back by tens, then by ones.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 133

Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-133

Another Key:
Use the count on strategy. Start at 82. Count on until you reach 358.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 134
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-134

Show and Grow

Question 1.
Use the count back strategy to find 273 – 36.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 135

Answer:
Use the count back strategy. Start at 273. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-135

Question 2.
Use the strategy to find 124 – 45.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 136

Answer:
Use the count on strategy. Start at 45. Count on until you reach 124.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-136

Apply and Grow: Practice

Question 3.
Use the count back strategy to find 961 – 38.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 137

Answer:
Use the count back strategy. Start at 961. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-137

Question 4.
Use the count back strategy to find 853 – 77.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 138

Answer:
Use the count back strategy. Start at 853. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-138

Find the difference.

Question 5.
316 – 24 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 139

Answer:
Use the count back strategy. Start at 316. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-139

Question 6.
548 – 113 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 140

Answer:
Use the count back strategy. Start at 548. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-140

Question 7.
Your friend knows 154 words in Italian. You want to know just as many words as your friend. So far, you have learned 73 words. How many words do you have left to learn?

Answer:
Given,
Your friend knows 154 words in Italian. You want to know just as many words as your friend. So far, you have learned 73 words.
154 – 73 = 81
Thus 81 words are left to learn.

Question 8.
Structure
Write the equation shown by the number line.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 141

Answer:
By seeing the above number line we can find the subtraction equation.
Use the count on strategy. Start at 36. Count on until you reach 407.
407 – 36 = 371

Think and Grow: Modeling Real Life

Each member of a marching band and a football team is awarded a ribbon. The marching band has 123 members. The football team has 66 members. How many more ribbons are needed for the marching band than for the football team?
Subtraction equation:
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 142
___ more ribbons are needed for the marching band.

Answer:
Given,
Each member of a marching band and a football team is awarded a ribbon. The marching band has 123 members. The football team has 66 members.
123 – 66 = 57
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-142
Thus 57 ribbons are needed for the marching band than for the football team.

Show and Grow

Question 9.
A marine biologist feeds 435 pounds of fish to an orca and 50 pounds of fish to a sea lion. How many more pounds did the orca eat than the sea lion?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 143

Answer:
Given,
A marine biologist feeds 435 pounds of fish to an orca and 50 pounds of fish to a sea lion.
435 – 50 = 385
Thus 385 more pounds the orca eat than the sea lion.

Question 10.
There are 620 paper lanterns for a festival. Some are let go. There are 42 left. How many paper lanterns were let go?

Answer:
Given,
There are 620 paper lanterns for a festival. Some are let go. There are 42 left.
620 – 42 = 578
Therefore 578 paper lanterns were let go.

Question 11.
DIG DEEPER!
There are some guests at an amusement park. 387 of them leave when it rains. 474 of them stay. How many guests were there before it rained?

Answer:
Given that,
There are some guests at an amusement park. 387 of them leave when it rains. 474 of them stay.
387 + 474 = 861
Thus 861 guests were there before it rained.

Use Number Lines to Subtract Homework & Practice 8.7

Question 1.
Use the count back strategy to find 232 – 53.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 144

Answer:
Use the count back strategy. Start at 232. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-144

Question 2.
Use the count back strategy to find 796 – 81.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 145

Answer:
Use the count back strategy. Start at 796. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-145

Find the difference

Question 3.
474 – 19 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 146

Answer:
Use the count back strategy. Start at 474. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-146

Question 4.
615 – 204 = ___
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 147

Answer:
Use the count back strategy. Start at 615. Count back by tens, then by ones.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-147

Question 5.
Writing
Write and solve a subtraction word problem using 995 and 238.

Answer:
There are 995 guests at an amusement park. 238 of them leave when it rains. How many guests were there before it rained?
995 – 238
We can solve the problem by using the number line.
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-146

Question 6.
DIG DEEPER!
Which number lines can you use to find 734 – 308?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 148

Answer:
Among all the number lines option ii is used to find 734 – 308.

Question 7.
Modeling Real Life
You take 107 pictures on a field trip to a zoo. Your friend takes 73 pictures. How many more pictures do you take than your friend?
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 149

Answer:
Given that,
You take 107 pictures on a field trip to a zoo. Your friend takes 73 pictures.
107 – 73 = 34
You take 34 pictures than your friend.

Question 8.
Modeling Real Life
An author has 350 copies, not of her book. Some are signed. 115 copies are signed. How many copies are signed?

Answer:
Given,
An author has 350 copies, not of her book. Some are signed. 115 copies are signed.
350 – 115 = 235 copies
Therefore 235 copies are to be signed.

Review & Refresh

Find the quotient.

Question 9.
Divide 25 by 5.

Answer: 5

Explanation:
Divide the two numbers 25 and 5.
25/5 = 5
Thus the quotient is 5.

Question 10.
Divide 40 by 4.

Answer: 10

Explanation:
Divide the two numbers 40 and 4.
40/4 = 10
Thus the quotient is 10.

Question 11.
Divide 72 by 8.

Answer: 9

Explanation:
Divide the two numbers 72 and 8.
72/8 = 9
Thus the quotient is 9.

Lesson 8.8 Use Mental Math to Subtract

Explore and Grow

750 – 300 = ___
650 – 200 = ____
750 – 300 = ___
700 – 250 = ____
750 – 300 = ___
740 – 290 = ____

What patterns do you notice? Explain.

Answer:
750 – 300 = 450
650 – 200 = 450
750 – 300 = 450
700 – 250 = 450
750 – 300 = 450
740 – 290 = 450
Here you notice that all the answers are 450 for different patterns.

Think and Grow: Mental Math Strategies for Subtraction

Example
Find 433 – 198.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 150
One Way: Use compensation to change both numbers.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 151
433 + 2 = 435
198 + 2 = 200
435
-200
235
So, 433 – 198 = 235

Another Way: Use compensation to change one number.
Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 152
198 is close to 200 and it is easier to subtract 200.
433
-198 + 2 = 200
233
233 + 2 = 235
So, 433 – 198 = 235

Show and Grow

Use compensation to find the difference

Question 1.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 153

Answer:
Use compensation to change both numbers.
Add 4 to both numbers and use mental math strategy.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-153

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 154

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-154

Question 3.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 155

Answer:
Use compensation to change one number.
219 is close to 220 and it is easier to subtract 220.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-155

Apply and Grow: Practice

Use compensation to find the difference

Question 4.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 156

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-156

Question 5.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 157

Answer:
117 is close to 120. So it is easier to subtract 120.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-157

Use mental math to find the difference

Question 6.
643 – 115 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
643 + 5 = 648
115 + 5 = 120
648
-120
528
So, 643 – 115 = 528

Question 7.
863 – 257 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
863 + 3 = 866
257 + 3 = 260
866
-260
606

Question 8.
768 – 543 = ___

Answer:
Use compensation to change both numbers.
Add 2 to both numbers and use mental math strategy.
768 + 2 = 770
543 + 2 = 545
770
-545
225

Question 9.
688 – 414 = ___

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
688 + 1 = 689
414 + 1 = 415
689
-415
274

Question 10.
499 – 106 = ___

Answer:
Use compensation to change both numbers.
Add 4 to both numbers and use mental math strategy.
499 + 4 = 503
106 + 4 = 110
503
-110
393

Question 11.
495 – 162 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
495 + 3 = 498
162 + 3 = 165
498
-165
333

Question 12.
874 – 515 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
874 + 5 = 879
515 + 5 = 520
879
-520
359

Question 13.
637 – 228 = ___

Answer:
Use compensation to change both numbers.
Add 2 to both numbers and use mental math strategy.
637 + 2 = 639
228 + 2 = 230
639
-230
409

Question 14.
986 – 432 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
986 + 3 = 989
432 + 3 = 435
989
-435
554

Question 15.
A movie theater has 225 seats. 108 seats are not taken. How many seats are not taken?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 158

Answer:
Given that,
A movie theater has 225 seats. 108 seats are not taken.
225 – 108 = 117
Therefore 117 seats are not taken.

Question 16.
Reasoning
Your friend starts to find 741 – 295. What is the next step? Explain.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 159

Answer:
The next step is to subtract the original numbers.
741
-295
446

Think and Grow: Modeling Real Life

A softball coach has $325 for new equipment. She buys the catching gear. Does she have enough money left to buy the bat?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 402

Subtraction equation:
Compare:
The coach ___ has enough money to buy the bat.

Answer:
Given A softball coach has $325 for new equipment. She buys the catching gear.
325 + 1 = 326
219 + 1 = 220
326
-220
106
The cost of the bat is $109.
Thus she does not have enough money.

Show and Grow

Question 17.
A store owner has 550 T-shirts. He sells 333 of them. Then he receives an order for 168 T-shirts. Does he have enough T-shirts to complete the order?

Answer:
Given that,
A store owner has 550 T-shirts. He sells 333 of them.
550 – 333 = 217
Then he receives an order for 168 T-shirts.
217 – 168 = 49
Yes, he has enough T-shirts to complete the order.

Question 18.
The manager of a gaming center has $700 for new electronics. She buys the game system. Does she have enough money left for either of the other two items? If so, which one?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 160

Answer:
Given that,
The manager of a gaming center has $700 for new electronics. She buys the game system.
$399 + $169 = $568
700 – 568 = $132
Thus she has enough money left for either of the other two items.
She can buy the game system and a bundle of games.

DIG DEEPER!
How much more money does the manager need to buy both the television and the bundle of games?

Answer:
The cost of television is $379
The cost of the bundle of games = $169
379 + 169 = 548
Thus the manager need $548 to buy both the television and the bundle of games

Use Mental Math to Subtract Homework & Practice 8.8

Use compensation to find the difference.

Question 1.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 161

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
596 + 3 = 599
317 + 3 = 320
599
-320
279
So, 596 – 317 = 279

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 162

Answer:
Add 6 to 214 to make the subtraction easier.
685
-214 + 6 = 220
685
-220
465
So, 685 – 214 = 465

Use mental math to find the difference.

Question 3.
782 – 489 = ___

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
782 + 1 = 783
489 + 1 = 490
783
-490
293
So, 782 – 489 = 293

Question 4.
672 – 266 = ___

Answer:
Use compensation to change both numbers.
Add 4 to both numbers and use mental math strategy.
672 + 4 = 676
266 + 4 = 270
676
-270
406
So, 672 – 266 = 406

Question 5.
983 – 155 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
983 + 5 = 988
155 + 5 = 160
988
-160
820
So, 983 – 155 = 820

Question 6.
744 – 125 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
744 + 5 = 749
125 + 5 = 130
749
130
619
So, 744 – 125 = 619

Question 7.
967 – 619 = ___

Answer:
Use compensation to change both numbers.
Add 1 to both numbers and use mental math strategy.
967 + 1 = 968
619 + 1 = 620
968
-620
348
So, 967 – 619 = 348

Question 8.
854 – 517 = ___

Answer:
Use compensation to change both numbers.
Add 3 to both numbers and use mental math strategy.
854 + 3 = 857
517 + 3 = 520
857
-520
337
So, 854 – 517 = 337

Question 9.
472 – 215 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
472 + 5 = 479
215 + 5 = 220
479
220
259
So, 472 – 215 = 259

Question 10.
883 – 335 = ___

Answer:
Use compensation to change both numbers.
Add 5 to both numbers and use mental math strategy.
883 + 5 = 888
335 + 5 = 340
888
-340
548
So, 883 – 335 = 548

Question 11.
575 – 198 = ___

Answer:
Use compensation to change both numbers.
Add 2 to both numbers and use mental math strategy.
575 + 2 = 577
198 + 2 = 200
577
-200
377
So, 575 – 198 = 377

Question 12.
Reasoning
To find 765 – 246, Newton adds 5 to each number and then subtracts. To find the difference, Descartes adds 4 to each number, and then subtracts. Will they both get the correct answer? Explain.

Answer:
Given,
To find 765 – 246, Newton adds 5 to each number and then subtracts.
765 + 5 = 770
246 + 5 = 251
770 – 251 = 519
To find the difference, Descartes adds 4 to each number, and then subtracts.
765 + 4 = 769
246 + 4 = 250
769 – 250 = 519
Yes they both get the correct answer.

Question 13.
Modeling Real Life
A custodian has 350 desks to clean. She cleans 124 desks on the first floor and 147 desks on the second floor. Does she clean all of the desks?

Answer:
Given that,
A custodian has 350 desks to clean.
She cleans 124 desks on the first floor and 147 desks on the second floor.
124 + 147 = 271
350
-271
79
No, she did not clean all of the desks.

Question 14.
Modeling Real Life
A fashion designer has $ 725 to spend on new supplies. She buys the sewing machine. Does she have enough money left for either of the other two items? If so, which one?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 163

Answer:
Given,
A fashion designer has $ 725 to spend on new supplies. She buys the sewing machine.
725 – 495 = 230
Yes she has enough money left for either of the other two items
She can buy a Mannequin.

DIG DEEPER!
How much more money does the fashion designer need to buy both the mannequin and the fashion design software?

Answer:
Given,
The cost of the Mannequin is $129
The cost of the fashion design software is $329
129 + 329 = 458
458 – 230 = 228
Thus she needs 228 to buy both the mannequin and the fashion design software.

Review & Refresh

Draw equal groups. Then complete the equations.

Question 15.
3 groups of 6
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 164

Answer:
3 groups of 6 means 3 times 6.
6 + 6 + 6 = 18
3 × 6 = 18

Question 16.
4 groups of 9
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 165

Answer:
4 groups of 9 means 4 times 9.
9 + 9 + 9 + 9 = 36
4 × 9 = 36

Lesson 8.9 Subtract Three-Digit Numbers

Explore and Grow

Model the equation. Draw to show your model. Then find the difference.
694 – 418 = ___

Reasoning
How can you use an estimate to check whether your answer is reasonable?

Answer:
Find 694 – 418. Check whether your answer is reasonable.
694 = 700
418 = 400
700
-400
300
The difference is about 300.
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup
Big-ideas-math-answers-grade-3-chapter-8-img-1
Step 3: Check 276 is close to 300, so the answer is reasonable.

Think and Grow: Subtract Three-Digit Numbers

Example
Find 604 – 215. Check whether your answer is reasonable.
Step 1: Estimate. Round each number to the nearest hundred.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 166
600
-200
400
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 167
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-167
Step 3: Check 389 is close to 400, so the answer is reasonable.

Show and Grow

Find the difference. Check whether your answer is reasonable.

Question 1.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 168

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
300
200
100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Grade-3-Chapter-8-Add-and-subtract-digit-numbers-img-2
Step 3: Check 136 is close to 100, so the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 169

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
538 = 500
371 = 400
500 – 400 = 100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers img-3
Step 3: Check 167 is close to 100, so the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 170

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
500 – 300 = 200
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Bigideas-math-answer-key-3rd-grade-chapter-8-add-and-subtract-multi-digit-numbers-img-4
Step 3: Check 238 is close to 200, so the answer is reasonable.

Question 4.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 171

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
963 = 1000
429 = 400
1000 – 400 = 600
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-answers-3rd-grade-chapter-8-add-and-subtract-multi-digit-numbers-img-5
Step 3: Check 534 is close to 600, so the answer is reasonable.

Question 5.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 172

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
641 = 600
287 = 300
600 – 300 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Book-Grade-3-chapter-8-add-and-subtract-multi-digit-numbers-img-6
Step 3: Check 354 is close to 300, so the answer is reasonable.

Question 6.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 173

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
832 = 800
359 = 300
800 – 300 = 500
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-answers-3rd-grade-chapter-8-img-7
Step 3: Check 473 is close to 500, so the answer is reasonable.

Apply and Grow: Practice

Find the difference. Check whether your answer is reasonable.

Question 7.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 174

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
518 = 500
232 = 200
500 – 200 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-14
Step 3: Check 286 is close to 300, so the answer is reasonable.

Question 8.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 175

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
971 = 1000
320 = 300
1000 – 300 = 700
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-15
Step 3: Check 651 is close to 700, so the answer is reasonable.

Question 9.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 176

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
565 = 600
289 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-16
Step 3: Check 276 is close to 300, so the answer is reasonable.

Question 10.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 177

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
546 = 500
341 = 300
500 – 300 = 200
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-17
Step 3: Check 205 is close to 200, so the answer is reasonable.

Question 11.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 178

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
707 = 700
453 = 400
700 – 400 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-amswers-chapter-8-add-and-subtract-multi-digit-numbers-img-18
Step 3: Check 254 is close to 300, so the answer is reasonable.

Question 12.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 179

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
406 = 400
77 = 100
400 – 100 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-19
Step 3: Check 329 is close to 300, so the answer is reasonable.

Question 13.
Estimate: ____
552 – 381 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
552 = 600
381 = 400
600 – 400 = 200
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-20
Step 3: Check 171 is close to 200, so the answer is reasonable.

Question 14.
Estimate: ___
725 – 146 = ____

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
725 = 700
146 = 100
700 – 100 = 600
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-21
Step 3: Check 579 is close to 600, so the answer is reasonable.

Question 15.
Estimate: ___
800 – 486 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
486 = 500
800 – 500 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-book-grade-3-answers-chapter-8-add-and-subtract-multi-digit-numbers-img-22
Step 3: Check 314 is close to 300, so the answer is reasonable.

Question 16.
The number of rings on a tree is equal to its age. A redwood tree has 473 rings. A bristlecone pine tree has 806 rings. How much older is the bristlecone pine tree than the redwood tree?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 180

Answer:
Given that,
The number of rings on a tree is equal to its age. A redwood tree has 473 rings. A bristlecone pine tree has 806 rings
806
-473
333
Thus 333 older is the bristlecone pine tree than the redwood tree.

Question 17.
Writing
Explain how to regroup 408 to subtract 259.

Answer:
Working each column from right to left
9 is greater than 8 so you must regroup:
Take 1 from 4, so 4 becomes 3.
Add 10 to 0, so 0 becomes 10.
Take 1 from 10, so 10 becomes 9.
Add 10 to 8, so 8 becomes 18.
18 minus 9 is 9.
9 minus 5 is 4.
3 minus 2 is 1.
408
-259
149

Think and Grow: Modeling Real Life

How many more pennies does the class need to reach the goal?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 500
Subtraction equation:
The class needs to collect ___ more pennies.

Answer:
800
-444
356
The class needs to collect 356 more pennies.

Show and Grow

Question 18.
How many more campers can attend the summer camp?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 501

Answer:
Total summer camp openings = 405
Filled = 316
400 – 316 = 84
84 more campers can attend the summer camp.

Question 19.
A musician wants to buy a set of speakers that costs $672. She saves $224 each month for 2 months. How much money does she still need to save?

Answer:
Given that,
A musician wants to buy a set of speakers that costs $672.
She saves $224 each month for 2 months.
224 + 224 = 448
672 – 448 = 224
Thus she still need to save $224.

Question 20.
DIG DEEPER!
Newton has 442 packages to deliver. Descartes has 464. Newton delivers 174 packages, and Descartes delivers 188. Who is closer to finishing his deliveries?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 502

Answer:
Given,
Newton has 442 packages to deliver. Descartes has 464. Newton delivers 174 packages, and Descartes delivers 188.
442 – 174 = 268
464 – 188 = 276
Thus Newton is closer to finishing his deliveries.

Subtract Three-Digit Numbers Homework & Practice 8.9

Find the difference. Check whether your answer is reasonable.

Question 1.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 503

Answer:
The estimated number for 571 is 600.
The estimated number for 220 is 200.
600
-200
400
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Grade-3-Answer-Key-Chapter-8-Add-&-Subtract-Multi-Digit-Numbers-img-8
352 is close to 400. So the answer is reasonable.

Question 2.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 504

Answer:
The estimated number for 421 is 400.
The estimated number for 277 is 300.
400
-300
100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-3rd-Grade-Solutions-Chapter-8-Add-and-subtract-multi-digit-numbers-img-9
144 is close to 100. So the answer is reasonable.

Question 3.
Estimate: ___
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 505

Answer:
The estimated number for 534 is 500.
The estimated number for 186 is 200
500
-200
300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Bigideas-math-answers-3rd-grade-chapter-8-img-10
348 is close to 300. So the answer is reasonable.

Question 4.
Estimate: ___
690 – 298 = ___

Answer:
The estimated number for 690 is 700
The estimated number for 298 is 300.
700
-300
400
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-grade-3-solution-key-chapter-8-add-and-subtract-multi-digit-numbers-img-11
392 is close to 400. So the answer is reasonable.

Question 5.
Estimate: ___
613 – 472 = ___

Answer:
The estimated number for 613 is 600.
The estimated number for 472 is 500
600
-500
100
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
Big-ideas-math-answer-key-grade-3-8th-chapter-add-&-subtract-multi-digit-numbers-img-12
141 is close to 100. So the answer is reasonable.

Question 6.
Estimate: ___
835 – 189 = ___

Answer:
The estimated number for 835 is 800.
The estimated number for 189 is 200.
800
-200
600
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
BIM-Solution-Key-Grade-3-Chapter-8-img-13
646 is close to 600. So the answer is reasonable.

Question 7.
YOU BE THE TEACHER
Your friend says you have to regroup every time you subtract from a number that has a zero. Is your friend correct? Explain.

Answer:
Your friend says you have to regroup every time you subtract from a number that has a zero.
Yes, your friend is correct. Because whenever you subtract a number from 0 you have to regroup.

Question 8.
Modeling Real Life
How many more soup can labels does the school need to reach the goal?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 506

Answer:
To find how many more soup can label does the school need to reach the goal you have to subtract label collected from school soup.
1000 – 638 = 362

Question 9.
Modeling Real Life
Newton wants to buy a couch that costs $594. He saves $198 each month for 2 months. How much money does he still need to save?

Answer:
Given,
Newton wants to buy a couch that costs $594. He saves $198 each month for 2 months.
198 × 2 = $396
594 – 396 = -198
Therefore Newton need to save $198 to buy the couch.

Question 10.
DIG DEEPER!
Find the missing digits.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 507

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-507

Review & Refresh

Round the number to the nearest ten and to the nearest hundred.

Question 11.
64
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 508

Answer:
The number 64 nearest ten is 60.
The number nearest hundred to 64 is 60.

Question 12.
411
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 509

Answer:
The number 411 nearest ten is 410
The number nearest hundred to 411 is 400.

Lesson 8.10 Relate Addition and Subtraction

Explore and Grow

How can you find the missing number? How do you know you are correct?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 510

Reasoning
How are addition and subtraction related?

Answer:
497 – 358 = 136
Check the answer by using the addition model.
136
+358
497
So, the answer is reasonable.

Think and Grow: Relate Addition and Subtraction

Inverse operations are operations that “undo” each other. Addition and subtraction are inverse operations.
Example
Find 846 – 283. Use the inverse operation to check.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 511

Answer:
First, subtract 283 from 846 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-511

Example
Find 355 + 437. Use the inverse operation to check.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 512

Answer:
First, add 355 from 437 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-512

Show and Grow

Find the sum or difference. Use the inverse operation to check.

Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 183

Answer:
First, subtract 682 from 419 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-183
So, the answer is reasonable.

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 184

Answer:
First, add 169 from 745 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-184
So, the answer is reasonable.

Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 513

Answer:
First, add 376 from 238 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-513
So, the answer is reasonable.

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 514

Answer:
First, subtract 547 from 285 and then use the inverse operation to check the solution.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-514
So, the answer is reasonable.

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 515

Answer:
First, add 463 from 349 and then use the inverse operation to check the solution.
463
+349
812
Use the inverse operation to check the solution
812
-349
463
So, the answer is reasonable.

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 516

Answer:
First, subtract 790 from 317 and then use the inverse operation to check the solution.
790
-317
473
Use the inverse operation to check the solution
473
+317
790
So, the answer is reasonable.

Apply and Grow: Practice

Find the sum or difference. Use the inverse operation to check.

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 517

Answer:
First, subtract 857 from 567 and then use the inverse operation to check the solution.
857
-567
290
Use the inverse operation to check the solution
290
+567
857
So, the answer is reasonable.

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 518

Answer:
First, add 762 from 143 and then use the inverse operation to check the solution.
762
+143
905
Use the inverse operation to check the solution
905
-143
762
So, the answer is reasonable.

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 519

Answer:
First, add 653 from 217 and then use the inverse operation to check the solution.
653
+217
870
Use the inverse operation to check the solution
870
-217
653
So, the answer is reasonable.

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 520

Answer:
First, subtract 294 from 156 and then use the inverse operation to check the solution.
294
-156
138
Use the inverse operation to check the solution
138
+156
294
So, the answer is reasonable.

Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 521

Answer:
First, add 475 from 438 and then use the inverse operation to check the solution.
475
+438
913
Use the inverse operation to check the solution
913
-438
475
So, the answer is reasonable.

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 522

Answer:
First, subtract 514 from 386 and then use the inverse operation to check the solution.
514
-386
128
Use the inverse operation to check the solution
128
+386
514
So, the answer is reasonable.

Question 13.
Which one Doesn’t Belong? Which equation does not belong with the other three?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 523

Answer:
The fourth equation does not belong with the other three. Because it is not using the inverse operation of addition and subtraction.

Question 14.
Open-Ended
Write a subtraction equation that has a difference of 381.

Answer: 901 – 520 = 381
Take the number on your own and write the subtraction equation with the difference of 381.

Think and Grow: Modeling Real Life

A kayak costs $321. A customer pays $196 for the kayak after using a gift card. How much money is the gift card worth?
The gift card is worth $___.
Check:
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 524

Answer:
Given that,
A kayak costs $321. A customer pays $196 for the kayak after using a gift card.
321 – 196 = 125
Thus the gift card worth $125.

Show and Grow

Question 15.
You print 600 flyers for an event. You hand out some of them. There are 237 left. How many flyers did you hand out?

Answer:
Given that,
You print 600 flyers for an event. You hand out some of them. There are 237 left.
600 – 237 = 363
Thus you hand out 363 flyers.
363 + 237 = 600

Question 16.
A building has 163 floors. You start on the 28th floor. You go up in the elevator 126 floors. Then you go down 145 floors. On which floor do you end?

Answer:
Given that,
A building has 163 floors. You start on the 28th floor. You go up in the elevator 126 floors. Then you go down 145 floors.
126 – 28 = 98 floors
145 – 98 = 47
Thus you end at 47th floor.

Question 17.
DIG DEEPER!
A bus travels from Boston to Washington, D.C. On the way back, the bus stops in New York City. How many miles has the bus traveled in all? How many miles does the bus have left to travel?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 525

Answer:
Given that,
A bus travels from Boston to Washington, D.C. On the way back, the bus stops in New York City.
We have to find How many miles has the bus traveled in all
441 + 225 = 666 miles
Thus a bus travels 666 miles.
666 – 225 = 441 miles

Relate Addition and Subtraction Homework & Practice 8.10

Find the sum or difference. Use the inverse operation to check.

Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 526

Answer:
931
-544
387
Now you have to do the inverse operation.
387
+544
931

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 527

Answer:
623
+285
908
Now you have to do the inverse operation.
908
-285
623

Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 528

Answer:
523
+237
760
Now you have to do the inverse operation.
760
-237
523

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 529

Answer:
403
-252
151
Now you have to do the inverse operation.
151
+252
403

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 530

Answer:
612
+387
999
Now you have to do the inverse operation.
999
-387
612

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 531

Answer:
511
-371
140
Now you have to do the inverse operation.
140
+371
511

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 532

Answer:
437
+156
593
Now you have to do the inverse operation.
593
-156
437

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 533

Answer:
726
-362
364
Now you have to do the inverse operation.
364
+362
726

Question 9.
YOU BE THE TEACHER
Your friend uses an inverse operation to check her answer. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 510

Answer:
No, your friend is incorrect.
380
-159
221
Now you have to do the inverse operation.
221
+159
380

Question 10.
Which One Doesn’t Belong? Which does not belong with the other three?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 535

Answer:
208 + 475 = 683
Now you have to do the inverse operation.
683 – 475 = 208
The second figure does not belong to the other three expressions.

Question 11.
Modeling Real Life
A telescope costs $169. A customer pays $119 for the telescope after using a gift card. How much money is the gift card worth?

Answer:
Given that,
A telescope costs $169. A customer pays $119 for the telescope after using a gift card.
169
-119
50
The cost of the gift card is $50.

Question 12.
DIG DEEPER!
A train travels from Dallas to San Antonio. On the way back, the train stops in Austin. How many miles has the train travel? How many miles does the train have left to travel?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 536

Answer:
Given,
A train travels from Dallas to San Antonio. On the way back, the train stops in Austin.
274 – 79 = 195 miles
The train has traveled 195 miles.
79 miles left to travel from Austin to San Antonio.

Review & Refresh

Question 13.
Use the Distributive Propertytofind the area of the rectangle.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 537

Answer:
4 × 8 = 4 × (4 + 4)
4 × 8 = (4 × 4) + (4 × 4)
4 × 8 = 16 +16
4 × 8 = 32
Thus the area of the rectangle = 32 square foot.

Lesson 8.11 Problem Solving: Addition and Subtraction

Explore and Grow

You read 150 pages in three weeks.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 538

what does p represent?
p = ___

Answer:
Let 9 be the number of pages read
56 + 47 + p = 150
103 + p = 150
p = 150 – 103
p = 47

Construct Arguments
Explain to your partner how to find what n represents.
250 + n = 580

Answer:
Given the expression 250 + n = 580
n = 580 – 250
n = 330

Think and Grow: Using the Problem-Solving Plan
Example
Newton has 368 baseball cards. He gives away 139 of them. He buys 26 more. How many cards does he have now?

Understand the Problem

  • Newton has ___ cards.
  • He gives away of them.
  • He buys ___ more.
  • You need to find how many ___ he has now.

Answer:

  • Newton has 368 cards.
  • He gives away of them.
  • He buys 26 more.
  • You need to find how many cards he has now.

Make a Plan

How will you solve?

  • Subtract ___ from ___ to find how many ___ he has left after he gives some away.
  • Then add ___ to the difference to find how many he has now.

Answer:

  • Subtract 139 from 368 to find how many cards he has left after he gives some away.
  • Then add 26 to the difference to find how many he has now.

Solve
Draw a part-part-whole model and write an equation.
Use a letter to represent the unknown number.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 539

Newton has __ cards now.

Answer:
Step 1:
c is the unknown difference.
368 – 139 = c
c = 229
Step 2:
c = 229
229 + 26 = n
n = 225
Thus the unknown sum is 225.

Show and Grow

Question 1.
Explain how you can check whether your answer above is reasonable.

Answer:
You can check the answer by using addition and subtraction.
368
-139
229
Now check whether the answer is correct or not.
229
+139
368
So, the answer is reasonable.

Apply and Grow: Practice

Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.

Question 2.
A baker makes 476 muffins. He sells 218 of them. Then he makes 390 more. How many muffins does the baker have now?

Answer:
Given,
A baker makes 476 muffins. He sells 218 of them.
476
-218
258
Then he makes 390 more.
390
+258
648
Thus the baker has 348 muffins now.

Question 3.
Newton knocks down 146 pins in his first bowling game. He knocks down 19 more pins in his second game than in his first game. How many pins does he knock down in all?

Answer:
Given that,
Newton knocks down 146 pins in his first bowling game. He knocks down 19 more pins in his second game than in his first game.
146
+19
165
He knocks down 165 pins in his second game.
To find the total number of pins we have to the points in the first game and second game.
146
+165
311
Thus he knockdowns 311 pins in all.

Question 4.
You are traveling to a campground that is 243 miles away. You travel 155 miles in the morning and 59 miles in the afternoon. How many more miles do you need to travel before you get to the campground?

Answer:
Given that,
You are traveling to a campground that is 243 miles away.
You travel 155 miles in the morning and 59 miles in the afternoon.
155
+59
214
243
-214
029
Thus you need to travel 29 miles to get to the campground.

Question 5.
There are 205 lawn tickets and 585 bleacher tickets sold for a concert. There are 680 fewer VIP tickets sold than lawn and bleacher tickets combined. How many VIP tickets are sold?

Answer:
Given,
There are 205 lawn tickets and 585 bleacher tickets sold for a concert.
205 + 585 = 790
There are 680 fewer VIP tickets sold than lawn and bleacher tickets combined.
790
-680
110
Thus 110 VIP tickets are sold.

Think and Grow: Modeling Real Life

How many more people went to see the movie on Friday than on Thursday and Saturday combined?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 542
Understand the problem:
Make a plan:
Solve:
__ more people went to see the movie on Friday than on Thursday and Saturday combined.

Answer:
Thursday – 346 people
Saturday – 512 people
346 + 512 = 858 people
897
-858
39
39 more people went to see the movie on Friday than on Thursday and Saturday combined.

Show and Grow

Question 6.
How many more people used the ferry on Friday than on Saturday and Sunday combined?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 543
Explain how you can check whether your answer is reasonable.

Answer:
Friday – 903 passengers
Saturday – 624 passengers
Sunday – 255 passengers
624
+255
879
Now subtract 879 from 903.
903
-879
024

Problem Solving: Addition and Subtraction Homework & Practice 8.11

Write equations to solve. Use letters to represent the unknown numbers. Check whether your answer is reasonable.

Question 1.
Newton has 387 tokens, and Descartes has 295. They use a total of 461 tokens. How many tokens do they have now?

Answer: 222 tokens

Explanation:
Given that,
Newton has 387 tokens, and Descartes has 295.
387 + 295 = 682
They use a total of 461 tokens.
682 – 461 = 222 tokens
Thus they have 222 tokens now.

Question 2.
There are 125-second graders and 118 third graders at a museum. There are 249 more adults than students at the museum. How many adults are at the museum?

Answer:
Given that,
There are 125-second graders and 118 third graders at a museum.
125 + 118 = 243 graders
There are 249 more adults than students at the museum.
243 + 249 = 492
Therefore 492 adults are at the museum.

Question 3.
You received 171 votes in a coloring contest. Your friend received 24 fewer votes than you. How many people voted for you and your friend in all?

Answer:
Given,
You received 171 votes in a coloring contest. Your friend received 24 fewer votes than you.
171 + 24 = 195 votes
The number of votes for your friend is 195.
171+ 195 = 366 votes
Thus 366 people voted for you and your friend.

Question 4.
Writing
Write and solve a two-step problem that can be solved using addition or subtraction.

Answer:
You bought 10 packs of sketches and your friend bought 4 packs less than you. Each pack contains 10 sketches. Find how many sketches you and your friend bought in all.
Sol: You bought 10 packs of sketches and your friend bought 4 packs less than you.
10 – 4 = 6 packs
Your friend bought 6 packs.
Each pack contains 10 sketches.
10 + 6 = 16
16 × 10 = 160
Thus you and your friend bought 160 sketches.

Question 5.
Modeling Real Life
How many more fish were caught on Sunday than on Friday and Saturday combined?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 544
Explain how you can check whether your answer is reasonable.

Answer:
Number of fish caught on Friday and Saturday = 127 + 244 = 371
Number of fish caught on Sunday = 564
564 – 371 = 193
193 more fish were caught on Sunday than on Friday and Saturday combined.

Review & Refresh

Question 6.
Use the multiplication table.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 545
Describe the pattern in the shaded row and column.
What property explains this pattern?

Answer: The pattern shows that it is the multiple of 5.
5 × 1 = 5
5 × 2 = 10
5 × 3 = 15
5 × 4 = 20
5 × 5 = 25

Add and Subtract Multi-Digit Numbers Performance Task

Your school holds a talent show.

Question 1.
You and your friend hand out programs to guests before the show. You each start with 250 programs. There are 114 programs left. How many programs did you and your friend hand out?

Answer:
250 – 114 = 136 programs
Thus you and your friend hand out 136 programs.
136 + 114 = 250 programs

Question 2.
75 students wait backstage to perform in the show. There are 336 children, 125 adults, and 14 teachers in the audience.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 216
a. How many people are at the talent show in all? Explain how to use addition properties to find the sum.

Answer:
Given,
75 students wait backstage to perform in the show. There are 336 children, 125 adults, and 14 teachers in the audience.
75 + 336 + 125 + 14 = 550
Thus there are 550 people are in the talent show.

b. Four students perform in each of the first 4 acts. How many students still need to perform?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 217

Answer: 16 students

Explanation:
Given,
Four students perform in each of the first 4 acts.
4 × 4 = 16
Thus 16 students need to perform.

c. Each performer is given a juice box backstage. Juice boxes come in packages of 10. How many packages did the teachers buy? How many juice boxes are left?

Answer:
Given,
Each performer is given a juice box backstage. Juice boxes come in packages of 10.
1 box – 10 packages
16 × 1 = 16 boxes
16 × 10 = 160 packages

Add and Subtract Multi-Digit Numbers Activity

Three in a Row: Addition and Subtraction

Directions:
1. Players take turns.
2. On your turn, spin both spinners. Add or subtract the two numbers. Cover the sum or difference.
3. If the sum or difference is already covered, then you lose your turn.
4. The first player to get three counters in a row, horizontally, vertically, or diagonally, wins!
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 218

Answer:
Sums:
547 + 107 = 654
547 + 338 = 885
547 + 262 = 809
I got the three counters vertically.

Add and Subtract Multi-Digit Numbers Chapter Practice

8.1 Identify Addition Properties

Identify the property.

Question 1.
59 + 0 = 59

Answer:
It shows the Addition Property of Zero. The Addition Property of Zero defines the sum of any number and 0 is that number.

Question 2.
(14 + 32) + 6 = 14 + (32 + 6)

Answer:
It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

Question 3.
27 + 51 = 51 + 27

Answer:
Commutative Property of addition Changing the grouping of addends does not change the sum.

Question 4.
Structure
Which equations show the Commutative Property of Addition?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 219

Answer:
64 + 12 = 12 + 64 – Commutative Property of addition changing the grouping of addends does not change the sum.
71 + 0 = 71 – It shows the Addition Property of Zero. The Addition Property of Zero defines the sum of any number and 0 is that number.
(56 + 21) + 34 = 56 + (21 + 34) – It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.
26 + (41 + 4) = 4 + (26 + 41) – It satisfies the Associative Property of Addition. It is defined as changing the grouping of addends does not change the sum.

8.2 Use Number Lines to Add

Question 5.
Find 648 + 37.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 220

Answer:
Use the count on strategy. Start at 648. Count on by tens, then by ones.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-220

8.3 Use Mental Math to Add

Use mental math to find the sum.

Question 6.
192 + 107 = ___

Answer:
You can find the sum of 192 and 107 by using mental math strategies.
192 – 2 = 190
107 + 2 = 109
190
+109
299

Question 7.
676 + 114 = ___

Answer:
You can find the sum of 676 and 114 by using mental math strategies.
676 + 4 = 680
114 – 4 = 110
680
+110
790

Question 8.
716 + 279 = ___

Answer:
You can find the sum of 716 and 279 by using mental math strategies.
716 – 1 = 715
279 + 1 = 280
715
+280
995

Question 9.
501 + 468 = ___

Answer:
You can find the sum of 501 and 468 by using mental math strategies.
501 – 1 = 500
468 + 1 = 469
500
+469
969

Question 10.
527 + 343 = ___

Answer:
You can find the sum of 527 and 343 by using mental math strategies.
527 + 3 = 530
343 – 3 = 340
530
+340
870

Question 11.
441 + 189 = ___

Answer:
You can find the sum of 441 and 189 by using mental math strategies.
441 – 1 = 440
189 + 1 = 190
440
+190
630

8.4 Use Partial Sums to Add

Use partial sums to add.

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 221

Answer:
586 = 500 + 80 + 6
107 = 100 + 00 + 7
693 = 600 + 80 + 13

Question 13.
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 222

Answer:
647 = 600 + 40 + 7
293 = 200 + 90 + 3
940 = 800 + 130 + 10

Question 14.
Modeling Real Life
On Earth, your cousin weighs 207 pounds less than he would on Jupiter. Your cousin weighs 135 pounds on Earth. How much would he weigh on Jupiter?
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 223

Answer:
Given,
On Earth, your cousin weighs 207 pounds less than he would on Jupiter. Your cousin weighs 135 pounds on Earth.
207 + 135 = 342
Thus your cousin would weigh 342 pounds on Jupiter.

8.5 Add Three-Digit Numbers

Find the sum. Check whether your answer is reasonable.

Question 15.
Estimate: ___
326 + 490 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
326 = 300
490 = 500
300 + 500 = 800
The sum is about 800.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
326
+490
816
816 is close to 800. So, the answer is reasonable.

Question 16.
Estimate: ___
657 + 189 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
657 = 700
189 = 200
600 + 200 = 800
The sum is about 800.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
657 + 189 = 846
846 is close to 800. So, the answer is reasonable.

Question 17.
Estimate: ___
543 + 261 = ___

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
543 = 500
261 = 300
500 + 300 = 800
The sum is about 800.
Step 2: Find the sum. Add the ones, tens, then the hundreds.
543 + 261 = 804
804 is close to 800. So, the answer is reasonable.

8.6 Add Three or More Numbers

Question 18.
Estimate: ___
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 224

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
78 = 100
433 = 400
367 = 400
100 + 400 + 400 = 900
The sum is about 900.

Question 19.
Estimate: ___
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 225

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
194 = 200
151 = 200
244 = 200
231 = 200
200 + 200 + 200 + 200 = 800
The sum is about 800.

Question 20.
Estimate: ___
Big Ideas Math Answer Key Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 226

Answer:
Step 1: Estimate. Round each addend to the nearest hundred.
373 = 400
329 = 300
118 = 100
61 = 100
400 + 300 + 100 + 100 = 900
The sum is about 900.

8.7 Use Number Lines to Subtract

Question 21.
Find 856 – 29.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 511

Answer:
Use the count back strategy. Start at 856. Count back by tens, then by ones.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-511

Question 22.
Structure
Write the equation shown by the number line.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 512

Answer:
By seeing the above number we can find the subtraction equation.
764 – 50 = 714
714 – 4 = 710
710 – 3 = 707
The subtraction equation is 764 – 57 = 707

8.8 Use Mental Math to Subtract

Use mental math to find the difference.

Question 23.
957 – 619 = ___

Answer:
957 – 7 = 950
619 + 7 = 626
950
-626
324
The difference is 324.

Question 24.
831 – 415 = ___

Answer:
831 – 1 = 830
415 + 1 = 416
830
-416
414
The difference is 414.

Question 25.
876 – 366 = ___

Answer:
876 – 6 = 870
366 + 6 = 372
870 – 372 = 498

Question 26.
636 – 317 = ___

Answer:
636 + 6 = 642
317 – 6 = 311
642 – 311 = 331
The difference is 331.

Question 27.
965 – 528 = ___

Answer:
528 – 8 = 520
965 + 8 = 973
973 – 520 = 453
The difference is 453.

Question 28.
384 – 118 = ____

Answer:
684 + 4 = 688
118 – 4 = 114
688 – 114 = 574
The difference is 574.

8.9 Subtract Three-Digit Numbers

Find the difference. Check whether your answer is reasonable.

Question 29.
Estimate: ___
963 – 51 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
963 = 1000
51 = 100
1000 – 100 = 900
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
963 -51 = 912
912 is close to 900. So, the answer is reasonable.

Question 30.
Estimate: ___
878 – 594 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
878 = 900
594 = 600
900 – 600 = 300
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
878 – 594 = 284
284 is close to 300. So, the answer is reasonable.

Question 31.
Estimate: ___
766 – 297 = ___

Answer:
Step 1: Estimate. Round each number to the nearest hundred.
766 = 800
297 = 300
800 – 300 = 500
Step 2: Find the difference.
Subtract the ones, then the tens, then the hundreds. There are not enough ones or tens to subtract, so regroup.
766 – 297 = 469
469 is close to 500. So, the answer is reasonable.

Question 32.
YOU BE THE TEACHER
Your friend finds 760 – 482. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 513

Answer: No your friend is not correct.
BIM Answer Key for Grade 3 Chapter 8 img-13

8.10 Relate Addition and Subtraction

Find the sum or difference. Use the inverse operation to check.

Question 33.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 514

Answer:
649
+227
876
Now you have to do the inverse operation of the sum.
876
-227
649

Question 34.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 515

Answer:
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 515
288
Now you have to do the inverse operation of the difference.
288
+517
805

8.11 Problem Solving: Addition and Subtraction

Question 35.
There are 532 dogs enrolled in police academies. 246 dogs graduate in July, and 187 dogs graduate in August. How many dogs still need to graduate?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 516

Answer:
Given that,
There are 532 dogs enrolled in police academies.
246 dogs graduate in July, and 187 dogs graduate in August.
246 + 187 = 433
532
-433
99
Thus 99 dogs are still needed to graduate.

Add and Subtract Multi-Digit Numbers Cumulative Practice 1 – 8

Question 1.
Which numbers round to 300 when rounded to the nearest hundred?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 517

Answer: The numbers round to 300 when rounded to the nearest hundred is
298, 309, 347
Thus the correct answer is the option a, b, c.

Question 2.
You buy 18 cups of yogurt. The yogurt is sold in packs of 6 cups. How many packs of yogurt do you buy?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 519

Answer:
Given,
You buy 18 cups of yogurt. The yogurt is sold in packs of 6 cups.
18/6 = 3 packs
Thus the correct answer is option c.

Question 3.
A bedroom floor is 9 feet long and 8 feet wide. What is the area of the bedroom floor?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 520

Answer:
Given,
A bedroom floor is 9 feet long and 8 feet wide.
The area of the rectangle = l × w
A = 9ft × 8ft
A = 72 sq. ft
Thus the area of the bedroom floor is 72 sq. ft.

Question 4.
Your friend says 458 – 298 = 160. How can you use inverse operations to check your friend’s answer? Is your friend correct?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 544

Answer:
Your friend says 458 – 298 = 160.
Your friend is correct.
The correct answer is option a.

Question 5.
Your friend says she needs (9 × 3) + (3 × 9) = 27 × 27 = 54 tiles to make the design. Why is her thinking incorrect?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 545

Answer:
Your friend says she needs (9 × 3) + (3 × 9) = 27 × 27 = 54 tiles to make the design.
Her thinking is incorrect because she needs to add 27 and 27 but she multiplied.
(9 × 3) + (3 × 9) = 27 + 27 = 54

Question 6.
Part A
What is the least number that can be made with the digits 7, 9, and 8 using each digit only once?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 546
Your friend says the greatest number he can make with the digits 7, 9, and 8 using each digit only once is 879. Is he correct? If not, correct his answer. Explain.

Answer:
Given,
Your friend says the greatest number he can make with the digits 7, 9, and 8 using each digit only once is 879
Your friend is incorrect because the greatest number with the digits 7, 9, and 8 is 987.

Question 7.
Which equation is shown by the number line?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 547

Answer:
The count starts from 0.
The count jumps from 0 and skips for every 3s.
3 × 7 = 21
Thus the correct answer is option c.

Question 8.
Find the sum.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 548

Answer:
548
+372
920

Question 9.
Which equations show the Associative Property of Addition?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 549

Answer:
According to the associative property of addition, the sum of three or more numbers remains the same regardless of how the numbers are grouped.
Options A and D show the equation for the Associative Property of Addition.

Question 10.
A teacher takes 7 students on a field trip. Each student pays $5. How much money does the teacher collect in all?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 550

Answer:
Given,
A teacher takes 7 students on a field trip. Each student pays$5.
7 × $5 = $35
Thus the teacher collects $35 in all.
The correct answer is option c.

Question 11.
What is the area of the shape?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 551

Answer:
You can divide the figure into two parts.
Figure 1:
l = 4m
b = 3m
A = l × b
A = 4 × 3 = 12 sq. m
Figure 2:
l = 8m
b = 3m
A = l × b
A = 8 × 3 = 24 sq. m
Add the area of both the figures 12 + 24 = 36 sq.m
Thus the correct answer is option D.

Question 12.
There are 459 girls and 552 boys in a school. How many more boys are there than girls?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 552

Answer:
Given that,
There are 459 girls and 552 boys in school.
Subtract the number of girls from the number of boys.
552
-459
93
Thus the correct answer is option b.

Question 13.
Look at the pattern. What rule was used to make the pattern?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 553

Answer:
1, 3, 9, 27 are the multiples of 3.
Thus the correct answer is option c.

Question 14.
A smoothie shop sells 368 smoothies in July and 205 smoothies in August. About how many more smoothies did the shop sell in July than in August?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 554

Answer:
Given that,
A smoothie shop sells 368 smoothies in July and 205 smoothies in August.
368
-205
163
Thus the shop sell 163 smoothies in July than in August.

Question 15.
Which shape does not have an area of 16 square units?
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 555

Answer:
There are 16 counters in the first figure
2 × 8 = 16
There are 16 counters in the second figure
4 × 4 = 16
There are 15 counters in the third figure
5 × 3 = 15
Thus the correct answer is option c.

Complete the table

Question 16.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 556

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-556
You can complete the table by multiplying the rows and columns.

Question 17.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 557

Answer:
You can complete the table by multiplying the rows and columns.
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-557

Add and Subtract Multi-Digit Numbers Steam Performance Task 1-8

Question 1.
The carpeting in the third-grade classrooms of an elementary school is being replaced. One roll of carpet covers 100 square yards. The map shows the classrooms that will receive the new carpet.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 560
a. Explain two different ways to find the area of Classroom A.

Answer:
You can find the area of classroom A by using the composite figure.
The shape of classroom A is square.
a = 10 yd
Area of the classroom is a × a
A = 10 × 10 = 100 sq.yd
Thus the area of the classroom A is 100 sq. yd
Another way:
l = 10 yd
w = 3 yd
A = 10 × 3 = 30 sq. yd
l = 10 yd
A = 8 × 4 = 32 sq. yd
l = 10 yd
w = 3 yd
A = 10 × 3 = 30 sq. yd

b. Find the total area of all of the classrooms in square yards.

Answer:
Area of the classroom is a × a
A = 10 × 10 = 100 sq.yd
Thus the area of the classroom A is 100 sq. yd
Area of Classroom D = 10 × 7 = 70 sq.yd
Area of Classroom B = 11 × 7 = 77 sq.yd
Area of Classroom C = 12 × 7 = 84 sq. yd, 8 × 3 = 24 sq.yd
Area of Classroom C = 84 + 24 = 108 sq. yd
Total area of the classrooms = 100 + 70 + 77 + 108 = 355 sq. yd

c. Estimate the number of rolls of carpet needed for the classrooms. Explain.

Answer:
There are 4 classrooms. So, the estimated number of rolls of carpet is 4.

d. Find the area of the hallway in square yards. Is there enough carpet for the hallway?

Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 561
Answer:
First, divide the hallway into 3 parts.
i. It is in the form of a square.
a = 3 yd
A = 3 × 3 = 9 sq.yd
ii. It is in the form of rectangle
A = l × b
A = 5 × 4 = 20 sq.yd
iii. It is in the form of rectangle
A = l × b
A = 5 × 3 = 15 sq.yd
The area of the hallway in square yards = 9 + 20 + 15 = 44 sq.yd

Question 2.
Each school keeps a record of the total number of students in each class and grade.
a. Use the number of students in your class to estimate the total number of students in your grade. Explain.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 562

Answer:
Let the number of students in your class is 47.
By this, we can estimate the total number of students in your grade i.e., 50.
The total number of students in your grade is 50.

b. Use the table to write the number of students in each grade of your school.
Big Ideas Math Solutions Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers 563

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-563
You can prepare the table by estimating the number of students in your school.

c. How does your estimate compare to the actual number of students in your grade? Explain.

Answer: You can compare the number of students in the above table with the actual number of students in your grade.

d. What is the total number of students in your school?

Answer:
Add the number of students of all the grades
40+ 50 + 52 + 47 + 47 + 50 = 286
Thus there are 286 students in your school.

e. Write and answer a question using the information from the table above.

Answer:
Compare the number of students in grade 3 with the actual number of students in your school in grade 3?
The estimated number of students in grade 3 in the above table is 50.
The actual number of students in your school is 48.
50 – 48 = 2

f. What is one reason your principal may want to know the total number of students in your class, grade, or school?

Answer: Shaping a vision of academic success for all students.

Conclusion:

I wish that the information given here regarding Big Ideas Math Answers Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers is helpful for you. This pdf will help to score good marks in the exam. Get the solutions of BIM Book Grade 3 Chapter 8 Add and Subtract Multi-Digit Numbers from here. Stay tuned to our page to know the solution and brief explanation for other chapters of grade 3.

Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

High School Students who are seeking homework help can refer to this Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume Ultimate Guide. By using this ultimate preparation material, you all can easily understand and learn the concepts covered in chapter 11 Circumference, Area, and Volume. Enhance your subject knowledge and score high in various exams after referring and practicing the questions and solutions provided in the BIM Geometry Ch 11 Circumference, Area, and volume Textbook Answer Key Pdf. In this article, you will find the aligned Lesson-wise Big Ideas Geometry Answers Ch 11 in pdf format to download & access offline too for free of cost.

Big Ideas Math Book Geometry Answer Key Chapter 11 Circumference, Area, and Volume

Access the Lesson-wise Big Ideas Math Geometry Chapter 11 Solution Key and solve all easy & complex questions of Circumference, Area, and Volume with ease. Also, you can excel in math concepts by practicing the BIM Geometry Textbook Answer Key. Ace up your preparation with BIM Geometry Ch 11 Circumference, Area, and Volume Answer Key and check your knowledge accordingly for better scores and math skills.

Circumference, Area, and Volume Maintaining Mathematical Proficiency

Find the surface area of the prism.

Question 1.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 1

Answer:
The surface area of the prism = 158.

Explanation:
In the above-given question,
given that,
l = 5 ft, w = 8 ft, and h = 3 ft.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(5×8 + 5×3 + 8×3).
surface area = 2(40 + 15 + 24).
surface area = 2(79).
surface area = 158.

Question 2.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 2

Answer:
The surface area of the triangular prism = 68m.

Explanation:
In the above-given question,
given that,
l = 10 m, p = 4 m, and h = 10.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(6 + 8) + 4(10).
surface area = 2(14) + 40.
surface area = 28 + 40.
surface area = 68m.

Question 3.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 3

Answer:
The surface area of the triangular prism = 42m.

Explanation:
In the above-given question,
given that,
w = 10 cm, p = 4 cm, and h = 5 cm, l = 6 cm.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(6 + 5) + 4(5).
surface area = 2(11) + 20.
surface area = 22 + 20.
surface area = 42cm.

Find the missing dimension.

Question 4.
A rectangle has a perimeter 0f 28 inches and a width of 5 inches. What is the length of the rectangle?

Answer:
The length of the rectangle = 9 in.

Explanation:
In the above-given question,
given that,
A rectangle has a perimeter of 28 inches and a width of 5 inches.
length of the rectangle = p/2 – w.
length = 28/2 – 5.
where perimeter = 28 in, and w = 5 in.
length = 14 – 5.
length = 9.
so the length of the rectangle = 9 in.

Question 5.
A triangle has an area of 12 square centimeters and a height of 12 centimeters. What is the base of the triangle?

Answer:
The base of the triangle = 2 cm.

Explanation:
In the above-given question,
given that,
A triangle has an area of 12 sq cm and a height of 12 cm.
The base of the triangle = 2(A)/h.
base = 2(12)/12.
base = 24/12.
base = 2cm.
so the base of the triangle = 2 cm.

Question 6.
A rectangle has an area of 84 square feet and a width of 7 feet. What is the length of the rectangle?

Answer:
The length of the rectangle = 12 ft.

Explanation:
In the above-given question,
given that,
A rectangle has an area of 84 sq ft and a width of 7 feet.
area of the rectangle = l x b.
84 = l x 7.
l = 84/7.
l = 12.
so the length of the rectangle = 12 ft.

Question 7.
ABSTRACT REASONING
Write an equation for the surface area of a Prism with a length, width, and height of x inches. What solid figure does the prism represent?

Answer:
The surface area of a prism = 2(lw + wh + lh).

Explanation:
In the above-given question,
given that,
length = l, width = w, and height = x inches.
the surface area of the prism = 2(lw + wh + lh).
the solid figure does the prism represent the rectangular prism.

Circumference, Area, and Volume Monitoring Progress

Draw a net of the three-dimensional figure. Label the dimensions.

Question 1.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 4

Answer:
The surface area of the prism = 64 cm.

Explanation:
In the above-given question,
given that,
l = 2 ft, w = 4 ft, and h = 4 ft.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(2×4 + 4×4 + 4×2).
surface area = 2(8 + 16 + 8).
surface area = 2(32).
surface area = 64.

Question 2.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 5

Answer:
The surface area of the prism = 392 m.

Explanation:
In the above-given question,
given that,
l =8 m, w = 12 m, and h = 5 m.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(8×12 + 12×5 + 5×8).
surface area = 2(96 + 60 + 40).
surface area = 2(196).
surface area = 392.

Question 3.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 6

Answer:
The surface area of the triangular prism = 170 in.

Explanation:
In the above-given question,
given that,
B = 10 in, p = 15 in, and h = 15 in, l = 10 in.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(10) + 15(10).
surface area = 2(10) + 150.
surface area = 20 + 150.
surface area = 170 in.

11.1 Circumference and Arc Length

Exploration 1

Finding the Length of a Circular Arc

Work with a partner: Find the length of each red circular arc.

a. entire circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 7

Answer:

b. one-fourth of a circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 8
Answer:

c. one-third of a circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 9
Answer:

d. five-eights of a circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 10
Answer:

Exploration 2

Using Arc Length

Work with a partner: The rider is attempting to stop with the front tire of the motorcycle in the painted rectangular box for a skills test. The front tire makes exactly one-half additional revolution before stopping. The diameter of the tire is 25 inches. Is the front tire still in contact with the painted box? Explain.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11

Answer:

Communicate Your Answer

Question 3.
How can you find the length of a circular arc?
Answer:
The length of a circular arc = 2
LOOKING FOR REGULARITY IN REPEATED REASONING
To be proficient in math, you need to notice if calculations are repeated and look both for general methods and for shortcuts.
Answer:

Question 4.
A motorcycle tire has a diameter of 24 inches. Approximately how many inches does the motorcycle travel when its front tire makes three-fourths of a revolution?
Answer:

Lesson 11.1 Circumference and Arc Length

Monitoring Progress

Question 1.
Find the circumference of a circle with a diameter of 5 inches.

Answer:
Circumference C = πd
C = 3.14x 5 = 15.7 in

Question 2.
Find the diameter of a circle with a circumference of 17 feet.

Answer:
Diameter d = C/π
d = 17/π = 5.41 ft

Find the indicated measure.

Question 3.
arc length of \(\widehat{P Q}\)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 12

Answer:
arc length of \(\widehat{P Q}\) is 5.887

Explanation:
\(\widehat{P Q}\) = \(\frac { 75 }{ 360 } \) . π(9)
= 5.887

Question 4.
circumference of ⊙N
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 13

Answer:
arc length of LM/C = LM/360
61.26/C = 270/360
C = 81.68

Question 5.
radius of ⊙G
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 14

Answer:
arc length of EF = \(\frac { 60 }{ 360 } \) • 2πr
10.5 = \(\frac { 1 }{ 6 } \) • 2πr
r = 10.02

Question 6.
A car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet?

Answer:
The car tire have to make 69 revolutions to travel 500 ft.

Explanation:
Circumference C = 2πr = πd
C = 28π
Distance travelled = number of revolutions x C
500 x 12 = number of revolutions x 28π
number of revolutions = 68.2

Question 7.
In Example 4. the radius of the arc for a runner on the blue path is 44.02 meters, as shown in the diagram. About how far does this runner travel to go once around the track? Round to the nearest tenth of a meter.

Answer:

Question 8.
Convert 15° to radians.

Answer:
15° = 15 . \(\frac { π radians }{ 180° } \) = \(\frac { π }{ 12 } \) radians

Question 9.
Convert \(\frac{4 \pi}{3}\) radians to degrees.

Answer:
\(\frac{4 \pi}{3}\) radians = \(\frac{4 \pi}{3}\) radians . \(\frac { 180° }{ π radians } \) = 240 degrees

Exercise 11.1 Circumference and Arc Length

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The circumference of a circle with diameter d is C = _______ .
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 1

Question 2.
WRITING
Describe the difference between an arc measure and an arc length.

Answer:
An arc measure is measured in degrees while an arc length is the distance along an arc measured in linear units.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 10, find the indicated measure.

Question 3.
circumference of a circle with a radius of 6 inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 3

Question 4.
diameter of a circle with a circumference of 63 feet

Answer:
C = 63 ft
πd = 63
d = 20.05

Question 5.
radius of a circle with a circumference of 28π
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 5

Question 6.
exact circumference of a circle with a diameter of 5 inches

Answer:
C = πd
C = 5π = 15.707

Question 7.
arc length of \(\widehat{A B}\)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 15
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 7

Question 8.
m\(\widehat{D E}\)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 16

Answer:
\(\frac { arc length of DE }{ 2πr } \) = \(\frac { DE }{ 360 } \)
\(\frac { 8.73 }{ 2π(10) } \) = \(\frac { DE }{ 360 } \)
DE = 50.01°

Question 9.
circumference of ⊙C
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 17
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 9

Question 10.
radius of ⊙R
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 18
Answer:
\(\frac { arc length of LM }{ 2πr } \) = \(\frac { LM }{ 360 } \)
\(\frac { 38.95 }{ 2πr } \) = \(\frac { 260 }{ 360 } \)
r = 8.583

Question 11.
ERROR ANALYSIS
Describe and correct the error in finding the circumference of ⊙C.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 19
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 11

Question 12.
ERROR ANALYSIS
Describe and correct the error in finding the length of \(\widehat{G H}\).
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 20

Answer:
\(\frac { arc length of GH }{ 2πr } \) = \(\frac { m GH }{ 360 } \)
\(\widehat{G H}\). = \(\frac { 5 }{ 24 } \) . 2π(10)
= 13.08

Question 13.
PROBLEM SOLVING
A measuring wheel is used to calculate the length of a path. The diameter of the wheel is 8 inches. The wheel makes 87 complete revolutions along the length of the path. To the nearest foot, how long is the path?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 21
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 13

Question 14.
PROBLEM SOLVING
You ride your bicycle 40 meters. How many complete revolutions does the front wheel make?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 22

Answer:
Circumference of the front wheel = 2π(32.5)
= 65π cm
Distance covered = 40 m = 40 x 100 = 4000 cm
Number of revolutions = \(\frac { 4000 }{ 65π } \) = 19.58

In Exercises 15-18 find the perimeter of the shaded region.

Question 15.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 23
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 15

Question 16.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 24
Answer:
Two horizontal edges are 2 . 3 = 6
Circumference of circle = 2π(3) = 6π
The perimeter of the shaded region = 6 + 6π

Question 17.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 25
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 17

Question 18.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 26
Answer:

In Exercises 19 – 22, convert the angle measure.

Question 19.
Convert 70° to radians.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 19

Question 20.
Convert 300° to radians.

Answer:
300 • (\(\frac { π }{ 180 } \)) = \(\frac { 5π }{ 3 } \) radian

Question 21.
Convert \(\frac{11 \pi}{12}\) radians to degrees.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 21

Question 22.
Convert \(\frac{\pi}{8}\) radian to degrees.

Answer:
\(\frac { π }{ 8 } \) • \(\frac { 180 }{ π } \)) = 22.5°

Question 23.
PROBLEM SOLVING
The London Eye is a Ferris wheel in London, England, that travels at a speed of 0.26 meter per second. How many minutes does it take the London Eye to complete one full revolution?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 27
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 23

Question 24.
PROBLEM SOLVING
You are planning to plant a circular garden adjacent to one of the corners of a building, as shown. You can use up to 38 feet of fence to make a border around the garden. What radius (in feet) can the garden have? Choose all that apply. Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 28
(A) 7
(B) 8
(C) 9
(D) 10

Answer:
C = 38 ft
2πr = 38
r = 6.04

In Exercises 25 and 26, find the circumference of the circle with the given equation. Write the circumference in terms of π

Question 25.
x2 + y2 = 16
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 25

Question 26.
(x + 2)2 + (y – 3)2 = 9

Answer:
The radius of circle (x + 2)² + (y – 3)² = 9 is 3
C = 2πr = 2π(3) = 6π
The circumference of the circle is 6π units.

Question 27.
USING STRUCTURE
A semicircle has endpoints (- 2, 5) and (2, 8). Find the arc length of the semicircle.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 27

Question 28.
REASONING
\(\widehat{E F}\) is an arc on a circle with radius r. Let x° be the measure of \(\widehat{E F}\). Describe the effect on the length of \(\widehat{E F}\) if you (a) double the radius of the circle, and (b) double the measure of \(\widehat{E F}\).
Answer:

Question 29.
MAKING AN ARGUMENT
Your friend claims that it is possible for two arcs with the same measure to have different arc lengths. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 29

Question 30.
PROBLEM SOLVING
Over 2000 years ago, the Greek scholar Eratosthenes estimated Earth’s circumference by assuming that the Sun’s rays were Parallel. He chose a day when the Sun shone straight down into a well in the city of Syene. At noon, he measured the angle the Sun’s rays made with a vertical stick in the city of Alexandria. Eratosthenes assumed that the distance from Syene to Alexandria was equal to about 575 miles. Explain how Eratosthenes was able to use this information to estimate Earth’s circumference. Then estimate Earth’s circumference.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 29
Answer:

Question 31.
ANALYZING RELATIONSHIPS
In ⊙C the ratio of the length of \(\widehat{P Q}\) to the length of \(\widehat{R S}\) is 2 to 1. What is the ratio of m∠PCQ to m∠RCS?
(A) 4 to 1
(B) 2 to 1
(C) 1 to 4
(D) 1 to 2
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 31

Question 32.
ANALYZING RELATIONSHIPS
A 45° arc in ⊙C and a 30° arc in ⊙P have the same length. What is the ratio of the radius r1 of ⊙C to the radius r2 of ⊙P? Explain your reasoning.
Answer:

Question 33.
PROBLEM SOLVING
How many revolutions does the smaller gear complete during a single revolution of the larger gear?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 30
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 33

Question 34.
USING STRUCTURE
Find the circumference of each circle.
a. a circle circumscribed about a right triangle whose legs are 12 inches and 16 inches long
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 1
c² = a² + b²
c² = 12²+ 16²= 400
c = 20 in
circumference c = dπ
= 20π = 62.83 in

b. a circle circumscribed about a square with a side length of 6 centimeters
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 2
d² = 6²+ 6² = 72
d = 8.49 cm
C = dπ
C = 8.49π
C = 26.67 cm

c. a circle inscribed in an equilateral triangle with a side length of 9 inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 3
r = \(\frac { a√3 }{ 3 } \)
r = \(\frac { 9√3 }{ 3 } \)
r = 3√3 = 5.2
C = 2πr
C = 2π (5.2) = 32.67 in

Question 35.
REWRITING A FORMULA
Write a formula in terms of the measure θ (theta) of the central angle in radians) that can he used to find the length of an arc of a circle. Then use this formula to find the length of an arc of a circle with a radius of 4 inches and a central angle of \(\frac{3 \pi}{4}\) radians.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 35

Question 36.
HOW DO YOU SEE IT?
Compare the circumference of ⊙P to the length of \(\widehat{D E}\). Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 31
Answer:

Question 37.
MAKING AN ARGUMENT
In the diagram. the measure of the red shaded angle is 30°. The arc length a is 2. Your classmate claims that it is possible to find the circumference of the blue circle without finding the radius of either circle. Is your classmate correct? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 32
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 37

Question 38.
MODELING WITH MATHEMATICS
What is the measure (in radians) of the angle formed by the hands of a clock at each time? Explain your reasoning.
a. 1 : 30 P.M.

Answer:
3π/4

b. 3:15 P.M.

Answer:
π/24

Question 39.
MATHEMATICAL CONNECTIONS
The sum of the circumferences of circles A, B, and C is 63π. Find AC.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 33
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 39

Question 40.
THOUGHT PROVOKING
Is π a rational number? Compare the rational number \(\frac{355}{113}\) to π. Find a different rational number that is even closer π.

Answer:
π is not a rational number as it can not be represented as an equivalent fraction. π = 3.14 and 355/113 = 3.14. This fraction resembles that value of π. Therefore a more accurate fraction will be starting by the value of 7 decimla places of π, therefore 3.1415926 x x = a.

Question 41.
PROOF
The circles in the diagram are concentric and \(\overrightarrow{F G}\) ≅ \(\overrightarrow{G H}\) Prove that \(\widehat{J K}\) and \(\widehat{N G}\) have the same length.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 34
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 41.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 41.2

Question 42.
REPEATED REASONING
\(\overline{A B}\) is divided into four congruent segments, and semicircles with radius r are drawn.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 35
a. What is the sum of the four arc lengths?
Answer:

b. What would the sum of the arc lengths be if \(\overline{A B}\) was divided into 8 congruent segments? 16 congruent segments? n congruent segments? Explain your reasoning.
Answer:

Maintaining Mathematical Proficiency

Find the area of the polygon with the given vertices.

Question 43.
X(2, 4), Y(8, – 1), Z(2, – 1)
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 43.1

Question 44.
L(- 3, 1), M(4, 1), N(4, – 5), P(- 3, – 5)

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 4
LP = √(-3 + 3)² + (-5 – 1)² = 6
PN = √(4 + 3)² + (-5 + 5)² = 7
MN = √(4 – 4)² + (-5 – 1)² = 6
LM = √(4 + 3)² + (1 – 1)²= 7
Area = 6 x 7 = 42 units

11.2 Areas of Circles and Sectors

Exploration 1

Finding the Area of a Sector of a Circle

Work with a partner: A sector of a circle is the region bounded by two radii of the circle and their intercepted arc. Find the area of each shaded circle or sector of a circle.

a. entire circle
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 36
Answer:

b. one – fourth of a circle
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 37
Answer:

c. seven – eighths of a circle.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 38
Answer:

d. two – thirds of a circle
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 39
Answer:

Exploration 2

Finding the Area of a Circular Sector

Work with a partner: A center pivot irrigation system consists of 400 meters of sprinkler equipment that rotates around a central pivot point at a rate of once every 3 days to irrigate a circular region with a diameter of 800 meters. Find the area of the sector that is irrigated by this system in one day.
REASONING ABSTRACTLY
To be proficient in math, you need to explain to yourself the meaning of a problem and look for entry points to its solution.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 40
Answer:

Communicate Your Answer

Question 3.
How can you find the area of a sector of a circle?
Answer:

Question 4.
In Exploration 2, find the area of the sector that is irrigated in 2 hours.
Answer:

Lesson 11.2 Areas of Circles and Sectors

Monitoring progress

Question 1.
Find the area of a circle with a radius of 4.5 meters.

Answer:
Circle area = πr²
A = π(4.5)² = 20.25π

Question 2.
Find the radius of a circle with an area of 176.7 square feet.

Answer:
Circle area = πr²
176.7 = πr²
r² = 56.24
r = 7.499

Question 3.
About 58,000 people live in a region with a 2-mile radius. Find the population density in people per square mile.

Answer:
The population density is about 4615.49 people per square mile.

Explanation:
A = πr² = π • 2² = 4π
Population density = \(\frac { number of people }{ area of land } \)
= \(\frac { 58000 }{ 4π } \) = 4615.49

Question 4.
A region with a 3-mile radius has a population density of about 1000 people per square mile. Find the number of people who live in the region.

Answer:
The number of people who live in the region are 28274.

Explanation:
A = πr² = π • 3² = 9π
Population density = \(\frac { number of people }{ area of land } \)
Number of people = 1000 x 9π = 28274

Find the indicated measure

Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 41

Question 5.
area of red sector

Answer:
The area of red sector = 205.25

Explanation:
m∠FDE = 120°, FE = 120° and FGE = 360° – 120° = 240°
Area of red sector = \(\frac { FE }{ 360° } \) • πr²
= \(\frac { 120 }{ 360° } \) • π(14²)
= 205.25

Question 6.
area of blue sector

Answer:
Area of blue sector = 410.5

Explanation:
Area of blue sector = \(\frac { FGE }{ 360° } \) • πr²
= \(\frac { 240 }{ 360° } \) • π(14²) = 410.5

Question 7.
Find the area of ⊙H.

Answer:
Area of ⊙H = 907.92 sq cm

Explanation:
Area of sector FHG =\(\frac { FG }{ 360° } \) • Area of ⊙H
214.37 = \(\frac { 85 }{ 360° } \) • Area of ⊙H
Area of ⊙H = 907.92 sq cm

Question 8.
Find the area of the figure.

Answer:
Area of triangle = \(\frac { 1 }{ 2 } \) • 7 • 7
= 24.5 sq m
Area of semi circle = πr²/2
= π(3.5)²/2
= 19.242255
Area of the figure = 24.5 + 19.24 = 43.74 sq m

Question 9.
If you know the area and radius of a sector of a circle, can you find the measure of the intercepted arc? Explain.

Answer:

Exercise 11.2 Areas of Circles and Sectors

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
A(n) ____________ of a circle is the region bounded by two radii of the circle and their intercepted arc.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 1

Question 2.
WRITING
The arc measure of a sector in a given circle is doubled. will the area of the sector also be doubled? Explain our reasoning.

Answer:
Yes

Explanation:
Area of sector with arc measure x and radius r is s = π/180(xr)
If x becomes doube, then s1 = π/180(2xr) = 2s
This means that if the arc measure doubles, area of the sector also doubles.

Monitoring Progress and Modeling with Mathematics

In Exercise 3 – 10, find the indicated measure,

Question 3.
area of ⊙C
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 42
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 3

Question 4.
area of ⊙C
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 43

Answer:
Area A = πr²
A = π(10)² = 100π sq in

Question 5.
area of a circle with a radius of 5 inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 5

Question 6.
area of a circle with a diameter of 16 feet

Answer:
d = 2r
Circle area = πr² = (π/4)d²
= (π/4)16² = 64π

Question 7.
radius of a circle with an area of 89 square feet
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 7

Question 8.
radius of a circle with an area of 380 square inches

Answer:
A = πr²
380 = πr²
r = 10.99

Question 9.
diameter of a circle with an area of 12.6 square inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 9

Question 10.
diameter of a circle with an area of 676π square centimeters

Answer:
Area A = 676π square centimeters
(π/4)d² = 676π
d² = 2704
d = 52

In Exercises 11 – 14, find the indicated measure.

Question 11.
About 210,000 people live in a region with a 12-mile radius. Find the population density in people per square mile.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 11

Question 12.
About 650,000 people live in a region with a 6-mile radius. Find the population density in people per square mile.

Answer:
The population density is about 5747 people per square mile.

Explanation:
Area of region = π(6)² = 36π
Population density = \(\frac { Number of people }{ area of land } \)
= \(\frac { 650,000 }{ 36π } \) = 5747.2

Question 13.
A region with a 4-mile radius has a population density of about 6366 people per square mile. Find the number of people who live in the region.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 13

Question 14.
About 79,000 people live in a circular region with a population density of about 513 people per square mile. Find the radius of the region.

Answer:
The radius of the region is 7

Explanation:
Population density = \(\frac { Number of people }{ area of land } \)
513 = \(\frac { 79,000 }{ πr² } \)
πr² = 153.99
r = 7

In Exercises 15-18 find the areas of the sectors formed by∠DFE.

Question 15.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 44
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 15

Question 16.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 45

Answer:
Area of sector = \(\frac { 104° }{ 360° } \) • π(14)²
= 177.88
Area of red region is 177.88 sq cm
Area of blue region = \(\frac { 256° }{ 360° } \) • π(14)²
= 437.86 sq cm

Question 17.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 46
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 17

Question 18.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 47

Answer:
Area of red region is 10.471 sq ft
Area of the blue region is 39.79 sq ft

Explanation:
Area of sector = \(\frac { 75° }{ 360° } \) • π(4)²
= 10.471
Area of red region is 10.471 sq ft
Area of blue region = \(\frac { 285° }{ 360° } \) • π(4)²
= 39.79 sq ft

Question 19.
ERROR ANALYSIS
Describe and correct the error in finding. the area of the circle.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 48
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 19

Question 20.
ERROR ANALYSIS
Describe and correct the error in finding the area of sector XZY when the area of ⊙Z is 255 square feet.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 49

Answer:
Area of ⊙Z is 255 square feet
πr² = 255
r = 9
Area of sector XZY = \(\frac { 115 }{ 360 } \) • 255
n = 81.458 sq ft

In Exercises 21 and 22, the area of the shaded sector is show. Find the indicated measure.

Question 21.
area of ⊙M
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 50
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 21

Question 22.
radius of ⊙M
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 51

Answer:
radius of ⊙M = 3.98

Explanation:
Area of region = \(\frac { 89 }{ 360 } \) . Area of ⊙M
12.36 = \(\frac { 89 }{ 360 } \) . Area of ⊙M
Area of ⊙M = 49.99
πr² = 49.99
r = 3.98

In Exercises 23 – 28, find the area of the shaded region.

Question 23.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 52
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 23

Question 24.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 53

Answer:
The area of the shaded region is 85.840 sq in.

Explanation:
Area of square = 20² = 400
Diameter of one circle = 10
radius of one circle = 5 in
Area of one circle = π(5)² = 78.53
Areas of four circle = 314.159
Area of shaded region = 400 – 314.159 = 85.840

Question 25.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 54
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 25.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 25.2

Question 26.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 55

Answer:
The area of shaded region is 301.59

Explanation:
The radius of smaller circle is 8 cm
The radius of bigger circle is 16 cm
Area of smaller semicircle = \(\frac { 1 }{ 2 } \)(π(8)²) = 100.53
Area of lager semicircle = \(\frac { 1 }{ 2 } \)(π(16)²) = 402.123
Area of shaded region = 402.123 – 100.53 = 301.59

Question 27.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 56
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 27.1

Question 28.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 57
Answer:
Area of shaded region = 7.63

Explanation:
c² = 3² + 4² = 25
c = 5
Radius = 2.5
Circle area = π(2.5)² = 19.63
Area of triangle = (3 x 4)/2 = 6
Area of shaded region = 19.63 – 12 = 7.63

Question 29.
PROBLEM SOLVING
The diagram shows the shape of a putting green at a miniature golf course. One part of the green is a sector of a circle. Find the area of the putting green.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 58
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 29.1

Question 30.
MAKING AN ARGUMENT
Your friend claims that if the radius of a circle is doubled, then its area doubles. Is your friend correct? Explain your reasoning.

Answer:
The friend is not correct. doubling the radius quadruples the area.

Explanation:
Area of circle with radius r = πr²
Area of circle with radius 2r = π(2r)² = 4πr²

Question 31.
MODELING WITH MATHEMATICS
The diagram shows the area of a lawn covered by a water sprinkler.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 59
a. What is the area of the lawn that is covered by the sprinkler?
b. The water pressure is weakened so that the radius is 12 feet. What is the area of the lawn that will be covered?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 31

Question 32.
MODELING WITH MATHEMATICS
The diagram shows a projected beam of light from a lighthouse.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 60
a. What is the area of water that can be covered by the light from the lighthouse?
Answer:
Area = \(\frac { 115 }{ 360 } \) x π(18)²
= 325.15 sq mi

b. What is the area of land that can be covered by the light from the lighthouse?
Answer:
Area = \(\frac { 245 }{ 360 } \) x π(18)²
= 692.72 sq mi

Question 33.
ANALYZING RELATIONSHIPS
Look back at the Perimeters of Similar Polygons Theorem (Theorem 8.1) and the Areas of Similar PoIyons Theorem (Theorem 8.2) in Section 8.1. How would you rewrite these theorems to apply to circles? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 33

Question 34.
ANALYZING RELATIONSHIPS
A square is inscribed in a circle. The same square is also circumscribed about a smaller circle. Draw a diagram that represents this situation. Then find the ratio of the area of the larger circle to the area of the smaller circle.

Answer:
We start by assigning a variable to the radius of the inner circle. It is r, therefore the area of the circle is πr²
It can be seen that the side length of square is twice this radius. Therefore it can be said that the side length of this square is 2r.
Next, it can be seen that the diagonal of the square is diameter of outer circle. Therefore, length of the diagonal of the circle d = 2r√2. outer circle radius = r√2
Area of outer circle 2πr²
The ratio of the area of larger circle to the smaller circle = 2.

Question 35.
CONSTRUCTION
The table shows how students get to school.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 61
a. Explain why a circle graph is appropriate for the data.
b. You will represent each method by a sector of a circle graph. Find the central angle to use for each sector. Then construct the graph using a radius of 2 inches.
c. Find the area of each sector in your graph.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 35.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 35.2

Question 36.
HOW DO YOU SEE IT?
The outermost edges of the pattern shown form a square. If you know the dimensions of the other square, is it possible to compute the total colored area? Explain.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 62
Answer:

Question 37.
ABSTRACT REASONING
A circular pizza with a 12-inch diameter is enough for you and 2 friends. You want to buy pizzas for yourself and 7 friends. A 10-inch diameter pizza with one topping Costs $6.99 and a 14-inch diameter pizza with one topping Costs $12.99. How many 10-inch and 14-inch pizzas should you buy in each situation? Explain.
a. You want to spend as little money as possible.
b. You want to have three pizzas. each with a different topping, and spend as little money as possible.
C. You want to have as much of the thick outer crust as possible.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 37.1

Question 38.
THOUGHT PROVOKING
You know that the area of a circle is πr2. Find the formula for the area of an ellipse, shown below.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 63

Answer:
Ellipse area = πab

Question 39.
MULTIPLE REPRESENTATIONS
Consider a circle with a radius of 3 inches.
a. Complete the table, where x is the measure of the arc and is the area of the corresponding sector. Round your answers to the nearest tenth.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 64
b. Graph the data in the table.
c. Is the relationship between x and y linear? Explain.
d. If parts (a) – (c) were repeated using a circle with a radius of 5 inches, would the areas in the table change? Would your answer to part (c) change? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 39.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 39.2
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 39.3

Question 40.
CRITICAL THINKING
Find the area between the three congruent tangent circles. The radius of each circle is 6 inches.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 65
Answer:

Question 41.
PROOF
Semicircles with diameters equal to three sides of a right triangle are drawn, as shown. Prove that the sum of the areas of the two shaded crescents equals the area of the triangle.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 66
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 41

Maintaining Mathematical proficiency

Find the area of the figure.

Question 42.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 67

Answer:
Area = \(\frac { 1 }{ 2 } \)(base x height)
Area = \(\frac { 1 }{ 2 } \)(18 x 6) = 54 sq in

Question 43.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 68
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 43

Question 44.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 69
Answer:
Area = \(\frac { 1 }{ 2 } \)(base x height)
A = \(\frac { 1 }{ 2 } \)(13 x 9) = 58.5 sq in

Question 45.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 70
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 45

11.3 Areas of Polygons

Exploration 1

Finding the Area of a Regular Polygon

Work with a partner: Use dynamic geometry software to construct each regular polygon with side lengths of 4, as shown. Find the apothem and use it to find the area of the polygon. Describe the steps that you used.
a.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 71
Answer:

b.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 72
Answer:

c.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 73
Answer:

d.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 74
Answer:

Exploration 2

Writing a Formula for Area

Work with a partner: Generalize the steps you used in Exploration 1 to develop a formula for the area of a regular polygon.
REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.
Answer:

Communicate Your Answer

Question 3.
How can you find the area of a regular polygon?
Answer:

Question 4.
Regular pentagon ABCDE has side lengths of 6 meters and an apothem of approximately 4.13 meters. Find the area of ABCDE.
Answer:

Lesson 11.3 Areas of Polygons

Monitoring Progress

Question 1.
Find the area of a rhombus with diagonals d1 = 4 feet and d2 = 5 feet.

Answer:
Area of rhombus = \(\frac { 1 }{ 4 } \)(d₁d₂)
= \(\frac { 1 }{ 4 } \)(4 x 5) = 5 sq ft

Question 2.
Find the area of a kite with diagonals d1 = 12 inches and d1 = 9 inches.

Answer:
Area of kite = \(\frac { 1 }{ 4 } \)(d₁d₂)
= \(\frac { 1 }{ 4 } \)(12 x 9) = 27 sq in

In the diagram. WXYZ is a square inscribed in ⊙P.

Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 75

Question 3.
Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer:
P is the center, PY or PX is the radius, PQ is apothem, ∠XPY is the central angle.

Question 4.
Find m∠XPY, m∠XPQ, and m∠PXQ.

Answer:
m∠XPY = \(\frac { 360 }{ 4 } \) = 90
m∠XPQ = 90/2 = 45
m∠PXQ = 180 – (90 + 45) = 45

Find the area of regular polygon

Question 5.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 76

Answer:
c = √(8² + 6.5²) = 10.3
a = 20.61
Area = 0.25(√5(5+2√5) a²
Area = 0.25(√5(5+2√5) 20.61² = 730.8

Question 6.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 77
Answer:
Area = \(\frac { 5a² }{ 2 } \)√(5+2√5)
= \(\frac { 5(7²) }{ 2 } \)√(5+2√5)
Area = 55377

Exercise 11.3 Areas of Polygons

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain how to find the measure of a central angle of a regular polygon.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 78
Find the radius of ⊙F. Answer:

Answer:
EF = radius = 6.8

Find the apothem of polygon ABCDE.

Answer:
GF = apothem = 5.5

Find AF.

Answer:
AF = √4² + 5.5²
AF = 6.8

Find the radius of polygon ABCDE.
Answer:
AF = radius = 6.8

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the area of the kite or rhombus.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 79
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 80

Answer:
d₁ = 6 + 6 = 12
d₂ = 2 + 10 = 12
area A = \(\frac { 1 }{ 4 } \)(d₁d₂)
= \(\frac { 1 }{ 4 } \)(12 x 12)
= 36

Question 5.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 81
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 82
Answer:
area A = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(5 x 6) = 7.5

In Exercises 7 – 10, use the diagram

Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 83

Question 7.
Identify the center of polygon JKLMN?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 7

Question 8.
Identify a central angle of polygon JKLMN.

Answer:
∠NPM is the central angle of polygon JKLMN

Question 9.
What is the radius of polygon JKLMN?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 9

Question 10.
What is the apothem of polygon JKLMN?
Answer:
QP is the apothem of polygon JKLMN

In Exercises 11 – 14, find the measure of a central angle of a regular polygon with the given number of sides. Round answers to the nearest tenth of a degree, if necessary.

Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 84

Question 11.
10 sides
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 11

Question 12.
18 sides

Answer:
The measure of central angle = \(\frac { 360 }{ 18 } \) = 20

Question 13.
24 sides
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 13

Question 14.
7 sides

Answer:
The measure of central angle = \(\frac { 360 }{ 7 } \) = 51.42

In Exercises 15 – 18, find the given angle measure for regular octagon ABCDEFGH.

Question 15.
m∠GJH
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 15

Question 16.
m∠GJK

Answer:
m∠GJK = m∠GJH/2
m∠GJK = 22.5

Question 17.
m∠KGJ
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 17

Question 18.
m∠EJH
Answer:
m∠EJH = 3(45) = 135

In Exercises 19 – 24, find the area of the regular polygon.

Question 19.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 85
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 19

Question 20.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 86
Answer:
Area = \(\frac { 5a² }{ 2 } \)√(5+2√5)
Area = \(\frac { 5(6.84)² }{ 2 } \)√(5+2√5)
A = 359.9784

Question 21.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 87
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 21

Question 22.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 88
Answer:
Area = \(\frac { 3√3 a²}{ 2 } \)
A = \(\frac { 3√3 (7)²}{ 2 } \)
A = 127.30

Question 23.
an octagon with a radius of 11 units
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 23

Question 24.
a pentagon with an apothem of 5 units

Answer:
A = 90.75

Explanation:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 6

We know apothem a = and it divides pentagon into triangles, the central angle is divided into 360/5 = 72
After that, we halved this angle and got 2 right triangles with x = 44 and y = 36. Since we know one side and all three angles of the triangle, we can calculate p with the tangent function.
tan y = p/a
tan 36 = p/5
p = 3.63
Since p is just half of the length of the side, we have to multiply it by 2
2 . p = 2 . 3.63 = 7.26 = s
Area = \(\frac { a . s. n }{ 2 } \)
A = \(\frac { 15 x 7.26 x 5 }{ 2 } \)
A = 90.75

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding the area of the kite.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 89
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 25

Question 26.
ERROR ANALYSIS
Describe and correct the error in finding. the area of the regular hexagon.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 90

Answer:
s = √15² – 13² = 7.48
Area = \(\frac { 1 }{ 2 } \)(a . ns)
A = \(\frac { 1 }{ 2 } \)(13 x 6 x 7.48)
A = 291.72

In Exercises 27 – 30, find the area of the shaded region.

Question 27.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 91
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 27

Question 28.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 92

Answer:
Area of the shaded region = 223.75

Explanation:
Square side = diagonal/√2
= 28/√2 = 19.79
Area of square = 19.79² = 392
Circle area = π(14)² = 615.75
Area of the shaded region = 615.75 – 392 = 223.75

Question 29.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 93
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 29

Question 30.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 94
Answer:

Question 31.
MODELING WITH MATHEMATICS
Basaltic columns arc geological formations that result from rapidly cooling lava. Giant’s Causeway in Ireland contains many hexagonal basaltic columns. Suppose the top of one of the columns is in the shape of a regular hexagon with a radius of 8 inches. Find the area of the top of the column to the nearest square inch.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 95
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 31

Question 32.
MODELING WITH MATHEMATICS
A watch has a circular surface on a background that is a regular octagon. Find the area of the octagon. Then find the area of the silver border around the circular face.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 96
Answer:

CRITICAL THINKING
In Exercises 33 – 35, tell whether the statement is true or false. Explain your reasoning

Question 33.
The area of a regular n-gon of a fixed radius r increases as n increases.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 33

Question 34.
The apothem of a regular polygon is always less than the radius.

Answer:
true, the radius always reaches the end of the circle but the apothem never does

Question 35.
The radius of a regular polygon is always less than the side length.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 35

Question 36.
REASONING
Predict which figure has the greatest area and which has the least area. Explain your reasoning. Check by finding the area of each figure.
(A) Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 97
(B) Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 98
(C) Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 99
Answer:
(B) has the highest area and C has the lowest area.

Explanation:
(A) area = π(6.5)² = 132.73
(B) area = 139.25
(C) area = \(\frac { 1 }{ 2 } \)(18 x 15) = 135

Question 37.
USING EQUATIONS
Find the area of a regular
pentagon inscribed in a circle whose equation is given by (x – 4)2 + y + 2)2 = 25.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 37.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 37.2

Question 38.
REASONING
What happens to the area of a kite if you double the length of one of the diagonals? if you double the length of both diagonals? Justify your answer.

Answer:
Area of a kite = \(\frac { 1 }{ 4 } \)(d₁d₂)
If you double the length of one diagonal, then d₁ = 2d₁
Area of kite = \(\frac { 1 }{ 2 } \)(d₁d₂)
If you double length of both diagonals
Area = \(\frac { 1 }{ 4 } \)(2d₁2d₂) = d₁d₂
If you double the length of one diagonal, then the area becomes halve. If you double length of both diagonals, then area becomes 4 times.

MATHEMATICAL CONNECTIONS
In Exercises 39 and 40, write and solve an equation to find the indicated lengths. Round decimal answers to the nearest tenth.

Question 39.
The area of a kite is 324 square inches. One diagonal is twice as long as the other diagonal. Find the length of each diagonal.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 39

Question 40.
One diagonal of a rhombus is four times the length of the other diagonal. The area of the rhombus is 98 square feet. Find the length of each diagonal.

Answer:
The length of each diagonal is 9.89, 2.47.

Explanation:
One diagonal of a rhombus is four times the length of the other diagonal.
d₁ = 4d₂
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
98 = \(\frac { 1 }{ 4 } \)(d₁(4d₁))
d₁ = 9.89
d₂ = 2.47

Question 41.
REASONING
The perimeter of a regular nonagon. or 9-gon, is 18 inches. Is this enough information to find the area? If so, find the area and explain your reasoning. If not, explain why not.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 41

Question 42.
MAKING AN ARGUMENT
Your friend claims that it is possible to find the area of any rhombus if you only know the perimeter of the rhombus. Is your friend correct? Explain your reasoning.

Answer:
No; A rhombus is not a regular polygon.

Question 43.
PROOF
Prove that the area of any quadrilateral with perpendicular diagonals is A = \(\frac{1}{2}\)d1d2, where d1 and d2 are the lengths of the diagonals.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 100
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 43.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 43.2

Question 44.
HOW DO YOU SEE IT?
Explain how to find the area of the regular hexagon by dividing the hexagon into equilateral triangles.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 101

Answer:

Question 45.
REWRITING A FORMULA
Rewrite the formula for the area of a rhombus for the special case of a square with side length s. Show that this is the same as the formula for the area of a square, A = s2.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 45

Question 46.
REWRITING A FORMULA
Use the formula for the area of a regular polygon to show that the area of an equilateral triangle can be found by using the formula A = \(\frac{1}{4}\)s2√3 where s is the side length.
Answer:

Question 47.
CRITICAL THINKING
The area of a regular pentagon is 72 square centimeters. Find the length of one side.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 47

Question 48.
CRITICAL THINKING
The area of a dodecagon, or 12-gon, is 140 square inches. Find the apothem of the polygon.

Answer:
Let the side length of dodecagon be 2x. The measure of each interior angle of a regular decagon is 150. This implies that the base angle C and A of the resulting isosceles triangle formed by the red sides is equal to 150/2 = 75. The adjacent to this angle is the length 2x/2 = x inches, while the opposite to it is the blue apothem in the right triangle BDC formed. Therefore a = x tan 75. Therefore, area of dodecagon is
140 = 1/2 (x tan75)(12 . 2x)
140 = 44.785 x²
x² = 3.126
x = 1.768

Question 49.
USING STRUCTURE
In the figure, an equilateral triangle lies inside a square inside a regular pentagon inside a regular hexagon. Find the approximate area of the entire shaded region to the nearest whole number.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 102
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 49.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 49.2

Question 50.
THOUGHT PROVOKING
The area of a regular n-gon is given by A = \(\frac{1}{2}\)aP. As n approaches infinity, what does the n-gon approach? What does P approach? What does a approach? What can you conclude from your three answers? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 103
Answer:

Question 51.
COMPARING METHODS
Find the area of regular pentagon ABCDE by using the formula A = \(\frac{1}{2}\)aP, or A = \(\frac{1}{2}\)a • ns. Then find the area by adding the areas of smaller polygons. Check that both methods yield the same area. Which method do you prefer? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 104
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 51.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 51.2
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 51.3

Question 52.
USING STRUCTURE
Two regular polygons both have n sides. One of the polygons is inscribed in, and the other is circumscribed about, a circle of radius r. Find the area between the two polygons in terms of n and r.

Answer:
The radius of the smaller polygon is equal to the apothem of the larger polygon. The central angle is 360/n, therefore the apothem makes an angle of 180/n. Use sine and cosine to find the apothem and side length of the smaller polygon.
asmall = r sin\(\frac { 180 }{ n } \)
ssmall = 2r cos\(\frac { 180 }{ n } \)
Use tangent to find the side length of the large polygon.
Slarge = 2r tan\(\frac { 180 }{ n } \)
Use the formula to find the area of the smaller polygon.
Asmall = 1/2 . asmall . n . ssmall
Asmall = 1/2 . r sin\(\frac { 180 }{ n } \) . n . 2r cos\(\frac { 180 }{ n } \)
Asmall = nr² sin \(\frac { 180 }{ n } \) cos\(\frac { 180 }{ n } \)
Use the formula to find the area of the larger polygon.
ALarge = 1/2 . alarge . n . slarge
= nr² tan\(\frac { 180 }{ n } \)
The area between the polygons is equal to the area of the larger polygon minus the area of the smaller polygon. Use some trig identities to simplify the expression.
A = Alarge – Asmall
A = nr² tan\(\frac { 180 }{ n } sin²[latex]\frac { 180 }{ n }

Maintaining Mathematical Proficiency

Determine whether the figure has line symmetry, rotational symmetry, both, or neither. If the
figure has line symmetry. determine the number of lines of symmetry. It the figure has rotational
symmetry, describe any rotations that map the figure onto itself.

Question 53.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 105
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 53

Question 54.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 106
Answer:
The figure has rotational symmetry.

Question 55.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 107
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 55

Question 56.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 108
Answer:
The figure has one line symmetry.

11.4 Three-Dimensional Figures

Exploration 1

Analyzing a Property of Polyhedra

Work with a partner: The five Platonic solids are shown below. Each of these solids has congruent regular polygons as faces. Complete the table by listing the numbers of vertices, edges, and faces of each Platonic solid.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 109
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 110
Answer:

Communicate Your Answer

Question 2.
What is the relationship between the numbers of vertices V, edges E, and faces F of a polyhedron? (Note: Swiss mathematician Leonhard Euler (1707 – 1783) discovered a formula that relates these quantities.)
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to reason inductively about data.
Answer:

Question 3.
Draw three polyhedra that are different from the Platonic solids given in Exploration 1. Count the numbers of vertices, edges, and faces of each polyhedron Then verify that the relationship you found in Question 2 is valid for each polyhedron.
Answer:

Lesson 11.4 Three-Dimensional Figures

Monitoring progress

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 1.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 111

Answer:
The solid is formed by polygons, so it is a polyhedron. The base is a square, it is a square pyramid.

Question 2.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 112
Answer:
The solid have curved faces. So it is not a polyhedron.

Question 3.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 113
Answer:
The solid is formed by polygons, so it is a polyhedron. iT has two triangles, one rectangle and two squares.

Describe the shape formed by the intersection of the plane and the solid.

Question 4.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 114
Answer:
The cross-section is a pentagon.

Question 5.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 115
Answer:
The cross-section is a hexagon.

Question 6.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 116
Answer:
The cross-section is a circle.

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 7.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 117
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 8

Question 8.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 118
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 9

Question 9.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 119
Answer:

Exercise 11.4 Three-Dimensional Figures

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
A(n) ___________ is a solid that is bounded by polygons.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which solid does not belong with the other three? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 120

Answer:
Cone does not belong with the other three as it has a curved surface and others not.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, match the polyhedron with its name.

3. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 121A. triangular Prism
4. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 122B. rectangular pyramid
5. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 123C. hexagonal pyramid
6. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 124D. Pentagonal prism

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 5

In Exercises 7 – 10, tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 7.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 125
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 7

Question 8.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 126
Answer:
Yes, it is a polyhedron. It is a hexagonal prism.

Question 9.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 127
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 128
Answer:
Yes, it is a polyhedron, truncated square pyramid.

In Exercises 11 – 14, describe the cross section formed by the intersection of the plane and the solid.

Question 11.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 129
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 11

Question 12.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 130
Answer:
The cross-section is a square.

Question 13.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 131
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 13

Question 14.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 132
Answer:
The cross-section is a hexagon

In Exercises 15 – 18, sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 15.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 133
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 15

Question 16.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 134
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

Question 17.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 135
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 17

Question 18.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 136
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 9

Question 19.
ERROR ANALYSIS
Describe and correct the error in identifying the solid.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 137
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 19

Question 20.
HOW DO YOU SEE IT?
Is the swimming pool shown a polyhedron? If it is, name the polyhedron. If not, explain why not.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 138
Answer:
It is an octagonal polyhedron.

In Exercises 21 – 26, sketch the polyhedron.

Question 21.
triangular prism
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 21

Question 22.
rectangular prism

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11

Question 23.
pentagonal prism
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 23

Question 24.
hexagonal prism
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 12

Question 25.
square pyramid
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 25

Question 26.
pentagonal pyramid

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 13

Question 27.
MAKING AN ARGUMENT
Your Friend says that the polyhedron shown is a triangular prism. Your cousin says that it is a triangular pyramid. Who is correct? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 139
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 27

Question 28.
ATTENDING TO PRECISION
The figure shows a plane intersection a cube through four of its vertices. The edge length of the cube is 6 inches.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 140
a. Describe the shape of the cross section.
Answer:
The cross-section is a rectangle.

b. What is the perimeter of the cross section?
Answer:
The perimeter is 2(l + b)

c. What is the area of the cross section?
Answer:
Area is lb.

REASONING
In Exercises 29 – 34, tell whether it is possible for a cross section of a cube to have the given shape. If it is, describe or sketch how the plane could intersect the cube.

Question 29.
circle
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 29

Question 30.
pentagon
Answer:
yes, cross-section of the cube can be a pentagon.

Question 31.
rhombus
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 31

Question 32.
isosceles triangle
Answer:
Yes, the cross-section can be an isosceles triangle.

Question 33.
hexagon
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 33

Question 34.
scalene triangle
Answer:
Yes, the cross-section can be scalene triangle.

Question 35.
REASONING
Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
a.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 141
b.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 142
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 35

Question 36.
THOUGHT PROVOKING
Describe how Plato might have argued that there are precisely five Platonic Solids (see page 617). (Hint: Consider the angles that meet at a vertex.)
Answer:

Maintaining Mathematical proficiency

Decide whether enough information is given to prove that the triangles are congruent. It so, state the theorem you would use.

Question 37.
∆ABD, ∆ CDB
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 143
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 37

Question 38.
∆JLK, ∆JLM
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 144

Answer:
∆JLK ≅ ∆JLM by SAS congruence theorem.

Question 39.
∆RQP, ∆RTS
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 145
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 39

11.1 – 11.4 Quiz

Find the indicated measure.

Question 1.
m[latex]\widehat{E F}\)
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 146
Answer:
13.7 = \(\frac { m[latex]\widehat{E F}\) }{ 360 } [/latex] • 2π(7)
m\(\widehat{E F}\) = 112.13

Question 2.
arc length of \(\widehat{Q S}\)
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 147
Answer:
arc length of \(\widehat{Q S}\) = \(\frac { 83 }{ 360 } \) • 2π(4)
= 5.79

Question 3.
circumference of ⊙N
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 148
Answer:
8 = \(\frac { 48 }{ 360 } \) • 2πr
C = 60

Question 4.
Convert 26° to radians and \(\frac{5 \pi}{9}\) radians to degrees.

Answer:
26° = 26 . \(\frac { π }{ 180 } \) = \(\frac { 13π }{ 90 } \) radians
\(\frac{5 \pi}{9}\) = \(\frac{5 \pi}{9}\) . \(\frac { 180 }{ π } \) = 100°

Use the figure to find the indicated measure.

Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 149

Question 5.
area of red sector

Answer:
area of red sector = \(\frac { 100 }{ 360 } \) . π(12)²
= 125.66

Question 6.
area of blue sector

Answer:
area of blue sector = \(\frac { 260 }{ 360 } \) . π(12)²
= 326.72

In the diagram, RSTUVWXY is a reuIar octagon inscribed in ⊙C.

Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 150

Question 7.
Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer:
C is center, CY is radius, CZ is apothem, ∠YCR is central angle of the polygon

Question 8.
Find m∠RCY, m∠RCZ, and m∠ZRC.

Answer:
m∠RCY = 360/8 = 45
m∠RCZ = 45/2 = 22.5
m∠ZRC = 180 – (22.5 + 90) = 67.5

Question 9.
The radius of the circle is 8 units. Find the area of the octagon.

Answer:
Area of octagon = 0.5 x 8 x 8 sin 45 = 22.62

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 10.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 151
Answer:
It is not a polyhedron.

Question 11.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 152
Answer:
The solid is a polyhedron. It is an octagonal pyramid.

Question 12.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 153
Answer:
The solid is a polyhedron. It is a pentagonal prism.

Question 13.
Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 154

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 14

Question 14.
The two white congruent circles just fit into the blue circle. What is the area of the blue region?
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 155
Answer:
White circle diameter = radius of the blue circle.
6 = radius of the blue circle.
Area of blue circle = π6² = 113.09
Area of white circle = π3² = 28.27
Area of blue region = 113.09 – 2(28.27) = 56.541

Question 15.
Find the area of each rhombus tile. Then find the area of the pattern.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 156

Answer:
Area of yellow tile = \(\frac { 1 }{ 4 } \)(15.7 x 11.4) = 44.745
area of red tile = \(\frac { 1 }{ 4 } \)(18.5 x 6) = 27.75
Area of pattern = 32(44.745) + 23(27.75) = 2070.09 sq mm

11.5 Volumes of Prisms and Cylinders

Exploration 1

Finding volume

Work with a partner: Consider a stack of square papers that is in the form of a right prism.

Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 157

a. What is the volume the prism?
Answer:

b. When you twist the slack of papers, as shown at the right, do you change the volume? Explain your reasoning.
Answer:

c. Write a carefully worded conjecture that describes the conclusion you reached in part (b).
ATTENDING TO PRECISION
To be proficient in math, you need to communicate precisely to others.
Answer:

d. Use your conjecture to find the volume of the twisted stack of papers.
Answer:

Exploration 2

Finding volume

Work with a partner: Use the conjecture you wrote in Exploration I to find the volume of the cylinder.

a.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 158
Answer:

b.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 159
Answer:

Communicate Your Answer

Question 3.
How can you find the volume of a prism or cylinder that is not a right prism or right cylinder?
Answer:

Question 4.
In Exploration 1, would the conjecture you wrote change if the papers in each stack were not squares? Explain your reasoning.
Answer:

Lesson 11.5 Volumes of Prisms and Cylinders

Monitoring Progress

Find the volume of the solid.

Question 1.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 160
Answer:
Volume = Area * height
Area = \(\frac { 1 }{ 2 } \)(5 x 9) = 22.5
Volume = 22. 5 x 8 = 180 cubic m.

Question 2.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 161

Answer:
Area of circle = πr² = π(8)² = 64π
Volume = 64π x 14 = 2814.86 cubic ft

Question 3.
The diagram shows the dimensions of a concrete cylinder. Concrete has a density of 2.3 grams per cubic centimeter. Find the mass of the concrete cylinder to the nearest gram.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 162
Answer:
Mass of the concrete cylinder = 32 x π(24)² = 18432π cubic in

Question 4.
WHAT IF?
In Example 4, you want the length to be 5 meters, the width to be 3 meters. and the volume to be 60 cubic meters. What should the height be?
Answer:
volume = lbh
60 = 5 x 3 x h
h = 4 m

Question 5.
WHAT IF?
In Example 5, you want the height to be 5 meters and the volume to be 75 cubic meters. What should the area of the base be? Give a possible length and width.
Answer:
volume V = base x height
75 = base x 5
Base = 15 sq m

Question 6.
Prism C and prism D are similar. Find the volume of prism D.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 163
Answer:
\(\frac { 12 }{ 3 } \) = \(\frac { 1536 }{ v } \)
v = 384
Volume of prism D = 384 cubic m

Question 7.
Find the volume of the composite solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 164
Answer:
Volume = area x height
Volume = \(\frac { 1 }{ 2 } \)(10 x 3 x 6)
= 90 cubic ft

Exercise 11.5 Volumes of Prisms and Cylinders

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
In what type of units is the volume of a solid measured?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 1

Question 2.
COMPLETE THE SENTENCE
Density is the amount of _______ that an object has in a given unit of __________ .

Answer:
Density is the mass of the object divided by its volume.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the volume of the prism.

Question 3.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 165
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 3

Question 4.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 166
Answer:
Volume V = lbh
V = 1.5 x 2 x 4 = 12 m³

Question 5.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 167
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 5

Question 6.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 168
Answer:
Volume = 6 x 11 x 14
V = 924 m³

In Exercises 7 – 10. find the volume of the cylinder.

Question 7.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 169
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 7

Question 8.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 170
Answer:
Volume = πr²h
V = π(13.4)² x 9.8
V = 1759.6π cm³

Question 9.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 171
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 9

Question 10.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 172
Answer:
Volume V = πr²h
Shoter side = 18/2 = 9
Height h = √18² – 9² = 15.588
V = π6² x 15.588
= 1763 m³

In Exercises 11 and 12. make a sketch of the solid and find its volume. Round your answer to the nearest hundredth.

Question 11.
A prism has a height of 11.2 centimeters and an equilateral triangle for a base, where each base edge is 8 centimeters.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 11

Question 12.
A pentagonal prism has a height of 9 feet and each base edge is 3 feet.

Answer:
volume is 139.32 ft³

explanation:
Pentagon area = 15.48
Height h = 9 ft
Volume V = area x height
= 15.48 x 9 = 139.32

Question 13.
PROBLEM SOLVING
A piece of copper with a volume of 8.25 cubic centimeters has a mass of 73.92 grams. A piece of iron with a volume of 5 cubic centimeters has a mass of 39.35 grams. Which metal has the greater density?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 173
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 13

Question 14.
PROBLEM SOLVING
The United States has minted one-dollar silver coins called the American Eagle Silver Bullion Coin since 1986. Each coin has a diameter of 40.6 millimeters and is 2.98 millimeters thick. The density of silver is 10.5 grams per cubic centimeter. What is the mass of an American Eagle Silver Bullion Coin to the nearest grain?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 174
Answer:
V = πr²h
V =π(20.3)²  x (2.98)
V = 3856 mm³

Question 15.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the cylinder.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 175
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 15

Question 16.
ERROR ANALYSIS
Describe and correct the error in finding the density of an object that has a mass of 24 grams and a volume of 28.3 cubic centimeters.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 176

Answer:
Density = mass / volume
Density = \(\frac { 24 }{ 28.3 } \)
Density = 0.8480

In Exercises 17 – 22, find the missing dimension of the prism or cylinder.

Question 17.
Volume = 500 ft3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 177
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 17

Question 18.
Volume = 2700 yd3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 178

Answer:
Volume = 2700 yd³
12 x 5 x v = 2700
v = 15 yd

Question 19.
Volume = 80 cm3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 179
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 19

Question 20.
Volume = 72.66 in.3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 180
Answer:
Volume = 72.66 in.3
Area . x = 72.66
10.39 x = 72.66
x = 6.9 in

Question 21.
Volume = 3000 ft3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 181
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 21

Question 22.
Volume = 1696.5 m3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 182
Answer:
Volume = 1696.5
πr²h = 1696.5
πz² x 15 = 1696.5
z² = 36.00
z = 6

In Exercises 23 and 24, find the area of the base of the rectangular prism with the given volume and height. Then give a possible length and width.

Question 23.
V= 154 in.3, h = 11 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 23

Question 24.
V = 27 m3,h = 3m

Answer:
V = Bh
27 = B x 3
B = 9

In Exercises 25 and 26, the solids are similar. Find the volume of solid B.

Question 25.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 183
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 25

Question 26.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 184
Answer:
\(\frac { 12 }{ 15 } \) = \(\frac { 4608π }{ V } \)
V = 5760π
Volume of cylinder B = 5760π

In Exercises 27 and 28, the solids are similar. Find the indicated measure.

Question 27.
height x of the base of prism A
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 185
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 27

Question 28.
height h of cylinder B
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 186
Answer:
height h of cylinder B is 40 ft

Explanation:
\(\frac { 7π }{ 5 } \) = \(\frac { 56π }{ h } \)
h = 40

In Exercises 29 – 32. find the volume of the composite solid.

Question 29.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 187
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 29

Question 30.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 188
Answer:
Volume V = 89.32

Explanation:
Volume of square = 4³ = 64
Volume of semicircle = π(2)² x 4 = 8π

Question 31.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 189
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 31

Question 32.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 190

Answer:
The volume of composite solid is 35 cubic ft

Explanation:
Volume of larger prism = 4 x 2 x 5 = 40
Volume of the smaller prism = 1 x 1 x 5 = 5
Volume of larger prism – volume of the smaller prism = 40 – 5 = 35 cubic ft

Question 33.
MODELING WITH MATHEMATICS
The Great Blue Hole is a cylindrical trench located off the coast of Belize. It is approximately 1000 feet wide and 400 feet deep. About how many gallons of water does the Great Blue Hole contain? (1 ft3 ≈ 7.48 gallons)
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 191
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 33

Question 34.
COMPARING METHODS
The Volume Addition Postulate states that the volume of a solid is the sum of the volumes of all its non overlapping parts. Use this postulate to find the volume of the block of concrete in Example 7 by subtracting the volume of each hole from the volume of the large rectangular prism. Which method do you prefer? Explain your reasoning.
Answer:

REASONING
In Exercises 35 and 36, you are melting a rectangular block of wax to make candles. how many candles of the given shape can be made using a block that measures 10 centimeters by 9 centimeters by 20 centimeters?

Question 35.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 192
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 35

Question 36.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 193

Answer:
7 triangular prism candles with the given measures can be made.

Explanation:
Volume of block = 1800
The volume of triangular prism = 4 x 6 x 10 = 240
1800/240 = 7.5

Question 37.
PROBLEM SOLVING
An aquarium shaped like a rectangular prism has a length of 30 inches, a width of 10 inches, and a height of 20 inches. You fill the aquarium \(\frac{3}{4}\) fill with water. When you submerge a rock in the aquarium, the water level rises 0.25 inch.
a. Find the volume of the rock.
b. How many rocks of this size can you place in the aquarium before water spills out?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 37

Question 38.
PROBLEM SOLVING
You drop an irregular piece of metal into a container partially filled with water and measure that the waler level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?

Answer:
The density of metal is 1.4648

Explanation:
Density = \(\frac { Mass }{ Volume } \)
Volume V = 4.8 x 64 = 307.2
Density = \(\frac { 450 }{ 307.2 } \) = 1.4648

Question 39.
WRITING
Both of the figures shown arc made up of the same number of congruent rectangles. Explain how Cavalieri’s Principle can be adapted to compare the areas of these figures.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 194
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 39

Question 40.
HOW DO YOU SEE IT?
Each stack of memo papers contains 500 equally-sized sheets of paper. Compare their volumes. Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 195
Answer:

Question 41.
USING STRUCTURE
Sketch the solid formed by the net. Then find the volume of the solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 196
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 41

Question 42.
USING STRUCTURE
Sketch the solid with the given views. Then find the volume of the solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 197

Answer:
Volume = 2.5 x 3.5 x 6
Volume = 52.5

Question 43.
OPEN-ENDED
Sketch two rectangular prisms that have volumes of 1000 cubic inches hut different surface areas. Include dimensions in your sketches.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 43

Question 44.
MODELING WITH MATHEMATICS
Which box gives you more cereal for your money? Explain.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 199

Answer:
First one gives more cerel for your money.

Explanation:
Bigger one volume = 16 x 4 x 10 = 640
Smaller one volume = 2 x 8 x 10 = 160
6 – 640 means 1 – 106.66
2 – 160 means 1 – 80

Question 45.
CRITICAL THINKING
A 3-inch by 5-inch index card is rotated around a horizontal line and a vertical line to produce two different solids. Which solid has a greater volume? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 199
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 45

Question 46.
CRITICAL THINKING
The height of cylinder X is twice the height of cylinder Y. The radius of cylinder X is half the radius of cylinder Y. Compare the volumes of cylinder X and cylinder Y. Justify your answer.

Answer:
Let the height of cylinder X be h, radius be r and its volume is πr²h
So, the height of cylinder Y is h/2 and radius is 2r, then the volume is 2πr²h
From both expressions, it can be seen that the volume of cylinder y is twice that of cylinder X.

Question 47.
USING STRUCTURE
Find the volume of the solid shown. The bases of the solid are sectors of circles.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 200
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 47

Question 48.
MATHEMATICAL CONNECTIONS
You drill a circular hole of radius r through the base of a cylinder of radius R. Assume the hole is drilled completely through to the other base. You want the volume of the hole to be half the volume of the cylinder. Express r as a function of R.

Answer:
r = √R²/2

Explanation:
The radius of a solid cylinder without a hole is R. So its volume is πR²h
As per the given condition, the volume of the hole must be half of that of the solid cylinder, hole volume is πR²h/2
Volume of cylinder V = πr²h
πR²h/2 = πr²h
R²/2 = r²
r = √R²/2
r = \(\frac { R√2 }{ 2 } \)

Question 49.
ANALYZING RELATIONSHIPS
How can you change the height of a cylinder so that the volume is increased by 25% but the radius remains the same?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 49

Question 50.
ANALYZING RELATIONSHIPS
How can you change the edge length of a cube so that the volume is reduced by 40%?

Answer:
Write the equation of volume of rectangular prism which can be used to evaluate the cube volume
Volume = s x s x s
The above equation shows that the volume of a cube is directly proportional to one of its side length, therefore, if the volume is to be reduced by 40%, then its the length of one of its side must be reduced by 40%, without changing the 2 other of its sides.

Question 51.
MAKING AN ARGUMENT
You have two objects of equal V0lume. Your friend says you can compare the densities of the objects by comparing their mass, because the heavier object will have a greater density. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 51

Question 52.
THOUGHT PROVOKING
Cavalieri’s Principle states that the two solids shown below have the same volume. Do they also have the same surface area? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 201
Answer:

Question 53.
PROBLEM SOLVING
A barn is in the shape of a pentagonal prism with the dimensions shown, The volume of the barn is 9072 cubic feel. Find the dimensions of each half of the root.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 202
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 53

Question 54.
PROBLEM SOLVING
A wooden box is in the shape of a regular pentagonal prism. The sides, top, and bottom of the box are 1 centimeter thick. Approximate the volume of wood used to construct the box. Round your answer to the nearest tenth.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 203
Answer:
Pentagon area = \(\frac { 5 }{ 2 } \)(4)² sin 72
A = 38
Volume = Area x height
Volume = 38 x 6 = 228

Maintaining Mathematical Proficiency

Find the surface area of the regular pyramid.

Question 55.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 204
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 55

Question 56.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 205
Answer:
Surface area = base area + 3bs
Surface area = 166.3 + 3(8)(10)
= 406.3

Question 57.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 206
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 57

11.6 Volumes of Pyramids

Exploration 1

Finding the Volume of a Pyramid

Work with a partner: The pyramid and the prism have the same height and the same square base.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 207
When the pyramid is filled with sand and poured into the prism, it takes three pyramids to fill the prism.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 208
Use this information to write a formula for the volume V of a pyramid.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Exploration 2

Finding the Volume of a Pyramid
Work with a partner: Use the formula you wrote in Exploration 1 to find the volume of the hexagonal pyramid.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 209
Answer:

Communicate Your Answer

Question 3.
How can you find the volume of a pyramid?
Answer:

Question 4.
In Section 11 .7, you will study volumes of cones. How do you think you could use a method similar to the one presented in Exploration 1 to write a formula for the volume of a cone? Explain your reasoning.
Answer:

Lesson 11.6 Volumes of Pyramids

Monitoring Progress

Find the volume of the pyramid.

Question 1.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 210
Answer:
The volume of the pyramid is 400 cm³

Explanation:
Volume V = \(\frac { 1 }{ 3 } \)Bh
V = \(\frac { 1 }{ 3 } \)(10 x 10 x 12)
V = 400

Question 2.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 211

Answer:
The volume of the pyramid is 2494.13 cm³

Explanation:
Volume V = \(\frac { 1 }{ 3 } \)Bh
V = \(\frac { 1 }{ 3 } \)(374.12 x 20)
V = 2494.13

Question 3.
The volume of a square pyramid is 75 cubic meters and the height is 9 meters. Find the side length of the square base.

Answer:
The side length of the square base is 5 m

Explanation:
Volume V = \(\frac { 1 }{ 3 } \)Bh = 75
\(\frac { 1 }{ 3 } \)B(9) = 75
B = 25
s = 5

Question 4.
Find the height of the triangular pyramid at the left.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 212

Answer:
The height of the triangular pyramid is 8 m

Explanation:
V = 24
\(\frac { 1 }{ 3 } \)Bh = 24
B = \(\frac { 1 }{ 2 } \)(3 x 6) = 9
\(\frac { 1 }{ 3 } \)(9)h = 24
h = 8

Question 5.
Pyramid C and pyramid D are similar. Find the volume of pyramid D.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 213

Answer:
The volume of pyramid D is 12 m³

Explanation:
\(\frac { volume of pyramid C }{ volume of pyramid D } \) = (\(\frac { pyramid C base }{ pyramid D base } \))³
\(\frac { 324 }{ V } \) = (\(\frac { 9 }{ 3 } \))³
V = 12

Question 6.
Find the volume of the composite solid.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 214

Answer:
the volume of solid = 96

Explanation:
Volume of prism = Bh
B = 8 x 2 = 16
V = 16 x 5 = 80
\(\frac { 1 }{ 3 } \)Bh
= \(\frac { 1 }{ 3 } \)(16 x 3) = 16
the volume of solid = 16 + 80 = 96

Exercise 11.6 Volumes of Pyramids

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Explain the difference between a triangular prism and a triangular pyramid.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 1

Question 2.
REASONING
A square pyramid and a cube have the same base and height. Compare the volume of the square pyramid to the volume of the cube.

Answer:
Square pyramid = 1/3 Bh
Cube = BH
So, the volume of the square pyramid is 1/3 of the volume of the cube.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, find the volume of the pyramid.

Question 3.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 215
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 3

Question 4.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 216

Answer:
V = 6 in³

Explanation:
V =  \(\frac { 1 }{ 3 } \)Bh
B = 2 x 3 = 6
V = \(\frac { 1 }{ 3 } \)(6 x 3)

In Exercises 5 – 8, find the indicated measure.

Question 5.
A pyramid with a square base has a volume of 120 cubic meters and a height of 10 meters. Find the side length of the square base.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 5

Question 6.
A pyramid with a square base has a volume of 912 cubic feet and a height of 19 feet. Find the side length of the square base.

Answer:
The side length of the square base is 12 ft

Explanation:
A pyramid with a square base has a volume of 912 cubic feet
h = 19
\(\frac { 1 }{ 3 } \)Bh = 912
\(\frac { 1 }{ 3 } \)B(19) = 912
B = 144
s = 12

Question 7.
A pyramid with a rectangular base has a volume of 480 cubic inches and a height of 10 inches. The width of the rectangular base is 9 inches. Find the length of the rectangular base.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 7

Question 8.
A pyramid with a rectangular base has a volume of 105 cubic centimeters and a height of 15 centimeters. The length of the rectangular base is 7 centimeters. Find the width of the rectangular base.

Answer:
The width of the rectangular base is 3 cm

Explanation:
A pyramid with a rectangular base has a volume of 105 cubic centimeters
h = 15
l = 7
\(\frac { 1 }{ 3 } \)Bh = 105
\(\frac { 1 }{ 3 } \)lbh = 105
\(\frac { 1 }{ 3 } \)(7 x 15 x b) = 105
b = 3

Question 9.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the pyramid.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 217
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 9

Question 10.
OPEN-ENDED
Give an example of a pyramid and a prism that have the same base and the same volume. Explain your reasoning.
Answer:
Let the rectangular prism have the base dimensions 4 x 2 nad a height of 5 so its volume is 4 x 2 x 5 = 40 cubic units
Therefore the base of the rectangular prism also have the dimensions of 4 x 2 and a height of 5 x 3 = 15 units so its volume V = 1/3 x 4 x 2 x 15 = 40 cubic units

In Exercises 11 – 14, find the height of the pyramid.

Question 11.
Volume = 15 ft3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 218
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 11

Question 12.
Volume = 224 in.3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 219

Answer:
The height of the pyramid is 10.5 in

Explanation:
Volume = 224
\(\frac { 1 }{ 3 } \)Bh = 224
B = 8² = 64
\(\frac { 1 }{ 3 } \)(64)h = 224
h = 10.5

Question 13.
Volume = 198 yd3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 220
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 13

Question 14.
Volume = 392 cm3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 221

Answer:
The height of the pyramid is 12 cm

Explanation:
Volume = 392
\(\frac { 1 }{ 3 } \)Bh = 392
B = 14 x 7 = 98
\(\frac { 1 }{ 3 } \)(98)h = 392
h = 12

In Exercises 15 and 16, the pyramids are similar. Find the volume of pyramid B.

Question 15.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 222
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 15

Question 16.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 223

Answer:
Volume of A = 80

Explanation:
\(\frac { Volume of B }{ Volume of A } \) = (\(\frac { Side of B }{ side of A } \))³
\(\frac { V }{ 10 } \) = (\(\frac { 6 }{ 3 } \))³
V = 8 x 10
Volume of A = 80

In Exercises 17 – 20, find the volume of the composite solid.

Question 17.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 224
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 17

Question 18.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 225

Answer:
Composite solid volume = 306

Explanation:
Base area = \(\frac { 1 }{ 2 } \)bh = \(\frac { 1 }{ 2 } \)(12 x 9) = 54
Bottom solid volume V = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(54 x 10)
V = 180
Top solid volume v = \(\frac { 1 }{ 3 } \)(54 x 7) = 126
Composite solid volume = 180 + 126 = 306

Question 19.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 226
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 19

Question 20.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 227

Answer:
Composite solid volume = 1152

Explanation:
Volume of Box = 12 x 12 x 12 = 1728
Square pyramid volume = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(144 x 12)
= 576
Composite solid volume = 1728 – 576 = 1152

Question 21.
ABSTRACT REASONING
A pyramid has a height of 8 feet and a square base with a side length of 6 feet.

a. How does the volume of the pyramid change when the base slays the same and the height is doubled?
b. How does the volume of the pyramid change when the height stays the same and the side length of the base is doubled?
C. Are your answers 10 parts (a) and (b) true for any square pyramid? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 21.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 21.2

Question 22.
HOW DO YOU SEE IT?
The cube shown is formed by three pyramids. each with the same square base and the same height. How could you use this to verify the formula for the volume of a pyramid?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 228
Answer:

Question 23.
CRITICAL THINKING
Find the volume of the regular pentagonal pyramid. Round your answer to the nearest hundredth. In the diagram. m∠ABC = 35°
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 229
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 23.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 23.2

Question 24.
THOUGHT PROVOKING
A frustum of a pyramid is the part of the pyramid that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a square pyramid in terms of a, b, and h. (Hint: Consider the “missing” top of the pyramid and use similar triangles.)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 230
Answer:

Question 25.
MODELING WITH MATHEMATICS
Nautical deck prisms were used as a safe way to illuminate decks on ships. The deck prism shown here is composed of the following three solids: a regular hexagonal prism with an edge length of 3.5 inches and a height of 1.5 inches, a regular hexagonal prism with an edge length of 3.25 inches arid a height of 0.25 inch, and a regular hexagonal pyramid with an edge length of 3 inches and a height of 3 inches. Find the volume of the deck prism.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 231
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 25.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 25.2

Maintaining Mathematical Proficiency

Find the value of X. Round your answer to the nearest tenth.

Question 26.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 232
Answer:
tan 35 = \(\frac { 9 }{ x } \)
0.7 = \(\frac { 9 }{ x } \)
x = 12.8

Question 27.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 233
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 27

Question 28.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 234
Answer:
tan 30 = \(\frac { x }{ 10 } \)
0.577 = \(\frac { x }{ 10 } \)
x = 5.77

Question 29.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 235
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 29

11.7 Surface Areas and Volumes of Cones

Exploration 1

Finding the Surface Area of a Cone

Work with a partner: Construct a circle with a radius of 3 inches. Mark the circumference of the circle into six equal parts, and label the length of each part. Then cut out one sector of the circle and make a cone.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 236
a. Explain why the base of the cone is a circle. What are the circumference and radius
of the base?
Answer:

b. What is the area of the original circle? What is the area with one sector missing?
Answer:

c. Describe the surface area of the cone, including the base. Use your description to find the surface area.
Answer:

Exploration 2

Finding the Volume of a Cone

Work with a partner: The cone and the cylinder have the same height and the same circular base.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 237
When the cone is filled with sand and poured into the cylinder. it takes three cones to fill the cylinder.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 238
Use this information to write a formula for the volume V of a cone.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
Answer:

Communicate Your Answer

Question 3.
How can you find the surface area and the volume of a cone?
Answer:

Question 4.
In Exploration 1, cut another sector from the circle and make a cone. Find the radius of the base and the surface area of the cone. Repeat this three times, recording your results in a table. Describe the pattern.
Answer:

Lesson 11.7 Surface Areas and Volumes of Cones

Monitoring progress

Question 1.
Find the surface area of the right cone.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 239

Answer:
The surface area of the right cone is 436.17 m²

Explanation:
r = 7.8
l = 10
S = πr² + πrl
S = π(7.8)² + π(7.8 x 10)
S = 436.17

Find the volume of the cone.

Question 2.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 240

Answer:
The volume of the cone is 2206.44 in³

Explanation:
r = 7, h = 13
l = √13² – 7²= 10.95
S = πr² + πrl
S = π7² + π(7 x 10.95)
S = 394.74
Volume V = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 7² x 13)
V = 2206.44

Question 3.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 241

Answer:
The volume of the cone is 163.4 m³

Explanation:
h = √8² – 5² = 6.24
Volume V = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 5² x 6.24)
V = 163.4

Question 4.
Cone C and cone D are similar. Find the volume of cone D.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 242

Answer:
Volume of cone D = 18.84 cm³

Explanation:
\(\frac { Volumeof cone C }{ Volume of cone D } \) = (\(\frac { height of C }{ height of D } \))³
\(\frac { 384π }{ Volume of cone D } \)= (\(\frac { 8 }{ 2 } \))³
Volume of cone D = 18.84

Question 5.
Find the volume of the composite solid.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 243

Answer:
Composite solid volume = 329.86 cm³

Explanation:
Volume of cylinder = πr²h = π(3)² x 10 = 90π
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
= \(\frac { 1 }{ 3 } \)(π x 3² x 5)
= 15π
Composite solid volume = 15π + 90π = 105π

Exercise 11.7 Surface Areas and Volumes of Cones

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe the differences between pyramids and cones. Describe their similarities.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 1

Question 2.
COMPLETE THE SENTENCE
The volume of a cone with radius r and height h is \(\frac{1}{3}\) the volume of a(n) __________ with radius r and height h.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the surface area of the right cone.

Question 3.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 244
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 3

Question 4.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 245

Answer:
The surface area of cone is 219.44 sq cm.

Explanation:
S = πr² + πrl
S = π(5.5)² + π(5.5 x 7.2)
S = 219.44

Question 5.
A right cone has a radius of 9 inches and a height of 12 inches.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 5

Question 6.
A right cone has a diameter of 11.2 feet and a height of 19.2 feet.

Answer:
The surface area is 421.52 sq ft.

Explanation:
r = 5.6
h = 19.2
l = √19.2² – 5.6² = 18.36
Surface area S = πr² + πrl
S = π(5.6)² + π(5.6 x 18.36)
S = 421.52

In Exercises 7 – 10, find the volume of the cone.

Question 7.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 246Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 246
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 7

Question 8.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 247

Answer:
The volume is 2.09 cubic meter

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π(1)² x 2)
V = 2.09

Question 9.
A cone has a diameter of 11.5 inches and a height of 15.2 inches.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 9

Question 10.
A right cone has a radius of 3 feet and a slant height of 6 feet.

Answer:
The volume is 56.54 cubic ft

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 3² x 6)
V = 56.54

In Exercises 11 and 12, find the missing dimension(s).

Question 11.
Surface area = 75.4 cm2
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 11

Question 12.
Volume = 216π in.3

Answer:
The radius is 6.13 in

Explanation:
Volume = 216π in.3
\(\frac { 1 }{ 3 } \)(πr²h) = 216
\(\frac { 1 }{ 3 } \)(πr² x 18) = 216
r = 6.13

In Exercises 13 and 14, the cones are similar. Find the volume of cone B.

Question 13.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 248
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 13

Question 14.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 249

Answer:
Volume of cone B = 24.127

Explanation:
\(\frac { Volumeof cone A }{ Volume of cone B } \) = (\(\frac { height of A }{ height of B } \))³
\(\frac { 120π }{ Volume of cone B } \) = (\(\frac { 10 }{ 4 } \))³
Volume of cone B = 24.127

In Exercises 15 and 16, find the volume of the composite solid.

Question 15.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 250
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 15

Question 16.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 251

Answer:
Volume of the composite solid = 97.93 cubic m

Explanation:
Volume of box = lbh
V = 51 x 5.1 x 5.1 = 132.651
Cone volume = \(\frac { 1 }{ 3 } \)(πr²h)
v = \(\frac { 1 }{ 3 } \)(π x 2.55² x 5.1)
v = 34.72
Volume of the composite solid = 132.651 – 34.72 = 97.93

Question 17.
ANALYZING RELATIONSHIPS
A cone has height h and a base with radius r. You warn to change the cone so its volume is doubled. What is the new height if you change only the height? What is the new radius if you change only the radius? Explain.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 17

Question 18.
HOW DO YOU SEE IT
A snack stand serves a small order of popcorn in a cone-shaped container and a large order of popcorn in a cylindrical container. Do not perform any calculations.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 252
a. How many small containers of popcorn do you have to buy to equal the amount of popcorn in a large container? Explain.

Answer:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 3² x 8) = 75.39
Volume of cylinder = πr²h = π x 3² x 8 = 226.19
Volume of cylinder / Volume of cone = \(\frac { 226.19 }{ 75.39 } \) = 3
You have to buy 3 small containers of popcorn to equal the amount of popcorn in a large containe.

b. Which container gives you more popcorn for your money? Explain.
Answer:
$1.25 -> 75.39 i.e $1 = 60.312
$2.50 -> 226.19 i.e $1 = 90.47
So, large containers gives you more popcorn for your money

In Exercises 19 and 20. find the volume of the right cone.

Question 19.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 253
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 19

Question 20.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 254

Answer:
Volume of cone is 575.62 cubic yd

Explanation:
tan 32 = \(\frac { 7 }{ h } \)
h = 11.21
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 7² x 11.21)
V = 575.62

Question 21.
MODELING WITH MATHEMATICS
A cat eats hail a cup of food, twice per day. Will the automatic pet feeder hold enough food for 10 days? Explain your reasoning. (1 cup ≈ 14.4 in.3)
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 255
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 21

Question 22.
MODELING WITH MATHEMATICS
During a chemistry lab, you use a funnel to pour a solvent into a flask. The radius of the funnel is 5 centimeters and its height is 10 centimeters. You pour the solvent into the funnel at a rate of 80 milliliters per second and the solvent flows out of the funnel at a rate of 65 milliliters per second. How long will it be before the funnel overflows? (1 mL = 1 cm3)
Answer:
17.45 seconds

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 5² x 10)
V = 261.8
\(\frac { 261.8 }{ 15 } \) = 17.45

Question 23.
REASONING
To make a paper drinking cup, start with a circular piece of paper that has a 3-inch radius, then follow the given steps. How does the surface area of the cup compare to the original paper circle? Find m∠ABC.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 256
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 23

Question 24.
THOUGHT PROVOKING
A frustum of a cone is the part of the cone that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a cone in terms of a, b, and h. (Hint: Consider the “missing” top of the cone and use similar triangles.)
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 257
Answer:
Volume V  = (1/3) * π * h * (r1² + r2² + (r1 * r2))

Question 25.
MAKING AN ARGUMENT
In the figure, the two cylinders are congruent The combined height of the two smaller cones equals the height of the larger cone. Your friend claims that this means the total volume of the two smaller cones is equal to the volume of the larger cone. Is your friend correct? Justify your answer.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 258
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 25

Question 26.
CRITICAL THINKING
When the given triangle is rotated around each of its sides. solids of revolution are formed. Describe the three solids and find their volumes. Give your answers in terms of π.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 259
Answer:

Maintaining Mathematical Proficiency

Find the indicated measure.

Question 27.
area of a circle with a radius of 7 feet
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 27

Question 28.
area of a circle with a diameter of 22 centimeters

Answer:
d = 11
A = πr²
A = 121π

Question 29.
diameter of a circle with an area of 256 square meters
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 29

Question 30.
radius of a circle with an area of 529 π square inches
Answer:
A = πr²
529π = πr²
r = 23

11.8 Surface Areas and Volumes of Spheres

Exploration 1

Finding the Surface Area of a Sphere

Work with a partner: Remove the covering from a baseball or softball.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 260
You will end up with two “figure 8” pieces of material, as shown above. From the amount of material it takes to cover the ball, what would you estimate the surface area S of the ball to be? Express your answer in terms of the radius r of the ball.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 261
Use the Internet or some other resource to confirm that the formula you wrote for the surface area of a sphere is correct.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to identify relevant external mathematical resources, such as content located on a website.
Answer:

Exploration 2

Finding the volume of a sphere

Work with a partner: A cylinder is circumscribed about a sphere, as shown. Write a formula for the volume V of the cylinder in terms of the radius r.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 262
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 263
When half of the sphere (a hemisphere) is filled with sand and poured into the cylinder, it takes three hemispheres to till the cylinder. Use this information to write a formula for the volume V of a sphere in terms of the radius r.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 264
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 265
Answer:

Communicate Your Answer

Question 3.
How can you find the surface area and the volume of a sphere?
Answer:

Question 4.
Use the results of Explorations 1 and 2 to find the surface area and the volume of a sphere with a radius of(a) 3 inches and (b) 2 centimeters.
Answer:

Lesson 11.8 Surface Areas and Volumes of Spheres

Monitoring Progress

Find the surface area of the sphere.

Question 1.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 266

Answer:
The surface area of the sphere is 5026.54 ft²

Explanation:
D = 40
r = 20
The surface area of the sphere = 4πr²
S = 4 x π x (20)²
S = 5026.54 ft²

Question 2.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 267

Answer:
The surface area of the sphere is 113.09 ft²

Explanation:
Circumference C = 6π
2πr = 6π
r = 3
The surface area of the sphere = 4πr²
S = 4π x 3²
S = 113.09

Question 3.
Find the radius of the sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 268

Answer:
The radius of the sphere is 2.73 m

Explanation:
The surface area of the sphere = 4πr²
30π = 4πr²
r = 2.73

Question 4.
The radius of a sphere is 5 yards. Find the volume of the sphere.

Answer:
The volume of the sphere is 523.59 yards³

Explanation:
r = 5
The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 5³
V = 523.59 yards³

Question 5.
The diameter of a sphere is 36 inches. Find the volume of the sphere.

Answer:
The volume of the sphere is 24429.02 in³

Explanation:
D = 36
r = 18
The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 18³
V = 24429.02

Question 6.
The surface area of a sphere is 576π square centimeters. Find the volume of the sphere.

Answer:
The volume of the sphere is 2304π cm³

Explanation:
The surface area of the sphere = 4πr²
576π = 4πr²
r = 12
The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 12³
V = 2304π

Question 7.
Find the volume of the composite solid at the left.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 269

Answer:
The volume of the composite solid is 7.324 m³

Explanation:
The volume of cone = πr²\(\frac { h }{ 3 } \)
= π x 1² x \(\frac { 5 }{ 3 } \) = 5.23
The volume of sphere = \(\frac { 4 }{ 3 } \)πr³
= \(\frac { 4 }{ 3 } \)π x 1³ = 4.188
The volume of the composite solid = The volume of cone + The volume of sphere/2
= 5.23 + 4.188/2
= 7.324 m³

Exercise 11.8 Surface Areas and Volumes of Spheres

Question 1.
VOCABULARY
When a plane intersects a sphere. what must be true for the intersection to be a great circle?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 1

Question 2.
WRITING
Explain the difference between a sphere and a hemisphere.

Answer:
Hemisphere is a related term of the sphere. Sphere and hemisphere are three-dimensional solids. The volume of sphere is \(\frac { 4 }{ 3 } \)πr³ and hemisphere volume is \(\frac { 2 }{ 3 } \)πr³. The surface area of the sphere is 4πr² and hemisphere surface area is 3πr².

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 6, find the surface area of the sphere.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 270
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 271

Answer:
The surface area of the sphere is 225π cm²

Explanation:
The surface area of the sphere = 4πr²
S = 4π x 7.5²
S = 225π

Question 5.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 272
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 273

Answer:
The surface area of the sphere is 8π ft²

Explanation:
C = 4π
2πr = 4π
r = 2
The surface area of the sphere = 4πr²
S = 4π x 2²
S = 8π

In Exercises 7 – 10. find the indicated measure.

Question 7.
Find the radius of a sphere with a surface area of 4π square feet.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 7

Question 8.
Find the radius of a sphere with a surface area of 1024π square inches.

Answer:
The radius of a sphere is 16 in

Explanation:
The surface area of the sphere = 1024π
4πr² = 1024π
r = 16

Question 9.
Find the diameter of a sphere with a surface area of 900π square meters.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 9

Question 10.
Find the diameter of a sphere with a surface area of 196π square centimeters.

Answer:
The diameter of a sphere is 14 cm

Explanation:
The surface area of the sphere = 196π
4πr² = 196π
r = 7
D = 2(7) = 14

In Exercises 11 and 12, find the surface area of the hemisphere.

Question 11.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 274
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 11

Question 12.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 275

Answer:
The surface area of the hemisphere is 108π in²

Explanation:
D = 12, r = 6
The surface area of the sphere = 3πr²
S = 3π x 6²
S = 108π

In Exercises 13 – 18. find the volume of the sphere.

Question 13.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 276
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 13

Question 14.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 277

Answer:
The volume of the sphere is 268.08 ft³

Explanation:
r = 4 ft
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 4³
V = 268.08 ft

Question 15.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 278
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 15

Question 16.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 279

Answer:
The volume of the sphere is 1436.75 ft³

Explanation:
D = 14 ft
r = 7 ft
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 7³
V = 1436.75 ft

Question 17.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 282
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 17

Question 18.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 280

Answer:
The volume of the sphere is 179.89 in³

Explanation:
C = 7π
2πr = 7π
r = 3.5
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 3.5³
V = 179.89 in

In Exercises 19 and 20, find the volume of the sphere with the given surface area.

Question 19.
Surface area = 16π ft2
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 19

Question 20.
Surface area = 484π cm2

Answer:
The volume of the sphere is 5575.27 cm³

Explanation:
Surface area = 484π
4πr² = 484π
r = 11
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 11³
V = 5575.27

Question 21.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 281
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 21

Question 22.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 283

Answer:
Diameter = 3
radius = 1.5
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x (1.5)³
V = 14.137 cubic in

In Exercises 23 – 26, find the volume of the composite solid.

Question 23.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 284
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 23

Question 24.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 285

Answer:
Volume is 288π ft³

Explanation:
Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³
= \(\frac { 2 }{ 3 } \)π x 6³ = 144π
volume of the cone = πr²\(\frac { h }{ 3 } \)
= π x 6² x \(\frac { 12 }{ 3 } \) = 144π
Area of circle = πr² = π x 6² = 36π
Volume of hemipshere + volume of the cone = 144π + 144π = 288π

Question 25.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 286
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 25

Question 26.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 287

Answer:
The volume of solid is 296π m³

Explanation:
Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³
= \(\frac { 2 }{ 3 } \)π x 6³ = 144π
Volume of cylinder = πr²h
= π x 6² x 14 = 504π
Volume of solid = 504π – 2(144π) = 296π

In Exercises 27 – 32, find the surface area and volume of the ball.

Question 27.
bowling ball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 289
d = 8.5 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 27

Question 28.
basketball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 289
C = 29.5 in.

Answer:
The surface area is 277 in², volume is 43212.27 in³

Explanation:
C = 29.5
2πr = 29.5
r = 4.69
Surface area = 4πr²
S = 4π x 4.69² = 277
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 4.69³
V = 43212.27

Question 29.
softball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 290
C = 12 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 29

Question 30.
golf ball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 291
d = 1.7 in.

Answer:
The surface area is 9.07 in², volume is 2.57 in³

Explanation:
d = 1.7
r = 0.85
Surface area = 4πr²
S = 4π x 0.85² = 9.07
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 0.85³
V = 2.57

Question 31.
volleyball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 292
C = 26 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 31

Question 32.
baseball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 293
C = 9 in.

Answer:
The surface area is 25.78 in², volume is 12.24 in³

Explanation:
C = 9
2πr = 9
r = 1.43
Surface area = 4πr²
S = 4π x 1.43²
S = 25.78
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 1.43³
V = 12.24

Question 33.
MAKING AN ARGUMENT
You friend claims that if the radius of a sphere is doubled, then the surface area of the sphere will also be doubled. Is our friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 33

Question 34.
REASONING
A semicircle with a diameter of 18 inches is rotated about its diameter. Find the surface area and the volume of the solid formed.

Answer:
The surface area is 1018 in², volume is 3054.02 in³

Explanation:
Diameter = 18
radius r = 9
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 9³
V = 3054.02
Surface area = 4πr²
S = 4π x 9²
S = 1018

Question 35.
MODELING WITH MATHEMATICS
A silo has the dimensions shown. The top of the silo is a hemispherical shape. Find the volume of the silo.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 294
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 35

Question 36.
MODELING WITH MATHEMATICS
Three tennis balls are stored in a cylindrical container with a height of 8 inches and a radius of 1.43 inches. The circumference of a tennis ball is 8 inches.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 295
a. Find the volume of a tennis ball.

Answer:
C = 8 in
2πr = 8
r = 1.27
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 1.27³ = 8.64
The volume of tennis ball = 8.64 in³

b. Find the amount of space within the cylinder not taken up by the tennis balls.
Answer:
The surface area of tennis ball S = 4πr²
S = 4π x 1.27² = 20.26
Area of cylinder s = 2πrh+2πr²
s = 2π x 1.43 x 8+2π x 1.43²
s = 84.72
Remaining space = 84.72 – 20.26 = 64.46 in²

Question 37.
ANALYZING RELATIONSHIPS
Use the table shown for a sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 296
a. Copy and complete the table. Leave your answers in terms of π.
b. What happens to the surface area of the sphere when the radius is doubled? tripled? quadrupled?
c. What happens to the volume of the sphere when the radius is doubled? tripled? quadrupled?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 37

Question 38.
MATHEMATICAL CONNECTIONS
A sphere has a diameter of 4(x + 3) centimeters and a surface area of 784 π square centimeters. Find the value of x.

Answer:
x =11

Explanation:
Surface area = 4πr²
784π = πr²
r = 28
2r = diameter = 4(x + 3)
r = 2(x + 3)
28 = 2(x + 3)
x = 11

Question 39.
MODELING WITH MATHEMATICS
The radius of Earth is about 3960 miles. The radius of the moon is about 1080 miles.
a. Find the surface area of Earth and the moon.
b. Compare the surface areas of Earth and the moon.
c. About 70% of the surface of Earth is water. How many square miles of water are on Earth’s surface?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 39

Question 40.
MODELING WITH MATHEMATICS
The Torrid Zone on Earth is the area between the Tropic of Cancer and the Tropic of Capricorn. The distance between these two tropics is about 3250 miles. You can estimate the distance as the height of a cylindrical belt around the Earth at the equator.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 297
a. Estimate the surface area of the Torrid Zone. (The radius of Earth is about 3960 miles.)
Answer:
Surface area of cylinder = 2πrh
S = 2π x 3960 x 3250 = 80875080
surface area of earth = 4πr²
= 4π x 3960² = 197086348.8

b. A meteorite is equally likely to hit anywhere on Earth. Estimate the probability that a meteorite will land in the Torrid Zone.
Answer:
Probability of meteorites hitting the torrid zone = 80875080/197086348.8 = 0.4104

Question 41.
ABSTRACT REASONING
A sphere is inscribed in a cube with a volume of 64 cubic inches. What is the surface area of the sphere? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 41

Question 42.
HOW DO YOU SEE IT?
The formula for the volume of a hemisphere and a Cone are shown. If each solid has the same radius and r = h, which solid will have a greater volume? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 298
Answer:
The hemisphere has the highest volume.

Explanation:
Volume of hemisphere v = \(\frac { 2 }{ 3 } \)πr³
Volume of cone V = \(\frac { 1 }{ 3 } \)πr²h
If r = h
Volume of cone V = \(\frac { 1 }{ 3 } \)πr² x r = \(\frac { 1 }{ 3 } \)πr³
So, the hemisphere has the highest volume.

Question 43.
CRITICAL THINKING
Let V be the volume of a sphere. S be the surface area of the sphere, and r be the radius of the sphere. Write an equation for V in terms of r and S. (Hint: Start with the ratio \(\frac{V}{S}\).)
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 43

Question 44.
THOUGHT PROVOKING
A spherical lune is the region between two great circles of a sphere. Find the formula for the area of a lune.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 299
Answer:

Question 45.
CRITICAL THINKING
The volume of a right cylinder is the same as the volume of a sphere. The radius of the sphere is 1 inch. Give three possibilities for the dimensions of the cylinder.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 45.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 45.2

Question 46.
PROBLEM SOLVING
A spherical cap is a portion of a sphere cut off by a plane. The formula for the volume of a spherical cap is V = \(\frac{\pi h}{6}\) (3a2 + h2), where a is the radius of the base of the cap and h is the height of the cap. Use the diagram and given information to find the volume of each spherical cap.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 300
a. r = 5ft, a = 4ft
Answer:

b. r = 34 cm, a = 30 cm
Answer:

c. r = 13 m, h = 8 m
Answer:

d. r=75 in., h = 54in.
Answer:

Question 47.
CRITICAL THINKING
A sphere with a radius of 2 inches is inscribed in a right cone with a height of 6 inches. Find the surface area and the volume of the cone.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 47.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 47.2
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 47.3

Maintaining Mathematical Proficiency

Solve the triangle. Round decimal answers to the nearest tenth.

Question 48.
A = 26°, C = 35°, b = 13

Answer:
B = 119°, a = 7.16, c = 9.5

Explanation:
B = 180 – (26 + 35) = 119
\(\frac { sin A }{ a } \) = \(\frac { sin B }{ b } \)
\(\frac { sin 26 }{ a } \) = \(\frac { sin 119 }{ 13 } \)
a = 7.16
\(\frac { sin C }{ c } \) = \(\frac { sin B }{ b } \)
\(\frac { sin 35 }{ c } \) = \(\frac { sin 119 }{ 13 } \)
c = 9.5

Question 49.
B = 102°, C = 43°, b = 21
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 49

Question 50.
a = 23, b = 24, c = 20

Answer:
A = 62.2, B = 65.5, C = 49.4

Explanation:
a² = b² + c² – 2bc cos A
23² = 24²+ 20² – 2(24 x 20) cos A
A = 62.2
\(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin B }{ 24 } \)
B = 65.5
\(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin C }{ 20 } \)
C = 49.4

Question 51.
A = 103°, b = 15, c = 24
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 51

Circumference, Area, and Volume Review

11.1 Circumference and Arc Length

Find the indicated measure.

Question 1.
diameter of ⊙P
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 301

Answer:
diameter of ⊙P is 29.99

Explanation:
Circumference = 94.24
πd = 94.24
d = 29.99

Question 2.
circumference of ⊙F
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 302

Answer:
circumference of ⊙F = 56.57

Explanation:
5.5 = \(\frac { 35 }{ 360 } \) . C
C = 56.57

Question 3.
arc length of \(\widehat{A B}\)
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 303

Answer:
arc length of \(\widehat{A B}\) = 26.09

Explanation:
arc length of \(\widehat{A B}\) = \(\frac { 115 }{ 360 } \) . 2π(13)
= 26.09

Question 4.
A mountain bike tire has a diameter of 26 inches. To the nearest foot, how far does the tire travel when it makes 32 revolutions?

Answer:
The tire travels 2613.80 inches.

Explanation:
D = 26 in
r = 13 in
Circumference C = 2π(13) = 81.68
32 revolutions = 32 x 81.68 = 2613.80

11.2 Areas of Circles and Sectors

Find the area of the blue shaded region.

Question 5.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 304

Answer:
Area = \(\frac { 240 }{ 360 } \) . π(9)²
= 169.64

Question 6.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 305

Answer:
Area of shaded region = 11.43

Explanation:
Area of rectangle = 6 x 4 = 24
Area of semicircle = π(2)² = 4π
Area of shaded region = 24 – 4π = 11.43

Question 7.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 306

Answer:
Area of shaded region = 173.13

Explanation:
Area of small region = 27.93 = \(\frac { 50 }{ 360 } \) . πr²
πr² = 201.096
r = 8
Area of shaded region = \(\frac { 310 }{ 360 } \) . π(8)²
= 173.13

11.3 Areas of Polygons

Find the area of the kite or rhombus.

Question 8.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 307
Answer:
Area = 65

Explanation:
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(13 x 20)
A = 65

Question 9.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 308
Answer:
Area = 48

Explanation:
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(16 x 12)
A = 48

Question 10.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 309
Answer:
Area = 52.5

Explanation:
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(14 x 15)
A = 52.5

Find the area of the regular polygon.

Question 11.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 310
Answer:
Area = \(\frac { 3√3 }{ 2 } \)a²
A = \(\frac { 3√3 }{ 2 } \)(8.8)²
A = 201.195

Question 12.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 311
Answer:
Area = \(\frac { 1 }{ 2 } \)(n . a. s)
A = \(\frac { 1 }{ 2 } \)(9 . 5.2 . 7.6)
A = 117.84

Question 13.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 312
Answer:
Area = \(\frac { 1 }{ 2 } \)(n . a. s)
A = \(\frac { 1 }{ 2 } \)(5 . 4 .  3.3)
A = 33

Question 14.
A platter is in the shape of a regular octagon with an apothem of 6 inches. Find the area of the platter.

Answer:
Area = \(\frac { 1 }{ 2 } \)(n . a. s)
A = \(\frac { 1 }{ 2 } \)(8 . 6 . sin 45)
A = 16.97

11.4 Three-Dimensional Figures

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 15.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 313
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

Question 16.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 314
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 15

Question 17.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 315
Answer:

Describe the cross section formed by the intersection of the plane and the solid.

Question 18.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 316
Answer:
The cross section is a rectangle.

Question 19.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 317
Answer:
The cross-section is a square.

Question 20.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 318
Answer:
the cross-section is a triangle.

11.5 Volumes of Prisms and Cylinders

Find the volume of the solid.

Question 21.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 319

Answer:
Volume = lbh
V = 3.6 x 2.1 x 1.5 = 113.4 m³

Question 22.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 320

Answer:
Volume = πr²h
V = π(2)² x 8 = 100.53 mm³

Question 23.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 321

Answer:
Pentagon area = 6.88
Volume = Area x height
V = 6.88 x 4 = 27.52 yd³

11.6 Volumes of Pyramids

Find the volume of the pyramid.

Question 24.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 322

Answer:
Volume V = Base area x height/3
Base Area = 9² = 81
V = 81 x 7/3 = 189 ft³

Question 25.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 323
Answer:
Volume V = Base area x height/3
Base Area = 4 x 15 = 60
V = 60x 20/3 = 400 yd³

Question 26.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 324

Answer:
Volume V = Base area x height/3
base area = 18 x 10 = 180
V = 180 x 5/3 = 300 m³

Question 27.
The volume of a square pyramid is 60 cubic inches and the height is 15 inches. Find the side length of the square base.

Answer:
The side length of the square base is 3.46 in

Explanation:
The volume of a square pyramid is 60 cubic inches
V = 60
s²h/3 = 60
s² x 15/3 = 60
s² = 12
s = 3.46

Question 28.
The volume of a square pyramid is 1024 cubic inches. The base has a side length of 16 inches. Find the height of the pyramid
Answer:
The volume of a square pyramid is 1024 cubic inches
s²h/3 = 1024
16²h = 3072
h = 12

11.7 Surface Areas and Volumes of Cones

Find the surface area and the volume of the cone.

Question 29.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 325
Answer:
Surface area is 678.58 cm²
volume is 1017.87 cm³

Explanation:
Surface area of cone S = πr² + πrl
S = π x 9² + π x 9 x 15
S = 678.58
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 9² x 12)
V = 1017.87

Question 30.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 326

Answer:
Surface area is 2513.27 cm²
volume is 8042.47 cm³

Explanation:
Surface area of cone S = πr² + πrl
S = π x 16² + π x 16 x 34
S = 2513.27
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 16² x 30)
V = 8042.47

Question 31.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 327

Answer:
Surface area is 439.82 m²
volume is 562.102 m³

Explanation:
Surface area of cone S = πr² + πrl
S = π x 7² + π x 7 x 13
S = 439.82
h = √13² – 7² = 10.95
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 7² x 10.95)
V = 562.102

Question 32.
A cone with a diameter of 16 centimeters has a volume of 320π cubic centimeters. Find the height of the cone.

Answer:
The height of the cone = 15 cm.

Explanation:
r = 8
Volume V = 320π
\(\frac { 1 }{ 3 } \)(πr²h) = 320π
\(\frac { 1 }{ 3 } \)(π x 8² x h) = 320π
h = 15

11.8 Surface Areas and Volumes of Spheres

Find the surface area and the volume of the sphere.

Question 33.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 328

Answer:
The surface area is 615.75 in², volume is 1436.75 in³

Explanation:
Surface area S = 4πr²
S = 4π x 7²
S = 615.75
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 7³
V = 1436.75

Question 34.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 329

Answer:
The surface area is 907.92 ft², volume is 2572.44 ft³

Explanation:
d = 17
r = 8.5
Surface area S = 4πr²
S = 4π x 8.5²
S = 907.92
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 7³
V = 2572.44

Question 35.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 330

Answer:
The surface area is 2827.43 ft², volume is 14137.16 ft³

Explanation:
C = 30π
2πr = 30π
r = 15
Surface area S = 4πr²
S = 4π x 15²
S = 2827.43
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 15³
V = 14137.16

Question 36.
The shape of Mercury can be approximated by a sphere with a diameter of 4880 kilometers. Find the surface area and the volume of Mercury.

Answer:
The surface area and the volume of Mercury is 23814400π, 19369045330π

Explanation:
d = 4880
r = 2440
Surface area S = 4πr²
S = 4π x 2440² = 23814400π
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 2440³
V = 19369045330π

Question 37.
A solid is composed of a cube with a side length of 6 meters and a hemisphere with a diameter of 6 meters. Find the volume of the composite solid.

Answer:
Volume of the composite solid = 272.52

Explanation:
Volume of cube = a³
= 6³ = 216
Volume of hemisphere = \(\frac { 4 }{ 6 } \)πr³
= \(\frac { 4 }{ 6 } \)π x 3³ = 18π
Volume of the composite solid = 216 + 18π = 272.52

Circumference, Area, and Volume Test

Find the volume of the solid.

Question 1.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 331

Answer:
Volume = 2577.29 m³

Explanation:
Volume = \(\frac { 3√3 }{ 2 } \)a²h
= \(\frac { 3√3 }{ 2 } \) x 8² x 15.5
= 2577.29

Question 2.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 332

Answer:
Volume is 17.157 ft³

Explanation:
d = 3.2
r = 1.6
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 1.6³
V = 17.157

Question 3.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 333

Answer:
Volume of sloid = 402.11 m³

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)πr²h
= \(\frac { 1 }{ 3 } \)π x 4² x 3
= 50.26
Volume of cylinder = πr²h
= π x 4² x 6 = 301.59
Volume of sloid = 2(50.26) + 301.59 = 402.11

Question 4.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 334

Answer:
Volume of solid = 106.66

Explanation:
Volume of rectangular box = 5 x 2 x 8 = 80
Volume of pyramid = 80/3 = 26.66
Volume of solid = 80 + 26.66 = 106.66

Find the indicated measure.

Question 5.
circumference of ⊙F
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 335

Answer:
circumference of ⊙F is 109.71 in

Explanation:
64 = \(\frac { 210 }{ 360 } \) • C
C = 109.7

Question 6.
m\(\widehat{G H}\)
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 336

Answer:
m\(\widehat{G H}\) = 74.27

Explanation:
35 = \(\frac { x }{ 360 } \) • 2π x 27
x = 74.27

Question 7.
area of shaded sector
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 337

Answer:
Area is 142.41 in²

Explanation:
Area = \(\frac { 360 – 105 }{ 360 } \) • π x 8²
Area = 142.41

Question 8.
Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 338
Answer:

Question 9.
Find the surface area of a right cone with a diameter of 10 feet and a height of 12 feet.

Answer:
The surface area is 486.7 sq ft

Explanation:
l² = r² + h²
l² = 5² + 12²
l = 13
Surface area S = πr² + 2πrl
S = π x 5² + 2π x 5 x 13
S = 486.7

Question 10.
You have a funnel with the dimensions shown.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 339
a. Find the approximate volume of the funnel.

Answer:
Volume = \(\frac { 1 }{ 3 } \)πr²h
= \(\frac { 1 }{ 3 } \)π x 6² x 10
= 376.99

b. You use the funnel to put oil in a ear. Oil flows out of the funnel at a rate of 45 milliliters per second. How long will it take to empty the funnel when it is full of oil? (1 mL = 1 cm3)
Answer:

c. How long would it take to empty a funnel with a radius of 10 centimeters and a height of 6 centimeters if oil flows out of the funnel at a rate of 45 milliliters per second?
Answer:

d. Explain why you can claim that the time calculated in part (c) is greater than the time calculated in part (b) without doing any calculations.
Answer:

Question 11.
A water bottle in the shape of a cylinder has a volume of 500 cubic centimeters. The diameter of a base is 7.5 centimeters. What is the height of the bottle? Justify your answer.

Answer:
The height of the bottle is 11.3 cm

Explanation:
Volume of cylinder = 500
πr²h = 500
π(3.75)²h = 500
h = 11.3 cm

Question 12.
Find the area of a dodecagon (12 sides) with a side length of 9 inches.

Answer:
Area is 237.31

Explanation:
Area = \(\frac { 1 }{ 4 } \)πa²cot(π/n)
= \(\frac { 1 }{ 4 } \)π x 9² x cot(π/12)
= 237.31

Question 13.
In general, a cardboard fan with a greater area does a better job of moving air and cooling you. The fan shown is a sector of a cardboard circle. Another fan has a radius of 6 centimeters and an intercepted are of 150°. Which fan does a better job of cooling you?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 340

Answer:

Explanation:

Circumference, Area, and Volume Cumulative Assessment

Question 1.
Identify the shape of the cross section formed by the intersection of the plane and the
solid.

a.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 341
Answer:
The cross-section is a trapezoid.

b.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 342
Answer:
The cross-section is a pentagon.

c.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 343
Answer:
The cross-section is a rectangle.

Question 2.
In the diagram, Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 344 is tangent to ⊙P at Q and \(\overline{P Q}\) is a radius of ⊙P? What must be true about Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 344 and \(\overline{P Q}\)? Select all that apply.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 345
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 346
Answer:
PQ is perpendicular to RS.

Question 3.
A crayon can be approximated by a composite solid made from a cylinder and a cone.
A crayon box is a rectangular prism. The dimensions of a crayon and a crayon box containing 24 crayons are shown.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 348
a. Find the volume of a crayon.
Answer:
The volume of a crayon = πr²h + \(\frac { 1 }{ 3 } \)πr²h
= π x 4.25² x 80 + \(\frac { 1 }{ 3 } \)π x 3.25² x 10
= 4539.6 + 110.61
= 4650.21 mm³

b. Find the amount of space within the crayon box not taken up by the crayons.
Answer:
Volume of box = 94 x 28 x 71 = 186872
The volume of a crayon = 4650.21
Remaining space = 186872 – 24 x 4650.21
= 75266.96

Question 4.
What is the equation ol the line passing through the point (2, 5) that is parallel to the line x + \(\frac{1}{2}\)y = – 1?
(A) y = – 2x + 9
(B) y = 2x + 1
(C) y = \(\frac{1}{2}\)x + 4
(D) y = –\(\frac{1}{2}\)x + 6
Answer:
(A) y = – 2x + 9

Explanation:
x + \(\frac{1}{2}\)y = – 1
y = -2 – 2x
The slope of the line is -2
The euation of line is y – 5 = -2(x – 2)
y – 5 = -2x + 4
y = -2x + 9

Question 5.
The top of the Washington Monument in Washington, D.C., is a square pyramid, called a pyramidion. What is the volume of the pyramidion?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 347
(A) 22,019.63 ft3
(B) 172,006.91 ft3
(C) 66,058.88 ft3
(D) 207,530.08 ft3

Answer:
(A) 22,019.63 ft3

Explanation:
Volume = a²\(\frac { h }{ 3 } \)
= 34.5² x \(\frac { 55.5 }{ 3 } \)
= 22019.62

Question 6.
Prove or disprove that the point (1, √3 ) lies on the circle centered at the origin and containing the point (0, 2).
Answer:
We consider the circle centered at the origin and containing the point (0, 2).
Therefore, we canconclude that rdaius is 2 and points be (0, 0), (1, √3)
distance = √(1 – 0)² + (√3 – 0)² = 2
As radius and distance are same. The point B(1, √3) lies on the circle.

Question 7.
Your friend claims that the house shown can be described as a composite solid made from a rectangular prism and a triangular prism. Do you support your friend’s claim? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 349
Answer:
Yes

Question 8.
The diagram shows a square pyramid and a cone. Both solids have the same height, h, and the base of the cone has radius r. According to Cavalieri’s Principle, the solids will have the same volume if the square base has sides of length ______ .
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 350
Answer:
2r/√2

Explanation:
Volume of square pyramid = a²\(\frac { h }{ 3 } \)
Square diagonal = √2a
radius = √2a/2
a = 2r/√2
Volume of cone = \(\frac { 1 }{ 3 } \)πr²h

Question 9.
About 19,400 people live in a region with a 5-mile radius. Find the population density in people per square mile.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 351

Answer:
The number of people per square mile is 247

Explanation:
S = πr²
= π x 5² = 78.5
Number of people per square mile = 19400/78.5 = 247

Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations

Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations

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Big Ideas Math Book Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations

Here is the complete list of topic-wise Big Ideas Math Book Algebra 1 Chapter 5 Solving Systems of Linear Equations Solution Key which covers the questions from the BIM Textbooks based on the latest Common Core Curriculum. Ch 5 Big Ideas Math Textbook Algebra 1 Answers material given here offers Questions from Exercises(5.1 to 5.7), Chapter Tests, Review Tests, Quiz, Assessment Tests, Cumulative Assessments, etc. Practice thoroughly and gain more subject knowledge on the concepts of BIM Math Book Algebra 1 Chapter 5 Solutions.

Solving Systems of Linear Equations Maintaining Mathematical Proficiency

Graph the equation.
Question 1.
y + 4 = x
Answer:

Question 2.
6x – y = -1
Answer:

Question 3.
4x + 5y = 20
Answer:

Question 4.
-2y + 12 = -3x
Answer:

Solve the inequality. Graph the solution.
Question 5.
m + 4 > 9
Answer:

Question 6.
24 ≤ -6t
Answer:

Question 7.
2a – 5 ≤ 13
Answer:

Question 8.
-5z + 1 < -14
Answer:

Question 9.
4k – 16 < k + 2
Answer:

Question 10.
7w + 12 ≥ 2w – 3
Answer:

Question 11.
ABSTRACT REASONING
The graphs of the linear functions g and h have different slopes. The value of both functions at x = a is b. When g and h are graphed in the same coordinate plane, what happens at the point (a, b)?
Answer:

Solving Systems of Linear Equations Mathematical Practices

Mathematically proficient students use technological tools to explore concepts.

Monitoring Progress

Use a graphing calculator to find the point of intersection of the graphs of the two linear equations.
Question 1.
y = -2x – 3
y = \(\frac{1}{2}\)x – 3
Answer:

Question 2.
y = -x + 1
y = x- 2
Answer:

Question 3.
3x – 2y = 2
2x – y = 2
Answer:

Lesson 5.1 Solving Systems of Linear Equations by Graphing

Essential Question How can you solve a system of linear equations?

EXPLORATION 1

Writing a System of Linear EquationsWork with a partner. Your family opens a bed-and-breakfast. They spend $600 preparing a bedroom to rent. The cost to your family for food and utilities is $15 per night. They charge $75 per night to rent the bedroom.
a. Write an equation that represents the costs.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 1
b. Write an equation that represents the revenue (income).
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 2
c. A set of two (or more) linear equations is called a system of linear equations. Write the system of linear equations for this problem.

EXPLORATION 2

Using a Table or Graph to Solve a System
Work with a partner. Use the cost and revenue equations from Exploration 1 to determine how many nights your family needs to rent the bedroom before recovering the cost of preparing the bedroom. This is the break-even point.
a. Copy and complete the table.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 3
b. How many nights does your family need to rent the bedroom before breaking even?
c. In the same coordinate plane, graph the cost equation and the revenue equation from Exploration 1.
d. Find the point of intersection of the two graphs. What does this point represent? How does this compare to the break-even point in part (b)? Explain.

Communicate Your Answer

Question 3.
How can you solve a system of linear equations? How can you check your solution?
Answer:

Question 4.
Solve each system by using a table or sketching a graph. Explain why you chose each method. Use a graphing calculator to check each solution.
a. y = -4.3x – 1.3
y = 1.7x + 4.7

b. y = x
y = -3x + 8

c. y = -x – 1
y = 3x + 5
Answer:

Monitoring Progress

Tell whether the ordered pair is a solution of the system of linear equations.
Question 1.
(1, -2); 2x + y = 0
-x + 2y = 5
Answer:

Question 2.
(1, 4); y = 3x + 1
y = -x + 5
Answer:

Solve the system of linear equations by graphing.
Question 3.
y = x – 2
y = -x + 4
Answer:

Question 4.
y = \(\frac{1}{2}\)x + 3
y = –\(\frac{3}{2}\)x – 5
Answer:

Question 5.
2x + y = 5
3x – 2y = 4
Answer:

Question 6.
You have a total of 18 math and science exercises for homework. You have six more math exercises than science exercises. How many exercises do you have in each subject?
Answer:

Solving Systems of Linear Equations by Graphing 5.1 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
Do the equations 5y – 2x = 18 and 6x = -4y – 10 form a system of linear equations? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Consider the system of linear equations -4x + 2y = 4 and 4x – y = -6. Which is different? Find “both” answers.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, tell whether the ordered pair is a solution of the system of linear equations.
Question 3.
(2, 6); x + y = 8
3x – y = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 3

Question 4.
(8, 2); x – y = 6
2x – 10y = 4
Answer:

Question 5.
(-1, 3); y = -7x – 4
y = 8x + 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 5

Question 6.
(-4, -2); y = 2x + 6
y = -3x – 14
Answer:

Question 7.
(-2, 1); 6x + 5y = -7
2x – 4y = -8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 7

Question 8.
(5, -6); 6x + 3y = 12
4x + y = 14
Answer:

In Exercises 9–12, use the graph to solve the system of linear equations. Check your solution.
Question 9.
x – y = 4
4x + y = 1
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 9

Question 10.
x + y = 5
y – 2x = -4
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 6
Answer:

Question 11.
6y + 3x = 18
-x + 2y = 24
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 11

Question 12.
2x – y = -2
2x + 4y = 8
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 8
Answer:

In Exercises 13–20, solve the system of linear equations by graphing.
Question 13.
y = -x + 7
y = x + 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 13

Question 14.
y = -x + 4
y = 2x – 8
Answer:

Question 15.
y = \(\frac{1}{3}\)x + 2
y = \(\frac{2}{3}\)x + 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 15

Question 16.
y = \(\frac{3}{4}\)x – 4
y = –\(\frac{1}{2}\)x + 11
Answer:

Question 17.
9x + 3y = -3
2x – y = -4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 17.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 17.2

Question 18.
4x – 4y = 20
y = -5
Answer:

Question 19.
x – 4y = -4
-3x – 4y = 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 19.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 19.2

Question 20.
3y + 4x = 3
x + 3y = -6
Answer:

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in solving the system of linear equations.
Question 21.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 21.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 21.2

Question 22.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 10
Answer:

USING TOOLS In Exercises 23–26, use a graphing calculator to solve the system of linear equations.
Question 23.
0.2x + 0.4y = 4
-0.6x + 0.6y = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 23

Question 24.
-1.6x – 3.2y = -24
2.6x + 2.6y = 26
Answer:

Question 25.
-7x + 6y = 0
0.5x + y = 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 25

Question 26.
4x – y = 1.5
2x + y = 1.5
Answer:

Question 27.
MODELING WITH MATHEMATICS
You have 40 minutes to exercise at the gym, and you want to burn 300 calories total using both machines. How much time should you spend on each machine?
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 11
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 27.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 27.3

Question 28.
MODELING WITH MATHEMATICS
You sell small and large candles at a craft fair. You collect $144 selling a total of 28 candles. How many of each type of candle did you sell?
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 12
Answer:

Question 29.
MATHEMATICAL CONNECTIONS
Write a linear equation that represents the area and a linear equation that represents the perimeter of the rectangle. Solve the system of linear equations by graphing. Interpret your solution.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 13
Answer:

Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 29.2

Question 30.
THOUGHT PROVOKING
Your friend’s bank account balance (in dollars) is represented by the equation y = 25x + 250, where x is the number of months. Graph this equation. After 6 months, you want to have the same account balance as your friend. Write a linear equation that represents your account balance. Interpret the slope and y-intercept of the line that represents your account balance.
Answer:

Question 31.
COMPARING METHODS
Consider the equation x + 2 = 3x – 4.
a. Solve the equation using algebra.
b. Solve the system of linear equations y = x + 2 and y = 3x – 4 by graphing.
c. How is the linear system and the solution in part (b) related to the original equation and the solution in part (a)?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 31.2

Question 32.
HOW DO YOU SEE IT?
A teacher is purchasing binders for students. The graph shows the total costs of ordering x binders from three different companies.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 14
a. For what numbers of binders are the costs the same at two different companies? Explain.
b. How do your answers in part (a) relate to systems of linear equations?
Answer:

Question 33.
MAKING AN ARGUMENT
You and a friend are going hiking but start at different locations. You start at the trailhead and walk 5 miles per hour. Your friend starts 3 miles from the trailhead and walks 3 miles per hour.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.1 15
a. Write and graph a system of linear equations that represents this situation.
b. Your friend says that after an hour of hiking you will both be at the same location on the trail. Is your friend correct? Use the graph from part (a) to explain your answer.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 33.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 33.2

Maintaining Mathematical Proficiency

Solve the literal equation for y.
Question 34.
10x + 5y = 5x + 20
Answer:

Question 35.
9x + 18 = 6y – 3x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.1 Question 35

Question 36.
\(\frac{3}{4}\) x + \(\frac{1}{4}\) y = 5
Answer:

Lesson 5.2 Solving Systems of Linear Equations by Substitution

Essential Question How can you use substitution to solve a system of linear equations?

EXPLORATION 1

Using Substitution to Solve Systems
Work with a partner. Solve each system of linear equations using two methods.
Method 1 Solve for x first.
Solve for x in one of the equations. Substitute the expression for x into the other equation to find y. Then substitute the value of y into one of the original equations to find x.

Method 2 Solve for y first.
Solve for y in one of the equations. Substitute the expression for y into the other equation to find x. Then substitute the value of x into one of the original equations to find y.
Is the solution the same using both methods? Explain which method you would prefer to use for each system
a. x + y = -7
-5x + y = 5

b. x – 6y = -11
3x + 2y = 7

c. 4x + y = -1
3x – 5y = -18

EXPLORATION 2

Writing and Solving a System of Equations
Work with a partner.
a. Write a random ordered pair with integer coordinates. One way to do this is to use a graphing calculator. The ordered pair generated at the right is (-2, -3).
b. Write a system of linear equations that has your ordered pair as its solution.
c. Exchange systems with your partner and use one of the methods from Exploration 1 to solve the system. Explain your choice of method.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 1

Communicate Your Answer

Question 3.
How can you use substitution to solve a system of linear equations?
Answer:

Question 4.
Use one of the methods from Exploration 1 to solve each system of linear equations. Explain your choice of method. Check your solutions.
a. x + 2y = -7
2x – y = -9

b.x – 2y = -6
2x + y = -2

c.-3x + 2y = -10
-2x + y = -6

d. 3x + 2y = 13
x – 3y = -3

e. 3x – 2y = 9
-x – 3y = 8

f. 3x – y = -6
4x + 5y = 11
Answer:

Monitoring Progress

Solve the system of linear equations by substitution. Check your solution.
Question 1.
y = 3x + 14
y = -4x
Answer:

Question 2.
3x + 2y = 0
y = \(\frac{1}{2}\)x – 1
Answer:

Question 3.
x = 6y – 7
4x + y = -3
Answer:

Solve the system of linear equations by substitution. Check your solution.
Question 4.
x + y = -2
-3x + y = 6
Answer:

Question 5.
-x + y = -4
4x – y = 10
Answer:

Question 6.
2x – y = -5
3x – y = 1
Answer:

Question 7.
x – 2y = 7
3x – 2y = 3
Answer:

Question 8.
There are a total of 64 students in a drama club and a yearbook club. The drama club has 10 more students than the yearbook club. Write a system of linear equations that represents this situation. How many students are in each club?
Answer:

Solving Systems of Linear Equations by Substitution 5.2 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Describe how to solve a system of linear equations by substitution.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 1

Question 2.
NUMBER SENSE
When solving a system of linear equations by substitution, how do you decide which variable to solve for in Step 1?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3−8, tell which equation you would choose to solve for one of the variables. Explain.
Question 3.
x + 4y = 30
x – 2y = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 3

Question 4.
3x – y = 0
2x + y = -10
Answer:

Question 5.
5x + 3y = 11
5x – y = 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 5

Question 6.
3x – 2y = 19
x + y = 8
Answer:

Question 7.
x – y = -3
4x + 3y = -5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 7

Question 8.
3x + 5y = 25
x – 2y = -6
Answer:

In Exercises 9–16, solve the system of linear equations by substitution. Check your solution.
Question 9.
x = 17 – 4y
y = x – 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 9.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 9.2

Question 10.
6x – 9 = y
y = -3x
Answer:

Question 11.
x = 16 – 4y
3x + 4y = 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 11.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 11.2

Question 12.
-5x + 3y = 51
y = 10x – 8
Answer:

Question 13.
2x = 12
x – 5y = -29
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 13

Question 14.
2x – y = 23
x – 9 = -1
Answer:

Question 15.
5x + 2y = 9
x + y = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 15.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 15.2

Question 16.
11x – 7y = -14
x – 2y = -4
Answer:

Question 17.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the linear system 8x + 2y = -12 and 5x – y = 4.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 17.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 17.2

Question 18.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the linear system 4x + 2y = 6 and 3x + y = 9.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 3
Answer:

Question 19.
MODELING WITH MATHEMATICS
A farmer plants corn and wheat on a 180-acre farm. The farmer wants to plant three times as many acres of corn as wheat. Write a system of linear equations that represents this situation. How many acres of each crop should the farmer plant?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 19.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 19.2

Question 20.
MODELING WITH MATHEMATICS
A company that offers tubing trips down a river rents tubes for a person to use and “cooler” tubes to carry food and water. A group spends $270 to rent a total of 15 tubes. Write a system of linear equations that represents this situation. How many of each type of tube does the group rent?
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 4
Answer:

In Exercises 21–24, write a system of linear equations that has the ordered pair as its solution.
Question 21.
(3, 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 21

Question 22.
(-2, 8)
Answer:

Question 23.
(-4, -12)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 23

Question 24.
(15, -25)
Answer:

Question 25.
PROBLEM SOLVING
A math test is worth 100 points and has 38 problems. Each problem is worth either 5 points or 2 points. How many problems of each point value are on the test?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 25.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 25.2

Question 26.
PROBLEM SOLVING
An investor owns shares of Stock A and Stock B. The investor owns a total of 200 shares with a total value of $4000. How many shares of each stock does the investor own?
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 5
Answer:

MATHEMATICAL CONNECTIONS In Exercises 27 and 28, (a) write an equation that represents the sum of the angle measures of the triangle and (b) use your equation and the equation shown to find the values of x and y.
Question 27.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 27.2

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 7
Answer:

Question 29.
REASONING
Find the values of a and b so that the solution of the linear system is (-9, 1).
ax + by = -31 Equation 1
ax – by = -41 Equation 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 29.2

Question 30.
MAKING AN ARGUMENT
Your friend says that given a linear system with an equation of a horizontal line and an equation of a vertical line, you cannot solve the system by substitution. Is your friend correct? Explain.
Answer:

Question 31.
OPEN-ENDED
Write a system of linear equations in which (3, -5) is a solution of Equation 1 but not a solution of Equation 2, and (-1, 7) is a solution of the system.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 31.2

Question 32.
HOW DO YOU SEE IT?
The graphs of two linear equations are shown.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 8
a. At what point do the lines appear to intersect?
b. Could you solve a system of linear equations by substitution to check your answer in part (a)? Explain.
Answer:

Question 33.
REPEATED REASONING
A radio station plays a total of 272 pop, rock, and hip-hop songs during a day. The number of pop songs is 3 times the number of rock songs. The number of hip-hop songs is 32 more than the number of rock songs. How many of each type of song does the radio station play?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 33.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 33.2

Question 34.
THOUGHT PROVOKING
You have $2.65 in coins. Write a system of equations that represents this situation. Use variables to represent the number of each type of coin.
Answer:

Question 35.
NUMBER SENSE
The sum of the digits of a two-digit number is 11. When the digits are reversed, the number increases by 27. Find the original number
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 35.2

Maintaining Mathematical Proficiency

Find the sum or difference.
Question 36.
(x – 4) + (2x – 7)
Answer:

Question 37.
(5y – 12) + (-5y – 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 37.1

Question 38.
(t – 8) – (t + 15)
Answer:

Question 39.
(6d + 2) – (3d – 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 39

Question 40.
4(m + 2) + 3(6m – 4)
Answer:

Question 41.
2(5v + 6) – 6(-9v + 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.2 Question 41

Lesson 5.3 Solving Systems of Linear Equations by Elimination

Essential Question How can you use elimination to solve a system of linear equations?

EXPLORATION 1

Writing and Solving a System of Equations
Work with a partner. You purchase a drink and a sandwich for $4.50. Your friend purchases a drink and five sandwiches for $16.50. You want to determine the price of a drink and the price of a sandwich.
a. Let x represent the price (in dollars) of one drink. Let y represent the price (in dollars) of one sandwich. Write a system of equations for the situation. Use the following verbal model.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 1
Label one of the equations Equation 1 and the other equation Equation 2.
b. Subtract Equation 1 from Equation 2. Explain how you can use the result to solve the system of equations. Then find and interpret the solution.

EXPLORATION 2

Using Elimination to Solve Systems
Work with a partner. Solve each system of linear equations using two methods.
Method 1 Subtract. Subtract Equation 2 from Equation 1. Then use the result to solve the system.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 2
Method 2 Add. Add the two equations. Then use the result to solve the system.
Is the solution the same using both methods? Which method do you prefer?
a. 3x – y = 6
3x + y = 0

b. 2x + y = 6
2x – y = 2

c. x – 2y = -7
x + 2y = 5

EXPLORATION 3

Using Elimination to Solve a System
Work with a partner.
2x + y = 7 Equation 1
x + 5y = 17 Equation 2
a. Can you eliminate a variable by adding or subtracting the equations as they are? If not, what do you need to do to one or both equations so that you can?
b. Solve the system individually. Then exchange solutions with your partner and compare and check the solutions.

Communicate Your Answer

Question 4.
How can you use elimination to solve a system of linear equations?
Answer:

Question 5.
When can you add or subtract the equations in a system to solve the system? When do you have to multiply first? Justify your answers with examples.
Answer:

Question 6.
In Exploration 3, why can you multiply an equation in the system by a constant and not change the solution of the system? Explain your reasoning.
Answer:

Monitoring Progress

Solve the system of linear equations by elimination. Check your solution.
Question 1.
3x + 2y = 7
-3x + 4y = 5
Answer:

Question 2.
x – 3y = 24
3x + y = 12
Answer:

Question 3.
x + 4y = 22
4x + y = 13
Answer:

Question 4.
Solve the system in Example 3 by eliminating x.
Answer:

Solving Systems of Linear Equations by Elimination 5.3 Exercises

Vocabulary and Core Concept Check
Question 1.
OPEN-ENDED
Give an example of a system of linear equations that can be solved by first adding the equations to eliminate one variable.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 1

Question 2.
WRITING
Explain how to solve the system of linear equations by elimination.
2x – 3y = -4 Equation 1
-5x + 9y = 7 Equation 2
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3−10, solve the system of linear equations by elimination. Check your solution.
Question 3.
x + 2y = 13
-x + y = 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 3.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 3.2

Question 4.
9x + y = 2
-4x – y = -17
Answer:

Question 5.
5x + 6y = 50
x – 6y = -26
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 5.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 5.2

Question 6.
-x + y = 4
x + 3y = 4
Answer:

Question 7.
-3x – 5y = -7
-4x + 5y = 14
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 7.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 7.2

Question 8.
4x – 9y = -21
-4x – 3y = 9
Answer:

Question 9.
-y – 10 = 6x
5x + y = -10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 9.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 9.2

Question 10.
3x – 30 = y
7y – 6 = 3x
Answer:

In Exercises 11–18, solve the system of linear equations by elimination. Check your solution.
Question 11.
x + y = 2
2x + 7y = 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 11.1

Question 12.
8x – 5y = 11
4x – 3y = 5
Answer:

Question 13.
11x – 20y = 28
3x + 4y = 36
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 13.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 13.2

Question 14.
10x – 9y = 46
-2x + 3y = 10
Answer:

Question 15.
4x – 3y = 8
5x – 2y = -11
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 15.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 15.2

Question 16.
-2x – 5y = 9
3x + 11y = 4
Answer:

Question 17.
9x + 2y = 39
6x + 13y = -9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 17.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 17.2

Question 18.
12x – 7y = -2
8x + 11y = 30
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the linear system 5x – 7y = 16 and x + 7y = 8.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 19

Question 20.
ERROR ANALYSIS
Describe and correct the error in solving for one of the variables in the linear system 4x + 3y = 8 and x – 2y = -13.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 4
Answer:

Question 21.
MODELING WITH MATHEMATICS
A service center charges a fee of x dollars for an oil change plus y dollars per quart of oil used. A sample of its sales record is shown. Write a system of linear equations that represents this situation. Find the fee and cost per quart of oil.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 21.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 21.2

Question 22.
MODELING WITH MATHEMATICS
A music website charges x dollars for individual songs and y dollars for entire albums. Person A pays $25.92to download 6 individual songs and 2 albums. Person B pays $33.93 to download 4 individual songs and 3 albums. Write a system of linear equations that represents this situation. How much does the website charge to download a song? an entire album?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 6
Answer:

In Exercises 23–26, solve the system of linear equations using any method. Explain why you chose the method.
Question 23.
3x + 2y = 4
2y = 8 – 5x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 23.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 23.2

Question 24.
-6y + 2 = -4x
y – 2 = x
Answer:

Question 25.
y – x = 2
y = – \(\frac{1}{4}\) x + 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 25

Question 26.
3x + y = \(\frac{1}{3}\)
2x – 3y = \(\frac{8}{3}\)
Answer:

Question 27.
WRITING
For what values of a can you solve the linear system ax + 3y = 2 and 4x + 5y = 6 by elimination without multiplying first? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 27

Question 28.
HOW DO YOU SEE IT?
The circle graph shows the results of a survey in which 50 students were asked about their favorite meal.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 7
a. Estimate the numbers of students who chose breakfast and lunch.
b. The number of students who chose lunch was 5 more than the number of students who chose breakfast. Write a system of linear equations that represents the numbers of students who chose breakfast and lunch.
c. Explain how you can solve the linear system in part (b) to check your answers in part (a).
Answer:

Question 29.
MAKING AN ARGUMENT
Your friend says that any system of equations that can be solved by elimination can be solved by substitution in an equal or fewer number of steps. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 29

Question 30.
THOUGHT PROVOKING
Write a system of linear equations that can be added to eliminate a variable or subtracted to eliminate a variable.
Answer:

Question 31.
MATHEMATICAL CONNECTIONS
A rectangle has a perimeter of 18 inches. A new rectangle is formed by doubling the width w and tripling the length ℓ, as shown. The new rectangle has a perimeter P of 46 inches.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 8
a. Write and solve a system of linear equations to find the length and width of the original rectangle.
b. Find the length and width of the new rectangle.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 31.2

Question 32.
CRITICAL THINKING
Refer to the discussion of System 1 and System 2 on page 248. Without solving, explain why the two systems shown have the same solution.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations 5.3 9
Answer:

Question 33.
PROBLEM SOLVING
You are making 6 quarts of fruit punch for a party. You have bottles of 100% fruit juice and 20% fruit juice. How many quarts of each type of juice should you mix to make 6 quarts of 80% fruit juice?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 33.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 33.2

Question 34.
PROBLEM SOLVING
A motorboat takes 40 minutes to travel 20 miles downstream. The return trip takes 60 minutes. What is the speed of the current?
Answer:

Question 35.
CRITICAL THINKING
Solve for x, y, and z in the system of equations. Explain your steps.
x + 7y + 3z = 29 Equation 1
3z + x – 2y = -7 Equation 2
5y = 10 – 2x Equation 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 35.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 35.3

Maintaining Mathematical Proficiency

Solve the equation. Determine whether the equation has one solution, no solution, or infinitely many solutions.
Question 36.
5d – 8 = 1 + 5d
Answer:

Question 37.
9 + 4t = 12 – 4t
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 37

Question 38.
3n + 2 = 2(n – 3)
Answer:

Question 39.
-3(4 – 2v) = 6v – 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 39

Write an equation of the line that passes through the given point and is parallel to the given line.
Question 40.
(4, -1); y = -2x + 7
Answer:

Question 41.
(0, 6); y = 5x – 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.3 Question 41

Question 42.
(-5, -2); y = \(\frac{2}{3}\)x + 1
Answer:

Lesson 5.4 Solving Special Systems of Linear Equations

Essential Question Can a system of linear equations have no solution or infinitely many solutions?

EXPLORATION 1

Using a Table to Solve a System
Work with a partner. You invest $450 for equipment to make skateboards. The materials for each skateboard cost $20. You sell each skateboard for $20.
a. Write the cost and revenue equations. Then copy and complete the table for your cost C and your revenue R.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 1
b. When will your company break even? What is wrong?
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 2

EXPLORATION 2

Writing and Analyzing a System
Work with a partner. A necklace and matching bracelet have two types of beads. The necklace has 40 small beads and 6 large beads and weighs 10 grams. The bracelet has 20 small beads and 3 large beads and weighs 5 grams. The threads holding the beads have no significant weight.
a. Write a system of linear equations that represents the situation. Let x be the weight (in grams) of a small bead and let y be the weight (in grams) of a large bead.
b. Graph the system in the coordinate plane shown. What do you notice about the two lines?
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 3
c. Can you find the weight of each type of bead? Explain your reasoning.

Communicate Your Answer

Question 3.
Can a system of linear equations have no solution or infinitely many solutions? Give examples to support your answers.
Answer:

Question 4.
Does the system of linear equations represented by each graph have no solution, one solution, or infinitely many solutions? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 4
Answer:

Monitoring Progress

Solve the system of linear equations.
Question 1.
x + y = 3
2x + 2y = 6
Answer:

Question 2.
y = -x + 3
2x + 2y = 4
Answer:

Question 3.
x + y = 3
x + 2y = 4
Answer:

Question 4.
y = -10x + 2
10x + y = 10
Answer:

Question 5.
WHAT IF?
What happens to the solution in Example 3 when the perimeter of the trapezoidal piece of land is 96 kilometers? Explain.
Answer:

Solving Special Systems of Linear Equations 5.4 Exercises

Vocabulary and Core Concept Check
Question 1.
REASONING
Is it possible for a system of linear equations to have exactly two solutions? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 1

Question 2.
WRITING
Compare the graph of a system of linear equations that has infinitely many solutions and the graph of a system of linear equations that has no solution.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3−8, match the system of linear equations with its graph. Then determine whether the system has one solution, no solution, or infinitely many solutions.
Question 3.
-x + y = 1
x – y = 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 2.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 2.2

Question 4.
2x – 2y = 4
-x + y = -2
Answer:

Question 5.
2x + y = 4
-4x – 2y = -8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 3.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 3.2

Question 6.
x – y = 0
5x – 2y = 6
Answer:

Question 7.
-2x + 4y = 1
3x – 6y = 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 7.1

Question 8.
5x + 3y = 17
x – 3y = -2
Answer:

Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 5

In Exercises 9–16, solve the system of linear equations.
Question 9.
y = -2x – 4
y = 2x – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 9

Question 10.
y = -6x – 8
y = -6x + 8
Answer:

Question 11.
3x – y = 6
-3x + y = -6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 11

Question 12.
-x + 2y = 7
x – 2y = 7
Answer:

Question 13.
4x + 4y = -8
-2x – 2y = 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 13

Question 14.
15x – 5y = -20
-3x + y = 4
Answer:

Question 15.
9x – 15y = 24
6x – 10y = -16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 15

Question 16.
3x – 2y = -5
4x + 5y = 47
Answer:

In Exercises 17–22, use only the slopes and y-intercepts of the graphs of the equations to determine whether the system of linear equations has one solution, no solution, or infinitely many solutions. Explain.
Question 17.
y = 7x + 13
-21x + 3y = 39
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 17.1

Question 18.
y = -6x – 2
12x + 2y = -6
Answer:

Question 19.
4x + 3y = 27
4x – 3y = -27
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 19

Question 20.
-7x + 7y = 1
2x – 2y = -18
Answer:

Question 21.
-18x + 6y = 24
3x – y = -2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 21

Question 22.
2x – 2y = 16
3x – 6y = 30
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in solving the system of linear equations.
Question 23.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 23.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 23.2

Question 24.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 7
Answer:

Question 25.
MODELING WITH MATHEMATICS
A small bag of trail mix contains 3 cups of dried fruit and 4 cups of almonds. A large bag contains 4\(\frac{1}{2}\) cups of dried fruit and 6 cups of almonds. Write and solve a system of linear equations to find the price of 1 cup of dried fruit and 1 cup of almonds.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 25.1

Question 26.
MODELING WITH MATHEMATICS
In a canoe race, Team A is traveling 6 miles per hour and is 2 miles ahead of Team B. Team B is also traveling 6 miles per hour. The teams continue traveling at their current rates for the remainder of the race. Write a system of linear equations that represents this situation. Will Team B catch up to Team A? Explain.
Answer:

Question 27.
PROBLEM SOLVING
A train travels from New York City to Washington, D.C., and then back to New York City. The table shows the number of tickets purchased for each leg of the trip. The cost per ticket is the same for each leg of the trip. Is there enough information to determine the cost of one coach ticket? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 27.2

Question 28.
THOUGHT PROVOKING
Write a system of three linear equations in two variables so that any two of the equations have exactly one solution, but the entire system of equations has no solution.
Answer:

Question 29.
REASONING
In a system of linear equations, one equation has a slope of 2 and the other equation has a slope of –\(\frac{1}{3}\). How many solutions does the system have? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 29

Question 30.
HOW DO YOU SEE IT?
The graph shows information about the last leg of a 4 × 200-meter relay for three relay teams. Team A’s runner ran about 7.8 meters per second, Team B’s runner ran about 7.8 meters per second, and Team C’s runner ran about 8.8 meters per second.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 10
a. Estimate the distance at which Team C’s runner passed Team B’s runner.
b. If the race was longer, could Team C’s runner have passed Team A’s runner? Explain.
c. If the race was longer, could Team B’s runner have passed Team A’s runner? Explain.
Answer:

Question 31.
ABSTRACT REASONING
Consider the system of linear equations y = ax + 4 and y = bx – 2, where a and b are real numbers. Determine whether each statement is always, sometimes, or never true. Explain your reasoning.
a. The system has infinitely many solutions.
b. The system has no solution.c. When a < b, the system has one solution.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 31

Question 32.
MAKING AN ARGUMENT
One admission to an ice skating rink costs x dollars, and renting a pair of ice skates costs y dollars. Your friend says she can determine the exact cost of one admission and one skate rental. Is your friend correct? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 11
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check your solutions.
Question 33.
|2x + 6| = |x|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 33

Question 34.
|3x – 45| = |12x|
Answer:

Question 35.
|x – 7| = |2x – 8|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.4 Question 35.2

Question 36.
|2x + 1| = |3x – 11|
Answer:

Solving Systems of Linear Equations Study Skills: Analyzing Your Errors

5.1 – 5.4 What Did You Learn?

Core Vocabulary
system of linear equations, p. 236
solution of a system of linear equations, p. 236

Core Concepts
Section 5.1
Solving a System of Linear Equations by Graphing, p. 237

Section 5.2
Solving a System of Linear Equations by Substitution, p. 242

Section 5.3
Solving a System of Linear Equations by Elimination, p. 248

Section 5.4
Solutions of Systems of Linear Equations, p. 254

Mathematical Practices

Question 1.
Describe the given information in Exercise 33 on page 246 and your plan for finding the solution.
Answer:

Question 2.
Describe another real-life situation similar to Exercise 22 on page 251 and the mathematics that you can apply to solve the problem.
Answer:

Question 3.
What question(s) can you ask your friend to help her understand the error in the statement she made in Exercise 32 on page 258?
Answer:

Study Skills: Analyzing Your Errors

Study Errors
What Happens: You do not study the right material or you do not learn it well enough to remember it on a test without resources such as notes.
How to Avoid This Error: Take a practice test. Work with a study group. Discuss the topics on the test with your teacher. Do not try to learn a whole chapter’s worth of material in one night.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.4 12

Solving Systems of Linear Equations 5.1–5.4 Quiz

Use the graph to solve the system of linear equations. Check your solution.
Question 1.
y = – \(\frac{1}{3}\)x + 2
y = x – 2
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 1
Answer:

Question 2.
y = \(\frac{1}{2}\)x – 1
y = 4x + 6
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 2
Answer:

Question 3.
y = 1
y = 2x + 1
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 3
Answer:

Solve the system of linear equations by substitution. Check your solution.
Question 4.
y = x – 4
-2x + y = 18
Answer:

Question 5.
2y + x = -4
y – x = -5
Answer:

Question 6.
3x – 5y = 13
x + 4y = 10
Answer:

Solve the system of linear equations by elimination. Check your solution.
Question 7.
x + y = 4
-3x – y = -8
Answer:

Question 8.
x + 3y = 1
5x + 6y = 14
Answer:

Question 9.
2x – 3y = -5
5x + 2y = 16
Answer:

Solve the system of linear equations.
Question 10.
x – y = 1
x – y = 6
Answer:

Question 11.
6x + 2y = 16
2x – y = 2
Answer:

Question 12.
3x – 3y = -2
-6x + 6y = 4
Answer:

Question 13.
You plant a spruce tree that grows 4 inches per year and a hemlock tree that grows 6 inches per year. The initial heights are shown.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 4
a. Write a system of linear equations that represents this situation.
b. Solve the system by graphing. Interpret your solution.
Answer:

Question 14.
It takes you 3 hours to drive to a concert 135 miles away. You drive 55 miles per hour on highways and 40 miles per hour on the rest of the roads.
a. How much time do you spend driving at each speed?
b. How many miles do you drive on highways? the rest of the roads?
Answer:

Question 15.
In a football game, all of the home team’s points are from 7-point touchdowns and 3-point field goals. The team scores six times. Write and solve a system of linear equations to find the numbers of touchdowns and field goals that the home team scores.
Big Ideas Math Answer Key Algebra 1 Chapter 5 Solving Systems of Linear Equations q 5
Answer:

Lesson 5.5 Solving Equations by Graphing

Essential Question How can you use a system of linear equations to solve an equation with variables on both sides?

Previously, you learned how to use algebra to solve equations with variables on both sides. Another way is to use a system of linear equations.

EXPLORATION 1

Solving an Equation by Graphing
Work with a partner. Solve 2x – 1 = – \(\frac{1}{2}\) x + 4 by graphing.
a. Use the left side to write a linear equation. Then use the right side to write another linear equation.
b. Graph the two linear equations from part (a). Find the x-value of the point of intersection. Check that the x-value is the solution of
2x – 1 = –\(\frac{1}{2}\)x + 4.
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 1
c. Explain why this “graphical method” works.

EXPLORATION 2

Solving Equations Algebraically and Graphically
Work with a partner. Solve each equation using two methods.
Method 1 Use an algebraic method.
Method 2 Use a graphical method.
Is the solution the same using both methods?
a. \(\frac{1}{2}\)x + 4 = –\(\frac{1}{4}\)x + 1
b. \(\frac{2}{3}\)x + 4 = \(\frac{1}{3}\)x + 3
c. –\(\frac{2}{3}\) x – 1 = \(\frac{1}{3}\)x – 4
d. \(\frac{4}{5}\) x + \(\frac{7}{5}\) = 3x – 3
e. -x + 2.5 = 2x – 0.5
f. – 3x + 1.5 = x + 1.5

Communicate Your Answer

Question 3.
How can you use a system of linear equations to solve an equation with variables on both sides?
Answer:

Question 4.
Compare the algebraic method and the graphical method for solving a linear equation with variables on both sides. Describe the advantages and disadvantages of each method.
Answer:

Monitoring Progress

Solve the equation by graphing. Check your solution.
Question 1.
\(\frac{1}{2}\)x – 3 = 2x
Answer:

Question 2.
-4 + 9x = -3x + 2
Answer:

Solve the equation by graphing. Check your solutions.
Question 3.
|2x + 2| = |x – 2|
Answer:

Question 4.
|x – 6| = |-x + 4|
Answer:

Question 5.
WHAT IF?
Company C charges $3.30 per mile plus a flat fee of $115 per week. After how many miles are the total costs the same at Company A and Company C?
Answer:

Solving Equations by Graphing 5.5 Exercises

Vocabulary and Core Concept Check
Question 1.
REASONING
The graphs of the equations y = 3x – 20 and y = -2x + 10 intersect at the point (6, −2). Without solving, find the solution of the equation 3x – 20 = -2x + 10.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 1

Question 2.
WRITING
Explain how to rewrite the absolute value equation |2x – 4| = |-5x + 1| as two systems of linear equations.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, use the graph to solve the equation. Check your solution.
Question 3.
-2x + 3 = x
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 3

Question 4.
-3 = 4x + 1
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 3
Answer:

Question 5.
-x – 1 = \(\frac{1}{3}\)x + 3
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 5

Question 6.
–\(\frac{3}{2}\)x – 2 = -4x + 3
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 5
Answer:

In Exercises 7−14, solve the equation by graphing. Check your solution.
Question 7.
x + 4 = -x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 7

Question 8.
4x = x + 3
Answer:

Question 9.
x + 5 = -2x – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 9

Question 10.
-2x + 6 = 5x – 1
Answer:

Question 11.
\(\frac{1}{2}\)x – 2 = 9 – 5x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 11

Question 12.
-5 + \(\frac{1}{4}\)x = 3x + 6
Answer:

Question 13.
5x – 7 = 2(x + 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 13

Question 14.
-6(x + 4) = -3x – 6
Answer:

In Exercises 15−20, solve the equation by graphing. Determine whether the equation has one solution, no solution, or infinitely many solutions.
Question 15.
3x – 1 = -x + 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 15

Question 16.
5x – 4 = 5x + 1
Answer:

Question 17.
-4(2 – x) = 4x – 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 17

Question 18.
-2x – 3 = 2(x – 2)
Answer:

Question 19.
-x – 5 = –\(\frac{1}{3}\) (3x + 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 19

Question 20.
\(\frac{1}{2}\)(8x + 3) = 4x + \(\frac{3}{2}\)
Answer:

In Exercises 21 and 22, use the graphs to solve the equation. Check your solutions.
Question 21.
|x – 4| = |3x|
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 21

Question 22.
|2x + 4| = |x – 1|
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 7
Answer:

In Exercises 23−30, solve the equation by graphing. Check your solutions.
Question 23.
|2x| = |x + 3|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 23.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 23.2

Question 24.
|2x – 6| = |x|
Answer:

Question 25.
|-x + 4| = |2x – 2|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 25.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 25.2

Question 26.
|x + 2| = |-3x + 6|
Answer:

Question 27.
|x + 1| = |x – 5|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 27.2

Question 28.
|2x + 5| = |-2x + 1|
Answer:

Question 29.
|x – 3| = 2|x|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 29.2

Question 30.
4|x + 2| = |2x + 7|
Answer:

USING TOOLS In Exercises 31 and 32, use a graphing calculator to solve the equation.
Question 31.
0.7x + 0.5 = -0.2x – 1.3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 31

Question 32.
2.1x + 0.6 = -1.4x + 6.9
Answer:

Question 33.
MODELING WITH MATHEMATICS
You need to hire a catering company to serve meals to guests at a wedding reception. Company A charges $500 plus $20 per guest. Company B charges $800 plus $16 per guest. For how many guests are the total costs the same at both companies?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 33

Question 34.
MODELING WITH MATHEMATICS
Your dog is 16 years old in dog years. Your cat is 28 years old in cat years. For every human year, your dog ages by 7 dog years and your cat ages by 4 cat years. In how many human years will both pets be the same age in their respective types of years?
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 8
Answer:

Question 35.
MODELING WITH MATHEMATICS
You and a friend race across a field to a fence and back. Your friend has a 50-meter head start. The equations shown represent you and your friend’s distances d (in meters) from the fence t seconds after the race begins. Find the time at which you catch up to your friend.
You: d = |-5t + 100|
Your friend: d = |-3\(\frac{1}{3}\)t + 50|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 35

Question 36.
MAKING AN ARGUMENT
The graphs of y = -x + 4 and y = 2x – 8 intersect at the point (4, 0). So, your friend says the solution of the equation -x + 4 = 2x – 8 is (4, 0). Is your friend correct? Explain.
Answer:

Question 37.
OPEN-ENDED
Find values for m and b so that the solution of the equation mx + b = – 2x – 1 is x = -3.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 37

Question 38.
HOW DO YOU SEE IT?
The graph shows the total revenue and expenses of a company x years after it opens for business.
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 9
a. Estimate the point of intersection of the graphs.
b. Interpret your answer in part (a).
Answer:

Question 39.
MATHEMATICAL CONNECTIONS
The value of the perimeter of the triangle (in feet) is equal to the value of the area of the triangle (in square feet). Use a graph to find x.
Big Ideas Math Answers Algebra 1 Chapter 5 Solving Systems of Linear Equations 5.5 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 39

Question 40.
THOUGHT PROVOKING
A car has an initial value of $20,000 and decreases in value at a rate of $1500 per year. Describe a different car that will be worth the same amount as this car in exactly 5 years. Specify the initial value and the rate at which the value decreases.
Answer:

Question 41.
ABSTRACT REASONING
Use a graph to determine the sign of the solution of the equation ax + b = cx + d in each situation.
a. 0 < b < d and a < c
b. d < b < 0 and a < c Answer: Maintaining Mathematical Proficiency Graph the inequality.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 41.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 41.2

Question 42.
y > 5
Answer:

Question 43.
x ≤ -2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 43

Question 44.
n ≥ 9
Answer:

Question 45.
c < -6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 45

Use the graphs of f and g to describe the transformation from the graph of f to the graph of g.
Question 46.
f(x) = x – 5; g(x) = f(x + 2)
Answer:

Question 47.
f(x) = 6x; g(x) = -f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 47

Question 48.
f(x) = -2x + 1; g(x) = f(4x)
Answer:

Question 49.
f(x) = \(\frac{1}{2}\)x – 2; g(x) = f(x – 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.5 Question 49

Lesson 5.6 Graphing Linear Inequalities in Two Variables

Essential Question How can you graph a linear inequality in two variables?
A solution of a linear inequality in two variables is an ordered pair (x, y) that makes the inequality true. The graph of a linear inequality in two variables shows all the solutions of the inequality in a coordinate plane.

EXPLORATION 1

Writing a Linear Inequality in Two Variables
Work with a partner.
a. Write an equation represented by the dashed line.
b. The solutions of an inequality are represented by the shaded region. In words, describe the solutions of the inequality.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 1
c. Write an inequality represented by the graph. Which inequality symbol did you use? Explain your reasoning.

EXPLORATION 2

Using a Graphing Calculator
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 2
Work with a partner. Use a graphing calculator to graph y ≥ \(\frac{1}{4}\)x – 3.
a. Enter the equation y = \(\frac{1}{4}\)x – 3 into your calculator.
b. The inequality has the symbol ≥. So, the region to be shaded is above the graph of y = \(\frac{1}{4}\)x – 3, as shown. Verify this by testing a point in this region, such as (0, 0), to make sure it is a solution of the inequality.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 3
Because the inequality symbol is greater than or equal to, the line is solid and not dashed. Some graphing calculators always use a solid line when graphing inequalities. In this case, you have to determine whether the line should be solid or dashed, based on the inequality symbol used in the original inequality

EXPLORATION 3

Graphing Linear Inequalities in Two Variables
Work with a partner. Graph each linear inequality in two variables. Explain your steps. Use a graphing calculator to check your graphs.
a. y > x + 5
b. y ≤ –\(\frac{1}{2}\)x + 1
c. y ≥ -x – 5

Communicate Your Answer

Question 4.
How can you graph a linear inequality in two variables?
Answer:

Question 5.
Give an example of a real-life situation that can be modeled using a linear inequality in two variables.
Answer:

Monitoring Progress

Tell whether the ordered pair is a solution of the inequality.
Question 1.
x + y > 0; (-2, 2)
Answer:

Question 2.
4x – y ≥ 5; (0, 0)
Answer:

Question 3.
5x – 2y ≤ -1; (-4, -1)
Answer:

Question 4.
-2x – 3y < 15; (5, -7)
Answer:

Graph the inequality in a coordinate plane.
Question 5.
y > -1
Answer:

Question 6.
x ≤ -4
Answer:

Question 7.
x + y ≤ -4
Answer:

Question 8.
x – 2y < 0
Answer:

Question 9.
You can spend at most $12 on red peppers and tomatoes for salsa. Red peppers cost $4 per pound, and tomatoes cost $3 per pound. Write and graph an inequality that represents the amounts of red peppers and tomatoes you can buy. Identify and interpret two solutions of the inequality.
Answer:

Graphing Linear Inequalities in Two Variables 5.6 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
How can you tell whether an ordered pair is a solution of a linear inequality?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 1

Question 2.
WRITING
Compare the graph of a linear inequality in two variables with the graph of a linear equation in two variables.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, tell whether the ordered pair is a solution of the inequality.
Question 3.
x + y < 7; (2, 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 3

Question 4.
x – y ≤ 0; (5, 2)
Answer:

Question 5.
x + 3y ≥ -2; (-9, 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 5

Question 6.
8x + y > -6; (-1, 2)
Answer:

Question 7.
-6x + 4y ≤ 6; (-3, -3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 7

Question 8.
3x – 5y ≥ 2; (-1, -1)
Answer:

Question 9.
-x – 6y > 12; (-8, 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 9

Question 10.
-4x – 8y < 15; (-6, 3)
Answer:

In Exercises 11−16, tell whether the ordered pair is a solution of the inequality whose graph is shown.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 4
Question 11.
(0, -1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 11

Question 12.
(-1, 3)
Answer:

Question 13.
(1, 4)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 13

Question 14.
(0, 0)
Answer:

Question 15.
(3, 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 15

Question 16.
(2, 1)
Answer:

Question 17.
MODELING WITH MATHEMATICS
A carpenter has at most $250 to spend on lumber. The inequality 8x + 12y ≤ 250 represents the numbers x of 2-by-8boards and the numbers y of 4-by-4 boards the carpenter can buy. Can the carpenter buy twelve 2-by-8 boards and fourteen 4-by-4 boards? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 17

Question 18.
MODELING WITH MATHEMATICS
The inequality 3x + 2y ≥ 93 represents the numbers x of multiple- choice questions and the numbers y of matching questions you can answer correctly to receive an A on a test. You answer 20 multiple-choice questions and 18 matching questions correctly. Do you receive an A on the test? Explain.
Answer:

In Exercises 19–24, graph the inequality in a coordinate plane.
Question 19.
y ≤ 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 19

Question 20.
y > 6
Answer:

Question 21.
x < 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 21

Question 22.
x ≥ -3
Answer:

Question 23.
y > -7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 23

Question 24.
x < 9
Answer:

In Exercises 25−30, graph the inequality in a coordinate plane.
Question 25.
y > -2x – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 25

Question 26.
y ≤ 3x – 1
Answer:

Question 27.
-4x + y < -7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 27

Question 28.
3x – y ≥ 5
Answer:

Question 29.
5x – 2y ≤ 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 29

Question 30.
-x + 4y > -12
Answer:

ERROR ANALYSIS In Exercises 31 and 32, describe and correct the error in graphing the inequality.
Question 31.
y < -x + 1
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 31

Question 32.
y ≤ 3x – 2
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 7
Answer:

Question 33.
MODELING WITH MATHEMATICS
You have at most $20 to spend at an arcade. Arcade games cost $0.75 each, and snacks cost $2.25 each. Write and graph an inequality that represents the numbers of games you can play and snacks you can buy. Identify and interpret two solutions of the inequality.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 33.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 33.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 33.3

Question 34.
MODELING WITH MATHEMATICS
A drama club must sell at least $1500 worth of tickets to cover the expenses of producing a play. Write and graph an inequality that represents how many adult and student tickets the club must sell. Identify and interpret two solutions of the inequality.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 8
Answer:

In Exercises 35–38, write an inequality that represents the graph.
Question 35.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 35

Question 36.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 10
Answer:

Question 37.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 11
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 37

Question 38.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 12
Answer:

Question 39.
PROBLEM SOLVING
Large boxes weigh 75 pounds, and small boxes weigh 40 pounds.
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 13
a. Write and graph an inequality that represents the numbers of large and small boxes a 200-pound delivery person can take on the elevator.
b. Explain why some solutions of the inequality might not be practical in real life.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 39.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 39.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 39.3

Question 40.
HOW DO YOU SEE IT?
Match each inequality with its graph.
a. 3x – 2y ≤ 6
b. 3x – 2y < 6 c. 3x – 2y > 6
d. 3x – 2y ≥ 6
Big Ideas Math Algebra 1 Answer Key Chapter 5 Solving Systems of Linear Equations 5.6 14
Answer:

Question 41.
REASONING
When graphing a linear inequality in two variables, why must you choose a test point that is not on the boundary line?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 41

Question 42.
THOUGHT PROVOKING
Write a linear inequality in two variables that has the following two properties.
• (0, 0), (0, -1), and (0, 1) are not solutions.
• (1, 1), (3, -1), and (-1, 3) are solutions.
Answer:

Question 43.
WRITING
Can you always use (0, 0) as a test point when graphing an inequality? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 43

CRITICAL THINKING
In Exercises 44 and 45, write and graph an inequality whose graph is described by the given information.
Question 44.
The points (2, 5) and (−3, −5) lie on the boundary line. The points (6, 5) and (−2, −3) are solutions of the inequality.
Answer:

Question 45.
The points (−7, −16) and (1, 8) lie on the boundary line. The points (−7, 0) and (3, 14) are not solutions of the inequality.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 45.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 45.2

Maintaining Mathematical Proficiency

Write the next three terms of the arithmetic sequence.
Question 46.
0, 8, 16, 24, 32, . . .
Answer:

Question 47.
-5, -8, -11, -14, -17, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.6 Question 47.1

Question 48.
\(-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \ldots\)

Lesson 5.7 Systems of Linear Inequalities

Essential Question How can you graph a system of linear inequalities?

EXPLORATION 1

Graphing Linear Inequalities
Work with a partner. Match each linear inequality with its graph. Explain your reasoning.
2x + y ≤ 4 Inequality 1
2x – y ≤ 0 Inequality 2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 1

EXPLORATION 2

Graphing a System of Linear Inequalities
Work with a partner. Consider the linear inequalities given in Exploration 1.
2x + y ≤ 4 Inequality 1
2x – y ≤ 0 Inequality 2
a. Use two different colors to graph the inequalities in the same coordinate plane. What is the result?
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 2
b. Describe each of the shaded regions of the graph. What does the unshaded region represent?

Communicate Your Answer

Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 3
Question 3.
How can you graph a system of linear inequalities?
Answer:

Question 4.
When graphing a system of linear inequalities, which region represents the solution of the system?
Answer:

Question 5.
Do you think all systems of linear inequalities have a solution? Explain your reasoning.
Answer:

Question 6.
Write a system of linear inequalities represented by the graph.
Answer:

Monitoring Progress

Tell whether the ordered pair is a solution of the system of linear inequalities.
Question 1.
(-1, 5); y < 5 y > x – 4
Answer:

Question 2.
(1, 4); y ≥ 3x + 1
y > x – 1
Answer:

Graph the system of linear inequalities.
Question 3.
y ≥ -x + 4
x + y ≤ 0
Answer:

Question 4.
y > 2x – 3
y ≥ \(\frac{1}{2}\)x + 1
Answer:

Question 5.
-2x + y < 4 2x + y > 4
Answer:

Write a system of linear inequalities represented by the graph.
Question 6.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 4
Answer:

Question 7.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 5
Answer:

Question 8.
Name another solution of Example 6.
Answer:

Question 9.
WHAT IF?
You want to spend at least 3 hours at the mall. How does this change the system? Is (2.5, 5) still a solution? Explain.
Answer:

Systems of Linear Inequalities 5.7 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
How can you verify that an ordered pair is a solution of a system of linear inequalities?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 1

Question 2.
WHICH ONE DOESN’T BELONG?
Use the graph shown. Which of the ordered pairs does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 6
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3−6, tell whether the ordered pair is a solution of the system of linear inequalities.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 7
Question 3.
(-4, 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 3

Question 4.
(-3, -1)
Answer:

Question 5.
(-2, 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 5

Question 6.
(1, 1)
Answer:

In Exercises 7−10, tell whether the ordered pair is a solution of the system of linear inequalities.
Question 7.
(-5, 2); y < 4
y > x + 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 7

Question 8.
(1, -1); y > -2
y > x – 5
Answer:

Question 9.
(0, 0); y ≤ x + 7
y ≥ 2x + 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 9

Question 10.
(4, -3); y ≤ -x + 1
y ≤ 5x – 2
Answer:

In Exercises 11−20, graph the system of linear inequalities.
Question 11.
y > -3
y ≥ 5x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 11

Question 12.
y < -1
x > 4
Answer:

Question 13.
y < -2 y > 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 13

Question 14.
y < x – 1
y ≥ x + 1
Answer:

Question 15.
y ≥ -5
y – 1 < 3x Answer: Question 16. x + y > 4
y ≥ \(\frac{3}{2}\)x – 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 15

Question 17.
x + y > 1
-x – y < -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 17

Question 18.
2x + y ≤ 5
y + 2 ≥ -2x
Answer:

Question 19.
x < 4
y > 1
y ≥ -x + 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 19

Question 20.
x + y ≤ 10
x – y ≥ 2
y > 2
Answer:

In Exercises 21−26, write a system of linear inequalities represented by the graph.
Question 21.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 21

Question 22.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 9
Answer:

Question 23.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 23

Question 24.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 11
Answer:

Question 25.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 25

Question 26.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 13
Answer:

ERROR ANALYSIS In Exercises 27 and 28, describe and correct the error in graphing the system of linear inequalities.
Question 27.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 14
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 27

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 15
Answer:

Question 29.
MODELING WITH MATHEMATICS
You can spend at most $21 on fruit. Blueberries cost $4 per pound, and strawberries cost $3 per pound. You need at least 3 pounds of fruit to make muffins.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 16
a. Write and graph a system of linear inequalities that represents the situation.
b. Identify and interpret a solution of the system.
c. Use the graph to determine whether you can buy 4 pounds of blueberries and 1 pound of strawberries.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 29.2
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 29.3

Question 30.
MODELING WITH MATHEMATICS
You earn $10 per hour working as a manager at a grocery store. You are required to work at the grocery store at least 8 hours per week. You also teach music lessons for $15 per hour. You need to earn at least $120 per week, but you do not want to work more than 20 hours per week.
a. Write and graph a system of linear inequalities that represents the situation.
b. Identify and interpret a solution of the system.
c. Use the graph to determine whether you can work 8 hours at the grocery store and teach 1 hour of music lessons.
Answer:

Question 31.
MODELING WITH MATHEMATICS
You are fishing for surfperch and rockfish, which are species of bottomfish. Gaming laws allow you to catch no more than 15 surfperch per day, no more than 10 rockfish per day, and no more than 20 total bottomfish per day.
a. Write and graph a system of linear inequalities that represents the situation.
b. Use the graph to determine whether you can catch 11 surfperch and 9 rockfish in 1 day.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 17
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 31.2

Question 32.
REASONING
Describe the intersection of the half-planes of the system shown.
x – y ≤ 4
x – y ≥ 4
Answer:

Question 33.
MATHEMATICAL CONNECTIONS
The following points are the vertices of a shaded rectangle.
(-1, 1), (6, 1), (6, -3), (-1, -3)
a. Write a system of linear inequalities represented by the shaded rectangle.
b. Find the area of the rectangle.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 33

Question 34.
MATHEMATICAL CONNECTIONS
The following points are the vertices of a shaded triangle.
(2, 5), (6, -3), (-2, -3)
a. Write a system of linear inequalities represented by the shaded triangle.
b. Find the area of the triangle.
Answer:

Question 35.
PROBLEM SOLVING
You plan to spend less than half of your monthly $2000 paycheck on housing and savings. You want to spend at least 10% of your paycheck on savings and at most 30% of it on housing. How much money can you spend on savings and housing?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 35.2

Question 36.
PROBLEM SOLVING
On a road trip with a friend, you drive about 70 miles per hour, and your friend drives about 60 miles per hour. The plan is to drive less than 15 hours and at least 600 miles each day. Your friend will drive more hours than you. How many hours can you and your friend each drive in 1 day?
Answer:

Question 37.
WRITING
How are solving systems of linear inequalities and solving systems of linear equations similar? How are they different?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 37

Question 38.
HOW DO YOU SEE IT?
The graphs of two linear equations are shown.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 18
Replace the equal signs with inequality symbols to create a system of linear inequalities that has point C as a solution, but not points A, B, and D. Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 19
Answer:

Question 39.
USING STRUCTURE
Write a system of linear inequalities that is equivalent to |y| < x, where x > 0. Graph the system.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 39

Question 40.
MAKING AN ARGUMENT
Your friend says that a system of linear inequalities in which the boundary lines are parallel must have no solution. Is your friend correct? Explain.
Answer:

Question 41.
CRITICAL THINKING
Is it possible for the solution set of a system of linear inequalities to be all real numbers? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 41

OPEN-ENDED In Exercises 42−44, write a system of linear inequalities with the given characteristic.
Question 42.
All solutions are in Quadrant I.
Answer:

Question 43.
All solutions have one positive coordinate and one negative coordinate.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 43

Question 44.
There are no solutions.
Answer:

Question 45.
OPEN-ENDED
One inequality in a system is -4x + 2y > 6. Write another inequality so the system has (a) no solution and (b) infinitely many solutions.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 45

Question 46.
THOUGHT PROVOKING
You receive a gift certificate for a clothing store and plan to use it to buy T-shirts and sweatshirts. Describe a situation in which you can buy 9 T-shirts and 1 sweatshirt, but you cannot buy 3 T-shirts and 8 sweatshirts. Write and graph a system of linear inequalities that represents the situation.
Answer:

Question 47.
CRITICAL THINKING
Write a system of linear inequalities that has exactly one solution.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 47

Question 48.
MODELING WITH MATHEMATICS
You make necklaces and key chains to sell at a craft fair. The table shows the amounts of time and money it takes to make a necklace and a key chain, and the amounts of time and money you have available for making them.
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 20
a. Write and graph a system of four linear inequalities that represents the number x of necklaces and the number y of key chains that you can make.
b. Find the vertices (corner points) of the graph of the system.
c. You sell each necklace for $10 and each key chain for $8. The revenue R is given by the equation R = 10x + 8y. Find the revenue corresponding to each ordered pair in part (b). Which vertex results in the maximum revenue?
Answer:

Maintaining Mathematical Proficiency

Write the product using exponents.
Question 49.
4 • 4 • 4 • 4 • 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 49

Question 50.
(-13) • (-13) • (-13)
Answer:

Question 51.
x • x • x • x • x • x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 51

Write an equation of the line with the given slope and y-intercept.
Question 52.
slope: 1
y-intercept: -6
Answer:

Question 53.
slope: -3
y-intercept: 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 53

Question 54.
slope: –\(\frac{1}{4}\)
y-intercept: -1
Answer:

Question 55.
slope: \(\frac{4}{3}\)
y-intercept: 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 Question 55

Solving Systems of Linear Equations Performance Task: Prize Patrol

5.5–5.7 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 21

Core Concepts
Section 5.5
Solving Linear Equations by Graphing, p. 262
Solving Absolute Value Equations by Graphing, p. 263

Section 5.6
Graphing a Linear Inequality in Two Variables, p. 269

Section 5.7
Graphing a System of Linear Inequalities, p. 275
Writing a System of Linear Inequalities, p. 276

Mathematical Practices

Question 1.
Why do the equations in Exercise 35 on page 266 contain absolute value expressions?
Answer:

Question 2.
Why is it important to be precise when answering part (a) of Exercise 39 on page 272?
Answer:

Question 3.
Describe the overall step-by-step process you used to solve Exercise 35 on page 279.
Answer:

Performance Task Prize Patrol

You have been selected to drive a prize patrol cart and place prizes on the competing teams’ predetermined paths. You know the teams’ routes and you can only make one pass. Where will you place the prizes so that each team will have a chance to find a prize on their route?
To explore the answers to these questions and more, go to
Big Ideas Math Algebra 1 Answers Chapter 5 Solving Systems of Linear Equations 5.7 22

Solving Systems of Linear Equations Chapter Review

5.1 Solving Systems of Linear Equations by Graphing (pp. 235–240)

Solve the system of linear equations by graphing.
Question 1.
y = -3x + 1
y = x – 7
Answer:

Question 2.
y = -4x + 3
4x – 2y = 6
Answer:

Question 3.
5x + 5y = 15
2x – 2y = 10
Answer:

5.2 Solving Systems of Linear Equations by Substitution (pp. 241–246)

Solve the system of linear equations by substitution. Check your solution.
Question 4.
3x + y = -9
y = 5x + 7
Answer:

Question 5.
x + 4y = 6
x – y = 1
Answer:

Question 6.
2x + 3y = 4
y + 3x = 6
Answer:

Question 7.
You spend $20 total on tubes of paint and disposable brushes for an art project. Tubes of paint cost $4.00 each and paintbrushes cost $0.50 each. You purchase twice as many brushes as tubes of paint. How many brushes and tubes of paint do you purchase?
Answer:

5.3 Solving Systems of Linear Equations by Elimination (pp. 247 – 252)

Solve the system of linear equations by elimination. Check your solution.
Question 8.
9x – 2y = 34
5x + 2y = -6
Answer:

Question 9.
x + 6y = 28
2x – 3y = -19
Answer:

Question 10.
8x – 7y = -3
6x – 5y = -1
Answer:

5.4 Solving Special Systems of Linear Equations (pp. 253–258)

Solve the system of linear equations.
Question 11.
x = y + 2
-3x + 3y = 6
Answer:

Question 12.
3x – 6y = -9
-5x + 10y = 10
Answer:

Question 13.
-4x + 4y = 32
3x + 24 = 3y
Answer:

5.5 Solving Equations by Graphing (pp. 261–26

Solve the equation by graphing. Check your solution(s).
Question 14.
\(\frac{1}{3}\)x + 5 = -2x – 2
Answer:

Question 15.
|x + 1| = |-x – 9|
Answer:

Question 16.
|2x – 8| = |x + 5|
Answer:

5.6 Graphing Linear Inequalities in Two Variables (pp. 267–272)

Graph the inequality in a coordinate plane.
Question 17.
y > -4
Answer:

Question 18.
-9x + 3y ≥ 3
Answer:

Question 19.
5x + 10y < 40 Answer: 5.7 Systems of Linear Inequalities (pp. 273–280) Graph the system of linear inequalities. Question 20. y ≤ x – 3 y ≥ x + 1 Answer: Question 21. y > -2x + 3
y ≥ \(\frac{1}{4}\)x – 1
Answer:

Question 22.
x + 3y > 6
2x + y < 7
Answer:

Solving Systems of Linear Equations Chapter Test

Solve the system of linear equations using any method. Explain why you chose the method.

Question 1.
8x + 3y = -9
-8x + y = 29
Answer:

Question 2.
\(\frac{1}{2}\)x + y = -6
y = \(\frac{3}{5}\)x + 5
Answer:

Question 3.
y = 4x + 4
-8x + 2y = 8
Answer:

Question 4.
x = y – 11
x – 3y = 1
Answer:

Question 5.
6x – 4y = 9
9x – 6y = 15
Answer:

Question 6.
y = 5x – 7
-4x + y = -1
Answer:

Question 7.
Write a system of linear inequalities so the points (1, 2) and (4, -3) are solutions of the system, but the point (-2, 8) is not a solution of the system.
Answer:

Question 8.
How is solving the equation |2x + 1| = |x – 7| by graphing similar to solving the equation 4x + 3 = -2x + 9 by graphing? How is it different?
Answer:

Graph the system of linear inequalities.
Question 9.
y > \(\frac{1}{2}\)x + 4
2y ≤ x + 4
Answer:

Question 10.
x + y < 1 5x + y > 4
Answer:

Question 11.
y ≥ – \(\frac{2}{3}\)x + 1
-3x + y > -2
Answer:

Question 12.
You pay $45.50 for 10 gallons of gasoline and 2 quarts of oil at a gas station. Your friend pays $22.75 for 5 gallons of the same gasoline and 1 quart of the same oil.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ct 1
a. Is there enough information to determine the cost of 1 gallon of gasoline and 1 quart of oil? Explain.
b. The receipt shown is for buying the same gasoline and same oil. Is there now enough information to determine the cost of 1 gallon of gasoline and 1 quart of oil? Explain.
c. Determine the cost of 1 gallon of gasoline and 1 quart of oil.
Answer:

Question 13.
Describe the advantages and disadvantages of solving a system of linear equations by graphing.
Answer:

Question 14.
You have at most $60 to spend on trophies and medals to give as prizes for a contest.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ct 2
a. Write and graph an inequality that represents the numbers of trophies and medals you can buy. Identify and interpret a solution of the inequality.
b. You want to purchase at least 6 items. Write and graph a system that represents the situation. How many of each item can you buy?
Answer:

Question 15.
Compare the slopes and y-intercepts of the graphs of the equations in the linear system 8x + 4y = 12 and 3y = -6x – 15 to determine whether the system has one solution, no solution, or infinitely many solutions. Explain.
Answer:

Solving Systems of Linear Equations Cumulative Assessment

Question 1.
The graph of which equation is shown?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 1
Answer:

Question 2.
A van rental company rents out 6-, 8-, 12-, and 16-passenger vans. The function C(x) = 100 + 5x represents the cost C (in dollars) of renting an x-passenger van for a day. Choose the numbers that are in the range of the function.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 2
Answer:

Question 3.
Fill in the system of linear inequalities with <, ≤, >, or ≥ so that the graph represents the system.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 3
Answer:

Question 4.
Your friend claims to be able to fill in each box with a constant so that when you set each side of the equation equal to y and graph the resulting equations, the lines will intersect exactly once. Do you support your friend’s claim? Explain.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 4
Answer:

Question 5.
Select the phrases you should use when describing the transformations from the graph of f to the graph of g.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 5
Answer:

Question 6.
Which two equations form a system of linear equations that has no solution?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 6
Answer:

Question 7.
Fill in a value for a so that each statement is true for the equation ax – 8 = 4 – x.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 7
Answer:

Question 8.
Which ordered pair is a solution of the linear inequality whose graph is shown?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 8
Answer:

Question 9.
Which of the systems of linear equations are equivalent?
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 9
Answer:

Question 10.
The value of x is more than 9. Which of the inequalities correctly describe the triangle? The perimeter (in feet) is represented by P, and the area (in square feet) is represented by A.
Big Ideas Math Algebra 1 Solutions Chapter 5 Solving Systems of Linear Equations ca 10
Answer:

Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions

Big Ideas Math Answers Grade 4 Chapter 8

Are you looking online to get Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions? Then, you have reached the right place. Don’t bother as we have included the best preparation resources for all of the students here in detail. Make the most use of these math books, study materials, and Big Ideas Math Grade 4 Chapter 8 Add and Subtract Fractions to get the best preparation. Stand out from the rest of the crowd using our Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions. Download the Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions pdf PDF for free now and prepare from anywhere and anytime.

Big Ideas 4th Grade Math Book Chapter 8 Add and Subtract Fractions Answer Key

Big Ideas Math Chapter 8 Add and Subtract Fractions Questions are provided as per the updated syllabus guidelines. Also, the answers and explanations are given by highly subject expertise people. Furthermore, the Big Ideas Grade 4 Chapter 8 Math Answer key covers assignment tests, questions from exercises, practice tests, etc. Prepare the topics you wish to prepare and you feel the lag. The below quick links will help you to find each and every concept along with questions.

Lesson 1: Use Models to Add Fractions

Lesson 2: Decompose Fractions

Lesson 3: Add Fractions with Like Denominators

Lesson 4: Use Models to subtract Fractions

Lesson 5: Subtract Fractions with Like Denominators

Lesson 6: Model Fractions and Mixed Numbers

Lesson 7: Add Mixed Numbers

Lesson 8: Subtract Mixed Numbers

Lesson 9: Problem Solving: Fractions

Performance Task

Lesson 8.1 Use Models to Add Fractions

Explore and Grow

Draw models to show \(\frac{2}{8}\) and \(\frac{5}{8}\).

Answer:
You can add fractions by joining parts that refer to the same whole.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_1

Use your models to find \(\frac{2}{8}\) + \(\frac{5}{8}\). Explain your method.

Answer:
Combine the like terms
\(\frac{2}{8}\) + \(\frac{5}{8}\) = (2 + 5)/8 = 7/8

Repeated Reasoning
Write two fractions that have a sum of \(\frac{6}{8}\). Explain your reasoning.

Think and Grow: Use Models to Add Fractions

You can add fractions by joining parts that refer to the same whole.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 1
Answer:
The denominators of the fraction are the same so you have to add numerators.
\(\frac{1}{5}\) + \(\frac{3}{5}\) = \(\frac{4}{5}\)

Example
Use a number line to find \(\frac{5}{4}\) + \(\frac{2}{4}\).
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 2
Answer:
The denominators of the fraction are the same so you have to add numerators.
\(\frac{5}{4}\) + \(\frac{2}{4}\) = \(\frac{7}{4}\)

Show and Grow

Find the sum. Explain how you used the model to add.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 3

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-3
(3+4)/10 = 7/10

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 4

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-4

Apply and Grow: Practice

Find the sum. Use a model or a number line to help.

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 5

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_2

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 6

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_3
\(\frac{5}{12}\) + \(\frac{4}{12}\) = \(\frac{9}{12}\)

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 7

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_5

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 8

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_4

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 9

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_6

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 10

Answer:
Add a fraction to the whole number.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_7
5 + \(\frac{6}{8}\) = \(\frac{46}{8}\)

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 11

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_8

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 12

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Fractions-img_10

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 13

Answer:
The denominators of the fraction are the same so you have to add numerators.
Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_11

Question 12.
Structure
Write the addition equation represented by the models.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 14

Answer:
By seeing the above model we can find the addition equation.
\(\frac{4}{8}\) + \(\frac{3}{8}\) = \(\frac{7}{8}\)

Question 13.
Open-Ended
Write three fractions with different numerators that have a sum of 1.

Answer:
\(\frac{2}{8}\) + \(\frac{5}{8}\) + \(\frac{1}{8}\) = \(\frac{8}{8}\) = 1

Question 14.
Writing
Explain why \(\frac{1}{8}\) + \(\frac{4}{8}\) does not equal \(\frac{5}{16}\).

Answer:
In the above expressions, the denominators are the same but the numerators are different.
So, you have to add the numerators not denominators.
\(\frac{1}{8}\) + \(\frac{4}{8}\) = \(\frac{5}{8}\)

Think and Grow: Modeling Real Life

Example
You need \(\frac{2}{3}\) cup of hot water and \(\frac{4}{3}\) cups of cold water for a science experiment. How many cups of water do you need in all?
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 15
Because each fraction represents a part of the same whole you can join the parts.
Use a model to find \(\frac{2}{3}\) + \(\frac{4}{3}\).
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 16

Answer:
Given that,
You need \(\frac{2}{3}\) cup of hot water and \(\frac{4}{3}\) cups of cold water for a science experiment.

Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-16
Thus you need 2 cups of water in all.

Show and Grow

Question 15.
You cut a foam noodle for a craft. You use \(\frac{2}{4}\) of the noodle for one part of the craft and \(\frac{1}{4}\) of the noodle for another part. What fraction of the foam noodle do you use altogether?
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 17

Answer:
Given that,
You cut a foam noodle for a craft. You use \(\frac{2}{4}\) of the noodle for one part of the craft and \(\frac{1}{4}\) of the noodle for another part.
\(\frac{2}{4}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\)
Thus \(\frac{3}{4}\) of the foam noodle is used.

Question 16.
You make a fruit drink using \(\frac{4}{8}\) gallon of orange juice, \(\frac{2}{8}\) gallon of mango juice, and \(\frac{4}{8}\) gallon of pineapple juice. How much juice do you use in all?

Answer:
Given that,
You make a fruit drink using \(\frac{4}{8}\) gallon of orange juice, \(\frac{2}{8}\) gallon of mango juice, and \(\frac{4}{8}\) gallon of pineapple juice.
\(\frac{4}{8}\) + \(\frac{2}{8}\) = \(\frac{6}{8}\)
\(\frac{6}{8}\) + \(\frac{4}{8}\) = \(\frac{10}{8}\)
Thus you used \(\frac{10}{8}\) fraction of juice.

Question 17.
DIG DEEPER!
A community plants cucumbers in \(\frac{5}{12}\) of a garden, broccoli in \(\frac{3}{12}\) of the garden, and carrots in \(\frac{4}{12}\) of the garden. What fraction of the garden is planted with green vegetables?

Answer:
Given that,
A community plants cucumbers in \(\frac{5}{12}\) of a garden, broccoli in \(\frac{3}{12}\) of the garden, and carrots in \(\frac{4}{12}\) of the garden.
\(\frac{5}{12}\) + \(\frac{3}{12}\) + \(\frac{4}{12}\) = \(\frac{12}{12}\) = 1
\(\frac{12}{12}\) fraction of the garden is planted with green vegetables.

Use Models to Add Fractions Homework & Practice 8.1

Find the sum. Explain how you used the model to add.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 18

Answer: \(\frac{9}{6}\)
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-18

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 19

Answer: 1
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-19

Find the sum. Use a model or a number line to help.

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 20

Answer: \(\frac{7}{8}\)
You can add fractions by joining parts that refer to the same whole.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_1

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 21

Answer: 2
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Fractions-img_20

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 22

Answer: 3 \(\frac{1}{4}\)

Explanation:
Add fraction to the whole number.
\(\frac{1}{4}\) + 3
\(\frac{1}{4}\) + 3 × \(\frac{4}{4}\)
\(\frac{1}{4}\) + \(\frac{13}{4}\) = \(\frac{13}{4}\)

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 23

Answer: \(\frac{9}{12}\)
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_21

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 24

Answer: 2
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_22
\(\frac{12}{10}\) = \(\frac{10}{10}\) + \(\frac{2}{10}\)
\(\frac{10}{10}\) + \(\frac{2}{10}\) + \(\frac{8}{10}\) = \(\frac{20}{10}\) = 2

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 25

Answer:
4/8 = 1/2
6 + 1/2 = (12 + 1)/2 = 13/2

Find the sum. Use a model or a number line to help.

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 26

Answer:
Add all the three unit fractions.
BIM Grade 4 Chapter 8 Add and Subtract Fractions img_23

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 27

Answer:
Big Ideas Math Answers 4th Grade Chapter 8 img_24

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 28

Answer:
The denominators of all the fractions are the same. So you have to add the numerators of the fraction.
\(\frac{50}{100}\) + \(\frac{25}{100}\) + \(\frac{5}{100}\) = \(\frac{80}{100}\)

Question 12.
YOU BE THE TEACHER
Newton says \(\frac{3}{5}\) + \(\frac{1}{5}\) = \(\frac{4}{10}\). Descartes says the sum is \(\frac{4}{5}\). Who is correct? Explain.

Answer:
Newton says \(\frac{3}{5}\) + \(\frac{1}{5}\) = \(\frac{4}{10}\). Descartes says the sum is \(\frac{4}{5}\).
Descartes is correct.
\(\frac{3}{5}\) + \(\frac{1}{5}\) = \(\frac{4}{5}\)
You have to add numerators, not denominators.
So, Newton’s equation is not correct.

Question 13.
Make each statement true by writing two fractions whose denominators are the same and whose numerators are 3 and 2.
The sum of ___ and __ is greater than 1.
___________________________
The sum of ___ and ___ is less than 1.
___________________________
The sum of ___ and ___ is equal to 1.

Answer:
The sum of 3/2 and 2/2 is greater than 1.
3/2 + 2/2 = 5/2
5/2 > 1
The sum of 3/6 and 2/6 is less than 1.
3/6 + 2/6 = 5/6
5/6 < 1
The sum of 3/5 and 2/5 is equal to 1.
3/5 + 2/5 = 5/5 = 1

Question 14.
Modeling Real Life
Your teacher assigns 5 pages to read. You read \(\frac{3}{5}\) of the pages in class and \(\frac{1}{5}\) of the pages at home. What fraction of the reading assignment is complete?

Answer:
Given that,
Your teacher assigns 5 pages to read. You read \(\frac{3}{5}\) of the pages in class and \(\frac{1}{5}\) of the pages at home.
The denominators of all the fractions are the same. So you have to add the numerators of the fraction.
\(\frac{3}{5}\) + \(\frac{1}{5}\) = \(\frac{4}{5}\)

Question 15.
Modeling Real Life
In the Sahara Desert, it rains \(\frac{2}{10}\) inch in September, \(\frac{3}{10}\) inch in October, and \(\frac{5}{10}\) inch in November. How much does it rain in the 3 months?

Answer:
Given that,
In the Sahara Desert, it rains \(\frac{2}{10}\) inch in September, \(\frac{3}{10}\) inch in October, and \(\frac{5}{10}\) inch in November.
\(\frac{2}{10}\) + \(\frac{3}{10}\) + \(\frac{5}{10}\)
The denominators of all the fractions are the same. So you have to add the numerators of the fraction.
\(\frac{2}{10}\) + \(\frac{3}{10}\) + \(\frac{5}{10}\) = (2 + 3 + 5)/10 = 10/10 = 1
It rains 10/10 in the 3 months.

Review & Refresh

Tell whether the number is prime or composite. Explain.

Question 16.
37

Answer: 37 is a prime number.
A prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. A natural number greater than 1 that is not prime is called a composite number.

Question 17.
21

Answer: 21 is a composite number.
A composite number is a positive integer that can be formed by multiplying two smaller positive integers. Equivalently, it is a positive integer that has at least one divisor other than 1 and itself.

Question 18.
99

Answer: 99 is a composite number.
A composite number is a positive integer that can be formed by multiplying two smaller positive integers. Equivalently, it is a positive integer that has at least one divisor other than 1 and itself.

Lesson 8.2 Decompose Fractions

Explore and Grow

Use a model to find Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 29.

How can you write \(\frac{7}{10}\) as a sum of unit fractions? Explain your reasoning.

Answer: The sum of unit fraction of \(\frac{7}{10}\) is \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\)

Structure
Explain how you can write \(\frac{7}{10}\) as a sum of two fractions. Draw a model to support your answer.

Answer: You can write sum of \(\frac{7}{10}\) as \(\frac{2}{10}\) + \(\frac{5}{10}\)
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_23

Think and Grow: Decompose Fractions

A unit fraction represents one equal part of a whole. You can write a fraction as a sum of unit fractions.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 31
Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-31
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 31.1

Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-31.1

Show and Grow

Question 1.
Write \(\frac{4}{5}\) as a sum of unit fractions.

Answer: The unit fraction of \(\frac{4}{5}\) is \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\)
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 2.
Write \(\frac{5}{6}\) as a sum of fractions in two different ways.

Answer: The unit fraction of \(\frac{5}{6}\) is \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\)
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.
You can also write \(\frac{5}{6}\) as \(\frac{2}{6}\) + \(\frac{3}{6}\)
That means \(\frac{5}{6}\) can be written as 2 parts of \(\frac{1}{6}\) and 3 parts of \(\frac{1}{6}\)

Apply and Grow: Practice

Question 3.
\(\frac{4}{7}\)

Answer: The unit fraction of \(\frac{4}{7}\) is \(\frac{1}{7}\) + \(\frac{1}{7}\) + \(\frac{1}{7}\) + \(\frac{1}{7}\)
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 4.
\(\frac{7}{8}\)

Answer: The unit fraction of \(\frac{7}{8}\) is \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\)
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 5.
\(\frac{3}{10}\)

Answer: The unit fraction of \(\frac{3}{10}\) is  \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\)
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 6.
\(\frac{10}{100}\)

Answer: The unit fraction of \(\frac{10}{100}\) is \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\)
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 7.
\(\frac{6}{2}\)

Answer: 3
The unit fraction of \(\frac{6}{2}\) is \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\)
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 8.
\(\frac{9}{4}\)

Answer:
Break apart 9 parts of \(\frac{1}{4}\) into 5 parts of \(\frac{1}{4}\) and 4 parts of \(\frac{1}{4}\).

Question 9.
\(\frac{8}{12}\)

Answer: Break apart 8 parts of \(\frac{1}{12}\) into 5 parts of \(\frac{1}{12}\) and 3 parts of \(\frac{1}{12}\).

Question 10.
\(\frac{5}{3}\)

Answer: The unit fraction of \(\frac{5}{3}\) is \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\)
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.

Question 11.
Writing
You write \(\frac{4}{6}\) as a sum of unit fractions. Explain how the numerator of \(\frac{4}{6}\) is related to the number of addends.

Answer: The unit fraction of \(\frac{4}{6}\) is \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\)
A unit fraction is a rational number written as a fraction where the numerator is one and the denominator is a positive integer.
Also, you can write \(\frac{4}{6}\) as 4 parts of \(\frac{1}{6}\), 2 equal parts of \(\frac{1}{6}\) and 2 equal parts of \(\frac{1}{6}\).

Question 12.
DIG DEEPER!
Why is it important to be able to write a fraction as a sum of fractions in different ways?

Answer:
Asking students to write a fraction as a sum of unit fractions, or as a sum of other fractions, encourages students to make sense of quantities and their relationships. Students further develop their understandings about fractions and decomposing numbers through this process.

Question 13.
Precision
Match each fraction with an equivalent expression.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 32

Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-32

Think and Grow: Modeling Real Life

Example
A chef has \(\frac{8}{10}\) liter of soup. How can the chef pour all of the soup into 2 bowls?
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 33
Break apart \(\frac{8}{10}\) into any two fractions that have a sum of \(\frac{8}{10}\).
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 34

Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-34

Show and Grow

Question 14.
You have \(\frac{7}{3}\) pounds of almonds. What are two different ways you can put all of the almonds into 2 bags?

Answer:
Given that,
You have \(\frac{7}{3}\) pounds of almonds.
Break apart 7 parts of \(\frac{1}{3}\) into 5 parts of \(\frac{1}{3}\) and 2 parts of \(\frac{1}{3}\)
Thus you can put 5 parts of \(\frac{1}{3}\) and 2 parts of \(\frac{1}{3}\) of the almonds into 2 bags.

Question 15.
A 3-person painting crew has \(\frac{10}{12}\) of a fence left to paint. What is one way the crew can finish painting the fence when each person paints a fraction of the fence?
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 35

Answer:
Given that,
A 3-person painting crew has \(\frac{10}{12}\) of a fence left to paint.
\(\frac{10}{12}\) can be written as \(\frac{3}{12}\) + \(\frac{3}{12}\) + \(\frac{4}{12}\)
Thus each person paints \(\frac{3}{12}\) + \(\frac{3}{12}\) + \(\frac{4}{12}\) fraction of the fence.

Question 16.
DIG DEEPER!
Three teammates have to run a total of miles for a relay race. Can each team member run the same fraction of a mile, in fourths, to complete the race? Explain.

Answer:
Three teammates have to run a total of miles for a relay race.
No three members cannot run the same fraction of a mile, in fourths, to complete the race
\(\frac{10}{12}\) can be written as \(\frac{3}{12}\) + \(\frac{3}{12}\) + \(\frac{4}{12}\)

Decompose Fractions Homework & Practice 8.2

write the fraction as a sum of unit fractions.

Question 1.
\(\frac{2}{2}\)

Answer: 1
The sum of unit fractions of \(\frac{2}{2}\) is \(\frac{1}{2}\) + \(\frac{1}{2}\)

Question 2.
\(\frac{3}{5}\)

Answer: The sum of unit fractions of \(\frac{3}{5}\) is \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\)

Question 3.
\(\frac{4}{3}\)

Answer: The sum of unit fractions of \(\frac{4}{3}\) is \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\)

Question 4.
\(\frac{6}{4}\)

Answer: The sum of unit fractions of \(\frac{6}{4}\) is \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\)

write the fraction as a sum of fractions in two different ways.

Question 5.
\(\frac{8}{12}\)

Answer: The sum of unit fractions of \(\frac{8}{12}\) is \(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\)

Another Way:
Break apart \(\frac{8}{12}\) as 4 parts of \(\frac{1}{12}\) and 4 parts of \(\frac{1}{12}\)

Question 6.
\(\frac{10}{6}\)

Answer: The sum of unit fractions of \(\frac{10}{6}\) is \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\)

Another way:
Break apart \(\frac{10}{6}\) as 5 parts of \(\frac{1}{6}\) and 5 parts of \(\frac{11}{6}\)

Question 7.
\(\frac{11}{100}\)

Answer: The sum of unit fractions of \(\frac{11}{100}\) is \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\)

Another way:
Break apart \(\frac{11}{100}\) as 5 parts of \(\frac{1}{100}\), 4 parts of \(\frac{1}{100}\) and 2 parts of \(\frac{1}{100}\)

Question 8.
\(\frac{14}{8}\)

Answer: The sum of unit fractions of \(\frac{14}{8}\) is \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\)

Another way:
Break apart \(\frac{14}{8}\) as 5 parts of \(\frac{1}{8}\), 9 parts of \(\frac{1}{8}\)

Question 9.
Which One Doesn’t Belong? Which expression does belong with the other three?
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 36

Answer: The expression \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) does not belong to the other three.

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 37

Answer: Yes your friend is correct.
\(\frac{1}{10}\) + \(\frac{3}{10}\)  + \(\frac{5}{10}\)
Here the denominators are the same so you have to add the numerators.
\(\frac{1}{10}\) + \(\frac{3}{10}\)  + \(\frac{5}{10}\) = \(\frac{9}{10}\)
\(\frac{2}{10}\) + \(\frac{4}{10}\) + \(\frac{3}{10}\)
Here the denominators are the same so you have to add the numerators.
\(\frac{2}{10}\) + \(\frac{4}{10}\) + \(\frac{3}{10}\) = \(\frac{9}{10}\)

Question 11.
Number Sense
Is it possible to write \(\frac{7}{12}\) as the sum of three fractions with three different numerators and the same denominator? Explain.

Answer: Yes it is possible to write \(\frac{7}{12}\) as the sum of three fractions with three different numerators and the same denominator.
\(\frac{7}{12}\) = \(\frac{3}{12}\) + \(\frac{3}{12}\) + \(\frac{1}{12}\)

Question 12.
You have \(\frac{8}{4}\) pounds of dried pineapple. What are two different ways you can put all of the pineapples into 2 bags?

Answer:
Given that,
You have \(\frac{8}{4}\) pounds of dried pineapple.
Break apart \(\frac{8}{4}\) as 4 parts of \(\frac{1}{4}\) and 4 parts of \(\frac{1}{4}\).
The two different ways you can put all of the pineapples into 2 bags are 4 parts of \(\frac{1}{4}\).

Question 13.
DIG DEEPER!
A carpenter has 3 planks of wood. Each plank has a different thickness. When stacked, the thickness of the 3 planks is \(\frac{6}{8}\) inch. What are the possible thickness of each plank?
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 38

Answer:
Given that,
A carpenter has 3 planks of wood. Each plank has a different thickness.
When stacked, the thickness of the 3 planks is \(\frac{6}{8}\) inch.
\(\frac{6}{8}\) = \(\frac{2}{8}\) + \(\frac{3}{8}\) + \(\frac{1}{8}\)
The possible thickness of each plank are \(\frac{2}{8}\), \(\frac{3}{8}\), \(\frac{1}{8}\)

Review & Refresh

Find the product. Check whether your answer is reasonable.

Question 14.
Estimate: ___
608 × 5 = ___

Answer:
600 × 5 = 3000
The number close to 608 is 600.
Step 2:
608 × 5 = 3040
3040 is close to 3000. So, the answer is reasonable.

Question 15.
Estimate: ___
7 × 5,394 = ___

Answer:
7 × 5400 = 37,800
The number close to 5394 is 5400.
Step 2:
7 × 5394 = 37,758
37,758 is close to 37,800. So, the answer is reasonable.

Question 16.
Estimate: ___
927 × 3 = ___

Answer:
900 × 3 = 2700
The number close to 927 is 900.
Step 2:
927 × 3 = 2781
2781 is close to 2700. So, the answer is reasonable.

Lesson 8.3 Add Fractions with Like Denominators

Explore and Grow

Write each fraction as a sum of unit fractions. Use models to help.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 39

How many unit fractions did you use in all to rewrite the fractions above? How does this relate to the sum \(\frac{3}{6}+\frac{5}{6}\) ?

Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-39
\(\frac{3}{6}+\frac{5}{6}\) = \(\frac{8}{6}\)

Construct Arguments
How can you use the numerators and the denominators to add fractions with like denominators? Explain why your method makes sense.

Answer:
To add fractions with like denominators, add the numerators and keep the same denominator. Then simplify the sum. You know how to do this with numeric fractions.
\(\frac{3}{6}+\frac{5}{6}\) = \(\frac{8}{6}\)

Think and Grow: Add Fractions

To add fractions with like denominators, add the numerators.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 40
The denominator stays the same.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 41

Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-41
Add the numerators of the like denominators.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 42

Answer:
Add the numerators of the like denominators.
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-42

Show and Grow

Add.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 43

Answer:
Add the numerators of the like denominators.
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-43

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 44

Answer:
Add the numerators of the like denominators.
6 + 2 = 8
\(\frac{6}{5}+\frac{2}{5}\) = \(\frac{8}{5}\)

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 45

Answer:
Add the numerators of the like denominators.
4 + 4 = 8
\(\frac{4}{8}+\frac{4}{8}\) = \(\frac{8}{8\) = 1

Apply and Grow: Practice

Add.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 46

Answer:
Add the numerators of the like denominators.
3 + 2 = 5
\(\frac{3}{6}+\frac{2}{6}\) = \(\frac{5}{6}\)

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 47

Answer:
Add the numerators of the like denominators.
8 + 4 = 12
\(\frac{8}{2}+\frac{4}{2}\) = \(\frac{12}{2}\) = 6

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 48

Answer:
Add the numerators of the like denominators.
4 + 1 = 5
\(\frac{4}{5}+\frac{1}{5}\) = \(\frac{5}{5}\) = 1

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 49

Answer:
Add the numerators of the like denominators.
60 + 35 = 95
\(\frac{60}{100}+\frac{35}{100}\) = \(\frac{95}{100}\)

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 50

Answer:
The denominators are not the same. So first you have to make the common denominators and add the fraction with the number.
2 × 3/3 = 6/3
\(\frac{6}{3}+\frac{5}{3}\) = \(\frac{11}{3}\)

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 51

Answer:
The denominators are not the same. So first you have to make the common denominators and add the fraction with the number.
6 × 12/12 = 72/12
\(\frac{72}{12}+\frac{1}{12}\) = \(\frac{73}{12}\)

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 52

Answer:
Add the numerators of the like denominators.
3 + 1 + 1 = 5
3/4 + 1/4 + 1/4 = 5/4

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 53

Answer:
Add the numerators of the like denominators.
\(\frac{6}{8}\) + \(\frac{5}{8}\) + \(\frac{4}{8}\)
6 + 5 + 4 = 15
\(\frac{6}{8}\) + \(\frac{5}{8}\) + \(\frac{4}{8}\) = \(\frac{15}{8}\)

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 54

Answer:
Add the numerators of the like denominators.
43 + 16 + 10 = 69
\(\frac{43}{100}\) + \(\frac{16}{100}\) + \(\frac{10}{100}\) = \(\frac{69}{100}\)

Question 13.
You eat \(\frac{2}{10}\) of a vegetable pizza. Your friend eats \(\frac{3}{10}\) of the pizza. What fraction of the pizza do you and your friend eat together?

Answer:
Given that,
You eat \(\frac{2}{10}\) of a vegetable pizza. Your friend eats \(\frac{3}{10}\) of the pizza.
\(\frac{2}{10}\) + \(\frac{3}{10}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)
\(\frac{1}{2}\) fraction of the pizza do you and your friend eat together

Question 14.
Number Sense
A sum has 5 addends. Each addend is a unit fraction. The sum is 1. What are the addends?

Answer:
\(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) = \(\frac{5}{5}\) = 1

Question 15.
Writing
Explain how to add \(\frac{3}{4}\) and \(\frac{1}{4}\). Use a model to support your answer.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 55

Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-55
\(\frac{3}{4}\) + \(\frac{1}{4}\) = \(\frac{4}{4}\) = 1

Think and Grow: Modeling Real Life

Example
The table shows the natural hazards studied by 100 students for a science project. What fraction of the students studied a weather-based natural hazard?
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 56
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 57

Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-57

Show and Grow

Question 16.
Use the graph above to find what fraction of the students studied an Earth-based natural hazard.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 57.1

Answer:
We can find the fraction of the students who studied an Earth-based natural hazard
1 × 8 = 8
Half drop = 4
8 + 4 = 12
=  Number of students/Total number of students surveyed
= 12/100
Thus \(\frac{12}{100}\) fraction of the students who studied an Earth-based natural hazard.

Question 17.
DIG DEEPER!
A caterer needs at least 2 pounds of lunch meat to make a sandwich platter. She has \(\frac{6}{4}\) pounds of turkey and \(\frac{3}{4}\) pound of ham. Does the caterer have enough lunch meat to make a sandwich platter? Explain.

Answer:
Given that,
A caterer needs at least 2 pounds of lunch meat to make a sandwich platter. She has \(\frac{6}{4}\) pounds of turkey and \(\frac{3}{4}\) pound of ham.
\(\frac{6}{4}\) + \(\frac{3}{4}\) = \(\frac{9}{4}\)
Convert it into mixed fraction
\(\frac{9}{4}\) = 1 \(\frac{3}{4}\)
Thus the caterer does not have enough lunch meat to make a sandwich platter.

Add Fractions with Like Denominators Homework & Practice 8.3

Add

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 58

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-58

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 59

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
\(\frac{2}{2}\) + \(\frac{7}{2}\) = (2 + 7)/2 = \(\frac{9}{2}\)

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 60

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
\(\frac{2}{5}\) + \(\frac{2}{5}\) = (2 + 2)/5 = \(\frac{4}{5}\)

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 61

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
\(\frac{4}{10}\) + \(\frac{6}{10}\) = (4 + 6)/10 = \(\frac{10}{10}\) = 1

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 62

Answer:
The denominators are not the same. So first you have to make the common denominators and add the fraction with the number.
Take the denominator as common and add the numerators.
4 × 3/3 = 12/3
\(\frac{12}{3}\) + \(\frac{1}{3}\) = (12 + 1)/3 = \(\frac{13}{2}\)

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 63

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
\(\frac{27}{100}\) + \(\frac{460}{100}\) = (27 + 460)/100 = \(\frac{487}{100}\)

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 64

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
\(\frac{8}{4}\) + \(\frac{5}{4}\) = (8 + 5)/4 = \(\frac{13}{4}\)

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 65

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
\(\frac{4}{6}\) + \(\frac{1}{6}\) = (4 + 1)/6 = \(\frac{5}{6}\)

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 66

Answer:
The denominators are not the same. So first you have to make the common denominators and add the fraction with the number.
Take the denominator as common and add the numerators.
10 × 12/12 = 120/12
\(\frac{120}{12}\) + \(\frac{7}{12}\) = (120 + 7)/3 = \(\frac{127}{12}\)

Question 10.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 67

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
\(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{2}{5}\) = (1 + 1 + 2)/5 = \(\frac{4}{5}\)

Question 11.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 68

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
\(\frac{38}{100}\) + \(\frac{13}{100}\) + \(\frac{21}{100}\) = (38+ 13 + 21)/100 = \(\frac{72}{100}\)

Question 12.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 69

Answer:
Add the numerators of the like denominators.
Take the denominator as common and add the numerators.
\(\frac{8}{8}\) + \(\frac{4}{8}\) + \(\frac{2}{8}\) = (8 + 4 + 2)/8 = \(\frac{14 }{8}\)

Question 13.
You plant a sunflower seed. After 11 week, the plant is \(\frac{1}{2}\) inch tall. The next week your plant grows \(\frac{3}{2}\) inches. How tall is your plant after the second week?

Answer:
Given that,
You plant a sunflower seed. After 11 week, the plant is \(\frac{1}{2}\) inch tall. The next week your plant grows \(\frac{3}{2}\) inches.
\(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{4}{2}\) = 2
The plant is 2 inches tall after the second week.

Question 14.
Writing
Explain how to find the unknown addend.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 70

Answer:
1 can be written as \(\frac{10}{10}\)
\(\frac{7}{10}\) + ? = \(\frac{10}{10}\)
? = \(\frac{10}{10}\) – \(\frac{7}{10}\)
? = \(\frac{3}{10}\)
Thus the unknown addend is \(\frac{3}{10}\)

Question 15.
DIG DEEPER!
When you double me and add \(\frac{1}{6}\), you get \(\frac{5}{6}\). What fraction am I?

Answer: \(\frac{2}{6}\)

Explanation:
If you add \(\frac{2}{6}\) twice and add \(\frac{1}{6}\) to it you get \(\frac{5}{6}\).

Question 16.
Reasoning
You eat \(\frac{2}{8}\) of a large apple at lunch and another \(\frac{4}{8}\) of it as a snack. Your friend eats \(\frac{4}{8}\) of a small apple at lunch and another \(\frac{2}{8}\) of it as a snack. Do you each eat the same amount? Explain.

Answer:
Given that,
You eat \(\frac{2}{8}\) of a large apple at lunch and another \(\frac{4}{8}\) of it as a snack. Your friend eats \(\frac{4}{8}\) of a small apple at lunch and another \(\frac{2}{8}\) of it as a snack.
\(\frac{2}{8}\) + \(\frac{4}{8}\) = \(\frac{6}{8}\)
\(\frac{2}{8}\) + \(\frac{4}{8}\) = \(\frac{6}{8}\)
Yes you and your friend eat same amount of food.

Question 17.
Modeling Real Life
The graph shows the classification of 100 species of birds in North America according to their extinction rate. What fraction of the species are classified as near threatened or vulnerable?
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 71

Answer:
• = 4 species
Near threatened = 4 × 4 = 16 species
half • = 2 species
16 + 2 = 18 species
Fraction of near threatened = number of species/total number of species of birds in North America
= 18/100
Vulnerable = 5 × 4 = 20 species
half • = 2 species
20 + 2 = 22 species
Fraction of near threatened = number of species/total number of species of birds in North America
= 22/100

Question 18.
Modeling Real Life
Use the graph above to find what fraction of not the species are critically endangered.

Answer:
• = 4 species
critically endangered = 4 × 4 = 16 species
half • = 2 species
16 + 2 = 18 species
Fraction of near threatened = number of species/total number of species of birds in North America
= 18/100

Review & Refresh

Question 19.
A pet store has 25 tanks with 32 fish in each tank. A customer buys 7 fish. How many fish does the pet store have now?

Answer:
Given,
A pet store has 25 tanks with 32 fish in each tank. A customer buys 7 fish.
\(\frac{25}{32}\) – \(\frac{7}{32}\) = \(\frac{18}{32}\)
Thus there are 18 fishes in the pet store.

Lesson 8.4 Use Models to subtract Fractions

Explore and Grow

Draw a model to show \(\frac{9}{12}\).
Answer:
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_25

Use your model to find \(\frac{9}{12}\) – \(\frac{5}{12}\). Explain your method.
Answer:
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_26

Repeated Reasoning
Write two fractions that have a difference of \(\frac{7}{12}\). Explain your reasoning.

Answer:  \(\frac{9}{12}\) – \(\frac{2}{12}\) = \(\frac{7}{12}\)

Think and Grow: Use Models to Subtract Fractions

You can subtract fractions by taking away parts that refer to the same whole.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 72

Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-72
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 73

Answer:
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-73

Show and Grow

Find the difference. Explain how you used the model to subtract.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 74

Answer: 5/10 = 1/2
Take away a length of 4/10 from the length of 9/10.
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-74

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 75

Answer:
Take away a length of 6/4 from the length of 2/4.
Big-Ideas-Math-Solutions-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-75

Apply and Grow: Practice

Find the difference. Use a model or a number line to help.

Question 3.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 76

Answer: 4/8
Take away a length of 8/8 from the length of 4/8.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_7

Question 4.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 77

Answer: 8/12
Take away a length of 10/12 from the length of 2/12.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_27

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 8 Add and Subtract Fractions 78

Answer: 3/5
Take away a length of 4/5 from the length of 1/5.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-28

Question 6.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 79

Answer: 6/2 = 3
Take away a length of 9/2 from the length of 3/2.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-29

Question 7.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 80

Answer: 10/6
Take away a length of 15/6 from the length of 5/6.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-30

Question 8.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 81

Answer: \(\frac{26}{100}\)
The denominators of both the fractions are the same. So subtract the numerators.
\(\frac{76}{100}\) – \(\frac{50}{100}\) = \(\frac{26}{100}\)

Question 9.
You need to walk \(\frac{3}{4}\) mile for your physical education class. So far, you have walked \(\frac{2}{4}\) mile. How much farther do you need to walk?

Answer:
Given that,
You need to walk \(\frac{3}{4}\) mile for your physical education class. So far, you have walked \(\frac{2}{4}\) mile.
\(\frac{3}{4}\) – \(\frac{2}{4}\) = \(\frac{1}{4}\)
You need to walk \(\frac{1}{4}\) miles more.

Question 10.
Number Sense
Which expressions have a difference of \(\frac{4}{5}\) ?
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 82

Answer:
5/5 – 1/5 = 4/5
10/5 – 6/5 = 4/5
6/5 – 3/5 = 3/5
9/5 – 5/5 = 4/5
i, ii, iv has the difference of \(\frac{4}{5}\)

Question 11.
Structure
Write the subtraction equation represented by the model.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 83

Answer: \(\frac{7}{8}\) – \(\frac{4}{8}\) = \(\frac{3}{8}\)

Question 12.
Writing
Explain why the numerator changes when you subtract fractions with like denominators, but the denominator stays the same.

Answer:
The most simple fraction subtraction problems are those that have two proper fractions with a common denominator. That is, each denominator is the same. The process is just as it is for the addition of fractions with like denominators, except you subtract! You subtract the second numerator from the first and keep the denominator the same.

Think and Grow: Modeling Real Life

Example
A lizard’s tail is \(\frac{10}{12}\) foot long. It sheds a \(\frac{7}{12}\) foot long part of its tail to escape a predator. How long is the remaining part of the lizard’s tail?
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 84
Because each fraction represents a part of the same whole, you can take away a part.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 85

Answer:
Given that,
A lizard’s tail is \(\frac{10}{12}\) foot long. It sheds a \(\frac{7}{12}\) foot long part of its tail to escape a predator.
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-85

Show and Grow

Question 13.
You have \(\frac{9}{8}\) cups of raisins. You eat \(\frac{2}{8}\) cup. What fraction of a cup of raisins do you have left?
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 86

Answer:
Given that,
You have \(\frac{9}{8}\) cups of raisins. You eat \(\frac{2}{8}\) cup.
\(\frac{9}{8}\) – \(\frac{2}{8}\) = \(\frac{7}{8}\)
Thus \(\frac{7}{8}\) fraction of a cup of raisins is left.

Question 14.
A large bottle has \(\frac{7}{4}\) quarts of liquid soap. A small bottle has \(\frac{3}{4}\) quart of liquid soap. How much more soap is in the large bottle than in the small bottle?

Answer:
Given that,
A large bottle has \(\frac{7}{4}\) quarts of liquid soap. A small bottle has \(\frac{3}{4}\) quart of liquid soap.
\(\frac{7}{4}\) – \(\frac{3}{4}\) = \(\frac{4}{4}\) = 1
Thus 1 more soap is in the large bottle than in the small bottle.

Question 15.
DIG DEEPER!
You need 2 cups of milk for a recipe. You have cup of \(\frac{1}{3}\) milk. How much more milk do you need? Explain.

Answer:
Given,
You need 2 cups of milk for a recipe. You have cup of \(\frac{1}{3}\) milk.
2 × \(\frac{1}{3}\) = \(\frac{2}{3}\)
Thus \(\frac{2}{3}\) more milk you need.

Use Models to subtract Fractions Homework & Practice 8.4

Find the difference. Explain how you used the model to subtract.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 87

Answer:
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-87

Question 2.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 88

Answer:
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-88

Find the difference. Use a model or a number line to help.

Question 3.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 89

Answer: 15/10
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-25

Question 4.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 90

Answer: 10/5

BIM 3rd Grade Answer Key Grade 4 Chapter 8 Add & Subtract Fractions img_31

Question 5.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 91

Answer: 8/12

Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_32

Question 6.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 92

Answer:

Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-226

Question 7.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 93

Answer: 7/4

Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-146

Question 8.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 94

Answer:
The denominators of both the fractions are the same. So subtract the numerators.
\(\frac{70}{100}\) – \(\frac{6}{100}\) = \(\frac{64}{100}\)

Question 9.
You have \(\frac{2}{3}\) yard of ribbon. You cut off \(\frac{1}{3}\) yard of the ribbon. How much ribbon do you have left?

Answer:
Given that,
You have \(\frac{2}{3}\) yard of ribbon. You cut off \(\frac{1}{3}\) yard of the ribbon.
The denominators of both the fractions are the same. So subtract the numerators.
\(\frac{2}{3}\) – \(\frac{1}{3}\) = \(\frac{1}{3}\)
\(\frac{1}{3}\) ribbon has left.

Question 10.
Structure
When using circular models to find the difference of \(\frac{4}{2}\) and \(\frac{1}{2}\), why do you shade two circles to represent \(\frac{4}{2}\)?

Answer:
The denominators of both the fractions are the same. So subtract the numerators.
\(\frac{4}{2}\) – \(\frac{1}{2}\) = \(\frac{3}{2}\)

Question 11.
YOU BE THE TEACHER
In a box of pens, \(\frac{3}{4}\) of the pens are blue. Your friend takes \(\frac{1}{4}\) of the blue pens and says that now \(\frac{2}{4}\) of the pens in the box are blue. Is your friend correct? Explain.

Answer:
Given,
In a box of pens, \(\frac{3}{4}\) of the pens are blue. Your friend takes \(\frac{1}{4}\) of the blue pens and says that now \(\frac{2}{4}\) of the pens in the box are blue.
The denominators of both the fractions are the same. So subtract the numerators.
\(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{2}{4}\)
Yes, your friend is correct.

Question 12.
DIG DEEPER!
Using numerators that even number, write two different subtraction equations that each have a difference of 1.

Answer: \(\frac{6}{4}\) – \(\frac{2}{4}\) = \(\frac{4}{4}\) = 1

Question 13.
Modeling Real Life
In our solar system, \(\frac{6}{8}\) of the planets have moons, and \(\frac{4}{8}\) of the planets have moons and rings. What fraction of the planets in our solar system have moons, but do not have rings?

Answer:
Given,
In our solar system, \(\frac{6}{8}\) of the planets have moons, and \(\frac{4}{8}\) of the planets have moons and rings.
The denominators of both the fractions are the same. So subtract the numerators.
\(\frac{6}{8}\) – \(\frac{4}{8}\) = \(\frac{2}{8}\)

Question 14.
Modeling Real Life
A professional pumpkin carver carves a pumpkin that weighs \(\frac{7}{10}\) ton. He carves a second pumpkin that weighs \(\frac{6}{10}\) ton. How much heavier is the first pumpkin than the second pumpkin?
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 95

Answer:
Given that,
A professional pumpkin carver carves a pumpkin that weighs \(\frac{7}{10}\) ton. He carves a second pumpkin that weighs \(\frac{6}{10}\) ton.
The denominators of both the fractions are the same. So subtract the numerators.
\(\frac{7}{10}\) – \(\frac{6}{10}\) = \(\frac{1}{10}\)
The first pumpkin is \(\frac{1}{10}\) heavier than the second pumpkin.

Review & Refresh

Find an equivalent fraction.

Question 15.
\(\frac{7}{4}\)

Answer:
The equivalent fraction of \(\frac{7}{4}\) is given below,
\(\frac{7}{4}\) × \(\frac{2}{2}\) = \(\frac{14}{8}\)

Question 16.
\(\frac{3}{5}\)

Answer:
The equivalent fraction of \(\frac{3}{5}\) is given below,
\(\frac{3}{5}\) × \(\frac{3}{3}\) = \(\frac{9}{15}\)

Question 17.
\(\frac{2}{3}\)

Answer:
The equivalent fraction of \(\frac{2}{3}\) is given below,
\(\frac{2}{3}\) × \(\frac{2}{2}\) = \(\frac{4}{6}\)

Lesson 8.5 Subtract Fractions with Like Denominators

Explore and Grow

Write each fraction as a sum of unit fractions. Use models to help.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 96
How many more unit fractions did you use to rewrite \(\frac{4}{5}\) than \(\frac{3}{5}\)?
How does this relate to the difference \(\frac{4}{5}\) – \(\frac{3}{5}\) ?

Answer:
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-96
\(\frac{4}{5}\) – \(\frac{3}{5}\) = \(\frac{1}{5}\)

Construct Arguments
How can you use the numerators and the denominators to subtract fractions with like denominators? Explain why your method makes sense.

Answer: Steps on How to Add and Subtract Fractions with the Same Denominator. To add fractions with like or the same denominator, simply add the numerators then copy the common denominator. Always reduce your final answer to its lowest term.

Think and Grow: Subtract Fractions

To subtract fractions with like denominators, subtract the numerators. The denominator stays the same.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 97
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 98

Answer:
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-98

Show and Grow

Subtract.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 100

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.

Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-100

Question 2.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 101

Answer:
First, make the denominators common and then subtract the numerators
1 can be written as \(\frac{12}{12}\)
\(\frac{12}{12}\) – \(\frac{8}{12}\) = (12 – 8)/12
= \(\frac{4}{12}\) or \(\frac{1}{3}\)

Question 3.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 102

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{50}{100}\) – \(\frac{30}{100}\) = (50 – 30)/100
= \(\frac{20}{100}\) or \(\frac{1}{5}\)

Apply and Grow: Practice

Subtract.

Question 4.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 103

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
Big Ideas Math Book 4th Grade Answer Key Chapter 8 Add and Subtract Fractions img_37

Question 5.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 104

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_35

Question 6.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 105

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{12}{6}\) – \(\frac{7}{6}\) = (12- 7)/6
\(\frac{5}{6}\)

Question 7.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 106

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{4}{5}\) – \(\frac{3}{5}\) = (4- 3)/5
\(\frac{1}{5}\)

Question 8.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 107

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{60}{100}\) – \(\frac{43}{100}\) = (60 – 43)/100
\(\frac{17}{100}\)

Question 9.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 108

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{10}{2}\) – \(\frac{2}{2}\) = (10 – 2)/2
\(\frac{8}{4}\) = 2

Question 10.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 109

Answer:
First, make the denominators common and then subtract the numerators.
\(\frac{12}{12}\) – \(\frac{7}{12}\) = (12 – 7)/12
= \(\frac{5}{12}\)
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_34

Question 11.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 110

Answer:
First, make the denominators common and then subtract the numerators
1 can be written as \(\frac{8}{8}\)
\(\frac{8}{8}\) – \(\frac{5}{8}\) = \(\frac{3}{8}\)

Question 12.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 111

Answer:
First, make the denominators common and then subtract the numerators
2 can be written as \(\frac{8}{4}\)
\(\frac{8}{4}\) – \(\frac{1}{4}\) = \(\frac{7}{4}\)

Question 13.
You have 1 gallon of paint. You use \(\frac{2}{3}\) gallon to paint a wall. How much paint do you have left?
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 112

Answer:
Given that,
You have 1 gallon of paint. You use \(\frac{2}{3}\) gallon to paint a wall.
First, make the denominators common and then subtract the numerators
1 – \(\frac{2}{3}\)
1 can be written as \(\frac{3}{3}\)
\(\frac{3}{3}\) – \(\frac{2}{3}\) = \(\frac{1}{3}\)

Question 14.
Reasoning
Why is it unreasonable to get a difference of \(\frac{7}{8}\) when subtracting \(\frac{1}{8}\) from \(\frac{7}{8}\)? Use a model to support your answer.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 113

Answer:
The difference of \(\frac{7}{8}\) when subtracting \(\frac{1}{8}\) from \(\frac{7}{8}\) is,
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-113

Question 15.
Your friend says each difference is \(\frac{3}{10}\). Is your friend correct? Explain.
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 114

Answer:
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_36
Your friend is correct.
10/10 – 7/10 = 3/10
100/100 = 70/100 = 30/100 = 3/10

Think and Grow: Modeling Real Life

Example
A flock of geese has completed \(\frac{5}{12}\) of its total migration. What fraction of its migration does the flock of geese have left to complete?
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 115
Because the total migration is 1 whole, find 1 − \(\frac{5}{12}\).
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions 116

Answer:
Given,
A flock of geese has completed \(\frac{5}{12}\) of its total migration.
Because the total migration is 1 whole, find 1 − \(\frac{5}{12}\).
First, make the denominators common and then subtract the numerators.
Big-Ideas-Math-Answers-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-116

Show and Grow

Question 16.
A runner has completed \(\frac{6}{10}\) of a race. What fraction of the race does the runner have left to complete?

Answer:
Given that,
A runner has completed \(\frac{6}{10}\) of a race.
1 – \(\frac{6}{10}\)
1 can be written as \(\frac{10}{10}\)
\(\frac{10}{10}\) – \(\frac{6}{10}\) = \(\frac{4}{10}\)
The runner has left \(\frac{4}{10}\) fraction of the race to complete.

Question 17.
A pizza buffet serves pizzas of the same size with different toppings. There is \(\frac{7}{8}\) of a vegetable pizza and \(\frac{2}{8}\) of a pineapple pizza left. How much more vegetable pizza is left than pineapple pizza?

Answer:
Given,
A pizza buffet serves pizzas of the same size with different toppings.
There is \(\frac{7}{8}\) of a vegetable pizza and \(\frac{2}{8}\) of a pineapple pizza left.
\(\frac{7}{8}\) – \(\frac{2}{8}\) = \(\frac{5}{8}\)
\(\frac{5}{8}\) more vegetable pizza is left than pineapple pizza.

Question 18.
DIG DEEPER!
Baseball practice is 1 hour long. You stretch for 7 minutes and play catch for 8 minutes. What fraction of an hour do you have left to practice?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 117

Answer:
Given,
Baseball practice is 1 hour long. You stretch for 7 minutes and play catch for 8 minutes.
7 minutes + 8 minutes = 15 minutes
15 minutes = \(\frac{1}{4}\) hour
1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
Thus \(\frac{3}{4}\) fraction of an hour is left to practice.

Subtract Fractions with Like Denominators Homework & Practice 8.5

Subtract

Question 1.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 118

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{7}{8}\) – \(\frac{3}{8}\) = \(\frac{4}{8}\)
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-118

Question 2.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 119

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{5}{4}\) – \(\frac{3}{4}\) = \(\frac{2}{4}\)

Question 3.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 120

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{13}{5}\) – \(\frac{6}{5}\) = \(\frac{7}{6}\)

Question 4.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 121

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{5}{12}\) – \(\frac{1}{12}\) = \(\frac{4}{12}\)

Question 5.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 122

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{9}{6}\) – \(\frac{4}{6}\) = \(\frac{5}{6}\)

Question 6.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 123

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{11}{3}\) – \(\frac{7}{3}\) = \(\frac{4}{3}\)

Question 7.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 124

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{10}{10}\) – \(\frac{4}{10}\) = \(\frac{6}{10}\)

Question 8.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 125

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{20}{2}\) – \(\frac{8}{2}\) = \(\frac{12}{2}\)

Question 9.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 126

Answer:
The denominators of the above fraction are the same so you have to subtract the numerators.
\(\frac{36}{100}\) – \(\frac{21}{100}\) = \(\frac{15}{100}\)

Question 10.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 127

Answer:
First, make the denominators common and then subtract the numerators.
1 can be written as 5/5.
\(\frac{5}{5}\) – \(\frac{3}{5}\) = \(\frac{2}{5}\)

Question 11.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 128

Answer:
First, make the denominators common and then subtract the numerators.
2 can be written as 8/4
\(\frac{8}{4}\) – \(\frac{2}{4}\) = \(\frac{6}{4}\)

Question 12.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 129

Answer:
First, make the denominators common and then subtract the numerators
3 can be written as 24/8.
\(\frac{24}{8}\) – \(\frac{15}{8}\) = \(\frac{9}{8}\)

Question 13.
A family eats \(\frac{2}{3}\) of a tray of lasagna. What fraction of the tray of lasagna is left?

Answer:
Given,
A family eats \(\frac{2}{3}\) of a tray of lasagna.
1 – \(\frac{2}{3}\)
1 can be written as \(\frac{3}{3}\)
\(\frac{3}{3}\) – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Therefore \(\frac{1}{3}\) fraction of the tray of lasagna is left.

Question 14.
Writing
Explain how finding is \(\frac{7}{10}-\frac{4}{10}\) similar to finding 7 – 4.

Answer:
Yes \(\frac{7}{10}-\frac{4}{10}\) similar to finding 7 – 4. Because the denominators of the fractions are the same.
\(\frac{7}{10}-\frac{4}{10}\) = \(\frac{3}{10}\)

Question 15.
Open-Ended
The model shows equal parts of a 1 whole. Write a subtraction problem whose answer is shown.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 130

Answer:
By seeing the above figure we can write the subtraction problem.
1 – \(\frac{3}{8}\)

Question 16.
Modeling Real Life
you fill \(\frac{2}{4}\) of your plate with vegetables. What fraction of your plate does not contain vegetables?

Answer:
Given,
you fill \(\frac{2}{4}\) of your plate with vegetables.
1 – \(\frac{2}{4}\)
1 can be written as \(\frac{4}{4}\)
\(\frac{4}{4}\) – \(\frac{2}{4}\) = \(\frac{2}{4}\)
Thus \(\frac{2}{4}\) fraction of your plate does not contain vegetables.

Question 17.
Modeling Real Life
A group of students designs a rectangular playground. They use \(\frac{2}{8}\) of the playground for a basketball court and \(\frac{3}{8}\) of the playground for a soccer field. How much space is left?

Answer:
Given,
A group of students designs a rectangular playground.
They use \(\frac{2}{8}\) of the playground for a basketball court and \(\frac{3}{8}\) of the playground for a soccer field.
\(\frac{2}{8}\) + \(\frac{3}{8}\) = \(\frac{5}{8}\)
1 – \(\frac{5}{8}\) = \(\frac{3}{8}\)
Thus \(\frac{3}{8}\) space is left.

Review & Refresh

Find the quotient and the remainder

Question 18.
34 ÷ 7 = ___R___

Answer: 4R6

Explanation:
34 ÷ 7 = \(\frac{34}{7}\)
\(\frac{34}{7}\) = 4R6
Thus the quotient is 4 and the remainder is 6.

Question 19.
28 ÷ 3 = ___R___

Answer: 9 R1

Explanation:
28 ÷ 3 = \(\frac{28}{3}\)
\(\frac{28}{3}\) = 9 R1
Thus the quotient is 9 and the remainder is 1.

Lesson 8.6 Model Fractions and Mixed Numbers

Explore and Grow

Draw a model to show 1 + 1 + \(\frac{2}{3}\).

Use your model to write the sum as a fraction.

Answer:
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_37

Repeated Reasoning
How can you write a fraction greater than 1 as the sum of a whole number and a fraction less than 1? Explain.

Answer:
\(\frac{3}{2}\)  = 1 \(\frac{1}{2}\)
The fraction 1 \(\frac{1}{2}\) the whole fraction is greater than 1 and the fraction is less than 1.

Think and Grow: Write Fractions and Mixed Numbers

A mixed number represents the sum of a whole number and a fraction less than 1.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 132
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 133

Answer:
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-133

Example
Write \(\frac{5}{2}\) as a mixed number.

Find how many wholes are in \(\frac{5}{2}\) and how many halves are left over.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 134

Answer:
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-134

Show and Grow

Question 1.
Write 3\(\frac{1}{4}\) as a fraction. Use a model or a number line to help.

Answer:

Big Ideas math 4th grade answers chapter 8 img_38

Question 2.
Write \(\frac{9}{6}\) as a mixed number. Use a model or a number line to help.

Answer:
\(\frac{9}{6}\) can be written as \(\frac{3}{2}\)
Now convert \(\frac{3}{2}\) into the mixed fraction
\(\frac{3}{2}\) = 1 \(\frac{1}{2}\)
BIM 4th Grade Answer key Chapter 8 Add and subtract fractions img_39

Apply and Grow: Practice

Write the mixed number as a fraction.

Question 3.
3\(\frac{4}{5}\)

Answer: \(\frac{19}{5}\)

Explanation:
Step 1
Multiply the denominator by the whole number
5 × 3 = 15
Step 2
Add the answer from Step 1 to the numerator
15 + 4 = 19
Step 3
Write an answer from Step 2 over the denominator
19/5

Question 4.
2\(\frac{1}{3}\)

Answer: \(\frac{7}{3}\)

Explanation:
Step 1
Multiply the denominator by the whole number
3 × 2 = 6
Step 2
Add the answer from Step 1 to the numerator
6 + 1 = 7
Step 3
Write an answer from Step 2 over the denominator
7/3

Question 5.
6\(\frac{7}{12}\)

Answer: \(\frac{79}{12}\)

Explanation:
Step 1
Multiply the denominator by the whole number
12 × 6 = 72
Step 2
Add the answer from Step 1 to the numerator
72 + 7 = 79
Step 3
Write an answer from Step 2 over the denominator
79/12

Question 6.
1\(\frac{82}{100}\)

Answer: \(\frac{182}{100}\)

Explanation:
Step 1
Multiply the denominator by the whole number
100 × 1 = 100
Step 2
Add the answer from Step 1 to the numerator
100 + 82 = 182
Step 3
Write an answer from Step 2 over the denominator
\(\frac{182}{100}\)

Question 7.
11\(\frac{3}{8}\)

Answer: \(\frac{91}{8}\)

Explanation:
Step 1
Multiply the denominator by the whole number
8 × 11 = 88
Step 2
Add the answer from Step 1 to the numerator
88 + 3 = 91
Step 3
Write an answer from Step 2 over the denominator
91/8

Question 8.
9\(\frac{5}{10}\)

Answer: \(\frac{95}{10}\)

Explanation:
Step 1
Multiply the denominator by the whole number
10 × 9 = 90
Step 2
Add the answer from Step 1 to the numerator
90 + 5 = 95
Step 3
Write an answer from Step 2 over the denominator
95/10

Write the fraction as a mixed number or a whole number.

Question 9.
\(\frac{9}{8}\)

Answer: 1 \(\frac{1}{8}\)

Explanation:
9÷8=1R1
\(\frac{9}{8}\) = 1 \(\frac{1}{8}\)

Question 10.
\(\frac{19}{3}\)

Answer: 6 \(\frac{1}{3}\)

Explanation:
19÷3=6R1
\(\frac{19}{3}\) = 6 \(\frac{1}{3}\)

Question 11.
\(\frac{38}{5}\)

Answer: 7 \(\frac{3}{5}\)

Explanation:
38÷5=7R3
\(\frac{38}{5}\) = 7 \(\frac{3}{5}\)

Question 12.
\(\frac{22}{10}\)

Answer: 2 \(\frac{1}{5}\)

Explanation:
11÷5=2R1
\(\frac{22}{10}\) = 2 \(\frac{1}{5}\)

Question 13.
\(\frac{460}{100}\)

Answer: 4 \(\frac{3}{5}\)

Explanation:
23÷5=4R3
\(\frac{460}{100}\) = 4 \(\frac{3}{5}\)

Question 14.
\(\frac{20}{4}\)

Answer: 5

Explanation:
4 divides 20 five times.
\(\frac{20}{4}\) = 5

Compare

Question 15.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 135

Answer: =

Explanation:
\(\frac{3}{2}\) can be written as 1 \(\frac{1}{2}\)
1 × 2 + 1 = 3
So, 1 \(\frac{1}{2}\) = \(\frac{3}{2}\)

Question 16.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 136

Answer: >

Explanation:
3\(\frac{3}{12}\) can be written as \(\frac{39}{12}\)
\(\frac{39}{12}\) > \(\frac{15}{12}\)
So, 3\(\frac{3}{12}\) > \(\frac{15}{12}\)

Question 17.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 137

Answer: <

Explanation:
\(\frac{21}{6}\) can be written as 3 \(\frac{3}{6}\) or 3 \(\frac{1}{2}\)
So, 3 \(\frac{1}{2}\) < 4
\(\frac{21}{6}\) < 4

Question 18.
Which One Doesn’t Belong? Which expression does not Belong to the other three?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 138

Answer:
3 \(\frac{2}{3}\) = \(\frac{11}{3}\)
\(\frac{9}{3}\) + \(\frac{3}{3}\) = \(\frac{12}{3}\)
\(\frac{3}{3}\) + \(\frac{3}{3}\) +\(\frac{3}{3}\) + \(\frac{2}{3}\) = \(\frac{11}{3}\)
\(\frac{11}{3}\)
So, the second expression does not belong to the other three expressions.

DIG DEEPER!
Find the unknown Number

Question 19.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 139

Answer: 2

Explanation:
\(\frac{8}{6}\) is 4÷3=1R1
\(\frac{8}{6}\) = 1 \(\frac{2}{6}\)
So, the unknown number is 2.

Question 20.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 140

Answer: 3

Explanation:
\(\frac{35}{4}\) = 8 R 3
8 \(\frac{3}{4}\) = \(\frac{35}{4}\)
So, the unknown number is 3.

Question 21.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 141

Answer: 10

Explanation:
12 × 10 + 9 = 129
\(\frac{129}{12}\) = 10 \(\frac{9}{12}\)
So, the unknown number is 10.

Think and Grow: Modeling Real Life

Example
A construction worker needs nails that are \(\frac{9}{4}\) inches long. Which size of nails should the worker use?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 142
Write \(\frac{9}{4}\) as a mixed number.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 143

Answer:
Given,
A construction worker needs nails that are \(\frac{9}{4}\) inches long.
Convert from improper fraction to the mixed fraction.
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-143

Show and Grow

Question 22.
You need screws that are \(\frac{13}{8}\) inches long to build a birdhouse. Which size of screws should you use?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 144

Answer:
Given,
You need screws that are \(\frac{13}{8}\) inches long to build a birdhouse.
Convert from improper fraction to the mixed fraction.
\(\frac{13}{8}\) = 1 \(\frac{5}{8}\)
8 × 1 + 5 = 13
So, you should use 1 \(\frac{5}{8}\) inches of screws.

Question 23.
You and your friend each measure the distance between two bean bag toss boards. You record the distance as 3\(\frac{3}{5}\) meters. Your friend records the distance as \(\frac{18}{5}\) meters. Did you and your friend record the same distance? Explain.

Answer:
Given that,
You and your friend each measure the distance between two bean bag toss boards.
You record the distance as 3\(\frac{3}{5}\) meters. Your friend records the distance as \(\frac{18}{5}\) meters.
3\(\frac{3}{5}\)
5 × 3 + 3 = 18
3\(\frac{3}{5}\) = \(\frac{18}{5}\)
Yes you and your friend record the same distance.

Question 24.
DIG DEEPER!
You use a \(\frac{1}{3}\)-cup scoop to measure 3\(\frac{1}{3}\) cups of rice. How many times do you fill the scoop?

Answer:
Given,
You use a \(\frac{1}{3}\)-cup scoop to measure 3\(\frac{1}{3}\) cups of rice.
\(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) = \(\frac{10}{3}\)
You need to measure 10 times to fill the scoop.

Question 25.
DIG DEEPER!
A sunflower plant is \(\frac{127}{10}\) centimeters tall. A snapdragon plant is 8\(\frac{9}{10}\) centimeters tall. Which plant is taller? Explain.

Answer:
Given that,
A sunflower plant is \(\frac{127}{10}\) centimeters tall. A snapdragon plant is 8\(\frac{9}{10}\) centimeters tall.
8\(\frac{9}{10}\) = \(\frac{89}{10}\)
\(\frac{127}{10}\) is greater than \(\frac{89}{10}\)
So, sunflower plant is taller.

Model Fractions and Mixed Numbers Homework & Practice 8.6

Write a mixed number as a fraction.

Question 1.
1\(\frac{7}{10}\)

Answer: \(\frac{17}{10}\)

Explanation:
Step 1
Multiply the denominator by the whole number
10 × 1 = 10
Step 2
Add the answer from Step 1 to the numerator
10 + 7 = 17
Step 3
Write an answer from Step 2 over the denominator
=17/10

Question 2.
1\(\frac{5}{6}\)

Answer: \(\frac{11}{6}\)

Explanation:
Step 1
Multiply the denominator by the whole number
6 × 1 = 6
Step 2
Add the answer from Step 1 to the numerator
6 + 5 = 11
Step 3
Write an answer from Step 2 over the denominator
11/6

Question 3.
2\(\frac{2}{3}\)

Answer: \(\frac{8}{3}\)

Explanation:
Step 1
Multiply the denominator by the whole number
3 × 2 = 6
Step 2
Add the answer from Step 1 to the numerator
6 + 2 = 8
Step 3
Write an answer from Step 2 over the denominator
8/3

Question 4.
4\(\frac{1}{2}\)

Answer: \(\frac{9}{2}\)

Explanation:
Step 1
Multiply the denominator by the whole number
2 × 4 = 8
Step 2
Add the answer from Step 1 to the numerator
8 + 1 = 9
Step 3
Write an answer from Step 2 over the denominator
9/2

Question 5.
3\(\frac{2}{8}\)

Answer: \(\frac{13}{4}\)

Explanation:
Step 1
Multiply the denominator by the whole number
8 × 3 = 24
Step 2
Add the answer from Step 1 to the numerator
24 + 2 = 26
Step 3
Write an answer from Step 2 over the denominator
26/8

Question 6.
9\(\frac{8}{12}\).

Answer: \(\frac{29}{3}\)

Explanation:
Step 1
Multiply the denominator by the whole number
12 × 9 = 108
Step 2
Add the answer from Step 1 to the numerator
108 + 8 = 116
Step 3
Write an answer from Step 2 over the denominator
116/12 = \(\frac{29}{3}\)

write the fraction as a mixed number or a whole number.

Question 7.
\(\frac{7}{5}\)

Answer: 1 \(\frac{2}{5}\)

Explanation:
Given the expression \(\frac{7}{5}\)
We have to convert the improper fraction to the mixed fraction.
7 ÷ 5=1R2
\(\frac{7}{5}\) = 1 \(\frac{2}{5}\)

Question 8.
\(\frac{10}{3}\)

Answer: 3 \(\frac{1}{3}\)

Explanation:
Given the expression \(\frac{10}{3}\)
We have to convert the improper fraction to the mixed fraction.
10÷3=3R1
\(\frac{10}{3}\) = 3 \(\frac{1}{3}\)

Question 9.
\(\frac{15}{4}\)

Answer: 3 \(\frac{3}{4}\)

Explanation:
Given the expression \(\frac{15}{4}\)
We have to convert the improper fraction to the mixed fraction.
15÷4=3 R 3
\(\frac{15}{4}\) = 3 \(\frac{3}{4}\)

Question 10.
\(\frac{32}{6}\)

Answer: 5 \(\frac{1}{3}\)

Explanation:
Given the expression \(\frac{32}{6}\)
We have to convert the improper fraction to the mixed fraction.
\(\frac{32}{6}\) = \(\frac{16}{3}\)
16÷3=5R1
\(\frac{32}{6}\) = 5 \(\frac{1}{3}\)

Question 11.
\(\frac{75}{8}\)

Answer: 9 \(\frac{3}{8}\)

Explanation:
Given the expression \(\frac{75}{8}\)
We have to convert the improper fraction to the mixed fraction.
75÷8=9R3
\(\frac{75}{8}\) = 9 \(\frac{3}{8}\)

Question 12.
\(\frac{40}{10}\)

Answer: 4

Explanation:
Given the expression \(\frac{40}{10}\)
We have to convert the improper fraction to the mixed fraction.
\(\frac{40}{10}\) = \(\frac{4}{1}\) = 4
Thus \(\frac{40}{10}\) = 4

Compare

Question 13.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 145

Answer: <

Explanation:
We have to convert the improper fraction to the mixed fraction.
5 \(\frac{1}{2}\) = \(\frac{11}{2}\)
\(\frac{11}{2}\) < \(\frac{15}{2}\)

Question 14.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 146

Answer: =

Explanation:
We have to convert the improper fraction to the mixed fraction.
\(\frac{27}{12}\) = \(\frac{27}{12}\)

Question 15.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 147

Answer:

Explanation:
We have to convert the improper fraction to the mixed fraction.
6 \(\frac{7}{8}\) = \(\frac{55}{8}\)
\(\frac{55}{8}\) > \(\frac{50}{8}\)

Question 16.
Number Sense
Complete the number line.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 148

Answer:
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-148

Question 17.
Modeling Real Life
You need pencil lead that is \(\frac{12}{10}\) millimeters thick to complete an art project. Which size of pencil lead should you use?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 149

Answer:
Given,
You need pencil lead that is \(\frac{12}{10}\) millimeters thick to complete an art project.
1 \(\frac{1}{10}\) = \(\frac{11}{10}\)
1 \(\frac{2}{10}\) = \(\frac{12}{10}\)
1 \(\frac{4}{10}\) = \(\frac{14}{10}\)
You should use 2nd pencil lead.

Question 18.
DIG DEEPER!
You have a \(\frac{1}{4}\)-cup measuring cup and a \(\frac{1}{2}\)-cup measuring cup. What are two ways you can 2\(\frac{3}{4}\) cups of water?

Answer:
Given,
You have a \(\frac{1}{4}\)-cup measuring cup and a \(\frac{1}{2}\)-cup measuring cup.
\(\frac{1}{4}\) + \(\frac{1}{2}\) = \(\frac{3}{4}\)
2\(\frac{3}{4}\) – \(\frac{3}{4}\) = 2

Review & Refresh

Question 19.
67 × 31 = ___

Answer:
Multiply the two numbers 67 and 31.
67 × 31 = 2077

Question 20.
83 × 47 = ___

Answer:
Multiply the two numbers 83 and 47.
83 × 47 = 3901

Lesson 8.7 Add Mixed Numbers

Use model to find 2\(\frac{3}{8}\) + 1\(\frac{1}{8}\).

Answer: 3 \(\frac{1}{2}\)

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=2+3/8+1+1/8
Solving the whole number parts
2+1=3
Solving the fraction parts
3/8+1/8=4/8
Reducing the fraction part, 4/8,
4/8=1/2
Combining the whole and fraction parts
3+1/2=3 1/2

Construct Arguments
How can you use the whole number parts and the fractional parts to add mixed numbers with like denominators? Explain why your method makes sense.

Answer: To add mixed numbers, we first add the whole numbers together, and then the fractions. If the sum of the fractions is an improper fraction, then we change it to a mixed number.

Think and Grow: Add Mixed Numbers

To add mixed numbers, add the fractional parts and add the whole number parts. Another way to add mixed numbers is to rewrite each number as a fraction, then add.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 150
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 151

Answer:
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-151

Example
Find 4\(\frac{2}{8}\) + 2\(\frac{7}{8}\).
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 152

Answer:
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-152

Apply and Grow: Practice

Add.

Question 3.
5\(\frac{1}{3}\) + 3\(\frac{2}{3}\) = ___

Answer: 9

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=5+1/3+3+2/3
Solving the whole number parts
5+3=8
Solving the fraction parts
1/3+2/3=33
Reducing the fraction part, 3/3,
3/3=1/1
Simplifying the fraction part, 1/1,
1/1=1
Combining the whole and fraction parts
8+1=9

Question 4.
2\(\frac{8}{12}\) + 7\(\frac{5}{12}\) = ___

Answer: 10 \(\frac{1}{12}\)

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=2+8/12+7+5/12
Solving the whole number parts
2+7=9
Solving the fraction parts
8/12+5/12=13/12
Simplifying the fraction part, 13/12,
13/12=1 1/12
Combining the whole and fraction parts
9+1+1/12=10 1/12

Question 5.
4 + 1\(\frac{1}{2}\) = ___

Answer: 5 \(\frac{1}{2}\)

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=4+1+1/2
Solving the whole number parts
4+1=5
Combining the whole and fraction parts
5+1/2=5 1/2

Question 6.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 153

Answer: 5 \(\frac{2}{100}\)

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=78/100+124/100 + 3
Solving the fraction parts
78/100+124/100=202/100
3 + 202/100 = 5 \(\frac{2}{100}\)

Question 7.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 154

Answer: 16 \(\frac{3}{8}\)

Explanation:
Add the fractional parts and then the whole numbers.
8 + 5 + 2 = 15
\(\frac{4}{8}\) + \(\frac{3}{8}\) + \(\frac{4}{8}\) = (4 + 4 + 3)/8 = \(\frac{11}{8}\)
\(\frac{11}{8}\) = 1 \(\frac{3}{8}\)
15 + 1\(\frac{3}{8}\) = 16 \(\frac{3}{8}\)

Question 8.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 155

Answer: 24 \(\frac{2}{5}\)

Explanation:
Add the fractional parts and then the whole numbers.
10 + 9 + 4 = 23
\(\frac{4}{5}\) + \(\frac{2}{5}\) + \(\frac{1}{5}\) = \(\frac{7}{5}\)
Convert it into the mixed fraction.
\(\frac{7}{5}\) = 1 \(\frac{2}{5}\)
23 + 1 \(\frac{2}{5}\) = 24 \(\frac{2}{5}\)

Question 9.
Number Sense
Explain how to use the addition properties to find Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 156 mentally. Then find the sum.

Answer: 16 \(\frac{3}{4}\)

Explanation:
Add the fractional parts and then the whole numbers.
6 + 8 + 1 = 15
\(\frac{3}{4}\) + \(\frac{2}{4}\) + \(\frac{1}{4}\) = \(\frac{6}{4}\)
Convert it into the mixed fraction.
\(\frac{6}{4}\) = 1 \(\frac{2}{4}\)
15 + 1 \(\frac{2}{4}\) = 16 \(\frac{2}{4}\)

Question 10.
DIG DEEPER!
When adding mixed numbers, is it always necessary to write the sum as a mixed number? Explain.

Answer: To add mixed numbers, we first add the whole numbers together, and then the fractions. If the sum of the fractions is an improper fraction, then we change it to a mixed number.

Question 11.
DIG DEEPER!
Find the unknown number.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 156.1

Answer:
Let the unknown number be x.
4 \(\frac{5}{6}\) + x = 8 \(\frac{3}{6}\)
x = 8 \(\frac{3}{6}\) – 4 \(\frac{5}{6}\)
x = 3 \(\frac{2}{3}\)

Think and Grow: Modeling Real Life

Example
You pick 2\(\frac{3}{4}\) pounds of cherries. Your friend picks 1\(\frac{2}{4}\) pounds of cherries. How many pounds of cherries do you and your friend pick in all?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 157
Add the amounts of cherries you and your friend each pick.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 158

Answer:
You pick 2\(\frac{3}{4}\) pounds of cherries. Your friend picks 1\(\frac{2}{4}\) pounds of cherries.
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-158

Show and Grow

Question 12.
Before noon, 2\(\frac{3}{8}\) inches of snow falls in a city. Afternoon, 4\(\frac{6}{8}\) inches of snow falls. How many inches of snow falls in the city that day?

Answer:
Given that,
Before noon, 2\(\frac{3}{8}\) inches of snow falls in a city. Afternoon, 4\(\frac{6}{8}\) inches of snow falls.
2\(\frac{3}{8}\) + 4\(\frac{6}{8}\)
2 + 4 = 6
\(\frac{3}{8}\) + \(\frac{6}{8}\) = \(\frac{9}{8}\)
Convert it into the mixed fraction.
\(\frac{9}{8}\) = 1 \(\frac{1}{8}\)
1 \(\frac{1}{8}\) inches of snow falls in the city that day.

Question 13.
DIG DEEPER!
A student driver must practice driving at night for a total of at least 10 hours. Has the student met the nighttime driving requirement yet?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 159

Answer:
2 \(\frac{1}{2}\) + 3 \(\frac{1}{2}\) + 2 \(\frac{1}{2}\)
First add the whole numbers
2 + 3 + 2 = 7
\(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1 \(\frac{1}{2}\)
7 + 1 \(\frac{1}{2}\) = 8 \(\frac{1}{2}\)

Add Mixed Numbers Homework & Practice 8.7

Add.

Question 1.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 160

Answer: 12 \(\frac{4}{5}\)

Explanation:
Add the fractional parts and then the whole numbers.
4 + 8 = 12
Add the fractions
\(\frac{1}{5}\) + \(\frac{3}{5}\) = \(\frac{4}{5}\)
Now add the fractions and thw whole numbers
12 + \(\frac{4}{5}\) = 12 \(\frac{4}{5}\)

Question 2.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 161

Answer: 20 \(\frac{3}{8}\)

Explanation:
Add the fractional parts and then the whole numbers.
10 + 9 = 19
Add the fractions
\(\frac{5}{8}\) + \(\frac{6}{8}\) = \(\frac{11}{8}\)
Convert it into the mixed fraction.
\(\frac{11}{8}\) = 1 \(\frac{3}{8}\)
Now add the fractions and the whole numbers
19 + 1 \(\frac{3}{8}\) = 20 \(\frac{3}{8}\)

Question 3.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 162

Answer: 8 \(\frac{1}{3}\)

Explanation:
Add the fractional parts and then the whole numbers.
2 + 6 = 8
Now add the fractions and the whole numbers
\(\frac{1}{3}\) + 8 = 8 \(\frac{1}{3}\)

Question 4.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 163

Answer:

Explanation:
Add the fractional parts and then the whole numbers.
3 + 4 = 7
\(\frac{10}{12}\) + \(\frac{10}{12}\) = \(\frac{20}{12}\)
Convert it into the mixed fraction.
\(\frac{20}{12}\) = 1 \(\frac{8}{12}\)
Now add the fractions and the whole numbers
7 + 1 \(\frac{8}{12}\) = 8 \(\frac{8}{12}\)

Question 5.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 164

Answer: 9 \(\frac{2}{6}\)

Explanation:
Add the fractional parts and then the whole numbers.
Convert it into the mixed fraction.
\(\frac{11}{6}\) = 1 \(\frac{5}{6}\)
7 + 1 = 8
\(\frac{3}{6}\) + \(\frac{5}{6}\) = \(\frac{8}{6}\)
Now add the fractions and the whole numbers
8 + \(\frac{8}{6}\)
\(\frac{8}{6}\) = 1 \(\frac{2}{6}\)
8 + 1 \(\frac{2}{6}\) = 9 \(\frac{2}{6}\)

Question 6.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 165

Answer: 15 \(\frac{1}{4}\)

Explanation:
Add the fractional parts and then the whole numbers.
Rewriting our equation with parts separated
=8+70/100+6+55/100
Solving the whole number parts
8+6=14
Solving the fraction parts
70/100+55/100=125/100
Reducing the fraction part, 125/100,
125/100=5/4
Simplifying the fraction part, 5/4,
5/4=1 1/4
Combining the whole and fraction parts
14+1+1/4= 15 \(\frac{1}{4}\)

Add

Question 7.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 166

Answer: 11

Explanation:
Add the fractional parts and then the whole numbers.
5 + 3 + 2 = 10
Now add the fractional part,
3/4 + 1/4 = 1
10 + 1 = 11

Question 8.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 167

Answer: 12 \(\frac{1}{2}\)

Explanation:
Add the fractional parts and then the whole numbers.
Add the whole numbers
1 + 1 + 9 = 11
\(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1 \(\frac{1}{2}\)
Combining the whole and fraction parts
11 + 1 \(\frac{1}{2}\) = 12 \(\frac{1}{2}\)

Question 9.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 168

Answer: 9 \(\frac{2}{10}\)

Explanation:
Add the fractional parts and then the whole numbers.
3 + 5 = 8
\(\frac{4}{10}\) + \(\frac{2}{10}\) + \(\frac{6}{10}\) = \(\frac{12}{10}\)
Convert it into the mixed fraction.
\(\frac{12}{10}\) = 1 \(\frac{2}{10}\)
Combining the whole and fraction parts
8 + 1 \(\frac{2}{10}\) = 9 \(\frac{2}{10}\)

Question 10.
Structure
Find 7\(\frac{4}{5}\) + 8\(\frac{2}{5}\) two different ways. Which way do you prefer? Why?

Answer:

Explanation:
Add the fractional parts and then the whole numbers.
7\(\frac{4}{5}\) + 8\(\frac{2}{5}\)
Add the fractional parts and then the whole numbers.
7 + 8 = 15
\(\frac{4}{5}\) + \(\frac{2}{5}\) = \(\frac{6}{5}\)
Convert it into the mixed fraction.
\(\frac{6}{5}\) = 1 \(\frac{1}{5}\)
15 + 1 \(\frac{1}{5}\) = 16 \(\frac{1}{5}\)

Question 11.
Modeling Real Life
A homeowner has two strings of lights. One is 8\(\frac{1}{3}\) yards long. The other is 16\(\frac{2}{3}\) yards long. He connects the strings of lights. How long will the string of lights be in all?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 169

Answer:
Given,
A homeowner has two strings of lights. One is 8 \(\frac{1}{3}\) yards long. The other is 16 \(\frac{2}{3}\) yards long. He connects the strings of lights.
8 \(\frac{1}{3}\) + 16 \(\frac{2}{3}\) = 24 \(\frac{3}{3}\)
\(\frac{3}{3}\) = 1
24 + 1 = 25

Question 12.
DIG DEEPER!
You can play the song “Mary Had a Little Lamb” by striking three glasses filled with water to make the tone. The first glass needs 1\(\frac{3}{4}\) cups, the second glass needs 1\(\frac{1}{2}\) cups, and the third glass needs 1\(\frac{1}{4}\) cups of water. How much water do you need in all?

Answer:
Given that,
You can play the song “Mary Had a Little Lamb” by striking three glasses filled with water to make the tone. The first glass needs 1\(\frac{3}{4}\)cups, the second glass needs 1\(\frac{1}{2}\) cups, and the third glass needs 1\(\frac{1}{4}\) cups of water.
1\(\frac{1}{2}\) + 1\(\frac{1}{4}\) + 1\(\frac{3}{4}\)
1 + 1 + 1 = 3
\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{3}{4}\) = 1 \(\frac{1}{2}\)
3 + 1 \(\frac{1}{2}\) = 4 \(\frac{1}{2}\)
Therefore you need 4 \(\frac{1}{2}\) cups of water.

Review & Refresh

Write the first six numbers in the pattern. Then describe another feature of the pattern.

Question 13.
Rule: Add 11.
First number: 22

Answer:
The first number is 22 you need to add 11 to it.
22 + 11 = 33
33 + 11 = 44
44 + 11 = 55
55 + 11 = 66
66 + 11 = 77
77 + 11 = 88

Question 14.
Rule: Multiply by 4.
First number: 7

Answer:
The first number is 7. You need to multiply by 4.
7 × 4 = 28
28 × 4 = 112
112 × 4 = 448
448 × 4 = 1792
1792 × 4 = 7168
7168 × 4 = 28672

Lesson 8.8 Subtract Mixed Numbers

Explore and Grow

Use a model to find 2\(\frac{3}{8}\) – 1\(\frac{1}{8}\).

Answer:
2\(\frac{3}{8}\) – 1\(\frac{1}{8}\).
First subtract the whole numbers
2 – 1 = 1
\(\frac{3}{8}\) – \(\frac{1}{8}\) = \(\frac{2}{8}\)
Combine the whole numbers and fractions.
1 \(\frac{2}{8}\) = 1 \(\frac{1}{4}\)

Construct Arguments
How can you use the whole number parts and the fractional parts to subtract mixed numbers with like denominators? Explain why your method makes sense.

Answer:
First, you have to subtract the whole number parts and then subtract the fraction parts with like denominators.
You can subtract the mixed fractions by using the number line or model.

Think and Grow: subtract Mixed Numbers

To subtract mixed numbers, subtract the fractional parts and subtract the whole number parts. Another way to subtract mixed numbers is to rewrite each number as a fraction, then subtract.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 170

Answer:
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-170

Example
Find 5\(\frac{3}{6}\) – 4\(\frac{5}{6}\).
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 171

Answer:
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-171

Show and Grow

Subtract

Question 1.
5\(\frac{4}{5}\) – 1\(\frac{2}{5}\) = ____

Answer:
There are enough fifths.
Subtract the whole numbers
5 – 1 = 4
\(\frac{4}{5}\) – \(\frac{2}{5}\) =\(\frac{2}{5}\)
4 + \(\frac{2}{5}\) = 4 \(\frac{2}{5}\)
Thus, 5\(\frac{4}{5}\) – 1\(\frac{2}{5}\) = 4 \(\frac{2}{5}\)

Question 2.
7\(\frac{1}{3}\) – 2\(\frac{2}{3}\) = _____

Answer:
There are not enough thirds.
7\(\frac{1}{3}\) = 6 \(\frac{3}{3}\) + \(\frac{1}{3}\) = 6\(\frac{4}{3}\)
Subtract the whole numbers
6 – 2 = 4
\(\frac{4}{3}\) – \(\frac{2}{3}\) = \(\frac{2}{3}\)
4 + \(\frac{2}{3}\) = 4 \(\frac{2}{3}\)

Question 3.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 172

Answer:
There are enough twelfths.
Subtract the whole numbers
15 – 4 = 11
\(\frac{10}{12}\) – \(\frac{8}{12}\) = \(\frac{2}{12}\)
11 + \(\frac{2}{12}\) = 11 \(\frac{2}{12}\)

Question 4.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 173

Answer:
There are enough eighths.
\(\frac{6}{8}\) – \(\frac{6}{8}\) = 0
Subtract the whole numbers
6 – 3 = 3
3 + 0 = 3

Question 5.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 174

Answer:
There are not enough tenths.
5 \(\frac{7}{10}\) can be written as 4 \(\frac{17}{10}\)
4 \(\frac{17}{10}\) – 1 \(\frac{9}{10}\)
Subtract the whole numbers
4 – 1 = 3
Subtract the fractional parts
\(\frac{17}{10}\) – \(\frac{9}{10}\) = \(\frac{8}{10}\)
3 + \(\frac{8}{10}\) = 3 \(\frac{8}{10}\)

Question 6.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 175

Answer:
There are not enough hundreds.
11 \(\frac{50}{100}\) can be written as 10 \(\frac{150}{100}\)
Subtract the whole numbers
10 – 7 = 3
Subtract the fractional parts
\(\frac{150}{100}\) – \(\frac{85}{100}\) = \(\frac{65}{100}\)
3 + \(\frac{65}{100}\) = 3 \(\frac{65}{100}\)

Question 7.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 176

Answer:
There are not enough sixths.
8 can be written as 7 \(\frac{6}{6}\)
Subtract the whole numbers
7 – 1 = 6
Subtract the fractional parts
\(\frac{6}{6}\) – \(\frac{3}{6}\) = \(\frac{3}{6}\)
6 + \(\frac{3}{6}\) = 6 \(\frac{3}{6}\)

Question 8.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 177

Answer:
There are not enough fourths.
10 can be written as 9 \(\frac{4}{4}\)
Subtract the whole numbers
9 – 9 = 0
Subtract the fractional parts
\(\frac{4}{4}\) – \(\frac{3}{4}\) = \(\frac{1}{4}\)
0 + \(\frac{1}{4}\)  = \(\frac{1}{4}\)

Question 9.
YOU BE THE TEACHER
Your friend says the difference of 9 and 2\(\frac{3}{5}\) is 7\(\frac{3}{5}\). Is your friend correct? Explain.

Answer:
9 – 2\(\frac{3}{5}\) = 7 \(\frac{3}{5}\)
Thus by this we can say that your friend is correct.

Question 10.
Writing
Explain how adding and subtracting mixed numbers are similar and different.

Answer:
Any mixed number can also be written as an improper fraction, in which the numerator is larger than the denominator.
Subtracting mixed numbers is very similar to adding them.
Write both fractions as equivalent fractions with a denominator. Then subtract the fractions.

Question 11.
DIG DEEPER!
Write two mixed numbers with like denominators that have a sum of 5\(\frac{2}{3}\) and a difference of 1.

Answer:
5\(\frac{2}{3}\) = 3 \(\frac{1}{3}\) + 2\(\frac{1}{3}\)
Now if you subtract the same fraction you need to get the difference as 1.
3 \(\frac{1}{3}\) – 2\(\frac{1}{3}\)
3 – 2 = 1
\(\frac{1}{3}\) – \(\frac{1}{3}\) = 0
So, 3 \(\frac{1}{3}\) – 2\(\frac{1}{3}\) = 1

Think and Grow: Modeling Real Life

Example
A replica of the Eiffel Tower is 6 inches tall. It is 2\(\frac{2}{5}\) inches taller than a replica of the Space Needle. How tall is the replica of the Space Needle?
Find the difference between the height of the Eiffel Tower replica, 6 inches, and 2\(\frac{2}{5}\) inches.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 178
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 179

Answer:
Given,
A replica of the Eiffel Tower is 6 inches tall. It is 2\(\frac{2}{5}\) inches taller than a replica of the Space Needle.
Big-Ideas-Math-Answers-4th-Grade-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-179

Show and Grow

Question 12.
A cook has a 5-pound bag of potatoes. He uses 2\(\frac{1}{3}\) pounds of potatoes to make a casserole. How many pounds of potatoes are left?

Answer:
Given,
A cook has a 5-pound bag of potatoes. He uses 2\(\frac{1}{3}\) pounds of potatoes to make a casserole.
5 – 2\(\frac{1}{3}\)
4 \(\frac{3}{3}\) – 2\(\frac{1}{3}\)
Subtract the whole numbers
4 – 2 = 2
Subtract the fractional parts
\(\frac{3}{3}\) – \(\frac{1}{3}\) = \(\frac{2}{3}\)
2 \(\frac{2}{3}\)
Thus 2 \(\frac{2}{3}\) pounds of potatoes are left.

Question 13.
A half-marathon is 13\(\frac{1}{10}\) miles long. A competitor runs 9\(\frac{6}{10}\) miles. How many miles does the competitor have left to run?

Answer:
Given,
A half-marathon is 13\(\frac{1}{10}\) miles long. A competitor runs 9\(\frac{6}{10}\) miles.
13\(\frac{1}{10}\) – 9\(\frac{6}{10}\)
12 \(\frac{11}{10}\) – 9\(\frac{6}{10}\)
Subtract the whole numbers
12 – 9 = 3
\(\frac{11}{10}\) – \(\frac{6}{10}\) = \(\frac{5}{10}\)
3 + \(\frac{5}{10}\) = 3 \(\frac{1}{2}\)
The competitor has left 3 \(\frac{1}{2}\) miles to run.

Question 14.
DIG DEEPER!
You want to mail a package that weighs 18\(\frac{2}{4}\) ounces. The weight limit is 13 ounces, so you remove 4\(\frac{3}{4}\) ounces of items from the package. Does the lighter package meet the weight requirement? If not, how much more weight do you need to remove?

Answer:
Given that,
You want to mail a package that weighs 18\(\frac{2}{4}\) ounces.
The weight limit is 13 ounces, so you remove 4\(\frac{3}{4}\) ounces of items from the package.
18\(\frac{2}{4}\) – 4\(\frac{3}{4}\) = 13 \(\frac{3}{4}\)
13 \(\frac{3}{4}\) – 13 = \(\frac{3}{4}\)
Thus you need to remove \(\frac{3}{4}\) ounces more.

Subtract Mixed Numbers Homework & Practice 8.8

Subtract

Question 1.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 180

Answer: 5 \(\frac{1}{2}\)

Explanation:
Rewriting our equation with parts separated
=10+3/4−5−1/4
Solving the whole number parts
10−5=5
Solving the fraction parts
3/4−1/4=2/4
Reducing the fraction part, 2/4,
2/4=1/2
Combining the whole and fraction parts
5+1/2=5 1/2

Question 2.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 181

Answer: 6

Explanation:
Rewriting our equation with parts separated
9 + 1/3 – 3 – 1/3
9 – 3 = 6
So, 9 \(\frac{1}{3}\) – 3 \(\frac{1}{3}\) = 6

Question 3.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 182

Answer: 4 \(\frac{2}{3}\)

Explanation:
Rewriting our equation with parts separated
=6+7/12−1−11/12
Solving the whole number parts
6−1=5
Solving the fraction parts
7/12−11/12=−4/12
Reducing the fraction part, 4/12,
−4/12=−1/3
Combining the whole and fraction parts
5−1/3=4 2/3

Question 4.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 183

Answer: 6 \(\frac{43}{50}\)

Explanation:
Rewriting our equation with parts separated
=15+6/100−8−20/100
Solving the whole number parts
15−8=7
Solving the fraction parts
6/100−20/100=−14/100
Reducing the fraction part, 14/100,
−14/100=−7/50
Combining the whole and fraction parts
7−7/50=6 43/50

Question 5.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 184

Answer: 1 \(\frac{2}{3}\)

Explanation:
Rewriting our equation with parts separated
=4+3/6−2−5/6
Solving the whole number parts
4−2=2
Solving the fraction parts
3/6−5/6=−2/6
Reducing the fraction part, 2/6,
−2/6=−1/3
Combining the whole and fraction parts
2−1/3=1 2/3

Question 6.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 185

Answer: 1 \(\frac{3}{5}\)

Explanation:
20 – 19 = 1
\(\frac{4}{5}\) – \(\frac{1}{5}\) = \(\frac{3}{5}\)
1 + \(\frac{3}{5}\) = 1 \(\frac{3}{5}\)

Subtract.

Question 7.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 186

Answer: 2 \(\frac{3}{5}\)

Explanation:
Rewriting our equation with parts separated
=5+6/10−3
Solving the whole number parts
5−3=2
Combining the whole and fraction parts
2+6/10=2 6/10

Question 8.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 187

Answer: 10 \(\frac{1}{2}\)

Explanation:
Rewriting our equation with parts separated
=13−2−1/2
Solving the whole number parts
13−2=11
Combining the whole and fraction parts
11−1/2=10 1/2

Question 9.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 188

Answer: 3 \(\frac{1}{4}\)

Explanation:
Rewriting our equation with parts separated
=18−14−6/8
Solving the whole number parts
18−14=4
Combining the whole and fraction parts
4−6/8=3 2/8

Question 10.
Reasoning
Explain why you rename 4\(\frac{1}{3}\) when finding 4\(\frac{1}{3}\) – \(\frac{2}{3}\) .

Answer:
4\(\frac{1}{3}\) – \(\frac{2}{3}\)
4 can be written as 3 \(\frac{3}{3}\)
3 \(\frac{3}{3}\) – \(\frac{2}{3}\)
3 + \(\frac{3}{3}\) – \(\frac{2}{3}\)
3 + \(\frac{1}{3}\) = 3 \(\frac{1}{3}\)
So, 4\(\frac{1}{3}\) – \(\frac{2}{3}\) = 3 \(\frac{1}{3}\)

Question 11.
DIG DEEPER!
Find the unknown number.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 189

Answer:
Let the unknown number be x.
10 \(\frac{3}{12}\) – x = \(\frac{4}{12}\)
10 \(\frac{3}{12}\) –  \(\frac{4}{12}\) = x
9 \(\frac{15}{12}\) –  \(\frac{4}{12}\) = x
x = 9 \(\frac{11}{12}\)
Thus the unknown number is 9 \(\frac{11}{12}\).

Question 12.
Modeling Real Life
A rare flower found in Indonesian rain forests can grow wider than a car tire. How much wider is the flower than a car tire that is 1\(\frac{11}{12}\) feet wide?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 189.1

Answer:
Given,
A rare flower found in Indonesian rain forests can grow wider than a car tire.
3 – 1\(\frac{11}{12}\)
2 \(\frac{12}{12}\) – 1\(\frac{11}{12}\)
= 1 \(\frac{1}{12}\)

Question 13.
Modeling Real Life
Your tablet battery is fully charged. You use \(\frac{32}{100}\) of the charge listening to music, and \(\frac{13}{100}\) of the charge playing games. What fraction of the charge remains on your tablet battery?

Answer:
Given,
Your tablet battery is fully charged. You use \(\frac{32}{100}\) of the charge listening to music, and \(\frac{13}{100}\) of the charge playing games.
\(\frac{32}{100}\) – \(\frac{13}{100}\) = \(\frac{19}{100}\)
Thus \(\frac{19}{100}\) fraction of the charge remains on your tablet battery.

Review & Refresh

Divide. Then check your answer.

Question 14.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 190

Answer:
Divide 84 by 5
84/5 = 16.8

Question 15.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 191

Answer:
Divide 51 by 4.
51/4 = 12.75

Question 16.
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 192

Answer:
Divide 89 by 8.
89/8 = 11.125

Lesson 8.9 Problem Solving: Fractions

Explore and Grow

Make a plan to solve the problem.

The table shows the tusk lengths of two elephants. Which elephant’s tusks have a greater total length? How much greater?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 193
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 194

Answer:
Male Elephant = 4 1/12 + 4 3/12 = 8 4/12
Female Elephant = 4 + 3 7/12 = 7 7/12
The Right Tusk of a Male Elephant is greater than Female Elephant.
The left tusk of a Male Elephant is greater than Female Elephant.
Thus the total length of the Male Elephant is greater than Female Elephant.

Make Sense of Problems
A \(\frac{7}{12}\)-foot long piece of one of the male elephant’s tusks breaks off. Does this change your plan to solve the problem? Will this change the answer? Explain.

Answer:
A \(\frac{7}{12}\)-foot long piece of one of the male elephant’s tusks breaks off.
8 4/12 – 7 7/12 = 3/4
No, if \(\frac{7}{12}\)-foot long piece of one of the male elephant’s tusks breaks off it will not change the answer. Still, the Male Elephant is greater than Female Elephant.

Think and Grow: Problem Solving: Fractions

Example
A family spends 2\(\frac{2}{4}\) hours traveling to a theme park, 7\(\frac{1}{4}\) hours at the theme park, and 2\(\frac{3}{4}\) hours traveling home. How much more time does the family spend at the theme park than traveling?
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 195

Understand the Problem

What do you know?

  • The family spends 2\(\frac{2}{4}\) hours traveling to the theme park, 7\(\frac{1}{4}\) hours at the theme park, 2\(\frac{3}{4}\) hours traveling home.
    What do you need to find?
  • You need to find how much more time the family spends at the theme park than the traveling.

Make a plan

How will you solve it?

  • Add 2\(\frac{2}{4}\) and 2\(\frac{3}{4}\) to find how much time the family spends traveling.
  • Then subtract the sum from 7\(\frac{1}{4}\) to find how much more time they spend at the theme park.

Solve
So, the family spends ___ more hours at the theme park than traveling.

Show and Grow

Question 1.
Explain how you can check your answer in each step of the example above.

Answer:
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-196
So, the family spends 2 more hours at the theme park than traveling.

Apply any and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.

Answer:

  • The family spends 2\(\frac{2}{4}\) hours traveling to the theme park, 7\(\frac{1}{4}\) hours at the theme park, 2\(\frac{3}{4}\) hours traveling home.
    What do you need to find?
  • You need to find how much more time the family spends at the theme park than the traveling.

Question 2.
You are making a sand art bottle. You fill \(\frac{1}{6}\) of the bottle with pink sand, \(\frac{3}{6}\) with red sand, and \(\frac{2}{6}\) with white sand. How much of the bottle is filled?

Answer:
Given that,
You are making a sand art bottle. You fill \(\frac{1}{6}\) of the bottle with pink sand, \(\frac{3}{6}\) with red sand, and \(\frac{2}{6}\) with white sand.
\(\frac{1}{6}\) + \(\frac{3}{6}\) + \(\frac{2}{6}\) = \(\frac{1}{6}\)
Thus \(\frac{1}{6}\) of the bottle is filled.

Question 3.
Your friend has \(\frac{1}{8}\) of a photo album filled with beach photographs and \(\frac{4}{8}\) of the album filled with photos of friends. What fraction of the photo album is left?

Answer:
Given that,
Your friend has \(\frac{1}{8}\) of a photo album filled with beach photographs and \(\frac{4}{8}\) of the album filled with photos of friends.
\(\frac{1}{8}\) + \(\frac{4}{8}\) = \(\frac{5}{8}\)
\(\frac{8}{8}\) – \(\frac{5}{8}\) = \(\frac{3}{8}\)
Thus \(\frac{3}{8}\) fraction of the photo album is left.

Understand the problem. Then make a plan. How will you solve? Explain.

Question 4.
In Race A, an Olympic swimmer swims 100 meters in 62\(\frac{25}{100}\) seconds. In Race B, she cuts 2\(\frac{38}{100}\) seconds off her Race A time. How many seconds does she need to cut off her Race B time to swim 100 meters in 58\(\frac{45}{100}\) seconds?

Answer:
Given,
In Race A, an Olympic swimmer swims 100 meters in 62\(\frac{25}{100}\) seconds. In Race B, she cuts 2\(\frac{38}{100}\) seconds off her Race A time.
62\(\frac{25}{100}\) – 2\(\frac{38}{100}\) = 59 \(\frac{87}{100}\)
59 \(\frac{87}{100}\) – 58\(\frac{45}{100}\) = 1 \(\frac{42}{100}\)
She need 1 \(\frac{42}{100}\) to cut off her Race B time to swim 100 meters in 58\(\frac{45}{100}\) seconds.

Question 5.
A semi-truck has 2 fuel tanks that each hold the same amount of fuel. A truck driver fills up both tanks and uses \(\frac{3}{4}\) tank of gasoline driving to his first stop. He uses \(\frac{2}{4}\) tank of gasoline driving to his second stop. How much gasoline does he have left?

Answer:
Given that,
A semi-truck has 2 fuel tanks that each hold the same amount of fuel. A truck driver fills up both tanks and uses \(\frac{3}{4}\) tank of gasoline driving to his first stop. He uses \(\frac{2}{4}\) tank of gasoline driving to his second stop.
\(\frac{3}{4}\) + \(\frac{2}{4}\) = \(\frac{5}{4}\)
2 – \(\frac{5}{4}\) = \(\frac{3}{4}\)
Thus \(\frac{3}{4}\) gasoline has left.

Question 6.
A bootlace worm holds the record as the longest animal at 180 feet long. How much longer is it than 2 blue whales combined?
Big Ideas Math Answers 4th Grade Chapter 8 Add and Subtract Fractions 197

Answer:
Given,
A bootlace worm holds the record as the longest animal at 180 feet long.
1 blue whale = 85 \(\frac{8}{12}\)
2 blue whales = 85 \(\frac{8}{12}\) + 85 \(\frac{8}{12}\) = 171 \(\frac{1}{3}\)
180 – 171 \(\frac{1}{3}\)
179 \(\frac{3}{3}\) – 171 \(\frac{1}{3}\) = 8 \(\frac{2}{3}\)

Think and Grow: Modeling Real Life

Example
You walk \(\frac{1}{10}\) kilometer on Monday, \(\frac{3}{10}\) kilometer on Tuesday, and \(\frac{5}{10}\) kilometer on Wednesday. You continue the pattern on Thursday and Friday. How many kilometers do you walk in all?
Think: What do you know? What do you need to find? How will you solve?
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 198
Step 1: Identify the pattern.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 199
Step 2: Use the pattern to find the distances you walk on Thursday and Friday.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 200
Step 3: Add all of the distances.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 201
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 202.

Answer:
Step 1: Identify the pattern.
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-199
Step 2: Use the pattern to find the distances you walk on Thursday and Friday.
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-200
Step 3: Add all of the distances.
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-201
So, you walk 2 \(\frac{5}{10}\) kilometers in all.

Show and Grow

Question 7.
You save \(\frac{1}{4}\) dollar the first week, \(\frac{2}{4}\) dollar the next week, and dollar \(\frac{3}{4}\) dollar the following week. You continue the pattern for 3 more weeks. How much money do you save after 6 weeks?
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 203

Answer:
You save \(\frac{1}{4}\) dollar the first week, \(\frac{2}{4}\) dollar the next week, and dollar \(\frac{3}{4}\) dollar the following week. You continue the pattern for 3 more weeks.
\(\frac{1}{4}\), \(\frac{2}{4}\), \(\frac{3}{4}\), \(\frac{4}{4}\), \(\frac{5}{4}\), \(\frac{6}{4}\)
You save \(\frac{6}{4}\) dollar after 6 weeks.

Problem Solving: Fractions Homework & Practice 8.9

Question 1.
An older washing machine uses 170\(\frac{3}{10}\) liters of water per load. A new, high-efficiency, washing machine uses 75\(\frac{7}{10}\) fewer liters than the older washing machine. How many liters of water will the high-efficiency washing machine use for 2 loads of laundry?

Answer:
Given,
An older washing machine uses 170\(\frac{3}{10}\) liters of water per load. A new, high-efficiency, washing machine uses 75\(\frac{7}{10}\) fewer liters than the older washing machine.
75\(\frac{7}{10}\) + 75\(\frac{7}{10}\) = 151\(\frac{2}{5}\)
170\(\frac{3}{10}\) – 151\(\frac{2}{5}\) = 18 \(\frac{9}{10}\)

Question 2.
A student jumps 40 \(\frac{5}{12}\) inches for the high jump. On his second try, he jumps 1\(\frac{8}{12}\) inches higher. He can tie the school record if he raises the bar another 3\(\frac{10}{12}\) inches and successfully jumps over it. What is the school record for the high jump?

Answer:
Given,
A student jumps 40 \(\frac{5}{12}\) inches for the high jump.
On his second try, he jumps 1\(\frac{8}{12}\) inches higher.
He can tie the school record if he raises the bar another 3\(\frac{10}{12}\) inches and successfully jumps over it.
40 \(\frac{5}{12}\) + 1 \(\frac{8}{12}\) = 42 \(\frac{1}{12}\)
40 \(\frac{5}{12}\) + 3\(\frac{10}{12}\) = 44 \(\frac{3}{12}\)
44 \(\frac{3}{12}\) is the school record for the high jump.

Question 3.
You are shipping three care packages. The first package weighs 10\(\frac{1}{10}\) pounds. The second weighs 5\(\frac{7}{10}\) pounds, and the third weighs 25\(\frac{8}{10}\) pounds. What is the total weight of the packages?

Answer:
Given,
You are shipping three care packages. The first package weighs 10\(\frac{1}{10}\) pounds.
The second weighs 5\(\frac{7}{10}\) pounds, and the third weighs 25\(\frac{8}{10}\) pounds.
10\(\frac{1}{10}\) + 5\(\frac{7}{10}\) + 25\(\frac{8}{10}\) = 41 \(\frac{6}{10}\)
The total weight of the packages is 41 \(\frac{6}{10}\) pounds.

Question 4.
A person’s arm span is approximately equal to the person’s height. How tall is this fourth grader according to his arm span?
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 204

Answer:
Given,
A person’s arm span is approximately equal to the person’s height.
By using the pattern we can find the arm span of the fourth-grader i.e., 1 \(\frac{7}{12}\)

Question 5.
Writing
Write and solve a two-step word problem with mixed numbers that can be solved using addition or subtraction.

Answer:
I have 5 \(\frac{8}{12}\) episodes of my favorite series download onto my computer. I Downloaded some yesterday and \(\frac{7}{12}\) of the episodes this morning. The download speed was really slow. What fraction of the episodes did I download yesterday?
5 \(\frac{8}{12}\) – \(\frac{7}{12}\) = 5 \(\frac{1}{12}\) = \(\frac{61}{12}\)

Question 6.
Modeling Real Life
Your friend walks \(\frac{2}{10}\) mile to school each day. She walks the same distance home. How many miles does she walk to and from school in one 5-day school week?

Answer:
Given,
Your friend walks \(\frac{2}{10}\) mile to school each day. She walks the same distance home.
\(\frac{2}{10}\) + \(\frac{2}{10}\) = \(\frac{4}{10}\)
5 × \(\frac{4}{10}\) = \(\frac{20}{10}\) = 2
Thus she walk to and from school in one 5-day school week is 2 miles.

Question 7.
DIG DEEPER!
A store sells cashews in \(\frac{2}{3}\)-pound bags. You buy some bags and repackage the cashews into 1-pound bags. What is the least number of bags you should buy so that you do not have any cashews left over?

Answer:
Given,
A store sells cashews in \(\frac{2}{3}\)-pound bags. You buy some bags and repackage the cashews into 1-pound bags.
1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Thus \(\frac{1}{3}\) pound of cashews left over.

Review & Refresh

Compare.

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 205

Answer: >

Explanation:
\(\frac{8}{12}\) = \(\frac{4}{6}\)
\(\frac{4}{6}\) > \(\frac{1}{6}\)

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 206

Answer: <

Explanation:
First, make the denominators common.
\(\frac{9}{10}\) = \(\frac{18}{20}\)
\(\frac{14}{8}\) = \(\frac{35}{20}\)
\(\frac{18}{20}\) < \(\frac{35}{20}\)

Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 207

Answer: >

Explanation:
First, make the denominators common.
\(\frac{3}{4}\)
\(\frac{1}{2}\) × 2/2 = \(\frac{2}{4}\)
\(\frac{3}{4}\) > \(\frac{2}{4}\)

Add and Subtract Fractions Performance Task 8

The notes on sheet music tell you what note to play and how long to hold each note. The table shows how long you hold some notes compared to the length of one whole note.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 208
1. a. Complete the table by writing equivalent fractions.

Answer:
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-208
A whole note is nothing but 1 so the fraction is 8/8.
1/2 note is nothing but 4/8.
1/4 note is nothing but 2/8.

b. Each group of notes represents one measure. What is the sum of the values of the notes in each measure?
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 209

Answer:
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-209

c. Draw the missing note to complete each measure.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 210

Answer:
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-210

d. Draw one measure of notes where the sum of the values is 1. Show your work.
___________

Answer:
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-209 (1)

e. Write the fraction represented by the sum of the notes. Then write the fraction as a sum of fractions in two different ways.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 211

Answer:
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 211 = \(\frac{1}{8}\) + \(\frac{4}{8}\) + \(\frac{2}{8}\) = \(\frac{7}{8}\)

Add and Subtract Fractions Activity

Three In a Row: Fraction Add or Subtract

Directions:

  1. Players take turns.
  2. On your turn, spin both spinners. Choose whether to add or subtract.
  3. Add or subtract the mixed number and fraction. Cover the sum or difference.
  4. If the sum or difference is already covered, you lose your turn.
  5. The first player to get three in a row wins!

Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 212

Answer:
1 \(\frac{1}{8}\) + \(\frac{3}{8}\) = 1 + \(\frac{1}{8}\) + \(\frac{3}{8}\) = 1 \(\frac{4}{8}\)
3 \(\frac{7}{8}\) + \(\frac{8}{8}\) = 3 + \(\frac{7}{8}\) + 1 = 4 \(\frac{7}{8}\)
2 \(\frac{5}{8}\) + \(\frac{4}{8}\) = 2 + \(\frac{5}{8}\) + \(\frac{4}{8}\) = 3 \(\frac{1}{8}\)

Add and Subtract Fractions Chapter Practice 8

8.1 Use Models to Add Fractions

Find the sum. Explain how you used the model to add.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 214

Answer: 5/6
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-214

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 215

Answer:
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-215

Find the sum. Use a model or a number line to help.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 216

Answer:
BIM Grade 4 Chapter 8 add & subtract fractions img_25

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 217

Answer:
Big Ideas Math Answers Grade 4 Chapter 8 Add and Subtract Fractions img_216

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 218

Answer:
Denominators are the same so add the numerators.
\(\frac{45}{100}\) + \(\frac{10}{100}\) + \(\frac{9}{100}\) = \(\frac{64}{100}\)

8.2 Decompose Fractions

Write the fraction as a sum of unit fractions.

Question 6.
\(\frac{2}{12}\)

Answer:
The unit fraction for \(\frac{2}{12}\) is \(\frac{1}{12}\) + \(\frac{1}{12}\)

Question 7.
\(\frac{3}{3}\)

Answer: The unit fraction for \(\frac{3}{3}\) is \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\)

Write the fraction as a sum of fractions in two different ways.

Question 8.
\(\frac{5}{8}\)

Answer:
The unit fraction for \(\frac{5}{8}\) is \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\)

Question 9.
\(\frac{6}{100}\)

Answer:
The unit fraction for \(\frac{6}{100}\) is \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\)

Question 10.
\(\frac{90}{100}\)

Answer:
\(\frac{90}{100}\) = 9/10
The unit fraction for \(\frac{9}{10}\) is \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\)

Question 11.
\(\frac{4}{5}\)

Answer:
The unit fraction for \(\frac{4}{5}\) is \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\)

8.3 Add Fractions with Like Denominators

Add

Question 12.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 219

Answer:
Denominators are the same so add the numerators.
\(\frac{5}{10}\) + \(\frac{10}{10}\) = \(\frac{15}{10}\)

Question 13.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 220

Answer:
BIM Grade 4 Chapter 8 add & subtract fractions img_24
Denominators are the same so add the numerators.
\(\frac{1}{3}\) + \(\frac{1}{3}\) = \(\frac{2}{3}\)

Question 14.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 221

Answer:
Denominators are the same so add the numerators.
\(\frac{1}{8}\) + \(\frac{6}{8}\) = \(\frac{7}{8}\)

Question 15.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 222

Answer:
Denominators are the same so add the numerators.
\(\frac{7}{4}\) + \(\frac{3}{4}\) = \(\frac{11}{4}\)

Question 16.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 223

Answer:
Denominators are the same so add the numerators.
\(\frac{2}{6}\) + \(\frac{2}{6}\) = \(\frac{4}{6}\)

Question 17.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 224

Answer:
Denominators are the same so add the numerators.
\(\frac{8}{12}\) + \(\frac{4}{12}\) = \(\frac{12}{12}\) = 1

Question 18.
Logic
When you add two of me you get \(\frac{100}{100}\). What fraction am I?

Answer: \(\frac{50}{100}\)
If you add \(\frac{50}{100}\) two times you get \(\frac{100}{100}\)
\(\frac{50}{100}\) + \(\frac{50}{100}\) = \(\frac{100}{100}\)

8.4 Use Models to Subtract Fractions

Find the difference. Explain how you used the model to subtract.

Question 19.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 225

Answer:
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-225

Question 20.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 226

Answer:
Big-Ideas-Math-Answer-Key-Grade-4-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-226

Find the difference. Use a model or a number line to help.

Question 21.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 227

Answer: 3/2
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-139

Question 22.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 228

Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-8-Add-and-Subtract-Multi-Digit-Numbers-139

Question 23.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 229

Answer:
Denominators are the same so subtract the numerators.
\(\frac{30}{100}\) – \(\frac{21}{100}\) = (30 – 21)/100 = \(\frac{9}{100}\)

Question 24.
Modeling Real Life
A football team wins \(\frac{7}{10}\) of their games this season. They lose \(\frac{3}{10}\) of their games. How many more games does the team win than lose?

Answer:
Given that,
A football team wins \(\frac{7}{10}\) of their games this season. They lose \(\frac{3}{10}\) of their games.
\(\frac{7}{10}\) – \(\frac{3}{10}\) = \(\frac{4}{10}\)
Thus \(\frac{4}{10}\) more games the team win than lose.

8.5 Subtract Fractions with Like Denominators

Subtract.

Question 25.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 230

Answer: \(\frac{5}{10}\)

Explanation:
Denominators are the same so subtract the numerators.
\(\frac{9}{10}\) – \(\frac{4}{10}\) = \(\frac{5}{10}\)

Question 26.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 231

Answer: \(\frac{7}{12}\)

Explanation:
Denominators are the same so subtract the numerators.
\(\frac{14}{12}\) – \(\frac{7}{12}\) = \(\frac{7}{12}\)

Question 27.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 232

Answer: \(\frac{24}{100}\)

Explanation:
Denominators are the same so subtract the numerators.
\(\frac{80}{100}\) – \(\frac{56}{100}\) = \(\frac{24}{100}\)

Question 28.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 233

Answer: 3/8\(\frac{3}{8}\)

Explanation:
Denominators are the same so subtract the numerators.
1 can be written as \(\frac{8}{8}\)
\(\frac{8}{8}\) – \(\frac{5}{8}\) = \(\frac{3}{8}\)

Question 29
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 234

Answer: \(\frac{2}{3}\)

Explanation:
Denominators are the same so subtract the numerators.
1 can be written as \(\frac{3}{3}\)
\(\frac{3}{3}\) – \(\frac{1}{3}\) = \(\frac{2}{3}\)

Question 30.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 235

Answer: \(\frac{2}{6}\)

Explanation:
Denominators are the same so subtract the numerators.
2 can be written as \(\frac{12}{6}\)
\(\frac{12}{6}\) – \(\frac{10}{6}\) = \(\frac{2}{6}\)

8.6 Model Fractions and Mixed Numbers

Write the mixed number as a fraction.

Question 31.
1 \(\frac{6}{8}\)

Answer: \(\frac{7}{4}\)

Explanation:
Step 1
Multiply the denominator by the whole number
8 × 1 = 8
Step 2
Add the answer from Step 1 to the numerator
8 + 6 = 14
Step 3
Write an answer from Step 2 over the denominator
14/8 = \(\frac{7}{4}\)

Question 32.
4 \(\frac{1}{2}\)

Answer: \(\frac{9}{2}\)

Explanation:
Step 1
Multiply the denominator by the whole number
2 × 4 = 8
Step 2
Add the answer from Step 1 to the numerator
8 + 1 = 9
Step 3
Write an answer from Step 2 over the denominator
\(\frac{9}{2}\)

Question 33.
5 \(\frac{10}{12}\)

Answer: \(\frac{35}{6}\)

Explanation:
Step 1
Multiply the denominator by the whole number
12 × 5 = 60
Step 2
Add the answer from Step 1 to the numerator
60 + 10 = 70
Step 3
Write an answer from Step 2 over the denominator
70/12 = \(\frac{35}{6}\)

Write the fraction as a mixed number or a whole number.

Question 34.
\(\frac{17}{4}\)

Answer: 4 \(\frac{1}{4}\)

Explanation:
Converting from improper fraction to the mixed fraction.
\(\frac{17}{4}\) = 4 \(\frac{1}{4}\)

Question 35.
\(\frac{30}{6}\)

Answer: 5

Explanation:
Converting from improper fraction to the mixed fraction.
6 divides 30 five times.
So, \(\frac{30}{6}\) = 5

Question 36.
\(\frac{63}{10}\)

Answer: 6 \(\frac{3}{10}\)

Explanation:
Converting from improper fraction to the mixed fraction.
\(\frac{63}{10}\) = 63 ÷ 10
= 6 \(\frac{3}{10}\)

Compare.

Question 37.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 236

Answer: <

Explanation:
2 4/100
Step 1
Multiply the denominator by the whole number
100 × 2 = 200
Step 2
Add the answer from Step 1 to the numerator
200 + 4 = 204
Step 3
Write an answer from Step 2 over the denominator
204/100
240/100
Step 1
Multiply the denominator by the whole number
100 × 2 = 200
Step 2
Add the answer from Step 1 to the numerator
200 + 40 = 240
Step 3
Write an answer from Step 2 over the denominator
240/100
204/100 < 240/100

Question 38.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 237

Answer: >

Explanation:
Step 1
Multiply the denominator by the whole number
3 × 8 = 24
Step 2
Add the answer from Step 1 to the numerator
24 + 2 = 26
Step 3
Write an answer from Step 2 over the denominator
26/3
26/3 > 25/3

Question 39.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 238

Answer: =

Explanation:
25/5 = 5
5 = 5

Question 40.
Which One Doesn’t Belong? Which expression does not belong with the other three?
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 239

Answer: 20/8 does not belong with the other three.

8.7 Add Mixed Numbers

Add.

Question 41.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 240

Answer: 9

Explanation:
Rewriting our equation with parts separated
=5+1/2+3+1/2
Solving the whole number parts
5+3=8
Solving the fraction parts
1/2+1/2=2/2
Reducing the fraction part, 2/2,
2/2=1/1
Simplifying the fraction part, 1/1,
1/1=1
Combining the whole and fraction parts
8+1=9

Question 42.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 241

Answer: 4 1/3

Explanation:
Rewriting our equation with parts separated
=2+5/6+1+3/6
Solving the whole number parts
2+1=3
Solving the fraction parts
5/6+3/6 = 8/6
8/6 = 4/3
4/3 = 4 1/3

Question 43.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 242

Answer: 5 5/6

Explanation:
Rewriting our equation with parts separated
=4+1+10/12
Solving the whole number parts
4+1=5
Combining the whole and fraction parts
5+10/12= 5 10/12 = 5 5/6

Question 44.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 243

Answer: 19

Explanation:
Rewriting our equation with parts separated
=8+3/5+10+2/5
Solving the whole number parts
8+10=18
Solving the fraction parts
3/5+2/5=5/5
Simplifying the fraction part, 1/1,
1/1 = 1
Combining the whole and fraction parts
18+1=19

Question 45.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 244

Answer: 14 1/4

Explanation:
Rewriting our equation with parts separated
=7+2/4+1+2/4
Solving the whole number parts
7+1=8
Solving the fraction parts
2/4+2/4=4/4
Reducing the fraction part, 4/4,
4/4=1/1
Simplifying the fraction part, 1/1,
1/1=1
Combining the whole and fraction parts
8+1=9
9 + 5 1/4
Rewriting our equation with parts separated
=9+5+1/4
Solving the whole number parts
9+5=14
Combining the whole and fraction parts
14+1/4=14 1/4

Question 46.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 245

Answer: 19 5/100

Explanation:
Rewriting our equation with parts separated
=4+25/100+11+75/100
Solving the whole number parts
4+11=15
Solving the fraction parts
25/100+75/100=100/100
Reducing the fraction part, 100/100,
100/100=11
Simplifying the fraction part, 1/1,
1/1=1
Combining the whole and fraction parts
15+1=16
Rewriting our equation with parts separated
=16+3+5/100
Solving the whole number parts
16+3=19
Combining the whole and fraction parts
19+5/100=19 5/100

8.8 Subtract Mixed Numbers

Subtract

Question 47.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 246

Answer: 3

Explanation:
Rewriting our equation with parts separated
=9+2/3-6-2/3
Solving the whole number parts
9−6=3
Solving the fraction parts
2/3−2/3=0/3
Simplifying the fraction part, 0/3,
0/3=0
Combining the whole and fraction parts
3+0=3

Question 48.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 247

Answer: 5 2/5

Explanation:
Rewriting our equation with parts separated
=13+9/10−8−5/10
Solving the whole number parts
13−8=5
Solving the fraction parts
9/10−5/10=4/10
Reducing the fraction part, 4/10,
4/10=2/5
Combining the whole and fraction parts
5+2/5=5 2/5

Question 49.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 248

Answer: 1/2

Explanation:
Rewriting our equation with parts separated
=3+2/8−2−6/8
Solving the whole number parts
3−2=1
Solving the fraction parts
2/8−6/8=−4/8
Reducing the fraction part, 4/8,
−4/8=−1/2
Combining the whole and fraction parts
1−1/2=1/2

Question 50.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 249

Answer: 5 1/2

Explanation:
6 + 1/2 – 1 = 5 1/2

Question 51.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 250

Answer: 2 3/4

Explanation:
Rewriting our equation with parts separated
=7−4−1/4
Solving the whole number parts
7−4=3
Combining the whole and fraction parts
3−1/4=2 3/4

Question 52.
Big Ideas Math Answer Key Grade 4 Chapter 8 Add and Subtract Fractions 251

Answer: 1/6

Explanation:
Rewriting our equation with parts separated
=20−19−5/6
Solving the whole number parts
20−19=1
Combining the whole and fraction parts
1−5/6=1/6

8.9 Problem Solving: Fractions

Question 53.
You give \(\frac{3}{12}\) of your bag of grapes to one friend and \(\frac{5}{12}\) of your bag to another friend. What fraction of the bag of grapes do you have left?

Answer:
Given that,
You give \(\frac{3}{12}\) of your bag of grapes to one friend and \(\frac{5}{12}\) of your bag to another friend
\(\frac{3}{12}\) + \(\frac{5}{12}\) = \(\frac{8}{12}\)
\(\frac{12}{12}\) – \(\frac{8}{12}\) = \(\frac{4}{12}\)
Thus \(\frac{4}{12}\) fraction of the bag of grapes are left.

Final Words:

Hope you are all satisfied with the solutions provided in the BIM Grade 4 Chapter 8 Add and Subtract Fractions pdf. If you have any doubts regarding the problems you can ask your doubts in the below comment box. We are ready to clarify your doubts at any time. Stay with us to get the solutions of all 4th-grade chapters.

Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring

Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring

Looking for fun-learning ways to understand the concepts of algebra 1 chapter 7 Polynomial Equations and Factoring? Then, here is the best study resource that helps you during your exam preparation and also for homework help. Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring is the ultimate guide for students to grasp the topics covered under this chapter.

Also, it make you practice various questions and gain excellent knowledge on polynomial equations and factoring. BigIdeas Math Book Algebra 1 Ch 7 Polynomial Equations and Factoring includes 7.1 to 7.8 Exercises Questions, Chapter Review, Chapter Test, Cumulative Assessment, etc. for efficient preparation. So, make use of this material and score more marks in the examinations.

Big Ideas Math Textbook Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring

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Polynomial Equations and Factoring Maintaining Mathematical Proficiency

Simplify the expression.
Question 1.
3x – 7 + 2x
Answer:

Question 2.
4r + 6 – 9r – 1
Answer:

Question 3.
-5t + 3 – t – 4 + 8t
Answer:

Question 4.
3(s – 1) + 5
Answer:

Question 5.
2m – 7(3 – m)
Answer:

Question 6.
4(h + 6) – (h – 2)
Answer:

Find the greatest common factor.
Question 7.
20, 36
Answer:

Question 8.
42, 63
Answer:

Question 9.
54, 81
Answer:

Question 10.
72, 84
Answer:

Question 11.
28, 64
Answer:

Question 12.
30, 77
Answer:

Question 13.
ABSTRACT REASONING
Is it possible for two integers to have no common factors? Explain your reasoning.
Answer:

Polynomial Equations and Factoring Mathematical Practices

Mathematically proficient students consider concrete models when solving a mathematics problem.

Monitoring Progress

Write the algebraic expression modeled by the algebra tiles.
Question 1.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 1.1
Answer:

Question 2.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 1.2
Answer:

Question 3.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 1.3
Answer:

Question 4.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 1.4
Answer:

Question 5.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 1.5
Answer:

Question 6.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 1.6
Answer:

Question 7.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 1.7
Answer:

Question 8.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 1.8
Answer:

Question 9.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 1.9
Answer:

Lesson 7.1 Adding and Subtracting Polynomials

Essential Question How can you add and subtract polynomials?

EXPLORATION 1

Adding Polynomials
Work with a partner. Write the expression modeled by the algebra tiles in each step.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 1

EXPLORATION 2

Subtracting Polynomials
Work with a partner. Write the expression modeled by the algebra tiles in each step.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 2

Communicate Your Answer

Question 3.
How can you add and subtract polynomials?
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 3
Answer:

Question 4.
Use your methods in Question 3 to find each sum or difference.
a. (x2 + 2x – 1) + (2x2 – 2x + 1)
b. (4x + 3) + (x – 2)
c. (x2 + 2) – (3x2 + 2x + 5)
d. (2x – 3x) – (x2 – 2x + 4)
Answer:

Monitoring Progress

Find the degree of the monomial.
Question 1.
-3x4
Answer:

Question 2.
7c3d2
Answer:

Question 3.
\(\frac{5}{3}\)y
Answer:

Question 4.
-20.5
Answer:

Write the polynomial in standard form. Identify the degree and leading coefficient of the polynomial. Then classify the polynomial by the number of terms.
Question 5.
4 – 9z
Answer:

Question 6.
t2 – t2 – 10t
Answer:

Question 7.
2.8x + x3
Answer:

Find the sum or difference.
Question 8.
(b – 10) + (4b – 3)
Answer:

Question 9.
(x2 – x – 2) + (7x2 – x)
Answer:

Question 10.
(p2 + p + 3) – (-4p2 – p + 3)
Answer:

Question 11.
(-k + 5) – (3k2 – 6)
Answer:

Question 12.
WHAT IF?
The polynomial -16t2 – 25t + 200 represents the height of the penny after t seconds.
a. Write a polynomial that represents the distance between the penny and the paintbrush after t seconds.
b. Interpret the coefficients of the polynomial in part (a).
Answer:

Adding and Subtracting Polynomials 7.1 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
When is a polynomial in one variable in standard form?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 1

Question 2.
OPEN-ENDED
Write a trinomial in one variable of degree 5 in standard form.
Answer:

Question 3.
VOCABULARY
How can you determine whether a set of numbers is closed under an operation?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 3

Question 4.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 4
Answer:

In Exercises 5–12, find the degree of the monomial.
Question 5.
4g
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 5

Question 6.
23x4
Answer:

Question 7.
-1.75k2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 7

Question 8.
–\(\frac{4}{9}\)
Answer:

Question 9.
s8t
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 9

Question 10.
8m2n4
Answer:

Question 11.
9xy3z7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 11

Question 12.
-3q4rs6
Answer:

In Exercises 13–20, write the polynomial in standard form. Identify the degree and leading coefficient of the polynomial. Then classify the polynomial by the number of terms.
Question 13.
6c2 + 2c4 – c
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 13

Question 14.
4w11 – w12
Answer:

Question 15.
7 + 3p2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 15

Question 16.
8d – 2 – 4d3
Answer:

Question 17.
3t8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 17

Question 18.
5z + 2z3 + 3z4
Answer:

Question 19.
πr2 – \(\frac{5}{7}\)r8 + 2r5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 19

Question 20.
\(\sqrt{7}\)n4
Answer:

Question 21.
MODELING WITH MATHEMATICS
The expression \(\frac{4}{3}\) πr3 represents the volume of a sphere with radius r. Why is this expression a monomial? What is its degree?
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 21

Question 22.
MODELING WITH MATHEMATICS
The amount of money you have after investing $400 for 8 years and $600 for 6 years at the same interest rate is represented by 400x8 + 600x6, where x is the growth factor. Classify the polynomial by the number of terms. What is its degree?
Answer:

In Exercises 23–30, find the sum.
Question 23.
(5y + 4) + (-2y + 6)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 23

Question 24.
(-8x – 12) + (9x + 4)
Answer:

Question 25.
(2n2 – 5n – 6) + (-n2 – 3n + 11)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 25

Question 26.
(-3p3 + 5p2 – 2p) + (-p3 – 8p2 – 15p)
Answer:

Question 27.
(3g2 – g) + (3g2 – 8g + 4)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 27

Question 28.
(9r2 + 4r – 7) + (3r2 – 3r)
Answer:

Question 29.
(4a – a3 – 3) + (2a3 – 5a2 + 8)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 29

Question 30.
(s3 – 2s – 9) + (2s2 – 6s3 + s)
Answer:

In Exercises 31–38, find the difference.
Question 31.
(d – 9) – (3d – 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 31

Question 32.
(6x + 9) – (7x + 1)
Answer:

Question 33.
(y2 – 4y + 9) – (3y2 – 6y – 9)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 33

Question 34.
(4m2 – m + 2) – (-3m2 + 10m + 4)
Answer:

Question 35.
(k3 – 7k + 2) – (k2 – 12)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 35

Question 36.
(-r – 10) – (-4r3 + r2 + 7r)
Answer:

Question 37.
(t4 – t2 + t) – (12 – 9t2 – 7t)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 37

Question 38.
(4d – 6d3 + 3d2) – (10d3 + 7d – 2)
Answer:

ERROR ANALYSIS In Exercises 39 and 40, describe and correct the error in finding the sum or difference.
Question 39.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 39

Question 40.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 7
Answer:

Question 41.
MODELING WITH MATHEMATICS
The cost (in dollars)of making b bracelets is represented by 4 + 5b. The cost (in dollars) of making b necklaces is represented by 8b + 6. Write a polynomial that represents how much more it costs to make b necklaces than b bracelets.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 41

Question 42.
MODELING WITH MATHEMATICS
The number of individual memberships at a fitness center in m months is represented by 142 + 12m. The number of family memberships at the fitness center in m months is represented by 52 + 6m. Write a polynomial that represents the total number of memberships at the fitness center.
Answer:

In Exercises 43–46, find the sum or difference.
Question 43.
(2s2 – 5st – t2) – (s2 + 7st – t2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 43

Question 44.
(a2 – 3ab + 2b2) + (-4a2 + 5ab – b2)
Answer:

Question 45.
(c2 – 6d2) + (c2 – 2cd + 2d2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 45

Question 46.
(-x2 + 9xy) – (x2 + 6xy – 8y2)
Answer:

REASONING In Exercises 47–50, complete the statement with always, sometimes, or never. Explain your reasoning.
Question 47.
The terms of a polynomial are ________ monomials.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 47

Question 48.
The difference of two trinomials is _________ a trinomial.
Answer:

Question 49.
A binomial is ________ a polynomial of degree 2.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 49

Question 50.
The sum of two polynomials is _________ a polynomial.
Answer:

MODELING WITH MATHEMATICS The polynomial −16t2 – v0t – s0 represents the height (in feet) of an object, where v0 is the initial vertical velocity (in feet per second), s0 is the initial height of the object (in feet), and t is the time (in seconds). In Exercises 51 and 52, write a polynomial that represents the height of the object. Then  nd the height of the object after 1 second.
Question 51.
You throw a water balloon from a building.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 51

Question 52.
You bounce a tennis ball on a racket.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 10
Answer:

Question 53.
MODELING WITH MATHEMATICS
You drop a ball from a height of 98 feet. At the same time, your friend throws a ball upward. The polynomials represent the heights (in feet) of the balls after t seconds.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 11
a. Write a polynomial that represents the distance between your ball and your friend’s ball after t seconds.
b. Interpret the coefficients of the polynomial in part (a).
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 53

Question 54.
MODELING WITH MATHEMATICS
During a 7-year period, the amounts (in millions of dollars) spent each year on buying new vehicles N and used vehicles U by United States residents are modeled by the equations
N = -0.028t3 + 0.06t2+ 0.1t + 17
U = -0.38t2 + 1.5t + 42
where t = 1 represents the first year in the 7-year period.
a. Write a polynomial that represents the total amount spent each year on buying new and used vehicles in the 7-year period.
b. How much is spent on buying new and used vehicles in the fifth year?
Answer:

Question 55.
MATHEMATICAL CONNECTIONS
Write the polynomial in standard form that represents the perimeter of the quadrilateral.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 55

Question 56.
HOW DO YOU SEE IT?
The right side of the equation of each line is a polynomial.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 13
a. The absolute value of the difference of the two polynomials represents the vertical distance between points on the lines with the same x-value. Write this expression.
b. When does the expression in part (a) equal 0? How does this value relate to the graph?
Answer:

Question 57.
MAKING AN ARGUMENT
Your friend says that when adding polynomials, the order in which you add does not matter. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 57

Question 58.
THOUGHT PROVOKING
Write two polynomials whose sum is x2 and whose difference is 1.
Answer:

Question 59.
REASONING
Determine whether the set is closed under the given operation. Explain.
a. the set of negative integers; multiplication
b. the set of whole numbers; addition
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 59

Question 60.
PROBLEM SOLVING
You are building a multi-level deck.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 14
a. For each level, write a polynomial in standard form that represents the area of that level. Then write the polynomial in standard form that represents the total area of the deck.
b. What is the total area of the deck when x = 20?
c. A gallon of deck sealant covers 400 square feet. How many gallons of sealant do you need to cover the deck in part (b) once? Explain.
Answer:

Question 61.
PROBLEM SOLVING
A hotel installs a new swimming pool and a new hot tub.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.1 15
a. Write the polynomial in standard form that represents the area of the patio.
b. The patio will cost $10 per square foot. Determine the cost of the patio when x = 9.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 61

Maintaining Mathematical Proficiency

Simplify the expression.
Question 62.
2(x – 1) + 3(x + 2)
Answer:

Question 63.
8(4y – 3) + 2(y – 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.1 Question 63

Question 64.
5(2r + 1) – 3(-4r + 2)
Answer:

Lesson 7.2 Multiplying Polynomials

Essential Question How can you multiply two polynomials?

EXPLORATION 1

Multiplying Monomials Using Algebra Tiles
Work with a partner. Write each product. Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 1

EXPLORATION 2

Multiplying Binomials Using Algebra Tiles
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 2
Work with a partner. Write the product of two binomials modeled by each rectangular array of algebra tiles. In parts (c) and (d), first draw the rectangular array of algebra tiles that models each product.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 3

Communicate Your Answer

Question 3.
How can you multiply two polynomials?
Answer:

Question 4.
Give another example of multiplying two binomials using algebra tiles that is similar to those in Exploration 2.
Answer:

Monitoring Progress

Use the Distributive Property to find the product.
Question 1.
(y + 4)(y + 1)
Answer:

Question 2.
(z – 2)(z + 6)
Answer:

Use a table to find the product.
Question 3.
(p + 3)(p – 8)
Answer:

Question 4.
(r – 5)(2r – 1)
Answer:

Use the FOIL Method to find the product.
Question 5.
(m – 3)(m – 7)
Answer:

Question 6.
(x – 4)(x + 2)
Answer:

Question 7.
( 2u + \(\frac{1}{2}\))( u – \(\frac{3}{2}\))
Answer:

Question 8.
(n + 2)(n2 + 3)
Answer:

Find the product.
Question 9.
(x + 1)(x2 + 5x + 8)
Answer:

Question 10.
(n – 3)(n2 – 2n + 4)
Answer:

Question 11.
WHAT IF?
In Example 5(a), how does the polynomial change when the longer base is extended by 1 foot? Explain.
Answer:

Multiplying Polynomials 7.2 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Describe two ways to find the product of two binomials.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 1

Question 2.
WRITING
Explain how the letters of the word FOIL can help you to remember how to multiply two binomials.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, use the Distributive Property to find the product.
Question 3.
(x + 1)(x + 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 3

Question 4.
(y + 6)(y + 4)
Answer:

Question 5.
(z – 5)(z + 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 5

Question 6.
(a + 8)(a – 3)
Answer:

Question 7.
(g – 7)(g – 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 7

Question 8.
(n – 6)(n – 4)
Answer:

Question 9.
(3m + 1)(m + 9)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 9

Question 10.
(5s + 6)(s – 2)
Answer:

In Exercises 11–18, use a table to find the product.
Question 11.
(x + 3)(x + 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 11

Question 12.
(y + 10)(y – 5)
Answer:

Question 13.
(h – 8)(h – 9)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 13

Question 14.
(c – 6)(c – 5)
Answer:

Question 15.
(3k – 1)(4k + 9)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 15

Question 16.
(5g + 3)(g + 8)
Answer:

Question 17.
(-3 + 2j)(4j – 7)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 17

Question 18.
(5d – 12)(-7 + 3d)
Answer:

ERROR ANALYSIS
In Exercises 19 and 20, describe and correct the error in finding the product of the binomials.
Question 19.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 19

Question 20.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 5
Answer:

In Exercises 21–30, use the FOIL Method to find the product.
Question 21.
(b + 3)(b + 7)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 21

Question 22.
(w + 9)(w + 6)
Answer:

Question 23.
(k + 5)(k – 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 23

Question 24.
(x – 4)(x + 8)
Answer:

Question 25.
(q – \(\frac{3}{4}\)) (q + \(\frac{1}{4}\))
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 25

Question 26.
(z – \(\frac{5}{3}\)) (z – \(\frac{2}{3}\))
Answer:

Question 27.
(9 – r)(2 – 3r)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 27

Question 28.
(8 – 4x)(2x + 6)
Answer:

Question 29.
(w + 5)(w2 + 3w)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 29

Question 30.
(v – 3)(v2 + 8v)
Answer:

MATHEMATICAL CONNECTIONS In Exercises 31– 34, write a polynomial that represents the area of the shaded region.
Question 31.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 31

Question 32.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 7
Answer:

Question 33.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 33

Question 34.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 9
Answer:

In Exercises 35–42, find the product.
Question 35.
(x + 4)(x2 + 3x + 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 35

Question 36.
(f + 1)(f2 + 4f + 8)
Answer:

Question 37.
(y + 3)( y2 + 8y – 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 37

Question 38.
(t – 2)(t2 – 5t + 1)
Answer:

Question 39.
(4 – b)(5b2 + 5b – 4)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 39

Question 40.
(d + 6)(2d2 – d + 7)
Answer:

Question 41.
(3e2 – 5e + 7)(6e + 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 41

Question 42.
(6v2 + 2v – 9)(4 – 5v)
Answer:

Question 43.
MODELING WITH MATHEMATICS
The football field is rectangular.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 10
a. Write a polynomial that represents the area of the football field.
b. Find the area of the football field when the width is 160 feet.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 43.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 43.2

Question 44.
MODELING WITH MATHEMATICS
You design a frame to surround a rectangular photo. The width of the frame is the same on every side, as shown.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 11
a. Write a polynomial that represents the combined area of the photo and the frame.
b. Find the combined area of the photo and the frame when the width of the frame is 4 inches.
Answer:

Question 45.
WRITING
When multiplying two binomials, explain how the degree of the product is related to the degree of each binomial.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 45

Question 46.
THOUGHT PROVOKING
Write two polynomials that are not monomials whose product is a trinomial of degree 3.
Answer:

Question 47.
MAKING AN ARGUMENT
Your friend says the FOIL Method can be used to multiply two trinomials. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 47

Question 48.
HOW DO YOU SEE IT?
The table shows one method of finding the product of two binomials.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 12
a. Write the two binomials being multiplied.
b. Determine whether a, b, c, and d will be positive or negative when x > 0.
Answer:

Question 49.
COMPARING METHODS
You use the Distributive Property to multiply (x + 3)(x – 5). Your friend uses the FOIL Method to multiply (x – 5)(x + 3). Should your answers be equivalent? Justify your answer.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 49

Question 50.
USING STRUCTURE
The shipping container is a rectangular prism. Write a polynomial that represents the volume of the container.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 13
Answer:

Question 51.
ABSTRACT REASONING
The product of (x + m)(x + n) is x2 + bx + c.
a. What do you know about m and n when c > 0?
b. What do you know about m and n when c < 0?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 51

Maintaining Mathematical Proficiency

Write the absolute value function as a piecewise function.
Question 52.
y = |x| + 4
Answer:

Question 53.
y = 6|x – 3|
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 53

Question 54.
y = -4|x + 2|
Answer:

Simplify the expression. Write your answer using only positive exponents.
Question 55.
102 • 109
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 55

Question 56.
\(\frac{x^{5} \cdot x}{x^{8}}\)
Answer:

Question 57.
(3z6)-3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.2 Question 57

Question 58.
\(\left(\frac{2 y^{4}}{y^{3}}\right)^{-2}\)
Answer:

Lesson 7.3 Special Products of Polynomials

Essential Question What are the patterns in the special products (a+ b)(a – b), (a + b)2, and (a – b)2?

EXPLORATION 1

Finding a Sum and Difference Pattern
Work with a partner. Write the product of two binomials modeled by each rectangular array of algebra tiles.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 1

EXPLORATION 2

Finding the Square of a Binomial Pattern
Work with a partner. Draw the rectangular array of algebra tiles that models each product of two binomials. Write the product.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 2

Communicate Your Answer

Question 3.
What are the patterns in the special products (a + b)(a – b), (a + b)2, and (a – b)2?
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 3
Answer:

Question 4.
Use the appropriate special product pattern to find each product. Check your answers using algebra tiles.
a. (x + 3)(x – 3)
b. (x – 4)(x + 4)
c. (3x + 1)(3x – 1)
d. (x + 3)2
e. (x – 2)2
f. (3x + 1)2
Answer:

Monitoring Progress

Find the product.
Question 1.
(x + 7)2
Answer:

Question 2.
(7x – 3)2
Answer:

Question 3.
(4x – y)2
Answer:

Question 4.
(3m + n)2
Answer:

Find the product.
Question 5.
(x + 10)(x – 10)
Answer:

Question 6.
(2x + 1)(2x – 1)
Answer:

Question 7.
(x + 3y)(x – 3y)
Answer:

Question 8.
Describe how to use special product patterns to find 212.
Answer:

Question 9.
Each of two dogs has one black gene (B) and one white gene (W). The Punnett square shows the possible gene combinations of an offspring and the resulting colors.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 4
a. What percent of the possible gene combinations result in black?
b. Show how you could use a polynomial to model the possible gene combinations of the offspring.
Answer:

Special Products of Polynomials 7.3 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain how to use the square of a binomial pattern.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 5
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, find the product.
Question 3.
(x + 8)2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 3

Question 4.
(a – 6)2
Answer:

Question 5.
(2f – 1)2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 5

Question 6.
(5p + 2)2
Answer:

Question 7.
(-7t + 4)2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 7

Question 8.
(-12 – n)2
Answer:

Question 9.
(2a + b)2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 9

Question 10.
(6x – 3y)2
Answer:

MATHEMATICAL CONNECTIONS In Exercises 11–14, write a polynomial that represents the area of the square.
Question 11.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 11

Question 12.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 7
Answer:

Question 13.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 13

Question 14.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 9
Answer:

In Exercises 15–24, find the product.
Question 15.
(t – 7)(t + 7)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 15

Question 16.
(m + 6)(m – 6)
Answer:

Question 17.
(4x + 1)(4x – 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 17

Question 18.
(2k – 4)(2k + 4)
Answer:

Question 19.
(8 + 3a)(8 – 3a)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 19

Question 20.
(\(\frac{1}{2}\) – c )(\(\frac{1}{2}\) + c )
Answer:

Question 21.
(p – 10q)(p + 10q)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 21

Question 22.
(7m + 8n)(7m – 8n)
Answer:

Question 23.
(-y + 4)(-y – 4)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 23

Question 24.
(-5g – 2h)(-5g + 2h)
Answer:

In Exercises 25–30, use special product patterns to find the product.
Question 25.
16 • 24
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 25

Question 26.
33 • 27
Answer:

Question 27.
422
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 27

Question 28.
292
Answer:

Question 29.
30.52
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 29

Question 30.
10\(\frac{1}{3}\) • 9\(\frac{2}{3}\)
Answer:

ERROR ANALYSIS In Exercises 31 and 32, describe and correct the error in finding the product.
Question 31.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 31

Question 32.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 11
Answer:

Question 33.
MODELING WITH MATHEMATICS
A contractor extends a house on two sides.
a. The area of the house after the renovation is represented by (x + 50)2. Find this product.
b. Use the polynomial in part (a) to fond the area when x = 15. What is the area of the extension?
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 12.1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 33

Question 34.
MODELING WITH MATHEMATICS
A square-shaped parking lot with 100-foot sides is reduced by x feet on one side and extended by x feet on an adjacent side.
a. The area of the new parking lot is represented by (100 – x)(100 + x). Find this product.
b. Does the area of the parking lot increase, decrease, or stay the same? Explain.
c. Use the polynomial in part (a) to find the area of the new parking lot when x = 21.
Answer:

Question 35.
MODELING WITH MATHEMATICS
In deer, the gene N is for normal coloring and the gene a is for no coloring, or albino. Any gene combination with an N results in normal coloring. The Punnett square shows the possible gene combinations of an offspring and the resulting colors from parents that both have the gene combination Na.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 12
a. What percent of the possible gene combinations result in albino coloring?
b. Show how you could use a polynomial to model the possible gene combinations of the offspring.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 35

Question 36.
MODELING WITH MATHEMATICS
Your iris controls the amount of light that enters your eye by changing the size of your pupil.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 13
a. Write a polynomial that represents the area of your pupil. Write your answer in terms of π.
b. The width x of your iris decreases from 4millimeters to 2 millimeters when you enter a dark room. How many times greater is the area of your pupil after entering the room than before entering the room? Explain.
Answer:

Question 37.
CRITICAL THINKING
Write two binomials that have the product x2 – 121. Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 37

Question 38.
HOW DO YOU SEE IT?
In pea plants, any gene combination with a green gene (G) results in a green pod. The Punnett square shows the possible gene combinations of the offspring of two Gy pea plants and the resulting pod colors.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.3 14
A polynomial that models the possible gene combinations of the offspring is
(0.5G + 0.5y)2 – 0.25G2 + 0.5Gy + 0.25y2.
Describe two ways to determine the percent of possible gene combinations that result in green pods.
Answer:

In Exercises 39–42, find the product.
Question 39.
(x2 + 1)(x2 – 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 39

Question 40.
(y3 + 4)2
Answer:

Question 41.
(2m2 – 5n2)2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 41

Question 42.
(r3 – 6t4)(r3 + 6t4)
Answer:

Question 43.
MAKING AN ARGUMENT
Your friend claims to be able to use a special product pattern to determine that (4\(\frac{1}{3}\))2 is equal to 16\(\frac{1}{9}\). Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 43

Question 44.
THOUGHT PROVOKING
The area (in square meters) of the surface of an artificial lake is represented by x2. Describe three ways to modify the dimensions of the lake so that the new area can be represented by the three types of special product patterns discussed in this section.
Answer:

Question 45.
REASONING
Find k so that 9x2 – 48x + k is the square of a binomial.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 45

Question 46.
REPEATED REASONING
Find (x + 1)3 and (x + 2)3. Find a pattern in the terms and use it to write a pattern for the cube of a binomial (a + b)3.
Answer:

Question 47.
PROBLEM SOLVING
Find two numbers a and b such that (a + b)(a – b) < (a – b)2 < (a + b)2.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 47

Maintaining Mathematical Proficiency

Factor the expression using the GCF.
Question 48.
12y – 18
Answer:

Question 49.
9r + 27
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 49

Question 50.
49s + 35t
Answer:

Question 51.
15x – 10y
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.3 Question 51

Lesson 7.4 Solving Polynomial Equations in Factored Form

Essential Question How can you solve a polynomial equation?

EXPLORATION 1

Matching Equivalent Forms of an Equation
Work with a partner. An equation is considered to be in factored form when the product of the factors is equal to 0. Match each factored form of the equation with its equivalent standard form and nonstandard form. Factored Form Standard Form Nonstandard Form
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 1

EXPLORATION 2

Writing a Conjecture
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 2
Work with a partner. Substitute 1, 2, 3, 4, 5, and 6 for x in each equation and determine whether the equation is true. Organize your results in a table. Write a conjecture describing what you discovered.
a. (x – 1)(x – 2) = 0
b. (x – 2)(x – 3) = 0
c. (x – 3)(x – 4) = 0
d. (x – 4)(x – 5) = 0
e. (x – 5)(x – 6) = 0
f. (x – 6)(x – 1) = 0

EXPLORATION 3

Special Properties of 0 and 1
Work with a partner. The numbers 0 and 1 have special properties that are shared by no other numbers. For each of the following, decide whether the property is true for 0, 1, both, or neither. Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 3

Communicate Your Answer

Question 4.
How can you solve a polynomial equation?
Answer:

Question 5.
One of the properties in Exploration 3 is called the Zero-Product Property. It is one of the most important properties in all of algebra. Which property is it? Why do you think it is called the Zero-Product Property? Explain how it is used in algebra and why it is so important.
Answer:

Monitoring Progress

Solve the equation. Check your solutions.
Question 1.
x(x – 1) = 0
Answer:

Question 2.
3t(t + 2) = 0
Answer:

Question 3.
(z – 4)(z – 6) = 0
Answer:

Solve the equation. Check your solutions.
Question 4.
(3s + 5)(5s + 8) = 0
Answer:

Question 5.
(b + 7)2 = 0
Answer:

Question 6.
(d – 2)(d + 6)(d + 8) = 0
Answer:

Question 7.
Factor out the greatest common monomial factor from 8y2 – 24y.
Answer:

Solve the equation. Check your solutions.
Question 8.
a2 + 5a = 0
Answer:

Question 9.
3s2 – 9s = 0
Answer:

Question 10.
4x2 = 2x
Answer:

Question 11.
You can model the entrance to a mine shaft using the equation y = – \(\frac{1}{2}\)(x + 4)(x – 4), where x and y are measured in feet. The x-axis represents the ground. Find the width of the entrance at ground level.
Answer:

Solving Polynomial Equations in Factored Form 7.4 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain how to use the Zero-Product Property to find the solutions of the equation 3x(x – 6) = 0.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find both answers.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, solve the equation.
Question 3.
x(x + 7) = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 3

Question 4.
r(r – 10) = 0
Answer:

Question 5.
12t(t – 5) = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 5

Question 6.
-2v(v + 1) = 0
Answer:

Question 7.
(s – 9)(s – 1) = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 7

Question 8.
(y + 2)(y – 6) = 0
Answer:

In Exercises 9–20, solve the equation.
Question 9.
(2a – 6)(3a + 15) = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 9

Question 10.
(4q + 3)(q + 2) = 0
Answer:

Question 11.
(5m + 4)2 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 11

Question 12.
(h – 8)2 = 0
Answer:

Question 13.
(3 – 2g)(7 – g) = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 13

Question 14.
(2 – 4d )(2 + 4d ) = 0
Answer:

Question 15.
z(z + 2)(z – 1) = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 15

Question 16.
5p(2p – 3)(p + 7) = 0
Answer:

Question 17.
(r – 4)2(r + 8) = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 17

Question 18.
w(w – 6)2 = 0
Answer:

Question 19.
(15 – 5c)(5c + 5)(-c + 6) = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 19

Question 20.
(2 – n) ( 6 + \(\frac{2}{3}\)n ) (n – 2) = 0
Answer:

In Exercises 21–24, find the x-coordinates of the points where the graph crosses the x-axis.
Question 21.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 21

Question 22.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 6
Answer:

Question 23.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 23

Question 24.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 8
Answer:

In Exercises 25–30, factor the polynomial.
Question 25.
5z2 + 45z
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 25

Question 26.
6d2 – 21d
Answer:

Question 27.
3y3 – 9y2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 27

Question 28.
20x3 + 30x2
Answer:

Question 29.
5n6 + 2n5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 29

Question 30.
12a4 + 8a
Answer:

In Exercises 31–36, solve the equation.
Question 31.
4p2 – p = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 31

Question 32.
6m2 + 12m = 0
Answer:

Question 33.
25c + 10c2 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 33

Question 34.
18q – 2q2 = 0
Answer:

Question 35.
3n2 = 9n
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 35

Question 36.
-28r = 4r2
Answer:

Question 37.
ERROR ANALYSIS
Describe and correct the error in solving the equation.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 37

Question 38.
ERROR ANALYSIS
Describe and correct the error in solving the equation.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 10
Answer:

Question 39.
MODELING WITH MATHEMATICS
The entrance of a tunnel can be modeled by y = – \(\frac{11}{50}\)(x – 4)(x – 24), where x and y are measured in feet. The x-axis represents the ground. Find the width of the tunnel at ground level.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 11
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 39

Question 40.
MODELING WITH MATHEMATICS
The Gateway Arch in St. Louis can be modeled by y = – \(\frac{2}{315}\)(x + 315)(x – 315), where x and y are measured in feet. The x-axis represents the ground.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 12
a. Find the width of the arch at ground level.
b. How tall is the arch?
Answer:

Question 41.
MODELING WITH MATHEMATICS
A penguin leaps out of the water while swimming. This action is called porpoising. The height y (in feet) of a porpoising penguin can be modeled by y = -16x2 + 4.8x, where x is the time (in seconds) since the penguin leaped out of the water. Find the roots of the equation when y = 0. Explain what the roots mean in this situation.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 41

Question 42.
HOW DO YOU SEE IT?
Use the graph to fill in each blank in the equation with the symbol + or -. Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 13
Answer:

Question 43.
CRITICAL THINKING
How many x-intercepts does the graph of y = (2x + 5)(x – 9)2 have? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 43

Question 44.
MAKING AN ARGUMENT
Your friend says that the graph of the equation y = (x – a)(x – b) always has two x-intercepts for any values of a and b. Is your friend correct? Explain.
Answer:

Question 45.
CRITICAL THINKING
Does the equation (x2 + 3)(x4+ 1) = 0 have any real roots? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 45

Question 46.
THOUGHT PROVOKING
Write a polynomial equation of degree 4 whose only roots are x = 1, x = 2, and x = 3.
Answer:

Question 47.
REASONING
Find the values of x in terms of y that are solutions of each equation.
a. (x + y)(2x – y) = 0
b. (x2 – y2)(4x + 16y) = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 47

Question 48.
PROBLEM SOLVING
Solve the equation (4x-5 – 16)(3x – 81) = 0.
Answer:

Maintaining Mathematical Proficiency

List the factor pairs of the number.
Question 49.
10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 49

Question 50.
18
Answer:

Question 51.
30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.4 Question 51

Question 52.
48
Answer:

Polynomial Equations and Factoring Study Skills: Preparing for a Test

7.1–7.4 What Did You Learn?

Core Vocabulary
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 14

Core Concepts
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 15

Mathematical Practices

Question 1.
Explain how you wrote the polynomial in Exercise 11 on page 375. Is there another method you can use to write the same polynomial?
Answer:

Question 2.
Find a shortcut for exercises like Exercise 7 on page 381 when the variable has a coefficient of 1. Does your shortcut work when the coefficient is not 1?
Answer:

Study Skills: Preparing for a Test

  • Review examples of each type of problem that could appear on the test.
  • Review the homework problems your teacher assigned.
  • Take a practice test
    Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.4 16

Polynomial Equations and Factoring 7.1–7.4 Quiz

Write the polynomial in standard form. Identify the degree and leading coefficient of the polynomial. Then classify the polynomial by the number of terms.
Question 1.
-8q3
Answer:

Question 2.
9 + d2 – 3d
Answer:

Question 3.
\(\frac{2}{3}\)m4 – \(\frac{5}{6}\)m6
Answer:

Question 4.
-1.3z + 3z4 + 7.4z2
Answer:

Find the sum or difference.
Question 5.
(2x2 + 5) + (-x2 + 4)
Answer:

Question 6.
(-3n2 + n) – (2n2 – 7)
Answer:

Question 7.
(-p2 + 4p) – (p2 – 3p + 15)
Answer:

Question 8.
(a2 – 3ab + b2) + (-a2 + ab + b2)
Answer:

Find the product.
Question 9.
(w + 6)(w + 7)
Answer:

Question 10.
(3 – 4d )(2d – 5)
Answer:

Question 11.
(y + 9)(y2 + 2y – 3)
Answer:

Question 12.
(3z – 5)(3z + 5)
Answer:

Question 13.
(t + 5)2
Answer:

Question 14.
(2q – 6)2
Answer:

Solve the equation.
Question 15.
5x2 – 15x = 0
Answer:

Question 16.
(8 – g)(8 – g) = 0
Answer:

Question 17.
(3p + 7)(3p – 7)( p + 8) = 0
Answer:

Question 18.
-3y( y – 8)(2y + 1) = 0
Answer:

Question 19.
You are making a blanket with a fringe border of equal width on each side.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring q 1
a. Write a polynomial that represents the perimeter of the blanket including the fringe.
b. Write a polynomial that represents the area of the blanket including the fringe.
c. Find the perimeter and the area of the blanket including the fringe when the width of the fringe is 4 inches.
Answer:

Question 20.
You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The expression 1000(1 + r )2 represents the balance after 2 years, where r is the annual interest rate in decimal form.
a. Write the polynomial in a standard form that represents the balance of your account after 2 years.
b. The interest rate is 3%. What is the balance of your account after 2 years?
c. The guitar costs $1100. Do you have enough money in your account after 3 years? Explain.
Answer:

Question 21.
The front of a storage bunker can be modeled by y = – \(\frac{5}{216}\)(x – 72)(x + 72), where x and y are measured in inches. The x-axis represents the ground. Find the width of the bunker at ground level.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring q 2
Answer:

Lesson 7.5 Factoring x2 + bx + c

Essential Question How can you use algebra tiles to factor the trinomial x2 + bx + c into the product of two binomials?

EXPLORATION 1

Finding Binomial Factors
Work with a partner. Use algebra tiles to write each polynomial as the product of two binomials. Check your answer by multiplying.
Sample x2 + 5x + 6
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 1

Communicate Your Answer

Question 2.
How can you use algebra tiles to factor the trinomial x2 + bx + c into the product of two binomials?
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 2
Answer:

Question 3.
Describe a strategy for factoring the trinomial x2 + bx + c that does not use algebra tiles.
Answer:

Factor the polynomial.
Question 1.
x2 + 7x + 6
Answer:

Question 2.
x2 + 9x + 8
Answer:

Factor the polynomial.
Question 3.
w2 – 4w + 3
Answer:

Question 4.
n2 – 12n + 35
Answer:

Question 5.
x2 – 14x + 24
Answer:

Question 6.
x2 + 2x – 15
Answer:

Question 7.
y2 + 13y – 30
Answer:

Question 8.
v2 – v – 42
Answer:

Question 9.
WHAT IF?
The area of the pumpkin patch is 200 square meters. What is the area of the square plot of land?
Answer:

Factoring x2 + bx + c 7.5 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
You are factoring x2 + 11x = 26. What do the signs of the terms tell you about the factors? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 1

Question 2.
OPEN-ENDED
Write a trinomial that can be factored as (x + p)(x + q), where p and q are positive.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, factor the polynomial.
Question 3.
x2 + 8x + 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 3

Question 4.
z2 + 10z + 21
Answer:

Question 5.
n2 + 9n + 20
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 5

Question 6.
s2 + 11s + 30
Answer:

Question 7.
h2 + 11h + 18
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 7

Question 8.
y2 + 13y + 40
Answer:

In Exercises 9–14, factor the polynomial.
Question 9.
v2 – 5v + 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 9

Question 10.
x2 – 13x + 22
Answer:

Question 11.
d2 – 5d + 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 11

Question 12.
k2 – 10k + 24
Answer:

Question 13.
w2 – 17w + 72
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 13

Question 14.
j2 – 13j + 42
Answer:

In Exercises 15–24, factor the polynomial.
Question 15.
x2 + 3x – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 15

Question 16.
z2 + 7z – 18
Answer:

Question 17.
n2 + 4n – 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 17

Question 18.
s2 + 3s – 40
Answer:

Question 19.
y2 + 2y – 48
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 19

Question 20.
h2 + 6h – 27
Answer:

Question 21.
x2 – x – 20
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 21

Question 22.
m2 – 6m – 7
Answer:

Question 23.
-6t – 16 + t2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 23

Question 24.
-7y + y2 – 30
Answer:

Question 25.
MODELING WITH MATHEMATICS
A projector displays an image on a wall. The area (in square feet) of the projection is represented by x2 – 8x + 15.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 3
a. Write a binomial that represents the height of the projection.
b. Find the perimeter of the projection when the height of the wall is 8 feet.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 25

Question 26.
MODELING WITH MATHEMATICS
A dentist’s office and parking lot are on a rectangular piece of land. The area (in square meters) of the land is represented by x2 + x – 30.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 4
a. Write a binomial that represents the width of the land.
b. Find the area of the land when the length of the dentist’s office is 20 meters.
Answer:

ERROR ANALYSIS In Exercises 27 and 28, describe and correct the error in factoring the polynomial.
Question 27.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 27

Question 28.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 6
Answer:

In Exercises 29–38, solve the equation.
Question 29.
m2 + 3m + 2 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 29

Question 30.
n2 – 9n + 18 = 0
Answer:

Question 31.
x2 + 5x – 14 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 31

Question 32.
v2 + 11v – 26 = 0
Answer:

Question 33.
t2 + 15t = -36
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 33

Question 34.
n2 – 5n = 24
Answer:

Question 35.
a2 + 5a – 20 = 30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 35

Question 36.
y2 – 2y – 8 = 7
Answer:

Question 37.
m2 + 10 = 15m – 34
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 37

Question 38.
b2 + 5 = 8b – 10
Answer:

Question 39.
MODELING WITH MATHEMATICS
You trimmed a large square picture so that you could fit it into a frame. The area of the cut picture is 20 square inches. What is the area of the original picture?
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 39.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 39.2

Question 40.
MODELING WITH MATHEMATICS
A web browser is open on your computer screen.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 8
a. The area of the browser window is 24 square inches. Find the length of the browser window x.
b. The browser covers \(\frac{3}{13}\) of the screen. What are the dimensions of the screen?
Answer:

Question 41.
MAKING AN ARGUMENT
Your friend says there are six integer values of b for which the trinomial x2 + bx – 12 has two binomial factors of the form (x + p) and (x + q). Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 41

Question 42.
THOUGHT PROVOKING
Use algebra tiles to factor each polynomial modeled by the tiles. Show your work.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 9
Answer:

MATHEMATICAL CONNECTIONS In Exercises 43 and 44, find the dimensions of the polygon with the given area.
Question 43.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 43

Question 44.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 11
Answer:

Question 45.
REASONING
Write an equation of the form x2 + bx + c = 0 that has the solutions x = -4 and x = 6. Explain how you found your answer.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 45

Question 46.
HOW DO YOU SEE IT?
The graph of y = x2 + x – 6 is shown.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 12
a. Explain how you can use the graph to factor the polynomial x2 + x – 6.
b. Factor the polynomial.
Answer:

Question 47.
PROBLEM SOLVING
Road construction workers are paving the area shown.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring 7.5 13
a. Write an expression that represents the area being paved.
b. The area being paved is 280 square meters. Write and solve an equation to find the width of the road x.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 47.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 47.2
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 47.3

USING STRUCTURE In Exercises 48–51, factor the polynomial.
Question 48.
x2 + 6xy + 8y2
Answer:

Question 49.
r2 + 7rs + 12s2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 49

Question 50.
a2 + 11ab – 26b2
Answer:

Question 51.
x2 – 2xy – 35y2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 51

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.
Question 52.
p – 9 = 0
Answer:

Question 53.
z + 12 = -5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 53

Question 54.
6 = \(\frac{c}{-7}\)
Answer:

Question 55.
4k = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.5 Question 55

Lesson 7.6 Factoring ax2 + bx + c

Essential Question How can you use algebra tiles to factor the trinomial ax2 + bx + c into the product of two binomials?

EXPLORATION 1

Finding Binomial Factors
Work with a partner. Use algebra tiles to write each polynomial as the product of two binomials. Check your answer by multiplying.
Sample 2x2 + 5x + 2
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 1

Communicate Your Answer

Question 2.
How can you use algebra tiles to factor the trinomial ax2 + bx + c into the product of two binomials?
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 2
Answer:

Question 3.
Is it possible to factor the trinomial 2x2 + 2x + 1? Explain your reasoning.
Answer:

Monitoring Progress

Factor the polynomial.
Question 1.
8x2 – 56x + 48
Answer:

Question 2.
14x2 + 31x + 15
Answer:

Question 3.
2x2 – 7x + 5
Answer:

Question 4.
3x2 – 14x + 8
Answer:

Question 5.
4x2 – 19x – 5
Answer:

Question 6.
6x2 + x – 12
Answer:

Question 7.
-2y2 – 5y – 3
Answer:

Question 8.
-5m2 + 6m – 1
Answer:

Question 9.
-3x2 – x + 2
Answer:

Question 10.
WHAT IF?
The area of the reserve is 136 square miles. How wide is the reserve?
Answer:

Factoring ax2 + bx + c 7.6 Exercises

Vocabulary and Core Concept Check

Question 1.
REASONING
What is the greatest common factor of the terms of 3y2 – 21y + 36?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 1

Question 2.
WRITING
Compare factoring 6x2 – x – 2 with factoring x2 – x – 2.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, factor the polynomial.
Question 3.
3x2 + 3x – 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 3

Question 4.
8v2 + 8v – 48
Answer:

Question 5.
4k2 + 28k + 48
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 5

Question 6.
6y2 – 24y + 18
Answer:

Question 7.
7b2 – 63b + 140
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 7

Question 8.
9r2 – 36r – 45
Answer:

In Exercises 9–16, factor the polynomial.
Question 9.
3h2 + 11h + 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 9

Question 10.
8m2 + 30m + 7
Answer:

Question 11.
6x2 – 5x + 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 11

Question 12.
10w2 – 31w + 15
Answer:

Question 13.
3n2 + 5n – 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 13

Question 14.
4z2 + 4z – 3
Answer:

Question 15.
8g2 – 10g – 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 15.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 15.2
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 15.3

Question 16.
18v2 – 15v – 18
Answer:

In Exercises 17–22, factor the polynomial.
Question 17.
-3t2 + 11t – 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 17

Question 18.
-7v2 – 25v – 12
Answer:

Question 19.
-4c2 + 19c + 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 19

Question 20.
-8h2 – 13h + 6
Answer:

Question 21.
-15w2 – w + 28
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 21.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 21.2
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 21.3

Question 22.
-22d2 + 29d – 9
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in factoring the polynomial.
Question 23.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 23

Question 24.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 4
Answer:

In Exercises 25–28, solve the equation.
Question 25.
5x2 – 5x – 30 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 25

Question 26.
2k2 – 5k – 18 = 0
Answer:

Question 27.
-12n2 – 11n = -15
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 27.2
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 27.3

Question 28.
14b2 – 2 = -3b
Answer:

In Exercises 29–32, find the x-coordinates of the points where the graph crosses the x-axis.
Question 29.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 29.2

Question 30.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 6
Answer:

Question 31.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 31.2

Question 32.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 8
Answer:

Question 33.
MODELING WITH MATHEMATICS
The area (in square feet) of the school sign can be represented by 15x2 – x – 2.
a. Write an expression that represents the length of the sign.
b. Describe two ways to find the area of the sign when x = 3.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 33

Question 34.
MODELING WITH MATHEMATICS
The height h (in feet) above the water of a cliff diver is modeled by h = -16t2 + 8t + 80, where t is the time (in seconds). How long is the diver in the air?
Answer:

Question 35.
MODELING WITH MATHEMATICS
The Parthenon in Athens, Greece, is an ancient structure that has a rectangular base. The length of the base of the Parthenon is 8 meters more than twice its width. The area of the base is about 2170 square meters. Find the length and width of the base.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 35.2

Question 36.
MODELING WITH MATHEMATICS
The length of a rectangular birthday party invitation is 1 inch less than twice its width. The area of the invitation is 15 square inches. Will the invitation fi t in the envelope shown without being folded? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 10
Answer:

Question 37.
OPEN-ENDED
Write a binomial whose terms have a GCF of 3x.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 37

Question 38.
HOW DO YOU SEE IT?
Without factoring, determine which of the graphs represents the function g(x) = 21x2 + 37x + 12 and which represents the function h(x) = 21x2 – 37x + 12. Explain your reasoning.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 11
Answer:

Question 39.
REASONING
When is it not possible to factor ax2 + bx + c, where a ≠ 1? Give an example.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 39

Question 40.
MAKING AN ARGUMENT
Your friend says that to solve the equation 5x2 + x – 4 = 2, you should start by factoring the left side as (5x – 4)(x + 1). Is your friend correct? Explain.
Answer:

Question 41.
REASONING
For what values of t can 2x2 + tx + 10 be written as the product of two binomials?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 41.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 41.2

Question 42.
THOUGHT PROVOKING
Use algebra tiles to factor each polynomial modeled by the tiles. Show your work.
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 12
Answer:

Question 43.
MATHEMATICAL CONNECTIONS
The length of a rectangle is 1 inch more than twice its width. The value of the area of the rectangle (in square inches) is 5 more than the value of the perimeter of the rectangle (in inches). Find the width.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 43.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 43.2
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 43.3

Question 44.
PROBLEM SOLVING
A rectangular swimming pool is bordered by a concrete patio. The width of the patio is the same on every side. The area of the surface of the pool is equal to the area of the patio. What is the width of the patio?
Big Ideas Math Algebra 1 Answer Key Chapter 7 Polynomial Equations and Factoring 7.6 13
Answer:

In Exercises 45–48, factor the polynomial.
Question 45.
4k2 + 7jk – 2j2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 45

Question 46.
6x2 + 5xy – 4y2
Answer:

Question 47.
-6a2 + 19ab – 14b2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 47.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 47.2

Question 48.
18m3 + 39m2n – 15mn2
Answer:

Maintaining Mathematical Proficiency

Find the square root(s).
Question 49.
± \(\sqrt{64}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 49

Question 50.
\(\sqrt{4}\)
Answer:

Question 51.
– \(\sqrt{225}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 51

Question 52.
± \(\sqrt{81}\)
Answer:

Solve the system of linear equations by substitution. Check your solution.
Question 53.
y = 3 + 7x
y – x = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 53.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 53.2

Question 54.
2x = y + 2
-x + 3y = 14
Answer:

Question 55.
5x – 2y = 14
-7 = -2x + y
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 55.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.6 Question 55.2

Question 56.
-x – 8 = -y
9y – 12 + 3x = 0
Answer:

Lesson 7.7 Factoring Special Products

Essential Question How can you recognize and factor special products?

EXPLORATION 1

Factoring Special ProductsWork with a partner. Use algebra tiles to write each polynomial as the product of two binomials. Check your answer by multiplying. State whether the product is a “special product” that you studied in Section 7.3.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 1

EXPLORATION 2

Factoring Special Products
Work with a partner. Use algebra tiles to complete the rectangular array at the left in three different ways, so that each way represents a different special product. Write each special product in standard form and in factored form.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 2

Communicate Your Answer

Question 3.
How can you recognize and factor special products? Describe a strategy for recognizing which polynomials can be factored as special products.
Answer:

Question 4.
Use the strategy you described in Question 3 to factor each polynomial.
a. 25x2 + 10x + 1
b. 25x2 – 10x + 1
c. 25x2 – 1
Answer:

Monitoring Progress

Factor the polynomial.
Question 1.
x2 – 36
Answer:

Question 2.
100 – m2
Answer:

Question 3.
9n2 – 16
Answer:

Question 4.
16h2 – 49
Answer:

Use a special product pattern to evaluate the expression.
Question 5.
362 – 342
Answer:

Question 6.
472 – 442
Answer:

Question 7.
552 – 502
Answer:

Question 8.
282 – 242
Answer:

Factor the polynomial.
Question 9.
m2 – 2m + 1
Answer:

Question 10.
d2 – 10d + 25
Answer:

Question 11.
9z2 + 36z + 36
Answer:

Solve the equation.
Question 12.
a2 + 6a + 9 = 0
Answer:

Question 13.
w2 – \(\frac{7}{3}\)w + \(\frac{49}{36}\) = 0
Answer:

Question 14.
n2 – 81 = 0
Answer:

Question 15.
WHAT IF?
The golf ball does not hit the pine tree. After how many seconds does the ball hit the ground?
Answer:

Factoring Special Products 7.7 Exercises

Vocabulary and Core Concept Check

Question 1.
REASONING
Can you use the perfect square trinomial pattern to factor y2 + 16y + 64? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which polynomial does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 3
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, factor the polynomial.
Question 3.
m2 – 49
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 3

Question 4.
z2 – 81
Answer:

Question 5.
64 – 81d2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 5

Question 6.
25 – 4x2
Answer:

Question 7.
225a2 – 36b2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 7

Question 8.
16x2 – 169y2
Answer:

In Exercises 9–14, use a special product pattern to evaluate the expression.
Question 9.
122 – 92
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 9

Question 10.
192 – 112
Answer:

Question 11.
782 – 722
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 11

Question 12.
542 – 522
Answer:

Question 13.
532 – 472
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 13

Question 14.
392 – 362
Answer:

In Exercises 15–22, factor the polynomial.
Question 15.
h2 + 12h + 36
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 15

Question 16.
p2 + 30p + 225
Answer:

Question 17.
y2 – 22y + 121
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 17

Question 18.
x2 – 4x + 4
Answer:

Question 19.
a2 – 28a + 196
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 19

Question 20.
m2 + 24m + 144
Answer:

Question 21.
25n2 + 20n + 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 21

Question 22.
49a2 – 14a + 1
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in factoring the polynomial.
Question 23.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 23

Question 24.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 5
Answer:

Question 25.
MODELING WITH MATHEMATICS
The area (in square centimeters) of a square coaster can be represented by d2 + 8d + 16.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 6
a. Write an expression that represents the side length of the coaster.
b. Write an expression for the perimeter of the coaster.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 25

Question 26.
MODELING WITH MATHEMATICS
The polynomial represents the area (in square feet) of the square playground.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 7
a. Write a polynomial that represents the side length of the playground.
b. Write an expression for the perimeter of the playground.
Answer:

In Exercises 27–34, solve the equation.
Question 27.
z2 – 4 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 27

Question 28.
4x2 = 49
Answer:

Question 29.
k2 – 16k + 64 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 29

Question 30.
s2 + 20s + 100 = 0
Answer:

Question 31.
n2 + 9 = 6n
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 31

Question 32.
y2 = 12y – 36
Answer:

Question 33.
y2 + \(\frac{1}{2}\)y = – 1 — 16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 33

Question 34.
– \(\frac{4}{3}\)x + \(\frac{4}{9}\) = -x2
Answer:

In Exercises 35–40, factor the polynomial.
Question 35.
3z2 – 27
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 35

Question 36.
2m2 – 50
Answer:

Question 37.
4y2 – 16y + 16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 37

Question 38.
8k2 + 80k + 200
Answer:

Question 39.
50y2 + 120y + 72
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 39

Question 40.
27m2 – 36m + 12
Answer:

Question 41.
MODELING WITH MATHEMATICS
While standing on a ladder, you drop a paintbrush. The function represents the height y (in feet) of the paintbrush t seconds after it is dropped. After how many seconds does the paintbrush land on the ground?
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 41

Question 42.
MODELING WITH MATHEMATICS
The function represents the height y (in feet) of a grasshopper jumping straight up from the ground t seconds after the start of the jump. After how many seconds is the grasshopper 1 foot off the ground?
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 9
Answer:

Question 43.
REASONING
Tell whether the polynomial can be factored. If not, change the constant term so that the polynomial is a perfect square trinomial.
a. w2 + 18w + 84
b. y2 – 10y + 23
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 43.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 43.2

Question 44.
THOUGHT PROVOKING
Use algebra tiles to factor each polynomial modeled by the tiles. Show your work.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 10
Answer:

Question 45.
COMPARING METHODS
Describe two methods you can use to simplify (2x – 5)2 – (x – 4)2. Which one would you use? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 45

Question 46.
HOW DO YOU SEE IT?
The figure shows a large square with an area of a2 that contains a smaller square with an area of b2.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 11
a. Describe the regions that represent a2 – b2. How can you rearrange these regions to show that a2 – b2 = (a + b)(a – b)?
b. How can you use the figure to show that (a – b)2 = a2 – 2ab + b2?
Answer:

Question 47.
PROBLEM SOLVING
You hang nine identical square picture frames on a wall.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 12
a. Write a polynomial that represents the area of the picture frames, not including the pictures.
b. The area in part (a) is 81 square inches. What is the side length of one of the picture frames? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 47.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 47.2

Question 48.
MATHEMATICAL CONNECTIONS
The composite solid is made up of a cube and a rectangular prism.
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 13
a. Write a polynomial that represents the volume of the composite solid.
b. The volume of the composite solid is equal to 25x. What is the value of x? Explain your reasoning.
Answer:

Maintaining Mathematical Proficiency

Write the prime factorization of the number.
Question 49.
50
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 49

Question 50.
44
Answer:

Question 51.
85
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 51

Question 52.
96
Answer:

Graph the inequality in a coordinate plane.
Question 53.
y ≤ 4x – 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 53

Question 54.
y > – \(\frac{1}{2}\)x + 3
Answer:

Question 55.
4y – 12 ≥ 8x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.7 Question 55

Question 56.
3y + 3 < x
Answer:

Lesson 7.8 Factoring Polynomials Completely

Essential Question How can you factor a polynomial completely?

EXPLORATION 1

Writing a Product of Linear Factors
Work with a partner. Write the product represented by the algebra tiles. Then multiply to write the polynomial in standard form.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 1

EXPLORATION 2

Matching Standard and Factored Forms
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 2
Work with a partner. Match the standard form of the polynomial with the equivalent factored form. Explain your strategy.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 3

Communicate Your Answer

Question 3.
How can you factor a polynomial completely?
Answer:

Question 4.
Use your answer to Question 3 to factor each polynomial completely.
a. x3 + 4x2 + 3x
b. x3 – 6x2 + 9x
c. x3 + 6x2 + 9x
Answer:

Factor the polynomial by grouping.
Question 1.
a3 + 3a2 + a + 3
Answer:

Question 2.
y2 + 2x + yx + 2y
Answer:

Factor the polynomial completely.
Question 3.
3x3 – 12x
Answer:

Question 4.
2y3 – 12y2 + 18y
Answer:

Question 5.
m3 – 2m2 – 8m
Answer:

Solve the equation.
Question 6.
w3 – 8w2 + 16w = 0
Answer:

Question 7.
x3 – 25x = 0
Answer:

Question 8.
c3 – 7c2 + 12c = 0
Answer:

Question 9.
A box in the shape of a rectangular prism has a volume of 72 cubic feet. The box has a length of x feet, a width of (x – 1) feet, and a height of (x + 9) feet. Find the dimensions of the box.
Answer:

Factoring Polynomials Completely 7.8 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What does it mean for a polynomial to be factored completely?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 1

Question 2.
WRITING
Explain how to choose which terms to group together when factoring by grouping.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, factor the polynomial by grouping.
Question 3.
x3 + x2 + 2x + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 3

Question 4.
y3 – 9y2+ y – 9
Answer:

Question 5.
3z3 + 2z – 12z2 – 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 5

Question 6.
2s3 – 27 – 18s + 3s2
Answer:

Question 7.
x2 + xy + 8x + 8y
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 7

Question 8.
q2 + q + 5pq + 5p
Answer:

Question 9.
m2 – 3m + mn – 3n
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 9

Question 10.
2a2 + 8ab – 3a – 12b
Answer:

In Exercises 11–22, factor the polynomial completely.
Question 11.
2x3 – 2x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 11

Question 12.
36a4 – 4a2
Answer:

Question 13.
2c2 – 7c + 19
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 13

Question 14.
m2 – 5m – 35
Answer:

Question 15.
6g3 – 24g2 + 24g
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 15

Question 16.
-15d3 + 21d2 – 6d
Answer:

Question 17.
3r5 + 3r4 – 90r3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 17

Question 18.
5w4 – 40w3 + 80w2
Answer:

Question 19.
-4c4 + 8c3 – 28c2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 19

Question 20.
8t2 + 8t – 72
Answer:

Question 21.
b3 – 5b2 – 4b + 20
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 21

Question 22.
h3 + 4h2 – 25h – 100
Answer:

In Exercises 23–28, solve the equation.
Question 23.
5n3 – 30n2 + 40n = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 23

Question 24.
k4 – 100k2 = 0
Answer:

Question 25.
x3 + x2 = 4x + 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 25

Question 26.
2t5 + 2t4 – 144t3 = 0
Answer:

Question 27.
12s – 3s3 = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 27

Question 28.
4y3 – 7y2 + 28 = 16y
Answer:

In Exercises 29–32, find the x-coordinates of the points where the graph crosses the x-axis.
Question 29.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 29

Question 30.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 5
Answer:

Question 31.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 31

Question 32.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 7
Answer:

ERROR ANALYSIS In Exercises 33 and 34, describe and correct the error in factoring the polynomial completely.
Question 33.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 33

Question 34.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 9
Answer:

Question 35.
MODELING WITH MATHEMATICS
You are building a birdhouse in the shape of a rectangular prism that has a volume of 128 cubic inches. The dimensions of the birdhouse in terms of its width are shown.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 10
a. Write a polynomial that represents the volume of the birdhouse.
b. What are the dimensions of the birdhouse?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 35.2

Question 36.
MODELING WITH MATHEMATICS
A gift bag shaped like a rectangular prism has a volume of 1152 cubic inches. The dimensions of the gift bag in terms of its width are shown. The height is greater than the width. What are the dimensions of the gift bag?
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 11
Answer:

In Exercises 37–40, factor the polynomial completely.
Question 37.
x3 + 2x2y – x – 2y
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 37

Question 38.
8b3 – 4b2a – 18b + 9a
Answer:

Question 39.
4s2 – s + 12st – 3t
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 39

Question 40.
6m3 – 12mn + m2n – 2n2
Answer:

Question 41.
WRITING
Is it possible to find three real solutions of the equation x3 + 2x2 + 3x + 6 = 0? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 41

Question 42.
HOW DO YOU SEE IT?
How can you use the factored form of the polynomial x4 – 2x3 – 9x2 + 18x = x(x – 3)(x + 3)(x – 2) to find the x-intercepts of the graph of the function?
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 12
Answer:

Question 43.
OPEN-ENDED
Write a polynomial of degree 3 that satisfies each of the given conditions.
a. is not factorable
b. can be factored by grouping
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 43

Question 44.
MAKING AN ARGUMENT
Your friend says that if a trinomial cannot be factored as the product of two binomials, then the trinomial is factored completely. Is your friend correct? Explain.
Answer:

Question 45.
PROBLEM SOLVING
The volume (in cubic feet) of a room in the shape of a rectangular prism is represented by 12z3 – 27z. Find expressions that could represent the dimensions of the room.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 45

Question 46.
MATHEMATICAL CONNECTIONS
The width of a box in the shape of a rectangular prism is 4 inches more than the height h. The length is the difference of 9 inches and the height.
a. Write a polynomial that represents the volume of the box in terms of its height (in inches).
b. The volume of the box is 180 cubic inches. What are the possible dimensions of the box?
c. Which dimensions result in a box with the least possible surface area? Explain your reasoning.
Answer:

Question 47.
MATHEMATICAL CONNECTIONS
The volume of a cylinder is given by V = πr2h, where r is the radius of the base of the cylinder and h is the height of the cylinder. Find the dimensions of the cylinder.
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 13
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 47

Question 48.
THOUGHT PROVOKING
Factor the polynomial x5 – x4 – 5x3 + 5x2 + 4x – 4 completely.
Answer:

Question 49.
REASONING
Find a value for w so that the equation has (a) two solutions and (b) three solutions. Explain your reasoning. 5x3 + wx2 + 80x = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 49.1
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 49.2

Maintaining Mathematical Proficiency

Solve the system of linear equations by graphing.
Question 50.
y = x – 4
y = -2x + 2
Answer:

Question 51.
y = \(\frac{1}{2}\)x + 2
y = 3x – 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 51

Question 52.
5x – y = 12
\(\frac{1}{4}\)x + y = 9
Answer:

Question 53.
x = 3y
y – 10 = 2x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 53

Graph the function. Describe the domain and range.
Question 54.
f(x) = 5x
Answer:

Question 55.
y = 9 (\(\frac{1}{3}\))x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 55

Question 56.
y = -3(0.5)x
Answer:

Question 57.
f(x) = -3(4)x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 7 Polynomial Equations and Factoring 7.8 Question 57

Polynomial Equations and Factoring Performance Task: The View Matters

7.5 – 7.8 What Did You Learn?

Core Vocabulary
factoring by grouping, p. 404
factored completely, p. 404

Core Concepts
Section 7.5
Factoring x2 + bx + c When c Is Positive, p. 386
Factoring x2 + bx + c When cIs Negative, p. 387

Section 7.6
Factoring ax2 + bx + c When ac Is Positive, p. 392
Factoring ax2 + bx + c When ac Is Negative, p. 393

Section 7.7
Difference of Two Squares Pattern, p. 398
Perfect SquareTrinomial Pattern, p. 399

Section 7.8
Factoring by Grouping, p. 404
Factoring Polynomials Completely, p. 404

Mathematical Practices

Question 1.
How are the solutions of Exercise 29 on page 389 related to the graph of y = m2 + 3m + 2?
Answer:

Question 2.
The equation in part (b) of Exercise 47 on page 390 has two solutions. Are both solutions of the equation reasonable in the context of the problem? Explain your reasoning.
Answer:

Performance Task: The View Matters

The way an equation or expression is written can help you interpret and solve problems. Which representation would you rather have when trying to solve for specific information? Why?
To explore the answers to these questions and more, go to
Big Ideas Math Algebra 1 Solutions Chapter 7 Polynomial Equations and Factoring 7.8 14

Polynomial Equations and Factoring Chapter Review

7.1 Adding and Subtracting Polynomials

Write the polynomial in standard form. Identify the degree and leading coefficient of the polynomial. Then classify the polynomial by the number of terms.
Question 1.
6 + 2x2
Answer:

Question 2.
-3p3 + 5p6 – 4
Answer:

Question 3.
9x7 – 6x2 + 13x5
Answer:

Question 4.
-12y + 8y3
Answer:

Find the sum or difference.
Question 5.
(3a + 7) + (a – 1)
Answer:

Question 6.
(x2 + 6x – 5) + (2x2 + 15)
Answer:

Question 7.
(-y2 + y + 2) – (y2 – 5y – 2)
Answer:

Question 8.
(p + 7) – (6p2 + 13p)
Answer:

7.2 Multiplying Polynomials

Find the product.
Question 9.
(x + 6)(x – 4)
Answer:

Question 10.
(y – 5)(3y + 8)
Answer:

Question 11.
(x + 4)(x2 + 7x)
Answer:

Question 12.
(-3y + 1)(4y2 – y – 7)
Answer:

7.3 Special Products of Polynomials

Find the product.
Question 13.
(x + 9)(x – 9)
Answer:

Question 14.
(2y + 4)(2y – 4)
Answer:

Question 15.
( p + 4)2
Answer:

Question 16.
(-1+ 2d )2
Answer:

7.4 Solving Polynomial Equations in Factored Form (pp. 377–382)

Solve the equation.
Question 17.
x2 + 5x = 0
Answer:

Question 18.
(z + 3)(z – 7) = 0
Answer:

Question 19.
(b + 13)2 = 0
Answer:

Question 20.
2y(y – 9)(y + 4) = 0
Answer:

7.5 Factoring x2 + bx + c(pp. 385–390)

Factor the polynomial.
Question 21.
p2 + 2p – 35
Answer:

Question 22.
b2 + 18b + 80
Answer:

Question 23.
z2 – 4z – 21
Answer:

Question 24.
x2 – 11x + 28
Answer:

7.6 Factoring ax2 + bx + c (pp. 391–396)

Factor the polynomial.
Question 25.
3t2 + 16t – 12
Answer:

Question 26.
-5y2 – 22y – 8
Answer:

Question 27.
6x2 + 17x + 7
Answer:

Question 28.
-2y2 + 7y – 6
Answer:

Question 29.
3z2 + 26z – 9
Answer:

Question 30.
10a2 – 13a – 3
Answer:

7.7 Factoring Special Products (pp. 397–402)

Factor the polynomial.
Question 31.
x2 – 9
Answer:

Question 32.
y2 – 100
Answer:

Question 33.
z2 – 6z + 9
Answer:

Question 34.
m2 + 16m + 64
Answer:

7.8 Factoring Polynomials Completely (pp. 403–408)

Factor the polynomial completely.
Question 35.
n3 – 9n
Answer:

Question 36.
x2 – 3x + 4ax – 12a
Answer:

Question 37.
2x4 + 2x3 – 20x2
Answer:

Solve the equation.
Question 38.
3x3 – 9x2 – 54x = 0
Answer:

Question 39.
16x2 – 36 = 0
Answer:

Question 40.
z3 + 3z2 – 25z – 75 = 0
Answer:

Question 41.
A box in the shape of a rectangular prism has a volume of 96 cubic feet. The box has a length of (x + 8) feet, a width of x feet, and a height of (x – 2) feet. Find the dimensions of the box.
Answer:

Polynomial Equations and Factoring Chapter Test

Find the sum or difference. Then identify the degree of the sum or difference and classify it by the number of terms.
Question 1.
(-2p + 4) – (p2 – 6p + 8)
Answer:

Question 2.
(9c6 – 5b4) – (4c6 – 5b4)
Answer:

Question 3.
(4s4 + 2st + t) + (2s4 – 2st – 4t)
Answer:

Find the product.
Question 4.
(h – 5)(h – 8)
Answer:

Question 5.
(2w – 3)(3w + 5)
Answer:

Question 6.
(z + 11)(z – 11)
Answer:

Question 7.
Explain how you can determine whether a polynomial is a perfect square trinomial.
Answer:

Question 8.
Is 18 a polynomial? Explain your reasoning.
Answer:

Factor the polynomial completely.
Question 9.
s2 – 15s + 50
Answer:

Question 10.
h3 + 2h2 – 9h – 18
Answer:

Question 11.
-5k2 – 22k + 15
Answer:

Solve the equation.
Question 12.
(n – 1)(n + 6)(n + 5) = 0
Answer:

Question 13.
d2 + 14d + 49 = 0
Answer:

Question 14.
6x4 + 8x2 = 26x3
Answer:

Question 15.
The expression π(r – 3)2 represents the area covered by the hour hand on a clock in one rotation, where r is the radius of the entire clock. Write a polynomial in standard form that represents the area covered by the hour hand in one rotation.
Answer:

Question 16.
A magician’s stage has a trapdoor.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring ct 1
a. The total area (in square feet) of the stage can be represented by x2 + 27x + 176. Write an expression for the width of the stage.
b. Write an expression for the perimeter of the stage.
c. The area of the trapdoor is 10 square feet. Find the value of x.
d. The magician wishes to have the area of the stage be at least 20 times the area of the trapdoor. Does this stage satisfy his requirement? Explain.
Answer:

Question 17.
Write a polynomial equation in factored form that has three positive roots.
Answer:

Question 18.
You are jumping on a trampoline. For one jump, your height y (in feet) above the trampoline after t seconds can be represented by y = -16t2 + 24t. How many seconds are you in the air?
Answer:

Question 19.
A cardboard box in the shape of a rectangular prism has the dimensions shown.
Big Ideas Math Answer Key Algebra 1 Chapter 7 Polynomial Equations and Factoring ct 2
a. Write a polynomial that represents the volume of the box.
b. The volume of the box is 60 cubic inches. What are the length, width, and height of the box?
Answer:

Polynomial Equations and Factoring Cumulative Assessment

Question 1.
Classify each polynomial by the number of terms. Then order the polynomials by degree from least to greatest.
a. -4x3
b. 6y – 3y5
c. c2 + 2 + c
d. -10d4 + 7d2
e. -5z11 + 8z12
f. 3b6 – 12b8 + 4b4
Answer:

Question 2.
Which exponential function is increasing the fastest over the interval x = 0 to x = 2 ?
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring ca 2
Answer:

Question 3.
Find all solutions of the equation x3 + 6x2 – 4x = 24.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring ca 3
Answer:

Question 4.
The table shows the distances you travel over a 6-hour period. Create an equation that models the distance traveled as a function of the number of hours.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring ca 4
Answer:

Question 5.
Consider the equation y = – \(\frac{1}{3}\)x + 2.
a. Graph the equation in a coordinate plane.
b. Does the equation represent a linear or nonlinear function?
c. Is the domain discrete or continuous?
Answer:

Question 6.
Which expressions are equivalent to -2x + 15x2 – 8?
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring ca 6
Answer:

Question 7.
The graph shows the function f(x) = 2(3)x.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring ca 7
a. Is the function increasing or decreasing for increasing values of x?
b. Identify any x- and y-intercepts.
Answer:

Question 8.
Which polynomial represents the product of 2x – 4 and x2 + 6x – 2?
A. 2x3 + 8x2 – 4x + 8
B. 2x3 + 8x2 – 28x + 8
C. 2x3 + 8
D. 2x3 – 24x – 2
Answer:

Question 9.
You are playing miniature golf on the hole shown.
Big Ideas Math Answers Algebra 1 Chapter 7 Polynomial Equations and Factoring ca 9
a. Write a polynomial that represents the area of the golf hole.
b. Write a polynomial that represents the perimeter of the golf hole.
c. Find the perimeter of the golf hole when the area is 216 square feet.
Answer:

Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions

Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions

Need instant homework help or assignment help while solving the Algebra 1 Questions? Then, Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions guide is the perfect one to overcome your hurdles. Based on the common core standards, our subject experts prepared this BIM Math Algebra 1 Solution key for Chapter 4 Writing Linear Functions. You can observe step-by-step explanations for all questions covered on Ch 5 Big Ideas Math Algebra 1 Answers Pdf. With the help of BIM Algebra 1 Ch 5 Writing Linear Functions Solutions pdf, students can easily understand the concepts & learn efficiently for their examinations.

Big Ideas Math Book Algebra 1 Solutions for Chapter 4 Writing Linear Functions

Utilize the Big Ideas Math Algebra 1 Textbook Solutions for Ch 4 Writing Linear Functions concepts and practice to the next level. Students can attempt any type of questions asked from chapter 4 in the examination after practicing the questions covered in the Big Ideas Math Algebra 1 Chapter 4 Writing Linear Functions Answer Key. Access the links provided below and solve the respective exercise of Ch 4 Algebra 1 BIM Textbook Solutions. Also, you can test your knowledge of the 4th chapter writing linear functions topics by referring to this ultimate Big Ideas Math Algebra 1 Chapter 4 Writing Linear Functions Answers Guide.

Writing Linear Functions Maintaining Mathematical Proficiency

Use the graph to answer the question.

Question 1.
What ordered pair corresponds to point G?

Question 2.
What ordered pair corresponds to point D?

question 3.
Which point is located in Quadrant I?

Question 4.
Which point is located in Quadrant IV?

Solve the equation for y.

Question 5.
x – y = 5

Question 6.
6x + 3y = -1

Question 7.
0 = 2y – 8x + 10

Question 8.
-x + 4y – 28 = 0

Question 9.
2y + 1 – x = 7x

Question 10.
y – 4 = 3x + 5y

Question 11.
ABSTRACT REASONING
Both coordinates of the point (x, y) are multiplied by a negative number. How does this change the location of the point? Be sure to consider points originally located in all four quadrants.

Writing Linear Functions Mathematical Practices

Monitoring Progress

Question 1.
You work 37\(\frac{1}{2}\) hours and earn $352.50. What is your hourly wage?

Question 2.
You drive 1244.5 miles and use 47.5 gallons of gasoline. What is your car’s gas mileage (in miles per gallon)?

Question 3.
You drive 236 miles in 4.6 hours. At the same rate, how long will it take you to drive 450 miles?

Lesson 4.1 Writing Equations in Slope-Intercept Form

Essential Question

Given the graph of a linear function, how can you write an equation of the line?

EXPLORATION 1
Writing Equations in Slope-Intercept Form
Work with a partner.

  • Find the slope and y-intercept of each line.
  • Write an equation of each line in slope-intercept form.
  • Use a graphing calculator to verify your equation.

Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 1

EXPLORATION 2
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 2
Mathematical Modeling
Work with a partner. The graph shows the cost of a smartphone plan.
a. What is the y-intercept of the line? Interpret the y-intercept in the context of the problem.
b. Approximate the slope of the line. Interpret the slope in the context of the problem.
c. Write an equation that represents the cost as a function of data usage.
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 3

Communicate Your Answer

Question 3.
Given the graph of a linear function, how can you write an equation of the line?

Question 4.
Give an example of a graph of a linear function that is different from those above. Then use the graph to write an equation of the line.

4.1 Lesson

Monitoring Progress

Write an equation of the line with the given slope and y-intercept.

Question 1.
slope = 7; y-intercept = 2

Question 2.
slope = \(\frac{1}{3}\) ; y-intercept = -1

Write an equation of the line in slope-intercept form.

Question 3.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 4

Question 4.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 5

Question 5.
Write an equation of the line that passes through (0, -2) and (4, 10).

Question 6.
Write a linear function g with the values g(0) = 9 and g(8) = 7.

Monitoring Progress

Question 7.
The corresponding data for electricity generated by hydropower are 248 million megawatt hours in 2007 and 277 million megawatt hours in 2012. Write a linear model that represents the number of megawatt hours generated by hydropower as a function of the number of years since 2007.

Writing Equations in Slope-Intercept Form 4.1 Exercises

Question 1.
COMPLETE THE SENTENCE
A linear function that models a real-life situation is called a __________.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 1

Question 2.
WRITING
Explain how you can use slope-intercept form to write an equation of a line given its slope and y-intercept.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, write an equation of the line with the given slope and y-intercept.

Question 3.
slope: 2
y-intercept: 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 3

Question 4.
slope: 0
y-intercept: 5
Answer:

Question 5.
slope: -3
y-intercept: 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 5

Question 6.
slope: -7
y-intercept: 1
Answer:

Question 7.
slope: \(\frac{2}{3}\)
y-intercept: -8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 7

Question 8.
slope: –\(\frac{3}{4}\)
y-intercept: -6
Answer:

In Exercises 9–12, write an equation of the line in slope-intercept form.

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 9

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 7
Answer:

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 11

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 9
Answer:

In Exercises 13–18, write an equation of the line that passes through the given points.

Question 13.
(3, 1), (0, 10)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 13

Question 14.
(2, 7), (0, -5)
Answer:

Question 15.
(2, -4), (0, -4)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 15

Question 16.
(-6, 0), (0, -24)
Answer:

Question 17.
(0, 5), (-1.5, 1)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 17

Question 18.
(0, 3), (-5, 2.5)
Answer:

In Exercises 19–24, write a linear function f with the given values.

Question 19.
f(0) = 2, f(2) = 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 19

Question 20.
f(0) = 7, f(3) = 1
Answer:

Question 21.
f(4) = -3, f(0) = -2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 21

Question 22.
f(5) = -1, f(0) = -5
Answer:

Question 23.
f(-2) = 6, f(0) = -4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 23

Question 24.
f(0) = 3, f(-6) = 3
Answer:

In Exercises 25 and 26, write a linear function f with the given values.

Question 25.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 25

Question 26.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 11
Answer:

Question 27.
ERROR ANALYSIS
Describe and correct the error in writing an equation of the line with a slope of 2 and a y-intercept of 7.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 27

Question 28.
ERROR ANALYSIS
Describe and correct the error in writing an equation of the line shown.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 13
Answer:

Question 29.
MODELING WITH MATHEMATICS
In 1960, the world record for the men’s mile was 3.91 minutes. In 1980, the record time was 3.81 minutes.
a. Write a linear model that represents the world record (in minutes) for the men’s mile as a function of the number of years since 1960.
b. Use the model to estimate the record time in 2000 and predict the record time in 2020.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 29

Question 30.
MODELING WITH MATHEMATICS
A recording studio charges musicians an initial fee of $50 to record an album. Studio time costs an additional $75 per hour.
a. Write a linear model that represents the total cost of recording an album as a function of studio time (in hours).
b. Is it less expensive to purchase 12 hours of recording time at the studio or a $750 music software program that you can use to record on your own computer? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 14
Answer:

Question 31.
WRITING
A line passes through the points (0, -2) and (0, 5). Is it possible to write an equation of the line in slope-intercept form? Justify your answer.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 31

Question 32.
THOUGHT PROVOKING
Describe a real-life situation involving a linear function whose graph passes through the points.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 15
Answer:

Question 33.
REASONING
Recall that the standard form of a linear equation is Ax + By = C. Rewrite this equation in slope-intercept form. Use your answer to find the slope and y-intercept of the graph of the equation -6x + 5y = 9.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 33

Question 34.
MAKING AN ARGUMENT
Your friend claims that given f(0) and any other value of a linear function f, you can write an equation in slope-intercept form that represents the function. Your cousin disagrees, claiming that the two points could lie on a vertical line. Who is correct? Explain.
Answer:

Question 35.
ANALYZING A GRAPH
Line ℓ is a reflection in the x-axis of line k. Write an equation that represents line k.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 35

Question 36.
HOW DO YOU SEE IT?
The graph shows the approximate U.S. box office revenues (in billions of dollars) from 2000 to 2012, where x = 0 represents the year 2000.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 17
a. Estimate the slope and y-intercept of the graph.
b. Interpret your answers in part (a) in the context of the problem.
c. How can you use your answers in part (a) to predict the U.S. box office revenue in 2018?
Answer:

Question 37.
ABSTRACT REASONING
Show that the equation of the line that passes through the points (0, b) and (1, b + m) is y = mx + b. Explain how you can be sure that the point (-1, b – m) also lies on the line.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 37

Maintaining Mathematical Proficiency

Solve the equation. (Section 1.3)

Question 38.
3(x – 15) = x + 11
Answer:

Question 39.
-4y – 10 = 4(y – 3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 39

Question 40.
2(3d + 3) = 7 + 6d
Answer:

Question 41.
-5(4 – 3n) = 10(n – 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 41

Use intercepts to graph the linear equation. (Section 3.4)

Question 42.
-4x + 2y = 16
Answer:

Question 43.
3x + 5y = -15
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 43

Question 44.
x – 6y = 24
Answer:

Question 45.
-7x – 2y = -21
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.1 Question 45

Lesson 4.2 Writing Equations in Point-Slope Form

Essential Question
How can you write an equation of a line when you are given the slope and a point on the line?

EXPLORATION 1
Writing Equations of Lines
Work with a partner.

  • Sketch the line that has the given slope and passes through the given point.
  • Find the y-intercept of the line.
  • Write an equation of the line.

Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 18

EXPLORATION 2
Writing a Formula
Work with a partner.

The point (x1, y1) is a given point on a nonvertical line. The point (x, y) is any other point on the line. Write an equation that represents the slope m of the line. Then rewrite this equation by multiplying each side by the difference of the x-coordinates to obtain the point-slope form of a linear equation.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 19

EXPLORATION 3
Writing an Equation
Work with a partner.

For four months, you have saved $25 per month. You now have $175 in your savings account.
a. Use your result from Exploration 2 to write an equation that represents the balance A after t months.
b. Use a graphing calculator to verify your equation.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 20
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 21

Communicate Your Answer

Question 4.
How can you write an equation of a line when you are given the slope and a point on the line?

Question 5.
Give an example of how to write an equation of a line when you are given the slope and a point on the line. Your example should be different from those above.

4.2 Lesson

Monitoring Progress

Write an equation in point-slope form of the line that passes through the given point and has the given slope.

Question 1.
(3, -1); m = -2

Question 2.
(4, 0); m = – \(\frac{2}{3}\)

Write an equation in slope-intercept form of the line that passes through the given points.

Question 3.
(1, 4), (3, 10)

Question 4.
(-4, -1), (8, -4)

Question 5.
Write a linear function g with the values g(2) = 3 and g(6) = 5.

Question 6.
You pay an installation fee and a monthly fee for Internet service. The table shows the total cost for different numbers of months. Can the situation be modeled by a linear equation? Explain. If possible, write a linear model that represents the total cost as a function of the number of months.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 22

Writing Equations in Point-Slope Form 4.2 Exercises

Vocabulary and Core Concept Check

Question 1.
USING STRUCTURE
Without simplifying, identify the slope of the line given by the equation y – 5 = -2(x + 5). Then identify one point on the line.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 1

Question 2.
WRITING
Explain how you can use the slope formula to write an equation of the line that passes through (3, -2) and has a slope of 4.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3−10, write an equation in point-slope form of the line that passes through the given point and has the given slope.

Question 3.
(2, 1); m = 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 3

Question 4.
(3, 5); m = -1
Answer:

Question 5.
(7, -4); m = -6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 5

Question 6.
(-8, -2); m = 5
Answer:

Question 7.
(9, 0); m = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 7

Question 8.
(0, 2); m = 4
Answer:

Question 9.
(-6, 6); m – \(\frac{3}{2}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 9

Question 10.
(5, -12); m = –\(\frac{2}{5}\)
Answer:

In Exercises 11−14, write an equation in slope-intercept form of the line shown.

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 23
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 11

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 24
Answer:

Question 13.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 25
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 13

Question 14.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 26
Answer:

In Exercises 15−20, write an equation in slope-intercept form of the line that passes through the given points.

Question 15.
(7, 2), (2, 12)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 15

Question 16.
(6, -2), (12, 1)
Answer:

Question 17.
(6, -1), (3, -7)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 17

Question 18.
(-2, 5), (-4, -5)
Answer:

Question 19.
(1, -9), (-3, -9)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 19

Question 20.
(-5, 19), (5, 13)
Answer:

In Exercises 21−26, write a linear function f with the given values.

Question 21.
f(2) = -2, f(1) = 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 21

Question 22.
f(5) = 7, f(-2) = 0
Answer:

Question 23.
f(-4) = 2, f(6) = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 23

Question 24.
f(-10) = 4, f(-2) = 4
Answer:

Question 25.
f(-3) = 1, f(13) = 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 25

Question 26.
f(-9) = 10, f(-1) = -2
Answer:

In Exercises 27−30, tell whether the data in the table can be modeled by a linear equation. Explain. If possible, write a linear equation that represents y as a function of x.

Question 27.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 27
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 27

Question 28.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 28
Answer:

Question 29.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 29
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 29

Question 30.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 30
Answer:

Question 31.
ERROR ANALYSIS
Describe and correct the error in writing a linear function g with the values g(5) = 4 and g(3) = 10.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 31
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 31

Question 32.
ERROR ANALYSIS
Describe and correct the error in writing an equation of the line that passes through the points (1, 2) and (4, 3).
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 32
Answer:

Question 33.
MODELING WITH MATHEMATICS
You are designing a sticker to advertise your band. A company charges $225 for the first 1000 stickers and $80 for each additional 1000 stickers.
a. Write an equation that represents the total cost (in dollars) of the stickers as a function of the number (in thousands) of stickers ordered.
b. Find the total cost of 9000 stickers.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 33

Question 34.
MODELING WITH MATHEMATICS
You pay a processing fee and a daily fee to rent a beach house. The table shows the total cost of renting the beach house for different numbers of days.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 33
a. Can the situation be modeled by a linear equation? Explain.
b. What is the processing fee? the daily fee?
c. You can spend no more than $1200 on the beach house rental. What is the maximum number of days you can rent the beach house?
Answer:

Question 35.
WRITING
Describe two ways to graph the equation y – 1 = \(\frac{3}{2}\)(x – 4).
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 35.2

Question 36.
THOUGHT PROVOKING
The graph of a linear function passes through the point (12, -5) and has a slope of \(\frac{2}{5}\). Represent this function in two other ways.
Answer:

Question 37.
REASONING
You are writing an equation of the line that passes through two points that are not on the y-axis. Would you use slope-intercept form or point-slope form to write the equation? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 37

Question 38.
HOW DO YOU SEE IT? The graph shows two points that lie on the graph of a linear function.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 34
a. Does the y-intercept of the graph of the linear function appear to be positive or negative? Explain.
b. Estimate the coordinates of the two points. How can you use your estimates to confirm your answer in part (a)?
Answer:

Question 39.
CONNECTION TO TRANSFORMATIONS
Compare the graph of y = 2x to the graph of y – 1 = 2(x + 3). Make a conjecture about the graphs of y = mx and y – k = m(x – h).
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 39

Question 40.
COMPARING FUNCTIONS
Three siblings each receive money for a holiday and then spend it at a constant weekly rate. The graph describes Sibling A’s spending, the table describes Sibling B’s spending, and the equation y = -22.5x + 90 describes Sibling C’s spending. The variable y represents the amount of money left after x weeks.
Big Ideas Math Answer Key Algebra 1 Chapter 4 Writing Linear Functions 35
a. Which sibling received the most money? the least money?
b. Which sibling spends money at the fastest rate? the slowest rate?
c. Which sibling runs out of money first? last?
Answer:

Maintaining Mathematical Proficiency

Write the reciprocal of the number.

Question 41.
5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 41

Question 42.
-8
Answer:

Question 43.
–\(\frac{2}{7}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.2 Question 43

Question 44.
\(\frac{3}{2}\)
Answer:

Lesson 4.3 Writing Equations of Parallel and Perpendicular Lines

Essential Equation
How can you recognize lines that are parallel or perpendicular?

EXPLORATION 1
Recognizing Parallel Lines
Work with a partner.
Write each linear equation in slope-intercept form. Then use a graphing calculator to graph the three equations in the same square viewing window. (The graph of the first equation is shown.) Which two lines appear parallel? How can you tell?
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 36

EXPLORATION 2
Recognizing Perpendicular Lines
Work with a partner.
Write each linear equation in slope-intercept form. Then use a graphing calculator to graph the three equations in the same square viewing window. (The graph of the first equation is shown.) Which two lines appear perpendicular? How can you tell?
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 37

Communicate Your Answer

Question 3.
How can you recognize lines that are parallel or perpendicular?

Question 4.
Compare the slopes of the lines in Exploration 1. How can you use slope to determine whether two lines are parallel? Explain your reasoning.

Question 5.
Compare the slopes of the lines in Exploration 2. How can you use slope to determine whether two lines are perpendicular? Explain your reasoning.

4.3 Lesson

Question 1.
Line a passes through (-5, 3) and (-6, -1). Line b passes through (3, -2) and (2, -7). Are the lines parallel? Explain.

Question 2.
Write an equation of the line that passes through (-4, 2) and is parallel to the line y = \(\frac{1}{4}\)x + 1

Monitoring Progress

Question 3.
Determine which of the lines, if any, are parallel or perpendicular. Explain.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 38

Question 4.
Write an equation of the line that passes through (-3, 5) and is perpendicular to the line y = -3x – 1.

Question 5.
In Example 5, a boat is traveling parallel to the shoreline and passes through (9, 3). Write an equation that represents the path of the boat.

Writing Equations of Parallel and Perpendicular Lines 4.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Two distinct nonvertical lines that have the same slope are ____.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 1

Question 2.
VOCABULARY
Two lines are perpendicular. The slope of one line is –\(\frac{5}{7}\). What is the slope of the other line? Justify your answer.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, determine which of the lines, if any, are parallel. Explain.

Question 3.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 39
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 3

Question 4.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 40
Answer:

Question 5.
Line a passes through (-1, -2) and (1, 0).
Line b passes through (4, 2) and (2, -2).
Line c passes through (0, 2) and (-1, 1).
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 5

Question 6.
Line a passes through (-1, 3) and (1, 9).
Line b passes through (-2, 12) and (-1, 14).
Line c passes through (3, 8) and (6, 10).
Answer:

Question 7.
Line a: 4y + x = 8
Line b: 2y + x = 4
Line c: 2y = -3x + 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 7.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 7.2

Question 8.
Line a: 3y – x = 6
Line b: 3y = x + 18
Line c: 3y – 2x = 9
Answer:

In Exercises 9–12, write an equation of the line that passes through the given point and is parallel to the given line.

Question 9.
(-1, 3); y = 2x + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 9

Question 10.
(1, 2); y = -5x + 4
Answer:

Question 11.
(18, 2); 3y – x = -12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 11

Question 12.
(2, -5); 2y = 3x + 10
Answer:

In Exercises 13–18, determine which of the lines, if any, are parallel or perpendicular. Explain.

Question 13.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 41
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 13

Question 14.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 42
Answer:

Question 15.
Line a passes through (-2, 1) and (0, 3).
Line b passes through (4, 1) and (6, 4).
Line c passes through (1, 3) and (4, 1).
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 15

Question 16.
Line a passes through (2, 10) and (4, 13).
Line b passes through (4, 9) and (6, 12).
Line c passes through (2, 10) and (4, 9).
Answer:

Question 17.
Line a: 4x – 3y = 2
Line b: y = \(\frac{4}{3}\)x + 2
Line c: 4y + 3x = 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 17

Question 18.
Line a: y = 6x – 2
Line b: 6y = -x
Line c: y + 6x = 1
Answer:

In Exercises 19–22, write an equation of the line that passes through the given point and is perpendicular to the given line.

Question 19.
(7, 10); y = \(\frac{1}{2}\)x – 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 19

Question 20.
(-4, -1); y = \(\frac{4}{3}\)x + 6
Answer:

Question 21.
(-3, 3); 2y = 8x – 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 21

Question 22.
(8, 1); 2y + 4x = 12
Answer:

In Exercises 23 and 24, write an equation of the line that passes through the given point and is (a) parallel and(b) perpendicular to the given line.

Question 23.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 43
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 23

Question 24.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 44
Answer:

Question 25.
ERROR ANALYSIS
Describe and correct the error in writing an equation of the line that passes through (1, 3) and is parallel to the line y = \(\frac{1}{4}\)x + 2.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 45
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 25

Question 26.
ERROR ANALYSIS
Describe and correct the error in writing an equation of the line that passes through (4, -5) and is perpendicular to the line y = \(\frac{1}{3}\)x + 5.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 46
Answer:

Question 27.
MODELING WITH MATHEMATICS
A city water department is proposing the construction of a new water pipe, as shown. The new pipe will be perpendicular to the old pipe. Write an equation that represents the new pipe.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 47
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 27

Question 28.
MODELING WITH MATHEMATICS
A parks and recreation department is constructing a new bike path. The path will be parallel to the railroad tracks shown and pass through the parking area at the point (4, 5). Write an equation that represents the path.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 48
Answer:

Question 29.
MATHEMATICAL CONNECTIONS
The vertices of a quadrilateral are A(2, 2), B(6, 4), C(8, 10), and D(4, 8).
a. Is quadrilateral ABCD a parallelogram? Explain.
b. Is quadrilateral ABCD a rectangle? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 29.2

Question 30.
USING STRUCTURE
For what value of a are the graphs of 6y = -2x + 4 and 2y = ax – 5 parallel? perpendicular?
Answer:

Question 31.
MAKING AN ARGUMENT
A hockey puck leaves the blade of a hockey stick, bounces off a wall, and travels in a new direction, as shown. Your friend claims the path of the puck forms a right angle. Is your friend correct? Explain.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 49
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 31

Question 32.
HOW DO YOU SEE IT?
A softball academy charges students an initial registration fee plus a monthly fee. The graph shows the total amounts paid by two students over a 4-month period. The lines are parallel.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 50
a. Did one of the students pay a greater registration fee? Explain.
b. Did one of the students pay a greater monthly fee? Explain.
Answer:

REASONING
In Exercises 33–35, determine whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 33.
Two lines with positive slopes are perpendicular.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 33

Question 34.
A vertical line is parallel to the y-axis.
Answer:

Question 35.
Two lines with the same y-intercept are perpendicular.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 35

Question 36.
THOUGHT PROVOKING
You are designing a new logo for your math club. Your teacher asks you to include at least one pair of parallel lines and at least one pair of perpendicular lines. Sketch your logo in a coordinate plane. Write the equations of the parallel and perpendicular lines.
Answer:

Maintaining Mathematical Proficiency

Determine whether the relation is a function. Explain. (Section 3.1)

Question 37.
(3, 6), (4, 8), (5, 10), (6, 10), (7, 14)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.3 Question 37

Question 38.
(-1, 6), (1, 4), (-1, 2), (1, 6), (-1, 5)
Answer:

Writing Linear Functions Study Skills: Getting Actively Involved in Class

4.1–4.3 What Did You Learn

Core Vocabulary
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 51

Core Concepts

Section 4.1
Using Slope-Intercept Form, p. 176

Section 4.2
Using Point-Slope Form, p. 182

Section 4.3
Parallel Lines and Slopes, p. 188
Perpendicular Lines and Slopes, p. 189

Mathematical Practices

Question 1.
How can you explain to yourself the meaning of the graph in Exercise 36 on page 180?

Question 2.
How did you use the structure of the equations in Exercise 39 on page 186 to make a conjecture?

Question 3.
How did you use the diagram in Exercise 31 on page 192 to determine whether your friend was correct?

Study Skills

Getting Actively Involved in Class

If you do not understand something at all and do not even know how to phrase a question, just ask for clarification. You might say something like, “Could you please explain the steps in this problem one more time?”If your teacher asks for someone to go up to the board, volunteer. The student at the board often receives additional attention and instruction to complete the problem.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 52

Writing Linear Functions 4.1 – 4.3

4.1 – 4.3 Quiz

Write an equation of the line in slope-intercept form. (Section 4.1)

Question 1.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 53

Question 2.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 54

Question 3.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 55

Write an equation in point-slope form of the line that passes through the given points. (Section 4.2)

Question 4.
(-2, 5), (1, -1)

Question 5.
(-3, -2), (2, -1)

Question 6.
(1, 0), (4, 4)

Write a linear function f with the given values. (Section 4.1 and Section 4.2)

Question 7.
f(0) = 2, f(5) = -3

Question 8.
f(-1) = -6, f(4) = -6

Question 9.
f(-3) = -2, f(-2) = 3

Determine which of the lines, if any, are parallel or perpendicular. Explain. (Section 4.3)

Question 10.
Line a passes through (-2, 2) and (2, 1).
Line b passes through (1, -8) and (3, 0).
Line c passes through (-4, -3) and (0, -2).

Question 11.
Line a: 2x + 6y = -12
Line b: y = \(\frac{3}{2}\)x – 5
Line c : 3x – 2y = -4

Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line. (Section 4.3)

Question 12.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 56

Question 13.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 57

Question 14.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 58

Question 15.
A website hosting company charges an initial fee of $48 to set up a website. The company charges $44 per month to maintain the website. (Section 4.1)
a. Write a linear model that represents the total cost of setting up and maintaining a website as a function of the number of months it is maintained.
b. Find the total cost of setting up a website and maintaining it for 6 months.
c. A different website hosting company charges $62 per month to maintain a website, but there is no initial set-up fee. You have $620. At which company can you set up and maintain a website for the greatest amount of time? Explain.

Question 16.
The table shows the amount of water remaining in a water tank as it drains. Can the situation be modeled by a linear equation? Explain. If possible, write a linear model that represents the amount of water remaining in the tank as a function of time. (Section 4.2)
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 59

Lesson 4.4 Scatter Plots and Lines of Fit

Essential Question

How can you use a scatter plot and a line of fit to make conclusions about data?
A scatter plot is a graph that shows the relationship between two data sets. The two data sets are graphed as ordered pairs in a coordinate plane.

EXPLORATION 1
Finding a Line of Fit
Work with a partner.
A survey was taken of 179 married couples. Each person was asked his or her age. The scatter plot shows the results.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 60
a. Draw a line that approximates the data. Write an equation of the line. Explain the method you used.
b. What conclusions can you make from the equation you wrote? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 61

EXPLORATION 2
Work with a partner.
The scatter plot shows the median ages of American women at their first marriage for selected years from 1960 through 2010.
a. Draw a line that approximates the data. Write an equation of the line. Explain the method you used.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 62
b. What conclusions can you make from the equation you wrote?
c. Use your equation to predict the median age of American women at their first marriage in the year 2020.

Communicate Your Answer

Question 3.
How can you use a scatter plot and a line of fit to make conclusions about data?

Question 4.
Use the Internet or some other reference to find a scatter plot of real-life data that is different from those given above. Then draw a line that approximates the data and write an equation of the line. Explain the method you used.

4.4 Lesson

Monitoring Progress

Question 1.
How many calories are in the smoothie that contains 51 grams of sugar?

Question 2.
How many grams of sugar are in the smoothie that contains 250 calories?

Make a scatter plot of the data. Tell whether the data show a positive, a negative, or no correlation.

Question 3.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 63

Question 4.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 64

Question 5.
The following data pairs show the monthly income x (in dollars) and the monthly car payment y (in dollars) of six people: (2100, 410), (1650, 315), (1950, 405), (1500, 295), (2250, 440), and (1800, 375). Write an equation that models the monthly car payment as a function of the monthly income. Interpret the slope and y-intercept of the line of fit.

Scatter Plots and Lines of Fit 4.4 Exercises

Question 1.
COMPLETE THE SENTENCE
When data show a positive correlation, the dependent variable tends to ____________ as the independent variable increases.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 1

Question 2.
VOCABULARY
What is a line of fit?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, use the scatter plot to fill in the missing coordinate of the ordered pair.

Question 3.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 65
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 3

Question 4.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 203
Answer:

Question 5.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 67
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 5

Question 6.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 69
Answer:

Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 70

Question 7.
INTERPRETING A SCATTER PLOT
The scatter plot shows the hard drive capacities (in gigabytes) and the prices (in dollars) of 10 laptops.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 71
a. What is the price of the laptop with a hard drive capacity of 8 gigabytes?
b. What is the hard drive capacity of the $1200 laptop?
c. What tends to happen to the price as the hard drive capacity increases?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 7

Question 8.
INTERPRETING A SCATTER PLOT
The scatter plot shows the earned run averages and the winning percentages of eight pitchers on a baseball team.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 72
a. What is the winning percentage of the pitcher with an earned run average of 4.2?
b. What is the earned run average of the pitcher with a winning percentage of 0.33?
c. What tends to happen to the winning percentage as the earned run average increases?
Answer:

In Exercises 9–12, tell whether x and y show a positive, a negative, or no correlation.

Question 9.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 73
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 9

Question 10.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 74
Answer:

Question 11.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 75
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 11

Question 12.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 76
Answer:

In Exercises 13 and 14, make a scatter plot of the data. Tell whether x and y show a positive, a negative, or no correlation.

Question 13.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 77
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 13

Question 14.
Big Ideas Math Algebra 1 Solutions Chapter 4 Writing Linear Functions 78
Answer:

Question 15.
MODELING WITH MATHEMATICS
The table shows the world birth rates y (number of births per 1000 people) x years since 1960.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 79
a. Write an equation that models the birthrate as a function of the number of years since 1960.
b. Interpret the slope and y-intercept of the line of fit.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 15.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 15.2

Question 16.
MODELING WITH MATHEMATICS
The table shows the total earnings y (in dollars) of a food server who works x hours.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 80
a. Write an equation that models the server’s earnings as a function of the number of hours the server works.
b. Interpret the slope and y-intercept of the line of fit.
Answer:

Question 17.
OPEN-ENDED
Give an example of a real-life data set that shows a negative correlation.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 17

Question 18.
MAKING AN ARGUMENT
Your friend says that the data in the table show a negative correlation because the dependent variable y is decreasing. Is your friend correct? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 81
Answer:

Question 19.
USING TOOLS
Use a ruler or a yardstick to find the heights and arm spans of five people.
a. Make a scatter plot using the data you collected. Then draw a line of fit for the data.
b. Interpret the slope and y-intercept of the line of fit.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 19.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 19.2

Question 20.
THOUGHT PROVOKING
A line of fit for a scatter plot is given by the equation y = 5x + 20. Describe a real-life data set that could be represented by the scatter plot.
Answer:

Question 21.
WRITING
When is data best displayed in a scatter plot, rather than another type of display, such as a bar graph or circle graph?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 21

Question 22.
HOW DO YOU SEE IT?
The scatter plot shows part of a data set and a line of fit for the data set. Four data points are missing. Choose possible coordinates for these data points.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 82
Answer:

Question 23.
REASONING
A data set has no correlation. Is it possible to find a line of fit for the data? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 23

Question 24.
ANALYZING RELATIONSHIPS
Make a scatter plot of the data in the tables. Describe the relationship between the variables. Is it possible to fit a line to the data? If so, write an equation of the line. If not, explain why.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 83
Answer:

Maintaining Mathematical Proficiency

Evaluate the function when x = −3, 0, and 4. (Section 3.3)

Question 25.
g(x) = 6x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 25

Question 26.
h(x) = -10x
Answer:

Question 27.
f(x) = 5x – 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.4 Question 27

Question 28.
v(x) = 14 – 3x
Answer:

Lesson 4.5 Analyzing Lines of Fit

Essential Question
How can you analytically find a line of best fit for a scatter plot?

EXPLORATION 1
Finding a Line of Best Fit
Work with a partner.

The scatter plot shows the median ages of American women at their first marriage for selected years from 1960 through 2010. In Exploration 2 in Section 4.4, you approximated a line of fit graphically. To find the line of best fit, you can use a computer, spreadsheet, or graphing calculator that has a linear regression feature.
a. The data from the scatter plot is shown in the table. Note that 0, 5, 10, and so on represent the numbers of years since 1960. What does the ordered pair (25, 23.3) represent?
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 84
b. Use the linear regression feature to find an equation of the line of best fit. You should obtain results such as those shown below.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 85
c. Write an equation of the line of best fit. Compare your result with the equation you obtained in Exploration 2 in Section 4.4.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 85.1

Communicate Your Answer

Question 2.
How can you analytically find a line of best fit for a scatter plot?

Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 86
Question 3.
The data set relates the number of chirps per second for striped ground crickets and the outside temperature in degrees Fahrenheit. Make a scatter plot of the data. Then find an equation of the line of best fit. Use your result to estimate the outside temperature when there are 19 chirps per second.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 87

4.5 Lesson

Monitoring Progress

Question 1.
The table shows the attendances y (in thousands) at an amusement park from 2005 to 2014, where x = 0 represents the year 2005. The equation y = -9.8x + 850 models the data. Is the model a good fit?
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 88

Question 2.
Use the data in Monitoring Progress Question 1.
(a) Use a graphing calculator to find an equation of the line of best fit. Then plot the data and graph the equation in the same viewing window.
(b) Identify and interpret the correlation coefficient.
(c) Interpret the slope and y-intercept of the line of best fit.

Question 3.
Refer to Monitoring Progress Question 2. Use the equation of the line of best fit to predict the attendance at the amusement park in 2017.

Question 4.
Is there a correlation between time spent playing video games and grade point average? If so, is there a causal relationship? Explain your reasoning.

Analyzing Lines of Fit 4.5 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
When is a residual positive? When is it negative?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 1

Question 2.
WRITING
Explain how you can use residuals to determine how well a line of fit models a data set.
Answer:

Question 3.
VOCABULARY
Compare interpolation and extrapolation.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 3

Question 4.
WHICH ONE DOESN’T BELONG?
Which correlation coefficient does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 89
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, use residuals to determine whether the model is a good fit for the data in the table. Explain.

Question 5.
y = 4x – 5
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 90
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 5.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 5.2

Question 6.
y = 6x + 4
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 91
Answer:

Question 7.
y = -1.3x + 1
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 92
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 7.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 7.2

Question 8.
y = -0.5x – 2
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 93
Answer:

Question 9.
ANALYZING RESIDUALS
The table shows the growth y (in inches) of an elk’s antlers during week x. The equation y = -0.7x + 6.8 models the data. Is the model a good fit? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 94
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 9

Question 10.
ANALYZING RESIDUALS
The table shows the approximate numbers y (in thousands) of movie tickets sold from January to June for a theater. In the table, x = 1 represents January. The equation y = 1.3x + 27 models the data. Is the model a good fit? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 95
Answer:

In Exercises 11–14, use a graphing calculator to find an equation of the line of best fit for the data. Identify and interpret the correlation coefficient.

Question 11.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 96
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 11

Question 12.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 97
Answer:

Question 13.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 98
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 13

Question 14.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 99
Answer:

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in interpreting the graphing calculator display.

Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 100

Question 15.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 101
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 15

Question 16.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 102
Answer:

Question 17.
MODELING WITH MATHEMATICS
The table shows the total numbers y of people who reported an earthquake x minutes after it ended.
a. Use a graphing calculator to find an equation of the line of best fit. Then plot the data and graph the equation in the same viewing window.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 103
b. Identify and interpret the correlation coefficient.
c. Interpret the slope and y-intercept of the line of best fit.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 17

Question 18.
MODELING WITH MATHEMATICS
The table shows the numbers y of people who volunteer at an animal shelter on each day x.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 104
a. Use a graphing calculator to find an equation of the line of best fit. Then plot the data and graph the equation in the same viewing window.
b. Identify and interpret the correlation coefficient.
c. Interpret the slope and y-intercept of the line of best fit.
Answer:

Question 19.
MODELING WITH MATHEMATICS
The table shows the mileages x (in thousands of miles) and the selling prices y (in thousands of dollars) of several used automobiles of the same year and model.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 105
a. Use a graphing calculator to find an equation of the line of best fit.
b. Identify and interpret the correlation coefficient.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 106
c. Interpret the slope and y-intercept of the line of best fit.
d. Approximate the mileage of an automobile that costs $15,500. e. Predict the price of an automobile with 6000 miles.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 19.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 19.2

Question 20.
MODELING WITH MATHEMATICS
The table shows the lengths x and costs y of several sailboats.
a. Use a graphing calculator to find an equation of the line of best fit.
b. Identify and interpret the correlation coefficient.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 107
c. Interpret the slope and y-intercept of the line of best fit.
d. Approximate the cost of a sailboat that is 20 feet long.
e. Predict the length of a sailboat that costs $147,000.
Answer:

In Exercises 21–24, tell whether a correlation is likely in the situation. If so, tell whether there is a causal relationship. Explain your reasoning.

Question 21.
the amount of time spent talking on a cell phone and the remaining battery life
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 21

Question 22.
the height of a toddler and the size of the toddler’s vocabulary
Answer:

Question 23.
the number of hats you own and the size of your head
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 23

Question 24.
the weight of a dog and the length of its tail
Answer:

Question 25.
OPEN-ENDED
Describe a data set that has a strong correlation but does not have a causal relationship.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 25

Question 26.
HOW DO YOU SEE IT?
Match each graph with its correlation coefficient. Explain your reasoning.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 108
Answer:

Question 27.
ANALYZING RELATIONSHIPS
The table shows the grade point averages y of several students and the numbers x of hours they spend watching television each week.
a. Use a graphing calculator to find an equation of the line of best fit. Identify and interpret the correlation coefficient.
b. Interpret the slope and y-intercept of the line of best fit.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 109
c. Another student watches about 14 hours of television each week. Approximate the student’s grade point average.
d. Do you think there is a causal relationship between time spent watching television and grade point average? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 27.2

Question 28.
MAKING AN ARGUMENT
A student spends 2 hours watching television each week and has a grade point average of 2.4. Your friend says including this information in the data set in Exercise 27 will weaken the correlation. Is your friend correct? Explain.
Answer:

Question 29.
USING MODELS
Refer to Exercise 17.
a. Predict the total numbers of people who reported an earthquake 9 minutes and 15 minutes after it ended.
b. The table shows the actual data. Describe the accuracy of your extrapolations in part (a).
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 110
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 29.1

Question 30.
THOUGHT PROVOKING
A data set consists of the numbers x of people at Beach 1 and the numbers y of people at Beach 2 recorded daily for 1 week. Sketch a possible graph of the data set. Describe the situation shown in the graph and give a possible correlation coefficient. Determine whether there is a causal relationship. Explain.
Answer:

Question 31.
COMPARING METHODS
The table shows the numbers y (in billions) of text messages sent each year in a five-year period, where x = 1 represents the first year in the five-year period.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 111
a. Use a graphing calculator to find an equation of the line of best fit. Identify and interpret the correlation coefficient.
b. Is there a causal relationship? Explain your reasoning.
c. Calculate the residuals. Then make a scatter plot of the residuals and interpret the results.
d. Compare the methods you used in parts (a) and (c) to determine whether the model is a good fit. Which method do you prefer? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 31.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 31.2
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 31.3

Maintaining Mathematical Proficiency

Determine whether the table represents a linear or nonlinear function. Explain. (Section 3.2)

Question 32.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 112
Answer:

Question 33.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 113
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.5 Question 33

Lesson 4.6 Arithmetic Sequences

Essential Question
How can you use an arithmetic sequence to describe a pattern?

An arithmetic sequence is an ordered list of numbers in which the difference between each pair of consecutive terms, or numbers in the list, is the same.

EXPLORATION 1
Describing a Pattern
Work with a partner.
Use the figures to complete the table. Plot the points given by your completed table. Describe the pattern of the y-values.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 114

Communicate Your Answer

Question 2.
How can you use an arithmetic sequence to describe a pattern? Give an example from real life.

Question 3.
In chemistry, water is called H2O because each molecule of water has two hydrogen atoms and one oxygen atom. Describe the pattern shown below. Use the pattern to determine the number of atoms in 23 molecules.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 115

4.6 Lesson

Monitoring Progress

Write the next three terms of the arithmetic sequence.

Question 1.
-12, 0, 12, 24, . . .

Question 2.
0.2, 0.6, 1, 1.4, . . .

Question 3.
4, 3\(\frac{3}{4}\), 3\(\frac{1}{2}\), 3\(\frac{1}{4}\)

Graph the arithmetic sequence. What do you notice?

Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 116

Question 4.
3, 6, 9, 12, . . .

Question 5.
4, 2, 0, -2, . . .

Question 6.
1, 0.8, 0.6, 0.4, . . .

Question 7.
Does the graph shown represent an arithmetic sequence? Explain.

Write an equation for the nth term of the arithmetic sequence. Then find a25.

Question 8.
4, 5, 6, 7, . . .

Question 9.
8, 16, 24, 32, . . .

Question 10.
1, 0, -1, -2, . . .

Question 11.
A carnival charges $2 for each game after you pay a $5 entry fee.
a. Write a function that represents the arithmetic sequence.
b. Graph the function.
c. How many games can you play when you take $29 to the carnival?
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 117

Arithmetic Sequences 4.6 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe the graph of an arithmetic sequence.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Consider the arithmetic sequence represented by the graph. Which is different? Find “both” answers.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 118
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, write the next three terms of the arithmetic sequence.

Question 3.
First term: 2
Common difference: 13
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 3

Question 4.
First term: 18
Common difference: −6
Answer:

In Exercises 5−10, find the common difference of the arithmetic sequence.

Question 5.
13, 18, 23, 28, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 5

Question 6.
175, 150, 125, 100, . . .
Answer:

Question 7.
-16, -12, -8, -4, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 7

Question 8.
4, 3\(\frac{2}{3}\), 3\(\frac{1}{3}\), 3, . . .
Answer:

Question 9.
6.5, 5, 3.5, 2, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 9

Question 10.
-16, -7, 2, 11, . . .
Answer:

In Exercises 11−16, write the next three terms of the arithmetic sequence.

Question 11.
19, 22, 25, 28, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 11

Question 12.
1, 12, 23, 34, . . .
Answer:

Question 13.
16, 21, 26, 31, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 13

Question 14.
60, 30, 0, -30, . . .
Answer:

Question 15.
1.3, 1, 0.7, 0.4, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 15

Question 16.
\(\frac{5}{6}\), \(\frac{2}{3}\), \(\frac{1}{2}\), \(\frac{1}{3}\), . . .
Answer:

In Exercises 17−22, graph the arithmetic sequence.

Question 17.
4, 12, 20, 28, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 17

Question 18.
-15, 0, 15, 30, . . .
Answer:

Question 19.
−1, −3, −5, −7, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 19

Question 20.
2, 19, 36, 53, . . .
Answer:

Question 21.
0, 41\(\frac{1}{2}\), 9, 13\(\frac{1}{2}\), . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 21

Question 22.
6, 5.25, 4.5, 3.75, . . .
Answer:

In Exercises 23−26, determine whether the graph represents an arithmetic sequence. Explain.

Question 23.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 119
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 23

Question 24.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 120
Answer:

Question 25.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 121
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 25

Question 26.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 122
Answer:

In Exercises 27−30, determine whether the sequence is arithmetic. If so, find the common difference.

Question 27.
13, 26, 39, 52, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 27

Question 28.
5, 9, 14, 20, . . .29.
Answer:

Question 29.
48, 24, 12, 6, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 29

Question 30.
87, 81, 75, 69, . . .
Answer:

Question 31.
FINDING A PATTERN
Write a sequence that represents the number of smiley faces in each group. Is the sequence arithmetic? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 123
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 31

Question 32.
FINDING A PATTERN
Write a sequence that represents the sum of the numbers in each roll. Is the sequence arithmetic? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 124
Answer:

In Exercises 33−38, write an equation for the nth term of the arithmetic sequence. Then find a10.

Question 33.
-5, -4, -3, -2, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 33

Question 34.
-6, -9, -12, -15, . . .
Answer:

Question 35.
\(\frac{1}{2}\), 1, 1\(\frac{1}{2}\), 2, ……
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 35

Question 36.
100, 110, 120, 130, ….
Answer:

Question 37.
10, 0, -10, -20, ……
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 37

Question 38.
\(\frac{3}{7}\), \(\frac{4}{7}\), \(\frac{5}{7}\), \(\frac{6}{7}\), ………
Answer:

Question 39.
ERROR ANALYSIS
Describe and correct the error in finding the common difference of the arithmetic sequence.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 125
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 39

Question 40.
ERROR ANALYSIS
Describe and correct the error in writing an equation for the nth term of the arithmetic sequence.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 126
Answer:

Question 41.
NUMBER SENSE
The first term of an arithmetic sequence is 3. The common difference of the sequence is 1.5 times the first term. Write the next three terms of the sequence. Then graph the sequence.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 41

Question 42.
NUMBER SENSE
The first row of a dominoes display has 10 dominoes. Each row after the first has two more dominoes than the row before it. Write the first five terms of the sequence that represents the number of dominoes in each row. Then graph the sequence.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 127
Answer:

REPEATED REASONING
In Exercises 43 and 44, (a) draw the next three figures in the sequence and (b) describe the 20th figure in the sequence.

Question 43.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 128
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 43

Question 44.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 129
Answer:

Question 45.
MODELING WITH MATHEMATICS
The total number of babies born in a country each minute after midnight January 1st can be estimated by the sequence shown in the table. (See Example 5.)
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 130
a. Write a function that represents the arithmetic sequence.
b. Graph the function.
c. Estimate how many minutes after midnight January 1st it takes for 100 babies to be born.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 45.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 45.2

Question 46.
MODELING WITH MATHEMATICS
The amount of money a movie earns each week after its release can be approximated by the sequence shown in the graph.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 131
a. Write a function that represents the arithmetic sequence.
b. In what week does the movie earn $16 million?
c. How much money does the movie earn overall?
Answer:

MATHEMATICAL CONNECTIONS
In Exercises 47 and 48, each small square represents 1 square inch. Determine whether the areas of the figures form an arithmetic sequence. If so, write a function f that represents the arithmetic sequence and find f(30).

Question 47.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 132
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 47

Question 48.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 133
Answer:

Question 49.
REASONING
Is the domain of an arithmetic sequence discrete or continuous? Is the range of an arithmetic sequence discrete or continuous?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 49

Question 50.
MAKING AN ARGUMENT
Your friend says that the range of a function that represents an arithmetic sequence always contains only positive numbers or only negative numbers. Your friend claims this is true because the domain is the set of positive integers and the output values either constantly increase or constantly decrease. Is your friend correct? Explain.
Answer:

Question 51.
OPEN-ENDED
Write the first four terms of two different arithmetic sequences with a common difference of -3. Write an equation for the nth term of each sequence.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 51

Question 52.
THOUGHT PROVOKING
Describe an arithmetic sequence that models the numbers of people in a real-life situation.
Answer:

Question 53.
REPEATED REASONING
Firewood is stacked in a pile. The bottom row has 20 logs, and the top row has 14 logs. Each row has one more log than the row above it. How many logs are in the pile?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 53

Question 54.
HOW DO YOU SEE IT?
The bar graph shows the costs of advertising in a magazine.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 134
a. Does the graph represent an arithmetic sequence? Explain.
b. Explain how you would estimate the cost of a six-page advertisement in the magazine.
Answer:

Question 55.
REASONING
Write a function f that represents the arithmetic sequence shown in the mapping diagram.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 135
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 55

Question 56.
PROBLEM SOLVING
A train stops at a station every 12 minutes starting at 6:00 A.M. You arrive at the station at 7:29 A.M. How long must you wait for the train?
Answer:

Question 57.
ABSTRACT REASONING
Let x be a constant. Determine whether each sequence is an arithmetic sequence. Explain.
a. x + 6, 3x + 6, 5x + 6, 7x + 6, . . .
b. x + 1, 3x + 1, 9x + 1, 27x + 1, . . .
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 57.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 57.2

Maintaining Mathematical Proficiency

Solve the inequality. Graph the solution. (Section 2.2)

Question 58.
x + 8 ≥ -9
Answer:

Question 59.
15 < b – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 59

Question 60.
t – 21 < -12
Answer:

Question 61.
7 + y ≤ 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 61

Graph the function. Compare the graph to the graph of f(x) = | x |. Describe the domain and range. (Section 3.7)

Question 62.
h(x) = 3 | x |
Answer:

Question 63.
v(x) = | x – 5 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 63

Question 64.
g(x) = | x | + 1
Answer:

Question 65.
r(x) = -2 | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.6 Question 65

Lesson 4.7 Piecewise Functions

Essential Question 

How can you describe a function that is represented by more than on equation?

EXPLORATION 1
Writing Equations for a Function
Work with a partner.

a. Does the graph represent y as a function of x? Justify your conclusion.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 135.1
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 135.4
b. What is the value of the function when x = 0? How can you tell?
c. Write an equation that represents the values of the function when x ≤ 0.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 136
d. Write an equation that represents the values of the function when x > 0.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 137
e. Combine the results of parts (c) and (d) to write a single description of the function.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 138

EXPLORATION 2
Writing Equations for a Function
Work with a partner.

a. Does the graph represent y as a function of x? Justify your conclusion.
b. Describe the values of the function for the following intervals.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 139
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 140

Communicate Your Answer

Question 3.
How can you describe a function that is represented by more than one equation?

Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 141

Question 4.
Use two equations to describe the function represented by the graph.

4.7 Lesson

Monitoring Progress

Evaluate the function.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 142

Question 1.
f(-8)

Question 2.
f(-2)

Question 3.
f(0)

Question 4.
f(3)

Question 5.
f(5)

Question 6.
f(10)

Describe the domain and range.

Question 7.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 143

Question 8.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 144

Write a piecewise function for the graph.

Question 9.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 145

Question 10.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 146

Question 11.
A landscaper rents a wood chipper for 4 days. The rental company charges $100 for the first day and $50 for each additional day. Write and graph a step function that represents the relationship between the number x of days and the total cost y (in dollars) of renting the chipper.

Question 12.
WHAT IF? The reference beam originates at (3, 0) and reflects off a mirror at (5, 4).
a. Write an absolute value function that represents the path of the reference beam.
b. Write the function in part (a) as a piecewise function.

Piecewise Functions 4.7 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Compare piecewise functions and step functions.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 1

Question 2.
WRITING
Use a graph to explain why you can write the absolute value function y = | x | as a piecewise function.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, evaluate the function.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 500

Question 3.
f(-3)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 3

Question 4.
f(-2)
Answer:

Question 5.
f(0)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 5

Question 6.
f(5)
Answer:

Question 7.
g(-4)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 7

Question 8.
g(-1)
Answer:

Question 9.
g(0)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 9

Question 10.
g(1)
Answer:

Question 11.
g(2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 11

Question 12.
g(5)
Answer:

Question 13.
MODELING WITH MATHEMATICS
On a trip, the total distance (in miles) you travel in x hours is represented by the piecewise function
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 147
How far do you travel in 4 hours?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 13

Question 14.
MODELING WITH MATHEMATICS
The total cost (in dollars) of ordering x custom shirts is represented by the piecewise function
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 148
Determine the total cost of ordering 26 shirts.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 148.1
Answer:

In Exercises 15–20, graph the function. Describe the domain and range.

Question 15.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 149
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 15

Question 16.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 150
Answer:

Question 17.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 151
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 17

Question 18.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 152
Answer:

Question 19.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 153
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 19

Question 20.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 154
Answer:

Question 21.
ERROR ANALYSIS
Describe and correct the error in finding f(5) when Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 155.1
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 155
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 21

Question 22.
ERROR ANALYSIS
Describe and correct the error in graphing Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 157
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 158
Answer:

In Exercises 23–30, write a piecewise function for the graph.

Question 23.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 159
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 23

Question 24.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 160
Answer:

Question 25.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 161
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 25.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 25.2

Question 26.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 162
Answer:

Question 27.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 163
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 27.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 27.2

Question 28.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 164
Answer:

Question 29.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 165
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 29.1

Question 30.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 166
Answer:

In Exercises 31–34, graph the step function. Describe the domain and range.

Question 31.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 167
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 31

Question 32.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 168
Answer:

Question 33.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 169
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 33

Question 34.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 170
Answer:

Question 35.
MODELING WITH MATHEMATICS
The cost to join an intramural sports league is $180 per team and includes the first five team members. For each additional team member, there is a $30 fee. You plan to have nine people on your team. Write and graph a step function that represents the relationship between the number p of people on your team and the total cost of joining the league.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 35.2

Question 36.
MODELING WITH MATHEMATICS
The rates for a parking garage are shown. Write and graph a step function that represents the relationship between the number x of hours a car is parked in the garage and the total cost of parking in the garage for 1 day.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 171
Answer:

In Exercises 37–46, write the absolute value function as a piecewise function.

Question 37.
y = | x | + 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 37

Question 38.
y = | x | – 3
Answer:

Question 39.
y = | x – 2 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 39

Question 40.
y = | x + 5 |
Answer:

Question 41.
y = 2 | x + 3 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 41

Question 42.
y = 4 | x – 1 |
Answer:

Question 43.
y = -5 | x – 8 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 43

Question 44.
y = -3 | x + 6 |
Answer:

Question 45.
y = – | x – 3 | + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 45

Question 46.
y = 7| x + 1 | – 5
Answer:

Question 47.
MODELING WITH MATHEMATICS
You are sitting on a boat on a lake. You can get a sunburn from the sunlight that hits you directly and also from the sunlight that reflects off the water.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 172
a. Write an absolute value function that represents the path of the sunlight that reflects off the water.
b. Write the function in part (a) as a piecewise function.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 47.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 47.2

Question 48.
MODELING WITH MATHEMATICS
You are trying to make a hole in one on the miniature golf green.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 173
a. Write an absolute value function that represents the path of the golf ball.
b. Write the function in part (a) as a piecewise function.
Answer:

Question 49.
REASONING
The piecewise function f consists of two linear “pieces.” The graph of f is shown.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 174
a. What is the value of f(-10)?
b. What is the value of f(8)?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 49.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 49.2

Question 50.
CRITICAL THINKING
Describe how the graph of each piecewise function changes when < is replaced with ≤ and ≥ is replaced with >. Do the domain and range change? Explain.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 175
Answer:

Question 51.
USING STRUCTURE Graph
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 176
Describe the domain and range.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 51.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 51.2

Question 52.
HOW DO YOU SEE IT? The graph shows the total cost C of making x photocopies at a copy shop.
Big Ideas Math Algebra 1 Answer Key Chapter 4 Writing Linear Functions 177
a. Does it cost more money to make 100 photocopies or 101 photocopies? Explain.
b. You have $40 to make photocopies. Can you buy more than 500 photocopies? Explain.
Answer:

Question 53.
USING STRUCTURE
The output y of the greatest integer function is the greatest integer less than or equal to the input value x. This function is written as Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 204. Graph the function for -4 ≤ x < 4. Is it a piecewise function? a step function? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 53.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 53.2

Question 54.
THOUGHT PROVOKING
Explain why
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 205
does not represent a function. How can you redefine y so that it does represent a function?
Answer:

Question 55.
MAKING AN ARGUMENT
During a 9-hour snowstorm, it snows at a rate of 1 inch per hour for the first 2 hours, 2 inches per hour for the next 6 hours, and 1 inch per hour for the final hour.
a. Write and graph a piecewise function that represents the depth of the snow during the snowstorm.
b. Your friend says 12 inches of snow accumulated during the storm. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 55.1
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 55.2

Maintaining Mathematical Proficiency

Write the sentence as an inequality. Graph the inequality.(Section 2.5)

Question 56.
A number r is greater than -12 and no more than 13.
Answer:

Question 57.
A number t is less than or equal to 4 or no less than 18.Graph f and h. Describe the transformations from the graph of f to the graph of h.(Section 3.6)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 57

Question 58.
f(x) = x; h(x) = 4x + 3
Answer:

Question 59.
f(x) = x; h(x) = -x – 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 4 Writing Linear Functions 4.7 Question 59

Question 60.
f(x) = x; h(x) = –\(\frac{1}{2}\) + 5
Answer:

Writing Linear Functions Performance Task: Any Beginning

4.4–4.7 What Did You Learn?

Core Vocabulary

Section 4.4
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 179

Section 4.5
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 178

Section 4.6
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 181

Section 4.7
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 182

Mathematical Practices

Question 1.
What resources can you use to help you answer Exercise 17 on page 200?

Question 2.
What calculations are repeated in Exercises 11–16 on page 214? When finding a term such as a50, is there a general method or shortcut you can use instead of repeating calculations?

Question 3.
Describe the definitions you used when you explained your answer in Exercise 53 on page 224.

Performance Task

Any Beginning
With so many ways to represent a linear relationship, where do you start? Use what you know to move between equations, graphs, tables, and contexts.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 183
To explore the answers to this question and more, go to Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 206

Writing Linear Functions Chapter Review

Question 1.
Write an equation of the line in slope-intercept form.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 207

Question 2.
Write an equation in point-slope form of the line that passes through the point (4, 7) and has a slope of -1.

Write a linear function f with the given values.

Question 3.
f(10) = 5, f(2) = -3

Question 4.
f(3) = -4, f(5) = -4

Question 5.
f(6) = 8, f(9) = 3

Question 6.
Line a passes through (0, 4) and (4, 3).
Line b passes through (0, 1) and (4, 0).
Line c passes through (2, 0) and (4, 4).

Question 7.
Line a: 2x – 7y = 14
Line b: y = \(\frac{7}{2}\)x – 8
Line c: 2x + 7y = -21

Question 8.
Write an equation of the line that passes through (1, 5) and is parallel to the line y = -4x + 2.

Question 9.
Write an equation of the line that passes through (2, -3) and is perpendicular to the line y = -2x – 3

Question 10.
What is the roasting time for a 12-pound turkey?

Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 185

Question 11.
Write an equation that models the roasting time as a function of the weight of a turkey. Interpret the slope and y-intercept of the line of fit.

Question 12.
Make a scatter plot of the residuals to verify that the model in the example is a good fit.

Question 13.
Use the data in the example.
(a) Approximate the height of a student whose shoe size is 9.
(b) Predict the shoe size of a student whose height is 60 inches.

Question 14.
Is there a causal relationship in the data in the example? Explain.

write an equation for the nth term of the arithmetic sequence. Then find a30.

Question 15.
11, 10, 9, 8, …..

Question 16.
6, 12, 18, 24,….

Question 17.
-9, -6, -3, 0,….

Question 18.
Evaluate the function in the example when (a) x = 0 and (b) x = 5.

Graph the function. Describe the domain and range.

Question 19.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 186

Question 20.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 187

Write the absolute value function as a piecewise function.

Question 21.
y =| x | + 15

Question 22.
y = 4| x + 5 |

Question 23.
y = 2 | x + 2 | – 3

Question 24.
You are organizing a school fair and rent a popcorn machine for 3 days. The rental company charges $65 for the first day and $35 for each additional day. Write and graph a step function that represents the relationship between the number x of days and the total cost y (in dollars) of renting the popcorn machine.

Writing Linear Functions Chapter Test

Graph the function. Describe the domain and range.

Question 1.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 188

Question 2.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 189

Write an equation in slope-intercept form of the line with the given characteristics.

Question 3.
slope = \(\frac{2}{5}\) ; y-intercept = -7

Question 4.
passes through (0, 6) and (3, -3)

Question 5.
parallel to the line y = 3x – 1; passes through (-2, -8)

Question 6.
perpendicular to the line y = \(\frac{1}{4}\)x – 9; passes through (1, 1)

Write an equation in point-slope form of the line with the given characteristics.

Question 7.
slope = 10; passes through (6, 2)

Question 8.
passes through (-3, 2) and (6, -1)

Question 9.
The first row of an auditorium has 42 seats. Each row after the first has three more seats than the row before it.
a. Find the number of seats in Row 25.
b. Which row has 90 seats?

Question 10.
The table shows the amount x (in dollars) spent on advertising for a neighborhood festival and the attendance y of the festival for several years.
a. Make a scatter plot of the data. Describe the correlation.
b. Write an equation that models the attendance as a function of the amount spent on advertising.
c. Interpret the slope and y-intercept of the line of fit.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 190

Question 11.
Consider the data in the table in Exercise 10.
a. Use a graphing calculator to find an equation of the line of best fit.
b. Identify and interpret the correlation coefficient.
c. What would you expect the scatter plot of the residuals to look like?
d. Is there a causal relationship in the data? Explain your reasoning.
e. Predict the amount that must be spent on advertising to get 2000 people to attend the festival.

Question 12.
Let a, b, c, and d be constants. Determine which of the lines, if any, are parallel or perpendicular. Explain.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 191

Question 13.
Write a piecewise function defined by three equations that has a domain of all real numbers and a range of -3 < y ≤ 1.

Writing Linear Functions Cumulative Assessment

Question 1.
Which function represents the arithmetic sequence shown in the graph?
A. f(n) = 15 + 3n
B. f(n) = 4 – 3
C. f(n) = 27 – 3n
D. f(n) = 24 – 3n
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 192

Question 2.
Which of the inequalities are equivalent?
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 193

Question 3.
Complete the table for the four situations below. Explain your reasoning.
a. the price of a pair of pants and the number sold
b. the number of cell phones and the number of taxis in a city
c. a person’s IQ and the time it takes the person to run 50 meters
d. the amount of time spent studying and the score earned
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 194

Question 4.
Consider the function f(x) = x – 1. Select the functions that are shown in the graph. Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 195

Question 5.
Use the numbers to fill in values for m and b in the equation y = mx + b so that its graph passes through the points (6, 1) and (-2, -3).
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 196

Question 6.
Fill in the piecewise function with -, +, <, ≤, >, or ≥ so that the function is represented by the graph.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 197

Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 198

Question 7.
You claim that you can create a relation that is a function, and your friend claims that she can create a relation that is not a function. Using the given numbers, create a relation of five ordered pairs that supports your claim. What relation of five ordered pairs can your friend use to support her claim?
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 199

Question 8.
You have two coupons you can use at a restaurant. Write and solve an equation to determine how much your total bill must be for both coupons to save you the same amount of money.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 200

Question 9.
The table shows the daily high temperatures x (in degrees Fahrenheit) and the numbers y of frozen fruit bars sold on eight randomly selected days. The equation y = 3x – 50 models the data.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 201
a. Select the points that appear on a scatter plot of the residuals.
Big Ideas Math Answers Algebra 1 Chapter 4 Writing Linear Functions 202
b. Determine whether the model is a good fit for the data. Explain your reasoning.

Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry

Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry

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Big Ideas Math Book Geometry Answer Key Chapter 9 Right Triangles and Trigonometry

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Right Triangles and Trigonometry Maintaining Mathematical Proficiency

Simplify the expression.

Question 1.
√75

Answer:
square root of 75 = 5,625.

Explanation:
In the above-given question,
given that,
√75.
square root of 75 = 75 x 75.
75 x 75 = 5,625.
√75 = 5,625.

Question 2.
√270

Answer:
square root of 270 = 72,900.

Explanation:
In the above-given question,
given that,
√270.
square root of 270 = 270 x 270.
270 x 270 = 72900.
√270 = 72900.

Question 3.
√135

Answer:
square root of 135 = 18225.

Explanation:
In the above-given question,
given that,
√135.
square root of 135 = 135 x 135.
135 x 135 = 18225.
√135 = 18,225.

Question 4.
\(\frac{2}{\sqrt{7}}\)

Answer:
2/49 = 0.04.

Explanation:
In the above-given question,
given that,
square root of 7 = 7 x 7.
7 x 7 = 49.
\(\frac{2}{\sqrt{7}}\).
2/49 = 0.04.

Question 5.
\(\frac{5}{\sqrt{2}}\)

Answer:
5/4 = 1.25.

Explanation:
In the above-given question,
given that,
square root of 2 = 2 x 2.
2 x 2 = 4.
\(\frac{5}{\sqrt{2}}\).
5/4 = 1.25.

Question 6.
\(\frac{12}{\sqrt{6}}\)

Answer:
12/36 = 0.33.

Explanation:
In the above-given question,
given that,
square root of 6 = 6 x 6.
6 x 6 = 36.
\(\frac{12}{\sqrt{6}}\).
12/36 = 0.33.

Solve the proportion.

Question 7.
\(\frac{x}{12}=\frac{3}{4}\)

Answer:
x = 9.

Explanation:
In the above-given question,
given that,
\(\frac{x}{12}=\frac{3}{4}\)
x/12 = 3/4.
4x = 12 x 3.
4x = 36.
x = 36/4.
x = 9.

Question 8.
\(\frac{x}{3}=\frac{5}{2}\)

Answer:
x = 7.5.

Explanation:
In the above-given question,
given that,
\(\frac{x}{3}=\frac{5}{2}\)
x/3 = 5/2.
2x = 5 x 3.
2x = 15.
x = 15/2.
x = 7.5.

Question 9.
\(\frac{4}{x}=\frac{7}{56}\)

Answer:
x = 32.

Explanation:
In the above-given question,
given that,
\(\frac{4}{x}=\frac{7}{56}\)
4/x = 7/56.
7x = 56 x 4.
7x = 224.
x = 224/7.
x = 32.

Question 10.
\(\frac{10}{23}=\frac{4}{x}\)

Answer:
x = 9.2.

Explanation:
In the above-given question,
given that,
\(\frac{10}{23}=\frac{4}{x}\)
x/4 = 10/23.
10x = 23 x 4.
10x = 92.
x = 92/10.
x = 9.2.

Question 11.
\(\frac{x+1}{2}=\frac{21}{14}\)

Answer:
x = 135.

Explanation:
In the above-given question,
given that,
\(\frac{x + 1}{2}=\frac{21}{14}\)
x+12 x 2 = 21×14.
2x + 24= 294 .
2x = 294 – 24.
2x = 270.
x = 270/2.
x = 135.

Question 12.
\(\frac{9}{3 x-15}=\frac{3}{12}\)

Answer:
x = 6.33.

Explanation:
In the above-given question,
given that,
\(\frac{9}{3 x-15}=\frac{3}{12}\)
27x – 135 = 3×12.
27x = 36 + 135.
27x = 171.
x = 171/27.
x = 6.33.

Question 13.
ABSTRACT REASONING
The Product Property of Square Roots allows you to simplify the square root of a product. Are you able to simplify the square root of a sum? of a diffrence? Explain.

Answer:
Yes, I am able to simplify the square root of a sum.

Explanation:
In the above-given question,
given that,
The product property of square roots allows you to simplify the square root of a product.
√3 + 1 = √4.
√4 = 4 x 4.
16.
√3 – 1 = √2.
√2 = 2 x 2.
4.

Right Triangles and Trigonometry Mathematical practices

Monitoring progress

Question 1.
Use dynamic geometry software to construct a right triangle with acute angle measures of 30° and 60° in standard position. What are the exact coordinates of its vertices?

Answer:

Question 2.
Use dynamic geometry software to construct a right triangle with acute angle measures of 20° and 70° in standard position. What are the approximate coordinates of its vertices?
Answer:

9.1 The Pythagorean Theorem

Exploration 1

Proving the Pythagorean Theorem without Words

Work with a partner.

Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 1

a. Draw and cut out a right triangle with legs a and b, and hypotenuse c.

Answer:

Explanation:
In the above-given question,
given that,
proving the Pythagorean theorem without words.
a2 + b2 = c2.
Bid-Ideas-Math-Book-Geometry-Answer-Key-Chapter-9-Right Triangles and Trigonometry- 1

b. Make three copies of your right triangle. Arrange all tour triangles to form a large square, as shown.

Answer:
a2 + b2 = c2.

Explanation:
In the above-given question,
given that,
make three copies of your right triangle.
a2 + b2 = c2.

c. Find the area of the large square in terms of a, b, and c by summing the areas of the triangles and the small square.

Answer:
The area of the large square = a2 x b2.

Explanation:
In the above-given question,
given that,
the area of the square = l x b.
where l = length, and b = breadth.
the area of the square = a x b.
area = a2 x b2.

d. Copy the large square. Divide it into two smaller squares and two equally-sized rectangles, as shown.

Answer:

e. Find the area of the large square in terms of a and b by summing the areas of the rectangles and the smaller squares.
Answer:

f. Compare your answers to parts (c) and (e). Explain how this proves the Pythagorean Theorem.

Answer:
a2 + b2 = c2.

Explanation:
In the above-given question,
given that,
The length of the a and b is equal to the hypotenuse.
a2 + b2 = c2.
where a = one side and b = one side.

Exploration 2

Proving the Pythagorean Theorem

Work with a partner:

a. Draw a right triangle with legs a and b, and hypotenuse c, as shown. Draw the altitude from C to \(\overline{A B}\) Label the lengths, as shown.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 2

Answer:
a2 + b2 = c2.

Explanation:
In the above-given question,
given that,
a2 + b2 = c2.

Bid-Ideas-Math-Book-Geometry-Answer-Key-Chapter-9-Right Triangles and Trigonometry- 2

b. Explain why ∆ABC, ∆ACD, and ∆CBD are similar.

Answer:
In a right-angle triangle all the angle are equal.

Explanation:
In the above-given question,
given that,
∆ABC, ∆ACD, and ∆CBD are similar.
In a right-angle triangle all the angles are equal.

REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.
Answer:

c. Write a two-column proof using the similar triangles in part (b) to prove that a2 + b2 = c2

Answer:
a2 + b2 = c2

Explanation:
In the above-given question,
given that,
In the pythagorean theorem.
a2 + b2 = c2
the length of the hypotenuse ie equal to the two side lengths.
a2 + b2 = c2

Communicate Your Answer

Question 3.
How can you prove the Pythagorean Theorem?

Answer:
a2 + b2 = c2

Explanation:
In the above-given question,
given that,
In the pythagorean theorem,
the length of the hypotenuse is equal to the length of the other two sides.
hypotenuse = c.
length = a.
the breadth = b.
a2 + b2 = c2

Question 4.
Use the Internet or sonic other resource to find a way to prove the Pythagorean Theorem that is different from Explorations 1 and 2.
Answer:

Lesson 9.1 The Pythagorean Theorem

Monitoring Progress

Find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 1.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 3

Answer:
x = √52.

Explanation:
In the above-given question,
given that,
the side lengths are 6 and 4.
a2 + b2 = c2
6 x 6 + 4 x 4 = c2
36 + 16 = c2
52 = c2.
c = √52.

Question 2.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 4
Answer:
x = 4.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 5.
x2 + 3 x 3 = 5 x 5.
x2 + 9 = 25.
x2 = 25 – 9.
x2 = 16.
x = 4.

Question 3.
An anemometer is a device used to measure wind speed. The anemometer shown is attached to the top of a pole. Support wires are attached to the pole 5 feet above the ground. Each support wire is 6 feet long. How far from the base of the pole is each wire attached to the ground?
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 5

Answer:
x = √11.

Explanation:
In the above-given question,
given that,
An anemometer is a device used to measure wind speed.
support wires are attached to the pole 5 feet above the ground.
Each support wire is 6 feet long.
d2 + 6  x 6 = 5 x 5.
d2 + 36 = 25.
d2 = 25 – 36.
d2 = 11.
d
= √11.

Tell whether the triangle is a right triangle.

Question 4.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 6

Answer:
Yes the triangle is a right triangle.

Explanation:
In the above-given question,
given that,
the hypotenuse = 3 √34.
one side = 15.
the other side = 9.
so the triangle is a right triangle.

Question 5.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 7

Answer:
Yes, the triangle is a actute triangle.

Explanation:
In the above-given question,
given that,
the hypotenuse = 22.
one side = 26.
the other side = 14.
so the triangle is a acute triangle.

Question 6.
Verify that segments with lengths of 3, 4, and 6 form a triangle. Is the triangle acute, right, or obtuse?

Answer:
Yes the lengths of the triangle form a acute triangle.

Explanation:
In the above-given question,
given that,
the side lenths of 3, 4, and 6 form a triangle.
6 x 6 = 3 x 3 + 4 x 4.
36 = 9 + 16.
36 = 25.
so the length of the triangle forms a acute triangle.

Bid-Ideas-Math-Book-Geometry-Answer-Key-Chapter-9-Right Triangles and Trigonometry- 3

Question 7.
Verify that segments with lengths of 2, 1, 2, 8, and 3.5 form a triangle. Is the triangle acute, right, or obtuse?
Answer:

Exercise 9.1 The Pythagorean Theorem

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is a Pythagorean triple?
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 8
Find the length of the longest side.

Answer:
The length of the longest side = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Find the length of the hypotenuse

Answer:
The length of the hypotenuse = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the Pythagorean theorem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Find the length of the longest leg.

Answer:
The length of the longest leg = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Find the length of the side opposite the right angle.

Answer:
The length of the side opposite to the right angle = 5.

Explanation:
In the above-given question,
given that,
the side lengths are 3 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 3 x 3.
X x X = 16 + 9.
X x X = 25.
X = 5.

Monitoring progress and Modeling with Mathematics

In Exercises 3-6, find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 3.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 3

Question 4.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 10

Answer:
The length of the x = 34.

Explanation:
In the above-given question,
given that,
the side lengths are 30 and 16.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 16 x 16 + 30 x 30.
X x X = 256 + 900.
X x X = 1156.
X = 34.

Question 5.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 11
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 5

Question 6.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 12

Answer:
The length of the x = 7.2.

Explanation:
In the above-given question,
given that,
the side lengths are 6 and 4.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 4 x 4 + 6 x 6.
X x X = 16 + 36.
X x X = 52.
X = 7.2.

In Exercises 7 – 10, find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 7.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 13
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 7

Question 8.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 14

Answer:
The length of the X = 25.6.

Explanation:
In the above-given question,
given that,
the side lengths are 24 and 9.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 24 x 24 + 9 x 9.
X x X = 576 + 81.
X x X = 657.
X = 25.6.

Question 9.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 15
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 9

Question 10.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 16

Answer:
The length of the x = 11.4.

Explanation:
In the above-given question,
given that,
the side lengths are 7 and 9.
in the pythagorean threoem,
the longest side is equal to the side lengths.
X = 7 x 7 + 9 x 9.
X x X = 49 + 81.
X x X = 130.
X = 11.4.

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in using the Pythagorean Theorem (Theorem 9.1).

Question 11.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 17
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 11

Question 12.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 18

Answer:
x = 24.

Explanation:
In the above-given question,
given that,
the side lengths are 26 and 10.
26 x 26 = a x a + 10 x 10.
676 = a x a + 100.
676 – 100 = a x a.
576 = a x a.
a = 24.
x = 24.

Question 13.
MODELING WITH MATHEMATICS
The fire escape forms a right triangle, as shown. Use the Pythagorean Theorem (Theorem 9. 1) to approximate the distance between the two platforms. (See Example 3.)
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 19
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 13

Question 14.
MODELING WITH MATHEMATICS
The backboard of the basketball hoop forms a right triangle with the supporting rods, as shown. Use the Pythagorean Theorem (Theorem 9.1) to approximate the distance between the rods where the meet the backboard.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 20

Answer:
x = 9.1.

Explanation:
In the above-given question,
given that,
the side lengths are 13.4 and 9.8.
13.4 x 13.4 = X x X + 9.8 x 9.8.
179.56 = X x X + 96.04.
179.56 – 96.04 = X x X.
83.52 = X x X.
X = 9.1.
In Exercises 15 – 20, tell whether the triangle is a right triangle.

Question 15.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 21
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 15

Question 16.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 22

Answer:
No, the triangle is not a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 23 and 11.4.
the hypotenuse = 21.2.
21.2 x 21.2 = 23 x 23 + 11.4 x 11.4.
449.44 = 529 + 129.96.
449.44 = 658.96.
449 is not equal to 658.96.
so the triangle is not a right triangle.

Question 17.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 23
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 17

Question 18.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 24

Answer:
No, the triangle is not a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 5 and 1.
the hypotenuse = √26.
26 x 26 = 5 x 5 + 1 x 1.
676 = 25 + 1.
676 = 26.
676 is not equal to 26.
so the triangle is not a right triangle.

Question 19.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 25
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 19

Question 20.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 26

Answer:
Yes, the triangle forms a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 80 and 39.
the hypotenuse = 89.
89 x 89 = 80 x 80 + 39 x 39.
7921 = 6400 + 1521.
7921 = 7921.
7921 is equal to 7921.
so the triangle forms a right triangle.

In Exercises 21 – 28, verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?

Question 21.
10, 11, and 14
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 21

Question 22.
6, 8, and 10

Answer:
Yes, the triangle is forming a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 8 and 6.
the hypotenuse = 10.
10 x 10 = 8 x 8 + 6 x 6.
100 = 64 + 36.
100 = 100.
100 is  equal to 100.
so the triangle is forming a right triangle.

Question 23.
12, 16, and 20
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 23

Question 24.
15, 20, and 36

Answer:
Yes, the triangle is obtuse triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 15 and 20.
the hypotenuse = 36.
36 x 36 = 20 x 20 + 15 x 15.
1296 = 400 + 225.
1296 > 625.
1296 is greater than 625.
so the triangle is not a obtuse triangle.

Question 25.
5.3, 6.7, and 7.8
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 25

Question 26.
4.1, 8.2, and 12.2

Answer:
No, the triangle is obtuse triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 4.1 and 8.2.
the hypotenuse = 12.2.
12.2 x 12.2 = 4.1 x 4.1 + 8.2 x 8.2`.
148.84 = 16.81 + 67.24.
148.84 > 84.05.
148.84 is greater than 84.05.
so the triangle is obtuse triangle.

Question 27.
24, 30, and 6√43
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 27

Question 28.
10, 15 and 5√13

Answer:
Yes, the triangle is an acute triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 10 and 5√13.
the hypotenuse = 15.
15 x 15 = 10 x 10 + 5√13 x 5√13.
225 = 100 + 34.81.
225 < 134.81.
225 is less than 134.81.
so the triangle is acute triangle.

Question 29.
MODELING WITH MATHEMATICS
In baseball, the lengths of the paths between consecutive bases are 90 feet, and the paths form right angles. The player on first base tries to steal second base. How far does the ball need to travel from home plate to second base to get the player out?
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 29

Question 30.
REASONING
You are making a canvas frame for a painting using stretcher bars. The rectangular painting will be 10 inches long and 8 inches wide. Using a ruler, how can you be certain that the corners of the frame are 90°
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 27

Answer:
x = 12.8.

Explanation:
In the above-given question,
given that,
the side lengths are 10 and 8.
the hypotenuse = x.
X x X = 10 x 10 + 8 x 8.
X = 100 + 64.
X = 12.8.
In Exercises 31 – 34, find the area of the isosceles triangle.

Question 31.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 28
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 31

Question 32.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 29

Answer:
The area of the Isosceles triangle = 12 ft.

Explanation:
In the above-given question,
given that,
base = 32 ft.
hypotenuse = 20ft.
a2 + b2 = c2
h x h + 16 x 16 = 20 x 20.
h x h + 256 = 400.
h x h = 400 – 256.
h x h = 144.
h = 12 ft.
so the area of the isosceles triangle = 12 ft.

Question 33.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 30
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 33

Question 34.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 31

Answer:
The area of the Isosceles triangle = 48 m.

Explanation:
In the above-given question,
given that,
base = 28 m.
hypotenuse = 50 m.
a2 + b2 = c2
h x h + 14 x 14 = 50 x 50.
h x h + 196 = 2500.
h x h = 2500 – 196.
h x h = 2304.
h = 48 m.
so the area of the isosceles triangle = 48 m.

Question 35.
ANALYZING RELATIONSHIPS
Justify the Distance Formula using the Pythagorean Theorem (Thin. 9. 1).
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 35

Question 36.
HOW DO YOU SEE IT?
How do you know ∠C is a right angle without using the Pythagorean Theorem (Theorem 9.1) ?
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 32

Answer:
Yes, the triangle is forming a right triangle.

Explanation:
In the above-given question,
given that,
the side lengths are 8 and 6.
the hypotenuse = 10.
10 x 10 = 8 x 8 + 6 x 6.
100 = 64 + 36.
100 = 100.
100 is  equal to 100.
so the triangle is forming a right triangle.

Question 37.
PROBLEM SOLVING
You are making a kite and need to figure out how much binding to buy. You need the binding for the perimeter of the kite. The binding Comes in packages of two yards. How many packages should you buy?
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 33
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 37

Question 38.
PROVING A THEOREM
Use the Pythagorean Theorem (Theorem 9. 1) to prove the Hypotenuse-Leg (HL) Congruence Theorem (Theorem 5.9).
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 34
Answer:

Question 39.
PROVING A THEOREM
Prove the Converse of the Pythagorean Theorem (Theorem 9.2). (Hint: Draw ∆ABC with side lengths a, b, and c, where c is the length of the longest side. Then draw a right triangle with side lengths a, b, and x, where x is the length of the hypotenuse. Compare lengths c and x.)
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 39

Question 40.
THOUGHT PROVOKING
Consider two integers m and n. where m > n. Do the following expressions produce a Pythagorean triple? If yes, prove your answer. If no, give a counterexample.
2mn, m2 – n2, m2 + n2

Answer:

Question 41.
MAKING AN ARGUMENT
Your friend claims 72 and 75 Cannot be part of a pythagorean triple because 722 + 752 does not equal a positive integer squared. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 41

Question 42.
PROVING A THEOREM
Copy and complete the proof of the pythagorean Inequalities Theorem (Theorem 9.3) when c2 < a2 + b2.
Given In ∆ABC, c2 < a2 + b2 where c is the length
of the longest side.
∆PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and ∠R is a right angle.
Prove ∆ABC is an acute triangle.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 35

StatementsReasons
1. In ∆ABC, C2 < (12 + h2, where c is the length of the longest side. ∆PQR has side lengths a, b, and x, where x is the length of the hypotenuse, and ∠R is a right angle.1. _____________________________
2. a2 + b2 = x22. _____________________________
3. c2 < r23. _____________________________
4. c < x4. Take the positive square root of each side.
5. m ∠ R = 90°5. _____________________________
6. m ∠ C < m ∠ R6. Converse of the Hinge Theorem (Theorem 6.13)
7. m ∠ C < 90°7. _____________________________
8. ∠C is an acute angle.8. _____________________________
9. ∆ABC is an acute triangle.9. _____________________________

Answer:

Question 43.
PROVING A THEOREM
Prove the Pythagorean Inequalities Theorem (Theorem 9.3) when c2 > a2 + b2. (Hint: Look back at Exercise 42.)
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 43.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 43.2

Maintaining Mathematical Proficiency

Simplify the expression by rationalizing the denominator.

Question 44.
\(\frac{7}{\sqrt{2}}\)

Answer:
7/√2 = 7√2 /2.

Explanation:
In the above-given question,
given that,
\(\frac{7}{\sqrt{2}}\) = 7/√2.
7/√2 = 7/√2  x √2 /√2 .
7 √2 /√4.
7√2 /2.

Question 45.
\(\frac{14}{\sqrt{3}}\)
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 45

Question 46.
\(\frac{8}{\sqrt{2}}\)

Answer:
8/√2 = 8√2 /2.

Explanation:
In the above-given question,
given that,
\(\frac{8}{\sqrt{2}}\) = 8/√2.
8/√2 = 8/√2  x √2 /√2 .
8 √2 /√4.
8√2 /2.

Question 47.
\(\frac{12}{\sqrt{3}}\)
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.1 Ans 47

9.2 Special Right Triangles

Exploration 1

Side Ratios of an Isosceles Right Triangle

Work with a partner:

Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 36

a. Use dynamic geometry software to construct an isosceles right triangle with a leg length of 4 units.
Answer:

b. Find the acute angle measures. Explain why this triangle is called a 45° – 45° – 90° triangle.
Answer:

c. Find the exact ratios of the side lenghts (using square roots).
\(\frac{A B}{A C}\) = ____________
\(\frac{A B}{B C}\) = ____________
\(\frac{A B}{B C}\) = ____________
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.
Answer:

d. Repeat parts (a) and (c) for several other isosceles right triangles. Use your results to write a conjecture about the ratios of the side lengths of an isosceles right triangle.
Answer:

Exploration 2

Work with a partner.

a. Use dynamic geometry software to construct a right triangle with acute angle measures of 30° and 60° (a 30° – 60° – 90° triangle), where the shorter leg length is 3 units.

b. Find the exact ratios of the side lengths (using square roots).
\(\frac{A B}{A C}\) = ____________
\(\frac{A B}{B C}\) = ____________
\(\frac{A B}{B C}\) = ____________
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 37
Answer:

C. Repeat parts (a) and (b) for several other 30° – 60° – 90° triangles. Use your results to write a conjecture about the ratios of the side lengths of a 30° – 60° – 90° triangle.
Answer:

Communicate Your Answer

Question 3.
What is the relationship among the side lengths of 45°- 45° – 90° triangles? 30° – 60° – 90° triangles?
Answer:

Lesson 9.2 Special Right Triangles

Monitoring Progress

Find the value of the variable. Write your answer in simplest form.

Question 1.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 38

Answer:
x = 4

Explanation:
(2√2)² = x² + x²
8 = 2x²
x² = 4
x = 4

Question 2.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 39

Answer:
y = 2

Explanation:
y² = 2 + 2
y² = 4
y = 2

Question 3.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 40

Answer:
x = 3, y = 2√3

Explanation:
longer leg = shorter leg • √3
x = √3 • √3
x = 3
hypotenuse = shorter leg • 2
= √3 • 2 = 2√3

Question 4
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 41

Answer:
h = 2√3

Explanation:
longer leg = shorter leg • √3
h = 2√3

Question 5.
The logo on a recycling bin resembles an equilateral triangle with side lengths of 6 centimeters. Approximate the area of the logo.

Answer:
Area is \(\frac { 1 }{ 3√3 } \)

Explanation:
Area = \(\frac { √3 }{ 4 } \) a²
= \(\frac { √3 }{ 4 } \)(6)²
= \(\frac { 1 }{ 3√3 } \)

Question 6.
The body of a dump truck is raised to empty a load of sand. How high is the 14-foot-long body from the frame when it is tipped upward by a 60° angle?
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 42

Answer:
28/3 ft high is the 14-foot-long body from the frame when it is tipped upward by a 60° angle.

Explanation:
Height of body at 90 degrees = 14 ft
Height of body at 1 degree = 14/90
Height of body at 60 degrees = 14 x 60/90
= 14 x 2/3
= 28/3 ft

Exercise 9.2 Special Right Triangles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Name two special right triangles by their angle measures.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 1

Question 2.
WRITING
Explain why the acute angles in an isosceles right triangle always measure 45°.

Answer:
Because the acute angles of a right isosceles triangle must be congruent by the base angles theorem and complementary, their measures must be 90°/2 = 45°.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the value of x. Write your answer in simplest form.

Question 3.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 43
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 3

Question 4.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 44

Answer:
x = 10

Explanation:
hypotenuse = leg • √2
x = 5√2 • √2
x = 10

Question 5.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 46
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 5

Question 6.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 46

Answer:
x = \(\frac { 9 }{ √2 } \)

Explanation:
hypotenuse = leg • √2
9 = x • √2
x = \(\frac { 9 }{ √2 } \)

In Exercises 7 – 10, find the values of x and y. Write your answers in simplest form.

Question 7.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 47
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 7

Question 8.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 48

Answer:
x = 3, y = 6

Explanation:
hypotenuse = 2 • shorter leg
y = 2 • 3
y = 6
longer leg = √3 • shorter leg
3√3 = √3x
x = 3

Question 9.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 49
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 9

Question 10.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 50

Answer:
x = 18, y = 6√3

Explanation:
hypotenuse = 2 • shorter leg
12√3 = 2y
y = 6√3
longer leg = √3 • shorter leg
x = √3 . 6√3
x = 18

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in finding the length of the hypotenuse.

Question 11.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 51
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 11

Question 12.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 52

Answer:
hypotenuse = leg • √2 = √5 . √2 = √10

In Exercises 13 and 14. sketch the figure that is described. Find the indicated length. Round decimal answers to the nearest tenth.

Question 13.
The side length of an equilateral triangle is 5 centimeters. Find the length of an altitude.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 13

Question 14.
The perimeter of a square is 36 inches. Find the length of a diagonal.

Answer:
The length of a diagonal is 9√2

Explanation:
Side of the square = 36/4 = 9
square diagonal = √2a = √2(9) = 9√2

In Exercises 15 and 16, find the area of the figure. Round decimal answers to the nearest tenth.

Question 15.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 53
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 15

Question 16.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 54

Answer:
Area is 40√(1/3) sq m

Explanation:
longer leg = √3 • shorter leg
4 = √3 • shorter leg
shorter leg = 4/√3
h² = 16/3 + 16
h² = 16(4/3)
h = 8√(1/3)
Area of the parallelogram = 5(8√(1/3)) = 40√(1/3) sq m

Question 17.
PROBLEM SOLVING
Each half of the drawbridge is about 284 feet long. How high does the drawbridge rise when x is 30°? 45°? 60°?
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 55
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 17

Question 18.
MODELING WITH MATHEMATICS
A nut is shaped like a regular hexagon with side lengths of 1 centimeter. Find the value of x. (Hint: A regular hexagon can be divided into six congruent triangles.)
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 56

Answer:
Side length = 1cm
A regular hexagon has six equal the side length. A line drawn from the centre to any vertex will have the same length as any side.
This implies the radius is equal to the side length.
As a result, when lines are drawn from the centre to each of the vertexes, a
regular hexagon is said to be made of six equilateral triangles.
From the diagram, x = 2× apothem
Apothem is the distance from the centre of a regular polygon to the midpoint of side.
Using Pythagoras theorem, we would get the apothem
Hypotenuse² = opposite² + adjacent²
1² = apothem² + (½)²
Apothem = √(1² -(½)²)
= √(1-¼) = √¾
Apothem = ½√3
x = 2× Apothem = 2 × ½√3
x = √3

Question 19.
PROVING A THEOREM
Write a paragraph proof of the 45°- 45°- 90° Triangle Theorem (Theorem 9.4).
Given ∆DEF is a 45° – 45° – 90° triangle.
Prove The hypotenuse is √2 times as long as each leg.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 57
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 19

Question 20.
HOW DO YOU SEE IT?
The diagram shows part of the wheel of Theodorus.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 58
a. Which triangles, if any, are 45° – 45° – 90° triangles?

Answer:

b. Which triangles, if any, are 30° – 60° – 90° triangles?

Answer:

Question 21.
PROVING A THEOREM
Write a paragraph proof of the 30° – 60° – 90° Triangle Theorem (Theorem 9.5).
(Hint: Construct ∆JML congruent to ∆JKL.)
Given ∆JKL is a 30° 60° 9o° triangle.
Prove The hypotenuse is twice as long as the shorter leg, and the longer leg is √3 times as long as the shorter leg.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 59
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 21.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 21.2

Question 22.
THOUGHT PROVOKING
A special right triangle is a right triangle that has rational angle measures and each side length contains at most one square root. There are only three special right triangles. The diagram below is called the Ailles rectangle. Label the sides and angles in the diagram. Describe all three special right triangles.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 60

Answer:
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 9.2 1

Question 23.
WRITING
Describe two ways to show that all isosceles right triangles are similar to each other.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 23

Question 24.
MAKING AN ARGUMENT
Each triangle in the diagram is a 45° – 45° – 90° triangle. At Stage 0, the legs of the triangle are each 1 unit long. Your brother claims the lengths of the legs of the triangles added are halved at each stage. So, the length of a leg of a triangle added in Stage 8 will be \(\frac{1}{256}\) unit. Is your brother correct? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 61
Answer:

Question 25.
USING STRUCTURE
ΔTUV is a 30° – 60° – 90° triangle. where two vertices are U(3, – 1) and V( – 3, – 1), \(\overline{U V}\) is the hypotenuse. and point T is in Quadrant I. Find the coordinates of T.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 25.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 25.2

Maintaining Mathematical Proficiency

Find the Value of x.

Question 26.
ΔDEF ~ ΔLMN
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 62

Answer:
x = 18

Explanation:
\(\frac { DE }{ LM } \) = \(\frac { DF }{ LN } \)
\(\frac { 12 }{ x } \) = \(\frac { 20 }{ 30 } \)
\(\frac { 12 }{ x } \) = \(\frac { 2 }{ 3 } \)
x = 12(\(\frac { 3 }{ 2 } \)) = 18

Question 27.
ΔABC ~ ΔQRS
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 63
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.2 Ans 27

9.3 Similar Right Triangles

Exploration 1

Writing a Conjecture

a. Use dynamic geometry software to construct right ∆ABC, as shown. Draw \(\overline{C D}\) so that it is an altitude from the right angle to the hypotenuse of ∆ABC.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 64
Answer:

b. The geometric mean of two positive numbers a and b is the positive number x that satisfies
\(\frac{a}{x}=\frac{x}{b}\)
x is the geometric mean of a and b.
Write a proportion involving the side lengths of ∆CBD and ∆ACD so that CD is the geometric mean of two of the other side lengths. Use similar triangles to justify your steps.
Answer:

c. Use the proportion you wrote in part (b) to find CD.
Answer:

d. Generalize the proportion you wrote in part (b). Then write a conjecture about how the geometric mean is related to the altitude from the right angle to the hypotenuse of a right triangle.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
Answer:

Exploration 2

Comparing Geometric and Arithmetic Means

Work with a partner:
Use a spreadsheet to find the arithmetic mean and the geometric mean of several pairs of positive numbers. Compare the two means. What do you notice?
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 65
Answer:

Communicate Your Answer

Question 3.
How are altitudes and geometric means of right triangles related?
Answer:

Lesson 9.3 Similar Right Triangles

Monitoring progress

Identify the similar triangles.

Question 1.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 66

Answer:
△QRS ~ △ QST

Question 2.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 67
Answer:
△EFG ~ △ EHG

Question 3.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 68

Answer:
△EGH ~ △EFG
\(\frac { EF }{ EG } \) = \(\frac { GF }{ GH } \)
\(\frac { 5 }{ 3 } \) = \(\frac { 4 }{ x } \)
x = 2.4

Question 4.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 69
Answer:
△JLM ~ △LMK
\(\frac { JL }{ LM } \) = \(\frac { JM }{ KM } \)
\(\frac { 13 }{ 5 } \) = \(\frac { 12 }{ x } \)
x = 4.615

Find the geometric mean of the two numbers.

Question 5.
12 and 27

Answer:
x = √(12 x 27)
x = √324 = 18

Question 6.
18 and 54

Answer:
x = √(18 x 54) = √(972)
x = 31.17

Question 7.
16 and 18

Answer:
x = √(16 x 18) = √(288)
x = 16.970

Question 8.
Find the value of x in the triangle at the left.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 70

Answer:
x = √(9 x 4)
x = 6

Question 9.
WHAT IF?
In Example 5, the vertical distance from the ground to your eye is 5.5 feet and the distance from you to the gym wall is 9 feet. Approximate the height of the gym wall.

Answer:
9² = 5.5 x w
81 = 5.5 x w
w = 14.72
The height of the wall = 14.72 + 5.5 = 20.22

Exercise 9.3 Similar Right Triangles

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and ____________ .
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 1

Question 2.
WRITING
In your own words, explain geometric mean.

Answer:
The geometric mean is the average value or mean that signifies the central tendency of set of numbers by finding the product of their values.

Monitoring progress and Modeling with Mathematics

In Exercises 3 and 4, identify the similar triangles.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 71
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 72

Answer:
△LKM ~ △LMN ~ △MKN

In Exercises 5 – 10, find the value of x.

Question 5.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 73
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 74

Answer:
x = 9.6

Explanation:
\(\frac { QR }{ SR } \) = \(\frac { SR }{ TS } \)
\(\frac { 20 }{ 16 } \) = \(\frac { 12 }{ x } \)
1.25 = \(\frac { 12 }{ x } \)
x = 9.6

Question 7.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 75
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 7

Question 8.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 76

Answer:
x = 14.11

Explanation:
\(\frac { AB }{ AC } \) = \(\frac { BD }{ BC } \)
\(\frac { 16 }{ 34 } \) = \(\frac { x }{ 30 } \)
x = 14.11

Question 9.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 77
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 9

Question 10.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 78

Answer:
x = 2.77

Explanation:
\(\frac { 5.8 }{ 3.5 } \) = \(\frac { 4.6 }{ x } \)
x = 2.77

In Exercises 11 – 18, find the geometric mean of the two numbers.

Question 11.
8 and 32
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 11

Question 12.
9 and 16

Answer:
x = √(9 x 16)
x = 12

Question 13.
14 and 20
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 13

Question 14.
25 and 35

Answer:
x = √(25 x 35)
x = 29.5

Question 15.
16 and 25
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 15

Question 16.
8 and 28

Answer:
x = √(8 x 28)
x = 14.96

Question 17.
17 and 36
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 17

Question 18.
24 and 45

Answer:
x = √(24 x 45)
x = 32.86

In Exercises 19 – 26. find the value of the variable.

Question 19.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 79
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 19

Question 20.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 80

Answer:
y = √(5 x 8)
y = √40
y = 2√10

Question 21.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 81
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 21

Question 22.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 82

Answer:
10 • 10 = 25 • x
100 = 25x
x = 4

Question 23.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 83
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 23

Question 24.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 84

Answer:
b² = 16(16 + 6)
b² = 16(22) = 352
b = 18.76

Question 25.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 85
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 25

Question 26.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 86

Answer:
x² = 8(8 + 2)
x² = 8(10) = 80
x = 8.9

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in writing an equation for the given diagram.

Question 27.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 87
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 27

Question 28.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 88

Answer:
d² = g • e

MODELING WITH MATHEMATICS
In Exercises 29 and 30, use the diagram.

Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 89

Question 29.
You want to determine the height of a monument at a local park. You use a cardboard square to line up the top and bottom of the monument, as shown at the above left. Your friend measures the vertical distance from the ground to your eye and the horizontal distance from you to the monument. Approximate the height of the monument.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 29

Question 30.
Your classmate is standing on the other side of the monument. She has a piece of rope staked at the base of the monument. She extends the rope to the cardboard square she is holding lined up to the top and bottom of the monument. Use the information in the diagram above to approximate the height of the monument. Do you get the same answer as in Exercise 29? Explain your reasoning.

Answer:

MATHEMATICAL CONNECTIONS
In Exercises 31 – 34. find the value(s) of the variable(s).

Question 31.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 90
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 31

Question 32.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 91

Answer:
\(\frac { 6 }{ b + 3 } \) = \(\frac { 8 }{ 6 } \)
36 = 8(b + 3)
36 = 8b + 24
8b = 12
b = \(\frac { 3 }{ 2 } \)

Question 33.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 92
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 33

Question 34.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 93

Answer:
x = 42.66, y = 40, z = 53

Explanation:
\(\frac { 24 }{ 32 } \) = \(\frac { 32 }{ x } \)
0.75 = \(\frac { 32 }{ x } \)
x = 42.66
y = √24² + 32²
y = √576 + 1024 = 40
z = √42.66² + 32² = √1819.87 + 1024 = 53

Question 35.
REASONING
Use the diagram. Decide which proportions are true. Select all that apply.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 94
(A) \(\frac{D B}{D C}=\frac{D A}{D B}\)
(B) \(\frac{B A}{C B}=\frac{C B}{B D}\)
(C) \(\frac{C A}{B \Lambda}=\frac{B A}{C A}\)
(D) \(\frac{D B}{B C}=\frac{D A}{B A}\)
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 35

Question 36.
ANALYZING RELATIONSHIPS
You are designing a diamond-shaped kite. You know that AD = 44.8 centimeters, DC = 72 centimeters, and AC = 84.8 centimeters. You Want to use a straight crossbar \(\overline{B D}\). About how long should it be? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 95

Answer:
BD = 76.12

Explanation:
AD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm
Two disjoint pairs of consecutive sides are congruent.
So, AD = AB = 44.8 cm
DC = BC = 72 cm
The diagonals are perpendicular.
So, AC ⊥ BD
AC = AO + OC
AX = x + y = 84.8 — (i)
Perpendicular bisects the diagonal BD into equal parts let it be z.
BD = BO + OD
BD = z + z
Using pythagorean theorem
44.8² = x² + z² —- (ii)
72² = y² + z² —– (iii)
Subtract (ii) and (iii)
72² – 44.8² = y²+ z² – x² – z²
5184 – 2007.04 = (x + y) (x – y)
3176.96 = (84.8)(x – y)
37.464 = x – y —- (iv)
Add (i) & (iv)
x + y + x – y = 84.8 + 37.464
2x = 122.264
x = 61.132
x + y = 84.8
61.132 + y = 84.8
y = 23.668
44.8² = x² + z²
z = 38.06
BD = z + z
BD = 76.12

Question 37.
ANALYZING RELATIONSHIPS
Use the Geometric Mean Theorems (Theorems 9.7 and 9.8) to find AC and BD.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 96
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 37.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 37.2

Question 38.
HOW DO YOU SEE IT?
In which of the following triangles does the Geometric Mean (Altitude) Theorem (Theorem 9.7) apply?
(A)
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 97
(B)
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 98
(C)
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 99
(D)
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 100

Answer:

Question 39.
PROVING A THEOREM
Use the diagram of ∆ABC. Copy and complete the proof of the Pythagorean Theorem (Theorem 9. 1).
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 101
Given In ∆ABC, ∆BCA is a right angle.
Prove c2 = a2 + b2

StatementsReasons
1. In ∆ABC, ∠BCA is a right angle.1. ________________________________
2. Draw a perpendicular segment (altitude) from C to \(\overline{A B}\).2. Perpendicular Postulate (Postulate 3.2)
3. ce = a2 and cf = b23. ________________________________
4. ce + b2 = ___  + b24. Addition Property of Equality
5. ce + cf = a2 + b25. ________________________________
6. c(e + f) a2 + b26. ________________________________
7. e + f = ________7. Segment Addition Postulate (Postulate 1.2)
8. c  • c = a2 + b28. ________________________________
9. c2 = a2 + b29. Simplify.

Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 39.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 39.2

Question 40.
MAKING AN ARGUMENT
Your friend claims the geometric mean of 4 and 9 is 6. and then labels the triangle, as shown. Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 102

Answer:
G.M = √(4 x 9) = √36 = 6
My friend is correct.

In Exercises 41 and 42, use the given statements to prove the theorem.

Gien ∆ABC is a right triangle.
Altitude \(\overline{C D}\) is dravn to hypotenuse \(\overline{A B}\).

Question 41.
PROVING A THEOREM
Prove the Geometric Mean (Altitude) Theorem (Theorem 9.7) b showing that CD2 = AD • BD.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 41.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 41.2

Question 42.
PROVING A THEOREM
Prove the Geometric Mean ( Leg) Theorem (Theorem 9.8) b showing that CB2 = DB • AB and AC2 = AD • AB.

Answer:

Question 43.
CRITICAL THINKING
Draw a right isosceles triangle and label the two leg lengths x. Then draw the altitude to the hypotenuse and label its length y. Now, use the Right Triangle Similarity Theorem (Theorem 9.6) to draw the three similar triangles from the image and label an side length that is equal to either x or y. What can you conclude about the relationship between the two smaller triangles? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 43

Question 44.
THOUGHT PROVOKING
The arithmetic mean and geometric mean of two nonnegative numbers x and y are shown.
arithmetic mean = \(\frac{x+y}{2}\)
geometric mean = \(\sqrt{x y}\)
Write an inequality that relates these two means. Justify your answer.

Answer:

Question 45.
PROVING A THEOREM
Prove the Right Triangle Similarity Theorem (Theorem 9.6) by proving three similarity statements.
Given ∆ABC is a right triangle.
Altitude \(\overline{C D}\) is drawn to hvpotenuse \(\overline{A B}\).
Prove ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC,
∆CBD ~ ∆ACD
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 45.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 45.2
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 45.3

Maintaining Mathematical proficiency

Solve the equation for x.

Question 46.
13 = \(\frac{x}{5}\)

Answer:
13 = \(\frac{x}{5}\)
x = 65

Question 47.
29 = \(\frac{x}{4}\)
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 47

Question 48.
9 = \(\frac{78}{x}\)

Answer:
9 = \(\frac{78}{x}\)
9x = 78
x = 8.6

Question 49.
30 = \(\frac{115}{x}\)
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.3 Ans 49

9.1 to 9.3 Quiz

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 1.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 103
Answer:
x = 15

Explanation:
x² = 9² + 12²
x² = 81 + 144
x² = 225
x = 15

Question 2.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 104

Answer:
x = 10.63

Explanation:
x² = 7² + 8² = 49 + 64
x = √113
x = 10.63

Question 3.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 105
Answer:
x = 4√3

Explanation:
8² = x² + 4²
64 = x² + 16
x² = 48
x = 4√3

Verify that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?
(Section 9.1)
Question 4.
24, 32, and 40

Answer:
Triangle is a right angle trinagle.

Explanation:
40² = 1600
24² + 32² = 576 + 1024 = 1600
40² = 24² + 32²
So, the triangle is a right angle trinagle.

Question 5.
7, 9, and 13

Answer:
Triangle is an obtuse trinagle.

Explanation:
13² = 169
7² + 9² = 49 + 81 = 130
13² > 7² + 9²
So, the triangle is an obtuse trinagle.

Question 6.
12, 15, and 10√3

Answer:
Triangle is an acute trinagle.

Explanation:
15² = 225
12² + (10√3)² = 144 + 300 = 444
15² < 12² + (10√3)²
So, the triangle is an acute trinagle.

Find the values of x and y. Write your answers in the simplest form.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 106

Answer:
x = 6, y = 6√2

Explanation:
x = 6
hypotenuse = leg • √2
y = 6√2

Question 8.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 107

Answer:
y = 8√3, x = 16

Explanation:
longer leg = shorter leg • √3
y = 8√3
x² = 8² + (8√3)² = 64 + 192
x = 16

Question 9.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 108

Answer:
x = 5√2, y = 5√6

Explanation:
longer leg = shorter leg • √3
y = x√3
y = 5√6
hypotenuse = shorter leg • 2
10√2 = 2x
x = 5√2

Find the geometric mean of the two numbers.
Question 10.
6 and 12

Answer:
G.M = √(6 • 12) = 6√2

Question 11.
15 and 20

Answer:
G.M = √(15 • 20) = 10√3

Question 12.
18 and 26

Answer:
G.M = √(18 • 26) = 6√13

Identify the similar right triangles. Then find the value of the variable.

Question 13.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 109

Answer:
x = √(8 • 4)
x = 4√2

Question 14.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 110

Answer:
y = √(9 • 6) = 3√6

Question 15.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 111

Answer:

Question 16.
Television sizes are measured by the length of their diagonal. You want to purchase a television that is at least 40 inches. Should you purchase the television shown? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 112

Answer:
x² = 20.25² + 36²
x² = 410.0625 + 1296 = 1706.0625
x = 41.30
Yes, i will purchase the television.

Question 17.
Each triangle shown below is a right triangle.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 113
a. Are any of the triangles special right triangles? Explain your reasoning.
Answer:
A is a similar triangle.

b. List all similar triangles. if any.
Answer:
B, C and D, E are similar triangles.

c. Find the lengths of the altitudes of triangles B and C.
Answer:
B altitude = √(9 + 27) = 6
C altitude = √(36 + 72) = 6√3

9.4 The Tangent Ratio

Exploration 1

Calculating a Tangent Ratio

Work with a partner

a. Construct ∆ABC, as shown. Construct segments perpendicular to \(\overline{A C}\) to form right triangles that share vertex A and arc similar to ∆ABC with vertices, as shown.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 114
Answer:

b. Calculate each given ratio to complete the table for the decimal value of tan A for each right triangle. What can you Conclude?
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 115
Answer:

Exploration 2

Using a calculator

Work with a partner: Use a calculator that has a tangent key to calculate the tangent of 36.87°. Do you get the same result as in Exploration 1? Explain.
ATTENDING TO PRECISION
To be proficient in math, you need to express numerical answers with a degree of precision appropriate for the problem context.
Answer:

Communicate Your Answer

Question 3.
Repeat Exploration 1 for ∆ABC with vertices A(0, 0), B(8, 5), and C(8, 0). Construct the seven perpendicular segments so that not all of them intersect \(\overline{A C}\) at integer values of x. Discuss your results.
Answer:

Question 4.
How is a right triangle used to find the tangent of an acute angle? Is there a unique right triangle that must be used?
Answer:

Lesson 9.4 The Tangent Ratio

Monitoring progress

Find tan J and tan K. Write each answer as a fraction and as a decimal rounded to four places.

Question 1.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 116

Answer:
tan K = \(\frac { opposite side }{ adjacent side } \) = \(\frac { JL }{ KL } \)
= \(\frac { 32 }{ 24 } \) = \(\frac { 4 }{ 3 } \) = 1.33
tan J = \(\frac { KL }{ JL } \) = \(\frac { 24 }{ 32 } \) = \(\frac { 3 }{ 4 } \) = 0.75

Question 2.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 117

Answer:
tan K = \(\frac { LJ }{ LK } \) = \(\frac { 15 }{ 8 } \)
tan J = \(\frac { LK }{ LJ } \) = \(\frac { 8 }{ 15 } \)

Find the value of x. Round your answer to the nearest tenth.

Question 3.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 118

Answer:
Tan 61 = \(\frac { 22 }{ x } \)
1.804 = \(\frac { 22 }{ x } \)
x = 12.1951

Question 4.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 119

Answer:
tan 56 = \(\frac { x }{ 13 } \)
1.482 = \(\frac { x }{ 13 } \)
x = 19.266

Question 5.
WHAT IF?
In Example 3, the length of the shorter leg is 5 instead of 1. Show that the tangent of 60° is still equal to √3.

Answer:
longer leg = shorter leg • √3
= 5√3
tan 60 = \(\frac { 5√3 }{ 5 } \)
= √3

Question 6.
You are measuring the height of a lamppost. You stand 40 inches from the base of the lamppost. You measure the angle ot elevation from the ground to the top of the lamppost to be 70°. Find the height h of the lamppost to the nearest inch.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 120

Answer:
tan 70 = \(\frac { h }{ 40 } \)
2.7474 = \(\frac { h }{ 40 } \)
h = 109.896 in

Exercise 9.4 The Tangent Ratio

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The tangent ratio compares the length of _________ to the length of ___________ .
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 1

Question 2.
WRITING
Explain how you know the tangent ratio is constant for a given angle measure.

Answer:
When two triangles are similar, the corresponding sides are proportional which makes the ratio constant for a given acute angle measurement.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.

Question 3.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 121
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 3

Question 4.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 122

Answer:
tan F = \(\frac { DE }{ EF } \) = \(\frac { 24 }{ 7 } \)
tan D = \(\frac { EF }{ DE } \) = \(\frac { 7 }{ 24 } \)

Question 5.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 123
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 5

Question 6.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 124

Answer:
tan K = \(\frac { JL }{ LK } \) = \(\frac { 3 }{ 5 } \)
tan J = \(\frac { LK }{ JL } \) = \(\frac { 5 }{ 3 } \)

In Exercises 7 – 10, find the value of x. Round your answer to the nearest tenth.

Question 7.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 125
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 7

Question 8.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 126

Answer:
tan 27 = \(\frac { x }{ 15 } \)
0.509 = \(\frac { x }{ 15 } \)
x = 7.635

Question 9.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 127
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 128

Answer:
tan 37 = \(\frac { 6 }{ x } \)
0.753 = \(\frac { 6 }{ x } \)
x = 7.968

ERROR ANALYSIS
In Exercises 11 and 12, describe the error in the statement of the tangent ratio. Correct the error if possible. Otherwise, write not possible.

Question 11.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 129
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 11

Question 12.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 130

Answer:

In Exercises 13 and 14, use a special right triangle to find the tangent of the given angle measure.

Question 13.
45°
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 13

Question 14.
30°

Answer:
tan 30° = \(\frac { 1 }{ √3 } \)

Question 15.
MODELING WITH MATHEMATICS
A surveyor is standing 118 Feet from the base of the Washington Monument. The surveyor measures the angle of elevation from the ground to the top of the monument to be 78°. Find the height h of the Washington Monument to the nearest foot.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 131
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 15

Question 16.
MODELING WITH MATHEMATICS
Scientists can measure the depths of craters on the moon h looking at photos of shadows. The length of the shadow cast by the edge of a crater is 500 meters. The angle of elevation of the rays of the Sun is 55°. Estimate the depth d of the crater.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 132

Answer:
tan 55 = \(\frac { d }{ 500 } \)
1.428 = \(\frac { d }{ 500 } \)
d = 714 m
The depth of the crater is 714 m

Question 17.
USING STRUCTURE
Find the tangent of the smaller acute angle in a right triangle with side lengths 5, 12, and 13.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 17

Question 18.
USING STRUCTURE
Find the tangent 0f the larger acute angle in a right triangle with side lengths 3, 4, and 5.

Answer:
tan x = \(\frac { 4 }{ 3 } \)

Question 19.
REASONING
How does the tangent of an acute angle in a right triangle change as the angle measure increases? Justify your answer.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 19

Question 20.
CRITICAL THINKING
For what angle measure(s) is the tangent of an acute angle in a right triangle equal to 1? greater than 1? less than 1? Justify your answer.

Answer:
In order for the tangent of an angle to equal 1, the opposite and adjacent sides of a right triangle must be the same. This means the right triangle is an isosceles right triangle so the angles are 45 – 45 – 90. The acute angle must be 1. In order for the tangent to be greater than 1, the opposite side must be greater than the adjacent side. This means the angle must be between 45 and 90 degrees. If the tangent is less than 1, this means the opposite side must be smaller than the adjacent side. The acute angle must be between 0 and 45.

Question 21.
MAKING AN ARGUMENT
Your family room has a sliding-glass door. You want to buy an awning for the door that will be just long enough to keep the Sun out when it is at its highest point in the sky. The angle of elevation of the rays of the Sun at this points is 70°, and the height of the door is 8 feet. Your sister claims you can determine how far the overhang should extend by multiplying 8 by tan 70°. Is your sister correct? Explain.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 133
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 21

Question 22.
HOW DO YOU SEE IT?
Write expressions for the tangent of each acute angle in the right triangle. Explain how the tangent of one acute angle is related to the tangent of the other acute angle. What kind of angle pair is ∠A and ∠B?
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 134

Answer:
tan A = \(\frac { BC }{ AC } \) = \(\frac { a }{ b } \)
tan B = \(\frac { AC }{ BC } \) = \(\frac { b }{ a } \)

Question 23.
REASONING
Explain why it is not possible to find the tangent of a right angle or an obtuse angle.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 23

Question 24.
THOUGHT PROVOKING
To create the diagram below. you begin with an isosceles right triangle with legs 1 unit long. Then the hypotenuse of the first triangle becomes the leg of a second triangle, whose remaining leg is 1 unit long. Continue the diagram Until you have constructed an angle whose tangent is \(\frac{1}{\sqrt{6}}\). Approximate the measure of this angle.

Answer:

Question 25.
PROBLEM SOLVING
Your class is having a class picture taken on the lawn. The photographer is positioned 14 feet away from the center of the class. The photographer turns 50° to look at either end of the class.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 135
a. What is the distance between the ends of the class?
b. The photographer turns another 10° either way to see the end of the camera range. If each student needs 2 feet of space. about how many more students can fit at the end of each row? Explain.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 25

Question 26.
PROBLEM SOLVING
Find the perimeter of the figure. where AC = 26, AD = BF, and D is the midpoint of \(\overline{A C}\).
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 136

Answer:

Maintaining Mathematical proficiency

Find the value of x.

Question 27.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 137
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 27

Question 28.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 138
Answer:
longer side = shorter side • √3
7 = x√3
x = \(\frac { 7 }{ √3 } \)
x = 4.04

Question 29.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 139
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.4 Ans 29

9.5 The Sine and Cosine Ratios

Exploration 1

Work with a partner: Use dynamic geometry software.

a. Construct ∆ABC, as shown. Construct segments perpendicular to \(\overline{A C}\) to form right triangles that share vertex A arid are similar to ∆ABC with vertices, as shown.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 140
Answer:

b. Calculate each given ratio to complete the table for the decimal values of sin A and cos A for each right triangle. What can you conclude?
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 141
Answer:

Communicate Your Answer

Question 2.
How is a right triangle used to find the sine and cosine of an acute angle? Is there a unique right triangle that must be used?
Answer:

Question 3.
In Exploration 1, what is the relationship between ∠A and ∠B in terms of their measures’? Find sin B and cos B. How are these two values related to sin A and cos A? Explain why these relationships exist.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Lesson 9.5 The Sine and Cosine Ratios

Monitoring Progress

Question 1.
Find sin D, sin F, cos D, and cos F. Write each answer as a fraction and as a decimal rounded to four places.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 142
Answer:
sin D = \(\frac { 7 }{ 25 } \)
sin F = \(\frac { 24 }{ 25 } \)
cos D = \(\frac { 24 }{ 25 } \)
cos F = \(\frac { 7 }{ 25 } \)

Explanation:
sin D = \(\frac { EF }{ DF } \) = \(\frac { 7 }{ 25 } \)
sin F = \(\frac { DE }{ DF } \) = \(\frac { 24 }{ 25 } \)
cos D = \(\frac { DE }{ DF } \) = \(\frac { 24 }{ 25 } \)
cos F = \(\frac { EF }{ DF } \) = \(\frac { 7 }{ 25 } \)

Question 2.
Write cos 23° in terms of sine.

Answer:
cos X = sin(90 – X)
cos 23° = sin (90 – 23)
= sin(67)
So, cos 23° = sin 67°

Question 3.
Find the values of u and t using sine and cosine. Round your answers to the nearest tenth.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 143
Answer:
t = 7.2, u = 3.3

Explanation:
sin 65 = \(\frac { t }{ 8 } \)
0.906 = \(\frac { t }{ 8 } \)
t = 7.2
cos 65 = \(\frac { u }{ 8 } \)
0.422 x 8 = u
u = 3.3

Question 4.
Find the sine and cosine of a 60° angle.

Answer:
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 9.5 1
sin 60° = \(\frac { √3 }{ 2 } \)
cos 60° = \(\frac { 1 }{ 2 } \)

Question 5.
WHAT IF?
In Example 6, the angle of depression is 28°. Find the distance x you ski down the mountain to the nearest foot.

Answer:
sin 28° = \(\frac { 1200 }{ x } \)
x = \(\frac { 1200 }{ 0.469 } \)
x = 2558.6

Exercise 9.5 The Sine and Cosine Ratios

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The sine raio compares the length of ______________ to the length of _____________
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which ratio does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 144
sin B

Answer:
sin B = \(\frac { AC }{ BC } \)

cos C

Answer:
cos C = \(\frac { AC }{ BC } \)

tan B

Answer:
tan B = \(\frac { AC }{ AB } \)

\(\frac{A C}{B C}\)

Answer:
\(\frac{A C}{B C}\) = sin B

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, find sin D, sin E, cos D, and cos E. Write each answer as a Fraction and as a decimal rounded to four places.

Question 3.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 145
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 3

Question 4.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 146

Answer:
sin D = \(\frac { 35 }{ 37 } \)
sin E = \(\frac { 12 }{ 37 } \)
cos D = \(\frac { 12 }{ 37 } \)
cos E = \(\frac { 35 }{ 37 } \)

Question 5.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 147
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 5

Question 6.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 148

Answer:
sin D = \(\frac { 36 }{ 45 } \)
sin E = \(\frac { 27 }{ 45 } \)
cos D = \(\frac { 27 }{ 45 } \)
cos E = \(\frac { 36 }{ 45 } \)

Question 7.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 149
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 7

Question 8.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 150

Answer:
sin D = \(\frac { 8 }{ 17 } \)
sin E = \(\frac { 15 }{ 17 } \)
cos D = \(\frac { 15 }{ 17 } \)
cos E = \(\frac { 8 }{ 17 } \)

In Exercises 9 – 12. write the expression in terms of cosine.

Question 9.
sin 37°
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 9

Question 10.
sin 81°

Answer:
sin 81° = cos(90° – 81°) = cos9°

Question 11.
sin 29°
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 11

Question 12.
sin 64°

Answer:
sin 64° = cos(90° – 64°) = cos 26°

In Exercise 13 – 16, write the expression in terms of sine.

Question 13.
cos 59°
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 13

Question 14.
cos 42°

Answer:
cos 42° = sin(90° – 42°) = sin 48°

Question 15.
cos 73°
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 15

Question 16.
cos 18°

Answer:
cos 18° = sin(90° – 18°) = sin 72°

In Exercises 17 – 22, find the value of each variable using sine and cosine. Round your answers to the nearest tenth.

Question 17.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 151
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 17

Question 18.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 152

Answer:
p = 30.5, q = 14.8

Explanation:
sin 64° = \(\frac { p }{34 } \)
p = 0.898 x 34
p = 30.5
cos 64° = \(\frac { q }{ 34 } \)
q = 0.4383 x 34
q = 14.8

Question 19.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 153
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 19

Question 20.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 154

Answer:
s = 17.7, r = 19

Explanation:
sin 43° = \(\frac { s }{26 } \)
s = 0.681 x 26
s = 17.7
cos 43° = \(\frac { r }{ 26 } \)
r = 0.731 x 26
r = 19

Question 21.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 155
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 21

Question 22.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 156

Answer:
m = 6.7, n = 10.44

Explanation:
sin 50° = \(\frac { 8 }{n } \)
0.766 = \(\frac { 8 }{n } \)
n = 10.44
cos 50° = \(\frac { m }{ n } \)
0.642 = \(\frac { m }{ 10.44 } \)
m = 6.7

Question 23.
REASONING
Which ratios are equal? Select all that apply.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 157
sin X

cos X

sin Z

cos Z
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 23

Question 24.
REASONING
Which ratios arc equal to \(\frac{1}{2}\) Select all
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 158
sin L

Answer:
sin L = \(\frac { 2 }{ 4 } \) = \(\frac { 1 }{ 2 } \)

cos L

Answer:
cos L = \(\frac { 2√3 }{ 4 } \) = \(\frac { √3 }{ 2 } \)

sin J

Answer:
sin J = \(\frac { 2√3 }{ 4 } \) = \(\frac { √3 }{ 2 } \)

cos J

Answer:
cos J = \(\frac { 2 }{ 4 } \) = \(\frac { 1 }{ 2 } \)

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding sin A.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 159
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 25

Question 26.
WRITING
Explain how to tell which side of a right triangle is adjacent to an angle and which side is the hypotenuse.
Answer:

Question 27.
MODELING WITH MATHEMATICS
The top of the slide is 12 feet from the ground and has an angle of depression of 53°. What is the length of the slide?
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 160
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 27

Question 28.
MODELING WITH MATHEMATICS
Find the horizontal distance x the escalator covers.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 161

Answer:
cos 41 = \(\frac { x }{ 26 } \)
0.754 = \(\frac { x }{ 26 } \)
x = 19.6 ft

Question 29.
PROBLEM SOLVING
You are flying a kite with 20 feet of string extended. The angle of elevation from the spool of string to the kite is 67°.
a. Draw and label a diagram that represents the situation.
b. How far off the ground is the kite if you hold the spool 5 feet off the ground? Describe how the height where you hold the spool affects the height of the kite.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 29

Question 30.
MODELING WITH MATHEMATICS
Planes that fly at high speeds and low elevations have radar s sterns that can determine the range of an obstacle and the angle of elevation to the top of the obstacle. The radar of a plane flying at an altitude of 20,000 feet detects a tower that is 25,000 feet away. with an angle of elevation of 1°
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 162
a. How many feet must the plane rise to pass over the tower?

Answer:
sin 1 = \(\frac { h }{ 25000 } \)
0.017 = \(\frac { h }{ 25000 } \)
h = 425 ft
425 ft the plane rise to pass over the tower

b. PIanes Caillot come closer than 1000 feet vertically to any object. At what altitude must the plane fly in order to pass over the tower?

Answer:

Question 31.
MAKING AN ARGUMENT
Your friend uses che equation sin 49° = \(\frac{x}{16}\) to find BC. Your cousin uses the equation cos 41° = \(\frac{x}{16}\) to find BC. Who is correct? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 163
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 31

Question 32.
WRITING
Describe what you must know about a triangle in order to use the sine ratio and what you must know about a triangle in order to use the cosine ratio

Answer:
sin = \(\frac { opposite side }{ hypotenuse } \)
cos = \(\frac { adjacent side }{ hypotenuse } \)

Question 33.
MATHEMATICAL CONNECTIONS
If ∆EQU is equilateral and ∆RGT is a right triangle with RG = 2, RT = 1. and m ∠ T = 90°, show that sin E = cos G.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 33

Question 34.
MODELING WITH MATHEMATICS
Submarines use sonar systems, which are similar to radar systems, to detect obstacles, Sonar systems use sound to detect objects under water.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 164

a. You are traveling underwater in a submarine. The sonar system detects an iceberg 4000 meters a head, with an angle of depression of 34° to the bottom of the iceberg. How many meters must the submarine lower to pass under the iceberg?

Answer:
tan 34 = \(\frac { x }{ 4000 } \)
.674 = \(\frac { x }{ 4000 } \)
x = 2696

b. The sonar system then detects a sunken ship 1500 meters ahead. with an angle of elevation of 19° to the highest part of the sunken ship. How many meters must the submarine rise to pass over the sunken ship?

Answer:
tan 19 = \(\frac { x }{ 1500 } \)
0.344 = \(\frac { x }{ 1500 } \)
x = 516 m

Question 35.
ABSTRACT REASONING
Make a conjecture about how you could use trigonometric ratios to find angle measures in a triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 35

Question 36.
HOW DO YOU SEE IT?
Using only the given information, would you use a sine ratio or a cosine ratio to find the length of the hypotenuse? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 166

Answer:
sin 29 = \(\frac { 9 }{ x } \)
0.48 = \(\frac { 9 }{ x } \)
x = 18.75
The length of hypotenuse is 18.75

Question 37.
MULTIPLE REPRESENTATIONS
You are standing on a cliff above an ocean. You see a sailboat from your vantage point 30 feet above the ocean.

a. Draw and label a diagram of the situation.
b. Make a table showing the angle of depression and the length of your line of sight. Use the angles 40°, 50°, 60°, 70°, and 80°.
c. Graph the values you found in part (b), with the angle measures on the x-axis.
d. Predict the length of the line of sight when the angle of depression is 30°.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 37.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 37.2
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 37.3

Question 38.
THOUGHT PROVOKING
One of the following infinite series represents sin x and the other one represents cos x (where x is measured in radians). Which is which? Justify your answer. Then use each series to approximate the sine and cosine of \(\frac{\pi}{6}\).
(Hints: π = 180°; 5! = 5 • 4 • 3 • 2 • 1; Find the values that the sine and cosine ratios approach as the angle measure approaches zero).
a.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 166

Answer:
For x = 0
0 – \(\frac { 0³ }{ 3! } \) + \(\frac { 0⁵ }{ 5! } \) – \(\frac { 0⁷ }{ 7! } \) + . . . = 0
sin x = x – \(\frac { x³ }{ 3! } \) + \(\frac { x⁵ }{ 5! } \) – \(\frac { x⁷ }{ 7! } \) + . . .
sin \(\frac { π }{ 6 } \) = 0.5

b.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 167

Answer:
1 – \(\frac { 1² }{ 2! } \) + \(\frac { 1⁴ }{ 4! } \) – \(\frac { 1⁶ }{ 6! } \) + . .  = 1
cos x =x1 – \(\frac { x² }{ 2! } \) + \(\frac { x⁴ }{ 4! } \) – \(\frac { x⁶ }{ 6! } \) + . .
cos \(\frac { π }{ 6 } \) = 0.86

Question 39.
CRITICAL THINKING
Let A be any acute angle of a right triangle. Show that
(a) tan A = \(\frac{\sin A}{\cos A}\) and
(b) (sin A)2 + (cos A)2 = 1.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 39

Question 40.
CRITICAL THINKING
Explain why the area ∆ ABC in the diagram can be found using the formula Area = \(\frac{1}{2}\) ab sin C. Then calculate the area when a = 4, b = 7, and m∠C = 40°:
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 168
Answer:
Area = \(\frac{1}{2}\) ab sin C
= \(\frac{1}{2}\) (4 x 7) sin 40°
= 14 x 0.642
= 8.988

Maintaining Mathematical Proficiency

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 41.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 169
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 41

Question 42.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 170
Answer:
x = 12√2

Explanation:
c² = a² + b²
x² = 12² + 12²
x² = 144 + 144
x² = 288
x = 12√2

Question 43.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 171
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.5 Ans 43

Question 44.
Big Ideas Math Answer Key Geometry Chapter 9 Right Triangles and Trigonometry 172

Answer:
x = 6√2

Explanation:
c² = a² + b²
9² = x² + 3²
81 = x² + 9
x² = 81 – 9
x = 6√2

9.6 Solving Right Triangles

Exploration 1

Solving Special Right Triangles

Work with a partner. Use the figures to find the values of the sine and cosine of ∠A and ∠B. Use these values to find the measures of ∠A and ∠B. Use dynamic geometry software to verify your answers.
a.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 173
Answer:

b.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 174
Answer:

Exploration 2

Solving Right Triangles

Work with a partner: You can use a calculator to find the measure of an angle when you know the value of the sine, cosine, or tangent of the rule. Use the inverse sine, inverse cosine, 0r inverse tangent feature of your calculator to approximate the measures of ∠A and ∠B to the nearest tenth of a degree. Then use dynamic geometry software to verify your answers.
ATTENDING TO PRECISION
To be proficient in math, you need to calculate accurately and efficiently, expressing numerical answers with a degree of precision appropriate for the problem context.
a.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 175
Answer:

b.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 176
Answer:

Communicate Your Answer

Question 3.
When you know the lengths of the sides of a right triangle, how can you find the measures of the two acute angles?
Answer:

Question 4.
A ladder leaning against a building forms a right triangle with the building and the ground. The legs of the right triangle (in meters) form a 5-12-13 Pythagorean triple. Find the measures of the two acute angles to the nearest tenth of a degree.
Answer:

Lesson 9.6 Solving Right Triangles

Determine which of the two acute angles has the given trigonometric ratio.

Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 177

Question 1.
The sine of the angle is \(\frac{12}{13}\).

Answer:
Sin E = \(\frac{12}{13}\)
m∠E = sin-1(\(\frac{12}{13}\)) = 67.3°

Question 2.
The tangent of the angle is \(\frac{5}{12}\)

Answer:
tan F = \(\frac{5}{12}\)
m∠F = tan-1(\(\frac{5}{12}\)) = 22.6°

Let ∠G, ∠H, and ∠K be acute angles. Use a calculator to approximate the measures of ∠G, ∠H, and ∠K to the nearest tenth of a degree.

Question 3.
tan G = 0.43

Answer:
∠G = inverse tan of 0.43 = 23.3°

Question 4.
sin H = 0.68

Answer:
∠H = inverse sin of 0.68 = 42.8°

Question 5.
cos K = 0.94

Answer:
∠K = inverse cos of 0.94 = 19.9°

Solve the right triangle. Round decimal answers to the nearest tenth.

Question 6.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 178

Answer:
DE = 29, ∠D = 46.05°, ∠E = 42.84°

Explanation:
c² = a² + b²
x² = 20² + 21²
x² = 400 + 441
x² = 841
x = 29
sin D = \(\frac { 21 }{ 29 } \)
∠D = 46.05
sin E = \(\frac { 20 }{ 29 } \)
∠E = 42.84

Question 7.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 179

Answer:
GJ = 60, ∠G = 56.09°, ∠H = 33.3°

Explanation:
c² = a² + b²
109² = 91² + x²
x² = 11881 – 8281
x² = 3600
x = 60
sin G = \(\frac { 91 }{ 109 } \)
∠G = 56.09
sin H = \(\frac { 60 }{ 109 } \)
∠H = 33.3

Question 8.
Solve the right triangle. Round decimal answers to the nearest tenth.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 180

Answer:
XY = 13.82, YZ = 6.69, ∠Y = 37.5

Explanation:
cos 52 = \(\frac { 8.5 }{ XY } \)
0.615 = \(\frac { 8.5 }{ XY } \)
XY = 13.82
sin 52 = \(\frac { YZ }{ XY } \)
0.788 = \(\frac { YZ }{ 8.5 } \)
YZ = 6.69
sin Y = \(\frac { 8.5 }{ 13.82 } \)
∠Y = 37.5

Question 9.
WHAT IF?
In Example 5, suppose another raked stage is 20 feet long from front to back with a total rise of 2 feet. Is the raked stage within your desired range?

Answer:
x = inverse sine of \(\frac { 2 }{ 20 } \)
x = 5.7°

Exercise 9.6 Solving Right Triangles

Question 1.
COMPLETE THE SENTENCE
To solve a right triangle means to find the measures of all its ________ and _______ .
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 1

Question 2.
WRITING
Explain when you can use a trigonometric ratio to find a side length of a right triangle and when you can use the Pythagorean Theorem (Theorem 9.1 ).
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. determine which of the two acute angles has the given trigonometric ratio.

Question 3.
The cosine of the angle is \(\frac{4}{5}\)
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 181
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 3

Question 4.
The sine of the angle is \(\frac{5}{11}\)
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 182

Answer:
Sin(angle) = \(\frac { opposite }{ hypo } \)
sin A = \(\frac{5}{11}\)
The acute angle that has a sine of the angle is \(\frac{5}{11}\) is ∠A.

Question 5.
The sine of the angle is 0.95.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 183
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 5

Question 6.
The tangent of the angle is 1.5.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 184

Answer:
tan(angle) = \(\frac { opposite }{ adjacent } \)
1.5 = \(\frac { 18 }{ 12 } \)
tan C = 1.5
The acute angle that has a tangent of the angle is 1.5 ∠C.

In Exercises 7 – 12, let ∠D be an acute angle. Use a calculator to approximate the measure of ∠D to the nearest tenth of a degree.

Question 7.
sin D = 0.75
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 7

Question 8.
sin D = 0.19

Answer:
sin D = 0.19
∠D = inverse sine of 0.19
∠D = 10.9°

Question 9.
cos D = 0.33
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 9

Question 10.
cos D = 0.64

Answer:
cos D = 0.64
∠D = inverse cos of 0.64
∠D = 50.2°

Question 11.
tan D = 0.28
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 11

Question 12.
tan D = 0.72

Answer:
tan D = 0.72
∠D = inverse tan of 0.72
∠D = 35.8°

In Exercises 13 – 18. solve the right triangle. Round decimal answers to the nearest tenth.

Question 13.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 185
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 13

Question 14.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 186
Answer:
ED = 2√65, ∠E = 59.3, ∠D = 29.7

Explanation:
c² = 8² + 14²
x² = 64 + 196
x² = 260
x = 2√65
sin E = \(\frac { 14 }{ 2√65 } \)
∠E = 59.3
sin D = \(\frac { 8 }{ 2√65 } \)
∠D = 29.7

Question 15.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 187
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 15

Question 16.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 188
Answer:
HJ = 2√15, ∠G = 28.9, ∠J = 61

Explanation:
c² = a² + b²
16² = 14² + x²
x² = 256 – 196
x² = 60
x = 2√15
sin G = \(\frac { 2√15 }{ 16 } \)
∠G = 28.9
sin J = \(\frac { 14 }{ 16 } \)
∠J = 61

Question 17.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 189
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 17

Question 18.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 190

Answer:
RT = 17.8, RS = 9.68, ∠T = 32.8

Explanation:
sin 57 = \(\frac { 15 }{ x } \)
0.838 = \(\frac { 15 }{ x } \)
x = 17.899
RT = 17.8
cos 57 = \(\frac { x }{ 17.8 } \)
0.544 = \(\frac { x }{ 17.8 } \)
x = 9.68
RS = 9.68
sin T = \(\frac { 9.68 }{ 17.8 } \)
∠T = 32.8

Question 19.
ERROR ANALYSIS
Describe and correct the error in using an inverse trigonometric ratio.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 191
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 19

Question 20.
PROBLEM SOLVING
In order to unload clay easily. the body of a dump truck must be elevated to at least 45° The body of a dump truck that is 14 feet long has been raised 8 feet. Will the clay pour out easily? Explain your reasoning.

Answer:
Angle of elevation: sin x = \(\frac { 8 }{ 14 } \)
x = inverse sine of \(\frac { 8 }{ 14 } \) = 34.9
The clay will not pour out easily.

Question 21.
PROBLEM SOLVING
You are standing on a footbridge that is 12 feet above a lake. You look down and see a duck in the water. The duck is 7 feet away from the footbridge. What is the angle of elevation from the duck to you
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 192
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 21

Question 22.
HOW DO YOU SEE IT?
Write three expressions that can be used to approximate the measure of ∠A. Which expression would you choose? Explain your choice.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 193

Answer:
Three expressions are ∠A = inverse tan of (\(\frac { 15 }{ 22 } \)) = 34.2°
∠A = inverse sine of (\(\frac { 15 }{ BA } \))
∠A = inverse cos of (\(\frac { 22 }{ BA } \))

Question 23.
MODELING WITH MATHEMATICS
The Uniform Federal Accessibility Standards specify that awheel chair ramp may not have an incline greater than 4.76. You want to build a ramp with a vertical rise of 8 inches. you want to minimize the horizontal distance taken up by the ramp. Draw a diagram showing the approximate dimensions of your ramp.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 23

Question 24.
MODELING WITH MATHEMATICS
The horizontal part of a step is called the tread. The vertical part is called the riser. The recommended riser – to – tread ratio is 7 inches : 11 inches.

a. Find the value of x for stairs built using the recommended riser-to-tread ratio.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 194

Answer:

b. you want to build stairs that are less steep than the stairs in part (a). Give an example of a riser – to – tread ratio that you could use. Find the value of x for your stairs.
Answer:

Question 25.
USING TOOLS
Find the measure of ∠R without using a protractor. Justify your technique.
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 195
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 25

Question 26.
MAKING AN ARGUMENT
Your friend claims that tan-1x = \(\frac{1}{\tan x}\). Is your friend correct? Explain your reasoning.

Answer:
No
For example
tan-1(√3) = 60
\(\frac{1}{\tan √3}\) = 33.1

USING STRUCTURE
In Exercises 27 and 28, solve each triangle.

Question 27.
∆JKM and ∆LKM
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 196
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 27.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 27.2

Question 28.
∆TUS and ∆VTW
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 197

Answer:
TS = 8.2, UT = 7.3, ∠T = 28.6
TV = 13.2, TW = 9.6, ∠V = 46, ∠T = 42.84

Explanation:
tan 64 = \(\frac { TS }{ 4 } \)
TS = 2.05 x 4
TS = 8.2
sin 64 = \(\frac { UT }{ 8.2 } \)
0.898= \(\frac { UT }{ 8.2 } \)
UT = 7.3
sin T = \(\frac { 4 }{ 8.2 } \)
∠T = 28.6
TV = TS + SV
TV = 8.2 + 5 = 13.2
13.2² = TW² + 9²
TW² = 174.24 – 81
TW = 9.6
sin V = \(\frac { 9.6 }{ 13.2 } \)
∠V = 46
sin T = \(\frac { 9 }{ 13.2 } \)
∠T = 42.84

Question 29.
MATHEMATICAL CONNECTIONS
Write an expression that can be used to find the measure of the acute angle formed by each line and the x-axis. Then approximate the angle measure to the nearest tenth of a degree.
a. y = 3x
b. y = \(\frac{4}{3}\)x + 4
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 29

Question 30.
THOUGHT PROVOKING
Simplify each expression. Justify your answer.
a. sin-1 (sin x)

Answer:
sin-1 (sin x) = x

b. tan(tan-1 y)

Answer:
tan(tan-1 y) = y

C. cos(cos-1 z)

Answer:
cos(cos-1 z) = z

Question 31.
REASONING
Explain why the expression sin-1 (1.2) does not make sense.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 31

Question 32.
USING STRUCTURE
The perimeter of the rectangle ABCD is 16 centimeters. and the ratio of its width to its length is 1 : 3. Segment BD divides the rectangle into two congruent triangles. Find the side lengths and angle measures of these two triangles.

Answer:
The perimeter of the rectangle ABCD is 16 centimeters
2(l + b) = 16
l + b = 8
b : l = 1 : 3
4x = 8
x = 2
l = 6, b = 2
BD = √(6² + 2²) = √40 = 2√10
sin B = \(\frac { AD }{ BD } \) = \(\frac { 2 }{ 2√10 } \)
∠ABD = 18.4
∠CBD = 71.6
sin D = \(\frac { AB }{ BD } \) = \(\frac { 6 }{ 2√10 } \)
∠ADB = 71
∠CDB = 19

Maintaining Mathematical Proficiency

Solve the equation

Question 33.
\(\frac{12}{x}=\frac{3}{2}\)
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 33

Question 34.
\(\frac{13}{9}=\frac{x}{18}\)

Answer:
\(\frac{13}{9}=\frac{x}{18}\)
x = 1.44 x 18
x = 26

Question 35.
\(\frac{x}{2.1}=\frac{4.1}{3.5}\)
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.6 Ans 35

Question 36.
\(\frac{5.6}{12.7}=\frac{4.9}{x}\)

Answer:
\(\frac{5.6}{12.7}=\frac{4.9}{x}\)
0.44 = 4.9/x
x = 11.13

9.7 Law of Sines and Law of Cosines

Exploration 1

Discovering the Law of Sines

Work with a partner.

a. Copy and complete the table for the triangle shown. What can you conclude?
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 198
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 199
Answer:

b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about the relationship between the sines of the angles and the lengths of the sides of a triangle.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technology to compare predictions with data.
Answer:

Exploration 2

Discovering the Law of Cosines

Work with a partner:

a. Copy and complete the table for the triangle in Exploration 1 (a). What can you conclude?
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 200
Answer:

b. Use dynamic geometry software to draw two other triangles. Copy and complete the table in part (a) for each triangle. Use your results to write a conjecture about what you observe in the completed tables.
Answer:

Communicate Your Answer

Question 3.
What are the Law of Sines and the Law of Cosines?
Answer:

Question 4.
When would you use the Law of Sines to solve a triangle? When would you use the Law of Cosines to solve a triangle?
Answer:

Lesson 9.7 Law of Sines and Law of Cosines

Monitoring Progress

Use a calculator to find the trigonometric ratio. Round your answer to four decimal places.

Question 1.
tan 110°

Answer:
tan 110° = -2.7474

Question 2.
sin 97°

Answer:
sin 97° = 0.9925

Question 3.
cos 165°

Answer:
cos 165° = -0.9659

Find the area of ∆ABC with the given side lengths and included angle. Round your answer to the nearest tenth.

Question 4.
m ∠ B = 60°, a = 19, c = 14

Answer:
Area = 155.18

Explanation:
Area = \(\frac { 1 }{ 2 } \) ac sin B
= \(\frac { 1 }{ 2 } \) (19 x 14) sin 60°
= 133 x 0.866
= 155.18

Question 5.
m ∠ C = 29°, a = 38, b = 31

Answer:
Area = 282.72

Explanation:
Area = \(\frac { 1 }{ 2 } \) ab sin C
= \(\frac { 1 }{ 2 } \) (38 x 31) sin 29
= 598 x 0.48
= 282.72

Solve the triangle. Round decimal answers to the nearest tenth.

Question 6.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 201

Answer:
∠C = 46.6, ∠B = 82.4, AC = 23.57

Explanation:
Using law of sines
\(\frac { c }{ sin C } \) = \(\frac { a }{ sin A } \)
\(\frac { 17 }{ sin C } \) = \(\frac { 18 }{ sin 51 } \)
\(\frac { 17 }{ sin C } \) = \(\frac { 18 }{ 0.77 } \)
sin C = 0.7274
∠C = 46.6
∠A + ∠B + ∠C  = 180
51 + 46.6 + ∠B = 180
∠B = 82.4
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { 0.99 }{ b } \) = \(\frac { 0.77 }{ 18 } \)
b = 23.57

Question 7.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 202

Answer:
∠B = 31.3, ∠C  = 108.7, c = 23.6

Explanation:
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin B }{ 13 } \) = \(\frac { sin 40 }{ 16 } \)
sin B = \(\frac { .64 }{ 16 } \) x 13
sin B = 0.52
∠B = 31.3
∠A + ∠B + ∠C  = 180
40 + 31.3 + ∠C  = 180
∠C  = 108.7
\(\frac { c }{ sin C } \) = \(\frac { a }{ sin A } \)
\(\frac { c }{ sin 108.7 } \) = \(\frac {16 }{ sin 40 } \)
c = \(\frac {16 }{ .64 } \) x 0.947
c = 23.6

Solve the triangle. Round decimal answers to the nearest tenth.

Question 8.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 203

Answer:
∠C = 66, a = 4.36, c = 8.27

Explanation:
∠A + ∠B + ∠C = 180
29 + 85 + ∠C = 180
∠C = 66
\(\frac { a }{ sin A } \) = \(\frac { b }{ sin B } \) = \(\frac { c }{ sin C } \)
\(\frac { a }{ sin 29 } \) = \(\frac { 9 }{ sin 85 } \)
\(\frac { a }{ 0.48 } \) = \(\frac { 9 }{ 0.99 } \)
a = 4.36
\(\frac { b }{ sin B } \) = \(\frac { c }{ sin C } \)
\(\frac { 9 }{ sin 85 } \) = \(\frac { c }{ sin 66 } \)
\(\frac { 9 }{ 0.99 } \) = \(\frac { c }{ 0.91 } \)
c = 8.27

Question 9.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 204

Answer:
∠A = 29, b = 19.37, c = 20.41

Explanation:
∠A + ∠B + ∠C  = 180
∠A + 70 + 81 = 180
∠A = 29
\(\frac { a }{ sin A } \) = \(\frac { b }{ sin B } \) = \(\frac { c }{ sin C } \)
\(\frac { 10 }{ sin 29 } \) = \(\frac { b }{ sin 70 } \)
\(\frac { 10 }{ 0.48 } \) = \(\frac { b }{ 0.93 } \)
b = 19.37
\(\frac { a }{ sin A } \) = \(\frac { c }{ sin C } \)
\(\frac { 10 }{ sin 29 } \) = \(\frac { c }{ sin 81 } \)
\(\frac { 10 }{ 0.48 } \) = \(\frac { c }{ 0.98 } \)
c = 20.41

Question 10.
WHAT IF?
In Example 5, what would be the length of a bridge from the South Picnic Area to the East Picnic Area?

Answer:
The length of a bridge from the South Picnic Area to the East Picnic Area is 188 m.

Explanation:
\(\frac { a }{ sin A } \) = \(\frac { b }{ sin B } \) = \(\frac { c }{ sin C } \)
\(\frac { a }{ sin 71 } \) = \(\frac { 150 }{ sin 49 } \)
a = 188

Solve the triangle. Round decimal answers to the nearest tenth.

Question 11.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 205

Answer:
b = 61.3, ∠A = 46, ∠C  = 46

Explanation:
b² = a² + c² − 2ac cos B
b² = 45² + 43² – 2(45)(43) cos 88
b² = 2025 + 1849 – 3870 x 0.03 = 3757.9
b = 61.3
\(\frac { sin A }{ a } \) = \(\frac { sin B }{ b } \)
\(\frac { sin A }{ 45 } \) = \(\frac { sin 88 }{ 61.3 } \)
sin A = 0.72
∠A = 46
∠A + ∠B + ∠C  = 180
46 + 88 + ∠C  = 180
∠C  = 46

Question 12.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 206

Answer:
a = 41.1, ∠C  = 35.6, ∠B = 30.4

Explanation:
a² = b² + c² − 2bc cos A
a² = 23² + 26² – 2(23)(26) cos 114
a² = 529 + 676 – 1196 x -0.406
a² = 1690.5
a = 41.1
\(\frac { sin 114 }{ 41.1 } \) = \(\frac { sin B }{ 23 } \)
0.02 = \(\frac { sin B }{ 23 } \)
sin B = 0.507
∠B = 30.4
∠A + ∠B + ∠C  = 180
114 + 30.4 + ∠C  = 180
∠C  = 35.6

Question 13.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 207
Answer:
∠A = 41.4, ∠B = 81.8, ∠C  = 56.8

Explanation:
a² = b² + c² − 2bc cos A
4² = 6² + 5² – 2(6)(5) cos A
16 = 36 + 25 – 60 cos A
-45 = – 60 cos A
cos A = 0.75
∠A = 41.4
\(\frac { sin 41.4 }{ 4 } \) = \(\frac { sin B }{ 6 } \)
0.165 = \(\frac { sin B }{ 6 } \)
sin B = 0.99
∠B = 81.8
∠A + ∠B + ∠C  = 180
41.4 + 81.8 + ∠C  = 180
∠C  = 56.8

Question 14.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 208

Answer:
∠B = 81.8, ∠A = 58.6, ∠C  = 39.6

Explanation:
a² = b² + c² − 2bc cos A
23² = 27² + 16² – 2(27)(16) cos A
529 = 729 + 256 – 864 cos A
456 = 864 cos A
cos A = 0.52
∠A = 58.6
\(\frac { sin 58.6 }{ 23 } \) = \(\frac { sin B }{ 27 } \)
0.03 = \(\frac { sin B }{ 27 } \)
sin B = 0.99
∠B = 81.8
∠A + ∠B + ∠C  = 180
58.6 + 81.8 + ∠C  = 180
∠C  = 39.6

Exercise 9.7 Law of Sines and Law of Cosines

Vocabulary and Core Concept Check

Question 1.
WRITING
What type of triangle would you use the Law of Sines or the Law of Cosines to solve?
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 1

Question 2.
VOCABULARY
What information do you need to use the Law of Sines?

Answer:

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 8, use a calculator to find the trigonometric ratio, Round your answer to four decimal places.

Question 3.
sin 127°
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 3

Question 4.
sin 98°

Answer:
sin 98° = 0.9902

Question 5.
cos 139°
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 5

Question 6.
cos 108°

Answer:
cos 108° = -0.309

Question 7.
tan 165°
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 7

Question 8.
tan 116°

Answer:
tan 116° = -2.0503

In Exercises 9 – 12, find the area of the triangle. Round your answer to the nearest tenth.

Question 9.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 209
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 9

Question 10.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 210

Answer:
Area = \(\frac{1}{2}\)bc sin A
Area = \(\frac{1}{2}\)(28)(24) sin83
Area = 332.64

Question 11.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 211
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 11

Question 12.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 212

Answer:
Area = \(\frac{1}{2}\)ab sin C
Area = \(\frac{1}{2}\)(15)(7) sin 96
Area = 51.9

In Exercises 13 – 18. solve the triangle. Round decimal answers to the nearest tenth.

Question 13.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 213
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 13

Question 14.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 214

Answer:
∠B = 38.3, ∠A = 37.7, a = 15.7

Explanation:
\(\frac { sin B }{16 } \) = \(\frac { sin 104 }{ 25 } \)
sin B = 0.62
∠B = 38.3
∠A + ∠B + ∠C = 180
∠A + 38.3 + 104 = 180
∠A = 37.7
\(\frac { sin 37.7 }{ a } \) = \(\frac { sin 104 }{ 25 } \)
\(\frac { 0.61 }{ a } \) = 0.0388
a = 15.7

Question 15.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 215
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 15

Question 16.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 216

Answer:
∠B = 65, b = 33.55, a = 24.4

Explanation:
∠A + ∠B + ∠C = 180
42 + 73 + ∠B = 180
∠B = 65
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 65 }{ b } \) = \(\frac { sin 73 }{ 34 } \)
b = 33.55
\(\frac { sin A }{ a } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 42 }{ a } \) = \(\frac { sin 73 }{ 34 } \)
a = 24.4

Question 17.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 217
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 17

Question 18.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 218

Answer:
∠C = 90, b = 39.56, a = 17.6

Explanation:
∠A + ∠B + ∠C = 180
24 + 66 + ∠C = 180
∠C = 90
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 66 }{ b } \) = \(\frac { sin 90 }{ 43 } \)
\(\frac { .91 }{ b } \) = 0.023
b = 39.56
\(\frac { sin A }{ a } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 24 }{a } \) = \(\frac { sin 90 }{ 43 } \)
\(\frac { 0.406 }{a } \) = 0.023
a = 17.6

In Exercises 19 – 24, solve the triangle. Round decimal answers to the nearest tenth.

Question 19.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 219
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 19

Question 20.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 220

Answer:
b = 29.9, ∠A = 26.1, ∠C = 15.07

Explanation:
b² = a² + c² – 2ac cos B
b² = 20² + 12² – 2(12)(20) cos 138
b² = 400 + 144 – 480 (-0.74)
b² = 899.2
b = 29.9
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin 138 }{ 29.9 } \) = \(\frac { sin A }{ 20 } \)
\(\frac { 0.66 }{ 29.9 } \) = \(\frac { sin A }{ 20 } \)
sin A = 0.44
∠A = 26.1
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 138 }{ 29.9 } \) = \(\frac { sin C }{ 12 } \)
sin C = 0.26
∠C = 15.07

Question 21.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 221
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 21

Question 22.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 222

Answer:
∠A = 107.3, ∠B = 51.6, ∠C = 21.1

Explanation:
b² = a² + c² – 2ac cos B
28² = 18² + 13² – 2(18)(13) cos B
784 = 324 + 169 – 468 cos B
291 = 468 cos B
cos B = 0.62
∠B = 51.6
\(\frac { sin C }{ c } \) = \(\frac { sin B }{ b } \)
\(\frac { sin C }{ 13 } \) = \(\frac { sin 51.6 }{ 28 } \)
sin C = 0.36
∠C = 21.1
∠A + ∠B + ∠C = 180
51.6 + 21.1 + ∠A = 180
∠A = 107.3

Question 23.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 223
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 23

Question 24.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 224

Answer:
∠A = 23, ∠B = 132.1, ∠C = 24.9

Explanation:
b² = a² + c² – 2ac cos B
5² = 12² + 13² – 2(12)(13) cos B
25 = 144 + 169 – 312 cos B
288 = 312 cos B
cos B = 0.92
∠B = 23
\(\frac { sin C }{ c } \) = \(\frac { sin B }{ b } \)
\(\frac { sin C }{ 13 } \) = \(\frac { sin 23 }{ 5 } \)
sin C = 1.014
∠C = 24.9
∠A + ∠B + ∠C = 180
23 + 24.9 + ∠B = 180
∠B = 132.1

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding m ∠ C.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 225
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 25

Question 26.
ERROR ANALYSIS
Describe and correct the error in finding m ∠ A in ∆ABC when a = 19, b = 21, and c = 11.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 226

Answer:
a² = b² + c² – 2bc cos A
19² = 21² + 11² – 2(21)(11) cos A
361 = 441 + 121 – 462 cosA
201 = 462 cosA
cos A = 0.43
∠A = 64.5

COMPARING METHODS

In Exercise 27 – 32. tell whether you would use the Law of Sines, the Law of Cosines. or the Pythagorean Theorem (Theorem 9.1) and trigonometric ratios to solve the triangle with the given information. Explain your reasoning. Then solve the triangle.

Question 27.
m ∠ A = 72°, m ∠ B = 44°, b = 14
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 27

Question 28.
m ∠ B = 98°, m ∠ C = 37°, a = 18

Answer:
∠A = 45, b = 25.38, c = 15.38

Explanation:
∠A + ∠B + ∠C = 180
∠A + 98 + 37 = 180
∠A = 45
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin 98 }{ b } \) = \(\frac { sin 45 }{ 18 } \)
\(\frac { 0.99 }{ b } \) = 0.039
b = 25.38
\(\frac { sin A }{ a } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 45 }{ 18 } \) = \(\frac { sin 37 }{ c } \)
0.039 = \(\frac { sin 37 }{ c } \)
c = 15.38

Question 29.
m ∠ C = 65°, a = 12, b = 21
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 29

Question 30.
m ∠ B = 90°, a = 15, c = 6

Answer:
b = 3√29, ∠A = 66.9, ∠C = 23.1

Explanation:
b² = a² + c²- 2ac cos B
b² = 15² + 6² – 2(15)(6) cos 90
= 225 + 36 – 180(0)
b² = 261
b = 3√29
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin 90 }{ 3√29 } \) = \(\frac { sin A }{ 15 } \)
sin A = 0.92
∠A = 66.9
∠A + ∠B + ∠C = 180
66.9 + 90 + ∠C = 180
∠C = 23.1

Question 31.
m ∠ C = 40°, b = 27, c = 36
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 31

Question 32.
a = 34, b = 19, c = 27

Answer:
∠B = 33.9, ∠A = 78.5, ∠C = 67.6

Explanation:
b² = a² + c²- 2ac cos B
19² = 34² + 27²- 2(34)(27) cos B
361 = 1156 + 729 – 1836 cos B
cos B = 0.83
∠B = 33.9
\(\frac { sin 33.9 }{ 19 } \) = \(\frac { sin A }{ 34 } \)
sin A = 0.98
∠A = 78.5
∠A + ∠B + ∠C = 180
78.5 + 33.9 + ∠C = 180
∠C = 67.6

Question 33.
MODELING WITH MATHEMATICS
You and your friend are standing on the baseline of a basketball court. You bounce a basketball to your friend, as shown in the diagram. What is the distance between you and your friend?
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 227
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 33

Question 34.
MODELING WITH MATHEMATICS
A zip line is constructed across a valley, as shown in the diagram. What is the width w of the valley?
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 228

Answer:
w = 92.5 ft

Explanation:
w² = 25² + 84² – 2(25)(84) cos 102
w² = 7681 – 4200 cos 102
w = 92.5 ft

Question 35.
MODELING WITH MATHEMATICS
You are on the observation deck of the Empire State Building looking at the Chrysler Building. When you turn 145° clockwise, you see the Statue of Liberty. You know that the Chrysler Building and the Empire Slate Building arc about 0.6 mile apart and that the Chrysler Building and the Statue of Liberty are about 5.6 miles apart. Estimate the distance between the Empire State Building and the Statue of Liberty.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 35.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 35.2

Question 36.
MODELING WITH MATHEMATICS
The Leaning Tower of Pisa in Italy has a height of 183 feet and is 4° off vertical. Find the horizontal distance d that the top of the tower is off vertical.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 229

Answer:

Question 37.
MAKING AN ARGUMENT
Your friend says that the Law of Sines can be used to find JK. Your cousin says that the Law of Cosines can be used to find JK. Who is correct’? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 230
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 37

Question 38.
REASONING
Use ∆XYZ
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 231
a. Can you use the Law of Sines to solve ∆XYZ ? Explain your reasoning.
Answer:

b. Can you use another method to solve ∆XYZ ? Explain your reasoning.
Answer:

Question 39.
MAKING AN ARGUMENT
Your friend calculates the area of the triangle using the formula A = \(\frac{1}{2}\)qr sin S and says that the area is approximately 208.6 square units. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 232
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 39

Question 40.
MODELING WITH MATHEMATICS
You are fertilizing a triangular garden. One side of the garden is 62 feet long, and another side is 54 feet long. The angle opposite the 62-foot side is 58°.
a. Draw a diagram to represent this situation.
b. Use the Law of Sines to solve the triangle from part (a).
c. One bag of fertilizer covers an area of 200 square feet. How many bags of fertilizer will you need to cover the entire garden?

Answer:
C = 47.6, A = 74.4, a = 70.4
9 bags of fertilizer.

Question 41.
MODELING WITH MATHEMATICS
A golfer hits a drive 260 yards on a hole that is 400 yards long. The shot is 15° off target.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 233
a. What is the distance x from the golfer’s ball to the hole?
b. Assume the golfer is able to hit the ball precisely the distance found in part (a). What is the maximum angle θ (theta) by which the ball can be off target in order to land no more than 10 yards fr0m the hole?
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 41.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 41.2

Question 42.
COMPARING METHODS
A building is constructed on top of a cliff that is 300 meters high. A person standing on level ground below the cliff observes that the angle of elevation to the top of the building is 72° and the angle of elevation to the top of the cliff is 63°.
a. How far away is the person from the base of the cliff?

Answer:
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 9.7 1

b. Describe two different methods you can use to find the height of the building. Use one of these methods to find the building’s height.

Answer:
Consider △SYZ and evaluate d using tangent function
tan SYZ = \(\frac { 300 }{ d } \)
d = \(\frac { 300 }{ tan 63 } \)
d = 152.86
The person is standing 152.86 m away from the base of the cliff.
Consider △XYS and evaluate h + 300
tan XYZ = \(\frac { h + 300 }{ d } \)
h = 152.86 x tan 72 – 300
h = 170.45
The building is 170.45 m high.

Question 43.
MATHEMATICAL CONNECTIONS
Find the values of x and y.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 234
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 43.1

Question 44.
HOW DO YOU SEE IT?
Would you use the Law of Sines or the Law of Cosines to solve the triangle?
Answer:

Question 45.
REWRITING A FORMULA
A Simplify the Law of Cosines for when the given angle is a right angle.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 45

Question 46.
THOUGHT PROVOKING
Consider any triangle with side lengths of a, b, and c. Calculate the value of s, which is half the perimeter of the triangle. What measurement of the triangle is represented by \(\sqrt{s(s-a)(s-b)(s-c)} ?\)
Answer:

Question 47.
ANALYZING RELATIONSHIPS
The ambiguous case of the Law of Sines occurs when you are given the measure of one acute angle. the length of one adjacent side, and the length of the side opposite that angle, which is less than the length of the adjacent side. This results in two possible triangles. Using the given information, find two possible solutions for ∆ABC
Draw a diagram for each triangle.
(Hint: The inverse sine function gives only acute angle measures. so consider the acute angle and its supplement for ∠B.)
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 235
a. m ∠ A = 40°, a = 13, b = 16
b. m ∠ A = 21°, a = 17, b = 32
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 47.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 47.2
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 47.3

Question 48.
ABSTRACT REASONING
Use the Law of Cosines to show that the measure of each angle of an equilateral triangle is 60°. Explain your reasoning.

Answer:
a² = b² + c²- 2bc cos A
a² = a² + a² – 2 aa cos A
a² = 2a² coas A
cos A = 1/2
∠A = 60

Question 49.
CRITICAL THINKING
An airplane flies 55° east of north from City A to City B. a distance of 470 miles. Another airplane flies 7° north of east from City A to City C. a distance of 890 miles. What is the distance between Cities B and C?
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 49

Question 50.
REWRITING A FORMULA
Follow the steps to derive the formula for the area of a triangle.
Area = \(\frac{1}{2}\)ab sin C.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 236
a. Draw the altitude from vertex B to \(\overline{A C}\). Label the altitude as h. Write a formula for the area of the triangle using h.
Answer:

b. Write an equation for sin C
Answer:

c. Use the results of parts (a) and (b) to write a formula for the area of a triangle that does not include h.
Answer:

Question 51.
PROVING A THEOREM
Follow the steps to use the formula for the area of a triangle to prove the Law of Sines (Theorem 9.9).

a. Use the derivation in Exercise 50 to explain how to derive the three related formulas for the area of a triangle.
Area = \(\frac{1}{2}\)bc sin A,
Area = \(\frac{1}{2}\)ac sin B,
Area = \(\frac{1}{2}\)ab sin C
b. why can you use the formulas in part (a) to write the following statement?
\(\frac{1}{2}\)bc sin A = \(\frac{1}{2}\)ac sin B = \(\frac{1}{2}\)ab sin C
c. Show how to rewrite the statement in part (b) to prove the Law of Sines. Justify each step.
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 51.1
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 51.2

Question 52.
PROVING A THEOREM
Use the given information to complete the two – column proof of the Law of Cosines (Theorem 9.10).
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 237
Given \(\overline{B D}\) is an altitude of ∆ABC.
Prove a2 = b2 + c2 – 2bc cos A

StatementsReasons
1. \(\overline{B D}\) is an altitude of ∆ABC.1. Given
2. ∆ADB and ∆CDB are right triangles.2. _______________________
3. a2 = (b – x)2 + h23. _______________________
4. _______________________4. Expand binomial.
5. x2 + h2 = c25. _______________________
6. _______________________6. Substitution Property of Equality
7. cos A = \(\frac{x}{c}\)7. _______________________
8. x = c cos A8. _______________________
9. a2 = b2 + c2 – 2bc Cos A9. _______________________

Answer:

Maintaining Mathematical Proficiency

Find the radius and diameter of the circle.

Question 53.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 238
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 53

Question 54.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 239

Answer:
The radius is 10 in and the diameter is 20 in.

Question 55.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 240
Answer:
Big Ideas Math Geometry Answers Chapter 9 Right Triangles and Trigonometry 9.7 Ans 55

Question 56.
Big Ideas Math Answers Geometry Chapter 9 Right Triangles and Trigonometry 241

Answer:
The radius is 50 in and the diameter is 100 in.

Right Triangles and Trigonometry Review

9.1 The Pythagorean Theorem

Find the value of x. Then tell whether the side lengths form a Pythagorean triple.

Question 1.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 242

Answer:
x = 2√34
The sides will not form a Pythagorean triple.

Explanation:
x² = 6² + 10²
x²= 36 + 100
x = 2√34

Question 2.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 243

Answer:
x = 12
The sides form a Pythagorean triple.

Explanation:
20² = 16²+ x²
400 = 256 + x²
x² = 144
x = 12

Question 3.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 244

Answer:
x = 2√30
The sides will not form a Pythagorean triple.

Explanation:
13² = 7² + x²
169 = 49 + x²
x = 2√30

Verify that the segments lengths form a triangle. Is the triangle acute, right, or obtuse?

Question 4.
6, 8, and 9

Answer:
9² = 81
6² + 8² = 36 + 64 = 100
9² < 6² + 8²
So, the triangle is acute

Question 5.
10, 2√2, and 6√3

Answer:
10² = 100
(2√2)² + (6√3)² = 8 + 108 = 116
So, the triangle is acute.

Question 6.
13, 18 and 3√55

Answer:
18² = 324
13² + (3√55)² = 169 + 495 = 664
So, the triangle is acute.

9.2 Special Right Triangles

Find the value of x. Write your answer in simplest form.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 245

Answer:
hypotenuse = leg • √2
x = 6√2

Question 8.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 246

Answer:
longer leg = shorter leg • √3
14 = x • √3
x = 8.08

Question 9.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 247

Answer:
longer leg = shorter leg • √3
x = 8√3 • √3
x = 24

9.3 Similar Right Triangles

Identify the similar triangles. Then find the value of x.

Question 10.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 248

Answer:
9 = √(6x)
81 = 6x
x = \(\frac { 27 }{ 2 } \)

Question 11.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 249

Answer:
x = √(6 x 4)
x = 2√6

Question 12.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 250

Answer:
\(\frac { RP }{ RQ } \) = \(\frac { SP }{ RS } \)
\(\frac { 9 }{ x } \) = \(\frac { 6 }{ 3 } \)
x = 3.5

Question 13.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 251

Answer:
\(\frac { SU }{ ST } \) = \(\frac { SV }{ VU } \)
\(\frac { 16 }{ 20 } \) = \(\frac { x }{ (x – 16) } \)
4(x – 16) = 5x
x = 16

Find the geometric mean of the two numbers.

Question 14.
9 and 25

Answer:
mean = √(9 x 25)
= 15

Question 15.
36 and 48

Answer:
mean = √(36 x 48)
= 24√3

Question 16.
12 and 42

Answer:
mean = √(12 x 42)
= 6√14

9.4 The Tangent Ratio

Find the tangents of the acute angles in the right triangle. Write each answer as a fraction and as a decimal rounded to four decimal places.

Question 17.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 253

Answer:
tan J = \(\frac { Opposite side }{ Adjacent side } \)
tan J = \(\frac { LK }{ JK } \) = \(\frac { 11 }{ 60 } \)
tan L = \(\frac { JK }{ LK } \) = \(\frac { 60 }{ 11 } \)

Question 18.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 253

Answer:
tan P = \(\frac { MN }{ MP } \) = \(\frac { 35 }{ 12 } \)
tan N = \(\frac { MP }{ MN } \) = \(\frac { 12 }{ 35 } \)

Question 19.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 254

Answer:
tan A = \(\frac { BC }{ AC } \) = \(\frac { 7 }{ 4√2 } \)
tan B = \(\frac { AC }{ BC } \) = \(\frac { 4√2 }{ 7 } \)

Find the value of x. Round your answer to the nearest tenth.

Question 20.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 255

Answer:
tan 54 = \(\frac { x }{ 32 } \)
1.37 = \(\frac { x }{ 32 } \)
x = 43.8

Question 21.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 256

Answer:
tan 25 = \(\frac { x }{ 20 } \)
0.46 x 20 = x
x = 9.2

Question 22.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 257

Answer:
tan 38 = \(\frac { 10 }{ x } \)
x = 12.82

Question 23.
The angle between the bottom of a fence and the top of a tree is 75°. The tree is 4 let from the fence. How tall is the tree? Round your answer to the nearest foot.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 258

Answer:
tan 75 = \(\frac { x }{ 4 } \)
x = 14.92

9.5 The Sine and Cosine Ratios

Find sin X, sin Z, cos X, and cos Z. Write each answer as a fraction and as a decimal rounded to four decimal places.

Question 24.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 259

Answer:
sin X = \(\frac { 3 }{ 5 } \)
sin Z = \(\frac { 4 }{ 5 } \)
cos X = \(\frac { 4 }{ 5 } \)
cos Z = \(\frac { 3 }{ 5 } \)

Question 25.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 260

Answer:
sin X = \(\frac { 7 }{ √149 } \)
sin Z = \(\frac { 10 }{ √149 } \)
cos X = \(\frac { 10 }{ √149 } \)
cos Z = \(\frac { 7 }{ √149 } \)

Question 26.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 261

Answer:
sin X = \(\frac { 55 }{ 73 } \)
sin Z = \(\frac { 48 }{ 73 } \)
cos X = \(\frac { 48 }{ 73 } \)
cos Z = \(\frac { 55 }{ 73 } \)

Find the value of each variable using sine and cosine. Round your answers to the nearest tenth.

Question 27.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 262

Answer:
sin 23 = \(\frac { t }{ 34 } \)
t = 13.26
cos 23 = \(\frac { s }{ 34 } \)
s = 31.28

Question 28.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 263

Answer:
sin 36 = \(\frac { s }{ 5 } \)
s = 2.9
cos 36 = \(\frac { r }{ 5 } \)
r = 4

Question 29.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 264

Answer:
sin 70 = \(\frac { v }{ 10 } \)
v = 9.39
cos 70 = \(\frac { w }{ 10 } \)
w = 3.42

Question 30.
Write sin 72° in terms of cosine.

Answer:
sin 72 = cos(90 – 72)
= cos 18 = 0.95

Question 31.
Write cos 29° in terms of sine.

Answer:
sin 29 = cos(90 – 29)
= cos 61 = 0.48

9.6 Solving Right Triangles

Let ∠Q be an acute angle. Use a calculator to approximate the measure of ∠Q to the nearest tenth of a degree.

Question 32.
cos Q = 0.32

Answer:
cos Q = 0.32
∠Q = inverse cos of .32
∠Q = 71.3

Question 33.
sin Q = 0.91

Answer:
sin Q = 0.91
∠Q = inverse sin of 0.91
∠Q = 65.5

Question 34.
tan Q = 0.04

Answer:
tan Q = 0.04
∠Q = inverse tan of 0.04
∠Q = 2.29

Solve the right triangle. Round decimal answers to the nearest tenth.

Question 35.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 265

Answer:
a = 5√5, ∠A = 47.7, ∠B = 42.3

Explanation:
c² = a² + b²
15² = a² + 10²
a² = 125
a = 5√5
sin A = \(\frac { 5√5 }{ 15 } \) = 0.74
∠A = 47.7
∠A + ∠B + ∠C = 180
47.7 + ∠B + 90 = 180
∠B = 42.3

Question 36.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 266

Answer:
NL = 7.59, ∠L = 53, ML = 4.55

Explanation:
cos 37 = \(\frac { 6 }{ NL } \)
NL = 7.59
∠N + ∠M + ∠L = 180
37 + 90 + ∠L = 180
∠L = 53
sin 37 = \(\frac { ML }{ 7.59 } \)
ML = 4.55

Question 37.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 267

Answer:
XY = 17.34, ∠X = 46, ∠Z = 44

Explanation:
c² = a² + b²
25² = 18²+ b²
b² = 301
b = 17.34
sin X = \(\frac { 18 }{ 25 } \)
∠X = 46
sum of angles = 180
46 + 90 + ∠Z = 180
∠Z = 44

9.7 Law of Sines and Law of Cosines

Find the area of ∆ABC with the given side lengths and included angle.

Question 38.
m ∠ B = 124°, a = 9, c = 11

Answer:
Area = \(\frac { 1 }{ 2 } \) ac sin B
= \(\frac { 1 }{ 2 } \) (9 x 11) sin 124
= 40.59

Question 39.
m ∠ A = 68°, b = 13, c = 7

Answer:
Area = \(\frac { 1 }{ 2 } \) bc sin A
= \(\frac { 1 }{ 2 } \) (13 x 7) sin 68
= 41.86

Question 40.
m ∠ C = 79°, a = 25 b = 17

Answer:
Area = \(\frac { 1 }{ 2 } \) ab sin C
= \(\frac { 1 }{ 2 } \) (25 x 17) sin 79
= 208.25

Solve ∆ABC. Round decimal answers to the nearest tenth.

Question 41.
m ∠ A = 112°, a = 9, b = 4

Answer:
∠B = 24, ∠C = 44, c = 6.76

Explanation:
\(\frac { sin B }{ b } \) = \(\frac { sin A }{ a } \)
\(\frac { sin B }{ 4 } \) = \(\frac { sin 112 }{ 9 } \)
sin B = 0.408
∠B = 24
∠A + ∠B + ∠C = 180
112 + 24 + ∠C = 180
∠C = 44
\(\frac { sin 112 }{ 9 } \) = \(\frac { sin 44 }{ c } \)
c = 6.76

Question 42.
m ∠ 4 = 28°, m ∠ B = 64°, c = 55

Answer:
∠C = 88, b = 49.4, a = 25.5

Explanation:
∠A + ∠B + ∠C = 180
28 + 64 + ∠C = 180
∠C = 88
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin 64 }{ b } \) = \(\frac { sin 88 }{ 55 } \)
\(\frac { sin 64 }{ b } \) = 0.018
b = 49.4
\(\frac { sin 28 }{ a } \) = \(\frac { sin 88 }{ 55 } \)
a = 25.5

Question 43.
m ∠ C = 48°, b = 20, c = 28

Answer:
∠B = 31.3, ∠A = 100.7, a = 37.6

Explanation:
\(\frac { sin B }{ b } \) = \(\frac { sin C }{ c } \)
\(\frac { sin B }{ 20 } \) = \(\frac { sin 48 }{ 28 } \)
\(\frac { sin B }{ 20 } \) = 0.026
sin B = 0.52
∠B = 31.3
∠A + ∠B + ∠C = 180
∠A + 31.3 + 48 = 180
∠A = 100.7
\(\frac { sin 100.7 }{ a } \) = \(\frac { sin 48 }{ 28 } \)
a = 37.6

Question 44.
m ∠ B = 25°, a = 8, c = 3

Answer:
b = 5.45, ∠A = 37.5, ∠C = 117.5

Explanation:
b² = a² + c²- 2ac cos B
b² = 8² + 3² – 2(8 x 3) cos 25 = 73 – 43.2 = 29.8
b = 5.45
\(\frac { sin A }{ 8 } \) = \(\frac { sin 25 }{ 5.45 } \)
sin A = 0.61
∠A = 37.5
∠A + ∠B + ∠C = 180
37.5 + 25 + ∠C = 180
∠C = 117.5

Question 45.
m ∠ B = 102°, m ∠ C = 43°, b = 21

Answer:
∠A = 35, c = 14.72, a = 12.3

Explanation:
∠A + ∠B + ∠C = 180
∠A + 102 + 43 = 180
∠A = 35
\(\frac { sin 102 }{ 21 } \) = \(\frac { sin 43 }{ c } \)
0.046 = \(\frac { sin 43 }{ c } \)
c = 14.72
\(\frac { sin 102 }{ 21 } \) = \(\frac { sin 35 }{ a } \)
0.046 = \(\frac { sin 35 }{ a } \)
a = 12.3

Question 46.
a = 10, b = 3, c = 12

Answer:
∠B = 11.7, ∠C = 125.19, ∠A = 43.11

Explanation:
b² = a² + c²- 2ac cos B
3² = 10² + 12²- 2(10 x 12) cos B
9 = 100 + 144 – 240 cos B
cos B = 0.979
∠B = 11.7
a² = b² + c²- 2bc cos A
100 = 9 + 144 – 72 cos A
cos A = 0.73
∠A = 43.11
∠A + ∠B + ∠C = 180
43.11 + 11.7 + ∠C = 180
∠C = 125.19

Right Triangles and Trigonometry Test

Find the value of each variable. Round your answers to the nearest tenth.

Question 1.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 268
Answer:
sin 25 = \(\frac { t }{ 18 } \)
t = 7.5
cos 25 = \(\frac { s }{ 18 } \)
s = 16.2

Question 2.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 269
Answer:
sin 22 = \(\frac { 6 }{ x } \)
x = 16.21
cos 22 = \(\frac { y }{ 16.21 } \)
y = 14.91

Question 3.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 270
Answer:
tan 40 = \(\frac { k }{ 10 } \)
k = 8.3
cos 40 = \(\frac { 10 }{ j } \)
j = 13.15

Verity that the segment lengths form a triangle. Is the triangle acute, right, or obtuse?

Question 4.
16, 30, and 34

Answer:
34²= 16² + 30²
So, the triangle is a right-angled triangle.

Question 5.
4, √67, and 9

Answer:
9² = 81
4² + (√67)² = 83
So the triangle is acute

Question 6.
√5. 5. and 5.5

Answer:
5.5² = 30.25
√5² + 5² = 30
So the triangle is obtuse

Solve ∆ABC. Round decimal answers to the nearest tenth.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 271
Answer:
c = 12.08, ∠A = 24.22, ∠C = 65.78

Explanation:
tan A = \(\frac { 5 }{ 11 } \)
∠A = 24.22
c² = 11²+ 5²
c = 12.08
24.22 + 90 + ∠C = 180
∠C = 65.78

Question 8.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 272

Answer:
∠B = 35.4, ∠C = 71.6, c = 17.9

Explanation:
\(\frac { sin 73 }{ 18 } \) = \(\frac { sin B }{ 11 } \)
sin B = 0.58
∠B = 35.4
73 + 35.4 + ∠C = 180
∠C = 71.6
\(\frac { sin 73 }{ 18 } \) = \(\frac { sin 71.6 }{ c } \)
c = 17.9

Question 9.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 273

Answer:
BC = 4.54, ∠B = 59.3, ∠A = 30.7

Explanation:
9.2² = 8² + x²
x = 4.54
\(\frac { sin 90 }{ 9.2 } \) = \(\frac { sin B }{ 8 } \)
sin B = 0.86
∠B = 59.3
∠A + 59.3 + 90 = 180
∠A = 30.7

Question 10.
m ∠ A = 103°, b = 12, c = 24

Answer:
a = 29, ∠B = 53.5

Explanation:
a² = b² + c²- 2bc cos A
a² = 144 + 24² – 2(12 x 24) cos 103
a = 29
\(\frac { sin B }{ 12 } \) = \(\frac { sin 103 }{ 29 } \)
∠B = 23.5
∠C = 180 – (103 + 23.5) = 53.5

Question 11.
m ∠ A = 26°, m ∠ C = 35°, b = 13

Answer:
∠B = 119, a = 6.42, c = 8.5

Explanation:
∠B + 26 + 35 = 180
∠B = 119
\(\frac { sin 119 }{ 13 } \) = \(\frac { sin 26 }{ a } \)
a = 6.42
\(\frac { sin 119 }{ 13 } \) = \(\frac { sin 35 }{ c } \)
c = 8.5

Question 12.
a = 38, b = 31, c = 35

Answer:
∠B=50.2, ∠C = 59.8, ∠A = 70

Explanation:
b² = a² + c²- 2ac cos B
31² = 38²+ 35²- 2(35 x 38) cos B
cos B = 0.64
∠B=50.2
a² = b² + c²- 2bc cos A
38² = 31²+ 35²- 2(31 x 35) cos A
cos A = 0.341
∠A = 70
∠C = 59.8

Question 13.
Write cos 53° in terms of sine.

Answer:
cos 53° = sin (90 – 53) = sin 37

Find the value of each variable. Write your answers in simplest form.

Question 14.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 274
Answer:
sin 45 = \(\frac { 16 }{ q } \)
q = 22.6
cos 45 = \(\frac { r }{ q } \)
r = 16

Question 15.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 275
Answer:

Question 16.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 276
Answer:
sin 30 = \(\frac { f }{ 9.2 } \)
f = 4.6
cos 30 = \(\frac { 8 }{ h } \)
h = 9.2

Question 17.
In ∆QRS, m ∠ R = 57°, q = 9, and s = 5. Find the area of ∆QRS.

Answer:
Area = \(\frac { 1 }{ 2 } \) qs sin R
= \(\frac { 1 }{ 2 } \) (9 x 5) sin 57 = 18.675

Question 18.
You are given the measures of both acute angles of a right triangle. Can you determine the side lengths? Explain.

Answer:
No.

Question 19.
You are at a parade looking up at a large balloon floating directly above the street. You are 60 feet from a point on the street directly beneath the balloon. To see the top of the balloon, you look up at an angle of 53°. To see the bottom of the balloon, you look up at an angle of 29°. Estimate the height h of the balloon.
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 277

Answer:

Question 20.
You warn to take a picture of a statue on Easter Island, called a moai. The moai is about 13 feet tall. Your camera is on a tripod that is 5 feet tall. The vertical viewing angle of your camera is set at 90°. How far from the moai should you stand so that the entire height of the moai is perfectly framed in the photo?
Big Ideas Math Geometry Answer Key Chapter 9 Right Triangles and Trigonometry 278

Answer:

Right Triangles and Trigonometry Cummulative Assessment

Question 1.
The size of a laptop screen is measured by the length of its diagonal. you Want to purchase a laptop with the largest screen possible. Which laptop should you buy?
(A)
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 279
(B)
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 280
(C)
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 281
(D)
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 282
Answer:
(B)

Explanation:
(a) d = √9² + 12² = 15
(b) d = √11.25² + 20² = 22.94
(c) d = √12² + 6.75² = 13.76
(d) d = √8² + 6² = 10

Question 2.
In ∆PQR and ∆SQT, S is between P and Q, T is between R and Q, and \(\) What must be true about \(\overline{S T}\) and \(\overline{P R}\)? Select all that apply.
\(\overline{S T}\) ⊥ \(\overline{P R}\)      \(\overline{S T}\) || \(\overline{P R}\)    ST = PR        ST = \(\frac{1}{2}\)PR
Answer:

Question 3.
In the diagram, ∆JKL ~ ∆QRS. Choose the symbol that makes each statement true.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 283
<      =       >
sin J ___________ sin Q                 sin L ___________ cos J                   cos L ___________ tan Q
cos S ___________ cos                  J cos J ___________ sin S                 tan J ___________ tan Q
tan L ___________ tan Q               tan S ___________ cos Q                sin Q ___________ cos L
Answer:
sin J = sin Q                 sin L = cos J                   cos L = tan Q
cos S > cos J                  cos J > sin S                 tan J = tan Q
tan L < tan Q               tan S > cos Q                sin Q = cos L

Question 4.
A surveyor makes the measurements shown. What is the width of the river.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 284

Answer:
tan 34 = \(\frac { AB }{ 84 } \)
AB = 56.28

Question 5.
Create as many true equations as possible.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 285

____________ = ______________

sin X            cos X           tan x           \(\frac{X Y}{X Z}\)           \(\frac{Y Z}{X Z}\)

Sin Z            cos Z           tan Z           \(\frac{X Y}{Y Z}\)           \(\frac{Y Z}{X Y}\)

Answer:
sin X = \(\frac{Y Z}{X Z}\) = cos Z
cos X = \(\frac{X Y}{X Z}\) = sin Z
tan x = \(\frac{Y Z}{X Y}\)
tan Z = \(\frac{X Y}{Y Z}\)

Question 6.
Prove that quadrilateral DEFG is a kite.
Given \(\overline{H E} \cong \overline{H G}\), \(\overline{E G}\) ⊥ \(\overline{D F}\)
Prove \(\overline{F E} \cong \overline{F G}\), \(\overline{D E} \cong \overline{D G}\)
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 286
Answer:

Question 7.
What are the coordinates of the vertices of the image of ∆QRS after the composition of transformations show?
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 287
(A) Q’ (1, 2), R'(5, 4), S'(4, -1)
(B) Q'(- 1, – 2), R’ (- 5, – 4), S’ (- 4, 1)
(C) Q'(3, – 2), R’ (- 1, – 4), S’ (0, 1)
(D) Q’ (-2, 1), R'(- 4, 5), S'(1, 4)
Answer:

Question 8.
The Red Pyramid in Egypt has a square base. Each side of the base measures 722 feet. The height of the pyramid is 343 fee.
Big Ideas Math Geometry Solutions Chapter 9 Right Triangles and Trigonometry 288

a. Use the side length of the base, the height of the pyramid, and the Pythagorean Theorem to find the slant height, AB, of the pyramid.

Answer:
343² = h² + 722²
h = 635.3

b. Find AC.
Answer:
AC = 343

c. Name three possible ways of finding m ∠ 1. Then, find m ∠ 1.
Answer:
Three possible ways are sin 1, cos 1 and tan 1
tan 1 = \(\frac { 722 }{ 635.3 } \)
∠1 = 48

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions

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Big Ideas Math Book Algebra 2 Answer Key Chapter 5 Rational Exponents and Radical Functions

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Rational Exponents and Radical Functions Maintaining Mathematical Proficiency

Simplify the expression

Question 1.
y6 • y

Question 2.
\(\frac{n^{4}}{n^{3}}\)

Question 3.
\(\frac{x^{5}}{x^{6} \cdot x^{2}}\)

Question 4.
\(\frac{x^{-6}}{x^{5}}\) • 3x2/sup>

Question 5.
\(\left(\frac{4 w^{3}}{2 z^{2}}\right)^{3}\)

Question 6.
\(\left(\frac{m^{7} \cdot m}{z^{2} \cdot m^{3}}\right)^{2}\)

Solve the literal equation for y.

Question 7.
4x + y = 2

Question 8.
x − \(\frac{1}{3}\)y = −1

Question 9.
2y − 9 = 13x

Question 10.
2xy + 6y = 10

Question 11.
8x − 4xy = 3

Question 12.
6x + 7xy = 15

Question 13.
ABSTRACT REASONING Is the order in which you apply properties of exponents important? Explain your reasoning.

Rational Exponents and Radical Functions Mathematical Practices

Monitoring Progress

Question 1.
Use the Pythagorean Theorem to find the exact lengths of a, b, c, and d in the figure.

Question 2.
Use a calculator to approximate each length to the nearest tenth of an inch.

Question 3.
Use a ruler to check the reasonableness of your answers.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 1

Lesson 5.1 nth Roots and Rational Exponents

Essential Question

How can you use a rational exponent to represent a power involving a radical?
Previously, you learned that the nth root of a can be represented as
\(\sqrt[n]{a}\) = \(a^{1 / n}\) Definition of rational exponent
for any real number a and integer n greater than 1.

EXPLORATION 1
Exploring the Definition of a Rational Exponent
Work with a partner.
Use a calculator to show that each statement is true.
a. \(\sqrt{9}\) = \(9^{1 / 2}\)
b. \(\sqrt{2}\) = \(2^{1 / 2}\)
c. \(\sqrt[3]{8}\) = \(8^{1 / 3}\)
d. \(\sqrt[3]{3}\) = \(3^{1 / 3}\)
e. \(\sqrt[4]{16}\) = \(16^{1 / 4}\)
f. \(\sqrt[4]{12}\) = \(12^{1 / 4}\)

EXPLORATION 2
Writing Expressions in Rational Exponent Form

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 2
Work with a partner. Use the definition of a rational exponent and the properties of exponents to write each expression as a base with a single rational exponent. Then use a calculator to evaluate each expression. Round your answer to two decimal places.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 3
a. \((\sqrt{5})^{3}\)
b. \((\sqrt[4]{4})^{2}\)
c. \((\sqrt[3]{9})^{2}\)
d. \((\sqrt[5]{10})^{4}\)
e. \((\sqrt{15})^{3}\)
f. \((\sqrt[3]{27})^{4}\)

EXPLORATION 3
Writing Expressions in Radical Form
Work with a partner.
Use the properties of exponents and the definition of a rational exponent to write each expression as a radical raised to an exponent. Then use a calculator to evaluate each expression. Round your answer to two decimal places.
Sample \(5^{2 / 3}\) = (\left(5^{1 / 3}\right)\(\))2 = (\((\sqrt[3]{5})\))2 ≈ 2.92
a. \(8^{2 / 3}\)
b. \(6^{5 / 2}\)
c. \(12^{3 / 4}\)
d. \(10^{3 / 2}\)
e. \(16^{3 / 2}\)
f. \(20^{6 / 5}\)

Communicate Your Answer

Question 4.
How can you use a rational exponent to represent a power involving a radical?

Question 5.
Evaluate each expression without using a calculator. Explain your reasoning.
a. \(4^{3 / 2}\)
b. \(32^{4 / 5}\)
c. \(625^{3 / 4}\)
d. \(49^{3 / 2}\)
e. \(125^{4 / 3}\)
f. \(100^{6 / 3}\)

5.1 Lesson

Monitoring Progress

Question 1.
n = 4, a = 16

Question 2.
n = 2, a = −49

Question 3.
n = 3, a = −125

Question 4.
n = 5, a = 243

Evaluate the expression without using a calculator.

Question 5.
\(4^{5 / 2}\)

Question 6.
\(9^{-1 / 2}\)

Question 7.
\(81^{3 / 4}\)

Question 8.
\(1^{7 / 8}\)

Evaluate the expression using a calculator. Round your answer to two decimal places when appropriate.

Question 9.
\(6^{2 / 5}\)

Question 10.
\(64^{-2 / 3}\)

Question 11.
\((\sqrt[4]{16})^{5}\)

Question 12.
\((\sqrt[3]{-30})^{2}\)

Find the real solution(s) of the equation. Round your answer to two decimal places when appropriate.

Question 13.
8x3 = 64

Question 14.
\([\frac{1}{2}/latex]x5 = 512

Question 15.
(x + 5)4 = 16

Question 16.
(x − 2)3 = −14

Question 17.
WHAT IF? In Example 5, what is the annual depreciation rate when the salvage value is $6000?

nth Roots and Rational Exponents 5.1 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY Rewrite the expression [latex]a^{-s / t}\) in radical form. Then state the index of the radical.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 1

Question 2.
COMPLETE THE SENTENCE For an integer n greater than 1, if bn = a, then bis a(n) ___________ of a.
Answer:

Question 3.
WRITING Explain how to use the sign of a to determine the number of real fourth roots of a and the number of real fifth roots of a.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 3

Question 4.
WHICH ONE DOESN’T BELONG? Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, find the indicated real nth root(s) of a.

Question 5.
n = 3, a = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 5

Question 6.
n = 5, a = −1
Answer:

Question 7.
n = 2, a = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 7

Question 8.
n = 4, a = 256
Answer:

Question 9.
n = 5, a = −32
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 9

Question 10.
n = 6, a = −729
Answer:

In Exercises 11–18, evaluate the expression without using a calculator.

Question 11.
\(64^{1 / 6}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 11

Question 12.
\(8^{1 / 3}\)
Answer:

Question 13.
\(25^{3 / 2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 13

Question 14.
\(81^{3 / 4}\)
Answer:

Question 15.
\((-243)^{1 / 5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 15

Question 16.
\((-64)^{4 / 3}\)
Answer:

Question 17.
\(8^{-2 / 3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 17

Question 18.
\(16^{-7 / 4}\)
Answer:

ERROR ANALYSIS In Exercises 19 and 20, describe and correct the error in evaluating the expression.

Question 19.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 19

Question 20.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 6
Answer:

USING STRUCTURE In Exercises 21–24, match the equivalent expressions. Explain your reasoning.

Question 21.
\((\sqrt[3]{5})^{4}\) A. \(5^{-1 / 4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 21

Question 22.
\((\sqrt[4]{5})^{3}\) B. \(5^{4 / 3}\)
Answer:

Question 23.
\(\frac{1}{\sqrt[4]{5}}\) C. \(-5^{1 / 4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 23

Question 24.
\(-\sqrt[4]{5}\) D. \(5^{3 / 4}\)
Answer:

In Exercises 25–32, evaluate the expression using a calculator. Round your answer to two decimal places when appropriate.

Question 25.
\(\sqrt[5]{32,768}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 25

Question 26.
\(\sqrt[7]{1695}\)
Answer:

Question 27.
\(25^{-1 / 3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 27

Question 28.
\(85^{1 / 6}\)
Answer:

Question 29.
\(20,736^{4 / 5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 29

Question 30.
\(86^{-5 / 6}\)
Answer:

Question 31.
\((\sqrt[4]{187})^{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 31

Question 32.
\((\sqrt[5]{-8})^{8}\)
Answer:

MATHEMATICAL CONNECTIONS In Exercises 33 and 34, find the radius of the figure with the given volume.

Question 33.
V = 216 ft3
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 33

Question 34.
V = 1332 cm3
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 8
Answer:

In Exercises 35–44, find the real solution(s) of the equation. Round your answer to two decimal places when appropriate.

Question 35.
x3 = 125
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 35

Question 36.
5x3 = 1080
Answer:

Question 37.
(x + 10)5 = 70
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 37

Question 38.
(x − 5)4 = 256
Answer:

Question 39.
x5 = −48
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 39

Question 40.
7x4 = 56
Answer:

Question 41.
x6 + 36 = 100
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 41

Question 42.
x3 + 40 = 25
Answer:

Question 43.
\(\frac{1}{3}\)x4 = 27
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 43

Question 44.
\(\frac{1}{6}\)x3 = −36
Answer:

Question 45.
MODELING WITH MATHEMATICS When the average price of an item increases from p1 to p2 over a period of n years, the annual rate of inflation r (in decimal form) is given by r = \(\left(\frac{p_{2}}{p_{1}}\right)^{1 / n}\) − 1. Find the rate of inflation for each item in the table.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 45.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 45.2

Question 46.
HOW DO YOU SEE IT? The graph of y = xn is shown in red. What can you conclude about the value of n? Determine the number of real nth roots of a. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 10
Answer:

Question 47.
NUMBER SENSE Between which two consecutive integers does \(\sqrt[4]{125}\) lie? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 47

Question 48.
THOUGHT PROVOKING In 1619, Johannes Kepler published his third law, which can be given by d3 = t2, where d is the mean distance (in astronomical units) of a planet from the Sun and t is the time (in years) it takes the planet to orbit the Sun. It takes Mars 1.88 years to orbit the Sun. Graph a possible location of Mars. Justify your answer. (The diagram shows the Sun at the origin of the xy-plane and a possible location of Earth.)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 11
Answer:

Question 49.
PROBLEM SOLVING A weir is a dam that is built across a river to regulate the flow of water. The flow rate Q (in cubic feet per second) can be calculated using the formula Q= 3.367ℓ\(h^{3 / 2}\), where ℓ is the length (in feet) of the bottom of the spillway and his the depth (in feet) of the water on the spillway. Determine the flow rate of a weir with a spillway that is 20 feet long and has a water depth of 5 feet.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 49

Question 50.
REPEATED REASONING The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain river normally flows at a speed of 1 meter per second. What must its speed be in order to transport particles that are twice as massive as usual? 10 times as massive? 100 times as massive?
Answer:

Maintaining Mathematical Proficiency

Simplify the expression. Write your answer using only positive exponents. (Skills Review Handbook)

Question 51.
5 • 54

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 51

Question 52.
\(\frac{4^{2}}{4^{7}}\)
Answer:

Question 53.
\(\left(z^{2}\right)^{-3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 53

Question 54.
\(\left(\frac{3 x}{2}\right)^{4}\)
Answer:

Write the number in standard form. (Skills Review Handbook)

Question 55.
5 × 103

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 55

Question 56.
4 × 10−2

Answer:

Question 57.
8.2 × 10−1

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 57

Question 58.
6.93 × 106

Answer:

Lesson 5.2 Properties of Rational Exponents and Radicals

Essential Question
How can you use properties of exponents to simplify products and quotients of radicals?

EXPLORATION 1
Reviewing Properties of Exponents
Work with a partner.
Let a and b be real numbers. Use the properties of exponents to complete each statement. Then match each completed statement with the property it illustrates.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 13

EXPLORATION 2
Simplifying Expressions with Rational Exponents
Work with a partner.
Show that you can apply the properties of integer exponents to rational exponents by simplifying each expression. Use a calculator to check your answers.
a. \(5^{2 / 3}\) • \(5^{4 / 3}\)
b. \(3^{1 / 5}\) • \(3^{4 / 5}\)
c. \(\left(4^{2 / 3}\right)^{3}\)
d. \(\frac{\sqrt{98}}{\sqrt{2}}\)
e. \(\frac{\sqrt[4]{4}}{\sqrt[4]{1024}}\)
f. \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\)

EXPLORATION 3
Simplifying Products and Quotients of Radicals
Work with a partner.
Use the properties of exponents to write each expression as a single radical. Then evaluate each expression. Use a calculator to check your answers.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 14
a. \(\sqrt{3}\) • \(\sqrt{12}\)
b. \(\sqrt[3]{5}\) • \(\sqrt[3]{25}\)
c. \(\sqrt[4]{27}\) • \(\sqrt[4]{3}\)
d. \(\frac{\sqrt{98}}{\sqrt{2}}\)
e. \(\frac{\sqrt[4]{4}}{\sqrt[4]{1024}}\)
f. \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\)

Communicate Your Answer

Question 4.
How can you use properties of exponents to simplify products and quotients of radicals?

Question 5.
Simplify each expression.
a. \(\sqrt{27}\) • \(\sqrt{6}\)
b. \(\frac{\sqrt[3]{240}}{\sqrt[3]{15}}\)
c. (\(5^{1 / 2}\) \(16^{1 / 4}\))2

5.2 Lesson

Monitoring Progress

Simplify the expression.

Question 1.
\(2^{3 / 4}\) • \(2^{1 / 2}\)

Question 2.
\(\frac{3}{3^{1 / 4}}\)

Question 3.
\(\left(\frac{20^{1 / 2}}{5^{1 / 2}}\right)^{3}\)

Question 4.
(\(5^{1 / 3}\) • \(7^{1 / 4}\))3

Simplify the expression

Question 5.
\(\sqrt[4]{27}\) • \(\sqrt[4]{3}\)

Question 6.
\(\frac{\sqrt[3]{250}}{\sqrt[3]{2}}\)

Question 7.
\(\sqrt[3]{104}\)

Question 8.
\(\sqrt[5]{\frac{3}{4}}\)

Question 9.
\(\frac{3}{6-\sqrt{2}}\)

Question 10.
\(7 \sqrt[5]{12}\) – \(\sqrt[5]{12}\)

Question 11.
4(\(9^{2 / 3}\)) + (\(9^{2 / 3}\))

Question 12.
\(\sqrt[3]{5}\) + \(\sqrt[3]{40}\)

Simplify the expression. Assume all variables are positive.

Question 13.
\(\sqrt[3]{27 q^{9}}\)

Question 14.
\(\sqrt[5]{\frac{x^{10}}{y^{5}}}\)

Question 15.
\(\frac{6 x y^{3 / 4}}{3 x^{1 / 2} y^{1 / 2}}\)

Question 16.
\(\sqrt{9} w^{5}\) – \(w \sqrt{w^{3}}\)

Properties of Rational Exponents and Radicals 5.2 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING How do you know when a radical expression is in simplest form?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 1

Question 2.
WHICH ONE DOESN’T BELONG? Which radical expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 15
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, use the properties of rational exponents to simplify the expression.

Question 3.
\(\left(9^{2}\right)^{1 / 3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 3

Question 4.
\(\left(12^{2}\right)^{1 / 4}\)
Answer:

Question 5.
\(\frac{6}{6^{1 / 4}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 5

Question 6.
\(\frac{7}{7^{1 / 3}}\)
Answer:

Question 7.
\(\left(\frac{8^{4}}{10^{4}}\right)^{-1 / 4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 7

Question 8.
\(\left(\frac{9^{3}}{6^{3}}\right)^{-1 / 3}\)
Answer:

Question 9.
(\(3^{-2 / 3}\) • \(3^{1 / 3}\))-1

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 9

Question 10.
(\(5^{1 / 2}\) • \(5^{-3 / 2}\))-1/4

Answer:

Question 11.
\(\frac{2^{2 / 3} \cdot 16^{2 / 3}}{4^{2 / 3}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 11

Question 12.
\(\frac{49^{3 / 8} \cdot 49^{7 / 8}}{7^{5 / 4}}\)
Answer:

In Exercises 13–20, use the properties of radicals to simplify the expression.

Question 13.
\(\sqrt{2}\) • \(\sqrt{72}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 13

Question 14.
\(\sqrt[3]{16}\) • \(\sqrt[3]{32}\)
Answer:

Question 15.
\(\sqrt[4]{6}\) • \(\sqrt[4]{8}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 15

Question 16.
\(\sqrt[4]{8}\) • \(\sqrt[4]{8}\)
Answer:

Question 17.
\(\frac{\sqrt[5]{486}}{\sqrt[5]{2}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 17

Question 18.
\(\frac{\sqrt{2}}{\sqrt{32}}\)
Answer:

Question 19.
\(\frac{\sqrt[3]{6} \cdot \sqrt[3]{72}}{\sqrt[3]{2}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 19

Question 20.
\(\frac{\sqrt[3]{3} \cdot \sqrt[3]{18}}{\sqrt[6]{2} \cdot \sqrt[6]{2}}\)
Answer:

In Exercises 21–28, write the expression in simplest form.

Question 21.
\(\sqrt[4]{567}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 21

Question 22.
\(\sqrt[5]{288}\)
Answer:

Question 23.
\(\frac{\sqrt[3]{5}}{\sqrt[3]{4}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 23

Question 24.
\(\frac{\sqrt[4]{4}}{\sqrt[4]{27}}\)
Answer:

Question 25.
\(\sqrt{\frac{3}{8}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 25

Question 26.
\(\sqrt[3]{\frac{7}{4}}\)
Answer:

Question 27.
\(\sqrt[3]{\frac{64}{49}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 27

Question 28.
\(\sqrt[4]{\frac{1296}{25}}\)
Answer:

In Exercises 29–36, write the expression in simplest form.

Question 29.
\(\frac{1}{1+\sqrt{3}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 29

Question 30.
\(\frac{1}{2+\sqrt{5}}\)
Answer:

Question 31.
\(\frac{5}{3-\sqrt{2}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 31

Question 32.
\(\frac{11}{9-\sqrt{6}}\)
Answer:

Question 33.
\(\frac{9}{\sqrt{3}+\sqrt{7}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 33

Question 34.
\(\frac{2}{\sqrt{8}+\sqrt{7}}\)
Answer:

Question 35.
\(\frac{\sqrt{6}}{\sqrt{3}-\sqrt{5}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 35

Question 36.
\(\frac{\sqrt{7}}{\sqrt{10}-\sqrt{2}}\)
Answer:

In Exercises 37–46, simplify the expression.

Question 37.
\(9 \sqrt[3]{11}\) + \(3 \sqrt[3]{11}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 37

Question 38.
\(8 \sqrt[6]{5}\) – \(12 \sqrt[6]{5}\)
Answer:

Question 39.
\(3\left(11^{1 / 4}\right)\) + \(9\left(11^{1 / 4}\right)\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 39

Question 40.
\(13\left(8^{3 / 4}\right)\) – \(4\left(8^{3 / 4}\right)\)
Answer:

Question 41.
\(5 \sqrt{12}\) – \(19 \sqrt{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 41

Question 42.
\(27 \sqrt{6}\) + \(7 \sqrt{150}\)
Answer:

Question 43.
\(\sqrt[5]{224}\) + \(3 \sqrt[5]{7}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 43

Question 44.
\(7 \sqrt[3]{2}\) – \(\sqrt[3]{128}\)
Answer:

Question 45.
\(5\left(24^{1 / 3}\right)\) – 4(\(\left(3^{1 / 3}\right)\))
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 45

Question 46.
\(5^{1 / 4}\) + 6(\([405^{1 / 4}/latex])
Answer:

Question 47.
ERROR ANALYSIS Describe and correct the error in simplifying the expression.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 47

Question 48.
MULTIPLE REPRESENTATIONS Which radical expressions are like radicals?
A. [latex]\left(5^{2 / 9}\right)^{3 / 2}\)
B. \(\frac{5^{3}}{(\sqrt[3]{5})^{8}}\)
C. \(\sqrt[3]{625}\)
D. \(\sqrt[3]{5} 145\) – \(\sqrt[3]{875}\)
E. \(\sqrt[3]{5}\) + \(3 \sqrt[3]{5}\)
F. \(7 \sqrt[4]{80}\) – \(2 \sqrt[4]{405}\)
Answer:

In Exercises 49–54, simplify the expression.

Question 49.
\(\sqrt[4]{81 y^{8}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 49

Question 50.
\(\sqrt[3]{64 r^{3} t^{6}}\)
Answer:

Question 51.
\(\sqrt[5]{\frac{m^{10}}{n^{5}}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 51

Question 52.
\(\sqrt[4]{\frac{k^{16}}{16 z^{4}}}\)
Answer:

Question 53.
\(\sqrt[6]{\frac{g^{6} h}{h^{7}}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 53

Question 54.
\(\sqrt[8]{n^{2} p^{-1}}\)
Answer:

Question 55.
ERROR ANALYSIS Describe and correct the error in simplifying the expression.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 17
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 55

Question 56.
OPEN-ENDED Write two variable expressions involving radicals, one that needs absolute value in simplifying and one that does not need absolute value. Justify your answers.
Answer:

In Exercises 57–64, write the expression in simplest form. Assume all variables are positive.

Question 57.
\(\sqrt{81 a^{7} b^{12} c^{9}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 57

Question 58.
\(\sqrt[3]{125 r^{4} s^{9} t^{7}}\)
Answer:

Question 59.
\(\sqrt[5]{\frac{160 m^{6}}{n^{7}}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 59

Question 60.
\(\sqrt[4]{\frac{405 x^{3} y^{3}}{5 x^{-1} y}}\)
Answer:

Question 61.
\(\frac{\sqrt[3]{w} \cdot \sqrt{w^{5}}}{\sqrt{25 w^{16}}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 61

Question 62.
\(\frac{\sqrt[4]{v^{6}}}{\sqrt[7]{v^{5}}}\)
Answer:

Question 63.
\(\frac{18 w^{1 / 3} v^{5 / 4}}{27 w^{4 / 3} v^{1 / 2}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 63

Question 64.
\(\frac{7 x^{-3 / 4} y^{5 / 2} z^{-2 / 3}}{56 x^{-1 / 2} y^{1 / 4}}\)
Answer:

In Exercises 65–70, perform the indicated operation. Assume all variables are positive.

Question 65.
\(12 \sqrt[3]{y}\) + \(9 \sqrt[3]{y}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 65

Question 66.
\(11 \sqrt{2 z}\) – \(5 \sqrt{2 z}\)
Answer:

Question 67.
\(3 x^{7 / 2}\) – 5\(x^{7 / 2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 67

Question 68.
\(7 \sqrt[3]{m^{7}}\) + \(3 m^{7 / 3}\)
Answer:

Question 69.
\(\sqrt[4]{16 w^{10}}\) + \(2 w \sqrt[4]{w^{6}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 69

Question 70.
(p1/2 • p1/4) – \(\sqrt[4]{16 p^{3}}\)
Answer:

MATHEMATICAL CONNECTIONS In Exercises 71 and 72, find simplified expressions for the perimeter and area of the given figure.

Question 71.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 18
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 71

Question 72
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 19
Answer:

Question 73.
MODELING WITH MATHEMATICS The optimum diameter d (in millimeters) of the pinhole in a pinhole camera can be modeled by d = 1.9[(5.5 × 10−4)ℓ]1/2, where ℓ is the length (in millimeters) of the camera box. Find the optimum pinhole diameter for a camera box with a length of 10 centimeters.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 20
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 73

Question 74.
MODELING WITH MATHEMATICS The surface area S(in square centimeters) of a mammal can be modeled by S = km2/3, where m is the mass (in grams) of the mammal and k is a constant. The table shows the values of k for different mammals.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 21
a. Find the surface area of a bat whose mass is 32 grams.
b. Find the surface area of a rabbit whose mass is 3.4 kilograms (3.4 × 103 grams).
c. Find the surface area of a human whose mass is 59 kilograms.
Answer:

Question 75.
MAKING AN ARGUMENT Your friend claims it is not possible to simplify the expression 7\(\sqrt{11}\) − 9 \(\sqrt{44}\) because it does not contain like radicals. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 75

Question 76.
PROBLEM SOLVING The apparent magnitude of a star is a number that indicates how faint the star is in relation to other stars. The expression f(x) = \(\frac{2.512^{m_{1}}}{2.512^{m_{2}}}\) tells how many times fainter a star with apparent magnitude m1 is than a star with apparent magnitude m2.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 22
a. How many times fainter is Altair than Vega?
b. How many times fainter is Deneb than Altair?
c. How many times fainter is Deneb than Vega?
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 23
Answer:

Question 77.
CRITICAL THINKING Find a radical expression for the perimeter of the triangle inscribed in the square shown. Simplify the expression.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 24
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 77

Question 78.
HOW DO YOU SEE IT? Without finding points, match the functions f(x) = \(\sqrt{64 x^{2}}\) and g(x) = \(\sqrt[3]{64 x^{6}}\) with their graphs. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 25
Answer:

Question 79.
REWRITING A FORMULA You have filled two round balloons with water. One balloon contains twice as much water as the other balloon.
a. Solve the formula for the volume of a sphere, V = \(\frac{4}{3}\)πr3, for r.
b. Substitute the expression for r from part (a) into the formula for the surface area of a sphere, S = 4πr2. Simplify to show that S = (4π)1/3(3V)2/3.
c. Compare the surface areas of the two water balloons using the formula in part (b).
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 79.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 79.2

Question 80.
THOUGHT PROVOKING Determine whether the expressions (x2)1/6 and (x1/6)2 are equivalent for all values of x.
Answer:

Question 81.
DRAWING CONCLUSIONS Substitute different combinations of odd and even positive integers for m and n in the expression \(\sqrt[n]{x^{m}}\). When you cannot assume x is positive, explain when absolute value is needed in simplifying the expression.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 81
Maintaining Mathematical Proficiency

Identify the focus, directrix, and axis of symmetry of the parabola. Then graph the equation. (Section 2.3)

Question 82.
y = 2x2
Answer:

Question 83.
y2 = −x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 83.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 83.2

Question 84.
y2 = 4x
Answer:

Write a rule for g. Describe the graph of g as a transformation of the graph of f. (Section 4.7)

Question 85.
f(x) = x4 − 3x2 − 2x, g(x) = −f(x)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 85

Question 86.
f(x) = x3 − x, g(x) = f(x) − 3
Answer:

Question 87.
f(x) = x3 − 4, g(x) = f(x − 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 87

Question 88.
f(x) = x4 + 2x3 − 4x2, g(x) = f(2x)
Answer:

Lesson 5.3 Graphing Radical Functions

Essential Question

How can you identify the domain and range of a radical function?

EXPLORATION 1
Identifying Graphs of Radical Functions
Work with a partner.
Match each function with its graph. Explain your reasoning. Then identify the domain and range of each function.
a. f(x) = \(\sqrt{x}\)
b. f(x) = \(\sqrt[3]{x}\)
c. f(x) = \(\sqrt[4]{x}\)
d. f(x) = \(\sqrt[5]{x}\)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 26

EXPLORATION 2
Identifying Graphs of Transformations
Work with a partner.
Match each transformation of f(x) = \(\sqrt{x}\) with its graph. Explain your reasoning. Then identify the domain and range of each function.
a. g(x) = \(\sqrt{x+2}\)
b. g(x) = \(\sqrt{x-2}\)
c. g(x) = \(\sqrt{x}+2-2\)
d. g(x) = −\(\sqrt{x+2}\)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 27

Communicate Your Answer

Question 3.
How can you identify the domain and range of a radical function?

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 28
Question 4.
Use the results of Exploration 1 to describe how the domain and range of a radical function are related to the index of the radical.

5.3 Lesson

Monitoring Progress

Question 1.
Graph g(x) = \(\sqrt{x+1}\). Identify the domain and range of the function.

Question 2.
Describe the transformation of f(x) = \(\sqrt[3]{x}\) represented by g(x) = −\(\sqrt[3]{x}\) − 2. Then graph each function.

Question 3.
WHAT IF? In Example 3, the function N(d ) = 2.4 • E(d) approximates the number of seconds it takes a dropped object to fall d feet on the Moon. Write a rule for N. How long does it take a dropped object to fall 25 feet on the Moon?

Question 4.
In Example 4, is the transformed function the same when you perform the translation followed by the horizontal shrink? Explain your reasoning.

Question 5.
Use a graphing calculator to graph −4y2 = x + 1. Identify the vertex and the direction that the parabola opens.

Question 6.
Use a graphing calculator to graph x2 + y2 = 25. Identify the radius and the intercepts.

Graphing Radical Functions 5.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE Square root functions and cube root functions are examples of __________ functions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 1

Question 2.
COMPLETE THE SENTENCE When graphing y = a\(\sqrt[3]{x-h}\) + k, translate the graph of y = a\(\sqrt[3]{x}\)h units __________ and k units __________.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, match the function with its graph.

Question 3.
f(x) = \(\sqrt{x}+3\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 3

Question 4.
h(x) = \(\sqrt{x}\) + 3
Answer:

Question 5.
f(x) = \(\sqrt{x-3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 5

Question 6.
g(x) = \(\sqrt{x}\) − 3
Answer:

Question 7.
h(x) = \(\sqrt{x+3}\) − 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 7

Question 8.
f(x) = \(\sqrt{x-3}\) + 3
Answer:

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 28.1

In Exercises 9–18, graph the function. Identify the domain and range of the function.

Question 9.
h(x) = \(\sqrt{x}\) + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 9

Question 10.
g(x) = \(\sqrt{x}\) − 5
Answer:

Question 11.
g(x) = − \(\sqrt[3]{2 x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 11

Question 12.
f(x) = \(\sqrt[3]{-5 x}\)
Answer:

Question 13.
g(x) = \(\frac{1}{5} \sqrt{x}-3\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 13

Question 14.
f(x) = \(\frac{1}{2} \sqrt[3]{x}+6\)
Answer:

Question 15.
f(x) = \((6 x)^{1 / 2}\) + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 15

Question 16.
g(x) = −3(x + 1)1/3

Answer:

Question 17.
h(x) = −\(\sqrt[4]{x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 17

Question 18.
h(x) = \(\sqrt[5]{2 x}\)
Answer:

In Exercises 19–26, describe the transformation of f represented by g. Then graph each function.

Question 19.
f(x) = \(\sqrt{x}\), g(x) = \(\sqrt{x+1}\) + 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 19

Question 20.
f(x) = \(\sqrt{x}\), g(x) = 2\(\sqrt{x-1}\)
Answer:

Question 21.
f(x) = \(\sqrt[3]{x}\), g(x) = −\(\sqrt[3]{x}\) − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 21

Question 22.
f(x) = \(\sqrt[3]{x}\), g(x) = \(\sqrt[3]{x+4}\) − 5
Answer:

Question 23.
f(x) = \(x^{1 / 2}\), g(x) = \(\frac{1}{4}(-x)^{1 / 2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 23

Question 24.
f(x) = \(x^{1 / 3}\), g(x) = \(\frac{1}{3} x^{1 / 3}\) + 6
Answer:

Question 25.
f(x) = \(\sqrt[4]{x}\), g(x) = \(2 \sqrt[4]{x+5}\) − 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 25

Question 26.
f(x) = \(\sqrt[5]{x}\), g(x) = \(\sqrt[5]{-32 x}\) + 3
Answer:

Question 27.
ERROR ANALYSIS Describe and correct the error in graphing f(x) = \(\sqrt{x-2}\) − 2.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 29
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 27

Question 28.
ERROR ANALYSIS Describe and correct the error in describing the transformation of the parent square root function represented by g(x) = \(\sqrt{\frac{1}{2} x}\) + 3.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 30
Answer:

USING TOOLS In Exercises 29–34, use a graphing calculator to graph the function. Then identify the domain and range of the function.

Question 29.
g(x) = \(\sqrt{x^{2}+x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 29

Question 30.
h(x) = \(\sqrt{x^{2}-2 x}\)
Answer:

Question 31.
f(x) = \(\sqrt[3]{x^{2}+x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 31

Question 32.
f(x) = \(\sqrt[3]{3 x^{2}-x}\)
Answer:

Question 33.
f(x) = \(\sqrt{2 x^{2}+x+1}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 33

Question 34.
h(x) = \(\sqrt[3]{\frac{1}{2} x^{2}-3 x+4}\)
Answer:

ABSTRACT REASONING In Exercises 35–38, complete the statement with sometimes, always, or never.

Question 35.
The domain of the function y = a\(\sqrt{x}\) is ______ x ≥ 0.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 35

Question 36.
The range of the function y = a\(\sqrt{x}\) is ______ y ≥ 0.
Answer:

Question 37.
The domain and range of the function y = \(\sqrt[3]{x-h}\) + k are ________ all real numbers.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 37

Question 38.
The domain of the function y = a\(\sqrt{-x}\) + k is ________ x ≥ 0.
Answer:

Question 39.
PROBLEM SOLVING The distance (in miles) a pilot can see to the horizon can be approximated by E(n) = 1.22\(\sqrt{n}\), where n is the plane’s altitude (in feet above sea level) on Earth. The function M(n) = 0.75E(n) approximates the distance a pilot can see to the horizon n feet above the surface of Mars. Write a rule for M. What is the distance a pilot can see to the horizon from an altitude of 10,000 feet above Mars?
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 31
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 39

Question 40.
MODELING WITH MATHEMATICS The speed (in knots) of sound waves in air can be modeled by
v(K) = 643.855\(\sqrt{\frac{K}{273.15}}\)
where K is the air temperature (in kelvin). The speed (in meters per second) of sound waves in air can be modeled by s(K) = \(\frac{v(K)}{1.944}\)
Write a rule for s. What is the speed (in meters per second) of sound waves when the air temperature is 305 kelvin?
Answer:

In Exercises 41–44, write a rule for g described by the transformations of the graph of f.

Question 41.
Let g be a vertical stretch by a factor of 2, followed by a translation 2 units up of the graph of f(x) = \(\sqrt{x}\) + 3.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 41

Question 42.
Let g be a reflection in the y-axis, followed by a translation 1 unit right of the graph of f(x) = \(2 \sqrt[3]{x-1}\).
Answer:

Question 43.
Let g be a horizontal shrink by a factor of \(\frac{2}{3}\), followed by a translation 4 units left of the graph of f(x) = \(\sqrt{6 x}\).
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 43

Question 44.
Let g be a translation 1 unit down and 5 units right, followed by a reflection in the x-axis of the graph of f(x) = −\(-\frac{1}{2} \sqrt[4]{x}+\frac{3}{2}\)
Answer:

In Exercises 45 and 46, write a rule for g.

Question 45.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 32
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 45

Question 46.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 33
Answer:

In Exercises 47–50, write a rule for g that represents the indicated transformation of the graph of f.

Question 47.
f(x) = 2\(\sqrt{x}\), g(x) = f(x + 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 47

Question 48.
f(x) = \(\frac{1}{3} \sqrt{x-1}\), g(x) = −f(x) + 9
Answer:

Question 49.
f(x) = −\(\sqrt{x^{2}-2}\), g(x) = −2f(x + 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 49

Question 50.
f(x) = \(\sqrt[3]{x^{2}+10 x}\), g(x) =\(\frac{1}{4}\)f(−x) + 6
Answer:

In Exercises 51–56, use a graphing calculator to graph the equation of the parabola. Identify the vertex and the direction that the parabola opens.

Question 51.
\(\frac{1}{4}\)y2 = x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 51

Question 52.
3y2 = x
Answer:

Question 53.
−8y2 + 2 = x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 53

Question 54.
2y2 = x − 4
Answer:

Question 55.
x + 8 = \(\frac{1}{5}\)y2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 55

Question 56.
\(\frac{1}{2}\)x = y2 − 4
Answer:

In Exercises 57–62, use a graphing calculator to graph the equation of the circle. Identify the radius and the intercepts.

Question 57.
x2 + y2 = 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 57

Question 58.
x2 + y2 = 4
Answer:

Question 59.
1 − y2 = x2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 59

Question 60.
64 − x2 = y2

Answer:

Question 61.
−y2 = x2 − 36
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 61

Question 62.
x2 = 100 − y2

Answer:

Question 63.
MODELING WITH MATHEMATICS The period of a pendulum is the time the pendulum takes to complete one back-and-forth swing. The period T (in seconds) can be modeled by the function T = 1.11\(\sqrt{\ell}\), where ℓ is the length (in feet) of the pendulum. Graph the function. Estimate the length of a pendulum with a period of 2 seconds. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 34
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 63

Question 64.
HOW DO YOU SEE IT? Does the graph represent a square root function or a cube root function? Explain. What are the domain and range of the function?
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 35
Answer:

Question 65.
PROBLEM SOLVING For a drag race car with a total weight of 3500 pounds, the speed s(in miles per hour) at the end of a race can be modeled by s = 14.8\(\sqrt[3]{p}\), where p is the power (in horsepower). Graph the function.
a. Determine the power of a 3500-pound car that reaches a speed of 200 miles per hour.
b. What is the average rate of change in speed as the power changes from 1000 horsepower to 1500 horsepower?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 65

Question 66.
THOUGHT PROVOKING The graph of a radical function f passes through the points (3, 1) and (4, 0). Write two different functions that could represent f(x + 2) + 1. Explain.
Answer:

Question 67.
MULTIPLE REPRESENTATIONS The terminal velocity vt(in feet per second) of a skydiver who weighs 140 pounds is given by vt = 33.7\(\sqrt{\frac{140}{\Lambda}}\)
where A is the cross-sectional surface area (in square feet) of the skydiver. The table shows the terminal velocities (in feet per second) for various surface areas (in square feet) of a skydiver who weighs 165 pounds.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 36
a. Which skydiver has a greater terminal velocity for each value of A given in the table?
b. Describe how the different values of A given in the table relate to the possible positions of the falling skydiver.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 67

Question 68.
MATHEMATICAL CONNECTIONS The surface area S of a right circular cone with a slant height of 1 unit is given by S = πr + πr2, where r is the radius of the cone.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 37
a. Use completing the square to show that r = \(\frac{1}{\sqrt{\pi}} \sqrt{S+\frac{\pi}{4}}-\frac{1}{2}\).
b. Graph the equation in part (a) using a graphing calculator. Then find the radius of a right circular cone with a slant height of 1 unit and a surface area of \(\frac{3 \pi}{4}\) square units.
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check your solutions. (Skills Review Handbook)

Question 69.
| 3x + 2 | = 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 69

Question 70.
∣4x + 9 ∣ = −7
Answer:

Question 71.
| 2x − 6 ∣ = ∣ x ∣
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 71

Question 72.
| x + 8 | = | 2x + 2 |
Answer:

Solve the inequality. (Section 3.6)

Question 73.
x2 + 7x + 12 < 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 73

Question 74.
x2 − 10x + 25 ≥ 4
Answer:

Question 75.
2x2 + 6 > 13x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 75

Question 76.
\(\frac{1}{8}\)x2 + x ≤ −2
Answer:

Rational Exponents and Radical Functions Study Skills : Analyzing Your Errors

5.1–5.3 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 38

Core Concepts
Section 5.1
Real nth Roots of a, p. 238
Rational Exponents, p. 239

Section 5.2
Properties of Rational Exponents, p. 244
Properties of Radicals, p. 245

Section 5.3
Parent Functions for Square Root and Cube Root Functions, p. 252
Transformations of Radical Functions, p. 253

Mathematical Practices

Question 1.
How can you use definitions to explain your reasoning in Exercises 21–24 on page 241?

Question 2.
How did you use structure to solve Exercise 76 on page 250?

Question 3.
How can you check that your answer is reasonable in Exercise 39 on page 257?

Question 4.
How can you make sense of the terms of the surface area formula given in Exercise 68 on page 258?

Study Skills
Analyzing Your Errors
Application Errors
What Happens: You can do numerical problems, but you struggle with problems that have context.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 39
How to Avoid This Error: Do not just mimic the steps of solving an application problem. Explain out loud what the question is asking and why you are doing each step. After solving the problem, ask yourself, “Does my solution make sense?”

Rational Exponents and Radical Functions 5.1 – 5.3 Quiz

5.1–5.3 Quiz

Find the indicated real nth root(s) of a. (Section 5.1)

Question 1.
n = 4, a = 81

Question 2.
n = 5, a = −1024

Question 3.
Evaluate (a) 163/4 and (b) 1252/3 without using a calculator. Explain your reasoning. (Section 5.1)

Find the real solution(s) of the equation. Round your answer to two decimal places.(Section 5.1)

Question 4.
2x6 = 1458

Question 5.
(x + 6)3 = 28

Simplify the expression.(Section 5.2)

Question 6.
\(\left(\frac{48^{1 / 4}}{6^{1 / 4}}\right)^{6}\)

Question 7.
\(\sqrt[4]{3}\) • \(\sqrt[4]{432}\)

Question 8.
\(\frac{1}{3+\sqrt{2}}\)

Question 9.
\(\sqrt[3]{16}\) – \(5 \sqrt[3]{2}\)

Question 10.
Simplify \(\sqrt[8]{x^{9} y^{8} z^{16}}\). (Section 5.2)

Write the expression in simplest form. Assume all variables are positive.(Section 5.2)

Question 11.
\(\sqrt[3]{216 p^{9}}\)

Question 12.
\(\frac{\sqrt[5]{32}}{\sqrt[5]{m^{3}}}\)

Question 13.
\(\sqrt[4]{n^{4} q}\) + \(7 n \sqrt[4]{q}\)

Question 14.
Graph f(x) = 2\(2 \sqrt[3]{x}\) + 1. Identify the domain and range of the function. (Section 5.3)

Describe the transformation of the parent function represented by the graph of g. Then write a rule for g.(Section 5.3)

Question 15.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 40

Question 16.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 41

Question 17.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 42

Question 18.
Use a graphing calculator to graph x = 3y2 − 6. Identify the vertex and direction the parabola opens. (Section 5.3)

Question 19.
A jeweler is setting a stone cut in the shape of a regular octahedron. A regular octahedron is a solid with eight equilateral triangles as faces, as shown. The formula for the volume of the stone is V= 0.47s3, where s is the side length (in millimeters) of an edge of the stone. The volume of the stone is 161 cubic millimeters. Find the length of an edge of the stone. (Section 5.1)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 43

Question 20.
An investigator can determine how fast a car was traveling just prior to an accident using the model s = 4\(\sqrt{d}\), where s is the speed (in miles per hour) of the car and d is the length (in feet) of the skid marks. Graph the model. The length of the skid marks of a car is 90 feet. Was the car traveling at the posted speed limit prior to the accident? Explain your reasoning. (Section 5.3)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 44

Lesson 5.4 Solving Radical Equations and Inequalities

Essential Question
How can you solve a radical equation?

EXPLORATION 1
Solving Radical Equations
Work with a partner.
Match each radical equation with the graph of its related radical function. Explain your reasoning. Then use the graph to solve the equation, if possible. Check your solutions.
a. \(\sqrt{x-1}\) – 1 = 0
b. \(\sqrt{2 x+2}\) – \(\sqrt{x+4}\) = 0
c. \(\sqrt{9-x^{2}}\) = 0
d. \(\sqrt{x+2}\) – x = 0
e. \(\sqrt{-x+2}\) – x = 0
f. \(\sqrt{3 x^{2}+1}\) = 0
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 45

EXPLORATION 2
Solving Radical Equations
Work with a partner.
Look back at the radical equations in Exploration 1. Suppose that you did not know how to solve the equations using a graphical approach.
a. Show how you could use a numerical approach to solve one of the equations. For instance, you might use a spreadsheet to create a table of values.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 46
b. Show how you could use an analytical approach to solve one of the equations. For instance, look at the similarities between the equations in Exploration 1. What first step may be necessary so you could square each side to eliminate the radical(s)? How would you proceed to find the solution?

Communicate Your Answer

Question 3.
How can you solve a radical equation?

Question 4.
Would you prefer to use a graphical, numerical, or analytical approach to solve the given equation? Explain your reasoning. Then solve the equation.
\(\sqrt{x+3}\) – \(\sqrt{x-2}\) = 1

Monitoring Progress

Solve the equation. Check your solution.

Question 1.
\(\sqrt[3]{x}\) – 9 = -6

Question 2.
\(\sqrt{x+25}\) = 2

Question 3.
2\(\sqrt[3]{x-3}\) = 4

Question 4.
WHAT IF? Estimate the air pressure at the center of the hurricane when the mean sustained wind velocity is 48.3 meters per second.

Solve the equation. Check your solution(s).

Question 5.
\(\sqrt{10 x+9}\) = x + 3

Question 6.
\(\sqrt{2 x+5}\) = \(\sqrt{x+7}\)

Question 7.
\(\sqrt{x+6}\) – 2 = \(\sqrt{x-2}\)

Solve the equation. Check your solution(s).

Question 8.
(3x)1/3 = −3

Question 9.
(x + 6)1/2 = x

Question 10.
(x + 2)3/4 = 8

Question 11.
Solve
(a) 2\(\sqrt{x}\) − 3 ≥ 3 and
(b) 4\(\sqrt[3]{x+1}\) < 8

Solving Radical Equations and Inequalities 5.4 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY Is the equation 3x − \(\sqrt{2}\) = \(\sqrt{6}\) a radical equation? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 1

Question 2.
WRITING Explain the steps you should use to solve \(\sqrt{x}\) + 10 < 15
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation. Check your solution.

Question 3.
\(\sqrt{5 x+1}\) = 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 3

Question 4.
\(\sqrt{3 x+10}\) = 8
Answer:

Question 5.
\(\sqrt[3]{x-16}\) = 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 5

Question 6.
\(\sqrt[3]{x}\) − 10 = −7
Answer:

Question 7.
−2\(\sqrt{24 x}\) + 13 = −1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 7

Question 8.
8\(\sqrt[3]{10 x}\) − 15 = 17
Answer:

Question 9.
\(\frac{1}{5} \sqrt[3]{3 x}\) = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 9

Question 10.
\(\sqrt{2 x}-\frac{2}{3}\) = 0
Answer:

Question 11.
\(2 \sqrt[5]{x}\) + 7 = 15
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 11

Question 12.
\(\sqrt[4]{4 x}\) − 13 = −15
Answer:

Question 13.
MODELING WITH MATHEMATICS Biologists have discovered that the shoulder height h (in centimeters) of a male Asian elephant can be modeled by h = 62.5\(\sqrt[3]{t}\) + 75.8, where t is the age (in years) of the elephant. Determine the age of an elephant with a shoulder height of 250 centimeters.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 47
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 13

Question 14.
MODELING WITH MATHEMATICS In an amusement park ride, a rider suspended by cables swings back and forth from a tower. The maximum speed v (in meters per second) of the rider can be approximated by v = \(\sqrt{2 g h}\), where h is the height (in meters) at the top of each swing and g is the acceleration due to gravity (g ≈ 9.8 m/sec2). Determine the height at the top of the swing of a rider whose maximum speed is 15 meters per second.
Answer:

In Exercises 15–26, solve the equation. Check your solution(s).

Question 15.
x− 6 = \(\sqrt{3 x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 15

Question 16.
x − 10 = \(\sqrt{9 x}\)
Answer:

Question 17.
\(\sqrt{44-2 x}\) = x − 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 17

Question 18.
\(\sqrt{2 x+30}\) = x + 3
Answer:

Question 19.
\(\sqrt[3]{8 x^{3}-1}\) = 2x − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 19

Question 20.
\(\sqrt[4]{3-8 x^{2}}\) = 2x
Answer:

Question 21.
\(\sqrt{4 x+1}\) = \(\sqrt{x+10}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 21

Question 22.
\(\sqrt{3 x-3}\) – \(\sqrt{x+12}\) = 0
Answer:

Question 23.
\(\sqrt[3]{2 x-5}\) – \(\sqrt[3]{8 x+1}\) = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 23

Question 24.
\(\sqrt[3]{x+5}\) = 2\(\sqrt[3]{2 x+6}\)
Answer:

Question 25.
\(\sqrt{3 x-8}\) + 1 = \(\sqrt{x+5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 25.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 25.2

Question 26.
\(\sqrt{x+2}\) = 2 – \(\sqrt{x}\)
Answer:

In Exercises 27–34, solve the equation. Check your solution(s).

Question 27.
2x2/3 = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 27

Question 28.
4x3/2 = 32
Answer:

Question 29.
x1/4 + 3 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 29

Question 30.
2x3/4 − 14 = 40
Answer:

Question 31.
(x + 6)1/2 = x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 31

Question 32.
(5 − x)1/2 − 2x = 0
Answer:

Question 33.
2(x + 11)1/2 = x + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 33.1

Question 34.
(5x2 − 4)1/4 = x
Answer:

ERROR ANALYSIS In Exercises 35 and 36, describe and correct the error in solving the equation.

Question 35.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 48
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 35.1

Question 36.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 49
Answer:

In Exercises 37–44, solve the inequality.

Question 37.
\(2 \sqrt[3]{x}\) − 5 ≥ 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 37

Question 38.
\(\sqrt[3]{x-4}\) ≤ 5
Answer:

Question 39.
\(4 \sqrt{x-2}\) > 20
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 39

Question 40.
7\(\sqrt{x}\) + 1 < 9
Answer:

Question 41.
2\(\sqrt{x}\) + 3 ≤ 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 41

Question 42.
\(\sqrt[3]{x+7}\) ≥ 3
Answer:

Question 43.
\(-2 \sqrt[3]{x+4}\) < 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 43

Question 44.
−0.25\(\sqrt{x}\) − 6 ≤ −3
Answer:

Question 45.
MODELING WITH MATHEMATICS The length ℓ (in inches) of a standard nail can be modeled by ℓ = 54d3/2, where d is the diameter (in inches) of the nail. What is the diameter of a standard nail that is 3 inches long?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 45

Question 46.
DRAWING CONCLUSIONS “Hang time” is the time you are suspended in the air during a jump. Your hang time t (in seconds) is given by the function t = 0.5\(\sqrt{h}\), where h is the height (in feet) of the jump. Suppose a kangaroo and a snowboarder jump with the hang times shown.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 50
a. Find the heights that the snowboarder and the kangaroo jump.
b. Double the hang times of the snowboarder and the kangaroo and calculate the corresponding heights of each jump.
c. When the hang time doubles, does the height of the jump double? Explain.
Answer:

USING TOOLS In Exercises 47–52, solve the nonlinear system. Justify your answer with a graph.

Question 47.
y2 = x − 3
y = x − 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 47

Question 48.
y2 = 4x + 17
y = x + 5
Answer:

Question 49.
x2 + y2 = 4
y = x – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 49

Question 50.
x2 + y2 = 25
y = \(-\frac{3}{4}\)x + \(\frac{25}{4}\)
Answer:

Question 51.
x2 + y2 = 1
y = \(\frac{1}{2}\)x2 – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 51

Question 52.
x2 + y2 = 4
y2 = x + 2
Answer:

Question 53.
PROBLEM SOLVING The speed s (in miles per hour) of a car can be given by s = \(\sqrt{30 f d}\), where f is the coefficient of friction and d is the stopping distance (in feet). The table shows the coefficient of friction for different surfaces.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 51
a. Compare the stopping distances of a car traveling 45 miles per hour on the surfaces given in the table.
b. You are driving 35 miles per hour on an icy road when a deer jumps in front of your car. How far away must you begin to brake to avoid hitting the deer? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 53.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 53.2

Question 54.
MODELING WITH MATHEMATICS The Beaufort wind scale was devised to measure wind speed. The Beaufort numbers B, which range from 0 to 12, can be modeled by B = 1.69\(\sqrt{s+4.25}\) − 3.55, where s is the wind speed (in miles per hour).
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 52
a. What is the wind speed for B = 0? B = 3?
b. Write an inequality that describes the range of wind speeds represented by the Beaufort model.
Answer:

Question 55.
USING TOOLS Solve the equation x − 4 = \(\sqrt{2 x}\). Then solve the equation x − 4 = − \(\sqrt{2 x}\).
a. How does changing \(\sqrt{2 x}\) to −\(\sqrt{2 x}\) change the solution(s) of the equation?
b. Justify your answer in part (a) using graphs.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 55.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 55.2
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 55.3

Question 56.
MAKING AN ARGUMENT Your friend says it is impossible for a radical equation to have two extraneous solutions. Is your friend correct? Explain your reasoning.
Answer:

Question 57.
USING STRUCTURE Explain how you know the radical equation \(\sqrt{x+4}\) = −5 has no real solution without solving it.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 57

Question 58.
HOW DO YOU SEE IT? Use the graph to find the solution of the equation 2\(\sqrt{x-4}\) = −\(\sqrt{x-1}\) + 4. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 53
Answer:

Question 59.
WRITING A company determines that the price p of a product can be modeled by p = 70 − \(\sqrt{0.02 x+1}\), where x is the number of units of the product demanded per day. Describe the effect that raising the price has on the number of units demanded.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 59

Question 60.
THOUGHT PROVOKING City officials rope off a circular area to prepare for a concert in the park. They estimate that each person occupies 6 square feet. Describe how you can use a radical inequality to determine the possible radius of the region when P people are expected to attend the concert.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 54
Answer:

Question 61.
MATHEMATICAL CONNECTIONS The Moeraki Boulders along the coast of New Zealand are stone spheres with radii of approximately 3 feet. A formula for the radius of a sphere is
r = \(\frac{1}{2} \sqrt{\frac{S}{\pi}}\)
where S is the surface area of the sphere. Find the surface area of a Moeraki Boulder.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 61

Question 62.
PROBLEM SOLVING You are trying to determine the height of a truncated pyramid, which cannot be measured directly. The height h and slant heightℓof the truncated pyramid are related by the formula below.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 55
In the given formula, b1 and b2 are the side lengths of the upper and lower bases of the pyramid, respectively. When ℓ = 5, b1 = 2, and b2 = 4, what is the height of the pyramid?
Answer:

Question 63.
REWRITING A FORMULA A burning candle has a radius of r inches and was initially h0inches tall. After t minutes, the height of the candle has been reduced to h inches. These quantities are related by the formula
r = \(\sqrt{\frac{k t}{\pi\left(h_{0}-h\right)}}\)
where k is a constant. Suppose the radius of a candle is 0.875 inch, its initial height is 6.5 inches, and k = 0.04.
a. Rewrite the formula, solving for h in terms of t.
b. Use your formula in part (a) to determine the height of the candle after burning 45 minutes.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 63

Maintaining Mathematical Proficiency

Perform the indicated operation. (Section 4.2 and Section 4.3)

Question 64.
(x3 − 2x2 + 3x + 1) + (x4 − 7x)
Answer:

Question 65.
(2x5 + x4 − 4x2) − (x5 − 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 65

Question 66.
(x3 + 2x2 + 1)(x2 + 5)
Answer:

Question 67.
(x4 + 2x3 + 11x2 + 14x − 16) ÷ (x + 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 67

Let f(x) = x3 – 4x2 + 6. Write a rule for g. Describe the graph of g as a transformation of the graph of f.(Section 4.7)

Question 68.
g(x) = f(−x) + 4
Answer:

Question 69.
g(x) = \(\frac{1}{2}\)f(x) − 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 69

Question 70.
g(x) = −f(x − 1) + 6
Answer:

Lesson 5.5 Performing Function Operations

Essential Question
How can you use the graphs of two functions to sketch the graph of an arithmetic combination of the two functions?
Just as two real numbers can be combined by the operations of addition, subtraction, multiplication, and division to form other real numbers, two functions can be combined to form other functions. For example, the functions f(x) = 2x − 3 and g(x) = x2 − 1 can be combined to form the sum, difference, product, or quotient of f and g.
f(x) + g(x) = (2x − 3) + (x2 − 1) = x2 + 2x − 4 sum
f(x) − g(x) = (2x − 3) − (x2 − 1) = −x2 + 2x − 2 difference
f(x) • g(x) = (2x − 3)(x2 − 1) = 2x3 − 3x2 − 2x + 3 product
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 56

EXPLORATION 1
Graphing the Sum of Two Functions
Work with a partner.
Use the graphs of f and g to sketch the graph of f + g. Explain your steps.
Sample Choose a point on the graph of g. Use a compass or a ruler to measure its distance above or below the x-axis. If above, add the distance to the y-coordinate of the point with the same x-coordinate on the graph of f. If below, subtract the distance. Plot the new point. Repeat this process for several points. Finally, draw a smooth curve through the new points to obtain the graph of f + g.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 57
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 58

Communicate Your Answer

Question 2.
How can you use the graphs of two functions to sketch the graph of an arithmetic combination of the two functions?

Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 59
Question 3.
Check your answers in Exploration 1 by writing equations for f and g, adding the functions, and graphing the sum.

5.5 Lesson

Monitoring Progress

Question 1.
Let f(x) = −2x2/3 and g(x) = 7x2/3. Find (f + g)(x) and (f − g)(x) and state the domain of each. Then evaluate (f + g)(8) and (f − g)(8).

Question 2.
Let f(x) = 3x and g(x) = x1/5. Find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate (fg)(32) and (\(\frac{f}{g}\))(32).

Question 3.
Let f(x) = 8x and g(x) = 2x5/6. Use a graphing calculator to evaluate (f + g)(x), (f − g)(x), (fg)(x), and (\(\frac{f}{g}\)) (x) when x = 5. Round your answers to two decimal places.

Question 4.
In Example 5, explain why you can evaluate (f + g)(3), (f − g)(3), and (fg)(3) but not (\(\frac{f}{g}\))(3).

Question 5.
Use the answer in Example 6(a) to find the total number of heartbeats over the lifetime of a white rhino when its body mass is 1.7 × 105 kilograms.

Performing Function Operations 5.5 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING Let f and g be any two functions. Describe how you can use f, g, and the four basic operations to create new functions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 1

Question 2.
WRITING What x-values are not included in the domain of the quotient of two functions?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, find (f + g)(x) and (f – g)(x) and state the domain of each. Then evaluate f + g and f – g for the given value of x.

Question 3.
f(x) = \(-5 \sqrt[4]{x}\), g(x) = 19\(\sqrt[4]{x}\); x = 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 3

Question 4.
f(x) = \(\sqrt[3]{2 x}\), g(x) = −11\(\sqrt[3]{2 x}\) ; x = −4
Answer:

Question 5.
f(x) = 6x − 4x2− 7x3, g(x) = 9x2− 5x; x = −1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 5

Question 6.
f(x) = 11x + 2x2, g(x) = −7x − 3x2 + 4; x = 2
Answer:

In Exercises 7–12, find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate fg and \(\frac{f}{g}\) for the given value of x.

Question 7.
f(x) = 2x3, g(x) = \(\sqrt[3]{x}\) ; x = −27
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 7

Question 8.
f(x) = x4, g(x) = \(3 \sqrt{x}\) ; x = 4
Answer:

Question 9.
f(x) = 4x, g(x) = 9x1/2; x = 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 9

Question 10.
f(x) = 11x3, g(x) = 7x7/3; x = −8
Answer:

Question 11.
f(x) = 7x3/2, g(x) =−14x1/3; x = 64
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 11

Question 12.
f(x) = 4x5/4, g(x) = 2x1/2; x = 16
Answer:

USING TOOLS In Exercises 13–16, use a graphing calculator to evaluate (f + g)(x), (f − g)(x), (fg)(x), and (\(\frac{f}{g}\))(x) when x = 5. Round your answers to two decimal places.

Question 13.
f(x) = 4x4; g(x) = 24x1/3

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 13

Question 14.
f(x) = 7x5/3; g(x) = 49x2/3

Answer:

Question 15.
f(x) =−2x1/3; g(x) = 5x1/2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 15

Question 16.
f(x) = 4x1/2; g(x) = 6x3/4

Answer:

ERROR ANALYSIS In Exercises 17 and 18, describe and correct the error in stating the domain.

Question 17.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 60
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 17

Question 18.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 61
Answer:

Question 19.
MODELING WITH MATHEMATICS From 1990 to 2010, the numbers (in millions) of female F and male M employees from the ages of 16 to 19 in the United States can be modeled by F(t) =−0.007t2 + 0.10t + 3.7 and M(t) = 0.0001t3 − 0.009t2 + 0.11t + 3.7, where t is the number of years since 1990.
a. Find (F + M)(t).
b. Explain what (F + M)(t) represents.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 19

Question 20.
MODELING WITH MATHEMATICS From 2005 to 2009, the numbers of cruise ship departures (in thousands) from around the world W and Florida F can be modeled by the equations
W(t) = −5.8333t3 + 17.43t2 + 509.1t + 11496
F(t) = 12.5t3 − 60.29t2 + 136.6t + 4881
where t is the number of years since 2005.
a. Find (W − F )(t).
b. Explain what (W − F )(t) represents.
Answer:

Question 21.
MAKING AN ARGUMENT Your friend claims that the addition of functions and the multiplication of functions are commutative. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 21

Question 22.
HOW DO YOU SEE IT? The graphs of the functions f(x) = 3x2 − 2x − 1 and g(x) = 3x + 4 are shown. Which graph represents the function f + g? the function f − g? Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 62
Answer:

Question 23.
REASONING The table shows the outputs of the two functions f and g. Use the table to evaluate (f + g)(3), (f − g)(1), (fg)(2), and (\(\frac{f}{g}\))(0).
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 63
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 23

Question 24.
THOUGHT PROVOKING Is it possible to write two functions whose sum contains radicals, but whose product does not? Justify your answers.
Answer:

Question 25.
MATHEMATICAL CONNECTIONS A triangle is inscribed in a square, as shown. Write and simplify a function r in terms of x that represents the area of the shaded region.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 64
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 25

Question 26.
REWRITING A FORMULA For a mammal that weighs w grams, the volume b (in milliliters) of air breathed in and the volume d (in milliliters) of “dead space” (the portion of the lungs not filled with air) can be modeled by
b(w) = 0.007w and d(w) = 0.002w.
The breathing rate r (in breaths per minute) of a mammal that weighs w grams can be modeled by
r(w) = \(\frac{1.1 w^{0.734}}{b(w)-d(w)}\).
Answer:

Question 27.
PROBLEM SOLVING A mathematician at a lake throws a tennis ball from point A along the water’s edge to point B in the water, as shown. His dog, Elvis, first runs along the beach from point A to point D and then swims to fetch the ball at point B.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 65
a. Elvis runs at a speed of about 6.4 meters per second. Write a function r in terms of x that represents the time he spends running from point A to point D. Elvis swims at a speed of about 0.9 meter per second. Write a function s in terms of x that represents the time he spends swimming from point D to point B.
b. Write a function t in terms of x that represents the total time Elvis spends traveling from point A to point D to point B.
c. Use a graphing calculator to graph t. Find the value of x that minimizes t. Explain the meaning of this value.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 27

Maintaining Mathematical Proficiency

Solve the literal equation for n. (Skills Review Handbook)

Question 28.
3xn − 9 = 6y
Answer:

Question 29.
5z = 7n + 8nz
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 29

Question 30.
3nb = 5n − 6z
Answer:

Question 31.
\(\frac{3+4 n}{n}\) = 7b
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 31

Determine whether the relation is a function. Explain. (Skills Review Handbook)

Question 32.
(3, 4), (4, 6), (1, 4), (2, −1)
Answer:

Question 33.
(−1, 2), (3, 7), (0, 2), (−1, −1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 33

Question 34.
(1, 6), (7, −3), (4, 0), (3, 0)
Answer:

Question 35.
(3, 8), (2, 5), (9, 5), (2, −3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 35

Lesson 5.6 Inverse of a Function

Essential Question
How can you sketch the graph of the inverse of a function?

EXPLORATION 1
Graphing Functions and Their Inverses
Work with a partner.
Each pair of functions are inverses of each other. Use a graphing calculator to graph f and g in the same viewing window. What do you notice about the graphs?
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 65.1
a. f(x) = 4x + 3
g(x) = \(\frac{x-3}{4}\)
b. f(x) = x3 + 1
g(x) = \(\sqrt[3]{x-1}\)
c. f(x) = \(\sqrt{x-3}[latex]
g(x) = x2 + 3, x ≥ 0
d. f(x) = [latex]\frac{4 x+4}{x+5}\)
g(x) = \(\frac{4-5 x}{x-4}\)

EXPLORATION 2
Sketching Graphs of Inverse Functions
Work with a partner.
Use the graph of f to sketch the graph of g, the inverse function of f, on the same set of coordinate axes. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 66

Communicate Your Answer

Question 3.
How can you sketch the graph of the inverse of a function?

Question 4.
In Exploration 1, what do you notice about the relationship between the equations of f and g? Use your answer to find g, the inverse function of
f(x) = 2x − 3.
Use a graph to check your answer.

5.6 Lesson

Monitoring Progress

Solve y = f(x) for x. Then find the input(s) when the output is 2.

Question 1.
f(x) = x − 2

Question 2.
f(x) = 2x2

Question 3.
f(x) = −x3 + 3

Find the inverse of the function. Then graph the function and its inverse.

Question 4.
f(x) = 2x

Question 5.
f(x) = −x + 1

Question 6.
f(x) = \(\frac{1}{3}\)x − 2

Find the inverse of the function. Then graph the function and its inverse.

Question 7.
f(x) = −x2, x ≤ 0

Question 8.
f(x) = −x3 + 4

Question 9.
f(x) = \(\sqrt{x+2}\)

Determine whether the functions are inverse functions.

Question 10.
f(x) = x + 5, g(x) = x − 5

Question 11.
f(x) = 8x3, g(x) = \(\sqrt[3]{2 x}\)

Question 12.
The distance d (in meters) that a dropped object falls in t seconds on Earth is represented by d = 4.9t2. Find the inverse of the function. How long does it take an object to fall 50 meters?

Inverse of a Function 5.6 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY In your own words, state the definition of inverse functions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 1

Question 2.
WRITING Explain how to determine whether the inverse of a function is also a function.
Answer:

Question 3.
COMPLETE THE SENTENCE Functions f and g are inverses of each other provided that f(g(x)) = ____ and g(f(x)) = ____.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 3

Question 4.
DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 67
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, solve y = f(x) for x. Then find the input(s) when the output is -3.

Question 5.
f(x) = 3x + 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 5

Question 6.
f(x) = −7x − 2
Answer:

Question 7.
f(x) = \(\frac{1}{2}\)x − 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 7

Question 8.
f(x) = −\(\frac{2}{3}\)x + 1
Answer:

Question 9.
f(x) = 3x3

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 9

Question 10.
f(x) = 2x4 − 5
Answer:

Question 11.
f(x) = (x − 2)2 − 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 11

Question 12.
f(x) = (x − 5)3 − 1
Answer:

In Exercises 13–20, find the inverse of the function. Then graph the function and its inverse.

Question 13.
f(x) = 6x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 13.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 13.2

Question 14.
f(x) = −3x
Answer:

Question 15.
f(x) = −2x + 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 15.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 15.2

Question 16.
f(x) = 6x − 3
Answer:

Question 17.
f(x) = −\(\frac{1}{2}\)x + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 17.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 17.2

Question 18.
f(x) = \(\frac{1}{3}\)x − 1
Answer:

Question 19.
f(x) = \(\frac{2}{3}\)x − \(\frac{1}{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 19.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 19.2

Question 20.
f(x) = −\(\frac{4}{5}\)x + \(\frac{1}{5}\)
Answer:

Question 21.
COMPARING METHODS Find the inverse of the function f(x) = −3x + 4 by switching the roles of x and y and solving for y. Then find the inverse of the function f by using inverse operations in the reverse order. Which method do you prefer? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 21

Question 22.
REASONING Determine whether each pair of functions f and g are inverses. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 68
Answer:

In Exercises 23–28, find the inverse of the function. Then graph the function and its inverse.

Question 23.
f(x) = 4x2, x ≤ 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 23

Question 24.
f(x) = 9x2, x ≤ 0
Answer:

Question 25.
f(x) = (x − 3)3

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 25

Question 26.
f(x) = (x + 4)3

Answer:

Question 27.
f(x) = 2x4, x ≥ 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 27

Question 28.
f(x) = −x6, x ≥ 0
Answer:

ERROR ANALYSIS In Exercises 29 and 30, describe and correct the error in finding the inverse of the function.

Question 29.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 69
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 29

Question 30.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 70
Answer:

USING TOOLS In Exercises 31–34, use the graph to determine whether the inverse of f is a function. Explain your reasoning.

Question 31.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 71
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 31

Question 32.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 72
Answer:

Question 33.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 73
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 33

Question 34.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 74
Answer:

In Exercises 35–46, determine whether the inverse of f is a function. Then find the inverse.

Question 35.
f(x) = x3 − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 35

Question 36.
f(x) = −x3 + 3
Answer:

Question 37.
f(x) = \(\sqrt{x+4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 37

Question 38.
f(x) = \(\sqrt{x-6}\)
Answer:

Question 39.
f(x) = \(2 \sqrt[3]{x-5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 39

Question 40.
f(x) = 2x2 − 5
Answer:

Question 41.
f(x) = x4 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 41

Question 42.
f(x) = 2x3 − 5
Answer:

Question 43.
f(x) = \(3 \sqrt[3]{x+1}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 43

Question 44.
f(x) = \(-\sqrt[3]{\frac{2 x+4}{3}}\)
Answer:

Question 45.
f(x) = \(\frac{1}{2}\)x5

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 45

Question 46.
f(x) = \(-3 \sqrt{\frac{4 x-7}{3}}\)
Answer:

Question 47.
WRITING EQUATIONS What is the inverse of the function whose graph is shown?
A. g(x) = \(\frac{3}{2}\)x − 6
B. g(x) = \(\frac{3}{2}\)x + 6
C. g(x) = \(\frac{2}{3}\)x − 6
D. g(x) = \(\frac{2}{3}\)x + 12
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 75
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 47

Question 48.
WRITING EQUATIONS What is the inverse of f(x) = −\(-\frac{1}{64}\)x3?
A. g(x) = −4x3
B. g(x) = 4\(\sqrt[3]{x}\)
C. g(x) = −4\(\sqrt[3]{x}\)
D. g(x) = \(\sqrt[3]{-4 x}\)
Answer:

In Exercises 49–52, determine whether the functions are inverses.

Question 49.
f(x) = 2x − 9, g(x) = x — 2 + 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 49

Question 50.
f(x) = \(\frac{x-3}{4}\), g(x) = 4x + 3
Answer:

Question 51.
f(x) = \(\sqrt[5]{\frac{x+9}{5}}\), g(x) = 5x5 − 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 51.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 51.2

Question 52.
f(x) = 7x3/2 − 4, g(x) = (\(\frac{x+4}{7}\))3/2

Answer:

Question 53.
MODELING WITH MATHEMATICS The maximum hull speed v (in knots) of a boat with a displacement hull can be approximated by v = 1.34\(\sqrt{\ell}\), where ℓ is the waterline length (in feet) of the boat. Find the inverse function. What waterline length is needed to achieve a maximum speed of 7.5 knots?
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 76
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 53.1

Question 54.
MODELING WITH MATHEMATICS Elastic bands can be used for exercising to provide a range of resistance. The resistance R (in pounds) of a band can be modeled by R = \(\frac{3}{8}\)L − 5, where L is the total length (in inches) of the stretched band. Find the inverse function. What length of the stretched band provides 19 pounds of resistance?
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 77
Answer:

ANALYZING RELATIONSHIPS In Exercises 55–58, match the graph of the function with the graph of its inverse.

Question 55.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 78
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 55

Question 56.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 79
Answer:

Question 57.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 80
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 57

Question 58.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 81
Answer:

Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 82

Question 59.
REASONING You and a friend are playing a number-guessing game. You ask your friend to think of a positive number, square the number, multiply the result by 2, and then add 3. Your friend’s final answer is 53. What was the original number chosen? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 59

Question 60.
MAKING AN ARGUMENT Your friend claims that every quadratic function whose domain is restricted to nonnegative values has an inverse function. Is your friend correct? Explain your reasoning.
Answer:

Question 61.
PROBLEM SOLVING When calibrating a spring scale, you need to know how far the spring stretches for various weights. Hooke’s Law states that the length a spring stretches is proportional to the weight attached to it. A model for one scale is ℓ = 0.5w + 3, whereℓ is the total length (in inches) of the stretched spring and w is the weight (in pounds) of the object.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 83
a. Find the inverse function. Describe what it represents.
b. You place a melon on the scale, and the spring stretches to a total length of 5.5 inches. Determine the weight of the melon.
c. Verify that the function ℓ = 0.5w + 3 and the inverse model in part (a) are inverse functions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 61.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 61.2

Question 62.
THOUGHT PROVOKING Do functions of the form y = xm/n, where m and n are positive integers, have inverse functions? Justify your answer with examples.
Answer:

Question 63.
PROBLEM SOLVING At the start of a dog sled race in Anchorage, Alaska, the temperature was 5°C. By the end of the race, the temperature was −10°C. The formula for converting temperatures from degrees Fahrenheit F to degrees Celsius C is C = \(\frac{5}{9}\)(F − 32).
a. Find the inverse function. Describe what it represents.
b. Find the Fahrenheit temperatures at the start and end of the race.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 84
c. Use a graphing calculator to graph the original function and its inverse. Find the temperature that is the same on both temperature scales.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 63.1

Question 64.
PROBLEM SOLVING The surface area A (in square meters) of a person with a mass of 60 kilograms can be approximated by A = 0.2195h0.3964, where h is the height (in centimeters) of the person.
a. Find the inverse function. Then estimate the height of a 60-kilogram person who has a body surface area of 1.6 square meters.
b. Verify that function A and the inverse model in part (a) are inverse functions.
Answer:

USING STRUCTURE In Exercises 65–68, match the function with the graph of its inverse.

Question 65.
f(x) = \(\sqrt[3]{x-4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 65.1

Question 66.
f(x) = \(\sqrt[3]{x+4}\)
Answer:

Question 67.
f(x) = \(\sqrt{x+1}\) – 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 67.1

Question 68.
f(x) = \(\sqrt{x-1}\) + 3
Answer:

Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 85

Question 69.
DRAWING CONCLUSIONS Determine whether the statement is true or false. Explain your reasoning.
a. If f(x) = xn and n is a positive even integer, then the inverse of f is a function.
b. If f(x) = xn and n is a positive odd integer, then the inverse of f is a function.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 69.1

Question 70.
HOW DO YOU SEE IT? The graph of the function f is shown. Name three points that lie on the graph of the inverse of f. Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 86
Answer:

Question 71.
ABSTRACT REASONING Show that the inverse of any linear function f(x) = mx + b, where m ≠ 0, is also a linear function. Identify the slope and y-intercept of the graph of the inverse function in terms of m and b.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 71

Question 72.
CRITICAL THINKING Consider the function f(x) = −x.
a. Graph f(x) = −x and explain why it is its own inverse. Also, verify that f(x) = −x is its own inverse algebraically.
b. Graph other linear functions that are their own inverses. Write equations of the lines you graphed.
c. Use your results from part (b) to write a general equation describing the family of linear functions that are their own inverses.
Answer:

Maintaining Mathematical Proficiency

Simplify the expression. Write your answer using only positive exponents.(Skills Review Handbook)

Question 73.
(−3)−3

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 73

Question 74.
23 • 22

Answer:

Question 75.
\(\frac{4^{5}}{4^{3}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 75

Question 76.
(\(\frac{2}{3}\))4

Answer:

Describe the x-values for which the function is increasing, decreasing, positive, and negative.

Question 77.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 87
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 77

Question 78.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 88
Answer:

Question 79.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 89
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 79

Rational Exponents and Radical Functions Performance Task: Turning the Tables

5.4–5.6 What Did You Learn?

Core Vocabulary
radical equation, p. 262
extraneous solutions, p. 263
inverse functions, p. 277

Core Concepts
Section 5.4

Solving Radical Equations, p. 262
Solving Radical Inequalities, p. 265

Section 5.5
Operations on Functions, p. 270

Section 5.6
Exploring Inverses of Functions, p. 276
Inverses of Nonlinear Functions, p. 278
Horizontal Line Test, p. 278

Mathematical Practices

Question 1.
How did you find the endpoints of the range in part (b) of Exercise 54 on page 267?

Question 2.
How did you use structure in Exercise 57 on page 268?

Question 3.
How can you evaluate the reasonableness of the results in Exercise 27 on page 274?

Question 4.
How can you use a graphing calculator to check your answers in Exercises 49–52 on page 282?

Performance Task

Turning the Tables

In this chapter, you have used properties of rational exponents and functions to find an answer to the problem. Using those same properties, can you find a problem to the answer? How many problems can you find?
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 90
To explore the answers to these questions and more, go to BigIdeasMath.com

Rational Exponents and Radical Functions Chapter Review

Evaluate the expression without using a calculator.

Question 1.
87/3

Question 2.
95/2

Question 3.
(−27)−2/3

Find the real solution(s) of the equation. Round your answer to two decimal places when appropriate.

Question 4.
x5 + 17 = 35

Question 5.
7x3 = 189

Question 6.
(x + 8)4 = 16

Simplify the expression.

Question 7.
(\(\frac{6^{1 / 5}}{6^{2 / 5}}\))3

Question 8.
\(\sqrt[4]{32} \cdot \sqrt[4]{8}\)

Question 9.
\(\frac{1}{2-\sqrt[4]{9}}\)

Question 10.
\(4 \sqrt[5]{8}\) + \(3 \sqrt[5]{8}\)

Question 11.
\(2 \sqrt{48}\) – \(\sqrt{3}\)

Question 12.
(\(5^{2 / 3} \cdot 2^{3 / 2}\))1/2

Simplify the expression. Assume all variables are positive.

Question 13.
\(\sqrt[3]{125 z^{9}}\)

Question 14.
\(\frac{2^{1 / 4} z^{5 / 4}}{6 z}\)

Question 15.
\(\sqrt{10 z^{5}}-z^{2} \sqrt{40 z}\)

Describe the transformation of f represented by g. Then graph each function.

Question 16.
f(x) = \(\sqrt{x}\), g(x) = −2\(\sqrt{x}\)

Question 17.
f(x) = \(\sqrt[3]{x}\), g(x) = \(\sqrt[3]{-x}\) − 6

Question 18.
Let the graph of g be a reflection in the y-axis, followed by a translation 7 units to the right of the graph of f(x) = \(\sqrt[3]{x}\). Write a rule for g.

Question 19.
Use a graphing calculator to graph 2y2 = x − 8. Identify the vertex and the direction that the parabola opens.

Question 20.
Use a graphing calculator to graph x2 + y2 = 81. Identify the radius and the intercepts.

Solve the equation. Check your solution.

Question 21.
\(4 \sqrt[3]{2 x+1}\) = 20

Question 22.
\(\sqrt{4 x-4}\) = \(\sqrt{5 x-1}\) – 1

Question 23.
(6x)2/3 = 36

Solve the inequality.

Question 24.
\(5 \sqrt{x}\) + 2 > 17

Question 25.
\(2 \sqrt{x-8}\) < 24

Question 26.
7\(\sqrt[3]{x-3}\) ≥ 21

Question 27.
In a tsunami, the wave speeds (in meters per second) can be modeled by s(d ) = \(\sqrt{9.8 d}\), where d is the depth (in meters) of the water. Estimate the depth of the water when the wave speed is 200 meters per second.

Question 28.
Let f(x) = 2\(\sqrt{3-x}\) and g(x) = 4\(\sqrt[3]{3-x}\). Find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate(fg)(2) and (\(\frac{f}{g}\))(2).

Question 29.
Let f(x) = 3x2 + 1 and g(x) = x + 4. Find (f + g)(x) and (f − g)(x) and state the domain of each. Then evaluate (f + g)(−5) and (f − g)(−5).

Question 30.
f(x) = −\(\frac{1}{2}\)x + 10

Question 31.
f(x) = x2 + 8, x ≥ 0

Question 32.
f(x) = −x3 − 9

Question 33.
f(x) = 3\(\sqrt{x}\) + 5

Determine whether the functions are inverse functions.

Question 34.
f(x) = 4(x − 11)2, g(x) = \(\frac{1}{4}\)(x + 11)2

Question 35.
f(x) = −2x + 6, g(x) = −\(\frac{1}{2}\)x + 3

Question 36.
On a certain day, the function that gives U.S. dollars in terms of British pounds is d = 1.587p, where d represents U.S. dollars and p represents British pounds. Find the inverse function. Then find the number of British pounds equivalent to 100 U.S. dollars.

Rational Exponents and Radical Functions Chapter Test

Question 1.
Solve the inequality \(\) − 2 ≤ 13 and the equation 5\(\) − 2 = 13. Describe the similarities and differences in solving radical equations and radical inequalities.

Describe the transformation of f represented by g. Then write a rule for g.

Question 2.
f(x) = \(\sqrt{x}\)
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 92

Question 3.
f(x) = \(\sqrt[3]{x}\)
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 93

Question 4.
f(x) = \(\sqrt[5]{x}\)
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 94

Simplify the expression. Explain your reasoning.

Question 5.
642/3

Question 6.
(−27)5/3

Question 7.
\(\sqrt[4]{48 x y^{11} z^{3}}\)

Question 8.
\(\frac{\sqrt[3]{256}}{\sqrt[3]{32}}\)

Question 9.
Write two functions whose graphs are translations of the graph of y = \(\sqrt{x}\). The first function should have a domain of x ≥ 4. The second function should have a range of y ≥ −2.

Question 10.
In bowling, a handicap is a change in score to adjust for differences in the abilities of players. You belong to a bowling league in which your handicap h is determined using the formula h = 0.9(200 − a), where a is your average score. Find the inverse of the model. Then find the average for a bowler whose handicap is 36.

Question 11.
The basal metabolic rate of an animal is a measure of the amount of calories burned at rest for basic functioning. Kleiber’s law states that an animal’s basal metabolic rate R (in kilocalories per day) can be modeled by R = 73.3w3/4, where w is the mass (in kilograms) of the animal. Find the basal metabolic rates of each animal in the table.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 95

Question 12.
Let f(x) = 6x3/5 and g(x) = −x3/5. Find (f + g)(x) and (f − g)(x) and state the domain of each. Then evaluate (f + g)(32) and (f − g)(32).

Question 13.
Let f(x) = \(\frac{1}{2}\)x3/4 and g(x) = 8x. Find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate (fg)(16) and (\(\frac{f}{g}\))(16).

Question 14.
A football player jumps to catch a pass. The maximum height h (in feet) of the player above the ground is given by the function h = \(\frac{1}{64}\)s2, where s is the initial speed (in feet per second) of the player. Find the inverse of the function. Use the inverse to find the initial speed of the player shown. Verify that the functions are inverse functions.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 96

Rational Exponents and Radical Functions Cumulative Assessment

Question 1.
Identify three pairs of equivalent expressions. Assume all variables are positive. Justify your answer.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 97

Question 2.
The graph represents the function Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 98. Choose the correct values to complete the function.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 98.1

Question 3.
In rowing, the boat speed s (in meters per second) can be modeled by s= 4.62\(\sqrt[9]{n}\), where n is the number of rowers.
a. Find the boat speeds for crews of 2 people, 4 people, and 8 people.
b. Does the boat speed double when the number of rowers doubles? Explain.
c. Find the time (in minutes) it takes each crew in part (a) to complete a 2000-meter race.

Question 4.
A polynomial function fits the data in the table. Use finite differences to find the degree of the function and complete the table. Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 99

Question 5.
The area of the triangle is 42 square inches. Find the value of x.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 100

Question 6.
Which equations are represented by parabolas? Which equations are functions? Place check marks in the appropriate spaces. Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 101

Question 7.
What is the solution of the inequality 2\(\sqrt{x+3}\) − 1 < 3?
A. x < 1
B. −3 < x < 1
C. −3 ≤ x < 1
D. x ≥ −3

Question 8.
Which function does the graph represent? Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 102

Question 9.
Your friend releases a weather balloon 50 feet from you. The balloon rises vertically. When the balloon is at height h, the distance d between you and the balloon is given by d = \(\sqrt{2500+h^{2}}\), where h and d are measured in feet. Find the inverse of the function. What is the height of the balloon when the distance between you and the balloon is 100 feet?

Question 10.
The graphs of two functions f and g are shown. Are f and g inverse functions? Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 103

Big Ideas Math Geometry Answers Chapter 12 Probability

Big Ideas Math Geometry Answers Chapter 12

Big Ideas Math Book Geometry Answer Key Ch 12 Probability is aligned as per the BIM Geometry Textbooks. Refer to the BIM Geometry Solutions for quick guidance on the Topicwise Concepts of Probability. Big Ideas Math Geometry Solution Key can be downloaded for free of cost. Access the Big Ideas Math Geometry Chapter 12 Probability Answers created by subject experts adhering to the latest Common Core State Standards guidelines. Download the handy resources for Big Ideas Math Geometry Chapter 12 Probability Solutions and begin your preparation.

Big Ideas Math Book Geometry Answer Key Chapter 12 Probability

Make the most out of the Big Ideas Math Geometry Answer Key for Cha 12 Probability and score better grades in the exams. Simply tap on the quick links available and prepare the respective topics available as and when you need them. Big Ideas Math Geometry Answers Chapter 12 Probability available here covers questions from the Topics Sample Spaces, Independent and Dependent Events, Disjoint and Overlapping Events, Binomial Distributions, Probability Test, etc.

Lesson: 1 Sample Spaces and Probability

Lesson: 2 Independent and Dependent Events

Lesson: 3 Two-Way Tables and Probability

Quiz

Lesson: 4 Probability of Disjoint and Overlapping Events

Lesson: 5 Permutations and Combinations

Lesson: 6 Binomial Distributions

Chapter: 12 – Probability 

Probability Maintaining Mathematical Proficiency

Write and solve a proportion to answer the question.

Question 1.
What percent of 30 is 6?
Answer: 20

Explanation:
100% = 30
x% = 6
100% = 30(1)
x% = 6(2)
100%/x% = 30/6
Taking the inverse of both sides
x%/100% = 6/30
x = 20%
Thus 6 is 20% of 30.

Question 2.
What number is 68% of 25?
Answer: 17

Explanation:
68% × 25
(68 ÷ 100) × 25
(68 × 25) ÷ 100
1700 ÷ 100 = 17

Question 3.
34.4 is what percent of 86?
Answer: 40

Explanation:
100% = 86
x% = 34.4
100% = 86(1)
x% = 34.4(2)
100%/x% = 86/34.4
Taking the inverse of both sides
x%/100% = 34.4/86
x = 40%
Therefore, 34.4 is 40% of 86.

Display the data in a histogram.

Question 4.
Big Ideas Math Geometry Answers Chapter 12 Probability 1
Answer:
Big Ideas Math Answers Geometry Chapter 12 Probability img_1

Question 5.
ABSTRACT REASONING
You want to purchase either a sofa or an arm chair at a furniture store. Each item has the same retail price. The sofa is 20% off. The arm chair is 10% off. and you have a coupon to get an additional 10% off the discounted price of the chair. Are the items equally priced after the discounts arc applied? Explain.
Answer: Yes

Explanation:
You want to purchase either a sofa or an armchair at a furniture store.
Each item has the same retail price. The sofa is 20% off. The armchair is 10% off. and you have a coupon to get an additional 10% off the discounted price of the chair.
The price of armchair and sofa are the same.
If you add 10% to chair the discount for the chair and sofa will be the same.
10% + 10% = 20%

Probability Monitoring Progress

In Exercises 1 and 2, describe the event as unlikely, equally likely to happen or not happen, or likely. Explain your reasoning.

Question 1.
The oldest child in a family is a girl.
Answer:

Question 2.
The two oldest children in a family with three children are girls.
Answer:

Question 3.
Give an example of an event that is certain to occur.
Answer:
If A and B are independent event
P(A) = 1/2
P(B) = 1/5
P(A and B) = P(A) × P(B)
= 1/2 × 1/5
= 1/10

12.1 Sample Spaces and Probability

Exploration 1

Finding the Sample Space of an Experiment

Work with a partner: In an experiment, three coins are flipped. List the possible outcomes in the sample space of the experiment.
Big Ideas Math Geometry Answers Chapter 12 Probability 2
Answer:
The number of different outcomes when three coins are tossed is 2 × 2 × 2 = 8.
All 8 possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT.

Exploration 2

Finding the Sample Space of an Experiment

Work with a partner: List the possible outcomes in the sample space of the experiment.

a. One six-sided die is rolled.
Big Ideas Math Geometry Answers Chapter 12 Probability 3
Answer: 6 possible outcomes

b. Two six-sided die is rolled.
Big Ideas Math Geometry Answers Chapter 12 Probability 4
Answer:
Rolling two six-sided dice: Each die has 6 equally likely outcomes, so the sample space is 6 . 6 or 36 equally likely outcomes.

Exploration 3

Finding the Sample Space of an Experiment

Work with a partner: In an experiment, a spinner is spun.

Big Ideas Math Geometry Answers Chapter 12 Probability 5

a. How many ways can you spin a 1? 2? 3? 4? 5?
Answer: 1, 2, 3, 2, 4

b. List the sample space.
Answer: 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 5

c. What is the total number of outcomes?
Answer: 12

Exploration 4

Finding the Sample Space of an Experiment

Work with a partner: In an experiment, a bag contains 2 blue marbles and 5 red marbles. Two marbles arc drawn from the bag.

Big Ideas Math Geometry Answers Chapter 12 Probability 6

a. How many ways can you choose two blue? a red then blue? a blue then red? two red?
Answer: BB – 2, RB – 10, BR – 10, RR – 20

b. List the sample space.
Answer:

c. What is the total number of outcomes?
Answer: 42

Communicate Your Answer

Question 5.
How can you list the possible outcomes in the sample space of an experiment?
Answer:
There are four possible outcomes for each spin: red, blue, yellow, green. Then, multiply the number of outcomes by the number of spins. June flipped the coin three times. The answer is there are 12 outcomes in the sample space.

Question 6.
For Exploration 3, find the ratio of the number of each possible outcome to the total number of outcomes. Then find the sum of these ratios. Repeat for Exploration 4. What do you observe?
LOOKING FOR A PATTERN
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Lesson 12.1 Sample Spaces and Probability

Monitoring Progress

Find the number of possible outcomes in the sample space. Then list the possible outcomes.

Question 1.
You flip two coins.
Answer: 4

Explanation:
When we flip two coins simultaneoulsy then the possible outcomes will be (H, H), (T, T), (T, H), (H, T)
where H represents heads
T represents tails.
Thus the possible outcomes are 2² = 4

Question 2.
You flip two Coins and roll a six-sided die.
Answer: 6 × 2² = 24

Explanation:
We roll a die and flip two coins. We have to find the number of possible outcomes in this space. Also we have to list the possible outcomes.
1 = {When rolling the dice, the number 1 fell};
2 = {When rolling the dice, the number 2 fell};
3 = {When rolling the dice, the number 3 fell};
4 = {When rolling the dice, the number 4 fell};
5 = {When rolling the dice, the number 5 fell};
6 = {When rolling the dice, the number 6 fell};
On the other hand, using H for Heads and T for Tails we can list the outcomes.
(1, H, H), (2, H, H), (3, H, H), (4, H, H), (5, H, H), (6, H, H)
(1, T, H), (2, T, H), (3, T, H), (4, T, H), (5, T, H), (6, T, H)
(1, H, T), (2, H, T), (3, H, T), (4, H, T), (5, H, T), (6, H, T)
(1, T, T), (2, T, T), (3, T, T), (4, T, T), (5, T, T), (6, T, T)
Therefore, we can conclude that the number of all possible outcomes is
6 × 2² = 24

Question 3.
You flip a coin and roll a six-sided die. What is the probability that the coin shows tails and the die shows 4?
Answer:
The sample space has 12 possible outcomes.
Heads, 1
Heads, 2
Heads, 3
Heads, 4
Heads, 5
Heads, 6
Tails, 1
Tails, 2
Tails, 3
Tails, 4
Tails, 5
Tails, 6
Probability that the coin shows tails and the die shows 4 is 4/12 = 1/3

Find P(\(\bar{A}\)).

Question 4.
P(A) = 0.45
Answer:
P(\(\bar{A}\)) = 1 – P(A)
P(A) = 0.45
P(\(\bar{A}\)) = 1 – 0.45
P(\(\bar{A}\)) = 0.55

Question 5.
P(A) = \(\frac{1}{4}\)
Answer:
P(\(\bar{A}\)) = 1 – P(A)
P(A) = \(\frac{1}{4}\)
P(\(\bar{A}\)) = 1 – \(\frac{1}{4}\)
P(\(\bar{A}\)) = \(\frac{3}{4}\)

Question 6.
P(A) = 1
Answer:
P(\(\bar{A}\)) = 1 – P(A)
P(A) = 1
P(\(\bar{A}\)) = 1 – 1
P(\(\bar{A}\)) = 0

Question 7.
P(A) = 0.03
Answer:
P(\(\bar{A}\)) = 1 – P(A)
P(A) = 0.03
P(\(\bar{A}\)) = 1 – 0.03
P(\(\bar{A}\)) = 0.97

Question 8.
In Example 4, are you more likely to get 10 points or 5 points?
Answer:
10: 0.09
5: (5 – 10)/324
= (36π – 9π)/324
= 27π/324
= 0.26

Question 9.
In Example 4, are you more likely to score points (10, 5, or 2) or get 0 points?
Answer:
2: (2 – 5)/324
= (81π – 36π)/324
= 45π/324
= 0.43
0.09 + 0.26 + 0.43 = 0.78
More likely to get 2 points.

Question 10.
In Example 5, for which color is the experimental probability of stopping on the color greater than the theoretical probability?
Answer:
9/20 = 0.45

Question 11.
In Example 6, what is the probability that a pet-owning adult chosen at random owns a fish?
Answer:
146/1328 = 73/664 = 0.11

Exercise 12.1 Sample Spaces and Probability

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A number that describes the likelihood of an event is the ___________ of the event.
Answer:
A number that describes the likelihood of an event is the Probability of the event.

Question 2.
WRITING
Describe the difference between theoretical probability and experimental probability.
Answer: Experimental probability is the result of an experiment. Theoretical probability is what is expected to happen.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. find the number of possible outcomes in the sample space. Then list the possible outcomes.

Question 3.
You roll a die and flip three coins.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 3.1
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 3.2
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 3.3

Question 4.
You flip a coin and draw a marble at random from a hag containing two purple marbles and one while marble.
Answer:
Given data,
You flip a coin and draw a marble at random from a hag containing two purple marbles and one while marble.
the probability of getting a purple marble = 2/3
the probability of getting a white marble = 1/3

Question 5.
A bag contains four red cards numbered 1 through 4, four white cards numbered 1 through 4, and four black cards numbered 1 through 4. You choose a card at random.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 5

Question 6.
You draw two marbles without replacement from a bag containing three green marbles and four black marbles.
Answer:
In all there are 7 marbles when you first grab a marble, after that you take one marble away then you have 6 marbles to choose.
7 × 6 = 42
42 possible outcomes: GG, GG, GB, GB, GB, GB, GG, GG, GB, GB, GB, GB, GG, GG, GB, GB, GB, GB, BG, BG, BG, BB, BB, BB, BG, BG, BG, BB, BB, BB, BG, BG, BG, BB, BB, BB, BG, BG, BG, BB, BB, BB, BB.

Question 7.
PROBLEM SOLVING
A game show airs on television five days per week. Each day, a prize is randomly placed behind one of two doors. The contestant wins the prize by selecting the correct door. What is the probability that exactly two of the five contestants win a prize during a week?
Big Ideas Math Geometry Answers Chapter 12 Probability 7
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 7.1
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 7.2

Question 8.
PROBLEM SOLVING
Your friend has two standard decks of 52 playing cards and asks you to randomly draw one card from each deck. What is the probability that you will draw two spades?
Answer:
You have two decks of 52 cards and in a normal deck, there are 13 cards of each suit.
So there are 13 spades in the deck.
Therefore the probability of you drawing a spade is 13 out of all the 52 cards or \(\frac{13}{52}\), which can be reduced to \(\frac{1}{4}\). You do this two times with different decks that are exactly the same so you multiply \(\frac{1}{4}\) times \(\frac{1}{4}\). 1 times 1 is 1 and 4 times 4 is 16, so it is \(\frac{1}{16}\).
You can turn this into a percentage by dividing 1 by 16 and moving the decimal place to places to the right.
6.25% is the probability that you will draw two spades.

Question 9.
PROBLEM SOLVING
When two six-sided dice are rolled, there are 36 possible outcomes. Find the probability that
(a) the sum is not 4 and
(b) the sum is greater than 5.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 9.1

Question 10.
PROBLEM SOLVING
The age distribution of a population is shown. Find the probability of each event.
Big Ideas Math Geometry Answers Chapter 12 Probability 8
a. A person chosen at random is at least 15 years old.
Answer: 80%

b. A person chosen at random is from 25 to 44 years old.
Answer: 13% + 13% = 26%

Question 11.
ERROR ANALYSIS
A student randomly, guesses the answers to two true-false questions. Describe and correct the error in finding the probability of the student guessing both answers correctly.
Big Ideas Math Geometry Answers Chapter 12 Probability 9
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 11

Question 12.
ERROR ANALYSIS
A student randomly draws a number between 1 and 30. Describe and correct the error in finding the probability that the number drawn is greater than 4.
Big Ideas Math Geometry Answers Chapter 12 Probability 10
Answer:
The error is that the probability of the complement of the event is 4/30, not 3/30, because if you are looking for a sum greater than 4, than you subtract 1 by numbers less than or equal to 4 by the total amount of numbers, which is 30.
P(Sum is greater than 4)=1-P(Sum is less than or equal to 4)
1 – 2/15
= 13/15

Question 13.
MATHEMATICAL CONNECTIONS
You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. What is the probability your dart lands in the yellow region?
Big Ideas Math Geometry Answers Chapter 12 Probability 11
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 13

Question 14.
MATHEMATICAL CONNECTIONS
The map shows the length (in miles) of shoreline along the Gulf of Mexico for each state that borders the body of water. What is the probability that a ship coming ashore at a random point in the Gulf of Mexico lands in the given state?
Big Ideas Math Geometry Answers Chapter 12 Probability 12
a. Texas
Answer:
By using the above map we can solve the problem
Total length of the shoreline: 367 + 397 + 44 + 53 + 770 = 1631 miles
The probability of the ship landing in Texas: 367/1631 = 0.23

b. Alabama
Answer:
The probability of the ship landing in Alabama: 53/1631 = 0.03

c. Florida
Answer:
The probability of the ship landing in Florida: 770/1631 = 0.47

d. Louisiana
Answer:
The probability of the ship landing in Louisiana: 397/1631 = 0.24

Question 15.
DRAWING CONCLUSIONS
You roll a six-sided die 60 times. The table shows the results. For which number is the experimental probability of rolling the number the same as the theoretical probability?
Big Ideas Math Geometry Answers Chapter 12 Probability 13
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 15

Question 16.
DRAWING CONCLUSIONS
A bag contains 5 marbles that are each a different color. A marble is drawn, its color is recorded, and then the marble is placed back in the hag. This process is repeated until 30 marbles have been drawn. The table shows the results. For which marble is the experimental probability of drawing the marble the same as the theoretical probability?
Big Ideas Math Geometry Answers Chapter 12 Probability 14
Answer:
Total number of marbles = 30
6/30 = 1/5
For black marble the experimental probability of drawing the marble the same as the theoretical probability.

Question 17.
REASONING
Refer to the spinner shown. The spinner is divided into sections with the same area.
Big Ideas Math Geometry Answers Chapter 12 Probability 15
a. What is the theoretical probability that the spinner stops on a multiple of 3?
b. You spin the spinner 30 times. If stops on a multiple of 3 twenty times. What is the experimental probability of Stopping on a multiple of 3?
c. Explain why the probability you found in part (b) is different than the probability you found in part (a).
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 17

Question 18.
OPEN-ENDED
Describe a real-life event that has a probability of 0. Then describe a real-life event that has a probability of 1.
Answer:
The probability of rolling a 7 with a standard 6-sided die = 0
The probability of rolling a natural number with a standard 6-sided die = 1

Question 19.
DRAWING CONCLUSIONS
A survey of 2237 adults ages 18 and over asked which Sport 15 their favorite. The results are shown in the figure. What is the probability that an adult chosen at random prefers auto racing?
Big Ideas Math Geometry Answers Chapter 12 Probability 16
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 19

Question 20.
DRAWING CONCLUSIONS
A survey of 2392 adults ages 18 and over asked what type of food they Would be most likely to choose at a restaurant. The results are shown in the figure. What is the probability that an adult chosen at random prefers Italian food?
Big Ideas Math Geometry Answers Chapter 12 Probability 17
Answer:
P(Italian) = 526/1196
= 263/1196
P(Italian) ≈ 22%

Question 21.
ANALYZING RELATIONSHIPS
Refer to the board in Exercise 13. Order the likelihoods that the dart lands in the given region from least likely to most likely.
A. green
B. not blue
C. red
D. not yellow
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 21.1
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 21.2

Question 22.
ANALYZING RELATIONSHIPS
Refer to the chart below. Order the following events from least likely to most likely.
Big Ideas Math Geometry Answers Chapter 12 Probability 18
A. It rains on Sunday.
Answer:
80%
= 80/20 = 4/1
The ratio is 4:1

B. It does not rain on Saturday.
Answer:
100 – 30 = 70%
= 30/70 = 3 : 7

C. It rains on Monday.
Answer: 90%
= 90/10 = 9/1
= 9 : 1

D. It does not rain on Friday.
Answer:
100 – 5 = 95%
= 5/95 = 1/19
= 1 : 19

Question 23.
USING TOOLS
Use the figure in Example 3 to answer each question.
Big Ideas Math Geometry Answers Chapter 12 Probability 19
a. List the possible sums that result from rolling two six-sided dice.
b. Find the theoretical probability of rolling each sum.
c. The table below shows a simulation of rolling two six-sided dice three times. Use a random number generator to simulate rolling two six-sided dice 50 times. Compare the experimental probabilities of rolling each sum with the theoretical probabilities.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 23.1
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 23.2

Question 24.
MAKING AN ARGUMENT
You flip a coin three times. It lands on heads twice and on tails once. Your friend concludes that the theoretical probability of the coin landing heads up is P(heads up) = \(\frac{2}{3}\). Is your friend correct? Explain your reasoning.
Answer:
The friend is incorrect because the probability of heads here is \(\frac{2}{3}\), is the experimental probabiility of heads for this particular case, while its theoretical probability will always be \(\frac{1}{2}\).

Question 25.
MATHEMATICAL CONNECTIONS
A sphere fits inside a cube so that it touches each side, as shown. What is the probability a point chosen at random inside the cube is also inside the sphere ?
Big Ideas Math Geometry Answers Chapter 12 Probability 20
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 25.1

Question 26.
HOW DO YOU SEE IT?
Consider the graph of f shown. What is the probability that the graph of y = f(x) + c intersects the x-axis when c is a randomly chosen integer from 1 to 6? Explain.
Big Ideas Math Geometry Answers Chapter 12 Probability 21
Answer:

Question 27.
DRAWING CONCLUSIONS
A manufacturer tests 1200 computers and finds that 9 of them have defects. Find the probability that a computer chosen at random has a defect. Predict the number of computers with defects in a shipment of 15,000 computers. Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 27

Question 28.
THOUGHT PROVOKING
The tree diagram shows a sample space. Write a probability problem that can be represented by the sample space. Then write the answer(s) to the problem.
Big Ideas Math Geometry Answers Chapter 12 Probability 22
Answer:

Maintaining Mathematical Proficiency

Simplify the expression. Write your answer using only positive exponents.

Question 29.
\(\frac{2 x^{3}}{x^{2}}\)
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 29

Question 30.
\(\frac{2 x y}{8 y^{2}}\)
Answer:
\(\frac{2y}{8 y^{2}}\)
= \(\frac{2}{8y}\)
= \(\frac{1}{4y}\)

Question 31.
\(\frac{4 x^{9} y}{3 x^{3} y}\)
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 31

Question 32.
\(\frac{6 y^{0}}{3 x^{-6}}\)
Answer:
Given,
\(\frac{6 y^{0}}{3 x^{-6}}\)
= 6 . 1/3x-6
= 2 . x6

Question 33.
(3Pq)4
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.1 Qu 33

Question 34.
\(\left(\frac{y^{2}}{x}\right)^{-2}\)
Answer: (\(\frac{y}{x}\))²

12.2 Independent and Dependent Events

Exploration 1

Identifying Independent and Dependent Events

Work with a partner: Determine whether the events are independent or dependent. Explain your reasoning.
REASONING ABSTRACTLY
To be proficient in math, you need to make sense of quantities and their relationships in problem situations.
a. Two six-sided dice are rolled.
Big Ideas Math Answers Geometry Chapter 12 Probability 23
Answer:
P = Number of outcomes that satisfy the requirements/Total number of possible outcomes
Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die.
When two dice are thrown simultaneously, thus the number of event can be 62 = 36 because each die has 1 to 6 number on its faces.
This is independent event

b. Six pieces of paper, numbered 1 through 6, are in a bag, Two pieces of paper are selected one at a time without replacement.
Big Ideas Math Answers Geometry Chapter 12 Probability 24
Answer:
P = Number of outcomes that satisfy the requirements/Total number of possible outcomes
When we pick up any paper the probability result is equal of 1/6
This is dependent event

Exploration 2

Work with a partner:

a. In Exploration 1(a), experimentally estimate the probability that the sum of the two numbers rolled is 7. Describe your experiment.
Answer:

(b). In Exploration 1 (b), experimentally estimate the probability that the sum of the two numbers selected is 7. Describe your experiment.
Answer:

Exploration 3

Finding Theoretical Probabilities

Work with a partner:
a. In Exploration 1(a), find the theoretical probability that the sum of the two numbers rolled is 7. Then compare your answer with the experimental probability you found in Exploration 2(a).
Answer:
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.

b. In Exploration 1(b). find the theoretical probability that the sum of the two numbers selected is 7. Then compare your answer with the experimental probability you found in Exploration 2(b).
Answer:

C. Compare the probabilities you obtained in parts (a) and (b).
Answer:

Communicate Your Answer

Question 4.
How can you determine whether two events are independent or dependent?
Answer:
Two events A and B are said to be independent if the fact that one event has occurred does not affect the probability that the other event will occur.
If whether or not one event occurs does affect the probability that the other event will occur, then the two events are said to be dependent.

Question 5.
Determine whether the events are independent or dependent. Explain your reasoning.
a. You roil a 4 on a six-sided die and spin red on a spinner.
Answer: Independent

b. Your teacher chooses a student to lead a group. chooses another student to lead a second group. and chooses a third student to lead a third group.
Answer: Dependent

Lesson 12.2 Independent and Dependent Events

Monitoring progress

Question 1.
In Example 1, determine whether guessing Question 1 incorrectly and guessing Question 2 correctly are independent events.
Answer:

Question 2.
In Example 2, determine whether randomly selecting a girl first and randomly selecting a boy second are independent events.
Answer:

Question 3.
In Example 3, what is the probability that you spin an even number and then an odd number?
Answer:

Question 4.
In Example 4, what is the probability that both hills are $1 hills?
Answer:

Question 5.
In Example 5, what is the probability that none of the cards drawn are hearts when (a) you replace each card, and (b) you do not replace each card? Compare the probabilities.
Answer:
(a) you replace each card,
P(A and B and C) = P(A) P(B) P(C)
= 13/52 . 13/52 . 13/52 = 1/4 . 1/4 . 1/4 = 1/64 = 0.016

Question 6.
In Example 6, find (a) the probability that a non-defective part “passes” and (b) the probability that a defective part “fails.”
Answer:
(a) the probability that a non-defective part “passes”
P(P/D) = 3/39 = 1/13 = 0.077
(b) the probability that a defective part “fails.”
P(F/N) = 11/461 = 0.024

Question 7.
At a coffee shop. 80% of customers order coffee. Only 15% of customers order coffee and a bagel. What is the probability that a customer who orders coffee also orders a bagel?
Answer:
A: Customer order coffee
B: Customer order a bagel
P(B/A) = P(A and B)/P(A)
80% of customers order coffee and Only 15% of customers order coffee and a bagel.
P(A) = 80/100 = 0.8
P(A and B) = 15/100 = 0.15
P(B/A) = P(A and B)/P(A) = 0.15/0.8 = 0.1875 = 18.75%

Exercise 12.2 Independent and Dependent Events

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain the difference between dependent events and independent events, and give an example of each.
Answer:
When two events are dependent, the occurrence of one event affects the other. When two events are independent, the occurence of one event does not affect the other.

Question 2.
COMPLETE THE SENTENCE
The probability that event B will occur given that event A has occurred is called the _____________ of B given A and is written as _____________ .
Answer:
The probability that event B will occur given that event A has occurred is called the conditional probability of B given A and is written as P(B/A).

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, tell whether the events are independent or dependent. Explain your reasoning.

Question 3.
A box of granola bars contains an assortment of flavors. You randomly choose a granola bar and eat it. Then you randomly choose another bar.
Event A: You choose a coconut almond bar first.
Event B: You choose a cranberry almond bar second.
Answer:
The two events, which considered in this experiment are an example of dependent events.

Question 4.
You roll a six-sided die and flip a coin.
Event A: You get a 4 when rolling the die.
Event B: You get tails when flipping the coin
Big Ideas Math Answers Geometry Chapter 12 Probability 25
Answer: Independent, the events do not influence each other.

Question 5.
Your MP3 player contains hip-hop and rock songs. You randomly choose a song. Then you randomly choose another song without repeating song choices.
Event A: You choose a hip-hop song first.
Event B: You choose a rock song second.
Big Ideas Math Answers Geometry Chapter 12 Probability 26
Answer:
The events are dependent because the occurrence of event A affects the occurrence of event B.

Question 6.
There are 22 novels of various genres on a shell. You randomly choose a novel and put it back. Then you randomly choose another novel.
Event A: You choose a mystery novel.
Event B: You choose a science fiction novel.
Answer:
The 1st book chosen is put back so the second book picked has the same probability of being chosen a if the 1st book was never chosen to begin with the events are independent.

In Exercises 7 – 10. determine whether the events are independent.

Question 7.
You play a game that involves spinning a wheel. Each section of the wheel shown has the same area. Use a sample space to determine whether randomly spinning blue and then green are independent events.
Big Ideas Math Answers Geometry Chapter 12 Probability 28
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 7

Question 8.
You have one red apple and three green apples in a bowl. You randomly select one apple to eat now and another apple for your lunch. Use a sample space to determine whether randomly selecting a green apple first and randomly selecting a green apple second are independent events.
Answer:
Let R represent the red apple.
Let G1, G2, G3 represent the 3 green apples.
P(G first) = P(green apple first) = 9/12 = 3/4 = 0.75
P(G second) = P(green apple second) = 9/12 = 3/4 = 0.75
P(green apple first and second) = 6/12 = 1/2 = 0.5
Events are not independent.

Question 9.
A student is taking a multiple-choice test where each question has four choices. The student randomly guesses the answers to the five-question test. Use a sample space to determine whether guessing Question 1 correctly and Question 2 correctly are independent events.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 9

Question 10.
A vase contains four white roses and one red rose. You randomly select two roses to take home. Use a sample space to determine whether randomly selecting a white rose first and randomly selecting a white rose second are independent events.
Answer:
P(A and B) = P(A) and P(B)
A = {First randomly selected roses is white}
B = {Second randomly selected roses is white}
P(A) = Number of favorable outcomes/Total number of outcomes = 4/5
P(B) = 4/5
P(A) . P(B) = 4/5 . 4/5 = 16/25

Question 11.
PROBLEM SOLVING
You play a game that involves spinning the money wheel shown. You spin the wheel twice. Find the probability that you get more than $500 on your first spin and then go bankrupt on your second spin.
Big Ideas Math Answers Geometry Chapter 12 Probability 28
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 11

Question 12.
PROBLEM SOLVING
You play a game that involves drawing two numbers from a hat. There are 25 pieces of paper numbered from 1 to 25 in the hat. Each number is replaced after it is drawn. Find the probability that you will draw the 3 on your first draw and a number greater than 10 on your second draw.
Answer:
P(A) = 1/25
P(B) = 15/25
P(A and B) = P(A) . P(B)
= 1/25 . 15/25 = 3/125
Thus P(A and B) = 3/125

Question 13.
PROBLEM SOLVING
A drawer contains 12 white socks and 8 black socks. You randomly choose 1 sock and do not replace it. Then you randomly choose another sock. Find the probability that both events A and B will occur.
Event A: The first sock is white.
Event B: The second sock is white.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 13

Question 14.
PROBLEM SOLVING
A word game has 100 tiles. 98 of which are letters and 2 of which are blank. The numbers of tiles of each letter are shown. You randomly draw 1 tile, set it aside, and then randomly draw another tile. Find the probability that both events A and B will occur.
Big Ideas Math Answers Geometry Chapter 12 Probability 29
Answer:
P(A) = 56/100 = 0.56
P(B) = 42/(100 – 1) = 0.424
P(A) . P(B) = 0.56 × 0.424 = 0.2376

Question 15.
ERROR ANALYSIS
Events A and B are independent. Describe and correct the error in finding P(A and B).
Big Ideas Math Answers Geometry Chapter 12 Probability 30
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 15

Question 16.
ERROR ANALYSIS
A shelf contains 3 fashion magazines and 4 health magazines. You randomly choose one to read, set it aside, and randomly choose another for your friend to read. Describe and correct the error in finding the probability that both events A and B occur.
Event A: The first magazine is fashion.
Event B: The second magazine is health.
Big Ideas Math Answers Geometry Chapter 12 Probability 31
Answer:
P(A) = 3/7
P(B/A) = 4/(7 – 1) = 4/6
P(A and B) = P(A) × P(B/A)
P(A and B) = 3/7 × 4/6 = 2/7

Question 17.
NUMBER SENSE
Events A and B are independent. Suppose P(B) = 0.4 and P(A and B) = 0.13. Find P(A).
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 17

Question 18.
NUMBER SENSE
Events A and B are dependent. Suppose P(B/A) = 0.6 and P(A and B) = 0.15. Find P(A).
Answer:
P(A) = x
P(B/A) = 0.6
P(A and B) = 0.15
P(A and B) = P(A) × P(B/A)
0.15 = x × 0.6
x = 0.15/0.6
x = 0.25

Question 19.
ANALYZING RELATIONSHIPS
You randomly select three cards from a standard deck of 52 playing cards. What is the probability that all three cards are face cards when (a) you replace each card before selecting the next card, and (b) you do not replace each card before selecting the next card? Compare the probabilities.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 19

Question 20.
A bag contains 9 red marbles. 4 blue marbles, and 7 yellow marbles. You randomly select three marbles from the hag. what is the probability that all three marbles are red when (a) you replace each marble before selecting the next marble, and (b) you do not replace each marble before selecting the next marble? Compare the probabilities.
Answer:
a. There are a total of 9 + 4 + 7= 20 marbles.
Therefore, the probability of selecting a red marble in each attempt is 9/20 when the marble is replaced.
Therefore the probability of selecting a red marble in each of 3 of the attempt is
9/20 × 9/20 × 9/20 = 0.09125
The replacement makes these independent events.
b. There are total of 9 + 4 + 7 = 20 marbles.
The probability of selecting a red marble in the first attempt is 9/20, second attempt is 8/19 and the third attempt is 7/18 when the marbles are not replaced.
Therefore the probability of selecting a red marble in each of 3 of the attempts is 9/20 × 8/19 × 7/18 = 0.0737

Question 21.
ATTEND TO PRECISION
The table shows the number of species in the United States listed as endangered and threatened. Find (a) the probability that a randomly selected endangered species is a bird, and (b) the probability that a randomly selected mammal is endangered.
Big Ideas Math Answers Geometry Chapter 12 Probability 32
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 21

Question 22.
ATTEND TO PRECISION
The table shows the number of tropical cyclones that formed during the hurricane seasons over a 12-year period. Find (a) the probability to predict whether a Future tropical cyclone in the Northern Hemisphere is a hurricane, and (b) the probability to predict whether a hurricane is in the Southern Hemisphere.
Big Ideas Math Answers Geometry Chapter 12 Probability 33
Answer:

Question 23.
PROBLEM SOLVING
At a school, 43% of students attend the homecoming football game. Only 23% of students go to the game and the homecoming dance. What is the probability that a student who attends the football game also attends the dance?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 23

Question 24.
PROBLEM SOLVING
At a gas station. 84% of customers buy gasoline. Only 5% of customers buy gasoline and a beverage. What is the probability that a customer who buys gasoline also buys a beverage?
Answer:
Given,
P(A) = 84% = 0.84
P(A and B) = 5% = 0.05
P(A and B) = P(A) × P(B/A)
P(B/A) = 0.05/0.84
P(B/A) = 0.0595

Question 25.
PROBLEM SOLVING
You and 19 other students volunteer to present the “Best Teacher” award at a school banquet. One student volunteer will be chosen to present the award. Each student worked at least 1 hour in preparation for the banquet. You worked for 4 hours, and the group worked a combined total of 45 hours. For each situation, describe a process that gives you a “fair” chance to be chosen. and find the probability that you are chosen.
a. “Fair” means equally likely.
b. “Fair” means proportional to the number of hours each student worked in preparation.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 25

Question 26.
HOW DO YOU SEE IT?
A bag contains one red marble and one blue marble. The diagrams show the possible outcomes of randomly choosing two marbles using different methods. For each method. determine whether the marbles were selected with or without replacement.
a.
Big Ideas Math Answers Geometry Chapter 12 Probability 34
Answer:

b.
Big Ideas Math Answers Geometry Chapter 12 Probability 35
Answer:

Question 27.
MAKING AN ARGUMENT
A meteorologist claims that there is a 70% chance of rain. When it rains. there is a 75% chance that your softball game will be rescheduled. Your friend believes the game is more likely to be rescheduled than played. Is your friend correct? Explain your reasoning.
Answer:
The chance that the game will be rescheduled is (0.7)(0.75) = 0.525
which is 52.5 percent
making it greater than 50 percent.

Question 28.
THOUGHT PROVOKING
Two six-sided dice are rolled once. Events A and B are represented by the diagram. Describe each event. Are the two events dependent or independent? Justify your reasoning.
Big Ideas Math Answers Geometry Chapter 12 Probability 36
Answer:

Question 29.
MODELING WITH MATHEMATICS
A football team is losing by 14 points near the end of a game. The team scores two touchdowns (worth 6 points each) before the end of the game. After each touchdown, the coach must decide whether to go for 1 point with a kick (which is successful 99% of the time) or 2 points with a run or pass (which is successful 45% of the time).
Big Ideas Math Answers Geometry Chapter 12 Probability 37
a. If the team goes for 1 point after each touchdown, what is the probability that the team wins? loses? ties?
b. If the team goes for 2 points after each touchdown. what is the probability that the team wins? loses? ties?
c. Can you develop a strategy so that the coach’s team has a probability of winning the game that is greater than the probability of losing? If so, explain your strategy and calculate the probabilities of winning and losing the game.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 29

Question 30.
ABSTRACT REASONING
Assume that A and B are independent events.
a. Explain why P(B) = P(B/A) and P(A) = P(A/B).
Answer:
P(B) = P(B/A)
P(B) = P(A) . P(B/A)
P(A) = P(A/B).
P(A) = P(B) . P(A/B)

b. Can P(A and B) also be defined as P(B) • P(A/B)? Justify your reasoning.
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.

Question 31.
\(\frac{9}{10}\) x = 0.18
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 31

Question 32.
\(\frac{1}{4}\)x + 0.5x = 1.5
Answer:
Given,
\(\frac{1}{4}\)x + 0.5x = 1.5
0.25x + 0.50x = 1.5
0.75x = 1.5
x = 1.5/0.75
x = 2

Question 33.
0.3x – \(\frac{3}{5}\)x + 1.6 = 1.555
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.2 Qu 33

12.3 Two-Way Tables and Probability

Exploration 1

Completing and Using a Two-Way Table

Work with a partner: A two-way table displays the same information as a Venn diagram. In a two-way table, one category is represented by the rows and the other category is represented by the columns.

The Venn diagram shows the results of a survey in which 80 students were asked whether they play a musical instrument and whether they speak a foreign language. Use the Venn diagram to complete the two-way table. Then use the two-way table to answer each question.
Big Ideas Math Answers Geometry Chapter 12 Probability 38
Big Ideas Math Answers Geometry Chapter 12 Probability 39
a. How many students play an instrument?
Answer: 41

b. How many students speak a foreign language?
Answer: 46

c. How many students play an instrument and speak a foreign language?
Answer: 16

d. How many students do not play an instrument and do not speak a foreign language?
Answer: 9

e. How many students play an instrument and do not speak a foreign language?
Answer: 25

Exploration 2

Two – Way Tables and Probability

Work with a partner. In Exploration 1, one student is selected at random from the 80 students who took the survey. Find the probability that the student
a. plays an instrument.
Answer: 41/80

b. speaks a foreign language.
Answer: 46/80

c. plays an instrument and speaks a foreign language.
Answer: 16/80

d. does not play an instrument and does not speak a foreign language.
Answer: 9/80

e. plays an instrument and does not speak a foreign language.
Answer:

Exploration 3

Conducting a Survey

Work with your class. Conduct a survey of the students in your class. Choose two categories that are different from those given in Explorations 1 and 2. Then summarize the results in both a Venn diagram and a two-way table. Discuss the results.
MODELING WITH MATHEMATICS
To be proficient in math, you need to identify important quantities in a practical situation and map their relationships using such tools as diagrams and two-way tables.
Answer:

Communicate Your Answer

Question 4.
How can you construct and interpret a two-way table?
Answer:
Identify the variables. There are two variables of interest here: the commercial viewed and opinion.
Determine the possible values of each variable. For the two variables, we can identify the following possible values
Set up the table
Fill in the frequencies

Question 5.
How can you use a two-way table to determine probabilities?
Answer:

Lesson 12.3 Two-Way Tables and Probability

Monitoring Progress

Question 1.
You randomly survey students about whether they are in favor of planting a community garden at school. of 96 boys surveyed, 61 are in favor. 0f 88 girls surveyed, 17 are against. Organize the results in a two-way table. Then find and interpret the marginal frequencies.
Answer:
In order to find out how many boys are against you do 96 – 61. In order to find out how many girls are in favor you do 88 – 17.
In order to find the probability you make a proportion. It will be:
P/100 = number of girls against/total number of students
17/184 = p/100
17 × 100 = 184p
1700 = 184p
p = 9.23

Question 2.
Use the survey results in Monitoring Progress Question 1 to make a two-way table that shows the joint and marginal relative frequencies.
Answer:

Question 3.
Use the survey results in Example 1 to make a two-way table that shows the conditional relative frequencies based on the column totals. Interpret the conditional relative frequencies in the context of the problem.
Answer:

Question 4.
Use the survey results in Monitoring Progress Question 1 to make a two-way table that shows the conditional relative frequencies based on the row totals. Interpret the conditional relative frequencies in the context of the problem.
Answer:

Question 5.
In Example 4, what is the probability that a randomly selected customer who is located in Santa Monica will not recommend the provider to a friend?
Answer:

Question 6.
In Example 4, determine whether recommending the provider to a friend and living in Santa Monica are independent events. Explain your reasoning.
Answer:

Question 7.
A manager is assessing three employees in order to offer one of them a promotion. Over a period of time, the manager records whether the employees meet or exceed expectations on their assigned tasks. The table shows the managers results. Which employee should be offered the promotion? Explain.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 40
Answer:

Exercise 12.3 Two-Way Tables and Probability

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A(n) ______________ displays data collected from the same source that belongs to two different categories.
Answer:
A two-way table displays data collected from the same source that belongs to two different categories.

Question 2.
WRITING
Compare the definitions of joint relative frequency, marginal relative frequency, and conditional relative frequency.
Answer:
Joint relative frequency: joint relative frequency is the ratio of a frequency that is not in the total row or the total column to the total number of values.
Marginal relative frequency: marginal relative frequency is the sum of the joint relative frequencies in a given row or column.
Conditional relative frequency: It is the ratio of joint relative frequency to the marginal relative frequency.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, complete the two-way table.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 41
Answer:
Number of students who have passed the exam is 50 – 10 = 40
Of those 40 students, 6 did not study for the exam,
Number of the students who studied and have passed the exam is 40 – 6 = 34
Number of the students who did not study and did not pass the exam is 10 – 4 = 6
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 42
Answer:
Number of students who said no is 49 – 7 = 42
Total number of students is 56 + 42 = 98
Total number of people is 98 + 10 = 108
Out of the total number of people, 49 of them said no,
Total number of people who said yes is 108 – 49 = 59
Number of teachers who said yes is 59 – 56 = 3
Big-Ideas-Math-Geometry-Answer-Key-Chapter-12-Probability-42 (1)

Question 5.
MODELING WITH MATHEMATICS
You survey 171 males and 180 females at Grand Central Station in New York City. Of those, 132 males and 151 females wash their hands after using the public rest rooms. Organize these results in a two-way table. Then find and interpret the marginal frequencies.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 43
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 5.1
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 5.2

Question 6.
MODELING WITH MATHEMATICS
A survey asks 60 teachers and 48 parents whether school uniforms reduce distractions in school. Of those, 49 teachers and 18 parents say uniforms reduce distractions in school, Organize these results in a two-way table. Then find and interpret the marginal frequencies.
Answer:
Given,
A survey asks 60 teachers and 48 parents whether school uniforms reduce distractions in school. Of those, 49 teachers and 18 parents say uniforms reduce distractions in school
Number of teacher who said no is 60 – 49 = 11
Number of parents who said no = 48 – 18 = 30
Total number of people who said yes = 49 + 18 = 67
Total number of people who said no = 11 + 30 = 41
Big Ideas Math Answers Geometry Chapter 12 Probability img_15

USING STRUCTURE
In Exercises 7 and 8, use the two-way table to create a two-way table that shows the joint and marginal relative frequencies.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 44
Answer:
P1,1 = Number of favorable outcomes/Total number of outcomes
= 11/231 = 0.0476
P2,1 = 24/231 = 0.1039
P1,2 = 104/231 = 0.4502
P2,2 = 92/231 = 0.3983
The marginal relative frequencies we find as the sum of each row and each column.
P(A randomly chosen person is male) = 115/231 = 0.4978
P(A randomly chosen person is female) = 116/231 = 0.5022
P(A randomly chosen person have left dominant hand) = 35/231 = 0.1515
P(A randomly chosen person have left dominant hand) = 196/231 = 0.8484
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 7

Question 8.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 45
Answer:
P1,1 = Number of favorable outcomes/Total number of outcomes
= 62/410 = 0.1513
The marginal relative frequencies we find as the sum of each row and each column.
P(A randomly chosen person is male) = 377/410 = 0.9195
P(A randomly chosen person is female) = 33/410 = 0.0805

Question 9.
MODELING WITH MATHEMATICS
Use the survey results from Exercise 5 to make a two-way table that shows the joint and marginal relative frequencies.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 9

Question 10.
MODELING WITH MATHEMATICS
In a survey, 49 people received a flu vaccine before the flu season and 63 people did not receive the vaccine. Of those who receive the flu vaccine, 16 people got the flu. Of those who did not receive the vaccine, 17 got the flu. Make a two-way table that shows the joint and marginal relative frequencies.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 46
Answer:
We see that the total no. of people who received the vaccine is 49, of which 16 got a fly.
Number of people who received the vaccine and did not get a fly is 49 – 16 = 33
Number of people who did not received the vaccine and did not get a fly is 63 – 17 = 46
Total number of people who got a fly is 16 + 17 = 33
Total Number of people who did not get a fly is 33 + 46 = 79
We also know that the total number of people who were surveyed is 49 + 63 = 112
P1,1 = Number of favorable outcomes/Total number of outcomes
= 16/112 = 0.1428
The marginal relative frequencies we find as the sum of each row and each column.
P(A randomly chosen person got a fly) = 33/112 = 0.2946
P(A randomly chosen person did not get a fly) = 79/112 = 0.7053

Question 11.
MODELING WITH MATHEMATICS
A survey finds that 110 people ate breakfast and 30 people skipped breakfast. Of those who ate breakfast. 10 people felt tired. Of those who skipped breakfast. 10 people felt tired. Make a two-way table that shows the conditional relative frequencies based on the breakfast totals.
Answer:
Given,
A survey finds that 110 people ate breakfast and 30 people skipped breakfast.
Of those who ate breakfast. 10 people felt tired. Of those who skipped breakfast. 10 people felt tired.
Number of people who ate breakfast and not tired is 110 – 10 = 100
Number of people who did not eat breakfast and not tired is 30 – 10 = 20
Total number of people who felt tired is 10 + 10 = 20
Total number of people who did not get tired is 100 + 20 = 120
120 +20 = 140
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 11

Question 12.
MODELING WITH MATHEMATICS
Use the survey results from Exercise 10 to make a two-way table that shows the conditional relative frequencies based on the flu vaccine totals.
Answer:

Question 13.
PROBLEM SOLVING
Three different local hospitals in New York surveyed their patients. The survey asked whether the patients physician communicated efficiently. The results, given as joint relative frequencies. are shown in the two-way table.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 47
a. What is the probability that a randomly selected patient located in Saratoga was satisfied with the communication of the physician?
b. What is the probability that a randomly selected patient who was not satisfied with the physician’s communication is located in Glens Falls?
c. Determine whether being satisfied with the Communication of the physician and living in Saratoga are independent events.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 13

Question 14.
PROBLEM SOLVING
A researcher surveys a random sample of high school students in seven states. The survey asks whether students plan to stay in their home state after graduation. The results, given as joint relative frequencies, are shown in the two-way table.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 48
a. What is the probability that a randomly selected student who lives in Nebraska plans to stay in his or her home state after graduation?
Answer:
In this case we consider a event A = {arandomly chosen student lives in Nebraska}
B = {a randomly chosen student plans to stay in his or her home state after graduation}
P(A) = 0.044 + 0.4 = 0.444
P(B/A) = P(A and B)/P(A) = 0.044/0.444 = 0.099
About 1% students who lives Nebraska plans to stay in his or her home state after graduation.

b. What is the probability that a randomly selected student who does not plan to stay in his or her home state after graduation lives in North Carolina?
Answer:
C = {a randomly chosen student does not plan to stay in his or her home state after graduation}
D = {a randomly chosen student lives in North Carolina}
P(C) = 0.4 + 0.193 + 0.256 = 0.849
P(D/C) = P(C and D)/P(C)
= 0.193/0.849 = 0.227
Therefore about 22.7% students who does not plan to stay in his or her home state after graduation lives in North Carolina.

c. Determine whether planning to stay in their home state and living in Nebraska are independent events.
Answer:
P(B/A) = 0.099
P(B) = 0.044 + 0.05
1 + 0.056 = 0.151
P(B/A) ≠ P(B)
This events are independent.

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in finding the given conditional probability.

Big Ideas Math Geometry Answer Key Chapter 12 Probability 49

Question 15.
P(yes|Tokyo)
Big Ideas Math Geometry Answer Key Chapter 12 Probability 50
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 15

Question 16.
P(London|No)
Big Ideas Math Geometry Answer Key Chapter 12 Probability 51
Answer:
P(A and B) = P(A)P(B/A)
P(A) = 0.341 + 0.112+ 0.191 = 0.644
P(B/A) = P(A and B)/P(A) = 0.112/0.644 = 0.1739
In the denominator the probability P(B) = 0.248 is used instead of P(A), where P(B) is probability that a randomly chosen person live in London.

Question 17.
PROBLEM SOLVING
You want to find the quickest route to school. You map out three routes. Before school, you randomly select a route and record whether you are late or on time. The table shows your findings. Assuming you leave at the same time each morning, which route should you use? Explain.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 52
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 17.1
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 17.2

Question 18.
PROBLEM SOLVING
A teacher is assessing three groups of students in order to offer one group a prize. Over a period of time, the teacher records whether the groups meet or exceed expectations on their assigned tasks. The table shows the teacher’s results. Which group should be awarded the prize? Explain.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 53
Answer: Group 1 exceeded expectations 12 out of 16 times or 75% of the time.

Question 19.
OPEN-ENDED
Create and conduct a survey in your class. Organize the results in a two-way table. Then create a two-way table that shows the joint and marginal frequencies.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 19

Question 20.
HOW DO YOU SEE IT?
A research group surveys parents and Coaches of high school students about whether competitive sports are important in school. The two-way table shows the results of the survey.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 54
a. What does 120 represent?
Answer: 120 parents said that competitive sports are not important in school.

b. What does 1336 represent?
Answer: 1336 is the sum of parents and coaches who agree that competitive sports are important in school.

c. What does 1501 represent?
Answer: 1501 is the total number of people surveyed. Here this is the sum of the parents and the coaches who participated in the survey.

Question 21.
MAKING AN ARGUMENT
Your friend uses the table below to determine which workout routine is the best. Your friend decides that Routine B is the best option because it has the fewest tally marks in the “Docs Not Reach Goal” column. Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 55
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 21

Question 22.
MODELING WITH MATHEMATICS
A survey asks students whether they prefer math class or science class. Of the 150 male students surveyed, 62% prefer math class over science class. Of the female students surveyed, 74% prefer math. Construct a two-way table to show the number of students in each category if 350 students were surveyed.
Answer:
survey asks students whether they prefer math class or science class. Of the 150 male students surveyed, 62% prefer math class over science class.
62% = 0.62
0.62 = P(Math/Male)
= P(Math and Male)/P(Male)
= Number of male students who prefer math/150
Number of male students who prefer math = 0.62 × 150 = 93
0.74 = P(Math/Female)
= P(Math and Female)/P(Female)
= Number of female students who prefer math/200
Number of female students who prefer math = 0.74 × 200 = 148
So, the total number of students who prefer math class is 148 + 93 = 241
Number of male students who prefer science class = 150 -93 = 57
Number of female students who prefer science class = 200 – 148 = 52
Number of students who prefer science class = 57 + 52 = 109
Big Ideas Math Answers Geometry Chapter 12 probability img_16

Question 23.
MULTIPLE REPRESENTATIONS
Use the Venn diagram to construct a two-way table. Then use your table to answer the questions.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 56
a. What is the probability that a randomly selected person does not own either pet?
b. What is the probability that a randomly selected person who owns a dog also owns a cat?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 23

Question 24.
WRITING
Compare two-way tables and Venn diagrams. Then describe the advantages and disadvantages of each.
Answer:

Question 25.
PROBLEM SOLVING
A company creates a new snack, N, and tests it against its current leader, L. The table shows the results.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 57
The company is deciding whether it should try to improve the snack before marketing it, and to whom the snack should be marketed. Use probability to explain the decisions the company should make when the total size of the snack’s market is expected to (a) change very little, and (b) expand very rapidly.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 25

Question 26.
THOUGHT PROVOKING
Baye’s Theorem is given by
Big Ideas Math Geometry Answers Chapter 12 Probability 102
Use a two-way table to write an example of Baye’s Theorem.
Answer:
Big Ideas Math Answers Geometry Chapter 12 probability img_17
P(Cat owner) = 61/210 = 0.29
P(Dog owner) = 93/210 = 0.442
P(Cat Owner/Dog owner) = P(Dog owner and cat owner)/P(Dog owner) = 0.387
P(Dog owner/Cat owner) = P(Cat owner/Dog owner)P(Dog owner)/P(Cat owner)
= 0.387 × 0.442/0.29
= 0.5898

Maintaining Mathematical Proficiency

Draw a Venn diagram of the sets described.

Question 27.
Of the positive integers less than 15, set A consists of the factors of 15 and set B consists of all odd numbers.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 27

Question 28.
Of the positive integers less than 14, set A consists of all prime numbers and set B consists of all even numbers.
Answer:
Set A = {2. 3, 5, 7, 11, 13}
Set B = {2, 4, 6, 8, 10, 12}
It can be seen that here A and B are overlapping sets.
BIM Answers Geometry Chapter 12 Probability img_17

Question 29.
Of the positive integers less than 24, set A consists of the multiples of 2 and set B consists of all the multiples of 3.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.3 Qu 29

12.1 – 12.3 Quiz

Question 1.
You randomly draw a marble out of a bag containing 8 green marbles, 4 blue marbles 12 yellow marbles, and 10 red marbles. Find the probability of drawing a marble that is not yellow.
Answer:
Given,
You randomly draw a marble out of a bag containing 8 green marbles, 4 blue marbles 12 yellow marbles, and 10 red marbles.
Total number of outcomes here are 8 + 4 + 12 + 10 = 34
Thus the probability of obtaining a marble that is not yellow is (8 + 4 + 10)/34
= 22/34
= 11/17
= 0.647
= 64.7%
Thus the probability of obtaining a marble that is not yellow is 64.7%

Find P(\(\)(\bar{A}))

Question 2.
P(A) = 0.32
Answer:
P(\(\)(\bar{A}) = 1 – P(A)
1 – 0.32
= 0.68
P(\(\)(\bar{A}) = 0.68

Question 3.
P(A) = \(\frac{8}{9}\)
Answer:
P(\(\)(\bar{A}) = 1 – P(A)
1 – \(\frac{8}{9}\)
= \(\frac{1}{9}\)
P(\(\)(\bar{A}) = \(\frac{1}{9}\)

Question 4.
P(A) = 0.01
Answer:
P(\(\)(\bar{A}) = 1 – P(A)
1 – 0.01 = 0.99

Question 5.
You roll a six-sided die 30 times. A 5 is rolled 8 times. What is the theoretical probability of rolling a 5? What is the experimental probability of rolling a 5?
Answer:
Given,
You roll a six-sided die 30 times. A 5 is rolled 8 times.
The theoretical probability of rolling a 5 on a number cube is \(\frac{1}{6}\) while the experimental probability of rolling a 5 on a number cube is \(\frac{8}{30}\) = \(\frac{4}{15}\)

Question 6.
Events A and B are independent. Find the missing probability.
P(A) = 0.25
P(B) = ____
P(A and B) = 0.05
Answer:
Given,
P(A) = 0.25
P(A and B) = 0.05
P(A and B) = P(A) × P(B)
P(B) = P(A and B)/P(A)
P(B) = 0.05/0.25 = 0.2
P(B) = 0.2

Question 7.
Events A and B are dependent. Find the missing probability.
P(A) = 0.6
P(B/A) = 0.2
P(A and B) = ____
Answer:
Given,
P(A) = 0.6
P(B/A) = 0.2
P(A and B) = P(A) × P(B/A)
P(A and B) = 0.6 × 0.2
= 0.12
P(A and B) = 0.12

Question 8.
Find the probability that a dart thrown at the circular target Shown will hit the given region.
Assume the dart is equally likely to hit any point inside the target.
Big Ideas Math Geometry Answer Key Chapter 12 Probability 58
a. the center circle
Answer:
Total area of the given region is π × r² = π × 6² = 36π = 113.112 sq. units
Area of the center circle is π × r² = π × 2² = 4π sq. units.
Therefore the probability of hitting the center circle is 4π/36π = 1/9 = 0.11…

b. outside the square
Answer:
Area of the square is 6² = 36
So the region outside of it is equal to 36π – 36 = 77.112 sq. units
Thus the probability of hitting the region outside the square is 77.112/113.112 = 0.682

c. inside the square but outside the center circle
Answer:
Area of the center circle is π × r² = π × 2² = 4π sq. units.
Area of the square is 6² = 36
Thus the probability of hitting the region outside the center circle but inside the square is 36 – 4π = 23.432 sq. units
Thus the probability of hitting the region is 23.432/113.112 = 0.207

Question 9.
A survey asks 13-year-old and 15-year-old students about their eating habits. Four hundred students are surveyed, 100 male students and 100 female students from each age group. The bar graph shows the number of students who said they eat fruit every day.

Big Ideas Math Geometry Answer Key Chapter 12 Probability 59

a. Find the probability that a female student, chosen at random from the students surveyed, eats fruit every day.
Answer:
Total number of females who eat a fruit everyday are 61 + 58 = 119
Therefore the probability of randomly choosing a female who eats a fruit everyday is 119/400= 0.2975

b. Find the probability that a 15 – year – old student. chosen at random from the students surveyed, eats fruit every day.
Answer:
Total number of 15 year old student who eat a fruit everyday are 53 + 58 = 111
Therefore the probability of randomly choosing a 15 year old student who eats a fruit everyday is 111/200 = 0.555

Question 10.
There are 14 boys and 18 girls in a class. The teacher allows the students to vote whether they want to take a test on Friday or on Monday. A total of 6 boys and 10 girls vote to take the test on Friday. Organize the information in a two-way table. Then find and interpret the marginal frequencies.
Answer:
Given,
There are 14 boys and 18 girls in a class. The teacher allows the students to vote whether they want to take a test on Friday or on Monday.
14 + 18 = 32
Number of boys who vote to take the test on Monday is 14 – 6 = 8
Number of girls who vote to take the test on monday is 18 – 10 = 8
A total of 6 boys and 10 girls vote to take the test on Friday.
The total number of students who take the test on Friday is 10 + 6 = 16
The total number of students who vote to take the test on Friday is 8 + 8 = 16
BIM Answers Geometry Chapter 12 Probability img_14

Question 11.
Three schools compete in a cross country invitational. Of the 15 athletes on your team. 9 achieve their goal times. Of the 20 athletes on the home team. 6 achieve their goal times. On your rival’s team, 8 of the 13 athletes achieve their goal times. Organize the information in a two-way table. Then determine the probability that a randomly elected runner who achieves his or her goal time is from your school.
Answer:
Three schools compete in a cross country invitational. Of the 15 athletes on your team. 9 achieve their goal times.
Number of runners in your team who do not achieve their goal team is 15 – 9 = 6
Number of runners in home team who do not achieve their goal team is 20 – 6 = 14
Number of runners in rival’s team who do not achieve their goal team is 13 – 8 = 5
Total number of runners who achieve their goal team is 9 + 6 + 8 = 23
Total number of runners who do not achieve their goal team is 6 + 14 + 5 = 25
The total number of rubbers who was surveyed is 23 + 25 = 48
P = Your team ans achieve their goal team/P(Archive their goal team)
P = 9/23
P = 0.39

12.4 Probability of Disjoint and Overlapping Events

Exploration 1

Work with a partner: A six-sided die is rolled. Draw a Venn diagram that relates the two events. Then decide whether the cents are disjoint or overlapping.
MODELING WITH MATHEMATICS
To be proficient in math, you need to map the relationships between important quantities in a practical situation using such tools as diagrams.
Big Ideas Math Geometry Solutions Chapter 12 Probability 60
a. Event A: The result is an even number.
Event B: The result is a prime number.
Answer:
Bigideas Math Geometry Answers Chapter 12 Probability img_13

b. Event A: The result is 2 or 4.
Event B: The result is an odd number
Answer:
Bigideas Math Geometry Answers Chapter 12 Probability img_14

Exploration 2

Finding the Probability that Two Events Occur

Work with a partner: A six-sided die is roiled. For each pair of events. find (a) P(A), (b) P(B). (C) P(A) and (B). and (d) P(A or B).
Big Ideas Math Geometry Solutions Chapter 12 Probability 61
a. Event A: The result is an even number.
Event B: The result is a Prime number.
Answer:
P(A) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
P(B) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
P(A or B) = P(A) + P(B) – P(A and B)
P(A and B) = \(\frac{1}{6}\)
P(A or B) = \(\frac{5}{6}\)

b. Event A: The result is 2 or 4.
Event B: The result is an odd number.
Answer:
P(A) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
P(B) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
P(A or B) = P(A) + P(B) – P(A and B)
P(A and B) = 0
P(A or B) = \(\frac{5}{6}\)

Exploration 3

Discovering Probability Formulas

Work with a partner:
a. In general, if event A and event B arc disjoint, then what is the probability that event A or event B will occur? Use a Venn diagram to justify your conclusion.
Answer:
If event A and B are disjoint, there are no common outcomes.
So we add the probabilities that each event occurs:
P(A or B) = P(A) + P(B)
Big Ideas Math Geometry Chapter 12 Probability Answer Key img_11

b. In general, if event A and event B are overlapping, then what is the probability that event A or event B will occur? Use a Venn diagram to justify your conclusion.
Answer:
If event A and event B are overlapping, there are common outcomes.
So, we add the probabilities that each event occurs then subtract the probability of the common outcomes.
P(A or B) = P(A) + P(B) – P(A and B)
Big Ideas Math Geometry Chapter 12 Probability Answer Key img_12

c. Conduct an experiment using a six-sided die. Roll the die 50 times and record the results. Then use the results to find the probabilities described in Exploration 2. How closely do your experimental probabilities compare to the theoretical probabilities you found in Exploration 2?
Answer:
Big Ideas Math Geometry Chapter 12 Probability Answer Key img_12.1
a. P(A) = \(\frac{1}{2}\) = 50%
P(A) = \(\frac{21}{50}\) = 42%
P(B) = \(\frac{1}{2}\) = 50%
P(B) = \(\frac{32}{50}\) = 64%
P(A and B) = \(\frac{1}{6}\) ≈ 16.7%
P(A and B) = \(\frac{9}{50}\) ≈ 18%
P(A or B) = \(\frac{5}{6}\) ≈ 83.3%
P(A or B) = \(\frac{44}{50}\) ≈ 88%
P(A) = \(\frac{1}{3}\) ≈ 33.3%
P(A) = \(\frac{17}{50}\) = 34%
P(B) = \(\frac{1}{2}\) = 50%
P(B) = \(\frac{29}{50}\) = 58%
P(A and B) = 0 = 0%
P(A and B) = \(\frac{0}{50}\) = 0%
P(A or B) = \(\frac{5}{6}\) ≈ 83.3%
P(A or B) = \(\frac{46}{50}\) = 92%

Communicate Your Answer

Question 4.
How can you find probabilities of disjoint and overlapping events?
Answer:
If A and B are disjoint events, then the probability of A or B is P(A or B) = P(A) + P(B). If two events A and B are overlapping, then the outcomes in the intersection of A and B are counted twice when P(A) and P(B) are added.
P(A or B) = P(A) + P(B) – P(A and B)

Question 5.
Give examples of disjoint events and overlapping events that do not involve dice.
Answer:

a. Event A: The result is an even number.
Event B: The result is a prime number.
Answer:
Bigideas Math Geometry Answers Chapter 12 Probability img_13

b. Event A: The result is 2 or 4.
Event B: The result is an odd number
Answer:
Bigideas Math Geometry Answers Chapter 12 Probability img_14

Lesson 12.4 Probability of Disjoint and Overlapping Events

Monitoring Progress

A card is randomly selected from a standard deck of 52 playing cards. Find the probability of the event.

Question 1.
selecting an ace or an 8
Answer:
A: Selecting an ace
B: You select 8
We know that A has 4 outcomes and B also has 4 outcomes.
P(A or B) = P(A) + P(B)
= 4/52 + 4/52
= 8/52
= 2/13 ≈ 0.15

Question 2.
selecting a 10 or a diamond
Answer:
A: Selecting a 10
B: You select diamond
P(A or B) = P(A) + P(B) – P(A and B)
= 4/52 + 13/52 – 1/52
= 16/52
= 4/13

Question 3.
WHAT IF?
In Example 3, suppose 32 seniors are in the band and 64 seniors are in the band or on the honor roll. What is the Probability that a randomly selected senior is both in the band and on the honor roll?
Answer:

Question 4.
In Example 4, what is the probability that the diagnosis is incorrect?
Answer:

Question 5.
A high school basketball team leads at halftime in 60% of the games in a season. The team wins 80% of the time when the have the halftime lead, but only 10% of the time when the do not. What is the probability that the team wins a particular game during the season?
Answer:
Given,
A high school basketball team leads at halftime in 60% of the games in a season. The team wins 80% of the time when the have the halftime lead, but only 10% of the time when the do not.
Let event A be team leads on the halftime and event B be win.
When A occurs, P(B) = 0.8
When A does not occur, P(B) = 0.1
P(B) = P(A and B) + P(\(\bar{A}\) and B)
P(A) . P(B | A) + P(\(\bar{A}\)) . P(B | \(\bar{A}\))
= 0.6 × 0.8 + 0.4 × 0.1
= 0.52

Exercise 12.4 Probability of Disjoint and Overlapping Events

Vocabulary and Core Concept Check

Question 1.
WRITING
Are the events A and \(\bar{A}\) disjoint? Explain. Then give an example of a real-life event and its complement.
Answer:
Yes A and \(\bar{A}\) are disjoint events because they are the complement of one another and so can not occur together and hence the name disjoint.
Example:
We flip a coin.
So, A = (the head fell)
\(\bar{A}\) = {the tail falls} are disjoint events.

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Geometry Solutions Chapter 12 Probability 62
How many outcomes are in the intersection of A and B?
Answer: There are 2 outcomes in the intersection of A and B.

How many outcomes are shared by both A and B?
Answer: 2 outcomes are shared by both A and B.

How many outcomes are in the union of A and B?
Answer: There are 4 + 2 + 3 = 9 outcomes in the union of A and B.

How many outcomes in B are also in A?
Answer: There are 2 outcomes in B that are also in A.

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 6, events A and B are disjoint. Find P(A or B)

Question 3.
p(A) = 0.3, P(B) = 0.1
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 3

Question 4.
p(A) = 0.55, P(B) = 0.2
Answer:
Given,
p(A) = 0.55, P(B) = 0.2
P(A or B) = P(A) + P(B)
P(A or B) = 0.55 + 0.2 = 0.75

Question 5.
P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{4}\)
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 5

Question 6.
p(A) = \(\frac{2}{3}\), P(B) = \(\frac{1}{5}\)
Answer:
Given,
p(A) = \(\frac{2}{3}\), P(B) = \(\frac{1}{5}\)
P(A or B) = P(A) + P(B)
P(A or B) = \(\frac{2}{3}\) + \(\frac{1}{5}\)
P(A or B) = \(\frac{13}{15}\)

Question 7.
PROBLEM SOLVING
Your dart is equally likely to hit any point inside the board Shown. You throw a dart and pop a balloon. What is the probability that the balloon is red or blue?
Big Ideas Math Geometry Solutions Chapter 12 Probability 63
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 7

Question 8.
PROBLEM SOLVING
You and your friend are among several candidates running for class president. You estimate that there is a 45% chance you will win and a 25% chance your friend will win. What is the probability that you or your friend win the election?
Answer:
Let A be the event of you winning the election and B be of your friend winning the election
P(A) = 45% = 0.45
P(B) = 25% = 0.25
P(A or B) = P(A) + P(B)
P(A or B) = 0.45 + 0.25 = 0.70
Therefore the probability of you or your friend winning the election is 0.7

Question 9.
PROBLEM SOLVING
You are performing an experiment to determine how well plants grow under different light sources. 0f the 30 Plants in the experiment, 12 receive visible light, 15 receive ultraviolet light, and 6 receive both visible and ultraviolet light. What is the probability that a plant in the experiment receives visible or ultraviolet light?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 9

Question 10.
PROBLEM SOLVING
Of 162 students honored at an academic awards banquet, 48 won awards for mathematics and 78 won awards for English. There are 14 students who won awards for both mathematics and English. A newspaper chooses a student at random for an interview. What is the probability that the student interviewed won an award for English or mathematics?
Answer:
Given,
There are 162 students honored at an academic awards banquet, 48 won awards for mathematics and 78 won awards for English.
There are 14 students who won awards for both mathematics and English. A newspaper chooses a student at random for an interview.
P(A) = 48/162
P(B) = 78/162
P(A and B) = 14/162
P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = 48/162 + 78/162 – 14/162
P(A or B) = 112/162 = 56/81 = 0.691

ERROR ANALYSIS
In Exercises 11 and 12, describe and correct the error in finding the probability of randomly drawing the given card from a standard deck of 52 playing cards.

Question 11.
Big Ideas Math Geometry Solutions Chapter 12 Probability 64
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 11

Question 12.
Big Ideas Math Geometry Solutions Chapter 12 Probability 65
Answer:
These 2 events are overlapping events as there is 1 card that is both a club and 9, therefore write equation of P(A or B) for overlapping events
P(A or B) = P(A) + P(B) – P(A and B)
= 13/52 + 4/52 – 1/52
= 4/13

In Exercises 13 and 14, you roll a six-sided die. Find P(A or B).

Question 13.
Event A: Roll a 6.
Event B: Roll a prime number.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 13

Question 14.
Event A: Roll an odd number.
Event B: Roll a number less than 5.
Answer:
P(A) = 3/6
1, 3, 5 out of 6 possible outcomes
P(B) = 4/6
1, 2, 3, 4 out of 6 possible outcomes
P(A and B) = 2/6
3/4 + 4/6 – 2/6 = 5/6

Question 15.
DRAWING CONCLUSIONS
A group of 40 trees in a forest are not growing properly. A botanist determines that 34 of the trees have a disease or are being damaged by insects, with 18 trees having a disease and 20 being damaged by insects. What is the probability that a randomly selected tree has both a disease and is being damaged by insects?
Big Ideas Math Geometry Solutions Chapter 12 Probability 66
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 15

Question 16.
DRAWING CONCLUSIONS
A company paid overtime wages or hired temporary help during 9 months of the year. Overtime wages were paid during 7 months. and temporary help was hired during 4 months. At the end of the year, an auditor examines the accounting records and randomly selects one month to check the payroll. What is the probability that the auditor will select a month in which the company paid overtime wages and hired temporary help?
Answer:
P(A) = 7/12
P(B) = 4/12
P(A or B) = 9/12
P(A and B) = P(A) + P(B) – P(A or B)
P(A and B) = 7/12 + 4/12 – 9/12
P(A and B) = 2/12 = 1/6
The probability of randomly selecting a month in which overtime was paid and temporary help was hired is 1/6 = 0.166…

Question 17.
DRAWING CONCLUSIONS
A company is focus testing a new type of fruit drink. The focus group is 47% male. 0f the responses, 40% of the males and 54% of the females said they would buy the fruit drink. What is the probability that a randomly selected person would buy the fruit drink?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 17

Question 18.
DRAWING CONCLUSIONS
The Redbirds trail the Bluebirds by one goal with 1 minute left in the hockey game. The Redbirds coach must decide whether to remove the goalie and add a frontline player. The probabilities of each team scoring are shown in the table.
Big Ideas Math Geometry Solutions Chapter 12 Probability 67
a. Find the probability that the Redbirds score and the Bluebirds do not score when the coach leaves the goalie in.
Answer:
Redbirds 10% x Bluebirds 90% = 9%

b. Find the probability that the Redbirds score and the Bluebirds do not score when the coach takes the goalie out.
Answer:
Redbirds 30% x Bluebirds 40% = 12%

c. Based on parts (a) and (b), what should the coach do?
Answer: Looks to be a 3% better chance to tie it up in B – pull the goalie

Question 19.
PROBLEM SOLVING
You can win concert tickets from a radio station if you are the first person to call when the song of the day is played. or if you are the first person to correctly answer the trivia question. The song of the day is announced at a random time between 7:00 and 7:30 A.M. The trivia question is asked at a random Lime between 7:15 and 7:45 A.M. You begin listening to the radio station at 7:20. Find the probability that you miss the announcement of the song of the day or the trivia question.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 19

Question 20.
HOW DO YOU SEE IT?
Are events A and B disjoint events? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 12 Probability 68
Answer:
A and B are not disjoint events and in fact they are overlapping events with 1 overlapping outcome.
Disjoint events do not have any overlap and they are mutually exclusive from one another.

Question 21.
PROBLEM SOLVING
You take a bus from sour neighborhood to your school. The express bus arrives at your neighborhood at a random time between 7:30 and 7:36 AM. The local bus arrives at your neighborhood at a random time between 7:30 and 7:40 A.M. You arrive at the bus stop at 7:33 A.M. Find the probability that you missed both the express bus and the local bus.
Big Ideas Math Geometry Solutions Chapter 12 Probability 69
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 21

Question 22.
THOUGHT PROVOKING
Write a general rule for finding P(A or B or C) for (a) disjoint and (b) overlapping events A, B, and C.
Answer:
For 2 disjoint events, the equation becomes:
P(A or B) = P(A) + P(B), based on this it can be said that the equation of P(A or B or C) will be
P(A or B or C) = P(A) + P(B) + P(C)
For 2 overlapping events, the equation becomes:
P(A or B) = P(A) + P(B) – P(A and B)
P(A or B or C) = P(A) + P(B) + P(C) – P(A and B) – P(A and C) – P(B and C) + P(A and B and C)

Question 23.
MAKING AN ARGUMENT
A bag contains 40 cards numbered 1 through 40 that are either red or blue. A card is drawn at random and placed back in the bag. This is done four times. Two red cards are drawn. numbered 31 and 19, and two blue cards are drawn. numbered 22 and 7. Your friend concludes that red cards and even numbers must be mutually exclusive. Is your friend correct? Explain.
Answer:
Your friend is incorrect because we do not know all the number of cards. Also, from the given data we do not know all colors for cards. Therefore we can not conclude that red cards and even numbers be mutually exclusive.

Maintaining Mathematical Proficiency

Find the Product.

Question 24.
(n – 12)2
Answer:
We can solve the product by using the formula
(a – b)² = a² – 2ab + b²
(n – 12)2 = (n)² – 2(n)(12) + (12)²
n² – 24n + 144

Question 25.
(2x + 9)2
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.4 Qu 25

Question 26.
(- 5z + 6)2
Answer:
(- 5z + 6)2 = (6 – 5z)2
We can solve the product by using the formula
(a – b)² = a² + 2ab + b²
(6 – 5z)2 = (6)² – 2(6)(5z) + (5z)²
36 – 24n + 25z²

Question 27.
(3a – 7b)2
Answer:
We can solve the product by using the formula
(a – b)² = a² + 2ab + b²
(3a – 7b)2 = (3a)² – 2(3a)(7b) + (7b)²
9a² – 42ab + 49b²

12.5 Permutations and Combinations

Exploration 1

Reading a Tree Diagram

Work with a partner. Two coins are flipped and the spinner is spun. The tree diagram shows the possible outcomes.
Big Ideas Math Answer Key Geometry Chapter 12 Probability 71
Big Ideas Math Answer Key Geometry Chapter 12 Probability 72
a. How many outcomes are possible?
Answer:

b. List the possible outcomes.
Answer:

Exploration 2

Reading a Tree Diagram

Work with a partner: Consider the tree diagram below.
Big Ideas Math Answer Key Geometry Chapter 12 Probability 82
a. How many events are shown?
Answer:

b. What outcomes are possible for each event?
Answer:

c. How many outcomes are possible?
Answer:

d. List the possible outcomes.
Answer:

Exploration 3

Writing a Conjecture

Work with a partner:

a. Consider the following general problem: Event 1 can occur in in ways and event 2 can occur in n ways. Write a conjecture about the number of ways the two events can occur. Explain your reasoning.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math,
you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.
Answer:

b. Use the conjecture you wrote in part (a) to write a conjecture about the number of ways more than two events can occur. Explain your reasoning.
Answer:

c. Use the results of Explorations 1(a) and 2(c) to verify your conjectures.
Answer:

Communicate Your Answer

Question 4.
How can a tree diagram help you visualize the number of ways in which two or more events can occur?
Answer:

Question 5.
In Exploration 1, the spinner is spun a second time. How many outcomes are possible?
Answer:

Lesson 12.5 Permutations and Combinations

Monitoring Progress

Question 1.
In how many ways can you arrange the letters in the word HOUSE?
Answer:

Question 2.
In how many ways can you arrange 3 of the letters in the word MARCH?
Answer:

Question 3.
WHAT IF
In Example 2, suppose there are 8 horses in the race. In how many different ways can the horses finish first, second, and third? (Assume there are no ties.)
Answer:

Question 4.
WHAT IF?
In Example 3, suppose there are 14 floats in the parade. Find the probability that the soccer team is first and the chorus is second.
Answer:

Question 5.
Count the possible combinations of 3 letters chosen from the list A, B, C, D, E.
Answer:

Question 6.
WHAT IF?
In Example 5, suppose you can choose 3 side dishes out of the list of 8 side dishes. How many combinations are possible?
Answer:

Question 7.
WHAT IF?
In Example 6, suppose there are 20 photos in the collage. Find the probability that your photo and your friend’s photo are the 2 placed at the top of the page.
Answer:

Exercise 12.5 Permutations and Combinations

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
An arrangement of objects in which order is important is called a(n) _________ .
Answer:
An arrangement of objects in which order is important is called a Permutation.

Question 2.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.
\(\frac{7 !}{2 ! \cdot 5 !}\) 7C5 7C2 \(\frac{7 !}{(7-2) !}\)
Answer:
7C2 \(\frac{7 !}{(7-2) !}\) = \(\frac{7 !}{5!}\)
= 7C2
The expression \(\frac{7 !}{(7-2) !}\) does not belong with other three.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, find the number of ways you can arrange (a) all of the letters and (b) 2 of the letters in the given word.

Question 3.
AT
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 3

Question 4.
TRY
Answer:
a. In this case, we have to find the number of permutations all of the letters in a given word that will consist of 3 letters.
Number of permutations = (1st place can be one of three letters) × (2nd can be one of two letters that is left) × (3rd can be one letter that is left)
= 3 . 2 . 1 = 6
Therefore we have 6 ways for arrange all of the letters in given word, that is TRY, TYR, YTR, YRT, RTY and RYT.
Now, we have to find the number of permutation 2 of the letters in a given word that will consists of 3 letters
Number of permutations = (1st place can be one of three letters) × (2nd can be one of two letters that is left)
= 3 . 2 = 6
We have 6 ways for arrange 2 of the letters in given word, tthat is TR, TY, YT, YR, RT and RY.

Question 5.
ROCK
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 5

Question 6.
WATER
Answer:
In this case, we have to find the number of permutation all of the letters in a given word that will consist of 5 letters.
Number of permutations = (1st can be one of 5 letters) × (2nd place can be one of 4 letters that is left) × (3rd can be one of 3 letters that is left) × (4th can be two letters that is left) × (5th can be one letter that is left)
= 5 .4 . 3 . 2 . 1 = 120
Thus we have 120 ways to arrange all of the letters in given word, that WATER, WATRE, WARTE,…, RETAW.
Number of permutations = (1st can be one of 5 letters) × (2nd place can be one of 4 letters that is left)
= 5 . 4
= 20
Thus we have 20 ways to arrange 2 of the letters in given word, that WA, WT, WR, WE …, ER, RE.

Question 7.
FAMILY
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 7

Question 8.
FLOWERS
Answer:
We have to find the number of permutation all of the letters in a given word that will consist of 7 letters.
Number of permutations = (1st place can be one of 7 letters) × (2nd place can be one of 6 letters that is left) × (3rd place can be one of 5 letters that is left) × (4th place can be one of 4 letters that is left) × (5th place can be one of 3 letters that is left) × (6th place can be one of 2 letters that is left) × (7th place can be one of 1 letters that is left)
= 7 . 6 . 5 . 4 . 3 . 2 . 1 = 5040
Thus we have 5040 ways to arrange all of the letters in given word, that is FLOWERS, FLOWERS, FLOWSER… SREWOLF
Now we have to find the number of permutation 2 of the letters in a given word that will consist of 7 letters.
Number of permutations = (1st place can be one of 7 letters) × (2nd place can be one of 6 letters that is left)
= 7 . 6
= 42
Thus we have 42 ways to arrange all of the letters in given word, that is FL, FO, FW, FE…RS, RE.

In Exercises 9 – 16, evaluate the expression.

Question 9.
5P2
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 9

Question 10.
7P3
Answer:
7P3 = \(\frac{7 !}{(7-3) !}\) = \(\frac{7 !}{4!}\)
= 7 . 6 . 5 . 4 . 3 . 2 . 1/4 . 3 . 2 . 1
= 7 . 6 . 5
= 210
7P3 = 210

Question 11.
9P1
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 11

Question 12.
6P5
Answer:
6P5 = \(\frac{6 !}{(6-5) !}\) = \(\frac{6 !}{1!}\)
= 6 . 5 . 4 . 3 . 2 . 1/1
= 6 . 5 . 4 . 3 . 2
= 720
6P5 = 720

Question 13.
8P6
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 13

Question 14.
12P0
Answer:
12P0 = \(\frac{12 !}{(12-0) !}\)
= \(\frac{12 !}{12!}\)
= 1
12P0 = 1

Question 15.
30P2
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 15

Question 16.
25P5
Answer:
25P5 = \(\frac{25 !}{(25-5) !}\) = \(\frac{25 !}{20!}\)
= 25 . 24 . 23 . 22 . 21 . 20 . 19 . 18 . …. 6 . 5 . 4 . 3 . 2 . 1/20 . 19 . 18 . … 3 . 2 . 1
= 25 . 24 . 23 . 22 . 21
= 720
25P5 = 6375600

Question 17.
PROBLEM SOLVING
Eleven students are competing in an art contest. In how many different ways can the students finish first, second, and third?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 17

Question 18.
PROBLEM SOLVING
Six Friends go to a movie theater. In how many different ways can they sit together in a row of 6 empty seats?
Answer:
Given,
Six Friends go to a movie theater.
6P6 = \(\frac{6!}{(6-6) !}\)
= \(\frac{6!}{0!}\)
= 6!
= 6 . 5 . 4 . 3 . 2 . 1
6P6 = 720

Question 19.
PROBLEM SOLVING
You and your friend are 2 of 8 servers working a shill in a restaurant. At the beginning of the shill. the manager randomly assigns 0ne section to each server. Find the probability that you are assigned Section 1 and your friend is assigned Section 2.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 19

Question 20.
PROBLEM SOLVING
You make 6 posters to hold up at a basketball game. Each poster has a letter of the word TIGERS. You and 5 friends sit next to each other in a row. The posters are distributed at random. Find the probability that TIGERS is spelled correctly when you hold up the posters.
Big Ideas Math Answer Key Geometry Chapter 12 Probability 73
Answer:
Number of favorable outcomes = 1 (TIGERS)
Total number of outcomes = 6!
= 6 . 5 . 4 . 3 . 2 . 1
= 720
Number of favorable outcomes/Total number of outcomes = 1/720

In Exercises 21 – 24, count the possible combinations of r letters chosen from the given list.

Question 21.
A, B, C, D; r = 3
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 21

Question 22.
L, M, N, O; r = 2
Answer:
4P2 = \(\frac{4!}{(4-2) !}\)
= \(\frac{4!}{2!}\)
= 4 . 3 . 2 . 1/2 . 1
= 12
4P2 = 12
So, the possible permutations of two letters in the given list L, M, N, O is
LM, ML
LN, NL
LO, OL
MN, NM
MO, MO
NO, ON
Thus the number of possible combination of a = 2 letters chosen from the list L, M, N, O
4P2 = 12/2 = 6

Question 23.
U , V, W, X, Y, Z; r = 3
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 23.1
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 23.2

Question 24.
D, E, F, G, H; R = 4
Answer:
5P4 = \(\frac{5!}{(5-4) !}\)
= \(\frac{5!}{1!}\)
=5 . 4 . 3 . 2 . 1
= 120
5P4 = 120
Thus the possible permutations of four letters in the given list D, E, F, G, H is
DEFG, DEGF, DGEF, DGFE, DFGD…HEFG, EHFG, EHGF, FGEH.
Thus we conclude that the number of possible combination of a = 4 letters chosen from the list D, E, F, G, H is
5C4 = \(\frac{120}{(24) !}\) = 5

In Exercise 25 – 32, evaluate the expression

Question 25.
5C1
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 25

Question 26.
8C5
Answer:
8C5 = \(\frac{8!}{(8-5) !}\)
= \(\frac{8!}{3!}\) . \(\frac{1}{5!}\)
= 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(5 . 4 . 3 . 2 . 1) (3 . 2 . 1)
= 8 . 7
8C5 = 56

Question 27.
9C9
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 27

Question 28.
8C6
Answer:
8C6 = \(\frac{8!}{(8-6) !}\)
= \(\frac{8!}{2!}\) . \(\frac{1}{6!}\)
= 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(6. 5 . 4 . 3 . 2 . 1) (3 . 2 . 1)
= 8 . 7
8C5 = 56

Question 29.
12C3
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 29

Question 30.
11C4
Answer:
11C4 = \(\frac{11!}{(11-4) !}\)
= \(\frac{11!}{7!}\) . \(\frac{1}{4!}\)
= 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(7 . 6. 5 . 4 . 3 . 2 . 1) (4 . 3 . 2 . 1)
= 11 . 10 . 3
11C4 = 330

Question 31.
15C8
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 31

Question 32.
20C5
Answer:
20C5 = \(\frac{20!}{(20-5) !}\)
= \(\frac{20!}{5!}\) . \(\frac{1}{5!}\)
= 20 . 19 . 18 . 17 . 16 . 15 . … 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(5 . 4 . 3 . 2 . 1) (5 . 4 . 3 . 2 . 1)
= 19 . 3 . 17 . 16
20C5 = 15504

Question 33.
PROBLEM SOLVING
Each year, 64 golfers participate in a golf tournament. The golfers play in groups of 4. How many groups of 4 golfers are possible?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 33

Question 34.
PROBLEM SOLVING
You want to purchase vegetable dip for a party. A grocery store sells 7 different flavors of vegetable dip. You have enough money to purchase 2 flavors. How many combinations of 2 flavors of vegetable dip are possible?
Answer:
given that,
You want to purchase vegetable dip for a party. A grocery store sells 7 different flavors of vegetable dip. You have enough money to purchase 2 flavors.
7C2 = \(\frac{7!}{(7-2) !}\)
= \(\frac{7!}{5!}\) . \(\frac{1}{2!}\)
= 7 . 6 . 5 . 4 . 3 . 2 . 1/(5 . 4 . 3 . 2 . 1) (2 . 1)
= 7 . 3
7C2 = 21

ERROR ANALYSIS
In Exercises 35 and 36, describe and correct the error in evaluating the expression.

Question 35.
Big Ideas Math Answer Key Geometry Chapter 12 Probability 74
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 35

Question 36.
Big Ideas Math Answer Key Geometry Chapter 12 Probability 75
Answer:
The permutation formula was used instead of the combination formula.

REASONING
In Exercises 37 – 40, tell whether the question can be answered using permutations or combinations. Explain your reasoning. Then answer the question.

Question 37.
To complete an exam. u must answer 8 questions from a list of 10 questions. In how many ways can you complete the exam?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 37

Question 38.
Ten students are auditioning for 3 different roles in a play. In how many ways can the 3 roles be filled?

Answer:
As 10 students are auditioning for 3 roles, which are different from each other, the order in which the roles are assigned to the students is important and should be taken into account.
As the Permutations formula takes into account the order of distribution. Hence the number of ways to fill the 3 roles can be found by using the permutations formula in the chapter.
So, the number of permutations of assigning the 3 roles to 3 students chosen from 10, using the permutations formula
10P3 = \(\frac{10!}{(10-3) !}\)
= \(\frac{10!}{7!}\)
= 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(7 . 6 . 5 . 4 . 3 . 2 . 1)
=10 . 9 . 8
10P3 = 720

Question 39.
Fifty-two athletes arc competing in a bicycle race. In how many orders can the bicyclists finish first, second, and third? (Assume there are no ties.)
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 39

Question 40.
An employee at a pet store needs to catch 5 tetras in an aquarium containing 27 tetras. In how many groupings can the employee capture 5 tetras?

Answer:
Given,
An employee at a pet store needs to catch 5 tetras in an aquarium containing 27 tetras.
27C5 = \(\frac{27!}{(27-5) !}\)
= \(\frac{27!}{22!}\) . \(\frac{1}{5!}\)
= 27. 26 . 25 . 24 . … 7 . 6 . 5 . 4 . 3 . 2 . 1/(22 . 21. … 7 . 6 . 5 . 4 . 3 . 2 . 1)(5 . 4 . 3 . 2 . 1)
= 27 . 26 . 5 . 23
27C5 = 80730
Thus the number of combinations of 27 tetras taken 5 at the time is 80730.

Question 41.
CRITICAL THINKING
Compare the quantities 50C9 and 50C41 without performing an calculations. Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 41

Question 42.
CRITICAL THINKING
Show that each identity is true for any whole numbers r and n, where 0 ≤ r ≤ n.
a. nCn = 1
Answer:
nCn = n!/n!(n – n)!
nCn = n!/n!0!
= 1/0!
We know that,
0! = 1
= 1/1 = 1
Thus nCn = 1

b. nCr = nCn – r
Answer:
nCn-a= n!/a!(n – a)!
nCn-a = n!/a!(n-a)! = nCn-a

c. n + 1Cr = nCr + nCr – 1
Answer:
n+1Ca = n+1!/a!(n+1-a)!
nCr + nCa-1 = n!/a!(n – a)! + n!/(n – (a – 1))! (a – 1)!
= \(\frac{(n+1)!}{(n-a+1) !}\) . \(\frac{1}{a!}\)
Thus n + 1Ca = nCa + nCa – 1

Question 43.

REASONING
Complete the table for each given value of r. Then write an inequality relating nPr and nCr. Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 12 Probability 76
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 43

Question 44.
REASONING
Write an equation that relates nPr and nCr. Then use your equation to find and interpret the Value of \(\frac{182 P_{4}}{182 C_{4}}\).
Answer:
\(\frac{n P_{a}}{n C_{a}}\) = n!/(n – a)!/n!/a!(n – a)! = a!
\(\frac{182 P_{4}}{182 C_{4}}\) = 4! = 24

Question 45.
PROBLEM SOLVING
You and your friend are in the studio audience on a television game show. From an audience of 300 people, 2 people are randomly selected as contestants. What is the probability that you and your friend are chosen?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 45

Question 46.
PROBLEM SOLVING
You work 5 evenings each week at a bookstore. Your supervisor assigns you 5 evenings at random from the 7 possibilities. What is the probability that your schedule does not include working on the weekend?
Answer:
nCa = n!/a!(n – a)!
7C5 = \(\frac{7!}{(7-5) !}\)
= \(\frac{7!}{2!}\) . \(\frac{1}{5!}\)
=7 . 6 . 5 . 4 . 3 . 2 . 1/(2 . 1)(5 . 4 . 3 . 2 . 1)
= 7 . 3
7C5 = 21
Thus we see that there are 21 possible combinations of five days formed from 7 days.
Hence we see that one of 21 possible combination of 5 days does not contain saturday and sunday.
P = P(Are chosen one combination of 21 possible)
= Number of favorable outcomes/Number of possible outcomes
= 1/21

REASONING
In Exercises 47 and 48, find the probability of winning a lottery using the given rules. Assume that lottery numbers are selected at random.

Question 47.
You must correctly select 6 numbers, each an integer from 0 to 49. The order is not important.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 47

Question 48.
You must correctly select 4 numbers, each an integer from 0 to 9. The order is important.
Answer:
nCa = n!/a!(n – a)!
10C4 = \(\frac{10!}{(10-4) !}\)
= \(\frac{10!}{6!}\) . \(\frac{1}{4!}\)
=10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(6 . 5 . 4 . 3 . 2 . 1)(4 . 3 . 2 . 1)
= 10 . 3 . 7
10C4 = 210
There are 210 possible combination of 10 numbers taken 4 at a time.
P (You winning a lottery) = P(Are chosen one combination 210 possible)
= Number of favorable outcomes/Number of possible outcomes
= 1/210

Question 49.
MATHEMATICAL CONNECTIONS
A polygon is convex when no line that contains a side of the polygon contains a point in the interior of the polygon. Consider a convex polygon with n sides.
Big Ideas Math Answer Key Geometry Chapter 12 Probability 77
a. Use the combinations formula to write an expression for the number of diagonals in an n-sided polygon.
b. Use your result from part (a) to write a formula for the number of diagonals of an n-sided convex polygon.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 49

Question 50.
PROBLEM SOLVING
You are ordering a burrito with 2 main ingredients and 3 toppings. The menu below shows the possible choices. How many different burritos are possible
Big Ideas Math Answer Key Geometry Chapter 12 Probability 78
Answer:
You are ordering a burrito with 2 main ingredients and 3 toppings.
Total number of main ingredients = 6
As the order in which the ingredients are chosen is not important, the total number of ways to select 2 main ingredients out of 6 can be found by using the combinations formula.
nCa = n!/a!(n – a)!
6C2 = \(\frac{6!}{(6-2) !}\)
= \(\frac{6!}{4!}\) . \(\frac{1}{2!}\)
=6 . 5 . 4 . 3 . 2 . 1/(4 . 3 . 2 . 1)(2 . 1)
= 3 . 5
6C2 = 15
Total number of toppings = 8
Number of toppings to be chosen = 3
nCa = n!/a!(n – a)!
8C3 = \(\frac{8!}{(8-3) !}\)
= \(\frac{8!}{5!}\) . \(\frac{1}{3!}\)
=8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(5 . 4 . 3 . 2 . 1)(3 . 2 . 1)
= 8 . 7
8C3 = 56
Total possible selections = ways to select main ingredients × Ways to select toppings
= 15 × 56
= 840

Question 51.
PROBLEM SOLVING
You want to purchase 2 different types of contemporary music CDs and 1 classical music CD from the music collection shown. How many different sets of music types can you choose for your purchase?
Big Ideas Math Answer Key Geometry Chapter 12 Probability 79
a. How many combinations of three marbles can be drawn from the bag? Explain.
b. How many permutations of three marbles can be drawn from the bag? Explain.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 51

Question 52.
HOW DO YOU SEE IT?
A bag contains one green marble, one red marble, and one blue marble. The diagram shows the possible outcomes of randomly drawing three marbles from the hag without replacement.
Big Ideas Math Answer Key Geometry Chapter 12 Probability 80
a. How many combinations of three marbles can be drawn from the bag? Explain.
Answer:
nCa = n!/a!(n – a)!
3C3 = \(\frac{3!}{(3-3) !}\)
= \(\frac{3!}{0!}\) . \(\frac{1}{3!}\)
=3 . 2 . 1/(1)(3 . 2 . 1)
= 1
3C3 = 1

b. How many permutations of three marbles can be drawn from the bag? Explain.
Answer:
nPa = n!/(n – a)!
3P3 = \(\frac{3!}{(3-3) !}\)
= \(\frac{3!}{0!}\)
=3 . 2 . 1/1
= 6
3P3 = 3
There are 6 possible combinations.

Question 53.
PROBLEM SOLVING
Every student in your history class is required to present a project in front of the class. Each day, 4 students make their presentations in an order chosen at random by the teacher. you make your presentation on the first day.
a. What is the probability that you are chosen to be the first or second presenter on the first day ?
b. What is the probability that you are chosen to be the second or third presenter on the first day? Compare your answer with that in part (a).
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 53

Question 54.
PROBLEM SOLVING
The organizer of a cast party for a drama club asks each of the 6 cast members to bring 1 food item From a list of 10 items. Assuming each member randomly chooses a food item to bring. what is the probability that at least 2 of the 6 cast members bring the same item?
Answer:
Given,
The organizer of a cast party for a drama club asks each of the 6 cast members to bring 1 food item From a list of 10 items.
Assuming each member randomly chooses a food item to bring.
A = {At least 2 of the 6 cast members bring the same item}
\(\bar{A}\) = {All members bring different items}
First member can choose one of the 10 different products, the second can choose one of the 9 possible ways, and so on until the 6th member can choose one of the remaining 5 items.
10 . 9 . 8 . 7 . 6 . 5 = 151200
nPa = n!/(n – a)!
10P6 = \(\frac{10!}{(10-6) !}\)
= \(\frac{10!}{4!}\)
=10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/4 . 3 . 2 . 1
= 10 . 9 . 8 . 7 . 6 . 5 = 151200
10P6 = 151200
10 . 10 . 10 . 10 . 10 . 10 = 1000000
P(\(\bar{A}\)) = Number of favorable outcomes/Number of possible outcomes
= 151200/1000000
= 0.1512
P(A) = 1 – P(\(\bar{A}\))
P(A) = 1 – 0.1512 = 0.8488
The probability that at least 2 of the 6 members bring the same items is about 85%

Question 55.
PROBLEM SOLVING
You are one of 10 students performing in a school talent show. The order of the performances is determined at random. The first 5 performers go on stage before the intermission.
a. What is the probability that you are the last performer before the intermission and your rival performs immediately before you?
b. What is the probability that you are not the first performer?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 55

Question 56.
THOUGHT PROVOKING
How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3, and 4? Repetition of digits is allowed.
Answer:
Let fixed number 1 on the first place and in the second place can be one of 5 digits 0, 1, 2, 3 or 4. Because repetition of digits is allowed. In third and fourth place it can also be one of five digits.
Therefore we see that 5 . 5 . 5 = 125 integers which begin with 1, can be formed.
Second, If we fixed 2 on the first place, by the same logic as in the first part, we get that 5³ = 125 integers which begin with 2.
Next, if we fixed number 3 on the first place, we obtain 5³ = 125 integers which begin with 2.
5³ + 5³ + 5³ = 375
375 integers greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3, 4.

Question 57.
PROBLEM SOLVING
There are 30 students in your class. Your science teacher chooses 5 students at random to complete a group project. Find the probability that you and your 2 best friends in the science class are chosen to work in the group. Explain how you found your answer.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 57

Question 58.
PROBLEM SOLVING
Follow the steps below to explore a famous probability problem called the birthday problem. (Assume there are 365 equally likely birthdays possible.)

a. What is the probability that at least 2 people share the same birthday in a group of 6 randomly chosen people? in a group of 10 randomly chosen people?
Answer:
nPa = n!/(n – a)!
A = {At least  people share the same birthday in a group of 6 people}
So, we see that complement of event A is
\(\bar{A}\) = {All 6 people were born on different days}
nPa = n!/(n – a)!
365P6 = \(\frac{365!}{(365-6)!}\)
= \(\frac{365!}{(359)!}\)
= 365. 364 . 363 . 362 . 361 . 360 . 359 . 358 . ……. 2 . 1/(359 . 358. …. 2 . 1)
= 365. 364 . 363 . 362 . 361 . 360
For the first day we can also choose one of 365 possible for the second dat we can also choose one of 365 ways and so on.
For each of the 6 days we have the option to choose one day from 365 possible ways.
So, the number of possible outcomes is
365 . 365 . 365 . 365 . 365 . 365
P(\(\bar{A}\)) = Number of favorable outcomes/Number of possible outcomes
= 365. 364 . 363 . 362 . 361 . 360/365 . 365 . 365 . 365 . 365 . 365
= 0.959
P(A) = 1 – P(\(\bar{A}\))
P(A) = 1 – 0.959 = 0.04
P = 1 – 365P10/36510
1 – 0.883 = 0.117

b. Generalize the results from part (a) by writing a formula for the probability P(n) that at least 2 people in a group of n people share the same birthday. (Hint: Use nPr notation in your formula.)
Answer:
Based on the explanation in the part under a) we can conclude that the probability that at least two people share the same birthday in a group of n people is
P = 1 – 365Pn/365n

c. Enter the formula from part (b) into a graphing calculator. Use the table feature to make a table of values. For what group size does the probability that at least 2 people share the same birthday first exceed 50%?
Answer:
Big Ideas Math Answers Geometry Chapter 12 Probability img_11
Based on the above table we see that for a group of 23 people and more, the probability that at least tweople share the same birthday exceed 50%

Maintaining Mathematical Proficiency

Question 59.
A bag contains 12 white marbles and 3 black marbles. You pick 1 marble at random. What is the probability that you pick a black marble?
Answer:
Given,
A bag contains 12 white marbles and 3 black marbles. You pick 1 marble at random
Big Ideas Math Geometry Answers Chapter 12 Probability 12.5 Qu 59

Question 60.
The table shows the result of flipping two coins 12 times. For what outcome is the experimental probability the same as the theoretical probability?
Big Ideas Math Answer Key Geometry Chapter 12 Probability 81
Answer:
The table shows the result of flipping two coins 12 times
P(HH) = P(HT) = P(TH) = P(TT) = 1/2 . 1/2 = 1/4
P(HH) = Number of favorable outcomes/Number of possible outcomes
= 2/12 = 1/6
P(HT) = 6/12 = 1/2
P(TH) = 3/12 = 1/4
P(TT) = 1/12
The most likely fell first heads, and second tails. Also, with the least probability fell twice tails.

12.6 Binomial Distributions

Exploration 1

Analyzing Histograms

Work with a partner: The histograms show the results when n coins are flipped.
Big Ideas Math Geometry Answers Chapter 12 Probability 83
STUDY TIP
When 4 coins are flipped (n = 4), the possible outcomes are
TTTT TTTH TTHT TTHH
THTT THTH THHT THHH
HTTT HTTH HTHT HTHH
HHTT HHTH HHHT HHHH.
The histogram shows the numbers of outcomes having 0, 1, 2, 3, and 4 heads.
a. In how many ways can 3 heads occur when 5 coins are flipped?
Answer:

b. Draw a histogram that shows the numbers of heads that can occur when 6 coins are flipped.
Answer:

c. In how many ways can 3 heads occur when 6 coins are flipped?
Answer:

Exploration 2 

Determining the Number of Occurrences

Work with a partner:

a. Complete the table showing the numbers of ways in which 2 heads can occur when n coins are flipped.
Big Ideas Math Geometry Answers Chapter 12 Probability 84
Answer:

b. Determine the pattern shown in the table. Use your result to find the number of ways in which 2 heads can occur when 8 coins are flipped.
LOOKING FOR A PATTERN
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Communicate Your Answer

Question 3.
How can you determine the frequency of each outcome of an event?
Answer:

Question 4.
How can you use a histogram to find the probability of an event?
Answer:

Lesson 12.6 Binomial Distributions

Monitoring Progress

An octahedral die has eight sides numbered 1 through 8. Let x be a random variable that represents the sum when two such dice are rolled.

Big Ideas Math Geometry Answers Chapter 12 Probability 85

Question 1.
Make a table and draw a histogram showing the probability distribution tor x.
Answer:

Question 2.
What is the most likely sum when rolling the two dice?
Answer:

Question 3.
What is the probability that the sum of the two dice is at most 3?
Answer:

According to a survey, about 85% of people ages 18 and older in the U.S. use the Internet or e-mail. You ask 4 randomly chosen people (ages 18 and older) whether they use the Internet or email.

Question 4.
Draw a histogram of the binomial distribution for your survey.
Answer:

Question 5.
What is the most likely outcome of your survey?
Answer:

Question 6.
What is the probability that at most 2 people you survey use the Internet or e-mail?
Answer:

Exercise 12.6 Binomial Distributions

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is a random variable?
Answer:
A random variable is a variable whose value is determined by the outcomes of a probability experiment.

Question 2.
WRITING
Give an example of a binomial experiment and describe how it meets the conditions of a binomial experiment.
Answer:
We flipping a coin 10 times and register what fell.
We know that events, coin toss are independent.
Each trial has only two possible outcomes: H and T
The probabilities are P(H) = P(T) = 1/2 and are the same for each trial.
We can conclude that this experiment is a binomial experiment.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. make a table and draw a histogram showing the probability distribution for the random variable.

Question 3.
x = the number on a table tennis ball randomly chosen from a bag that contains 5 balls labeled “1,” 3 halls labeled “2,” and 2 balls labeled “3.”
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 3

Question 4.
c = 1 when a randomly chosen card out of a standard deck of 52 playing cards is a heart and c = 2 otherwise.
Answer:
Let C be a random variable that represents the randomly chosen card.
Standard desk have 52 playing cards.
P(C = 1) = Number of favorable outcomes/Total number of outcomes
= A randomly chosen card is a hard/Total number of cards
= 13/52
On the other hand,
P(C = 2) = A randomly chosen card is not a hard/Total number of cards
= 39/52

Question 5.
w = 1 when a randomly chosen letter from the English alphabet is a vowel and w = 2 otherwise.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 5

Question 6.
n = the number of digits in a random integer from O through 999.
Answer:
There are 10 outcomes for value 1, 90 outcomes for value 2, and 900 values for value 3.
P(N = 1) = Number of favorable outcomes/Total number of outcomes
= A randomly chosen integers in a one-digit number/Total number of integers
= 10/1000
= 1/100
P(N = 2) = 90/1000 = 9/100
P(N = 3) = 900/1000 = 9/10

In Exercises 7 and 8, use the probability distribution to determine (a) the number that is most likely to be spun on a spinner, and (b) the probability of spinning an even number.

Question 7.
Big Ideas Math Geometry Answers Chapter 12 Probability 86
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 7

Question 8.
Big Ideas Math Geometry Answers Chapter 12 Probability 87
Answer:
The most likely number to be spun on the spinner is the value of random variable X(Number of spinner) of which P(X) is greatest.
From given histogram we see that this probability is greatest for X = 5.
Hence the most likely number to be spun on the spinner is 5.
P(Spinning an even number) = P(X = 10) + P(X = 20) = 1/6 + 1/12 = 1/4

USING EQUATIONS
In Exercises 9 – 12, calculate the probability of flipping a coin 20 times and getting the given number of heads.
Question 9.
1
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 9

Question 10.
4
Answer:
P(Four success) = 20C4(1/2)4(1/2)20-4
= 20!/4!(20 – 4)!(1/2)20
= 20!/4!(16)!(1/2)20
= 0.0046
We see that the obtained probability is small, which is logical.

Question 11.
18
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 11

Question 12.
20
Answer:
P(Four success) = 20C20(1/2)20(1/2)20-20
= 20!/20!(20 – 20)!(1/2)20
=(1/2)20
= 0.00000095

Question 13.
MODELING WITH MATHEMATICS
According to a survey, 27% of high school students in the United States buy a class ring. You ask 6 randomly chosen high school students whether they own a class ring.
Big Ideas Math Geometry Answers Chapter 12 Probability 88
a. Draw a histogram of the binomial distribution for your survey.
b. What is the most likely outcome of your survey?
c. What is the probability that at most 2 people have a class ring?
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 13.1
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 13.2

Question 14.
MODELING WITH MATHEMATICS
According to a survey, 48% of adults in the United States believe that Unidentified Flying Objects (UFOs) are observing our planet. You ask 8 randomly chosen adults whether they believe UFOs are watching Earth.
a. Draw a histogram of the binomial distribution for your survey.
Answer:
p = P(the American is a sports fan) = 48% = 0.48
1 – p = P(the American is not a sports fan) = 1 – 0.48= 0.52
P (0 success) = 0C8 p0(1 – p)8-0
= 8!/0!(8 – 0)! 1 . 0.528
= 0.528
= 0.1513
P (One person believe that UFOs are watching Earth) =8C1 p¹(1 – p)8-1
=0.03948
P (Two person believe that UFOs are watching Earth) =8C2 p2(1 – p)8-2
=0.1275
P (Three person believe that UFOs are watching Earth) =8C3 p3(1 – p)8-3
=0.2355
P (Four person believe that UFOs are watching Earth) =8C4 p4(1 – p)8-4
=0.2717
P (Five person believe that UFOs are watching Earth) =8C5 p5(1 – p)8-5
=0.2006
P (Six person believe that UFOs are watching Earth) =8C6 p6(1 – p)8-6
=0.0926
P (Seven person believe that UFOs are watching Earth) =8C7 p7(1 – p)8-7
=0.0244
P (Eight person believe that UFOs are watching Earth) =8C8 p8(1 – p)8-8
=0.0028
Big Ideas Math Answers Geometry Chapter 12 probability img_10

b. What is the most likely outcome of your survey?
Answer:
P (Four person believe that UFOs are watching Earth) = 0.2717
This probability has the highest, so we conclude that the most likely outcome is that four of the eight adults believe that UFOs are watching Earth.

c. What is the probability that at most 3 people believe UFOs are watching Earth?
Answer:
P (At most 3 persons believe that UFOs are watching Earth) = P (One person believe that UFOs are watching Earth) + P (Two person believe that UFOs are watching Earth) + P (Three person believe that UFOs are watching Earth)
= 0.0053 + 0.0395 + 0.1275 + 0.2355
= 0.4078

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in calculating the probability of rolling a 1 exactly 3 times in 5 rolls of a six-sided die.

Question 15.
Big Ideas Math Geometry Answers Chapter 12 Probability 89
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 15

Question 16.
Big Ideas Math Geometry Answers Chapter 12 Probability 90
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 15

Question 17.
MATHEMATICAL CONNECTIONS
At most 7 gopher holes appear each week on the farm shown. Let x represent how many of the gopher holes appear in the carrot patch. Assume that a gopher hole has an equal chance of appearing at any point on the farm.
Big Ideas Math Geometry Answers Chapter 12 Probability 91
a. Find P(x) for x = 0, 1, 2 ….., 7.

Answer:
p = P(the gopher holes appear in the carrot patch)
= Area marked for carrot/Area of the whole farm
= Area of a square + Area of a triangle/Area of the whole farm
= 0.28125
1 – p = P(The gopher holes do not appear in the carrot patch)
= 1 – 0.28125
= 0.71875
P (0 success) = 0C7 p0(1 – p)7-0
= 7!/0!(7 – 0)! 1 . 0.727
= 0.727
= 0.099
P (There is one gopher hole in the carrot patch) =7C1 p¹(1 – p)7-1
=0.27143
P (There is two gopher hole in the carrot patch) =7C2 p¹(1 – p)7-2
=0.31863
P (There is three gopher hole in the carrot patch) =7C3 p¹(1 – p)7-3
=0.20781
P (There is four gopher hole in the carrot patch) =7C4 p¹(1 – p)7-4
=0.08131
P (There is five gopher hole in the carrot patch) =7C5 p¹(1 – p)7-5
=0.01909
P (There is six gopher hole in the carrot patch) =7C6 p¹(1 – p)7-6
=0.00249
P (There is seven gopher hole in the carrot patch) =7C7 p¹(1 – p)7-7
=0.00012

b. Make a table showing the probability distribution for x.
c. Make a histogram showing the probability distribution for x.
Answer:

Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 17.2

Question 18.
HOW DO YOU SEE IT?
Complete the probability distribution for the random variable x. What is the probability the value of x is greater than 2?
Big Ideas Math Geometry Answers Chapter 12 Probability 92
Answer:
P(X = 1) + P(X = 2) + … + P(X = n) = 1
P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.1 + 0.3 + 0.4 + P(X = 4)
= 1
P(X = 4) = 1 – 0.8 = 0.2
Now lets find a probability that value of X is greater then two, as
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4)
= 0.3 + 0.4 + 0.2
= 0.9
It is very likely that the random variable k will take a value greater than 2.

Question 19.
MAKING AN ARGUMENT
The binomial distribution Shows the results of a binomial experiment. Your friend claims that the probability p of a success must be greater than the probability 1 – p of a failure. Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 12 Probability 93
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 19

Question 20.
THOUGHT PROVOKING
There are 100 coins in a bag. Only one of them has a date of 2010. You choose a coin at random, check the date, and then put the coin back in the bag. You repeat this 100 times. Are you certain of choosing the 2010 coin at least once? Explain your reasoning.
Answer:
Given,
There are 100 coins in a bag. Only one of them has a date of 2010. You choose a coin at random, check the date, and then put the coin back in the bag. You repeat this 100 times
p = P(Selected coin has a date of 2010)
= Number of favorable outcomes/Total number of outcomes
= 1/100
1 – p = 1 – 1/100  = 99/100
P (0 success) = 100C0 p0(1 – p)100-0
= 100!/0!(100 – 0)! 1 . (99/100)100
= (99/100)100
= 0.366
P(1 or more success) = 1 – P(0 success) = 1 – 0.366 = 0.634
Hence with a probability of 0.634, we will choose a coin which has a date of 2010.
We are not certain of choosing the 2010 coin at least once.

Question 21.
MODELING WITH MATHEMATICS
Assume that having a male and having a female child are independent events, and that the probability of each is 0.5.
a. A couple has 4 male children. Evaluate the validity of this statement: “The first 4 kids were all boys, so the next one will probably be a girl.”
b. What is the probability of having 4 male children and then a female child?
c. Let x be a random variable that represents the number of children a couple already has when they have their first female child. Draw a histogram of the distribution of P(x) for 0 ≤ x ≤ 10. Describe the shape of the histogram.
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 21.1
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 21.2

Question 22.
CRITICAL THINKING
An entertainment system has n speakers. Each speaker will function properly with probability p. independent of whether the other speakers are functioning. The system will operate effectively when at least 50% of its speakers are functioning. For what values of p is a 5-speaker system more likely to operate than a 3-speaker system?
Answer:
Given,
An entertainment system has n speakers. Each speaker will function properly with probability p. independent of whether the other speakers are functioning.
The system will operate effectively when at least 50% of its speakers are functioning.
p = P(Speaker will function properly)
1 – 9 = P(Speaker will not function properly)
P(5-speaker system operate) = P(X = 3) + P(X = 4) + P(X = 5)
P(5-Speaker system operate) = P (0 success) = 5C3 p3(1 – p)5-3 + 5C4 p4(1 – p)5-4 + 5C5 p5(1 – p)5-5 = 10p3(1 – p)5-4 + 5p4(1 – p)5
P = P(X = 2) + P(X = 3)
= 5C2 p2(1 – p)5-2 + 5C3 p3(1 – p)5-3
= 10p2(1 – p)5-3 + 10p3(1 – p)2

10p3(1 – p)5-4 + 5p4(1 – p)5 + p5 > 10p2(1 – p)3 + 10p3(1 – p)2
5p2 – 4p3 – 10 + 30p – 30p2 + 10p3 > 0
A 5-speaker system operate more likely to operate than 3-speaker system when p ∈ (0.558, 1]

Maintaining Mathematical Proficiency

List the possible outcomes for the situation.

Question 23.
guessing the gender of three children
Answer:
Big Ideas Math Geometry Answers Chapter 12 Probability 12.6 Qu 23

Question 24.
picking one of two doors and one of three curtains
Answer:
If we denote by D1 and D2 first and second door and with C1, C2, C3 first, second and third curtain, then the possible outcomes are
D1C1, D1C2, D1C3, D2C1, D2C2, D2C3
Thus there are six possible outcomes.

Probability Review

12.1 Sample Spaces and Probability

Question 1.
A bag contains 9 tiles. one for each letter in the word HAPPINESS. You choose a tile at random. What is the probability that you choose a tile with the letter S? What is the probability that you choose a tile with a letter other than P?

Answer:
Let X be a random variable that represent the letter on tile.
We know that a bag contains tiles labeled with “H”, “A”, “P”, “I”, “N”, “E” and “S”
So we can conclude that the possible values for X are letters “H”, “A”, “P”, “I”, “N”, “E” and “S” and total number of outcomes is 9.

Question 2.
You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. Are you most likely to get 5 points, 10 points, or 20 points?
Big Ideas Math Geometry Answers Chapter 12 Probability 94
Answer: It is the most likely to get 20 points.

Explanation:
From given board, we can conclude that the probability that we get 5 points is
P(5 points) = Surface of red area/Surface of board = 4/36 = 1/9
On the other hand, we see that the probability we get 10 points is
P(10 points) = Surface of yellow area/Surface of board = (16 – 4)/36 = 1/3
and the probability that we get 20 points is
P(20 points) = Surface of blue area/Surface of board = (36 – 16)/36 = 5/9
From the obtained results we get that it is the most likely to get 20 points, which is logical because the surface of the blue area is the largest.

12.2 Independent and Dependent Events

Find the probability of randomly selecting the given marbles from a bag of 5 red, 8 green, and 3 blue marbles when (a) you replace the first marble before drawing the second, and (b) you do not replace the first marble. Compare the probabilities.

Question 3.
red, then green
Answer:
We selected two marbles from a bag of 5 red, 8 green and 3 blue. With R denote the event that the red marble is drawn, with P blue, and with G green. The draws are independent because we replace the first marble before drawing the second. Therefore, the probability that we selected first red marble and then green is
P(RG) = P(R)P(G) = 5/16 . 8/16 = k5/36 = 0.15625
b. In this case, we do not replace the first marble before drawing the second. So, the draws are not independent.
P(RG) = P(R)P(G|R) = 5/16 . 8/15 = 1/6 = 0.16667
It is more likely that we selected first red marble and then green when we not replace the first marble before drawing the second.

Question 4.
blue, then red
Answer:
We selected two marbles from a bag of 5 red, 8 green and 3 blue. With R denote the event that the red marble is drawn, with P blue, and with G green. The draws are independent because we replace the first marble before drawing the second. Therefore, the probability that we selected first red marble and then green is
P(BR) = P(B)P(R) = 3/16 . 5/16 = 15/256 = 0.05859
b. In this case, we do not replace the first marble before drawing the second. So, the draws are not independent.
P(BR) = P(B)P(R|B) = 3/16 . 5/15 = 1/16 = 0.0625
It is more likely that we selected first blue marble and then red when we not replace the first marble before drawing the second.

Question 5.
green, then green
Answer:
We selected two marbles from a bag of 5 red, 8 green and 3 blue. With R denote the event that the red marble is drawn, with P blue, and with G green. The draws are independent because we replace the first marble before drawing the second. Therefore, the probability that we selected first red marble and then green is
P(GG) = P(G)P(G) = 8/16 . 8/16 = 1/4 = 0.25
b. In this case, we do not replace the first marble before drawing the second. So, the draws are not independent.
P(GG) = P(G)P(G|G) = 8/16 . 7/15 = 0.23333
It is more likely that we selected first blue marble and then red when we not replace the first marble before drawing the second.

12.3 Two-Way Tables and Probability

Question 6.
What is the probability that a randomly selected resident who does not support the project in the example above is from the west side?
Answer:
P(West side|Does not support the project) = P(West side and Does not support the project)/P(Does not support the project)
= 0.09/(0.08 + 0.09)
= 0.529
The probability that random selected resident who does not support the project is from the west side is about 0.529

Question 7.
After a conference, 220 men and 270 women respond to a survey. Of those, 200 men and 230 women say the conference was impactful. Organize these results in a two-way table. Then find and interpret the marginal frequencies.
Answer:
After a conference, 220 men and 270 women respond to a survey. Of those, 200 men and 230 women say the conference was impactful.
Number of men who say the conference had not impact = 220 – 200 = 20
By the same method we come to the conclusion
Number of women who say the conference had not impact = 270 – 230 = 40
Now, we will find the marginal frequencies.
Total number of people who say the conference was impact is 200 + 230 = 430
Total Number of people who say the conference had not impact = 20 + 40 = 60
Also from given information we know that total number of people who was surveyed is 220 + 270 = 490
BIM Answers Geometry Chapter 12 Probability img_2

12.4 Probability of Disjoint and Overlapping Events

Question 8.
Let A and B be events such that P(A) = 0.32, P(B) = 0.48, and P(A and B) = 0.12. Find P(A or B).
Answer:
P(A or B) = P(A) + P(B) – P(A and B)
Given,
P(A) = 0.32, P(B) = 0.48, and P(A and B) = 0.12
P(A or B) = 0.32 + 0.48 – 0.12 = 0.68

Question 9.
Out of 100 employees at a company, 92 employees either work part time or work 5 days each week. There are 14 employees who work part time and 80 employees who work 5 days each week. What is the probability that a randomly selected employee works both part time and 5 days each week?
Answer:
A = {Employees either work part time},
B = {Employees either work 5 days}
Based on the given information we see that
P(A) = Number of favorable outcomes/Total Number of outcomes
= Number of Employees either work part time/Total number of employees
= 14/100
= 0.14
P(B) = Number of favorable outcomes/Total Number of outcomes
= Number of Employees either work 5 days/Total number of employees
= 80/100
= 0.8
Also, we know that 92 employees either work part time or 5 days each week
P(A or B) = Number of favorable outcomes/Total Number of outcomes
= 92/100 = 0.92
For given events A and B the probability of A or B is
P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = 0.14 + 0.8 – 0.92 = 0.02
Hence the probability that a randomly selected employee works part time and 5 days each week is 0.02

12.5 Permutations and Combinations

Evaluate the expression.

Question 10.
7P6
Answer:
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
=  7!/(7 – 6)!
= 7 . 6 . 5 . 4 . 3 . 2 . 1
= 5040
7P6 = 7! = 5040

Question 11.
13P10
Answer:
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
=  13!/(13 – 10)!
= 13!/3!
= 13. 12 . 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 = 94348800
13P10 = 94348800

Question 12.
6C2
Answer:
nCa = n!/a!(n – a)!
6C2 = 6!/2!(6 – 2)!
= 6!/2!(4)!
= 6 . 5 . 4 . 3 . 2 . 1/(2 . 1)(4 . 3 . 2 . 1)
15
6C2 = 15
nC2 = n(n – 1)/2
6C2 = 15

Question 13.
8C4
Answer:
nCa = n!/a!(n – a)!
8C4 = 8!/4!(8 – 4)!
= 68!/4!(4)!
= 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(4 . 3 . 2 . 1)(4 . 3 . 2 . 1)
= 2 . 7 . 5
6C2 = 70

Question 14.
Eight sprinters are competing in a race. How many different ways can they finish the race? (Assume there are no ties.)
Answer:
nPn = n!
We know thata 8 sprinters are participating in a race. It is important to us in what order each of them will reach the goal. This tells us that it is necessary to calculate the number of permutations of 8 sprinters.
8P8 = 8! = 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1 = 40320
Eight sprinters can finish a race in 40320 ways.

Question 15.
A random drawing will determine which 3 people in a group of 9 will win concert tickets. What is the probability that you and your 2 friends will win the tickets?
Answer:
nCa = n!/a!(n – a)!
9C3 = 9!/3!(9 – 3)!
= 9!/3!(6)!
= 9 . 8 . 7 6 . 5 . 4 . 3 . 2 . 1/(3 . 2 . 1)(6 . 5 .  4 . 3 . 2 . 1)
= 84
9C3 = 84
The probability that you and your 2 friends will win the tickets is equal to the probability that out of 84 possibilities, the trio in which you and your friend are in will be chosen.
P = Number of favorable outcomes/ Total number of outcomes = 1/84

12.6 Binomial Distributions

Question 16.
Find the Probability of flipping a coin 12 times and getting exactly 4 heads.
Answer:
The probabilities are P(H) = P(T) = 1/2, that is p = p – 1 = 1/2 and are the same for each trial.
n = 12
P = 12!/4!(12 – 4)! . (1/2)12

= 0.1208
Therefore the probability of flipping a coin 12 times and getting exactly 4 heads is about 0.12

Question 17.
A basketball player makes a free throw 82.6% of the time. The player attempts 5 free throws. Draw a histogram of the binomial distribution of the number of successful free throws. What is the most likely outcome?
Answer:
Given,
A basketball player makes a free throw 82.6% of the time. The player attempts 5 free throws.
p = P(Successful free throw) = 82.6% = 0.826
1 – p = P(Unsuccessful free throw) = 1 – 0.826 = 0.174
P (Out of 5 free throws one was successful) = 5C1 p¹(1 – p)5-1
P (Out of 5 free throws two was successful) =5C2 p²(1 – p)5-2
P(Out of 5 free throws three was successful) = 5C3p³(1 – p)5-3
P (Out of 5 free throws four was successful) =5C4 p4(1 – p)5-4
P (All 5 throws one was successful) = 5C5(1 – p)5-5
Bigideas Math Geometry Answers Chapter 12 Probability img_3

Probability Test

You roll a six-sided die. Find the probability of the event described. Explain your reasoning.

Question 1.
You roll a number less than 5.
Answer:
The die has 6 sides, thus the total number of possible outcomes is 6.
The favorable outcomes are
P(n<5) = Number of favorable outcomes/Total number of outcomes
Thus there are 4 favorable outcomes.
The probability to roll a number less than 5 is
= 4/6
= 2/3

Question 2.
You roll a multiple of 3.
Answer:
The die has 6 sides, thus the total number of possible outcomes is 6.
The favorable outcomes are
P(n = 3k) = Number of favorable outcomes/Total number of outcomes
Thus there are 2 favorable outcomes.
The probability to roll a number less than 3 is
= 2/6
= 1/3

Evaluate the expression.

Question 3.
7P2
Answer:
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
7P2 = 7!/(7 – 2)!
=  7!/5!
= (7 . 6 . 5 . 4 . 3 . 2 . 1)/(5 . 4 . 3 .  2 . 1)
= 7 . 6
= 42
7P2 = 42

Question 4.
8P3
Answer:
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
8P3 = 8!/(8 – 3)!
=  8!/5!
= (8 . 7 . 6 . 5 . 4 . 3 . 2 . 1)/(5 . 4 . 3 .  2 . 1)
= 8 . 7 . 6
= 336
8P3 = 336

Question 5.
6C3
Answer:
nCa = n!/a!(n – a)!
6C3 = 6!/3!(6 – 3)!
= 6!/3!(3)!
= 6 . 5 . 4 . 3 . 2 . 1/(3 . 2 . 1)(3 . 2 . 1)
= 2 . 5 . 2
= 20
6C3 = 20

Question 6.
12C7
Answer:
nCa = n!/a!(n – a)!
12C7 = 12!/7!(12 – 7)!
= 12!/7!(5)!
= 12 . 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(7 . 6 . 5 . 4 . 3 . 2 . 1)(5 . 4 . 3 . 2 . 1)
= 11 . 2 . 9 . 4 = 792
12C7 = 792

Question 7.
In the word PYRAMID, how many ways can you arrange
(a) all of the letters and
Answer:
We have to find the number of permutation all of the letters in a given word that will consist of 7 letters. We see that it matters that order of letters are important.
Number of permutations = (In the 1st place can be one of 7 letters) × (In the 2nd place can be one of 6 letters that is left)
(In the 3rd can be one of 5 letters that is left) × (In the 4th can be 4 letters that is left)
(In the 5th can be three letter that is left) × (In the 6th can be two letters that is left)
(In the 7th can be one letter that is left)
= 7 . 6 . 5 . 4 . 3 . 2 . 1 = 5040
Therefore, we have 5040 ways for arrange all of the letters in given word, that is PYRAMID, PYRAMDI, PYRADMI, ……., DIMARYP.

(b) 5 of the letters?
Answer:
We have to find the number of permutation all of the letters in a given word that will consist of 3 letters. We see that it matters that order of letters are important.
Number of permutations = (In the 1st place can be one of 7 letters) × (In the 2nd place can be one of 6 letters that is left)
(In the 3rd can be one of 5 letters that is left) × (In the 4th can be 4 letters that is left)
(In the 5th can be three letter that is left)
= 7 . 6 . 5 . 4 . 3
= 2520
Therefore, we have 2520 ways for arrange 5 of the letters in given word, that is PYRAM, PYRAI, PYRAD….

Question 8.
You find the probability P(A or B) by using hie equation P(A or B) = P(A) + P(B) – P(A and B). Describe why it is necessary to subtract P(A and B) when the events A and B are overlapping. Then describe why it is not necessary to subtract P(A and B) when the events A and B are disjoint.
Answer:
When the events are overlapping then P(A and B) ≠ 0. This means that the expression P(A) + P(B) already contains the probability P(A and B) and so P(A and B) is subtracted from their sum to evaluate P(A or B)
When the events are overlapping then P(A and B) = 0 and so the equation of P(A or B reduces to P(A) + P(B))

Question 9.
Is it possible to use the formula P(A and B) = P(A) • P(B/A) when events A and B are independent? Explain your reasoning.
Answer:
If events A and B are independent events, then
P(A and B) = P(A) . P(B)
Also, because event B is independent of event A then P(B|A) = P(A)
P(A and B) = P(A) . P(B) = P(A) . P(B|A)
where A and B are the independent events.

Question 10.
According to a survey, about 58% of families sit down tor a family dinner at least four times per week. You ask 5 randomly chosen families whether the have a family dinner at least four times per week.
a. Draw a histogram of the binomial distribution for the survey.
Answer:
p = P(Family have dinner four times per week) = 58% = 0.58
1 – p = P(Family have no dinner dour times per week) = 1 – 0.58 = 0.42
P (One Family have dinner four times per week) = 5C1 p¹(1 – p)5-1
= 0.09024
P (Two Family have dinner four times per week) =5C2 p²(1 – p)5-2
= 0.24923
P(Three Family have dinner four times per week) = 5C3p³(1 – p)5-3
= 0.34418
P (Four Family have dinner four times per week) =5C4 p4(1 – p)5-4
= 0.23765
P (Family have dinner four times per week) = 5C5(1 – p)5-5 = 0.06563

Big Ideas Math Answers Geometry Chapter 12 Probability img_4

b. What is the most likely outcome of the survey?
Answer:
P(Three families have dinner four times per week) = 0.34418
This probability is the highest, so we can conclude that the most likely outcome is that three of the five families have dinner four times per week.

c. What is the probability that at least 3 families have a family dinner four times per week?
Answer:
In this part we have to find the probability that,
P(At least 3 families have dinner four times per week)
= P(Three families have dinner four times per week) + P(Four families have dinner four times per week) + P(Five families have dinner four times per week)
= 0.34418 + 0.23765 + 0.06563
= 0.64746

Question 11.
You are choosing a cell phone company to sign with for the next 2 years. The three plans you consider are equally priced. You ask several of your neighbors whether they are satisfied with their current cell phone company. The table shows the results. According to this survey, which company should you choose?
Big Ideas Math Geometry Answers Chapter 12 Probability 95
Answer:
Big Ideas Math Answers Geometry Chapter 12 Probability img_5
To find the joint relative frequencies we divide each frequency by the total number of people in the survey. Also the marginal relative frequencies we find as the sum of each row and each column.
So, we can present a two way table that shows the joint and marginal relative frequencies.
Big Ideas Math Answers Geometry Chapter 12 Probability img_6
Finally, to get conditional relative frequencies we use the previous the marginal relative frequency of each row.
Big Ideas Math Answers Geometry Chapter 12 Probability img_7

Therefore we should choose company A.

Question 12.
The surface area of Earth is about 196.9 million square miles. The land area is about 57.5 million square miles, and the rest is water. What is the probability that a meteorite that reaches the surface of Earth will hit land? What is the probability that it will hit water?
Answer:
From the formula for geometric probability we get the probability that a meteorite that reaches the surface of Earth will hit lend is
P = The lend area/ The surface area of Earth = 57.5/196.9 = 0.292
Alos we know that the probability for complement of event A is
P(\(\bar{A}\)) = 1 – P(A)
In this case, complement of event {A meteorite will hit lend} is {A meteorite will hit water}.
P = 1 – 0.292 = 0.708

Question 13.
Consider a bag that contains all the chess pieces in a set, as shown in the diagram.
Big Ideas Math Geometry Answers Chapter 12 Probability 96
a. You choose one piece at random. Find the probability that you choose a black piece or a queen.
Answer:
The total number of pieces is 2 + 2 + 4 + 4 + 4 + 16 = 32.
Also we see that the number of black and white pieces are the same, 16 and there are one black queen.
A = {We choose a black piece}
B = {We choose a queen}
P(A) = Number of favorable outcomes/Total number of outcomes = Number of black pieces/Total number of pieces = 16/32 = 1/2
P(B) = Number of queens/Total number of pieces = 2/32 = 1/16
P(A and B) = Number of black queens/Total number of pieces = 1/32
P(A or B) = P(A) + P(B) – P(A and B)
= 1/2 + 1/16 – 1/32 = 17/32

b. You choose one piece at random, do not replace it, then choose a second piece at random. Find the probability that you choose a king, then a pawn.
Answer:
C = {We choose a king} and D = {We choose a pawn}
P(C) = Number of pawns/Total number of pieces = 16/32 = 1/2
P(C and D) = Number of pawns and king/Total number of pieces = (16 + 2)/32 = 9/16
We know that for two dependent events A and B probability that both occur is
P(A and B) = P(A)P(B|A)
P(D|C) = P(C and D)/P(C) = 1/2 × 16/9 = 8/9

Question 14.
Three volunteers are chosen at random from a group of 12 to help at a summer camp.
a. What is the probability that you, your brother, and your friend are chosen?
Answer:
nCa = n!/a!(n – a)!
12C3 = 12!/3!(12 – 3)!
= 12!/3!(9)!
= 12 . 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1/(3 . 2 . 1)(9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1)
= 12 . 11 . 10/3 . 2 . 1
= 2 . 11 . 10
12C3 = 220
P = Number of favorable outcomes/Total number of outcomes = 1/220

b. The first person chosen will be a counselor, the second will be a lifeguard, and the third will be a cook. What is the probability that you are the cook, your brother is the lifeguard, and your friend is the counselor?
Answer:
We know that number of permutations of n objects taken at a time (a ≤ n)
nPa = n!/(n – a)!
12P3 = 12!/(12 – 3)!
=  12!/9!
= (12 . 11 . 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1)/(9 . 8 . 7 . 6 . 5 . 4 . 3 .  2 . 1)
= 12 . 11 . 10
= 1320
12P3 = 1320
P = 1/1320

Probability Cumulative Assessment

Question 1.
According to a survey, 63% of Americans consider themselves sports fans. You randomly select 14 Americans to survey.
a. Draw a histogram of the binomial distribution of your survey.
Answer:
p = P(the American is a sports fan) = 63% = 0.63
1 – p = P(the American is not a sports fan) = 1 – 0.63= 0.37
P (0 success) = 12C0 p0(1 – p)12-0
= 12!/0!(12 – 0)! 1 . 0.3712
= 0.3712
P (One American is a sports fans) =12C1 p¹(1 – p)12-1
=0.000134506
P (Two Americans are sports fans) =12C2 p2(1 – p)12-2
=0.001259628
P (Three Americans are sports fans) =12C3 p3(1 – p)12-3
=0.007149239
P (Four American is a sports fans) =12C4 p4(1 – p)12-4
=0.027389316
P (Five American is a sports fans) =12C5 p5(1 – p)12-5
=0.07461738
P (Six American is a sports fans) =12C6 p6(1 – p)12-6
=0.148226418
P (Seven American is a sports fans) =12C7 p7(1 – p)12-7
=0.216330447
P (Eight American is a sports fans) =12C8 p8(1 – p)12-8
=0.230216523
P (Nine American is a sports fans) =12C9 p9(1 – p)12-9
=0.174217909
P (Ten American is a sports fans) =12C10p10(1 – p)12-10
=0.088992392
P (Eleven American is a sports fans) =12C11 p11(1 – p)12-11
=0.023755047
P (Twelve American is a sports fans) =12C1 p12(1 – p)12-12
=0.003909188
Big Ideas math Answers Geometry Chapter 12 Probability img_8

b. What is the most likely number of Americans who consider themselves Sports fans?
Answer:
From histogram in part a and based on the calculations in the first part
We know that,
P(Eight Americans are sports fans) = 0.23
This probability is highest, so we can conclude that the most likley outcome is that eight of the 12 selected Americans consider themselves sports fans.

c. What is the probability at least 7 Americans consider themselves sports fans?
Answer:
In this part we have to find dthe probability that
P(At least 7 Americans consider themselves sports fans)
= P(Seven American is a sports fans) + P(Eight American is a sports fans) + P(Nine American is a sports fans) + P(Ten American is a sports fans) + P(Eleven American is a sports fans) + P(Twelve American is a sports fans)
= 0.2163 + 0.2302 + 0.1742 + 0.08899 + 0.0275 + 0.0039
= 0.7412
Therefore, we can conclude that the probability that at least 7 Americans consider themselves sports fans is 0.7412

Question 2.
What is the arc length of \(\widehat{A B}\) ?
Big Ideas Math Geometry Answers Chapter 12 Probability 97
(A) 3.5 π cm
(B) 7 π cm
(C) 21 π cm
(D) 42 π cm
Answer:
Arc length of \(\widehat{A B}\) = 105/360 × 2π × 12 = 7π cm

Question 3.
you order a fruit smoothie made with 2 liquid ingredients and 3 fruit ingredients from the menu shown. How many different fruit smoothies can you order?
Big Ideas Math Geometry Answers Chapter 12 Probability 98
Answer:

Question 4.
The point (4, 3) is on a circle with center (- 2, – 5), What is the standard equation of the circle?
Answer:
Given,
The point (4, 3) is on a circle with center (- 2, – 5)
(x – h)² + (y – k)² = r²
Substitute the values in the given equation
(4 – (-2))² + (3 – (-5))² = r²
100 = r²
r = √100 = 10
General equation of circle
(x – h)² + (y – k)² = r²
Substitute the values in the given equation
(x – (-2))² + (y – (-5))² = 10²
(x + 2)² + (y + 5)² = 100

Question 5.
Find the length of each line segment with the given endpoints. Then order the line segments from shortest to longest.
a. A(1, – 5), B(4, 0)
Answer:
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dAB = √(4 – 1)² + (0 – (-5))²
= √34
= 5.83

b. C(- 4, 2), D(1, 4)
Answer:
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dCD = √(1 – (-4))² + (4 – 2)²
= √29
= 5.39

c. E(- 1, 1), F(- 2, 7)
Answer:
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dEF = √(-2 – (-1))² + (7 – 1)²
= √37
= 6.083

d. G(- 1.5, 0), H(4.5, 0)
Answer:
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dGH = √(4.5 – (-1.5))² + (0 – 0)²
= √36
= 6

e. J(- 7, – 8), K(- 3, – 5)
Answer:
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dJK = √(-3 – (-7))² + (-5 – (-8))²
= √25
= 5

f. L(10, – 2), M(9, 6)
Answer:
The general equation of distance between 2 points is
d = √(x2 – x1)² + (y2 – y1)²
dLM = √(9 – 10)² + (6 – (-2))²
= √65
= 8.06
Therefore, the line segments in ascending order are JK < CD < AB < GH < EF < LM

Question 6.
Use the diagram to explain why the equation is true.
P(A) + P(B) = P(A or B) + P(A and B)
Big Ideas Math Geometry Answers Chapter 12 Probability 99
Answer:
Given,
P(A) = 8/12
P(B) = 7/12
P(A or B) = 12/12
P(A and B) = 3/12
P(A) + P(B) = P(A or B) + P(A and B)
8/12 + 7/12 = 12/12 + 3/12
5/4 = 5/4

Question 7.
A plane intersects a cylinder. Which of the following cross sections cannot be formed by this intersection?
(A) line
(B) triangle
(C) rectangle
(D) circle
Answer:
If the plane is tangent to the curved surface of the cylinder, then the intersection is a line.
If the plane is cuts the cylinder parallel to its circular base, then the intersection is a circle.
If the plane cuts the cylinder perpendicular to its circular base and parallel to its curved surface, then the intersection is a rectangle.
The correct answer is option B.

Question 8.
A survey asked male and female students about whether the prefer to take gym class or choir. The table shows the results of the survey,
Big Ideas Math Geometry Answers Chapter 12 Probability 100
a. Complete the two-way table.
Answer:
From the given table we first find the joint frequencies.
Number of people who prefer gym class = 106 – 49 = 57
On the other hand, we know that 57 people prefer gym class, of which 23 was female.
Number of male who prefer the gym class = 57 – 23 = 34
Number of male who prefer the choir class = 50 – 34 = 16
Number of female who prefer the choir class = 49 – 16 = 33
Now, we will find the remaining marginal frequencies. Hence we know that 23 female prefer gym class and 33 female prefer choir class.
Total number of female is 23 + 33 = 56
BIM Answers Geometry Chapter 12 Probability img_9

b. What is the probability that a randomly selected student is female and prefers choir?
Answer:
P = Number of female who prefer choir class/Total number of students
= 33/106
= 0.31

c. What is the probability that a randomly selected male student prefers gym class?
Answer:
P(Prefer gym class|Male) = P(Male and prefer gym class)/P(Male)
= Number of male who prefer gym class/Number of male students
= 34/50
= 0.68

Question 9.
The owner of a lawn-mowing business has three mowers. As long as one of the mowers is working. the owner can stay productive. One of the mowers is unusable 10% of the time, one is unusable 8% of the time, and one is unusable 18% of the time.
a. Find the probability that all three mowers are unusable on a given day.
Answer:
P(A) = 10% = 0.1
P(B) = 8% = 0.08
P(C) = 18% = 0.18
\(\bar{A}\) = {The first mower is usable}
\(\bar{B}\) = {The second mower is usable}
\(\bar{C}\) = {The third mower is usable}
P(\(\bar{A}\)) = 1 – 0.1 = 0.9
P(\(\bar{B}\)) = 1 – 0.08 = 0.92
P(\(\bar{C}\)) = 1 – 0.18 = 0.82
Event that all three mowers are unusable on a given day is A and B and C. If we assume that the operation of the mowers are independent, then we get the probability that all three mowers are unusable on a given day as
P(A and B and C) = P(A) . P(B) . P(C) = 0.1 × 0.08 × 0.18 = 0.00144

b. Find the probability that at least one of the mowers is unusable on a given day.
Answer:
Event at least one of mowers is unusable on a given day is equivalent to event one of mowers is unusable on a given day or two of mowers is unusable on a given day.
A and \(\bar{B}\) and \(\bar{C}\) or \(\bar{A}\) and B and \(\bar{C}\) or \(\bar{A}\) and \(\bar{B}\) and C
On the other hand, event two of mowers is unusable on a given day is
A and B and \(\bar{C}\) or \(\bar{A}\) and B and C or A and \(\bar{B}\) and C
P(A and \(\bar{B}\) and \(\bar{C}\)) + P(\(\bar{A}\) and B and \(\bar{C}\)) + P(\(\bar{A}\) and \(\bar{B}\) and C)
= 0.1 × 0.92 × 0.82 + 0.9 × 0.08 × 0.82 + 0.9 × 0.92 × 0.18
= 0.28352
P(A and \(\bar{B}\) and \(\bar{C}\)) + P(\(\bar{A}\) and B and \(\bar{C}\)) + P(\(\bar{A}\) and \(\bar{B}\) and C)
= 0.1 × 0.08 × 0.82 + 0.9 × 0.08 × 0.18  + 0.1 × 0.92 × 0.18
= 0.03608
P(At least one of mowers is unusable) = P(One of mowers is unusable) + P(two of mowers is unusable)
= 0.28352 + 0.03608
= 0.3196

c. Suppose the least-reliable mower stops working completely. How does this affect the probability that the lawn-moving business can be productive on a given day?
Answer:
If the least-reliable mower stops working completely, this will not affect the probability that the lawn mowing business can be productive on a given day, because we know that as long as one of the mowers is working, the owner can stay productive.

Question 10.
You throw a dart at the board shown. Your dart is equally likely to hit any point inside the square board. What is the probability your dart lands in the yellow region?
Big Ideas Math Geometry Answers Chapter 12 Probability 101
(A) \(\frac{\pi}{36}\)
(B) \(\frac{\pi}{12}\)
(C) \(\frac{\pi}{9}\)
(D) \(\frac{\pi}{4}\)
Answer:
The total area of the given figure is (6(2))² = 144 sq. units
Area of the red circle = π × 2² = 4π
The area of the yellow ring is π × (4² – 2²) = 12π
The area of the blue ring is π × (6² – 4²) = 20π
Therefore, the probability of hitting the blue region is 12π/144 = π/12
Thus the correct answer is option B.

Conclusion:

Hope you are all satisfied with the given solutions. You can get free access to Download Big Ideas Math Geometry Answers Chapter 12 Probability pdf from here. Bookmark our Big Ideas Math Answers to get detailed solutions for all Geometry Chapters.

Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions

Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions

Are you struggling to solve the trigonometric ratios and functions in homework and assignments? Then, check out this Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions and solve your issues within seconds. The perfect way to learn the concepts of ch 9 is by practicing the questions given in the BIM Math Book Algebra 2 Chapter 9 Trigonometric Ratios and Functions Solution Key. Look no further and start your preparation by using this quick & ultimate guide. Constant practice can make you succeed in your math learning journey.

Big Ideas Math Book Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions

Enhance your math skills by taking the help of our provided Big Ideas Math Book Algebra 2 Ch 9 Trigonometric Ratios and Functions Solution key. After learning the concepts of the Algebra 2 9th chapters from the BIM Textbook Answers of Algebra 2, students can become math proficient. Access the Topicwise Big Ideas Math Book Algebra 2 Ch 9 Trigonometric Ratios and Functions Answers for free from the respective links presented below and prepare well for the exams.

Trigonometric Ratios and Functions Maintaining Mathematical Proficiency

Order the expressions by value from least to greatest.
Question 1.
log2 x
∣4∣, 2 − 9∣, ∣6 + 4∣, − ∣7∣
Answer:

Question 2.
∣9 − 3∣, ∣0∣, ∣−4∣, \(\frac{|-5|}{|2|}\)
Answer:

Question 3.
∣−83∣,∣−2 • 8 ∣, ∣9 − 1∣, ∣9∣ + ∣−2∣ − ∣1 ∣
Answer:

Question 4.
∣−4 + 20∣, −∣42∣, ∣5∣−∣3 • 2 ∣, ∣−15∣
Answer:

Find the missing side length of the triangle.
Question 5.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 1
Answer:

Question 6.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 2
Answer:

Question 7.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 3
Answer:

Question 8.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 4
Answer:

Question 9.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 5
Answer:

Question 10.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 6
Answer:

Question 11.
ABSTRACT REASONING
The line segments connecting the points (x1, y1), (x2, y1), and (x2, y2) form a triangle. Is the triangle a right triangle? Justify your answer.
Answer:

Trigonometric Ratios and Functions Mathematical Practices

Mathematically proficient students reason quantitatively by creating valid representations of problems.

Monitoring Progress

Find the exact coordinates of the point (x, y) on the unit circle.
Question 1.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 7.1 1
Answer:

Question 2.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 7.1 2
Answer:

Question 3.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 7.1 3
Answer:

Lesson 9.1 Right Triangle Trigonometry

Essential Question How can you find a trigonometric function of an acute angle θ?
Consider one of the acute angles θ of a right triangle. Ratios of a right triangle’s side lengths are used to define the six trigonometric functions, as shown.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 1

EXPLORATION 1

Trigonometric Functions of Special Angles
Work with a partner. Find the exact values of the sine, cosine, and tangent functions for the angles 30°, 45°, and 60° in the right triangles shown.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 2

EXPLORATION 2

Exploring Trigonometric Identities
Work with a partner.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 3
Use the definitions of the trigonometric functions to explain why each trigonometric identity is true.
a. sin θ = cos(90° − θ)
b. cos θ = sin(90° − θ)
c. sin θ =\(\frac{1}{\csc \theta}\)
d. tan θ = \(\frac{1}{\cot \theta}\)
Use the definitions of the trigonometric functions to complete each trigonometric identity.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 4

Communicate Your Answer

Question 3.
How can you find a trigonometric function of an acute angle θ?
Answer:

Question 4.
Use a calculator to find the lengths x and y of the legs of the right triangle shown.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 5
Answer:

Monitoring Progress

Evaluate the six trigonometric functions of the angle θ.
Question 1.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 6
Answer:

Question 2.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 7
Answer:

Question 3.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 8
Answer:

Question 4.
In a right triangle, θ is an acute angle and cos θ = \(\frac{7}{10}\). Evaluate the other five trigonometric functions of θ.
Answer:

Question 5.
Find the value of x for the right triangle shown.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 9
Answer:

Solve △ABC using the diagram at the left and the given measurements.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 10
Question 6.
B = 45°, c = 5
Answer:

Question 7.
A = 32°, b = 10
Answer:

Question 8.
A = 71°, c = 20
Answer:

Question 9.
B = 60°, a = 7
Answer:

Question 10.
In Example 5, find the distance between B and C.
Answer:

Question 11.
WHAT IF?
In Example 6, estimate the height of the parasailer above the boat when the angle of elevation is 38°.
Answer:

Right Triangle Trigonometry 9.1 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
In a right triangle, the two trigonometric functions of θ that are defined using the lengths of the hypotenuse and the side adjacent to θ are __________ and __________.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 1

Question 2.
VOCABULARY
Compare an angle of elevation to an angle of depression.
Answer:

Question 3.
WRITING
Explain what it means to solve a right triangle.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 3

Question 4.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 11
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, evaluate the six trigonometric functions of the angle θ.
Question 5.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 12
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 5

Question 6.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 13
Answer:

Question 7.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 14
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 7

Question 8.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 15
Answer:

Question 9.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 16
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 9

Question 10.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 17
Answer:

Question 11.
REASONING
Let θ be an acute angle of a right triangle. Use the two trigonometric functions tan θ = \(\frac{4}{9}\) and sec θ = \(\frac{\sqrt{97}}{9}\) to sketch and label the right triangle. Then evaluate the other four trigonometric functions of θ.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 11

Question 12.
ANALYZING RELATIONSHIPS
Evaluate the six trigonometric functions of the 90° − θ angle in Exercises 5–10. Describe the relationships you notice.
Answer:

In Exercises 13–18, let θ be an acute angle of a right triangle. Evaluate the other five trigonometric functions of θ.
Question 13.
sin θ = \(\frac{7}{11}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 13.1
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 13.2

Question 14.
cos θ = \(\frac{5}{12}\)
Answer:

Question 15.
tan θ = \(\frac{7}{6}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 15.1
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 15.2

Question 16.
csc θ = \(\frac{15}{8}\)
Answer:

Question 17.
sec θ = \(\frac{14}{9}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 17.1
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 17.2

Question 18.
cot θ = \(\frac{16}{11}\)
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in finding sin θ of the triangle below.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 18
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 19

Question 20.
ERROR ANALYSIS
Describe and correct the error in finding csc θ, given that θ is an acute angle of a right triangle and cos θ = \(\frac{7}{11}\).
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 19
Answer:

In Exercises 21–26, find the value of x for the right triangle.
Question 21.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 20
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 21

Question 22.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 21
Answer:

Question 23.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 22
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 23

Question 24.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 23
Answer:

Question 25.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 24
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 25

Question 26.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 25
Answer:

USING TOOLS In Exercises 27–32, evaluate the trigonometric function using a calculator. Round your answer to four decimal places.
Question 27.
cos 14°
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 27

Question 28.
tan 31°
Answer:

Question 29.
csc 59°
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 29

Question 30.
sin 23°
Answer:

Question 31.
cot 6°
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 31

Question 32.
sec 11°
Answer:

In Exercises 33–40, solve △ABC using the diagram and the given measurements.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 26.1
Question 33.
B = 36°, a = 23
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 33

Question 34.
A = 27°, b = 9
Answer:

Question 35.
A = 55°, a = 17
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 35

Question 36.
B = 16°, b = 14
Answer:

Question 37.
A = 43°, b = 31
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 37

Question 38.
B = 31°, a = 23
Answer:

Question 39.
B = 72°, c = 12.8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 39

Question 40.
A = 64°, a = 7.4
Answer:

Question 41.
MODELING WITH MATHEMATICS
To measure the width of a river, you plant a stake on one side of the river, directly across from a boulder. You then walk 100 meters to the right of the stake and measure a 79° angle between the stake and the boulder. What is the width w of the river?
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 26
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 41

Question 42.
MODELING WITH MATHEMATICS
Katoomba Scenic Railway in Australia is the steepest railway in the world. The railway makes an angle of about 52° with the ground. The railway extends horizontally about 458 feet. What is the height of the railway?
Answer:

Question 43.
MODELING WITH MATHEMATICS
A person whose eye level is 1.5 meters above the ground is standing 75 meters from the base of the Jin Mao Building in Shanghai, China. The person estimates the angle of elevation to the top of the building is about 80°. What is the approximate height of the building?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 43

Question 44.
MODELING WITH MATHEMATICS
The Duquesne Incline in Pittsburgh, Pennsylvania, has an angle of elevation of 30°. The track has a length of about 800 feet. Find the height of the incline.
Answer:

Question 45.
MODELING WITH MATHEMATICS
You are standing on the Grand View Terrace viewing platform at Mount Rushmore, 1000 feet from the base of the monument.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 27
a. You look up at the top of Mount Rushmore at an angle of 24°. How high is the top of the monument from where you are standing? Assume your eye level is 5.5 feet above the platform.
b. The elevation of the Grand View Terrace is 5280 feet. Use your answer in part (a) to find the elevation of the top of Mount Rushmore.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 45

Question 46.
WRITING
Write a real-life problem that can be solved using a right triangle. Then solve your problem.
Answer:

Question 47.
MATHEMATICAL CONNECTIONS
The Tropic of Cancer is the circle of latitude farthest north of the equator where the Sun can appear directly overhead. It lies 23.5° north of the equator, as shown.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 28
a. Find the circumference of the Tropic of Cancer using 3960 miles as the approximate radius of Earth.
b. What is the distance between two points on the Tropic of Cancer that lie directly across from each other?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 47

Question 48.
HOW DO YOU SEE IT?
Use the figure to answer each question.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 29
a. Which side is adjacent to θ?
b. Which side is opposite of θ?
c. Does cos θ = sin(90° − θ)? Explain.
Answer:

Question 49.
PROBLEM SOLVING
A passenger in an airplane sees two towns directly to the left of the plane.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 30
a. What is the distance d from the airplane to the first town?
b. What is the horizontal distance x from the airplane to the first town?
c. What is the distance y between the two towns? Explain the process you used to find your answer.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 49

Question 50.
PROBLEM SOLVING
You measure the angle of elevation from the ground to the top of a building as 32°. When you move 50 meters closer to the building, the angle of elevation is 53°. What is the height of the building?
Answer:

Question 51.
MAKING AN ARGUMENT
Your friend claims it is possible to draw a right triangle so the values of the cosine function of the acute angles are equal. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 51

Question 52.
THOUGHT PROVOKING
Consider a semicircle with a radius of 1 unit, as shown below. Write the values of the six trigonometric functions of the angle θ. Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 31
Answer:

Question 53.
CRITICAL THINKING
A procedure for approximating π based on the work of Archimedes is to inscribe a regular hexagon in a circle.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 32
a. Use the diagram to solve for x. What is the perimeter of the hexagon?
b. Show that a regular n-sided polygon inscribed in acircle of radius 1 has a perimeter of 2n • sin (\(\frac{180}{n}\))°.
c. Use the result from part (b) to find an expression in terms of n that approximates π. Then evaluate the expression when n= 50.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 53

Maintaining Mathematical Proficiency.

Perform the indicated conversion.
Question 54.
5 years to seconds
Answer:

Question 55.
12 pints to gallons
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 55

Question 56.
5.6 meters to millimeters
Answer:

Find the circumference and area of the circle with the given radius or diameter.
Question 57.
r = 6 centimeters
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 57

Question 58.
r = 11 inches
Answer:

Question 59.
d = 14 feet
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.1 a 59

Lesson 9.2 Angles and Radian Measure

Essential Question How can you find the measure of an angle in radians?
Let the vertex of an angle be at the origin, with one side of the angle on the positive x-axis. The radian measure of the angle is a measure of the intercepted arc length on a circle of radius 1. To convert between degree and radian measure, use the fact that \(\frac{\pi \text { radians }}{180^{\circ}}\) = 1.

EXPLORATION 1

Writing Radian Measures of Angles
Work with a partner. Write the radian measure of each angle with the given degree measure. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 1

EXPLORATION 2

Writing Degree Measures of Angles
Work with a partner. Write the degree measure of each angle with the given radian measure. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 2

Communicate Your Answer

Question 3.
How can you find the measure of an angle in radians?
Answer:

Question 4.
The figure shows an angle whose measure is 30 radians. What is the measure of the angle in degrees? How many times greater is 30 radians than 30 degrees? Justify your answers.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 3
Answer:

Monitoring Progress

Draw an angle with the given measure in standard position.
Question 1.
65°
Answer:

Question 2.
300°
Answer:

Question 3.
−120°
Answer:

Question 4.
−450°
Answer:

Find one positive angle and one negative angle that are coterminal with the given angle.
Question 5.
80°
Answer:

Question 6.
230°
Answer:

Question 7.
740°
Answer:

Question 8.
−135°
Answer:

Convert the degree measure to radians or the radian measure to degrees.
Question 9.
135°
Answer:

Question 10.
−40°
Answer:

Question 11.
\(\frac{5 \pi}{4}\)
Answer:

Question 12.
−6.28
Answer:

Question 13.
WHAT IF?
In Example 4, the outfield fence is 220 feet from home plate. Estimate the length of the outfield fence and the area of the field.
Answer:

Angles and Radian Measure 9.2 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
An angle is in standard position when its vertex is at the __________ and its __________ lies on the positive x-axis.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 1

Question 2.
WRITING
Explain how the sign of an angle measure determines its direction of rotation.
Answer:

Question 3.
VOCABULARY
In your own words, define a radian.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 3

Question 4.
WHICH ONE DOESN’T BELONG?
Which angle does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, draw an angle with the given measure in standard position.
Question 5.
110°
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 5

Question 6.
450°
Answer:

Question 7.
−900°
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 7

Question 8.
−10°
Answer:

In Exercises 9–12, find one positive angle and one negative angle that are coterminal with the given angle.
Question 9.
70°
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 9

Question 10.
255°
Answer:

Question 11.
−125°
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 11

Question 12.
−800°
Answer:

In Exercises 13–20, convert the degree measure to radians or the radian measure to degrees.
Question 13.
40°
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 13

Question 14.
315°
Answer:

Question 15.
−260°
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 15

Question 16.
−500°
Answer:

Question 17.
\(\frac{\pi}{9}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 17

Question 18.
\(\frac{3 \pi}{4}\)
Answer:

Question 19.
−5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 19

Question 20.
12
Answer:

Question 21.
WRITING
The terminal side of an angle in standard position rotates one-sixth of a revolution counterclockwise from the positive x-axis. Describe how to find the measure of the angle in both degree and radian measures.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 21

Question 22.
OPEN-ENDED
Using radian measure, give one positive angle and one negative angle that are coterminal with the angle shown. Justify your answers.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 5
Answer:

ANALYZING RELATIONSHIPS In Exercises 23–26, match the angle measure with the angle.
Question 23.
600°
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 23

Question 24.
\(-\frac{9 \pi}{4}\)
Answer:

Question 25.
\(\frac{5 \pi}{6}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 25

Question 26.
−240°
Answer:

Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 6

Question 27.
MODELING WITH MATHEMATICS
The observation deck of a building forms a sector with the dimensions shown. Find the length of the safety rail and the area of the deck.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 27

Question 28.
MODELING WITH MATHEMATICS
In the men’s shot put event at the 2012 Summer Olympic Games, the length of the winning shot was 21.89 meters. A shot put must land within a sector having a central angle of 34.92° to be considered fair.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 8
a. The officials draw an arc across the fair landing area, marking the farthest throw. Find the length of the arc.
b. All fair throws in the 2012 Olympics landed within a sector bounded by the arc in part (a). What is the area of this sector?
Answer:

Question 29.
ERROR ANALYSIS
Describe and correct the error in converting the degree measure to radians.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 29

Question 30.
ERROR ANALYSIS
Describe and correct the error in finding the area of a sector with a radius of 6 centimeters and a central angle of 40°.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 10
Answer:

Question 31.
PROBLEM SOLVING
When a CD player reads information from the outer edge of a CD, the CD spins about 200 revolutions per minute. At that speed, through what angle does a point on the CD spin in oneminute? Give your answer in both degree and radian measures.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 31

Question 32.
PROBLEM SOLVING
You work every Saturday from 9:00 A.M. to 5:00 P.M. Draw a diagram that shows the rotation completed by the hour hand of a clock during this time. Find the measure of the angle generated by the hour hand in both degrees and radians. Compare this angle with the angle generated by the minute hand from 9:00 A.M. to 5:00 P.M.
Answer:

USING TOOLS In Exercises 33–38, use a calculator to evaluate the trigonometric function.
Question 33.
cos \(\frac{4 \pi}{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 33

Question 34.
sin \(\frac{7 \pi}{8}\)
Answer:

Question 35.
csc \(\frac{10 \pi}{11}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 35

Question 36.
cot (− \(\frac{6 \pi}{5}\))
Answer:

Question 37.
cot(−14)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 37

Question 38.
cos 6
Answer:

Question 39.
MODELING WITH MATHEMATICS
The rear windshield wiper of a car rotates 120°, as shown. Find the area cleared by the wiper.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 39

Question 40.
MODELING WITH MATHEMATICS
A scientist performed an experiment to study the effects of gravitational force on humans. In order for humans to experience twice Earth’s gravity, they were placed in a centrifuge 58 feet long and spun at a rate of about 15 revolutions per minute.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 12
a. Through how many radians did the people rotate each second?
b. Find the length of the arc through which the people rotated each second.
Answer:

Question 41.
REASONING
In astronomy, the terminator is the day-night line on a planet that divides the planet into daytime and nighttime regions. The terminator moves across the surface of a planet as the planet rotates. It takes about 4 hours for Earth’s terminator to move across the continental United States. Through what angle has Earth rotated during this time? Give your answer in both degree and radian measures.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 13
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 41

Question 42.
HOW DO YOU SEE IT?
Use the graph to find the measure of θ. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 14
Answer:

Question 43.
MODELING WITH MATHEMATICS
A dartboard is divided into 20 sectors. Each sector is worth a point value from 1 to 20 and has shaded regions that double or triple this value. A sector is shown below. Find the areas of the entire sector, the double region, and the triple region.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 15
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 43

Question 44.
THOUGHT PROVOKING
π is an irrational number, which means that it cannot be written as the ratio of two whole numbers. π can, however, be written exactly as a continued fraction, as follows.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 16
Answer:

Question 45.
MAKING AN ARGUMENT
Your friend claims that when the arc length of a sector equals the radius, the area can be given by A = \(\frac{s^{2}}{2}\). Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 45

Question 46.
PROBLEM SOLVING
A spiral staircase has 15 steps. Each step is a sector with a radius of 42 inches and a central angle of \(\frac{\pi}{8}\).
a. What is the length of the arc formed by the outer edge of a step?
b. Through what angle would you rotate by climbing the stairs?
c. How many square inches of carpeting would you need to cover the 15 steps?
Answer:

Question 47.
MULTIPLE REPRESENTATIONS
There are 60 minutes in 1 degree of arc, and 60 seconds in 1 minute of arc. The notation 50° 30′ 10″ represents an angle with a measure of 50 degrees, 30 minutes, and 10 seconds.
a. Write the angle measure 70.55° using the notation above.
b. Write the angle measure 110° 45′ 30″ to the nearest hundredth of a degree. Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 47

Maintaining Mathematical Proficiency

Find the distance between the two points.
Question 48.
(1, 4), (3, 6)
Answer:

Question 49.
(−7, −13), (10, 8)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 49

Question 50.
(−3, 9), (−3, 16)
Answer:

Question 51.
(2, 12), (8, −5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 51

Question 52.
(−14, −22), (−20, −32)
Answer:

Question 53.
(4, 16), (−1, 34)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.2 a 53

Lesson 9.3 Trigonometric Functions of Any Angle

Essential Question How can you use the unit circle to define the trigonometric functions of any angle?
Let θ be an angle in standard position with (x, y) a point on the terminal side of θ and r = \(\sqrt{x^{2}+y^{2}}\) ≠ 0. The six trigonometric functions of θ are defined as shown.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 1

EXPLORATION 1

Writing Trigonometric Functions
Work with a partner. Find the sine, cosine, and tangent of the angle θ in standard position whose terminal side intersects the unit circle at the point (x, y) shown.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 2

Communicate Your Answer

Question 2.
How can you use the unit circle to define the trigonometric functions of any angle?
Answer:

Question 3.
For which angles are each function undefined? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 3.1
a. tangent
b. cotangent
c. secant
d. cosecant
Answer:

Monitoring Progress

Evaluate the six trigonometric functions of θ.
Question 1.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 3
Answer:

Question 2.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 4
Answer:

Question 3.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 5
Answer:

Question 4.
Use the unit circle to evaluate the six trigonometric functions of θ = 180º.
Answer:

Sketch the angle. Then find its reference angle.
Question 5.
210°
Answer:

Question 6.
−260°
Answer:

Question 7.
\(\frac{-7 \pi}{9}\)
Answer:

Question 8.
\(\frac{15 \pi}{4}\)
Answer:

Evaluate the function without using a calculator.
Question 9.
cos(−210º)
Answer:

Question 10.
sec \(\frac{11 \pi}{4}\)
Answer:

Question 11.
Use the model given in Example 5 to estimate the horizontal distance traveled by a track and field long jumper who jumps at an angle of 20° and with an initial speed of 27 feet per second.
Answer:

Trigonometric Functions of Any Angle 9.3 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
A(n) ___________ is an angle in standard position whose terminal side lies on an axis.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 1

Question 2.
WRITING
Given an angle θ in standard position with its terminal side in Quadrant III, explain how you can use a reference angle to find cos θ.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, evaluate the six trigonometric functions of θ.
Question 3.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 6
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 3

Question 4.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 7
Answer:

Question 5.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 8
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 5

Question 6.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 9
Answer:

Question 7.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 10
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 7

Question 8.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 11
Answer:

In Exercises 9–14, use the unit circle to evaluate the six trigonometric functions of θ.
Question 9.
θ = 0°
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 9

Question 10.
θ = 540°
Answer:

Question 11.
θ = \(\frac{\pi}{2}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 11

Question 12.
θ = \(\frac{7 \pi}{2}\)
Answer:

Question 13.
θ = −270°
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 13

Question 14.
θ = −2π
Answer:

In Exercises 15–22, sketch the angle. Then find its reference angle.
Question 15.
−100°
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 15

Question 16.
150°
Answer:

Question 17.
320°
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 17

Question 18.
−370°
Answer:

Question 19.
\(\frac{15 \pi}{4}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 19

Question 20.
\(\frac{8 \pi}{3}\)
Answer:

Question 21.
−\(\frac{5 \pi}{6}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 21

Question 22.
−\(\frac{13 \pi}{6}\)
Answer:

Question 23.
ERROR ANALYSIS
Let (−3, 2) be a point on the terminal side of an angle θ in standard position. Describe and correct the error in finding tan θ.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 12
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 23

Question 24.
ERROR ANALYSIS
Describe and correct the error in finding a reference angle θ′ for θ = 650°.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 13
Answer:

In Exercises 25–32, evaluate the function without using a calculator.
Question 25.
sec 135°
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 25

Question 26.
tan 240°
Answer:

Question 27.
sin(−150°)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 27

Question 28.
csc(−420°)
Answer:

Question 29.
tan (−\(\frac{3 \pi}{4}\))
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 29

Question 30.
cot (\(\frac{-8 \pi}{3}\))
Answer:

Question 31.
cos \(\frac{7 \pi}{4}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 31

Question 32.
sec \(\frac{11 \pi}{6}\)
Answer:

In Exercises 33–36, use the model for horizontal distance given in Example 5.
Question 33.
You kick a football at an angle of 60° with an initial speed of 49 feet per second. Estimate the horizontal distance traveled by the football.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 33

Question 34.
The “frogbot” is a robot designed for exploring rough terrain on other planets. It can jump at a 45° angle with an initial speed of 14 feet per second. Estimate the horizontal distance the frogbot can jump on Earth.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 14
Answer:

Question 35.
At what speed must the in-line skater launch himself off the ramp in order to land on the other side of the ramp?
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 15
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 35

Question 36.
To win a javelin throwing competition, your last throw must travel a horizontal distance of at least 100 feet. You release the javelin at a 40° angle with an initial speed of 71 feet per second. Do you win the competition? Justify your answer.
Answer:

Question 37.
MODELING WITH MATHEMATICS
A rock climber is using a rock climbing treadmill that is 10 feet long. The climber begins by lying horizontally on the treadmill, which is then rotated about its midpoint by 110° so that the rock climber is climbing toward the top. If the midpoint of the treadmill is 6 feet above the ground, how high above the ground is the top of the treadmill?
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 16
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 37

Question 38.
REASONING
A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet above the ground, and rotate 255°counterclockwise before the ride temporarily stops. How high above the ground are you when the ride stops? If the radius of the Ferris wheel is doubled, is your height above the ground doubled? Explain your reasoning.
Answer:

Question 39.
DRAWING CONCLUSIONS
A sprinkler at ground level is used to water a garden. The water leaving the sprinkler has an initial speed of 25 feet per second.
a. Use the model for horizontal distance given in Example 5 to complete the table.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 17
b. Which value of θ appears to maximize the horizontal distance traveled by the water? Use the model for horizontal distance and the unit circle to explain why your answer makes sense.
c. Compare the horizontal distance traveled by the water when θ = (45 − k)° with the distance when θ = (45 + k)°, for 0 < k < 45.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 39

Question 40.
MODELING WITH MATHEMATICS
Your school’s marching band is performing at halftime during a football game. In the last formation, the band members form a circle 100 feet wide in the center of the field. You start at a point on the circle 100 feet from the goal line, march 300° around the circle, and then walk toward the goal line to exit the field. How far from the goal line are you at the point where you leave the circle?
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 18
Answer:

Question 41.
ANALYZING RELATIONSHIPS
Use symmetry and the given information to label the coordinates of the other points corresponding to special angles on the unit circle.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 19
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 41

Question 42.
THOUGHT PROVOKING
Use the interactive unit circle tool at BigIdeasMath.com to describe all values of θ for each situation.
a. sin θ > 0, cos θ < 0, and tan θ > 0
b. sin θ > 0, cos θ < 0, and tan θ < 0
Answer:

Question 43.
CRITICAL THINKING
Write tan θ as the ratio of two other trigonometric functions. Use this ratio to explain why tan 90° is undefined but cot 90° = 0.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 43

Question 44.
HOW DO YOU SEE IT?
Determine whether each of the six trigonometric functions of θ is positive, negative, or zero. Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 20
Answer:

Question 45.
USING STRUCTURE
A line with slope m passes through the origin. An angle θ in standard position has a terminal side that coincides with the line. Use a trigonometric function to relate the slope of the line to the angle.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 45

Question 46.
MAKING AN ARGUMENT
Your friend claims that the only solution to the trigonometric equation tan θ = \(\sqrt{3}\) is θ= 60°. Is your friend correct? Explain your reasoning.
Answer:

Question 47.
PROBLEM SOLVING
When two atoms in a molecule are bonded to a common atom, chemists are interested in both the bond angle and the lengths of the bonds. An ozone molecule is made up of two oxygen atoms bonded to a third oxygen atom, as shown.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 21
a. In the diagram, coordinates are given in picometers (pm). (Note: 1 pm = 10−12 m) Find the coordinates (x, y) of the center of the oxygen atom in Quadrant II.
b. Find the distance d (in picometers) between the centers of the two unbonded oxygen atoms.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 47

Question 48.
MATHEMATICAL CONNECTIONS
The latitude of a point on Earth is the degree measure of the shortest arc from that point to the equator. For example, the latitude of point P in the diagram equals the degree measure of arc PE. At what latitude θ is the circumference of the circle of latitude at P half the distance around the equator?
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 22
Answer:

Maintaining Mathematical Proficiency

Find all real zeros of the polynomial function.
Question 49.
f (x) = x4 + 2x3 + x2 + 8x − 12
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 49

Question 50.
f(x) = x5 + 4x4 − 14x3 − 14x2 − 15x− 18
Answer:

Graph the function.
Question 51.
f(x) = 2(x+ 3)2 (x − 1)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 51.1
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 51.2

Question 52.
f(x) = \(\frac{1}{2}\) (x − 4)(x + 5)(x + 9)
Answer:

Question 53.
f(x) = x2(x + 1)3 (x − 2)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 53.1
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.3 a 53.2

Lesson 9.4 Graphing Sine and Cosine Functions

Essential Question What are the characteristics of the graphs of the sine and cosine functions?

EXPLORATION 1

Graphing the Sine FunctionWork with a partner.
a. Complete the table for y= sin x, where x is an angle measure in radians.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 1
b. Plot the points (x, y) from part (a). Draw a smooth curve through the points to sketch the graph of y = sin x.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 2
c. Use the graph to identify the x-intercepts, the x-values where the local maximums and minimums occur, and the intervals for which the function is increasing or decreasing over −2π ≤ x ≤ 2π. Is the sine function even, odd, or neither?

EXPLORATION 2

Graphing the Cosine Function
Work with a partner.
a. Complete a table for y= cos x using the same values of x as those used in Exploration 1.
b. Plot the points (x, y) from part (a) and sketch the graph of y= cos x.
c. Use the graph to identify the x-intercepts, the x-values where the local maximums and minimums occur, and the intervals for which the function is increasing or decreasing over −2π ≤ x ≤ 2π. Is the cosine function even, odd, or neither?

Communicate Your Answer

Question 3.
What are the characteristics of the graphs of the sine and cosine functions?
Answer:

Question 4.
Describe the end behavior of the graph of y = sin x
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 3
Answer:

Monitoring Progress

Identify the amplitude and period of the function. Then graph the function and describe the graph of g as a transformation of the graph of its parent function.
Question 1.
g(x) = \(\frac{1}{4}\)sin x
Answer:

Question 2.
g(x) = cos 2x
Answer:

Question 3.
g(x) = 2 sin πx
Answer:

Question 4.
g(x) = \(\frac{1}{3}\) cos \(\frac{1}{2}\)x
Answer:

Graph the function.
Question 5.
g(x) = cos x+ 4
Answer:

Question 6.
g(x) = \(\frac{1}{2}\)sin (x − \(\left.\frac{\pi}{2}\right\))
Answer:

Question 7.
g(x) = sin(x + π) − 1
Answer:

Graph the function.
Question 8.
g(x) = −cos (x + \(\left.\frac{\pi}{2}\right\))
Answer:

Question 9.
g(x) = −3 sin \(\frac{1}{2}\)x + 2
Answer:

Question 10.
g(x) = −2 cos 4x − 1
Answer:

Graphing Sine and Cosine Functions 9.4 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
The shortest repeating portion of the graph of a periodic function is called a(n) _________.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 1

Question 2.
WRITING
Compare the amplitudes and periods of the functions y = \(\frac{1}{2}\)cos x and y = 3 cos 2x.
Answer:

Question 3.
VOCABULARY
What is a phase shift? Give an example of a sine function that has a phase shift.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 3

Question 4.
VOCABULARY
What is the midline of the graph of the function y = 2 sin 3(x + 1) − 2?
Answer:

Monitoring Progress and Modeling with Mathematics

USING STRUCTURE In Exercises 5–8, determine whether the graph represents a periodic function. If so, identify the period.
Question 5.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 4
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 5

Question 6.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 5
Answer:

Question 7.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 6
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 7

Question 8.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 7
Answer:

In Exercises 9–12, identify the amplitude and period of the graph of the function.
Question 9.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 8
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 9

Question 10.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 9
Answer:

Question 11.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 10
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 11

Question 12.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 11
Answer:

In Exercises 13–20, identify the amplitude and period of the function. Then graph the function and describe the graph of g as a transformation of the graph of its parent function.
Question 13.
g(x) = 3 sin x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 13

Question 14.
g(x) = 2 sin x
Answer:

Question 15.
g(x) = cos 3x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 15

Question 16.
g(x) = cos 4x
Answer:

Question 17.
g(x) = sin 2πx
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 17

Question 18.
g(x) = 3 sin 2x
Answer:

Question 19.
g(x) = \(\frac{1}{3}\)cos 4x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 19

Question 20.
g(x) = \(\frac{1}{2}\)cos 4πx
Answer:

Question 21.
ANALYZING EQUATIONS
Which functions have an amplitude of 4 and a period of 2?
A. y = 4 cos 2x
B. y = −4 sin πx
C. y = 2 sin 4x
D. y = 4 cos πx
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 21

Question 22.
WRITING EQUATIONS
Write an equation of the form y = a sin bx, where a > 0 and b > 0, so that the graph has the given amplitude and period.
a. amplitude: 1
period: 5
b. amplitude: 10
period: 4
c. amplitude: 2
period: 2π
d. amplitude: \(\frac{1}{2}\)
period: 3π
Answer:

Question 23.
MODELING WITH MATHEMATICS
The motion of a pendulum can be modeled by the function d = 4 cos 8πt, where d is the horizontal displacement (in inches) of the pendulum relative to its position at rest and t is the time (in seconds). Find and interpret the period and amplitude in the context of this situation. Then graph the function.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 23

Question 24.
MODELING WITH MATHEMATICS
A buoy bobs up and down as waves go past. The vertical displacement y (in feet) of the buoy with respect to sea level can be modeled by y = 1.75 cos \(\frac{\pi}{3}\)t, where t is the time (in seconds). Find and interpret the period and amplitude in the context of the problem. Then graph the function.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 12
Answer:

In Exercises 25–34, graph the function.
Question 25.
g(x) = sin x + 2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 25

Question 26.
g(x) = cos x − 4
Answer:

Question 27.
g(x) = cos (x − \(\frac{\pi}{2}\))
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 27

Question 28.
g(x) = sin (x + \(\frac{\pi}{4}\))
Answer:

Question 29.
g(x) = 2 cos x − 1
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 29

Question 30.
g(x) = 3 sin x + 1
Answer:

Question 31.
g(x) = sin 2(x + π)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 31

Question 32.
g(x) = cos 2(x − π)
Answer:

Question 33.
g(x) = sin \(\frac{1}{2}\)(x + 2π) + 3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 33

Question 34.
g(x) = cos \(\frac{1}{2}\)(x − 3π) − 5
Answer:

Question 35.
ERROR ANALYSIS
Describe and correct the error in finding the period of the function y = sin \(\frac{2}{3}\)x.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 13
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 35

Question 36.
ERROR ANALYSIS
Describe and correct the error in determining the point where the maximum value of the function y = 2 sin (x − \(\frac{\pi}{4}\)) occurs.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 14
Answer:

USING STRUCTURE In Exercises 37–40, describe the transformation of the graph of f represented by the function g
Question 37.
f(x) = cos x, g(x) = 2 cos (x − \(\frac{\pi}{2}\)) + 1
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 37

Question 38.
f(x) = sin x, g(x) = 3 sin (x + \(\frac{\pi}{4}\)) − 2
Answer:

Question 39.
f(x) = sin x, g(x) = sin 3(x + 3π) − 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 39

Question 40.
f(x) = cos x, g(x) = cos 6(x − π) + 9
Answer:

In Exercises 41–48, graph the function.
Question 41.
g(x) = −cos x + 3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 41

Question 42.
g(x) = −sin x − 5
Answer:

Question 43.
g(x) = −sin \(\frac{1}{2}\)x − 2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 43

Question 44.
g(x) = −cos 2x + 1
Answer:

Question 45.
g(x) = −sin(x − π) + 4
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 45

Question 46.
g(x) = −cos(x + π) − 2
Answer:

Question 47.
g(x) = −4 cos (x + \(\frac{\pi}{4}\)) − 1
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 47

Question 48.
g(x) = −5 sin (x − \(\frac{\pi}{2}\)) + 3
Answer:

Question 49.
USING EQUATIONS
Which of the following is a point where the maximum value of the graph of y =−4 cos (x − \(\frac{\pi}{2}\))occurs?
A. (−\(\frac{\pi}{2}\), 4 )
B. (\(\frac{\pi}{2}\), 4 )
C. (0, 4)
D. (π, 4)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 49

Question 50.
ANALYZING RELATIONSHIPS
Match each function with its graph. Explain your reasoning.
a. y = 3 + sin x
b. y = −3 + cos x
c. y = sin 2(x − \(\frac{\pi}{2}\))
d. y = cos 2 (x − \(\frac{\pi}{2}\))
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 15
Answer:

WRITING EQUATIONS In Exercises 51–54, write a rule for g that represents the indicated transformations of the graph of f.
Question 51.
f(x) = 3 sin x; translation 2 units up and π units right
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 51

Question 52.
f(x) = cos 2πx; translation 4 units down and 3 units left
Answer:

Question 53.
f(x) = \(\frac{1}{3}\)cos πx; translation 1 unit down, followed by a reflection in the line y =−1
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 53

Question 54.
f(x) = \(\frac{1}{2}\) sin 6x; translation \(\frac{3}{2}\) units down and 1 unit right, followed by a reflection in the line y = −\(\frac{3}{2}\)
Answer:

Question 55.
MODELING WITH MATHEMATICS
The height h(in feet) of a swing above the ground can be modeled by the function h = −8 cos θ+ 10, where the pivot is 10 feet above the ground, the rope is 8 feet long, and θ is the angle that the rope makes with the vertical. Graph the function. What is the height of the swing when θ is 45°?
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 16
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 55

Question 56.
DRAWING A CONCLUSION
In a particular region, the population L (in thousands) of lynx (the predator) and the population H (in thousands) of hares (the prey) can be modeled by the equations
L = 11.5 + 6.5 sin\(\frac{\pi}{5}\)t
H = 27.5 + 17.5 cos \(\frac{\pi}{5}\)t
where t is the time in years.
a. Determine the ratio of hares to lynx when t = 0, 2.5, 5, and 7.5 years.
b. Use the figure to explain how the changes in the two populations appear to be related.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 17
Answer:

Question 57.
USING TOOLS
The average wind speed s (in miles per hour) in the Boston Harbor can be approximated by s = 3.38 sin\(\frac{\pi}{180}\)(t + 3) + 11.6 where t is the time in days and t = 0 represents January 1. Use a graphing calculator to graph the function. On which days of the year is the average wind speed 10 miles per hour? Explain your reasoning.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 57

Question 58.
USING TOOLS
The water depth d (in feet) for the Bay of Fundy can be modeled by d = 35 − 28 cos \(\frac{\pi}{6.2}\)t, where t is the time in hours and t = 0 represents midnight. Use a graphing calculator to graph the function. At what time(s) is the water depth 7 feet? Explain.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 18
Answer:

Question 59.
MULTIPLE REPRESENTATIONS
Find the average rate of change of each function over the interval 0 < x < π.
a. y = 2 cos x
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 19
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 59

Question 60.
REASONING
Consider the functions y = sin(−x) and y = cos(−x).
a. Construct a table of values for each equation using the quadrantal angles in the interval −2π ≤ x ≤ 2π.
b. Graph each function.
c. Describe the transformations of the graphs of the parent functions.
Answer:

Question 61.
MODELING WITH MATHEMATICS
You are riding a Ferris wheel that turns for 180 seconds. Your height h (in feet) above the ground at any time t (in seconds) can be modeled by the equation h = 85 sin\(\frac{\pi}{20}\)(t − 10) + 90.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 20
a. Graph the function.
b. How many cycles does the Ferris wheel make in 180 seconds?
c. What are your maximum and minimum heights?
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 61

Question 62.
HOW DO YOU SEE IT?
Use the graph to answer each question.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 21
a. Does the graph represent a function of the form f(x) = a sin bx or f(x) =a cos bx? Explain.
b. Identify the maximum value, minimum value, period, and amplitude of the function.
Answer:

Question 63.
FINDING A PATTERN
Write an expression in terms of the integer n that represents all the x-intercepts of the graph of the function y = cos 2x. Justify your answer.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 63

Question 64.
MAKING AN ARGUMENT
Your friend states that for functions of the form y = a sin bx and y = a cos bx, the values of a and b affect the x-intercepts of the graph of the function. Is your friend correct? Explain.
Answer:

Question 65.
CRITICAL THINKING
Describe a transformation of the graph of f(x) = sin x that results in the graph of g(x) = cos x.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 65

Question 66.
THOUGHT PROVOKING
Use a graphing calculator to find a function of the form y = sin b1x + cos b2xwhose graph matches that shown below.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 22
Answer:

Question 67.
PROBLEM SOLVING
For a person at rest, the blood pressure P (in millimeters of mercury) at time t (in seconds) is given by the function
P = 100 − 20 cos \(\frac{8 \pi}{3}\)t.
Graph the function. One cycle is equivalent to one heartbeat. What is the pulse rate (in heartbeats per minute) of the person?
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 23
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 67

Question 68.
PROBLEM SOLVING
The motion of a spring can be modeled by y = A cos kt, where y is the vertical displacement (in feet) of the spring relative to its position at rest, A is the initial displacement (in feet), k is a constant that measures the elasticity of the spring, and t is the time (in seconds).
a. You have a spring whose motion can be modeled by the function y= 0.2 cos 6t. Find the initial displacement and the period of the spring. Then graph the function.
b. When a damping force is applied to the spring, the motion of the spring can be modeled by the function y = 0.2e−4.5t cos 4t. Graph this function. What effect does damping have on the motion?
Answer:

Maintaining Mathematical Proficiency

Simplify the rational expression, if possible.
Question 69.
\(\frac{x^{2}+x-6}{x+3}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 69

Question 70.
\(\frac{x^{3}-2 x^{2}-24 x}{x^{2}-2 x-24}\)
Answer:

Question 71.
\(\frac{x^{2}-4 x-5}{x^{2}+4 x-5}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 71

Question 72.
\(\frac{x^{2}-16}{x^{2}+x-20}\)
Answer:

Find the least common multiple of the expressions.
Question 73.
2x, 2(x − 5)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 73

Question 74.
x2 − 4, x + 2
Answer:

Question 75.
x2 + 8x + 12, x + 6
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 a 75

Trigonometric Ratios and Functions Study Skills: Form a Final Exam Study Group

9.1–9.4 What Did You Learn?

Core Vocabulary
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 24

Core Concepts
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 25

Mathematical Practices
Question 1.
Make a conjecture about the horizontal distances traveled in part (c) of Exercise 39 on page 483.
Answer:

Question 2.
Explain why the quantities in part (a) of Exercise 56 on page 493 make sense in the context of the situation.
Answer:

Study Skills: Form a Final Exam Study Group

Form a study group several weeks before the final exam. The intent of this group is to review what you have already learned while continuing to learn new material.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.4 26

Trigonometric Ratios and Functions 9.1–9.4 Quiz

Question 1.
In a right triangle, θ is an acute angle and sin θ = \(\frac{1}{2}\). Evaluate the other five trigonometric functions of θ.
Answer:

Find the value of x for the right triangle.
Question 2.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions q 1
Answer:

Question 3.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions q 2
Answer:

Question 4.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions q 3
Answer:

Draw an angle with the given measure in standard position. Then find one positive angle and one negative angle that are coterminal with the given angle.
Question 5.
40°
Answer:

Question 6.
\(\frac{5 \pi}{6}\)
Answer:

Question 7.
−960°
Answer:

Convert the degree measure to radians or the radian measure to degrees.
Question 8.
\(\frac{3 \pi}{10}\)
Answer:

Question 9.
−60°
Answer:

Question 10.
72°
Answer:

Evaluate the six trigonometric functions of θ.
Question 11.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions q 4
Answer:

Question 12.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions q 5
Answer:

Question 13.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions q 6
Answer:

Question 14.
Identify the amplitude and period of g(x) = 3 sin x. Then graph the function and describe the graph of g as a transformation of the graph of f (x) = sin x.
Answer:

Question 15.
Identify the amplitude and period of g(x) = cos 5πx + 3. Then graph the function and describe the graph of g as a transformation of the graph of f(x) = cos x.
Answer:

Question 16.
You are flying a kite at an angle of 70°. You have let out a total of 400 feet of string and are holding the reel steady 4 feet above the ground.
Big Ideas Math Answer Key Algebra 2 Chapter 9 Trigonometric Ratios and Functions q 7
a. How high above the ground is the kite?
b. A friend watching the kite estimates that the angle of elevation to the kite is 85°. How far from your friend are you standing?
Answer:

Question 17.
The top of the Space Needle in Seattle, Washington, is a revolving, circular restaurant. The restaurant has a radius of 47.25 feet and makes one complete revolution in about an hour. You have dinner at a window table from 7:00 P.M. to 8:55 P.M. Compare the distance you revolve with the distance of a person seated 5 feet away from the windows.
Answer:

Lesson 9.5 Graphing Other Trigonometric Functions

Essential Question What are the characteristics of the graph of the tangent function?

EXPLORATION 1

Graphing the Tangent Function
Work with a partner. a. Complete the table for y = tan x, where x is an angle measure in radians.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 1
b. The graph of y = tan x has vertical asymptotes at x-values where tan x is undefined. Plot the points (x, y) from part (a). Then use the asymptotes to sketch the graph of y = tan x.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 2
c. For the graph of y = tan x, identify the asymptotes, the x-intercepts, and the intervals for which the function is increasing or decreasing over −\(\frac{\pi}{2}\) ≤ x ≤ \(\frac{3 \pi}{2}\). Is the tangent function even, odd, or neither?

Communicate Your Answer

Question 2.
What are the characteristics of the graph of the tangent function?
Answer:

Question 3.
Describe the asymptotes of the graph of y = cot x on the interval −\(\frac{\pi}{2}\) < x < \(\frac{3 \pi}{2}\).
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 3
Answer:

Monitoring Progress

Graph one period of the function. Describe the graph of g as a transformation of the graph of its parent function.
Question 1.
g(x) = tan 2x
Answer:

Question 2.
g(x) = \(\frac{1}{3}\)cot x
Answer:

Question 3.
g(x) = 2 cot 4x
Answer:

Question 4.
g(x) = 5 tan πx
Answer:

Graph one period of the function. Describe the graph of g as a transformation of the graph of its parent function.
Question 5.
g(x) = csc 3x
Answer:

Question 6.
g(x) = \(\frac{1}{2}\)sec x
Answer:

Question 7.
g(x) = 2 csc 2x
Answer:

Question 8.
g(x) = 2 sec πx
Answer:

Graphing Other Trigonometric Functions 9.5 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Explain why the graphs of the tangent, cotangent, secant, and cosecant functions do not have an amplitude.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 1

Question 2.
COMPLETE THE SENTENCE
The _______ and _______ functions are undefined for x-values at which sin x = 0.
Answer:

Question 3.
COMPLETE THE SENTENCE
The period of the function y = sec x is _____, and the period of y = cot x is _____.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 3

Question 4.
WRITING
Explain how to graph a function of the form y = a sec bx.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, graph one period of the function. Describe the graph of gas a transformation of the graph of its parent function.
Question 5.
g(x) = 2 tan x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 5

Question 6.
g(x) = 3 tan x
Answer:

Question 7.
g(x) = cot 3x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 7

Question 8.
g(x) = cot 2x
Answer:

Question 9.
g(x) = 3 cot \(\frac{1}{4}\)x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 9.1
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 9.2

Question 10.
g(x) = 4 cot\(\frac{1}{2}\)x
Answer:

Question 11.
g(x) = \(\frac{1}{2}\)tan πx
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 11

Question 12.
g(x) = \(\frac{1}{3}\) tan 2πx
Answer:

Question 13.
ERROR ANALYSIS
Describe and correct the error in finding the period of the function y = cot 3x.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 13

Question 14.
ERROR ANALYSIS
Describe and correct the error in describing the transformation of f(x) = tan x represented by g(x) = 2 tan 5x.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 5
Answer:

Question 15.
ANALYZING RELATIONSHIPS
Use the given graph to graph each function.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 6
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 15

Question 16.
USING EQUATIONS
Which of the following are asymptotes of the graph of y = 3 tan 4x?
A. x = \(\frac{\pi}{8}\)
B. x = \(\frac{\pi}{4}\)
C. x = 0
D. x = −\(\frac{5 \pi}{8}\)
Answer:

In Exercises 17–24, graph one period of the function. Describe the graph of gas a transformation of the graph of its parent function.
Question 17.
g(x) = 3 csc x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 17

Question 18.
g(x) = 2 csc x
Answer:

Question 19.
g(x) = sec 4x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 19

Question 20.
g(x) = sec 3x
Answer:

Question 21.
g(x) = \(\frac{1}{2}\)sec πx
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 21

Question 22.
g(x) = \(\frac{1}{4}\) sec 2πx
Answer:

Question 23.
g(x) = csc \(\frac{\pi}{2}\)x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 23

Question 24.
g(x) = csc \(\frac{\pi}{4}\)x
Answer:

ATTENDING TO PRECISION In Exercises 25–28, use the graph to write a function of the form y = a tan bx.
Question 25.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 7
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 25

Question 26.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 8
Answer:

Question 27.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 9
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 27

Question 28.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 10
Answer:

USING STRUCTURE In Exercises 29–34, match the equation with the correct graph. Explain your reasoning.
Question 29.
g(x) = 4 tan x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 29

Question 30.
g(x) = 4 cot x
Answer:

Question 31.
g(x) = 4 csc πx
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 31

Question 32.
g(x) = 4 sec πx
Answer:

Question 33.
g(x) = sec 2x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 33

Question 34.
g(x) = csc 2x
Answer:

Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 11

Question 35.
WRITING
Explain why there is more than one tangent function whose graph passes through the origin and has asymptotes at x = −π and x = π.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 35

Question 36.
USING EQUATIONS
Graph one period of each function. Describe the transformation of the graph of its parent function.
a. g(x) = sec x + 3
b. g(x) = csc x − 2
c. g(x) = cot(x − π)
d. g(x) = −tan x
Answer:

WRITING EQUATIONS In Exercises 37–40, write a rule for g that represents the indicated transformation of the graph of f.
Question 37.
f(x) = cot 2x; translation 3 units up and \(\frac{\pi}{2}\) units left
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 37

Question 38.
f(x) = 2 tan x; translation π units right, followed by a horizontal shrink by a factor of \(\frac{1}{3}\)
Answer:

Question 39.
f(x) = 5 sec (x − π); translation 2 units down, followed by a reflection in the x-axis
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 39

Question 40.
f(x) = 4 csc x; vertical stretch by a factor of 2 and a reflection in the x-axis
Answer:

Question 41.
MULTIPLE REPRESENTATIONS
Which function has a greater local maximum value? Which has a greater local minimum value? Explain.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 12
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 41

Question 42.
ANALYZING RELATIONSHIPS
Order the functions from the least average rate of change to the greatest average rate of change over the interval −\(\frac{\pi}{4}\) < x < \(\frac{\pi}{4}\).
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 13
Answer:

Question 43.
REASONING
You are standing on a bridge 140 feet above the ground. You look down at a car traveling away from the underpass. The distance d (in feet) the car is from the base of the bridge can be modeled by d= 140 tan θ. Graph the function. Describe what happens to θ as d increases
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 14
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 43

Question 44.
USING TOOLS
You use a video camera to pan up the Statue of Liberty. The height h (in feet) of the part of the Statue of Liberty that can be seen through your video camera after time t (in seconds) can be modeled by h= 100 tan \(\frac{\pi}{36}\)t. Graph the function using a graphing calculator. What viewing window did you use? Explain.
Answer:

Question 45.
MODELING WITH MATHEMATICS
You are standing 120 feet from the base of a 260-foot building. You watch your friend go down the side of the building in a glass elevator.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 15
a. Write an equation that gives the distance d (in feet) your friend is from the top of the building as a function of the angle of elevation θ.
b. Graph the function found in part (a). Explain how the graph relates to this situation.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 45

Question 46.
MODELING WITH MATHEMATICS
You are standing 300 feet from the base of a 200-foot cliff. Your friend is rappelling down the cliff.
a. Write an equation that gives the distance d(in feet) your friend is from the top of the cliff as a function of the angle of elevation θ.
b. Graph the function found in part (a).
c. Use a graphing calculator to determine the angle of elevation when your friend has rappelled halfway down the cliff.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 16
Answer:

Question 47.
MAKING AN ARGUMENT
Your friend states that it is not possible to write a cosecant function that has the same graph as y = sec x. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 47

Question 48.
HOW DO YOU SEE IT?
Use the graph to answer each question.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 17
a. What is the period of the graph?
b. What is the range of the function?
c. Is the function of the form f(x) = a csc bx or f(x) = a sec bx? Explain.
Answer:

Question 49.
ABSTRACT REASONING
Rewrite a sec bx in terms of cos bx. Use your results to explain the relationship between the local maximums and minimums of the cosine and secant functions.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 49

Question 50.
THOUGHT PROVOKING
A trigonometric equation that is true for all values of the variable for which both sides of the equation are defined is called a trigonometric identity.Use a graphing calculator to graph the function
y = \(\frac{1}{2}\)(tan \(\frac{x}{2}\) + cot \(\frac{x}{2}\)) .
Use your graph to write a trigonometric identity involving this function. Explain your reasoning.
Answer:

Question 51.
CRITICAL THINKING
Find a tangent function whose graph intersects the graph of y = 2 + 2 sin x only at minimum points of the sine function.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 51

Maintaining Mathematical Proficiency

Write a cubic function whose graph passes through the given points.
Question 52.
(−1, 0), (1, 0), (3, 0), (0, 3)
Answer:

Question 53.
(−2, 0), (1, 0), (3, 0), (0, −6)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 53

Question 54.
(−1, 0), (2, 0), (3, 0), (1, −2)
Answer:

Question 55.
(−3, 0), (−1, 0), (3, 0), (−2, 1)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 55

Find the amplitude and period of the graph of the function.
Question 56.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 18
Answer:

Question 57.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 19
Answer:
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 a 57

Question 58.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions 9.5 20
Answer:

Lesson 9.6 Modeling with Trigonometric Functions

Essential Question What are the characteristics of the real-life problems that can be modeled by trigonometric functions?

EXPLORATION 1

Modeling Electric Currents
Work with a partner. Find a sine function that models the electric current shown in each oscilloscope screen. State the amplitude and period of the graph.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 1

Communicate Your Answer

Question 2.
What are the characteristics of the real-life problems that can be modeled by trigonometric functions?
Answer:

Question 3.
Use the Internet or some other reference to find examples of real-life situations that can be modeled by trigonometric functions.
Answer:

Monitoring Progress

Question 1.
WHAT IF?
In Example 1, how would the function change when the audiometer produced a pure tone with a frequency of 1000 hertz?
Answer:

Write a function for the sinusoid.
Question 2.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 2
Answer:

Question 3.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 3
Answer:

Question 4.
WHAT IF?
Describe how the model in Example 3 changes when the lowest point of a rope is 5 inches above the ground and the highest point is 70 inches above the ground.
Answer:

Question 5.
The table shows the average daily temperature T (in degrees Fahrenheit) for a city each month, where m = 1 represents January. Write a model that gives T as a function of m and interpret the period of its graph.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 4
Answer:

Modeling with Trigonometric Functions 9.6 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
Graphs of sine and cosine functions are called __________.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 1

Question 2.
WRITING
Describe how to find the frequency of the function whose graph is shown.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 5
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, find the frequency of the function.
Question 3.
y = sin x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 3

Question 4.
y = sin 3x
Answer:

Question 5.
y = cos 4x + 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 5

Question 6.
y =−cos 2x
Answer:

Question 7.
y = sin 3πx
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 7

Question 8.
y = cos \(\frac{\pi x}{4}\)
Answer:

Question 9.
y = \(\frac{1}{2}\) cos 0.75x − 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 9

Question 10.
y = 3 sin 0.2x + 6
Answer:

Question 11.
MODELING WITH MATHEMATICS
The lowest frequency of sounds that can be heard by humans is 20 hertz. The maximum pressure P produced from a sound with a frequency of 20 hertz is 0.02 millipascal. Write and graph a sine model that gives the pressure P as a function of the time t(in seconds).
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 11.1
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 11.2

Question 12.
MODELING WITH MATHEMATICS
A middle-A tuning fork vibrates with a frequency f of 440 hertz (cycles per second). You strike a middle-A tuning fork with a force that produces a maximum pressure of 5 pascals. Write and graph a sine model that gives the pressure Pas a function of the time t (in seconds).
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 6
Answer:

In Exercises 13–16, write a function for the sinusoid.
Question 13.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 13

Question 14.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 8
Answer:

Question 15.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 15

Question 16.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 10
Answer:

Question 17.
ERROR ANALYSIS
Describe and correct the error in finding the amplitude of a sinusoid with a maximum point at (2, 10) and a minimum point at (4, −6).
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 11
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 17

Question 18.
ERROR ANALYSIS
Describe and correct the error in finding the vertical shift of a sinusoid with a maximum point at (3, −2) and a minimum point at (7, −8).
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 12
Answer:

Question 19.
MODELING WITH MATHEMATICS
One of the largest sewing machines in the world has a flywheel (which turns as the machine sews) that is 5 feet in diameter. The highest point of the handle at the edge of the flywheel is 9 feet above the ground, and the lowest point is 4 feet. The wheel makes a complete turn every 2 seconds. Write a model for the height h(in feet) of the handle as a function of the time t(in seconds) given that the handle is at its lowest point when t = 0.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 19

Question 20.
MODELING WITH MATHEMATICS
The Great LaxeyWheel, located on the Isle of Man, is the largest working water wheel in the world. The highest point of a bucket on the wheel is 70.5 feet above the viewing platform, and the lowest point is 2 feet below the viewing platform. The wheel makes a complete turn every 24 seconds. Write a model for the height h(in feet) of the bucket as a function of time t (in seconds) given that the bucket is at its lowest point when t = 0.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 13
Answer:

USING TOOLS In Exercises 21 and 22, the time t is measured in months, where t = 1 represents January. Write a model that gives the average monthly high temperature D as a function of t and interpret the period of the graph.
Question 21.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 14
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 21

Question 22.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 15
Answer:

Question 23.
MODELING WITH MATHEMATICS
A circuit has an alternating voltage of 100 volts that peaks every 0.5 second. Write a sinusoidal model for the voltage Vas a function of the time t (in seconds).
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 16
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 23

Question 24.
MULTIPLE REPRESENTATIONS
The graph shows the average daily temperature of Lexington, Kentucky. The average daily temperature of Louisville, Kentucky, is modeled by y =−22 cos \(\frac{\pi}{6}\)t + 57, where y is the temperature (in degrees Fahrenheit) and t is the number of months since January 1. Which city has the greater average daily temperature? Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 17
Answer:

Question 25.
USING TOOLS
The table shows the numbers of employees N (in thousands) at a sporting goods company each year for 11 years. The time t is measured in years, with t = 1 representing the first year.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 18
a. Use sinusoidal regression to find a model that gives N as a function of t.
b. Predict the number of employees at the company in the 12th year.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 25

Question 26.
THOUGHT PROVOKING
The figure shows a tangent line drawn to the graph of the function y = sin x. At several points on the graph, draw a tangent line to the graph and estimate its slope. Then plot the points (x, m), where m is the slope of the tangent line. What can you conclude?
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 19
Answer:

Question 27.
REASONING
Determine whether you would use a sine or cosine function to model each sinusoid with the y-intercept described. Explain your reasoning.
a. The y-intercept occurs at the maximum value of the function.
b. The y-intercept occurs at the minimum value of the function.
c. The y-intercept occurs halfway between the maximum and minimum values of the function.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 27

Question 28.
HOW DO YOU SEE IT?
What is the frequency of the function whose graph is shown? Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 20
Answer:

Question 29.
USING STRUCTURE
During one cycle, a sinusoid has a minimum at (\(\frac{\pi}{2}\), 3 ) and a maximum at (\(\frac{\pi}{4}\), 8 ). Write a sine function and a cosine function for the sinusoid. Use a graphing calculator to verify that your answers are correct.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 29

Question 30.
MAKING AN ARGUMENT
Your friend claims that a function with a frequency of 2 has a greater period than a function with a frequency of \(\frac{1}{2}\). Is your friend correct? Explain your reasoning.
Answer:

Question 31.
PROBLEM SOLVING
The low tide at a port is 3.5 feet and occurs at midnight. After 6 hours, the port is at high tide, which is 16.5 feet.
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 21
a. Write a sinusoidal model that gives the tide depth d(in feet) as a function of the time t(in hours). Let t = 0 represent midnight.
b. Find all the times when low and high tides occur in a 24-hour period.
c. Explain how the graph of the function you wrote in part (a) is related to a graph that shows the tide depth d at the port t hours after 3:00 A.M.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 31

Maintaining Mathematical Proficiency

Simplify the expression.
Question 32.
\(\frac{17}{\sqrt{2}}\)
Answer:

Question 33.
\(\frac{3}{\sqrt{6}-2}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 33

Question 34.
\(\frac{8}{\sqrt{10}+3}\)
Answer:

Question 35.
\(\frac{13}{\sqrt{3}+\sqrt{11}}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 35

Expand the logarithmic expression.
Question 36.
log8\(\frac{x}{7}\)
Answer:

Question 37.
ln 2x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 37

Question 38.
log3 5x3
Answer:

Question 39.
ln \(\frac{4 x^{6}}{y}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions 9.6 a 39

Lesson 9.7 Using Trigonometric Identities

Essential Question How can you verify a trigonometric identity?

EXPLORATION 1

Writing a Trigonometric Identity
Work with a partner. In the figure, the point (x, y) is on a circle of radius c with center at the origin.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 1
a. Write an equation that relates a, b, and c.
b. Write expressions for the sine and cosine ratios of angle θ.
c. Use the results from parts (a) and (b) to find the sum of sin2θ and cos2θ. What do you observe?
d. Complete the table to verify that the identity you wrote in part (c) is valid for angles (of your choice) in each of the four quadrants.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 2

EXPLORATION 2

Writing Other Trigonometric Identities
Work with a partner. The trigonometric identity you derived in Exploration 1 is called a Pythagorean identity. There are two other Pythagorean identities. To derive them, recall the four relationships:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 3
a. Divide each side of the Pythagorean identity you derived in Exploration 1 by cos2θ and simplify. What do you observe?
b. Divide each side of the Pythagorean identity you derived in Exploration 1 by sin2θ and simplify. What do you observe?

Communicate Your Answer

Question 3.
How can you verify a trigonometric identity?
Answer:

Question 4.
Is sin θ = cos θ a trigonometric identity? Explain your reasoning.
Answer:

Question 5.
Give some examples of trigonometric identities that are different than those in Explorations 1 and 2.
Answer:

Monitoring Progress

Question 1.
Given that cos θ = \(\frac{1}{6}\) and 0 < θ < \(\frac{\pi}{2}\), find the values of the other five trigonometric functions of θ.
Answer:

Simplify the expression.
Question 2.
sin x cot x sec x
Answer:

Question 3.
cos θ − cos θ sin2θ
Answer:

Question 4.
\(\frac{\tan x \csc x}{\sec x}\)
Answer:

Verify the identity.
Question 5.
cot(−θ) =−cot θ
Answer:

Question 6.
csc2x(1 − sin2x) = cot2x
Answer:

Question 7.
cos x csc x tan x = 1
Answer:

Question 8.
(tan2x + 1)(cos2x− 1) = −tan2x
Answer:

Using Trigonometric Identities 9.7 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Describe the difference between a trigonometric identity and a trigonometric equation.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 1

Question 2.
WRITING
Explain how to use trigonometric identities to determine whether sec(−θ) = sec θ or sec(−θ) = −sec θ.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, find the values of the other five trigonometric functions of θ.
Question 3.
sinθ = \(\frac{1}{3}\), 0 < θ < \(\frac{\pi}{2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 3

Question 4.
sin θ = −\(\frac{7}{10}\), π < θ < \(\frac{3 \pi}{2}\)
Answer:

Question 5.
tanθ = −\(\frac{3}{7}\), \(\frac{\pi}{2}\) < θ < π
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 5.1
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 5.2

Question 6.
cot θ = −\(\frac{2}{5}\), \(\frac{\pi}{2}\) < θ < π
Answer:

Question 7.
cos θ = −\(\frac{5}{6}\), π < θ < \(\frac{3 \pi}{2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 7.1
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 7.2

Question 8.
sec θ = \(\frac{9}{4}\), \(\frac{3 \pi}{2}\) < θ < 2π
Answer:

Question 9.
cot θ = −3, \(\frac{3 \pi}{2}\) < θ < 2π
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 9

Question 10.
csc θ = −\(\frac{5}{3}\), π < θ < \(\frac{3 \pi}{2}\)
Answer:

In Exercises 11–20, simplify the expression.
Question 11.
sin x cot x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 11

Question 12.
cos θ (1 + tan2θ)
Answer:

Question 13.
\(\frac{\sin (-\theta)}{\cos (-\theta)}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 13

Question 14.
\(\frac{\cos ^{2} x}{\cot ^{2} x}\)
Answer:

Question 15.
\(\frac{\cos \left(\frac{\pi}{2}-x\right)}{\csc x} \)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 15

Question 16.
sin(\(\frac{\pi}{2}\) – θ) sec θ
Answer:

Question 17.
\(\frac{\csc ^{2} x-\cot ^{2} x}{\sin (-x) \cot x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 17

Question 18.
\(\frac{\cos ^{2} x \tan ^{2}(-x)-1}{\cos ^{2} x}\)
Answer:

Question 19.
\(\frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\csc \theta}\) + cos2 θ
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 19

Question 20.
\(\frac{\sec x \sin x+\cos \left(\frac{\pi}{2}-x\right)}{1+\sec x}\)
Answer:

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in simplifying the expression.
Question 21.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 21

Question 22.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 5
Answer:

In Exercises 23–30, verify the identity.
Question 23.
sin x csc x = 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 23

Question 24.
tan θ csc θ cos θ = 1
Answer:

Question 25.
cos (\(\frac{3 \pi}{2}\) − x)cot x = cos x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 25

Question 26.
sin (\(\frac{3 \pi}{2}\) − x)tan x = sin x
Answer:

Question 27.
\(\frac{\cos \left(\frac{\pi}{2}-\theta\right)+1}{1-\sin (-\theta)}\) = 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 27

Question 28.
\(\frac{\sin ^{2}(-x)}{\tan ^{2} x}\) = cos2 x
Answer:

Question 29.
\(\frac{1+\cos x}{\sin x}+\frac{\sin x}{1+\cos x}\) = 2 csc x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 29

Question 30.
\(\frac{\sin x}{1-\cos (-x)}\) = csc x + cot x
Answer:

Question 31.
USING STRUCTURE
A function f is odd when f(−x) = −f(x). A function f is even when (−x) = f (x). Which of the six trigonometric functions are odd? Which are even? Justify your answers using identities and graphs.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 31.1
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 31.2

Question 32.
ANALYZING RELATIONSHIPS
As the value of cos θ increases, what happens to the value of sec θ? Explain your reasoning.
Answer:

Question 33.
MAKING AN ARGUMENT
Your friend simplifies an expression and obtains sec x tan x− sin x. You simplify the same expression and obtain sin x tan2x. Are your answers equivalent? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 33

Question 34.
HOW DO YOU SEE IT?
The figure shows the unit circle and the angle θ.
a. Is sin θ positive or negative? cos θ? tan θ?
b. In what quadrant does the terminal side of −θ lie?
c. Is sin(−θ) positive or negative? cos(−θ)? tan(−θ)?
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 6
Answer:

Question 35.
MODELING WITH MATHEMATICS
A vertical gnomon(the part of a sundial that projects a shadow) has height h. The length s of the shadow cast by the gnomon when the angle of the Sun above the horizon is θ can be modeled by the equation below. Show that the equation below is equivalent to s = h cot θ.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 35

Question 36.
THOUGHT PROVOKING
Explain how you can use a trigonometric identity to find all the values of x for which sin x = cos x.
Answer:

Question 37.
DRAWING CONCLUSIONS
Static friction is the amount of force necessary to keep a stationary object on a flat surface from moving. Suppose a book weighing W pounds is lying on a ramp inclined at an angle θ. The coefficient of static friction u for the book can be found using the equation uW cos θ = W sin θ.
a. Solve the equation for u and simplify the result.
b. Use the equation from part (a) to determine what happens to the value of u as the angle θ increases from 0° to 90°.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 37

Question 38.
PROBLEM SOLVING
When light traveling in a medium (such as air) strikes the surface of a second medium (such as water) at an angle θ1, the light begins to travel at a different angle θ2. This change of direction is defined by Snell’s law, n1 sin θ1 = n2 sin θ2, where n1 and n2 are the indices of refraction for the two mediums. Snell’s law can be derived from the equation.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 8
a. Simplify the equation to derive Snell’s law.
b. What is the value of n1 when θ1 = 55°, θ2 = 35°, and n2 = 2?
c. If θ1 = θ2, then what must be true about the values of n1 and n2? Explain when this situation would occur.
Answer:

Question 39.
WRITING
Explain how transformations of the graph of the parent function f(x) = sin x support the cofunction identity sin (\(\frac{\pi}{2}\) − θ) = cos θ.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 39

Question 40.
USING STRUCTURE
Verify each identity.
a. ln ∣sec θ∣= −ln ∣cos θ∣
b. ln ∣tan θ∣= ln ∣sin θ∣− ln ∣cos θ∣
Answer:

Maintaining Mathematical Proficiency

Find the value of x for the right triangle.
Question 41.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 41

Question 42.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 10
Answer:

Question 43.
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions 9.7 a 43

Lesson 9.8 Using Sum and Difference Formulas

Essential Question How can you evaluate trigonometric functions of the sum or difference of two angles?

EXPLORATION 1

Deriving a Difference Formula
Work with a partner.
a. Explain why the two triangles shown are congruent.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 1
b. Use the Distance Formula to write an expression for d in the first unit circle.
c. Use the Distance Formula to write an expression for d in the second unit circle.
d. Write an equation that relates the expressions in parts (b) and (c). Then simplify this equation to obtain a formula for cos(a − b).

EXPLORATION 2

Deriving a Sum Formula
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 2
Work with a partner. Use the difference formula you derived in Exploration 1 to write a formula for cos(a+b) in terms of sine and cosine of a and b. Hint: Use the fact that cos(a+b) = cos[a − (−b)].

EXPLORATION 3

Deriving Difference and Sum Formulas
Work with a partner. Use the formulas you derived in Explorations 1 and 2 to write formulas for sin(a − b) and sin(a + b) in terms of sine and cosine of a and b. Hint: Use the cofunction identities sin (\(\frac{\pi}{2}\) − a)= cos a and cos (\(\frac{\pi}{2}\) − a)= sin a and the fact that
cos [(\(\frac{\pi}{2}\) − a) + b ]= sin(a − b) and sin(a+b) = sin[a − (−b)].

Communicate Your Answer

Question 4.
How can you evaluate trigonometric functions of the sum or difference of two angles?
Answer:

Question 5.
a. Find the exact values of sin 75° and cos 75° using sum formulas. Explain your reasoning.
b. Find the exact values of sin 75° and cos 75° using difference formulas. Compare your answers to those in part (a).
Answer:

Monitoring Progress

Find the exact value of the expression.
Question 1.
sin 105°
Answer:

Question 2.
cos 15°
Answer:

Question 3.
tan \(\frac{5 \pi}{2}\)
Answer:

Question 4.
cos \(\frac{\pi}{12}\)
Answer:

Question 5.
Find sin(a−b) given that sin a = \(\frac{8}{17}\) with 0 < a < \(\frac{\pi}{2}\) and cos b = −\(\frac{24}{25}\) with π < b < \(\frac{3 \pi}{2}\) .
Answer:

Simplify the expression.
Question 6.
sin(x + π)
Answer:

Question 7.
cos(x − 2π)
Answer:

Question 8.
tan(x − π)
Answer:

Question 9.
Solve sin (\(\frac{\pi}{4}\) − x)− sin (x + \(\frac{\pi}{4}\))= 1 for 0 ≤ x < 2π.
Answer:

Using Sum and Difference Formulas 9.8 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
Write the expression cos 130° cos 40°− sin 130° sin 40° as the cosine of an angle.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 1

Question 2.
WRITING
Explain how to evaluate tan 75° using either the sum or difference formula for tangent.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, find the exact value of the expression.
Question 3.
tan(−15°)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 3

Question 4.
tan 195°
Answer:

Question 5.
sin \(\frac{23 \pi}{12}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 5

Question 6.
sin(−165°)
Answer:

Question 7.
cos 105°
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 7

Question 8.
cos \(\frac{11 \pi}{12}\)
Answer:

Question 9.
tan \(\frac{17 \pi}{12}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 9

Question 10.
sin (−\(\frac{7 \pi}{12}\))
Answer:

In Exercises 11–16, evaluate the expression given that cos a = \(\frac{4}{5}\) with 0 < a < \(\frac{\pi}{2}\) and sin b = –\(\frac{15}{17}\) with \(\frac{3 \pi}{2}\) < b < 2π.
Question 11.
sin(a + b)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 11.1
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 11.2

Question 12.
sin(a − b)
Answer:

Question 13.
cos(a − b)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 13.1
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 13.2

Question 14.
cos(a + b)
Answer:

Question 15.
tan(a + b)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 15.1
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 15.2

Question 16.
tan(a − b)
Answer:

In Exercises 17–22, simplify the expression.
Question 17.
tan(x + π)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 17

Question 18.
cos (x − \(\frac{\pi}{2}\))
Answer:

Question 19.
cos(x + 2π)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 19

Question 20.
tan(x − 2π)
Answer:

Question 21.
sin (x − \(\frac{3 \pi}{2}\))
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 21

Question 22.
tan (x + \(\frac{\pi}{2}\))
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in simplifying the expression.
Question 23.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 3
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 23

Question 24.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 4
Answer:

Question 25.
What are the solutions of the equation 2 sin x − 1 = 0 for 0 ≤ x < 2π?
A. \(\frac{\pi}{3}\)
B. \(\frac{\pi}{6}\)
C. \(\frac{2 \pi}{3}\)
D. \(\frac{5 \pi}{6}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 25

Question 26.
What are the solutions of the equation tan x + 1 = 0 for 0 ≤ x < 2π?
A. \(\frac{\pi}{4}\)
B. \(\frac{3 \pi}{4}\)
C. \(\frac{5 \pi}{4}\)
D. \(\frac{7 \pi}{4}\)
Answer:

In Exercises 27–32, solve the equation for 0 ≤ x < 2π.
Question 27.
sin (x + \(\frac{\pi}{2}\)) = \(\frac{1}{2}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 27

Question 28.
tan (x − \(\frac{\pi}{4}\)) = 0
Answer:

Question 29.
cos (x +\(\frac{\pi}{6}\)) − cos (x −\(\frac{\pi}{6}\)) = 1
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 29

Question 30.
sin (x + \(\frac{\pi}{4}\)) + sin (x − \(\frac{\pi}{4}\)) = 0
Answer:

Question 31.
tan(x + π) − tan(π − x) = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 31

Question 32.
sin(x + π) + cos(x + π) = 0
Answer:

Question 33.
USING EQUATIONS
Derive the cofunction identity sin (\(\frac{\pi}{2}\) − θ)= cos θ using the difference formula for sine.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 33

Question 34.
MAKING AN ARGUMENT
Your friend claims it is possible to use the difference formula for tangent to derive the cofunction identity tan (\(\frac{\pi}{2}\) − θ) = cot θ. Is your friend correct? Explain your reasoning.
Answer:

Question 35.
MODELING WITH MATHEMATICS
A photographer is at a height h taking aerial photographs with a 35-millimeter camera. The ratio of the image length WQ to the length NA of the actual object is given by the formula
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 5
where θ is the angle between the vertical line perpendicular to the ground and the line from the camera to point A and t is the tilt angle of the film. When t = 45°, show that the formula can be rewritten as \(\frac{W Q}{N A}=\frac{70}{h(1+\tan \theta)}\).
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 35

Question 36.
MODELING WITH MATHEMATICS
When a wave travels through a taut string, the displacement y of each point on the string depends on the time t and the point’s position x. The equation of a standing wave can be obtained by adding the displacements of two waves traveling in opposite directions. Suppose a standing wave can be modeled by the formula
y = A cos (\(\frac{2 \pi t}{3}-\frac{2 \pi x}{5}\)) + A cos (\(\frac{2 \pi t}{3}+\frac{2 \pi x}{5}\)) .When t= 1, show that the formula can be rewritten as y = −A cos \(\frac{2 \pi x}{5}\).
Answer:

Question 37.
MODELING WITH MATHEMATICS
The busy signal on a touch-tone phone is a combination of two tones with frequencies of 480 hertz and 620 hertz. The individual tones can be modeled by the equations:
480 hertz: y1 = cos 960πt
620 hertz: y2 = cos 1240πt
The sound of the busy signal can be modeled by y1 + y2. Show that y1 + y2 = 2 cos 1100πt cos 140πt.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 37

Question 38.
HOW DO YOU SEE IT?
Explain how to use the figure to solve the equation sin (x + \(\frac{\pi}{4}\)) − sin (\(\frac{\pi}{4}\) − x) = 0 for 0 ≤ x < 2π.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 6
Answer:

Question 39.
MATHEMATICAL CONNECTIONS
The figure shows the acute angle of intersection, θ2 − θ1, of two lines with slopes m1 and m2.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 7
a. Use the difference formula for tangent to write an equation for tan (θ2 − θ1) in terms of m1 and m2.
b. Use the equation from part (a) to find the acute angle of intersection of the lines y = x− 1 and y = \(\left(\frac{1}{\sqrt{3}-2}\right)\)x + \(\frac{4-\sqrt{3}}{2-\sqrt{3}}\).
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 39.1
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 39.2

Question 40.
THOUGHT PROVOKING
Rewrite each function. Justify your answers.
a. Write sin 3x as a function of sin x.
b. Write cos 3x as a function of cos x.
c. Write tan 3x as a function of tan x.
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check your solution(s).
Question 41.
1 − \(\frac{9}{x-2}\) = −\(\frac{7}{2}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 41

Question 42.
\(\frac{12}{x}\) + \(\frac{3}{4}\) = \(\frac{8}{x}\)
Answer:

Question 43.
\(\frac{2 x-3}{x+1}\) = \(\frac{10}{x^{2}-1}\) + 5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 43.1
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 a 43.2

Trigonometric Ratios and Functions Performance Task: Lightening the Load

9.5–9.8 What Did You Learn?

Core Vocabulary
frequency, p. 506
sinusoid, p. 507
trigonometric identity, p. 514

Core Concepts
Section 9.5
Characteristics of y = tan x and y = cot x, p. 498
Period and Vertical Asymptotes of y = a tan bx and y = a cot bx, p. 499
Characteristics of y = sec x and y = csc x, p. 500

Section 9.6
Frequency, p. 506
Writing Trigonometric Functions, p. 507
Using Technology to Find Trigonometric Models, p. 509

Section 9.7
Fundamental Trigonometric Identities, p. 514

Section 9.8
Sum and Difference Formulas, p. 520
Trigonometric Equations and Real-Life Formulas, p. 522

Mathematical Practices
Question 1.
Explain why the relationship between θ and d makes sense in the context of the situation in Exercise 43 on page 503.
Answer:

Question 2.
How can you use definitions to relate the slope of a line with the tangent of an angle in Exercise 39 on page 524?
Answer:

Performance Task: Lightening the Load

You need to move a heavy table across the room. What is the easiest way to move it? Should you push it? Should you tie a rope around one leg of the table and pull it? How can trigonometry help you make the right decision?
To explore the answers to these questions and more, go to BigIdeasMath.com.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions 9.8 8

Trigonometric Ratios and Functions Chapter Review

9.1 Right Triangle Trigonometry (pp. 461−468)

Question 1.
In a right triangle, θ is an acute angle and cos θ = \(\frac{6}{11}\). Evaluate the other five trigonometric functions of θ.
Answer:

Question 2.
The shadow of a tree measures 25 feet from its base. The angle of elevation to the Sun is 31°. How tall is the tree?
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions cr 2
Answer:

9.2 Angles and Radian Measure (pp. 469−476)

Question 3.
Find one positive angle and one negative angle that are coterminal with 382°.
Answer:

Convert the degree measure to radians or the radian measure to degrees.
Question 4.
30°
Answer:

Question 5.
225°
Answer:

Question 6.
\(\frac{3 \pi}{4}\)
Answer:

Question 7.
\(\frac{5 \pi}{3}\)
Answer:

Question 8.
A sprinkler system on a farm rotates 140°and sprays water up to 35 meters. Draw a diagram that shows the region that can be irrigated with the sprinkler. Then find the area of the region.
Answer:

9.3 Trigonometric Functions of Any Angle (pp. 477−484)

Evaluate the six trigonometric functions of θ.
Question 9.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions cr 9
Answer:

Question 10.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions cr 10
Answer:

Question 11.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions cr 11
Answer:

Evaluate the function without using a calculator.
Question 12.
tan 330°
Answer:

Question 13.
sec(−405°)
Answer:

Question 14.
sin \(\frac{13 \pi}{6}\)
Answer:

Question 15.
sec \(\frac{11 \pi}{3}\)
Answer:

9.4 Graphing Sine and Cosine Functions (pp. 485−494)

Identify the amplitude and period of the function. Then graph the function and describe the graph of g as a transformation of the graph of the parent function.
Question 16.
g(x) = 8 cos x
Answer:

Question 17.
g(x) = 6 sin πx
Answer:

Question 18.
g(x) = \(\frac{1}{4}\) cos 4x
Answer:

Graph the function.
Question 19.
g(x) = cos(x + π) + 2
Answer:

Question 20.
g(x) = −sin x − 4
Answer:

Question 21.
g(x) = 2 sin (x + \(\frac{\pi}{2}\))
Answer:

9.5 Graphing Other Trigonometric Functions (pp. 497−504)

Graph one period of the function. Describe the graph of g as a transformation of the graph of its parent function.
Question 22.
g(x) = tan \(\frac{1}{2}\)x
Answer:

Question 23.
g(x) = 2 cot x
Answer:

Question 24.
g(x) = 4 tan 3πx
Answer:

Graph the function.
Question 25.
g(x) = 5 csc x
Answer:

Question 26.
g(x) = sec \(\frac{1}{2}\)x
Answer:

Question 27.
g(x) = 5 sec πx
Answer:

Question 28.
g(x) = \(\frac{1}{2}\) csc \(\frac{\pi}{4}\)x
Answer:

9.6 Modeling with Trigonometric Functions (pp. 505−512)

Write a function for the sinusoid.
Question 29.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions cr 29
Answer:

Question 30.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions cr 30
Answer:

Question 31.
You put a reflector on a spoke of your bicycle wheel. The highest point of the reflector is 25 inches above the ground, and the lowest point is 2 inches. The reflector makes 1 revolution per second. Write a model for the height h (in inches) of a reflector as a function of time t (in seconds) given that the reflector is at its lowest point when t = 0.
Answer:

Question 32.
The table shows the monthly precipitation P (in inches) for Bismarck, North Dakota, where t = 1 represents January. Write a model that gives P as a function of t and interpret the period of its graph.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions cr 32
Answer:

9.7 Using Trigonometric Identities (pp. 513−518)

Simplify the expression.
Question 33.
cot2x − cot2x cos2x
Answer:

Question 34.
\(\frac{(\sec x+1)(\sec x-1)}{\tan x}\)
Answer:

Question 35.
sin (\(\frac{\pi}{2}\) − x)tan x
Answer:

Verify the identity.
Question 36.
\(\frac{\cos x \sec x}{1+\tan ^{2} x}\) = cos2x
Answer:

Question 37.
tan (\(\frac{\pi}{2}\) − x)cot x = csc2x − 1
Answer:

9.8 Using Sum and Difference Formulas (pp. 519−524)

Find the exact value of the expression.
Question 38.
sin 75°
Answer:

Question 39.
tan(−15°)
Answer:

Question 40.
cos \(\frac{\pi}{12}\)
Answer:

Question 41.
Find tan(a + b), given that tan a = \(\frac{1}{4}\) with π < a < \(\frac{3 \pi}{2}\) and tan b = \(\frac{3}{7}\) with 0 < b < \(\frac{\pi}{2}\) .
Answer:

Solve the equation for 0 ≤ x < 2π.
Question 42.
cos (x + \(\frac{3 \pi}{4}\)) + cos (x − \(\frac{3 \pi}{4}\)) = 1
Answer:

Question 43.
tan(x + π) + cos (x + \(\frac{\pi}{2}\))= 0
Answer:

Trigonometric Ratios and Functions Chapter Test

Verify the identity.
Question 1.
\(\frac{\cos ^{2} x+\sin ^{2} x}{1+\tan ^{2} x}\) = cos2x
Answer:

Question 2.
\(\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x}\) = 2 sec x
Answer:

Question 3.
cos (x + \(\frac{3 \pi}{2}\)) = sin x
Answer:

Question 4.
Evaluate sec(−300°) without using a calculator.
Answer:

Write a function for the sinusoid.
Question 5.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions ct 5
Answer:

Question 6.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions ct 6
Answer:

Graph the function. Then describe the graph of g as a transformation of the graph of its parent function.
Question 7.
g(x) = −4 tan 2x
Answer:

Question 8.
g(x) = −2 cos \(\frac{1}{3}\)x + 3
Answer:

Question 9.
g(x) = 3 csc πx
Answer:

Convert the degree measure to radians or the radian measure to degrees. Then find one positive angle and one negative angle that are coterminal with the given angle.
Question 10.
−50°
Answer:

Question 11.
\(\frac{4 \pi}{5}\)
Answer:

Question 12.
\(\frac{8 \pi}{3}\)
Answer:

Question 13.
Find the arc length and area of a sector with radius r = 13 inches and central angle θ = 40°.
Answer:

Evaluate the six trigonometric functions of the angle θ.
Question 14.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions ct 14
Answer:

Question 15.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions ct 15
Answer:

Question 16.
In which quadrant does the terminal side of θ lie when cos θ < 0 and tan θ > 0? Explain.
Answer:

Question 17.
How tall is the building? Justify your answer.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions ct 17
Answer:

Question 18.
The table shows the average daily high temperatures T (in degrees Fahrenheit) in Baltimore, Maryland, where m= 1 represents January. Write a model that gives T as a function of m and interpret the period of its graph.
Big Ideas Math Algebra 2 Solutions Chapter 9 Trigonometric Ratios and Functions ct 18
Answer:

Trigonometric Ratios and Functions Cumulative Assessment

Question 1.
Which expressions are equivalent to 1?
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions ca 1
Answer:

Question 2.
Which rational expression represents the ratio of the perimeter to the area of the playground shown in the diagram?
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions ca 2
Answer:

Question 3.
The chart shows the average monthly temperatures (in degrees Fahrenheit) and the gas usages (in cubic feet) of a household for 12 months.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions ca 3
a. Use a graphing calculator to find trigonometric models for the average temperature y1 as a function of time and the gas usage y2 (in thousands of cubic feet) as a function of time. Let t = 1 represent January.
b. Graph the two regression equations in the same coordinate plane on your graphing calculator. Describe the relationship between the graphs.
Answer:

Question 4.
Evaluate each logarithm using log2 5 ≈ 2.322 and log2 3 ≈ 1.585, if necessary. Then order the logarithms by value from least to greatest.
a. log 1000
b. log2 15
c. ln e
d. log2 9
e. log2\(\frac{5}{3}\)
f. log2 1
Answer:

Question 5.
Which function is not represented by the graph?
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions ca 5
Answer:

Question 6.
Complete each statement with < or > so that each statement is true.
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions ca 6
Answer:

Question 7.
Use the Rational Root Theorem and the graph to find all the real zeros of the function f(x) = 2x3 − x2 − 13x− 6. (HSA-APR.B.3)
Big Ideas Math Answers Algebra 2 Chapter 9 Trigonometric Ratios and Functions ca 7
Answer:

Question 8.
Your friend claims −210° is coterminal with the angle \(\frac{5 \pi}{6}\). Is your friend correct? Explain your reasoning.
Answer:

Question 9.
Company A and Company B offer the same starting annual salary of $20,000. Company A gives a $1000 raise each year. Company B gives a 4% raise each year.
a. Write rules giving the salaries an and bn for your nth year of employment at Company A and Company B, respectively. Tell whether the sequence represented by each rule is arithmetic, geometric, or neither.
b. Graph each sequence in the same coordinate plane.
c. Under what conditions would you choose to work for Company B?
d. After 20 years of employment, compare your total earnings.
Answer:

Big Ideas Math Geometry Answers Chapter 8 Similarity

Big Ideas Math Geometry Answers Chapter 8 Similarity

Do you want a perfect guide for a better understanding of Similarity concepts in High School? Then, Big Ideas Math Geometry Answers Chapter 8 Similarity is the one-stop destination for all your requirements during preparation. Practicing and Solving no. of questions from chapter 8 similarity BIM Geometry Textbook Solutions is the best way to understand the concepts easily and gain more subject knowledge. Kickstart your preparation by taking the help of BigIdeas math geometry ch 8 similarity Answer key pdf and test your skills up-to-date for better scoring in the exams and become math proficient.  

Big Ideas Math Book Geometry Answer Key Chapter 8 Similarity

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Similarity Maintaining Mathematical Proficiency

Tell whether the ratios form a proportion.

Question 1.
\(\frac{5}{3}, \frac{35}{21}\)

Answer:
Yes, the ratios \(\frac{5}{3}, \frac{35}{21}\) form a proportion.

Explanation:
A proportion means two ratios are equal.
So, cross product of \(\frac{5}{3}, \frac{35}{21}\) is 21 x 5 = 105 = 35 x 3
Therefore, \(\frac{5}{3}, \frac{35}{21}\) form a proportion.

Question 2.
\(\frac{9}{24}, \frac{24}{64}\)

Answer:
Yes, the ratios \(\frac{9}{24}, \frac{24}{64}\) form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 24 x 24 = 576 = 64 x 9
Therefore, the ratios \(\frac{9}{24}, \frac{24}{64}\) form a proportion.

Question 3.
\(\frac{8}{56}, \frac{6}{28}\)

Answer:
The ratios \(\frac{8}{56}, \frac{6}{28}\) do not form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 56 x 6 = 336, 28 x 8 = 224
Therefore, the ratios \(\frac{8}{56}, \frac{6}{28}\) do not form a proportion.

Question 4.
\(\frac{18}{4}, \frac{27}{9}\)

Answer:
The ratios \(\frac{18}{4}, \frac{27}{9}\) do not form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 9 x 18 = 162, 27 x 4 = 108
Therefore, the ratios \(\frac{18}{4}, \frac{27}{9}\) do not form a proportion.

Question 5.
\(\frac{15}{21}, \frac{55}{77}\)

Answer:
The ratios \(\frac{15}{21}, \frac{55}{77}\) form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 15 x 77 = 1155, 55 x 21 = 1155
Therefore, the ratios \(\frac{15}{21}, \frac{55}{77}\) form a proportion.

Question 6.
\(\frac{26}{8}, \frac{39}{12}\)

Answer:
The ratios \(\frac{26}{8}, \frac{39}{12}\) form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 26 x 12 = 312, 39 x 8 = 312
Therefore, the ratios \(\frac{26}{8}, \frac{39}{12}\) form a proportion.

Find the scale factor of the dilation.

Question 7.
Big Ideas Math Geometry Answers Chapter 8 Similarity 1

Answer:
k = \(\frac { 3 }{ 7 } \)

Explanation:
The scale factor k = \(\frac { CP’ }{ CP } \)
= \(\frac { 6 }{ 14 } \)
= \(\frac { 3 }{ 7 } \)

Question 8.
Big Ideas Math Geometry Answers Chapter 8 Similarity 2

Answer:
k = \(\frac { 3 }{ 8 } \)

Explanation:
The scale factor k = \(\frac { CP }{ CP’ } \)
= \(\frac { 9 }{ 24 } \)
= \(\frac { 3 }{ 8 } \)

Question 9.
Big Ideas Math Geometry Answers Chapter 8 Similarity 3

Answer:
k = \(\frac { 1 }{ 2 } \)

Explanation:
The scale factor k = \(\frac { MK }{ M’K’ } \)
= \(\frac { 14 }{ 28 } \)
= \(\frac { 1 }{ 2 } \)

Question 10.
ABSTRACT REASONING
If ratio X and ratio Y form a proportion and ratio Y and ratio Z form a proportion, do ratio X and ratio Z form a proportion? Explain our reasoning.

Answer:
Yes, ratio X and ratio Z form a proportion.

Explanation:
If ratios are proportional means they are equal.
So, ratio X and ratio Y form a proportion that means X = Y
ratio Y and ratio Z form a proportion that means Y = Z
From the above two equations, we can say that X = Z
So, ratio X and ratio Z also form a proportion.

Similarity Mathematical Practices

Monitoring Progress

Question 1.
Find the perimeter and area of the image when the trapezoid is dilated by a scale factor of
(a) 2, (b) 3, and (c) 4.
Big Ideas Math Geometry Answers Chapter 8 Similarity 4

Answer:
(a) Perimeter is 32 cm, area is 48 sq cm.
(b) Perimeter is 48 cm, the area is 108 sq cm.
(c) Perimeter is 64 cm, the area is 192 sq cm.

Explanation:
The perimeter of trapezoid P = 2 + 5 + 6 + 3 = 16 cm
Area of trapezoid A = \(\frac { (2 + 6)3 }{ 2 } \)
= \(\frac { 3(8) }{ 2 } \)
= \(\frac { 24 }{ 2 } \)
= 12 sq cm
(a) If scale factor k = 2, then
Perimeter = kP
= 2 x 16 = 32
Area = k²A
= 2² x 12
= 4 x 12
= 48
(b) If scale factor k = 3, then,
Perimeter = kP
= 3 x 16 = 48 cm
Area = k²A
= 3² x 12 = 9 x 12 = 108
(c) If scale factor k = 4, then
Perimeter = kP
= 4 x 16 = 64
Area = k²A
= 4² x 12 = 16 x 12 = 192

Question 2.
Find the perimeter and area of the image when the parallelogram is dilated by a scale factor of
(a) 2, (b) 3, and (c) \(\frac{1}{2}\)
Big Ideas Math Geometry Answers Chapter 8 Similarity 5

Answer:
(a) Perimeter is 28 ft, area is 32 sq ft
(b) Perimeter is 42 ft, area is 72 sq ft
(c) Perimeter is 7 ft, area is 2 sq ft

Explanation:
Perimeter of parallelogram P = 2(2 + 5) = 7 x 2 = 14 ft
Area of the parallelogram = 2 x 4 = 8 sq ft
(a) If scale factor k = 2, then
Perimeter = kP
= 2 x 14 = 28
Area = k²A
= 2² x 8 = 4 x 8 = 32
(b) If scale factor k = 3, then
Perimeter = kP
= 3 x 14 = 42
Area = k²A
= 3² x 8 = 72
(c) If scale factor k = \(\frac{1}{2}\), then
Perimeter = kP
= \(\frac{1}{2}\) x 14 = 7
Area = k²A
= \(\frac{1}{2²}\) x 8 = \(\frac{1}{4}\) x 8 = 2

Question 3.
A rectangular prism is 3 inches wide, 4 inches long, and 5 inches tall. Find the surface area and volume of the image of the prism when it is dilated by a scale factor of
(a) 2, (b) 3, and (c) 4.

Answer:
(a) Surface area is 376 sq in, volume is 480 cubic in
(b) Surface area is 846 sq in, volume is 1620 cubic in
(c) Surface area is 1504 sq in, volume is 3840 cubic in

Explanation:
The surface area of the rectangular prism A = 2(3 x 4 + 4 x 5 + 5 x 3)
= 2(12 + 20 + 15) = 2(47) = 94 in
Volume of the rectangular prism V = 3 x 4 x 5 = 60 in³
(a) If the scale factor k = 2, then
Surface Area = k²A
= 2² x 94 = 4 x 94 = 376 sq in
Volume = k³V
= 2³ x 60 = 8 x 60 = 480 cubic in
(b) If the scale factor k = 3, then
Surface Area = k²A
= 3² x 94 = 9 x 94 = 846
Volume = k³V
= 3³ x 60 = 27 x 60 = 1620
(c) If the scale factor k = 4, then
Surface Area = k²A
= 4² x 94 = 16 x 94 = 1504
Volume = k³V
= 4³ x 60 = 64 x 60 = 3840

8.1 Similar Polygons

Exploration 1

Comparing Triangles after a Dilation

Work with a partner: Use dynamic geometry software to draw any ∆ABC. Dilate ∆ABC to form a similar ∆A’B’C’ using an scale factor k and an center of dilation.

Big Ideas Math Geometry Answers Chapter 8 Similarity 6

a. Compare the corresponding angles of ∆A’B’C and ∆ABC.
Answer:

b. Find the ratios of the lengths of the sides of ∆A’B’C’ to the lengths of the corresponding sides of ∆ABC. What do you observe?
Answer:

c. Repeat parts (a) and (b) for several other triangles, scale factors, and centers of dilation. Do you obtain similar results?
Answer:

Exploration 2

Comparing Triangles after a Dilation

Work with a partner: Use dynamic geometry Software to draw any ∆ABC. Dilate ∆ABC to form a similar ∆A’B’C’ using any scale factor k and any center of dilation.

Big Ideas Math Geometry Answers Chapter 8 Similarity 7

a. Compare the perimeters of ∆A’B’C and ∆ABC. What do you observe?
Answer:

b. Compare the areas of ∆A’B’C’ and ∆ABC. What do you observe?
Answer:

c. Repeat parts (a) and (b) for several other triangles, scale factors, and centers of dilation. Do you obtain similar results?
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Communicate Your Answer

Question 3.
How are similar polygons related?

Answer:
if two polygons are similar means they have the same shape, corresponding angles are congruent and the ratios of lengths of their corresponding sides are equal. The common ratio is called the scale factor.

Question 4.
A ∆RST is dilated by a scale factor of 3 to form ∆R’S’T’. The area of ∆RST is 1 square inch. What is the area of ∆R’S’T’?

Answer:
Area of ∆R’S’T’ = 9 sq in

Explanation:
Given that,
Area of ∆RST = 1 sq inch
Scale factor k = 3
Area of ∆R’S’T’ = k² x Area of ∆RST
= 3² x 1 = 9 x 1 = 9

Lesson 8.1 Similar Polygons

Monitoring Progress

Question 1.
In the diagram, ∆JKL ~ ∆PQR. Find the scale factor from ∆JKL to ∆PQR. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality.
Big Ideas Math Geometry Answers Chapter 8 Similarity 8

Answer:
The pairs of congruent angles are ∠K = ∠Q, ∠J = ∠P, ∠ L = ∠R
The scale factor is \(\frac { 3 }{ 2 } \)
The ratios of the corresponding side lengths in a statement of proportionality are \(\frac { PQ }{ JK } = \frac { PR }{ JL } = \frac { QR }{ LK } \)

Explanation:
Given that,
∆JKL ~ ∆PQR
The pairs of congruent angles are ∠K = ∠Q, ∠J = ∠P, ∠ L = ∠R
To find the scale factor,
\(\frac { PQ }{ JK } = \frac { 9 }{ 6 } \) = \(\frac { 3 }{ 2 } \), \(\frac { PR }{ JL } = \frac { 12 }{ 8 } \) = \(\frac { 3 }{ 2 } \), \(\frac { QR }{ LK } = \frac { 6 }{ 4 } \) = \(\frac { 3 }{ 2 } \)
So, the scale factor is \(\frac { 3 }{ 2 } \)

Question 2.
Find the value of x.
Big Ideas Math Geometry Answers Chapter 8 Similarity 9
ABCD ~ QRST

Answer:
x = 2

Explanation:
The triangles are similar, so corresponding side lengths are proportional.
\(\frac { RS }{ BC } \) = \(\frac { AB }{ QR } \)
\(\frac { 4 }{ x } \) = \(\frac { 12 }{ 6 } \)
\(\frac { 4 }{ x } \) = 2
4 = 2x
x = 2

Question 3.
Find KM
Big Ideas Math Geometry Answers Chapter 8 Similarity 10
∆JKL ~ ∆EFG

Answer:
KM = 42

Explanation:
Scale factor = \(\frac { JM }{ GH } \)
= \(\frac { 48 }{ 40 } \)
= \(\frac { 6 }{ 5 } \)
Because the ratio of the lengths of the altitudes in similar triangles is equal to the scale factor, you can write the following proportion
\(\frac { KM }{ HF } \) = \(\frac { 6 }{ 5 } \)
\(\frac { KM }{ 35 } \) = \(\frac { 6 }{ 5 } \)
KM = \(\frac { 6 }{ 5 } \) x 35
KM = 42

Question 4.
The two gazebos shown are similar pentagons. Find the perimeter of Gazebo A.
Big Ideas Math Geometry Answers Chapter 8 Similarity 11

Answer:
Perimeter of Gazebo A = 46 m

Explanation:
Scale factor = \(\frac { AB }{ FG } \)
= \(\frac { 10 }{ 15 } \)
= \(\frac { 2 }{ 3 } \)
So, \(\frac { AE }{ FK } \) = \(\frac { 2 }{ 3 } \)
\(\frac { x }{ 18 } \) = \(\frac { 2 }{ 3 } \)
x = 12
\(\frac { ED }{ KJ } \) = \(\frac { 2 }{ 3 } \)
\(\frac { ED }{ 15 } \) = \(\frac { 2 }{ 3 } \)
ED = 10
\(\frac { DC }{ JH } \) = \(\frac { 2 }{ 3 } \)
\(\frac { DC }{ 12 } \) = \(\frac { 2 }{ 3 } \)
DC = 8
\(\frac { BC }{ GH } \) = \(\frac { 2 }{ 3 } \)
\(\frac { BC }{ 9 } \) = \(\frac { 2 }{ 3 } \)
BC = 6
Therefore, perimeter = 6 + 8 + 10 + 12 + 10 = 46

Question 5.
In the diagram, GHJK ~ LMNP. Find the area of LMNP.
Big Ideas Math Geometry Answers Chapter 8 Similarity 12
Area of GHJK = 84m2

Answer:
Area of LMNP = 756 m2

Explanation:
As shapes are similar, their corresponding side lengths are proportional.
Scale Factor k = \(\frac { NP }{ JK } \)
= \(\frac { 21 }{ 7 } \)
= 3
Area of LMNP = k² x Area of GHJK
= 3² x 84
= 756 m2

Question 6.
Decide whether the hexagons in Tile Design 1 are similar. Explain.
Big Ideas Math Geometry Answers Chapter 8 Similarity 13

Answer:

Question 7.
Decide whether the hexagons in Tile Design 2 are similar. Explain.
Big Ideas Math Geometry Answers Chapter 8 Similarity 14

Answer:

Exercise 8.1 Similar Polygons

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
For two figures to be similar, the corresponding angles must be ____________ . and the corresponding side lengths must be _____________ .

Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Geometry Answers Chapter 8 Similarity 15
What is the scale factor?
Answer:
Scale Factor = \(\frac{image-length}{actual-length}\)
= \(\frac{3}{12}\)= \(\frac{4}{16}\)= \(\frac{5}{20}\)
= \(\frac{1}{4}\)

What is the ratio of their areas?

Answer:
Big Ideas Math Geometry Answers Exercise 8.1 Similar Polygons 1

What is the ratio of their corresponding side lengths?
Answer:
Big Ideas Math Geometry Answers Exercise 8.1 Similar Polygons 2

What is the ratio of their perimeters?
Answer:
Big Ideas Math Geometry Answers Exercise 8.1 Similar Polygons 3

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, find the scale factor. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality.

Question 3.
∆ABC ~ ∆LMN
Big Ideas Math Geometry Answers Chapter 8 Similarity 16
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 3

Question 4.
DEFG ~ PQRS
Big Ideas Math Geometry Answers Chapter 8 Similarity 17
Answer:
Big Ideas Math Geometry Answers Exercise 8.1 Similar Polygons 5

In Exercises 5-8, the polygons are similar. Find the value of x.

Question 5.
Big Ideas Math Geometry Answers Chapter 8 Similarity 18
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 5

Question 6.
Big Ideas Math Geometry Answers Chapter 8 Similarity 19

Answer:
x = 20

Explanation:
\(\frac { DF }{ GJ } \) = \(\frac { DE }{ GH } \)
\(\frac { 16 }{ 12 } \) = \(\frac { x }{ 15 } \)
x = \(\frac { 16 x 15 }{ 12 } \) = \(\frac { 240 }{ 12 } \)
x = 20

Question 7.
Big Ideas Math Geometry Answers Chapter 8 Similarity 20
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 7

Question 8.
Big Ideas Math Geometry Answers Chapter 8 Similarity 21

Answer:
x = 12

Explanation:
\(\frac { PN }{ KJ } \) = \(\frac { MN }{ JH } \)
\(\frac { x }{ 8 } \) = \(\frac { 9 }{ 6 } \)
x = \(\frac { 9 x 8 }{ 6 } \) = \(\frac { 72 }{ 6 } \)
x = 12

In Exercises 9 and 10, the black triangles are similar. Identify the type of segment shown in blue and find the value of the variab1e.

Question 9.
Big Ideas Math Geometry Answers Chapter 8 Similarity 22
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 9

Question 10.
Big Ideas Math Geometry Answers Chapter 8 Similarity 23

Answer:

Explanation:
\(\frac { y }{ 18 } \) = \(\frac { y – 1 }{ 16 } \)
18(y – 1) = 16y
18y – 18 = 16y
18y – 16y = 18
2y = 18
y = 9

In Exercises 11 and 12, RSTU ~ ABCD. Find the ratio of their perimeters.

Question 11.
Big Ideas Math Geometry Answers Chapter 8 Similarity 24
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 11

Question 12.
Big Ideas Math Geometry Answers Chapter 8 Similarity 25
Answer:
\(\frac { RS + ST + TU + UR }{ AB + BC + CD + DA } \) = \(\frac { RS }{ AB } \) = \(\frac { 18 }{ 24 } \)
The ratio of perimeter is \(\frac { 3 }{ 4 } \).

In Exercises 13-16, two polygons are similar. The perimeter of one polygon and the ratio of the corresponding side lengths are given. Find the perimeter of the other polygon.

Question 13.
perimeter of smaller polygon: 48 cm: ratio: \(\frac{2}{3}\)
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 13

Question 14.
perimeter of smaller polygon: 66 ft: ratio: \(\frac{3}{4}\)

Answer:
The perimeter of larger polygon is 88 ft.

Explanation:
\(\frac { smaller }{ larger } \) = \(\frac { 66 }{ x } \) = \(\frac{3}{4}\)
66 x 4 = 3x
3x = 264
x = \(\frac { 264 }{ 3 } \) = 88

Question 15.
perimeter of larger polygon: 120 yd: rttio: \(\frac{1}{6}\)
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 15

Question 16.
perimeter of larger polygon: 85 m; ratio: \(\frac{2}{5}\)

Answer:
The perimeter of smaller polygon is 34 m.

Explanation:
\(\frac { smaller }{ larger } \) = \(\frac { x }{ 85 } \) = \(\frac{2}{5}\)
85 x 2 = 5x
5x = 170
x = \(\frac { 170 }{ 5 } \) = 34

Question 17.
MODELING WITH MATHEMATICS
A school gymnasium is being remodeled. The basketball court will be similar to an NCAA basketball court, which has a length of 94 feet and a width of 50 feet. The school plans to make the width of the new court 45 feet. Find the perimeters of ail NCAA court and of the new court in the school.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 17

Question 18.
MODELING WITH MATHEMATICS
Your family has decided to put a rectangular patio in your backyard. similar to the shape of your backyard. Your backyard has a length of 45 feet and a width of 20 feet. The length of your new patio is 18 feet. Find the perimeters of your backyard and of the patio.

Answer:
The perimeter of the backyard is 130 ft.
Perimeter of patio is 52 ft

Explanation:
Draw a rectangle to represent the patio and a larger rectangle to represent our backyard and its going to similar figures
Big Ideas Math Geometry Answers Chapter 8 1
Perimeter of backyard = 2(45 + 20) = 2(65)
= 130 ft
Scale factor = \(\frac { 18 }{ 45 } \) = \(\frac { 2 }{ 5 } \)
So, \(\frac { perimeter of patio }{ perimeter of backyard } \) = \(\frac { 2 }{ 5 } \)
\(\frac { perimeter of patio }{ 130 } \) = \(\frac { 2 }{ 5 } \)
Perimeter of patio = \(\frac { 260 }{ 5 } \) = 52 ft

In Exercises 19-22, the polygons are similar. The area of one polygon is given. Find the area of the other polygon.

Question 19.
Big Ideas Math Geometry Answers Chapter 8 Similarity 26
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 19

Question 20.
Big Ideas Math Geometry Answers Chapter 8 Similarity 27

Answer:
Area of the larger triangle is 90 cm²

Explanation:
\(\frac { 10 }{ A } \) = (\(\frac { 4 }{ 12 } \))²
\(\frac { 10 }{ A } \) = \(\frac { 1 }{ 9 } \)
A = 10 x 9
A = 90

Question 21.
Big Ideas Math Geometry Answers Chapter 8 Similarity 28
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 21

Question 22.
Big Ideas Math Geometry Answers Chapter 8 Similarity 29

Answer:
Area of smaller triangle = 6 sq cm

Explanation:
\(\frac { A }{ 96 } \) = (\(\frac { 3 }{ 12 } \))²
\(\frac { A }{ 96 } \) = \(\frac { 1 }{ 16 } \)
16A = 96
A = \(\frac { 96 }{ 16 } \)
A = 6

Question 23.
ERROR ANALYSIS
Describe and correct the error in finding the perimeter of triangle B. The triangles are similar.
Big Ideas Math Geometry Answers Chapter 8 Similarity 30
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 23

Question 24.
ERROR ANALYSIS
Describe and correct the error in finding the area of triangle B. The triangles are similar.
Big Ideas Math Geometry Answers Chapter 8 Similarity 31

Answer:
Because the first ratio has a side of A over the side length of B, the square of the second ratio should have the area of B over the area of A.
\(\frac { 24 }{ x } \) = (\(\frac { 6 }{ 18 } \))²
\(\frac { 24 }{ x } \) = \(\frac { 1 }{ 9 } \)
x = 24 x 9
x = 216

In Exercises 25 and 26, decide whether the red and blue polygons are similar.

Question 25.
Big Ideas Math Geometry Answers Chapter 8 Similarity 32
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 25

Question 26.
Big Ideas Math Geometry Answers Chapter 8 Similarity 33

Answer:
Yes
Because both shapes are apparent and their side lengths are proportional and their corresponding angles are congruent.

Question 27.
REASONING
Triangles ABC and DEF are similar. Which statement is correct? Select all that apply.
(A) \(\frac{B C}{E F}=\frac{A C}{D F}\)
(B) \(\frac{A B}{D E}=\frac{C A}{F E}\)
(C) \(\frac{A B}{E F}=\frac{B C}{D E}\)
(D) \(\frac{C A}{F D}=\frac{B C}{E F}\)
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 27

ANALYZING RELATIONSHIPS
In Exercises 28 – 34, JKLM ~ EFGH.

Big Ideas Math Geometry Answers Chapter 8 Similarity 45

Question 28.
Find the scale factor of JKLM to EFGH.

Answer:
scale factor = \(\frac { EF }{ JK } \) =\(\frac { 8 }{ 20 } \)
k = \(\frac { 2 }{ 5 } \)

Question 29.
Find the scale factor of EFGH to JKLM.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 29

Question 30.
Find the values of x, y, and z.

Answer:
x = \(\frac { 55 }{ 2 } \)
y = 12
z = 65°

Explanation:
\(\frac { KL }{ GF } \) = \(\frac { x }{ 11 } \) = \(\frac { 5 }{ 2 } \)
2x = 55
x = \(\frac { 55 }{ 2 } \)
\(\frac { MJ }{ HE } \) = \(\frac { 30 }{ y } \) = \(\frac { 5 }{ 2 } \)
5y = 60
y = 12

Question 31.
Find the perimeter of each polygon.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 31

Question 32.
Find the ratio of the perimeters of JKLM to EFGH.

Answer:
The perimeter of JKLM : Perimeter of EFGH = 85 : 34

Question 33.
Find the area of each polygon.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 33

Question 34.
Find the ratio of the areas of JKLM to EFGH.

Answer:
Area of JKLM : Area of EFGH = 378.125 : 60.5 = 25 : 4

Question 35.
USING STRUCTURE
Rectangle A is similar to rectangle B. Rectangle A has side lengths of 6 and 12. Rectangle B has a side length of 18. What are the possible values for the length of the other side of rectangle B? Select all that apply.
(A) 6
(B) 9
(C) 24
(D) 36
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 35

Question 36.
DRAWING CONCLUSIONS
In table tennis, the table is a rectangle 9 feet long and 5 feet wide. A tennis Court is a rectangle 78 feet long and 36 feet wide. Are the two surfaces similar? Explain. If so, find the scale factor of the tennis court to the table.
Big Ideas Math Geometry Answers Chapter 8 Similarity 34

Answer:
The tennis table and court are not similar

Explanation:
If two figures are similar then their angles are congruent and sides are proportional.
If the tennis court and table are similar, then
\(\frac { length of table }{ length of court } \) = \(\frac { width of table }{ width of court } \)
\(\frac { 9 }{ 78 } \) = \(\frac { 5 }{ 36 } \)
9 • 36 = 5 • 78
324 = 390
So, Table and court are not similar.

MATHEMATICAL CONNECTIONS
In Exercises 37 and 38, the two polygons are similar. Find the values of x and y.

Question 37.
Big Ideas Math Geometry Answers Chapter 8 Similarity 35
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 37

Question 38.
Big Ideas Math Geometry Answers Chapter 8 Similarity 36

Answer:
x = 7.5

Explanation:
\(\frac { x }{ 5 } \) = \(\frac { 6 }{ 4 } \)
x = \(\frac { 15 }{ 2 } \)

ATTENDING TO PRECISION
In Exercises 39 – 42. the figures are similar. Find the missing corresponding side length.

Question 39.
Figure A has a pen meter of 72 meters and one of the side lengths is 18 meters. Figure B has a perimeter of 120 meters.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 39

Question 40.
Figure A has a perimeter of 24 inches. Figure B has a perimeter of 36 inches and one of the side lengths is 12 inches.

Answer:
The corresponding side length of figure A is 8 in

Explanation:
\(\frac { Perimeter of A }{ Perimeter of B } \) = \(\frac { Side length of A }{ Side length of B } \)
\(\frac { 24 }{ 36 } \) = \(\frac { x }{ 12 } \)
\(\frac { 2 }{ 3 } \) = \(\frac { x }{ 12 } \)
12 • 2 = 3x
x = 8

Question 41.
Figure A has an area of 48 square feet and one of the side lengths is 6 feet. Figure B has an area of 75 square feet.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 41

Question 42.
Figure A has an area of 18 square feet. Figure B has an area of 98 square feet and one of the side lengths is 14 feet.

Answer:
The corresponding side length of figure A is 6 ft.

Explanation:
\(\frac { Area of A }{ Area of B } \) = (\(\frac { Side length of A }{ Side length of B } \))²
\(\frac { 18 }{ 98 } \) = (\(\frac { x }{ 14 } \))²
\(\frac { 9 }{ 49 } \) = \(\frac { x² }{ 196 } \)
x² = 36
x = 6

CRITICAL THINKING
In Exercises 43-48, tell whether the polygons are always, sometimes, or never similar.

Question 43.
two isosceles triangles
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 43

Question 44.
two isosceles trapezoids

Answer:
Two isosceles trapezoids are sometimes similar.

Question 45.
two rhombuses
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 45

Question 46.
two squares

Answer:
Two squares are always similar.

Question 47.
two regular polygons
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 47

Question 48.
a right triangle and an equilateral triangle

Answer:
A right triangle and an equilateral triangle are never similar.

Question 49.
MAKING AN ARGUMENT
Your sister claims that when the side lengths of two rectangles are proportional, the two rectangles must be similar. Is she correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 49

Question 50.
HOW DO YOU SEE IT?
You shine a flashlight directly on an object to project its image onto a parallel screen. Will the object and the image be similar? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 8 Similarity 37
Answer:
The object and image are similar.

Question 51.
MODELING WITH MATHEMATICS
During a total eclipse of the Sun, the moon is directly in line with the Sun and blocks the Sun’s rays. The distance DA between Earth and the Sun is 93,00,000 miles. the distance DE between Earth and the moon is 2,40,000 miles, and the radius AB of the Sun is 432,5000 miles. Use the diagram and the given measurements to estimate the radius EC of the moon.
Big Ideas Math Geometry Answers Chapter 8 Similarity 38
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 51

Question 52.
PROVING A THEOREM
Prove the Perimeters of Similar Polygons Theorem (Theorem 8.1) for similar rectangles. Include a diagram in your proof.

Answer:
Big Ideas Math Geometry Answers Chapter 8 2
\(\frac { PQ + QR + RS + SP }{ KL + LM + MN + NK } \) = \(\frac { PQ }{ KL } \) = \(\frac { QR }{ LM } \) = \(\frac { RS }{ MN } \) = \(\frac { SP }{ NK } \)

Question 53.
PROVING A THEOREM
Prove the Areas of Similar Polygons Theorem (Theorem 8.2) for similar rectangles. Include a diagram in our proof.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 53

Question 54.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry. all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. A plane is the surface of the sphere. In spherical geometry, is it possible that two triangles are similar but not congruent? Explain your reasoning.
Answer:

Question 55.
CRITICAL THINKING
In the diagram, PQRS is a square, and PLMS ~ LMRQ. Find the exact value of x. This value is called the golden ratio. Golden rectangles have their length and width in this ratio. Show that the similar rectangles in the diagram are golden rectangles.
Big Ideas Math Geometry Answers Chapter 8 Similarity 39
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 55

Question 56.
MATHEMATICAL CONNECTIONS
The equations of the lines shown are y = \(\frac{4}{3}\)x + 4 and y = \(\frac{4}{3}\)x – 8. Show that ∆AOB ~ ∆COD.
Big Ideas Math Geometry Answers Chapter 8 Similarity 40
Answer:
The two lines slopes are equal and triangles angles are congruent and side lengths are proportional. So, triangles are similar.

Maintaining Mathematical proficiency

Find the value of x.

Question 57.
Big Ideas Math Geometry Answers Chapter 8 Similarity 41
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 57

Question 58.
Big Ideas Math Geometry Answers Chapter 8 Similarity 42

Answer:
x = 66°

Explanation:
x + 24 + 90 = 180
x + 114 = 180
x = 180 – 114
x = 66

Question 59.
Big Ideas Math Geometry Answers Chapter 8 Similarity 43
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.1 Answ 59

Question 60.
Big Ideas Math Geometry Answers Chapter 8 Similarity 44

Answer:
x = 60°

Explanation:
x + x + x = 180
3x = 180
x = 60

8.2 Proving Triangle Similarity by AA

Exploration 1

Comparing Triangles

Work with a partner. Use dynamic geometry software.

a. Construct ∆ABC and ∆DEF So that m∠A = m∠D = 106°, m∠B = m∠E = 31°, and ∆DEF is not congruent to ∆ABC.
Big Ideas Math Answers Geometry Chapter 8 Similarity 46
Answer:
m∠C ≠ m∠F

b. Find the third angle measure and the side lengths of each triangle. Copy the table below and record our results in column 1.
Big Ideas Math Answers Geometry Chapter 8 Similarity 47
Answer:

c. Are the two triangles similar? Explain.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
Answer:

d. Repeat parts (a) – (c) to complete columns 2 and 3 of the table for the given angle measures.
Answer:

e. Complete each remaining column of the table using your own choice of two pairs of equal corresponding angle measures. Can you construct two triangles in this way that are not similar?
Answer:

f. Make a conjecture about any two triangles with two pairs of congruent corresponding angles.
Answer:

Communicate Your Answer

Big Ideas Math Answers Geometry Chapter 8 Similarity 48

Question 2.
What can you conclude about two triangles when you know that two pairs of corresponding angles are congruent?
Answer:

Question 3.
Find RS in the figure at the left.
Answer:

Lesson 8.2 Proving Triangle Similarity by AA

Monitoring Progress

Show that the triangles are similar. Write a similarity statement.

Question 1.
∆FGH and ∆RQS
Big Ideas Math Answers Geometry Chapter 8 Similarity 49
Answer:
∆FGH and ∆RQS are similar by the AA similarity theorem.

Question 2.
∆CDF and ∆DEF
Big Ideas Math Answers Geometry Chapter 8 Similarity 50

Answer:

Question 3.
WHAT IF?
Suppose that \(\overline{S R}\) Big Ideas Math Answers Geometry Chapter 8 Similarity 51 \(\overline{T U}\) in Example 2 part (b). Could the triangles still be similar? Explain.
Answer:

Question 4.
WHAT IF?
A child who is 58 inches tall is standing next to the woman in Example 3. How long is the child’s shadow’?
Answer:

Question 5.
You are standing outside, and you measure the lengths 0f the shadows cast by both you and a tree. Write a proportion showing how you could find the height of the tree.
Answer:

Exercise 8.2 Proving Triangle Similarity by AA

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
If two angles of one triangle are congruent to two angles of another triangle. then the triangles are __________ .
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 1

Question 2.
WRITING
Can you assume that corresponding sides and corresponding angles of any two similar triangles are congruent? Explain.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. determine whether the triangles are similar. If they are, write a similarity statement. Explain your reasoning.

Question 3.
Big Ideas Math Answers Geometry Chapter 8 Similarity 52
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 3

Question 4.
Big Ideas Math Answers Geometry Chapter 8 Similarity 53
Answer:

Question 5.
Big Ideas Math Answers Geometry Chapter 8 Similarity 54
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 5

Question 6.
Big Ideas Math Answers Geometry Chapter 8 Similarity 55
Answer:

In Exercises 7 – 10. show that the two triangles are similar.

Question 7.
Big Ideas Math Answers Geometry Chapter 8 Similarity 56
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 7

Question 8.
Big Ideas Math Answers Geometry Chapter 8 Similarity 57
Answer:

Question 9.
Big Ideas Math Answers Geometry Chapter 8 Similarity 58
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 9

Question 10.
Big Ideas Math Answers Geometry Chapter 8 Similarity 59
Answer:

In Exercises 11 – 18, use the diagram to copy and complete the statement.

Big Ideas Math Answers Geometry Chapter 8 Similarity 60

Question 11.
∆CAG ~ _________
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 11

Question 12.
∆DCF ~ _________
Answer:

Question 13.
∆ACB ~ _________
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 13

Question 14.
m∠ECF = _________
Answer:

Question 15.
m∠ECD = _________
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 15

Question 16.
CF = _________
Answer:

Question 17.
BC = _________
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 17

Question 18.
DE = _________
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in using the AA Similarity Theorem (Theorem 8.3).
Big Ideas Math Answers Geometry Chapter 8 Similarity 61
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 19

Question 20.
ERROR ANALYSIS
Describe and correct the error in finding the value of x.
Big Ideas Math Answers Geometry Chapter 8 Similarity 62
Answer:

Question 21.
MODELING WITH MATHEMATICS
You can measure the width of the lake using a surveying technique, as shown in the diagram. Find the width of the lake, WX. Justify your answer.
Big Ideas Math Answers Geometry Chapter 8 Similarity 63
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 21

Question 22.
MAKING AN ARGUMENT
You and your cousin are trying to determine the height of a telephone pole. Your cousin tells you to stand in the pole’s shadow so that the tip of your shadow coincides with the tip of the pole’s shadow. Your Cousin claims to be able to use the distance between the tips of the shadows and you, the distance between you and the pole, and your height to estimate the height of the telephone pole. Is this possible? Explain. Include a diagram in your answer.
Answer:

REASONING
In Exercises 23 – 26, is it possible for ∆JKL and ∆XYZ to be similar? Explain your reasoning.

Question 23.
m∠J = 71°, m∠K = 52°, m∠X = 71°, and m∠Z = 57°
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 23

Question 24.
∆JKL is a right triangle and m∠X + m∠Y= 150°.
Answer:

Question 25.
m∠L = 87° and m∠Y = 94°
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 25

Question 26.
m∠J + m∠K = 85° and m∠Y + m∠Z = 80°
Answer:

Question 27.
MATHEMATICAL CONNECTIONS
Explain how you can use similar triangles to show that any two points on a line can be used to find its slope.
Big Ideas Math Answers Geometry Chapter 8 Similarity 64
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 27

Question 28.
HOW DO YOU SEE IT?
In the diagram, which triangles would you use to find the distance x between the shoreline and the buoy? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 8 Similarity 65
Answer:

Question 29.
WRITING
Explain why all equilateral triangles are similar.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 29

Question 30.
THOUGHT PROVOKING
Decide whether each is a valid method of showing that two quadrilaterals are similar. Justify your answer.
a. AAA
Answer:

b. AAAA
Answer:

Question 31.
PROOF
Without using corresponding lengths in similar polygons. prove that the ratio of two corresponding angle bisectors in similar triangles is equal to the scale factor.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 31

Question 32.
PROOF
Prove that if the lengths of two sides of a triangle are a and b, respectively, then the lengths of the corresponding altitudes to those sides are in the ratio \(\frac{b}{a}\).
Answer:

Question 33.
MODELING WITH MATHEMATICS
A portion of an amusement park ride is shown. Find EF. Justify your answer.
Big Ideas Math Answers Geometry Chapter 8 Similarity 66
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 33.1
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 33.2

Maintaining Mathematical Practices

Determine whether there is enough information to prove that the triangles are congruent. Explain your reasoning.

Question 34.
Big Ideas Math Answers Geometry Chapter 8 Similarity 67
Answer:

Question 35.
Big Ideas Math Answers Geometry Chapter 8 Similarity 68
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.2 Answ 35

Question 36.
Big Ideas Math Answers Geometry Chapter 8 Similarity 69
Answer:

8.1 & 8.2 Quiz

List all pairs of congruent angles. Then write the ratios of the corresponding side lengths in a statement of proportionality.

Question 1.
∆BDG ~ ∆MPQ
Big Ideas Math Answers Geometry Chapter 8 Similarity 70
Answer:

Question 2.
DEFG ~ HJKL
Big Ideas Math Answers Geometry Chapter 8 Similarity 71
Answer:

The polygons are similar. Find the value of x.

Question 3.
Big Ideas Math Answers Geometry Chapter 8 Similarity 72
Answer:

Question 4.
Big Ideas Math Answers Geometry Chapter 8 Similarity 73
Answer:

Determine whether the polygons are similar. If they are, write a similarity statement. Explain your reasoning. (Section 8.1 and Section 8.2)

Question 5.
Big Ideas Math Answers Geometry Chapter 8 Similarity 74
Answer:

Question 6.
Big Ideas Math Answers Geometry Chapter 8 Similarity 75
Answer:

Question 7.
Big Ideas Math Answers Geometry Chapter 8 Similarity 76
Answer:

Show that the two triangles are similar.

Question 8.
Big Ideas Math Answers Geometry Chapter 8 Similarity 77
Answer:

Question 9.
Big Ideas Math Answers Geometry Chapter 8 Similarity 78
Answer:

Question 10.
Big Ideas Math Answers Geometry Chapter 8 Similarity 79
Answer:

Question 11.
The dimensions of an official hockey rink used by the National Hockey League (NHL) are 200 feet by 85 feet. The dimensions of an air hockey table are 96 inches by 408 inches. Assume corresponding angles are congruent. (Section 8.1)
a. Determine whether the two surfaces are similar.
Answer:

b. If the surfaces are similar, find the ratio of their perimeters and the ratio ol their areas. If not, find the dimensions of an air hockey table that are similar to an NHL hockey rink.
Answer:

Question 12.
you and a friend buy camping tents made by the same company but in different sizes and colors. Use the information given in the diagram to decide whether the triangular faces of the tents are similar. Explain your reasoning. (Section 8.2)
Big Ideas Math Answers Geometry Chapter 8 Similarity 80
Answer:

8.3 Proving Triangle Similarity by SSS and SAS

Exploration 1

Deciding Whether Triangles Are Similar

Work with a partner: Use dynamic geometry software.

a. Construct ∆ABC and ∆DEF with the side lengths given in column 1 of the table below.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 81
Answer:

b. Copy the table and complete column 1.
Answer:

c. Are the triangles similar? Explain your reasoning.
Answer:

d. Repeat parts (a) – (c) for columns 2 – 6 in the table.
Answer:

e. How are the corresponding side lengths related in each pair of triangles that are similar? Is this true for each pair of triangles that are not similar?
Answer:

f. Make a conjecture about the similarity of two triangles based on their corresponding side lengths.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to analyze situations by breaking them into cases and recognize and use counter examples.
Answer:

g. Use your conjecture to write another set of side lengths of two similar triangles. Use the side lengths to complete column 7 of the table.
Answer:

Exploration 2

Deciding Whether Triangles Are Similar

Work with a partner: Use dynamic geometry software. Construct any ∆ABC.
a. Find AB, AC, and m∠A. Choose any positive rational number k and construct ∆DEF so that DE = k • AB, DF = k • AC, and m∠D = m∠A.
Answer:

b. Is ∆DEF similar to ∆ABC? Explain your reasoning.
Answer:

c. Repeat parts (a) and (b) several times by changing ∆ABC and k. Describe your results.
Answer:

Communicate Your Answer

Question 3.
What are two ways to use corresponding sides of two triangles to determine that the triangles are similar?
Answer:

Lesson 8.3 Proving Triangle Similarity by SSS and SAS

Monitoring progress

Use the diagram.

Big Ideas Math Geometry Answer Key Chapter 8 Similarity 82

Question 1.
Which of the three triangles are similar? Write a similarity statement.

Answer:
The ratios are equal. So, △LMN, △XYZ are similar.
The ratios are not equal. So △LMN, △RST are not similar.

Explanation:
Compare △LMN, △XYZ by finding the ratios of corresponding side lengths
Shortest sides: \(\frac { LM }{ YZ } \) = \(\frac { 20 }{ 30 } \) = \(\frac { 2 }{ 3 } \)
Longest sides: \(\frac { LN }{ XY } \) = \(\frac { 26 }{ 39 } \) = \(\frac { 2 }{ 3 } \)
Remaining sides: \(\frac { MN }{ ZX } \) = \(\frac { 24 }{ 36 } \) = \(\frac { 2 }{ 3 } \)
The ratios are equal. So, △LMN, △XYZ are similar.
Compare △LMN, △RST by finding the ratios of corresponding side lengths
Shortest sides: \(\frac { LM }{ RS } \) = \(\frac { 20 }{ 24 } \) = \(\frac { 5 }{ 6 } \)
Longest sides: \(\frac { LN }{ ST } \) = \(\frac { 26 }{ 33 } \)
Remaining sides: \(\frac { MN }{ RT } \) = \(\frac { 24 }{ 30 } \) = \(\frac { 4 }{ 5 } \)
The ratios are not equal. So △LMN, △RST are not similar.

Question 2.
The shortest side of a triangle similar to ∆RST is 12 units long. Find the other side 1enths of the triangle.

Answer:
The other side lengths of the triangle are 15 units, 16.5 units.

Explanation:
The shortest side of a triangle similar to ∆RST is 12 units
Scale factor = \(\frac { 12 }{ 24 } \) = \(\frac { 1 }{ 2 } \)
So, other sides are 33 x \(\frac { 12 }{ 2 } \) = 16.5, 30 x \(\frac { 12 }{ 2 } \) = 15.

Explain how to show that the indicated triangles are similar.

Question 3.
∆SRT ~ ∆PNQ
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 83

Answer:
The shorter sides: \(\frac { 18 }{ 24 } \) = \(\frac { 3 }{ 4 } \)
Longer sides: \(\frac { 21 }{ 28 } \) = \(\frac { 3 }{ 4 } \)
The side lengths are proportional. So ∆SRT ~ ∆PNQ

Question 4.
∆XZW ~ ∆YZX
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 84

Answer:
∆XZW and ∆YZX are not proportional.

Explanation:
The shorter sides: \(\frac { 9 }{ 16 } \)
Longer sides: \(\frac { 15 }{ 20 } \) = \(\frac { 3 }{ 4 } \)
The side lengths are not proportional. So ∆XZW and ∆YZX are not proportional.

Exercise 8.3 Proving Triangle Similarity by SSS and SAS

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
You plan to show that ∆QRS is similar to ∆XYZ by the SSS Similarity Theorem (Theorem 8.4). Copy and complete the proportion that you will use:
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 85
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which triangle does not belong with the other three? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 86
Answer:

Monitoring progress and Modeling with Mathematics

In Exercises 3 and 4, determine whether ∆JKL or ∆RST is similar to ∆ABC.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 87
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 88
Answer:

In Exercises 5 and 6, find the value of x that makes ∆DEF ~ ∆XYZ.

Question 5.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 89
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 90
Answer:

In Exercises 7 and 8, verify that ∆ABC ~ ∆DEF Find the scale factor of ∆ABC to ∆DEF

Question 7.
∆ABC: BC = 18, AB = 15, AC = 12
∆DEF: EF = 12, DE = 10, DF = 8
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 7

Question 8.
∆ABC: AB = 10, BC = 16, CA = 20
∆DEF: DE = 25, EF = 40, FD =50
Answer:

In Exercises 9 and 10. determine whether the two triangles are similar. If they are similar, write a similarity statement and find the scale factor of triangle B to triangle A.

Question 9.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 91
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 9

Question 10.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 92
Answer:

In Exercises 11 and 12, sketch the triangles using the given description. Then determine whether the two triangles can be similar.

Question 11.
In ∆RST, RS = 20, ST = 32, and m∠S = 16°. In ∆FGH, GH = 30, HF = 48, and m∠H = 24°.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 11

Question 12.
The side lengths of ∆ABC are 24, 8x, and 48, and the side lengths of ∆DEF are 15, 25, and 6x.

Answer:
\(\frac { AB }{ DE } \) = \(\frac { AC }{ DF } \) = \(\frac { BC }{ EF } \)
\(\frac { 24 }{ 15 } \) = \(\frac { 8x }{ 25 } \)
x = 5

In Exercises 13 – 16. show that the triangles are similar and write a similarity statement. Explain your reasoning.

Question 13.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 93
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 13

Question 14.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 94
Answer:

Question 15.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 95
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 15

Question 16.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 96
Answer:

In Exercises 17 and 18, use ∆XYZ.

Big Ideas Math Geometry Answer Key Chapter 8 Similarity 97

Question 17.
The shortest side of a triangle similar to ∆XYZ is 20 units long. Find the other side lengths of the triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 17

Question 18.
The longest side of a triangle similar to ∆XYZ is 39 units long. Find the other side lengths of the triangle.
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in writing a similarity statement.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 98
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 19

Question 20.
MATHEMATICAL CONNECTIONS
Find the value of n that makes ∆DEF ~ ∆XYZ when DE = 4, EF = 5, XY = 4(n + 1), YZ = 7n – 1, and ∠E ≅ ∠Y. Include a sketch.

Answer:
\(\frac { DE }{ XY } \) = \(\frac { EF }{ YZ } \)
\(\frac { 4 }{ 4(n + 1) } \) = \(\frac { 5 }{ 7n – 1 } \)
cross multiply the fractions
4(7n – 1) = 20(n + 1)
28n – 4 = 20n + 20
28n – 20n = 20 + 4
8n = 24
n = \(\frac { 24 }{ 8 } \)
n = 3

ATTENDING TO PRECISION
In Exercises 21 – 26, use the diagram to copy and complete the statement.

Big Ideas Math Geometry Answer Key Chapter 8 Similarity 99

Question 21.
m∠LNS = ___________
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 21

Question 22.
m∠NRQ = ___________
Answer:
m∠NRQ = m∠NRP = 91° by the vertical congruence

Question 23.
m∠NQR = ___________
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 23

Question 24.
RQ = ___________

Answer:
RQ = 4√3

Explanation:
Using the pythogrean theorem
NQ² = NR² + RQ²
8² = 4² + RQ²
64 = 16 + RQ²
64 – 16 = RQ²
48 = RQ²
RQ = 4√3

Question 25.
m∠NSM = ___________
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 25

Question 26.
m∠NPR = ___________

Answer:
m∠NPR = 28°

Explanation:
m∠NPR + m∠NRP + m∠RNP = 180°
m∠NPR + 91° + 61° = 180°
m∠NPR + 152° = 180°
m∠NPR = 180° – 152°
m∠NPR = 28°

Question 27.
MAKING AN ARGUMENT
Your friend claims that ∆JKL ~ ∆MNO by the SAS Similarity Theorem (Theorem 8.5) when JK = 18, m∠K = 130° KL = 16, MN = 9, m∠N = 65°, and NO = 8, Do you support your friend’s claim? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 27

Question 28.
ANALYZING RELATIONSHIPS
Certain sections of stained glass are sold in triangular, beveled pieces. Which of the three beveled pieces, if any, are similar?
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 100
Answer:
Out of three triangles, violet and blue triangles are similar.

Explanation:
Check the similarity of maroon and violet triangles.
longest sides: \(\frac { 5 }{ 7 } \)
shortest sides: \(\frac { 3 }{ 4 } \)
remaining sides: \(\frac { 3 }{ 4 } \)
The ratios are not equal. So those traingles are not similar.
Check the similarity of blue and violet triangles.
longest sides: \(\frac { 5 }{ 5.25 } \) = 1
shortest sides: \(\frac { 3 }{ 3 } \) = 1
remaining sides: \(\frac { 3 }{ 3 } \) = 1
The ratios are equal. So those traingles are similar.

Question 29.
ATTENDING TO PRECISION
In the diagram, \(\frac{M N}{M R}=\frac{M P}{M Q}\) Which of the statements must be true?
Select all that apply. Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 101
(A) ∠1 ≅∠2
(B) \(\overline{Q R}\) || \(\overline{N P}\)
(C)∠1 ≅ ∠4
(D) ∆MNP ~ ∆MRQ
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 29

Question 30.
WRITING
Are any two right triangles similar? Explain.

Answer:
Yes, any two right triangles can be similar. If two right triangles are similar, then the ratio of their longest, smallest and remaining side lengths must be equal and their angles must be congruent.

Question 31.
MODELING WITH MATHEMATICS
In the portion of the shuffleboard court shown, \(\frac{B C}{A C}=\frac{B D}{A E}\)
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 102
a. What additional information do you need to show that ∆BED ~ ∆ACE using the SSS Similarity Theorem (Theorem 8.4)?
b. What additional information do, you need to show that ∆BCD ~ ∆ACE using the SAS Similarity Theorem (Theorem 8.5)?
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 31

Question 32.
PROOF
Given that ∆BAC is a right triangle and D, E, and F are midpoints. prove that m∠DEF = 90°.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 103

Answer:
By observing the triangle ABC, m∠BAC = 90°
Join the midpoints of the sides of the triangle.
m∠DEF = 90°

Question 33.
PROVING A THEOREM
Write a two-column proof of the SAS Similarity Theorem (Theorem 8.5).
Given ∠A ≅ ∠D, \(\frac{A B}{D E}=\frac{A C}{D F}\)
Prove ∆ABC ~ ∆DEF
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 104
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 33.1
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 33.2
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 33.3

Question 34.
CRITICAL THINKING
You are given two right triangles with one pair of corresponding legs and the pair of hypotenuses having the same length ratios.
a. The lengths of the given pair of corresponding legs are 6 and 18, and the lengths of the hypotenuses are 10 and 30. Use the Pythagorean Theorem to find the lengths of the other pair of corresponding legs. Draw a diagram.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 3

b. Write the ratio of the lengths of the second pair of corresponding legs.
Answer:
First find the length of AC using pythagorean theorem
AC² + AB² = BC²
AC² + 36 = 100
AC² = 64
AC = 8
Find the length of DF using pythagorean theorem
DF² + DE² = EF²
DF² + 18² = 30²
DF² = 900 – 324
DF² = 576
DF= 24

c. Are these triangles similar? Does this suggest a Hypotenuse-Leg Similarity Theorem for right triangles? Explain.
Answer:
k = \(\frac { AC }{ DF } \) = \(\frac { 8 }{ 24 } \) = \(\frac { 1 }{ 3 } \)
k = \(\frac { AB }{ DE } \) = \(\frac { 6 }{ 18 } \) = \(\frac { 1 }{ 3 } \)
So, triangles are similar.

Question 35.
WRITING
Can two triangles have all three ratios of corresponding angle measures equal to a value greater than 1 ? less than 1 ? Explain.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 35

Question 36.
HOW DO YOU SEE IT?
Which theorem could you use to show that ∆OPQ ~ ∆OMN in the portion of the Ferris wheel shown when PM = QN = 5 feet and MO = NO = 10 feet?
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 105
Answer:
The corresponding angle theorem states that ∆OPQ is similar to ∆OMN.

Question 37.
DRAWING CONCLUSIONS
Explain why it is not necessary to have an Angle-Side-Angle Similarity Theorem.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 37

Question 38.
THOUGHT PROVOKING
Decide whether each is a valid method of showing that two quadrilaterals are similar. Justify your answer.
a. SASA
Answer:
If an angle of one triangle is congruent to an angle of a second triangle and the lengths of the sides including these angles are proportional, then the triangles are similar.

b. SASAS
Answer:
If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

c. SSSS
Answer:
If the corresponding side lengths of two triangles are proportional, then the triangles are similar.

d. SASSS
Answer:
If two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then those two triangles are similar.

Question 39.
MULTIPLE REPRESENTATIONS
Use a diagram to show why there is no Side-Side-Angle Similarity Theorem.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 39

Question 40.
MODELING WITH MATHEMATICS
The dimensions of an actual swing set are shown. You want to create a scale model of the swing set for a dollhouse using similar triangles. Sketch a drawing of your swing set and label each side length. Write a similarity statement for each pair of similar triangles. State the scale factor you used to create the scale model.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 106
Answer:
Here we have to check the similarity statement for △ABC, △DEF.
The scale factor k = \(\frac { AB }{ DE } \) = \(\frac { 8 }{ 6 } \) = \(\frac { 4 }{ 3 } \)

Question 41.
PROVING A THEOREM
Copy and complete the paragraph proot of the second part of the Slopes of Parallel Lines Theorem (Theorern 3. 13) from page 439.
Given ml = mn, l and n are nonvertical.
Prove l || n
You are given that ml = mn. By the definition of slope. ml = \(\frac{B C}{A C}\) and mn = \(\frac{E F}{D F}\) By ____________, \(\frac{B C}{A C}=\frac{E F}{D F}\). Rewriting this proportion yields ___________,

By the Right Angles Congruence Theorem (Thin. 2.3), ___________, So. ∆ABC ~ ∆DEF by ___________ . Because corresponding angles of similar triangles are congruent, ∠BAC ≅∠EDF. By ___________, l || n.
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 107
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 41

Question 42.
PROVING A THEOREM
Copy and complete the two-column proof 0f the second part of the Slopes of Perpendicular Lines Theorem (Theorem 3.14)
Given ml mn = – 1, l and n are nonvertical.
Prove l ⊥ n
Big Ideas Math Geometry Answer Key Chapter 8 Similarity 108

StatementsReasons
1. mlmn = – 11. Given
2. ml = \(\frac{D E}{A D}\), mn = \(\frac{A B}{B C}\)2. Definition of slope
3. \(\frac{D E}{A D} \cdot-\frac{A B}{B C}\) = – 13. ________________________________
4. \(\frac{D E}{A D}=\frac{B C}{A B}\)4. Multiply each side of statement 3 by –\(\frac{B C}{A B}\).
5. \(\frac{D E}{B C}\) = ____________5. Rewrite proportion.
6. ________________________________6. Right Angles Congruence Theorem (Thm. 2.3)
7. ∆ABC ~ ∆ADE7. ________________________________
8. ∠BAC ≅ ∠DAE8. Corresponding angles of similar figures are congruent.
9. ∠BCA ≅ ∠CAD9. Alternate Interior Angles Theorem (Thm. 3.2)
10. m∠BAC = m∠DAE, m∠BCA = m∠CAD10. ________________________________
11. m∠BAC + m∠BCA + 90° = 180°11. ________________________________
12. ________________________________12. Subtraction Property of Equality
13. m∠CAD + m∠DAE = 90°13. Substitution Property of Equality
14. m∠CAE = m∠DAE + m∠CAD14. Angle Addition Postulate (Post. 1.4)
15. m∠CAE = 90°15. ________________________________
16. ________________________________16. Definition of perpendicular lines

Answer:

StatementsReasons
1. mlmn = – 11. Given
2. ml = \(\frac{D E}{A D}\), mn = \(\frac{A B}{B C}\)2. Definition of slope
3. \(\frac{D E}{A D} \cdot-\frac{A B}{B C}\) = – 13. Correspomsding sides are opposite
4. \(\frac{D E}{A D}=\frac{B C}{A B}\)4. Multiply each side of statement 3 by –\(\frac{B C}{A B}\).
5. \(\frac{D E}{B C}\) = \(\frac { AB }{ AD } \)5. Rewrite proportion.
6. Two right-angled triangles are said to be congruent to each other if the hypotenuse and one side of the right triangle are equal to the hypotenuse and the corresponding side of the other right-angled triangle.6. Right Angles Congruence Theorem (Thm. 2.3)
7. ∆ABC ~ ∆ADE7. According to the side angle side theorem.
8. ∠BAC ≅ ∠DAE8. Corresponding angles of similar figures are congruent.
9. ∠BCA ≅ ∠CAD9. Alternate Interior Angles Theorem (Thm. 3.2)
10. m∠BAC = m∠DAE, m∠BCA = m∠CAD10. Congruent angles
11. m∠BAC + m∠BCA + 90° = 180°11. △ABC is a right-angled triangle
12. m∠CAD + m∠DAE = 90°12. Subtraction Property of Equality
13. m∠CAD + m∠DAE = 90°13. Substitution Property of Equality
14. m∠CAE = m∠DAE + m∠CAD14. Angle Addition Postulate (Post. 1.4)
15. m∠CAE = 90°15. Right Angle
16. If two lines meet each other a an angle of 90°, then they are called the perpendicular lines.16. Definition of perpendicular lines

Maintaining Mathematical proficiency

Find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio.

Question 43.
A(- 3, 6), B(2, 1); 3 to 2
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 43

Question 44.
A(- 3, – 5), B(9, – 1); 1 to 3
Answer:

Question 45.
A(1, – 2), B(8, 12); 4 to 3
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.3 Answ 45

8.4 Proportionality Theorems

Exploration 1

Discovering a Proportionality Relationship

Work with a partner. Use dynamic geometry software to draw any ∆ABC.
a. Construct \(\overline{D E}\) parallel to \(\overline{B C}\) with endpoints on \(\overline{A B}\) and \(\overline{A C}\), respectively.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 109
Answer:

b. Compare the ratios of AD to BD and AE to CE.
Answer:

c. Move \(\overline{D E}\) to other locations Parallel to \(\overline{B C}\) with endpoints on \(\overline{A B}\) and \(\overline{A C}\), and repeat part (b).
Answer:

d. Change ∆ABC and repeat parts (a) – (c) several times. Write a conjecture that summarizes your results.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Exploration 2

Discovering a Proportionality Relationship

Work with a partner. Use dynamic geometry software to draw any AABC.

Big Ideas Math Geometry Solutions Chapter 8 Similarity 110

a. Bisect ∆B and plot point D at the intersection of the angle bisector and \(\overline{A C}\).
Answer:

b. Compare the ratios of AD to DC and BA to BC.
Answer:

c. Change ∆ABC and repeat parts (a) and (b) several times. Write a conjecture that summarizes your results.
Answer:

Communicate Your Answer

Question 3.
What proportionality relationships exist in a triangle intersected by an angle bisector or by a line parallel to one of the sides?
Big Ideas Math Geometry Solutions Chapter 8 Similarity 111
Answer:

Question 4.
Use the figure at the right to write a proportion.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 111
Answer:

Lesson 8.4 Proportionality Theorems

Monitoring Progress

Question 1.
Find the length of \(\overline{Y Z}\).
Big Ideas Math Geometry Solutions Chapter 8 Similarity 112
Answer:
YZ = \(\frac { 315 }{ 11 } \)

Explanation:
Triangle property thorem is \(\frac { XW }{ WV } \) = \(\frac { XY }{ YZ } \)
\(\frac { 44 }{ 35 } \) = \(\frac { 36 }{ YZ } \)
cross multiply the fractions
44 • YZ = 36 • 35
44 • YZ = 1260
YZ = \(\frac { 1260 }{ 44 } \)
YZ = \(\frac { 315 }{ 11 } \)

Question 2.
Determine whether \(\overline{P S}\) || \(\overline{Q R}\)
Big Ideas Math Geometry Solutions Chapter 8 Similarity 113
Answer:
\(\frac { PQ }{ PN } \) = \(\frac { 50 }{ 90 } \) = \(\frac { 5 }{ 9 } \)
\(\frac { SR }{ SN } \) = \(\frac { 40 }{ 72 } \) = \(\frac { 5 }{ 9 } \)
\(\frac { PQ }{ PN } \) = \(\frac { SR }{ SN } \) so PS is parallel to QR

Find the length of the given line segment.

Question 3.
\(\overline{B D}\)
Big Ideas Math Geometry Solutions Chapter 8 Similarity 114

Answer:
\(\overline{B D}\) = 12

Explanation:
All the angles are congruent. So, \(\overline{A B}\), \(\overline{C D}\), \(\overline{E F}\) are parallel.
using the three parallel lines theorem
\(\frac { BD }{ DF } \) = \(\frac { AC }{ CE } \)
\(\frac { [latex]\overline{B D}\) }{ 30 } [/latex] = \(\frac { 16 }{ 40 } \)
\(\overline{B D}\) = \(\frac { 16 }{ 40 } \) • 30
\(\overline{B D}\) = 12

Question 4.
\(\overline{J M}\)
Big Ideas Math Geometry Solutions Chapter 8 Similarity 115

Answer:
\(\overline{J M}\) = \(\frac { 96 }{ 5 } \)

Explanation:
All the angles are congruent. So, \(\overline{G H}\), \(\overline{J K}\), \(\overline{M N}\) are parallel.
using the three parallel lines theorem
\(\frac { HK }{ KN } \) = \(\frac { GJ }{ JM } \)
\(\frac { 15 }{ 18 } \) = \(\frac { 16 }{ [latex]\overline{J M}\) } [/latex]
Cross multiply
15 • \(\overline{J M}\) = 16 • 18 = 288
\(\overline{J M}\) = \(\frac { 288 }{ 15 } \)
\(\overline{J M}\) = \(\frac { 96 }{ 5 } \)

Find the value of the variable.

Question 5.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 116

Answer:
x = 28

Explanation:
\(\overline{T V}\) is the angle bisector
So, \(\frac { ST }{ TU } \) = \(\frac { SV }{ VU } \)
\(\frac { 14 }{ x } \) = \(\frac { 24 }{ 48 } \)
cross multiply
24x = 14 • 48 = 672
x = \(\frac { 672 }{ 24 } \)
x = 28

Question 6.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 117

Answer:
x = 4√2

Explanation:
\(\overline{W Z}\) is the angle bisector
So, \(\frac { YZ }{ ZX } \) = \(\frac { YW }{ WX } \)
\(\frac { 4 }{ 4 } \) = \(\frac { 4√2 }{ x } \)
cross multiply
4x = 4 • 4√2 = 16√2
x = 4√2

Exercise 8.4 Proportionality Theorems

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE STATEMENT
If a line divides two sides of a triangle proportionally, then it is ____________ to the third side. This theorem is knon as the ____________ .
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 1

Question 2.
VOCABULARY
In ∆ABC, point R lies on \(\overline{B C}\) and \(\vec{A}\)R bisects ∆CAB. Write the proportionality statement for the triangle that is based on the Triangle Angle Bisector Theorem (Theorem 8.9).

Answer:
According to the triangle angle bisector theorem \(\frac { CR }{ BR } \) = \(\frac { AC }{ AB } \)

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, find the length of \(\overline{A B}\) .

Question 3.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 118
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 3

Question 4.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 119

Answer:
\(\frac { AE }{ ED } \) = \(\frac { AB }{ BC } \)
\(\frac { 14 }{ 12 } \) = \(\frac { AB }{ 18 } \)
AB = \(\frac { 14 }{ 12 } \) • 18
AB = 21 units.

In Exercises 5 – 8, determine whether \(\overline{K M}\) || \(\overline{J N}\).

Question 5.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 120
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 5

Question 6.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 121
Answer:
If \(\frac { JK }{ KL } \) = \(\frac { NM }{ ML } \), then KM || JN
\(\frac { JK }{ KL } \) = latex]\frac { 22.5 }{ 25 } [/latex] = latex]\frac { 9 }{ 10 } [/latex]
\(\frac { NM }{ ML } \) = \(\frac { 18 }{ 20 } \) = latex]\frac { 9 }{ 10 } [/latex]
\(\frac { JK }{ KL } \) = \(\frac { NM }{ ML } \)
Hence KM || JN

Question 7.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 122
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 7

Question 8.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 123
Answer:
If \(\frac { JK }{ KL } \) = \(\frac { NM }{ ML } \), then KM || JN
\(\frac { JK }{ KL } \) = latex]\frac {35 }{ 16 } [/latex]
\(\frac { NM }{ ML } \) = \(\frac { 34 }{ 15 } \)
\(\frac { JK }{ KL } \) ≠ \(\frac { NM }{ ML } \)
So, KM is not parallel to JN

CONSTRUCTION
In Exercises 9 – 12, draw a segment with the given length. Construct the point that divides the segment in the given ratio.
Question 9.
3 in.; 1 to 4
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 9

Question 10.
2 in.; 2 to 3

Answer:
Construct a 2 inch segment and divide the segment into 2 + 3 or 5 congruent pieces. Point P is the point that is \(\frac { 1 }{ 5 } \) of the way from point A to point B.

Question 11.
12 cm; 1 to 3
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 11

Question 12.
9 cm ; 2 to 5
Answer:
Construct a 9 cm segment and divide the segment into 2 + 5 or 7 congruent pieces. Point p is the point that is \(\frac { 1 }{ 7 } \) of the way from point A to point B.

In Exercises 13 – 16, use the diagram to complete the proportion.

Big Ideas Math Geometry Solutions Chapter 8 Similarity 124

Question 13.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 125
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 13

Question 14.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 126
Answer:
\(\frac { CG }{ EG } \) = \(\frac { BF }{ DF } \)

Question 15.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 127
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 15

Question 16.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 128
Answer:
\(\frac { BF }{ BD } \) = \(\frac { CG }{ CE } \)

In Exercises 17 and 18, find the length of the indicated line segment.

Question 17.
\(\overline{V X}\)
Big Ideas Math Geometry Solutions Chapter 8 Similarity 129
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 17

Question 18.
\(\overline{S U}\)
Big Ideas Math Geometry Solutions Chapter 8 Similarity 130
Answer:
\(\frac { SU }{ NS } \) = \(\frac { RT }{ PR } \)
\(\frac { SU }{ 10 } \) = \(\frac { 12 }{ 8 } \)
SU = \(\frac { 12 }{ 8 } \) • 10
SU = 10

In Exercises 19 – 22, find the value of the variable.

Question 19.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 131
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 19

Question 20.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 132
Answer:
\(\frac { z }{ 1.5 } \) = \(\frac { 3 }{ 4.5 } \)
z = \(\frac { 3 }{ 4.5 } \) • 1.5
z = 1

Question 21.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 133
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 21

Question 22.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 134
Answer:
\(\frac { q }{ 16 – q } \) = \(\frac { 36 }{ 28 } \)
28q = 36 (16 – q)
28q = 576 – 36q
28q + 36q = 576
64q = 576
q = 9

Question 23.
ERROR ANALYSIS
Describe and correct the error in solving for x.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 135
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 23

Question 24.
ERROR ANALYSIS
Describe and correct the error in the students reasoning.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 136
Answer:
\(\frac { BD }{ CD } \) = \(\frac { AB }{ AC } \)
BD = CD
So, 1 = \(\frac { AB }{ AC } \)
AC = AB

MATHEMATICAL CONNECTIONS
In Exercises 25 and 26, find the value of x for which \(\overline{P Q}\) || \(\overline{R S}\).

Question 25.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 137
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 25

Question 26.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 138
Answer:
\(\frac { PR }{ RT } \) = \(\frac { QS }{ ST } \)
\(\frac { 12 }{ 2x – 2 } \) = \(\frac { 21 }{ 3x – 1 } \)
12(3x – 1) = 21(2x – 2)
36x – 12 = 42x – 42
42x – 36x = 42 – 12
6x = 30
x = 5

Question 27.
PROVING A THEOREM
Prove the Triangle Proportionality Theorem (Theorem 8.6).
Big Ideas Math Geometry Solutions Chapter 8 Similarity 139
Given \(\overline{Q S}\) || \(\overline{T U}\)
Prove \(\frac{Q T}{T R}=\frac{S U}{U R}\)
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 27.1
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 27.2

Question 28.
PROVING A THEOREM
Prove the Converse of the Triangle Proportionality Theorem (Theorem 8.7).
Big Ideas Math Geometry Solutions Chapter 8 Similarity 140
Given \(\frac{Z Y}{Y W}=\frac{Z X}{X V}\)
Prove \(\overline{Y X}\) || \(\overline{W V}\)
Answer:

Question 29.
MODELING WITH MATHEMATICS
The real estate term lake frontage refers to the distance along the edge of a piece of property that touches a lake.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 141
a. Find the lake frontage (to the nearest tenth) of each lot shown.
b. In general, the more lake frontage a lot has, the higher its selling price. Which lot(s) should be listed for the highest price?
c. Suppose that low prices are in the same ratio as lake frontages. If the least expensive lot is $250,000, what are the prices of the other lots? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 29.1
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 29.2

Question 30.
USING STRUCTURE
Use the diagram to find the values of x and y.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 142
Answer:
\(\frac { 5 }{ 2 } \) = \(\frac { x }{ 1.5 } \)
x = \(\frac { 5 }{ 2 } \) • 1.5
x = 3.75
\(\frac { 3 }{ 7 } \) = \(\frac { y }{ 5.25 } \)
y = \(\frac { 3 }{ 7 } \) • 5.25
y = 2.25

Question 31.
REASONING
In the construction on page 447, explain why you can apply the Triangle Proportionality Theorem (Theorem 86) in Step 3.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 31

Question 32.
PROVING A THEOREM
Use the diagram with the auxiliary line drawn to write a paragraph proof of the Three Parallel Lines Theorem (Theorem 8.8).
Given K1 || K2 || K3
Prove \(\frac{C B}{B A}=\frac{D E}{E F}\)
Big Ideas Math Geometry Solutions Chapter 8 Similarity 143
Answer:
From the diagram, we can see that K₁ || K₂ || K₃
Those three parallel lines interest two traversals t₁, t₂
So, \(\frac{C B}{B A}=\frac{D E}{E F}\)

Question 33.
CRITICAL THINKING
In ∆LMN, the angle bisector of ∠M also bisects \(\overline{L N}\). Classify ∆LMN as specifically as possible. Justify your answer.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 33

Question 34.
HOW DO YOU SEE IT?
During a football game, the quarterback throws the ball to the receiver. The receiver is between two defensive players, as shown. If Player 1 is closer to the quarterback when the ball is thrown and both defensive players move at the same speed, which player will reach the receiver first? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 144
Answer:
As per the image, player 1 is closer to the receiver. So, player 1 will reach the receiver first.

Question 35.
PROVING A THEOREM
Use the diagram with the auxiliary lines drawn to write a paragraph proof of the Triangle Angle Bisector Theorem (Theorem 8.9).
Given ∠YXW ≅ ∠WXZ
prove \(\frac{Y W}{W Z}=\frac{X Y}{X Z}\)
Big Ideas Math Geometry Solutions Chapter 8 Similarity 145
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 35

Question 36.
THOUGHT PROVOKING
Write the converse of the Triangle Angle Bisector Theorem (Theorem 8.9). Is the converse true? Justify your answer.
Answer:

Question 37.
REASONING
How is the Triangle Midsegment Theorem (Theorem 6.8) related to the Triangle Proportionality Theorem (Theorem 8.6)? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 37

Question 38.
MAKING AN ARGUMENT
Two people leave points A and B at the same time. They intend to meet at point C at the same time. The person who leaves point A walks at a speed of 3 miles per hour. You and a friend are trying to determine how fast the person who leaves point B must walk. Your friend claims you need to know the length of \(\overline{A C}\). Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 8 Similarity 146
Answer:
My self starts walking from point A with a speed of 3 miles per hour and reaches point C.
My friend starts walking from point B with x speed and reaches point C.
\(\frac { AD }{ DC } \) = \(\frac { BE }{ CE } \)
I have to travel from A to C. So, I need to know distance between AC.
Therefore, my friend is correct.

Question 39.
CONSTRUCTION
Given segments with lengths r, s, and t, construct a segment of length x, such that \(\frac{r}{s}=\frac{t}{x}\)
Big Ideas Math Geometry Solutions Chapter 8 Similarity 147
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 39

Question 40.
PROOF
Prove Ceva’s Theorem: If P is any point inside ∆ABC, then \(\frac{A Y}{Y C} \cdot \frac{C X}{X B} \cdot \frac{B Z}{Z A}\) = 1
Big Ideas Math Geometry Solutions Chapter 8 Similarity 148
(Hint: Draw segments parallel to \(\overline{B Y}\) through A and C, as shown. Apply the Triangle Proportionality Theorem (Theorem 8.6) to ∆ACM. Show that ∆APN ~ ∆MPC, ∆CXM ~ ∆BXP, and ∆BZP ~ ∆AZN.)
Answer:

Maintaining Mathematical Proficiency

Use the triangle.

Big Ideas Math Geometry Solutions Chapter 8 Similarity 149

Question 41.
Which sides are the legs?
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 41

Question 42.
Which side is the hypotenuse?
Answer:
The leg c is the hypotenuse.

Solve the equation.

Question 43.
x2 = 121
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 43

Question 44.
x2 + 16 = 25
Answer:
x² + 16 = 25
x² = 25 – 16
x² = 9
x = 3

Question 45.
36 + x2 = 85
Answer:
Big Ideas Math Geometry Answers Chapter 8 Similarity 8.4 Answ 45

Similarity Review

8.1 Similar Polygons

Find the scale factor. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality.

Question 1.
ABCD ~ EFGH
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 150
Answer:
\(\frac { BC }{ CD } \) = \(\frac { 8 }{ 12 } \) = \(\frac { 2 }{ 3 } \)
\(\frac { EH }{ GH } \) = \(\frac { 6 }{ 9 } \) = \(\frac { 2 }{ 3 } \)
So, scale factor = \(\frac { 2 }{ 3 } \)

Question 2.
∆XYZ ~ ∆RPQ
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 151
Answer:
longer sides: \(\frac { 10 }{ 25 } \) = \(\frac { 2 }{ 5 } \)
shorter sides: \(\frac { 6 }{ 15 } \) = \(\frac { 2 }{ 5 } \)
remaining sides: \(\frac { 8 }{ 20 } \) = \(\frac { 2 }{ 5 } \)
So, scale factor = \(\frac { 2 }{ 5 } \)

Question 3.
Two similar triangles have a scale factor of 3 : 5. The altitude of the larger triangle is 24 inches. What is the altitude of the smaller triangle?
Answer:
Scale factor of smaller triangle to larger traingle is \(\frac { 3 }{ 5 } \) and larger traingle altitude is 24 inches
Let x be the smaller triangle altitude
\(\frac { altitude of smaller triangle }{ altitude of larger traingle } \) = scale factor
\(\frac { x }{ 24 } \) = \(\frac { 3 }{ 5 } \)
x = \(\frac { 3 }{ 5 } \) • 24
x = 14.4

Question 4.
Two similar triangles have a pair of corresponding sides of length 12 meters and 8 meters. The larger triangle has a perimeter of 48 meters and an area of 180 square meters. Find the perimeter and area of the smaller triangle.
Answer:
Scale factor = \(\frac { 2 }{ 3 } \)
perimeter of smaller triangle = 32
Area of smaller triangle = 80

Explanation:
The scale factor of smaller to larger traingle = \(\frac { 8 }{ 12 } \) = \(\frac { 2 }{ 3 } \)
\(\frac { perimeter of smaller triangle }{ perimeter of larger triangle } \) = scale factor\(\frac { perimeter of smaller triangle }{ 48 } \) = \(\frac { 2 }{ 3 } \)
perimeter of smaller triangle = \(\frac { 2 }{ 3 } \) • 48
perimeter of smaller triangle = 32
\(\frac { Area of smaller triangle }{ Area of larger triangle } \) = (scale factor)²
\(\frac { Area of smaller triangle }{ 180 } \) = ( \(\frac { 2 }{ 3 } \))²
Area of smaller triangle = \(\frac { 4 }{ 9 } \) • 180
= 80

8.2 Proving Triangle Similarity by AA

Show that the triangles are similar. Write a similarity statement.

Question 5.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 152
Answer:
m∠RQS = m∠UTS = 30°.
△QRS and △STU are similar as per the AA similarity theorem.

Question 6.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 153
Answer:
m∠CAB = 60°, m∠DEF = 30°
△ABC and △DEF are not similar as per the AA similarity theorem.

Question 7.
A cellular telephone tower casts a shadow that is 72 feet long, while a nearby tree that is 27 feet tall casts a shadow that is 6 feet long. How tall is the tower?
Answer:
\(\frac { shadow of tree }{ shadow of tower } \) = \(\frac { height of tree }{ height of tower } \)
\(\frac { 6 }{ 72 } \) = \(\frac { 27 }{ x } \)
6x = 1944
x = 324 ft
The height of the tower is 324 ft.

8.3 Proving Triangle Similarity by SSS and SAS

Use the SSS Similarity Theorem (Theorem 8.4) or the SAS Similarity Theorem (Theorem 8.5) to show that the triangles are similar.

Question 8.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 154
Answer:
\(\frac { DE }{ CD } \) = \(\frac { 7 }{ 3.5 } \) = 2
\(\frac { AB }{ BC } \) = \(\frac { 8 }{ 4 } \) = 2
\(\frac { DE }{ CD } \) = \(\frac { AB }{ BC } \)
So, BD is parallel to AE.

Question 9.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 155
Answer:
\(\frac { QU }{ TU } \) = \(\frac { 9 }{ 4.5 } \) = 2
\(\frac { QR }{ SR } \) = \(\frac { 14 }{ 7 } \) = 2
\(\frac { QU }{ TU } \) = \(\frac { QR }{ SR } \)
So, ST is parallel to RU.

Question 10.
Find the value of x that makes ∆ABC ~ ∆DEF
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 156

Answer:
\(\frac { 24 }{ 6 } \) = 4
\(\frac { 32 }{ 2x } \) = 4
32 = 8x
x = \(\frac { 32 }{ 8 } \)
x = 4

8.4 Proportionality Theorems

Determine whether \(\overline{A B}\) || \(\overline{C D}\)

Question 11.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 157

Answer:
\(\frac { DB }{ BE } \) = \(\frac { 10 }{ 16 } \) = \(\frac { 5 }{ 8 } \)
\(\frac { CA }{ AE } \) = \(\frac { 20 }{ 28 } \) = \(\frac { 5 }{ 7 } \)
\(\frac { DB }{ BE } \) ≠ \(\frac { CA }{ AE } \)
So, CD and AB are not parallel.

Question 12.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 158

Answer:
\(\frac { DB }{ BE } \) = \(\frac { 12 }{ 20 } \) = \(\frac { 3 }{ 5 } \)
\(\frac { CA }{ AE } \) = \(\frac { 13.5 }{ 22.5 } \) = \(\frac { 3 }{ 5 } \)
\(\frac { DB }{ BE } \) = \(\frac { CA }{ AE } \)
So, AB and CD are parallel.

Question 13.
Find the length of \(\overline{Y B}\).
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 159

Answer:
\(\frac { ZC }{ AZ } \) = \(\frac { 24 }{ 15 } \) = \(\frac { 8 }{ 5 } \)
\(\frac { YB }{ AY } \) = \(\frac { ZC }{ AZ } \)
\(\frac { YB }{ 7 } \) = \(\frac { 8 }{ 5 } \)
YB = \(\frac { 8 }{ 5 } \) • 7
YB = \(\frac { 56 }{ 5 } \)
The length of \(\overline{Y B}\) is \(\frac { 56 }{ 5 } \)

Find the length of \(\overline{A B}\).

Question 14.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 160

Answer:
\(\frac { AB }{ 7 } \) = \(\frac { 6 }{ 4 } \)
\(\overline{A B}\) = \(\frac { 6 }{ 4 } \) • 7
\(\overline{A B}\) = \(\frac { 42 }{ 4 } \)
\(\overline{A B}\) = \(\frac { 21 }{ 2 } \)
The length of \(\overline{A B}\) is \(\frac { 21 }{ 2 } \).

Question 15.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 161

Answer:
\(\frac { DB }{ CD } \) = \(\frac { AB }{ AC } \)
\(\frac { 4 }{ 10 } \) = \(\frac { AB }{ 18 } \)
\(\frac { 2 }{ 5 } \) = \(\frac { AB }{ 18 } \)
AB = \(\frac { 36 }{ 5 } \)

Similarity Test

Determine whether the triangles are similar. If they are, write a similarity statement. Explain your reasoning.

Question 1.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 162

Answer:
Longer sides: \(\frac { 32 }{ 24 } \) = \(\frac { 4 }{ 3 } \)
shorter sides: \(\frac { 18 }{ 14 } \) = \(\frac { 9 }{ 7 } \)
remaining sides: \(\frac { 20 }{ 15 } \) = \(\frac { 4 }{ 3 } \)
Those are not equal.
So, triangles are not similar.

Question 2.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 163

Answer:
\(\frac { AC }{ KJ } \) = \(\frac { 6 }{ 8 } \) = \(\frac { 3 }{ 4 } \)
\(\frac { BC }{ JL } \) = \(\frac { 8 }{ [latex]\frac { 32 }{ 3 } \) } [/latex] = \(\frac { 3 }{ 4 } \)
\(\frac { AC }{ KJ } \) = \(\frac { BC }{ JL } \)
∠C = ∠J
So, △ABC and △JLK are similar.

Question 3.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 164

Answer:
\(\frac { XY }{ XW } \) = \(\frac { PZ }{ PW } \)

Find the value of the variable.

Question 4.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 165

Answer:
\(\frac { 9 }{ w } \) = \(\frac { 15 }{ 5 } \)
\(\frac { 9 }{ w } \) = 3
9 = 3w
w = 3

Question 5.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 166

Answer:
\(\frac { 17.5 }{ 21 } \) = \(\frac { q }{ 33 } \)
q = \(\frac { 17.5 }{ 21 } \) • 33
q = \(\frac { 55 }{ 2 } \)

Question 6.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 167

Answer:
\(\frac { 21 – p }{ p } \) = \(\frac { 12 }{ 24 } \)
\(\frac { 21 – p }{ p } \) = \(\frac { 1 }{ 2 } \)
cross multiply
2(21 – p) = p
42 – 2p = p
42 = p + 2p
42 = 3p
p = 14

Question 7.
Given ∆QRS ~ ∆MNP, list all pairs of congruent angles, Then write the ratios of the corresponding side lengths in a statement of proportionality.

Answer:
The pairs of congruent anglres are m∠QRS, m∠RSQ, m∠SQR, m∠MNP, m∠NPM, m∠PMN.
The ratios of side lengths are \(\frac { RQ }{ MN } \), \(\frac { QS }{ MP } \), \(\frac { RS }{ NP } \)

Use the diagram.

Big Ideas Math Answer Key Geometry Chapter 8 Similarity 168

Question 8.
Find the length of \(\overline{E F}\).

Answer:
\(\frac { DE }{ EF } \) = \(\frac { CD }{ BC } \)
\(\frac { 3.2 }{ EF } \) = \(\frac { 2.8 }{ 1.4 } \)
EF = 1.6
The length of \(\overline{E F}\) is 1.6

Question 9.
Find the length of \(\overline{F G}\).

Answer:
\(\frac { EF }{ FG } \) = \(\frac { BC }{ AB } \)
\(\frac { 1.6 }{ FG } \) = \(\frac { 1.4 }{ 4.2 } \)
FG = 4.8
The length of \(\overline{F G}\) is 4.8

Question 10.
Is quadrilateral FECB similar to quadrilateral GFBA? If so, what is the scale factor of the dilation that maps quadrilateral FECB to quadrilateral GFBA?

Answer:
The scale factor of dilation from quadrilateral FECB to quadrilateral GFBA is \(\frac { GF }{ FE } \)

Question 11.
You are visiting the Unisphere at Flushing Meadows Corona Park in New York. To estimate the height of the stainless steel model of Earth. you place a mirror on the ground and stand where you can see the top of the model in the mirror. Use the diagram to estimate the height of the model. Explain why this method works.

Answer:

Question 12.
You are making a scale model of a rectangular park for a school project. Your model has a length of 2 feet and a width of 1.4 feet. The actual park is 800 yards long. What are the perimeter and area of the actual park?
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 169

Answer:
we know that 1 yard = ft
As per the similarity theorem
\(\frac { AD }{ EH } \) = \(\frac { AB }{ EF } \)
\(\frac { 2 }{ 800.3 } \) = \(\frac { 1.4 }{ EF } \)
EF = 1200 • 1 • 4
EF = 1680
Perimete of the park P = 2(1680 + 2400)
= 2(4080) = 8160 ft
Area of the actual park = 1680 • 2400 = 4,032,000 sq ft
Therefore, perimeter of the actual park = 8160 ft
area of the actual park is 4,032,000 sq ft.

Question 13.
In a Perspective drawing, lines that are parallel in real life must meet at a vanishing point on the horizon. To make the train cars in the drawing appear equal in length, they are drawn so that the lines connecting the opposite corners of each car are parallel. Use the dimensions given and the yellow parallel lines to find the length of the bottom edge of the drawing of Car 2.
Big Ideas Math Answer Key Geometry Chapter 8 Similarity 170
Answer:
\(\frac { 5.4 }{ 10.6 } \) = \(\frac { 5.4 + C }{ 19 } \)
C = \(\frac { 45.36 }{ 10.6 } \)
\(\frac { (19 – x – 8.4) }{ (19 – 8.4) } \) = \(\frac { 5.4 }{ 5.4 + c } \)
\(\frac { 19 – 8.4 }{ 19 } \) = \(\frac { c2 + 5.4 }{ c1 + c2 + 5.4 } \)
\(\frac { 19 – 8.4 }{ 19 } \) = \(\frac { 5.4 }{ 5.4 + c2 } \)
c1 = 7.6, c2 = 4.2, x = 4.6
The length of car 2 is 4.2 cm.

Similarity Cumulative Assessment

Question 1.
Use the graph of quadrilaterals ABCD and QRST.
Big Ideas Math Geometry Answers Chapter 8 Similarity 171
a. Write a composition of transformations that maps quadrilateral ABCD to quadrilateral QRST.
Answer:
The scale factor = \(\frac { AD }{ QT } \) = \(\frac { 2 }{ 1.5 } \)

b. Are the quadrilaterals similar? Explain your reasoning.
Answer:
No.
\(\frac { AD }{ QT } \) = \(\frac { 2 }{ 1.5 } \)
\(\frac { CD }{ TS } \) = \(\frac { 2.8 }{ 1.4 } \) = 2
So, quadrilaterals are not similar.

Question 2.
In the diagram. ABCD is a parallelogram. Which congruence theorem(s) could you Use to show that ∆AED ≅ ∆CEB? Select all that apply.
Big Ideas Math Geometry Answers Chapter 8 Similarity 172
SAS Congruence Theorem (Theorem 5.5)
Answer:
It states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent.

SSS Congruence Theorem (Theorem 5.8)
Answer:
If all the three sides of one triangle are equivalent to the corresponding three sides of the second triangle, then the two triangles are said to be congruent by SSS rule.

HL Congruence Theorem (Theorem 5.9)
Answer:
A given set of triangles are congruent if the corresponding lengths of their hypotenuse and one leg are equal.

ASA Congruence Theorem (Theorem 5. 10)
Answer:
If any two angles and the side included between the angles of one triangle are equivalent to the corresponding two angles and side included between the angles of the second triangle, then the two triangles are said to be congruent by ASA rule

AAS Congruence Theorem (Theorem 5. 11)
Answer:
AAS stands for Angle-angle-side. When two angles and a non-included side of a triangle are equal to the corresponding angles and sides of another triangle, then the triangles are said to be congruent.

Question 3.
By the Triangle Proportionality Theorem (Theorem 8.6), \(\frac{V W}{W Y}=\frac{V X}{X Z}\) In the diagram, VX > VW and XZ > WY. List three possible values for VX and XZ.
Big Ideas Math Geometry Answers Chapter 8 Similarity 173
Answer:
\(\frac{V W}{W Y}=\frac{V X}{X Z}\)
\(\frac{ 4 }{6}=\frac{V X}{X Z}\)
The possible values of VX are greater than 4 means 5, 6, 7, . . .
The possible values of XZ are greater than 6 means 7, 8, 9, . .

Question 4.
The slope of line l is – \(\frac{3}{4}\). The slope of line n is \(\frac{4}{3}\) What must be true about lines l and n ?
(A) Lines l and n are parallel.
(B) Lines l and n arc perpendicular.
(C) Lines l and n are skew.
(D) Lines l and n are the same line.
Answer:
The slope of l = – \(\frac{3}{4}\)
Slope of n = \(\frac{4}{3}\)
lines slopes are reciprocal and opposite. So, they are perpendicular.

Question 5.
Enter a statement or reason in each blank to complete the two-column proof.
Big Ideas Math Geometry Answers Chapter 8 Similarity 174
Given \(\frac{K J}{K L}=\frac{K H}{K M}\)
Prove ∠LMN ≅ ∠JHG

StatementsReasons
1. \(\frac{K J}{K L}=\frac{K H}{K M}\)1. Given
2. ∠JKH ≅ ∠LKM2. ________________________
3. ∆JKH ~ ∆LKM3. ________________________
4. ∠KHJ ≅∠KML4. ________________________
5. _______________________5. Definition of congruent angles
6. m∠KHJ + m∠JHG = 180°6. Linear Pair Postulate (Post. 18)
7. m∠JHG = 180° – m∠KHJ7. ________________________
8. m∠KML + m∠LMN = 180°8. ________________________
9. ________________________9. Subtraction Property of Equality
10. m∠LMN = 180° – m∠KHJ10. ________________________
11. ________________________11. Transitive Property of Equality
12. ∠LMN ≅ ∠JHG12. ________________________

Answer:

Question 6.
The coordinates of the vertices of ∆DEF are D(- 8, 5), E(- 5, 8), and F(- 1, 4), The coordinates of the vertices of ∆JKL are J(16, – 10), K(10, – 16), and L(2, – 8), ∠D ≅ ∠J. Can you show that ∆DEF ∆JKL by using the AA Similarity Theorem (Theorem 8.3)? If so, do so by listing the congruent corresponding angles and writing a similarity transformation that maps ∆DEF to ∆JKL. If not, explain why not.
Answer:
AA similarity theorem states that ∠D = ∠J. So, ∆DEF and ∆JKL are similar.

Question 7.
Classify the quadrilateral using the most specific name.
Big Ideas Math Geometry Answers Chapter 8 Similarity 175
rectangle     square    parallelogram    rhombus
Answer:

Question 8.
‘Your friend makes the statement “Quadrilateral PQRS is similar to quadrilateral WXYZ.” Describe the relationships between corresponding angles and between corresponding sides that make this statement true.

Answer:
When 2 figures are similar, then their corresponding angles are congruent and their corresponding lengths are proportional. hence if PQRS is similar to wxyZ, then the following statements are true.
∠P = ∠W, ∠Q = ∠X, ∠R = ∠Y and ∠S = ∠Z and
\(\frac { PQ }{ WX } \) = \(\frac { QR }{ XY } \) = \(\frac { RS }{ YZ } \) = \(\frac { PS }{ WZ } \) = k
Here is a constant of proportionality.

Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies

Big Ideas Math Answers Grade 3 Chapter 3

Detailed explanation for all questions of Big Ideas Math Book 3rd Grade Chapter 3 More Multiplication Facts and Strategies is provided here. Students can find the exact solution for every question of BIM 3rd Grade 3rd Chapter More Multiplication Facts and Strategies Book. So one who wants to become an expert in solving multiplication problems can download the Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies pdf from this page.

Big Ideas Math 3rd Grade Answer Key Chapter 3 More Multiplication Facts and Strategies

Topic-wise quick links of Big Ideas Math Book Grade 3 Chapter 3 More Multiplication Facts and Strategies Solutions are mentioned below. The different lessons covered in grade 3 chapter 3 More Multiplication Facts and Strategies are multiply by 3, multiply by 4, multiply by 6, multiply by 7,  multiply by 8, multiply by 9, and multiply three factors. Interested students must download BIM Grade3 Chapter 3 More Multiplication Facts and Strategies Answer Key pdf and start their preparation.

In addition to the exercise problems, you can also find the answers for homework, assignments and practice problems. Tap on the concept, you want to explore and get the solutions for all questions from the below sections.

Lesson 1 – Multiply by 3

Lesson 2 Multiply by 4

Lesson 3 Multiply by 6

Lesson 4 Multiply by 7

Lesson 5 Multiply by 8

Lesson 6 Multiply by 9

Lesson 7 Practice Multiplication Strategies

Lesson 8 Multiply Three Factors

Lesson 9 More Problem Solving: Multiplication

Performance Task

Lesson 3.1 Multiply by 3

Explore and Grow

Use the number line to find the product.
4 × 3 = _____
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 1

Answer: 12
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.1-1

Repeated Reasoning
Explain how you can use the number line to find products greater than 20. Complete the table
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 2
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.1-2

Think and Grow: Multiply by 3

Example
Find 5 × 3.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 3
Answer:
Distribute 3 to 2 and 1.
5 × 3 = 5 × (2 + 1)
5 × 3 = (5 × 2) + (5 × 1)
5 × 3 = 10 + 5
5 × 3 = 15

Show and Grow

Find the product
Question 1.
6 × 3
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 4
Answer:
Distribute 3 to 2 and 1.
6 × 3 = 6 × (2 + 1)
6 × 3 = (6 × 2) + (6 × 1)
6 × 3 = 12 + 6
6 × 3 = 18

Question 2.
3 × 4
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 5
Answer:
Distribute 3 to 2 and 1.
3 × 4 = (2 + 1) × 4
3 × 4 = (2 × 4) + (1 × 4)
3 × 4 = 8 + 4
3 × 4 = 12

Question 3.
2 × 3
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 6
Answer:
Distribute 3 to 2 and 1.
2 × 3 = 2 × (2 + 1)
2 × 3 = (2 × 2) + (2 × 1)
2 × 3 = 4 + 2
2 × 3 = 6

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 7
Answer:
Distribute 3 to 2 and 1.
3 × 7 = (2 + 1) × 7
3 × 7 = (2 × 7) + (1 × 7)
3 × 7 = 14 + 7
3 × 7 = 21

Question 5.
3 × 8 = ____
Answer:
Distribute 3 to 2 and 1.
3 × 8 = (2 + 1) × 8
3 × 8 = (2 × 8) + (1 × 8)
3 × 8 = 16 + 8
3 × 8 = 24

Question 6.
3 × 1 = ____
Answer:
Distribute 3 to 2 and 1.
3 × 1 = (2 + 1) × 1
3 × 1 = (2 × 1) + (1 × 1)
3 × 1 = 2 + 1
3 × 1 = 3

Question 7.
3 × 3 = _____
Answer:
Distribute 3 to 2 and 1.
3 × 3 = (2 + 1) × 3
3 × 3 = (2 × 3) + (1 × 3)
3 × 3 = 6 + 3
3 × 3 = 9

Question 8.
3 × 5 = _____
Answer:
Distribute 3 to 2 and 1.
3 × 5 = (2 + 1) × 5
3 × 5 = (2 × 5) + (1 × 5)
3 × 5 = 10 + 5
3 × 5 = 15

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 8
Answer:
Multiply the two numbers 3 and 6.
3 × 6 = 18

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 9
Answer:
Multiply the two numbers 3 and 10.
Multiply 3 by 0 and 3 by 1
10 × 3 = 30

Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 10
Answer:
Multiply the two numbers 3 and 4.
4 × 3 = 12

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 11
Answer:
Multiply the two numbers 3 and 9.
3 × 9 =27

Compare
Question 13.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 12
Answer: >

Explanation:
8 × 3 = 24
3 × 6 = 18
24 is greater than 18
Thus 8 × 3 > 3 × 6

Question 14.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 13
Answer: <

Explanation:
3 × 1 = 3
3 × 10 = 30
3 is less than 30.
3 × 1 < 3 × 10

Question 15.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 14
Answer: =

Explanation:
Any number multiplied by 0 will be always 0.
3 × 0 = 0
0 × 3 = 0
Thus 3 × 0 = 0 × 3

Question 16.
A baseball game has 9 innings. Each team gets 3 outs every inning. How many outs does each team get in one game?
Answer:
Given that,
A baseball game has 9 innings. Each team gets 3 outs every inning.
9 × 3 = 27
Therefore each team gets 27 outs in one game.

Question 17.
YOU BE THE TEACHER
Your friend says 23 is a multiple of 3 because there is a 3 in the ones place. Is your friend correct? Explain.
Answer:
Given,
Your friend says 23 is a multiple of 3 because there is a 3 in the ones place.
No, my friend is not correct. Because 23 is a prime number it is not a multiple of 3.

Think and Grow: Modeling Real Life

Ten people want toride a camel. There are 3 camels. Twopeople can ride on each camel. Are there enough camels for all of the people to ride at the same time?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 15
Multiplication equation:
There _____ enough camels for all of the people.

Answer:
Given,
Ten people want to ride a camel. There are 3 camels. Two people can ride on each camel.
3 × 2 = 6
There are not enough camels for all the 10 people.

Show and Grow

Question 18.
You want to give each of your 10 friends a sticker. You buy 4 sheets. Each sheet has 3 stickers. Do you have enough stickers?
Answer:
Given,
You want to give each of your 10 friends a sticker. You buy 4 sheets. Each sheet has 3 stickers.
1 sheet – 3 stickers
4 sheets – 4 × 3 stickers = 12 stickers
12 – 10 = 2 stickers
Yes, I have enough stickers.

Question 19.
DIG DEEPER!
You buy 2 heads of cauliflower and 5 ears of corn. How much money do you spend?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 16
You buy 3 heads of broccoli and 1 head of cauliflower.How much money do you spend?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 17
Answer:
The cost of 1 Broccoli is $3
The cost of 1 Cauliflower is $3
3 broccoli = 3 × $3 = $9
1 cauliflower = 1 × $3 = $3
9 + 3 = $12
Therefore I spent $12.

Multiply by 3 Homework & Practice 3.1

Find the product
Question 1.
3 × 3
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 18
Answer:
Distribute 3 to 2 and 1.
3 × 3 = 3 × (2 + 1)
3 × 3 = (3 × 2) + (3 × 1)
3 × 3 = 6 + 3
3 × 3 = 9

Question 2.
3 × 9
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 19
Answer:
Distribute 3 to 2 and 1.
3 × 9 = (2 + 1) × 9
3 × 9 = (2 × 9) + (1 × 9)
3 × 9 = 18 + 9
3 × 9 = 27

Question 3.
3 × 2 = ___
Answer:
Distribute 3 to 2 and 1.
3 × 2 = (2 + 1) × 2
3 × 2 = (2 × 2) + (1 × 2)
3 × 2 = 4 + 2
3 × 2 = 6

Question 4.
3 × 5 = ____
Answer:
Distribute 3 to 2 and 1.
3 × 5 = (2 + 1) × 5
3 × 5 = (2 × 5) + (1 × 5)
3 × 5 = 10 + 5
3 × 5 = 15

Question 5.
7 × 3 = _____
Answer:
Distribute 3 to 2 and 1.
7 × 3 = 7 × (2 + 1)
7 × 3 = (7 × 2) + (7 × 1)
7 × 3 = 14 + 7
7 × 3 = 21

Question 6.
0 × 3 = ____
Answer:
Any number multiplied by 0 will be always 0.
0 × 3 = 0

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 20
Answer:
Multiply the two numbers 3 and 4.
3 × 4 = 12

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 21
Answer:
multiply the two numbers 3 and 10.
3 × 10 = 30

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 22
Answer:
Multiply the two numbers 6 and 3.
6 × 3 = 18

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 23
Answer:
Multiply the two numbers 3 and 1.
3 × 1 = 3

Compare
Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 24
Answer: <

Explanation:
0 × 3 = 0
3 × 3 = 9
0 is less than 9.
0 × 3 < 3 × 3

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 25
Answer: >

Explanation:
8 × 3 = 24
7 × 3 = 21
24 > 21
8 × 3 > 7 × 3

Question 13.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 26
Answer: =

Explanation:
6 × 3 = 18
18 = 18

Question 14.
A Russian guitar, called a , has 3 strings. A music teacher is replacing the strings on 5 balalaikas. How many strings does the teacher need?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 27
Answer:
Given that,
A Russian guitar, called a, has 3 strings. A music teacher is replacing the strings on 5 balalaikas.
3 × 5 = 15
Thus the teacher needs 15 strings.

Question 15.
Number Sense
Circle the multiples of 3.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 28
Answer: The multiples of 3 are 12, 15, 21.

Question 16.
Modeling Real Life
You need 24 tennis balls. You buy 6 packs of 3 tennis balls. Do you have enough tennis balls?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 29
Answer:
Given,
You need 24 tennis balls. You buy 6 packs of 3 tennis balls.
6 × 3 = 18
24 – 18 = 6 tennis balls
Thus I don’t have enough tennis balls.

Question 17.
DIG DEEPER!
Newton buys 3 packs of plates and 3 packs of cups. How much money does he spend?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 30
Answer:
Given,
Newton buys 3 packs of plates and 3 packs of cups.
Cost of 1 plate = $4
Cost of 1 cup = $2
3 plates = 3 × $4 = $12
3 cups = 3 × $2 = $6
12 + 6 = 18
Thus Newton spend $18.

Review & Refresh

Question 18.
Which shapes show thirds?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.1 31
Answer: The 2nd, 3rd, and 4th figures show thirds. Because the figures are divided into 3 parts.

Lesson 3.2 Multiply by 4

Explore and Grow

Use the tape diagram to find the product
3 × 4 = _____
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 1

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-3-More-Multiplication-Facts-and-Strategies-3.2-1

Repeated Reasoning
Explain how you can use a tape diagram to find the missing products in the table. Complete the table.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 2

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-3-More-Multiplication-Facts-and-Strategies-3.2-2

Think and Grow: Multiply by 4

Example
Find 5 × 4.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 3

Answer:
Rewrite 4 as 2 + 2.
5 × 4 = 5 × (2 + 2)
5 × 4 = (5 × 2) + (5 × 2)
5 × 4 = 10 + 10
5 × 4 = 20

Show and Grow

Find the product
Question 1.
6 × 4
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 4
Answer:
Rewrite 4 as 2 + 2.
6 × 4 = 6 × (2 + 2)
6 × 4 = (6 × 2) + (6 × 2)
6 × 4 = 12 + 12
6 × 4 = 24

Question 2.
4 × 8
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 5
Answer:
Rewrite 4 as 2 + 2.
4 × 8 = (2 + 2) × 8
4 × 8 = (2 × 8) + (2 × 8)
4 × 8 = 16 + 16
4 × 8 = 32

Apply and Grow:Practice

Find the product
Question 3.
4 × 4
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 6
Answer:
Rewrite 4 as 2 + 2.
4 × 4 = 4 × (2 + 2)
4 × 4 = (4 × 2) + (4 × 2)
4 × 4 = 8 + 8
4 × 4 = 16

Question 4.
4 × 2
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 7
Answer:
Rewrite 4 as 2 + 2.
4 × 2 = (2 + 2) × 2
4 × 2 = (2 × 2) + (2 × 2)
4 × 2 = 4 + 4
4 × 2 = 8

Question 5.
4 × 5 = ___
Answer:
Rewrite 4 as 2 + 2.
4 × 5 = (2 + 2) × 5
4 × 5 = (2 × 5) + (2 × 5)
4 × 5 = 10 + 10
4 × 5 = 20

Question 6.
4 × 7 = ___
Answer:
Rewrite 4 as 2 + 2.
4 × 7 = (2 + 2) × 7
4 × 7 = (2 × 7) + (2 × 7)
4 × 7 =  14 + 14
4 × 7 = 28

Question 7.
4 × 6 = ____
Answer:
Rewrite 4 as 2 + 2.
4 × 6 = (2 + 2) × 6
4 × 6 = (2 × 6) + (2 × 6)
4 × 6 = 12 + 12
4 × 6 = 24

Question 8.
4 × 1 = ____
Answer:
Rewrite 4 as 2 + 2.
4 × 1 = (2 + 2) × 1
4 × 1 = (2 × 1) + (2 × 1)
4 × 1 = 2 + 2
4 × 1 = 4

Question 9.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 8
Answer:
Rewrite 4 as 2 + 2.
10 × 4 = 10 × (2 + 2)
10 × 4 = (10 × 2) + (10 × 2)
10 × 4 = 20 + 20
10 × 4 = 40

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 9
Answer:
Rewrite 4 as 2 + 2.
3 × 4 = 3 × (2 + 2)
3 × 4 = (3 × 2) + (3 × 2)
3 × 4 = 6 + 6
3 × 4 = 12

Question 11.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 10
Answer:
Rewrite 4 as 2 + 2.
8 × 4 = 8 × (2 + 2)
8 × 4 = (8 × 2) + (8 × 2)
8 × 4 = 16 + 16
8 × 4 = 32

Question 12.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 11
Answer:
Rewrite 4 as 2 + 2.
4 × 9 = (2 + 2) × 9
4 × 9 = (2 × 9) + (2 × 9)
4 × 9 = 18 + 18
4 × 9 = 36

Find the missing factor.
Question 13.
4 × ___ = 0
Answer: 0

Explanation:
Let the missing factor be x.
4 × x = 0
x = 0/4
x = 0
Thus the missing factor is 0.

Question 14.
___ × 4 = 40
Answer: 10

Explanation:
Let the missing factor be y.
y × 4 = 40
y = 40/4
y = 10
Thus the missing factor is 10.

Question 15.
4 = ____ × 4
Answer:
Let the missing factor be t.
4 = t × 4
t = 4/4
t = 1
Thus the missing factor is 1.

Question 16.
Number Sense
You exchange some dollar bills for quarters. How many quarters might you receive?
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 12
Answer:
Convert from dollars to quarters
1 dollar = 4 quarters
Multiples of 4 are 12, 16, 24, 36, 40.

Question 17.
Explain how you can use 2 × 3 to find 4 × 3.
Answer:
Rewrite 4 as 2 + 2.
4 × 3 = (2 + 2) × 3
4 × 3 = (2 × 3) + (2 × 3)
4 × 3 = 6 + 6
4 × 3 = 12

Think and Grow: Modeling Real Life

The graph shows the number of cars scheduled to get a new set of tires installed. How many new tires are installed on the busiest day?
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 13
What is the busiest day?
Multiplication equation:
______ new tires are installed on the busiest day.

Answer:
Monday = 6 × 1 = 6 new tires
Tuesday = 7 × 1 = 7 new tires
Wednesday = 3 × 1 = 3 new tires
Thursday = 8 × 1 = 8 new tires
Friday = 4 × 1 = 4 new tires
8 new tires are installed on the busiest day.

Show and Grow

Question 18.
Use the graph above. How many new tires are installed on the least busy day?
Answer:
Monday = 6 × 1 = 6 new tires
Tuesday = 7 × 1 = 7 new tires
Wednesday = 3 × 1 = 3 new tires
Thursday = 8 × 1 = 8 new tires
Friday = 4 × 1 = 4 new tires
3 new tires are installed on the least busy day.

Question 19.
DIGD EEPER!
Descartes is packing bags for party favors. Each bag needs 1 container of bubbles, 5 stickers, and 3 balloons. How many of each item does Descartes need to make 4 bags?
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 14
Answer:
Given that,
Descartes is packing bags for party favors.
Each bag needs 1 container of bubbles, 5 stickers, and 3 balloons.
4 bags = 1 bubble × 4 + 5 stickers × 4 + 3 balloons × 4
4 bags = 4 + 20 + 12 = 36
Thus Descartes need 4 bubbles, 20 stickers and 12 balloons.

Multiply by 4 Homework & Practice 3.2

Fing the product
Question 1.
3 × 4
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 15
Answer:
Rewrite 4 as 2 + 2.
3 × 4 = 3 × (2 + 2)
3 × 4 = (3 × 2) + (3 × 2)
3 × 4 = 6 + 6
3 × 4 = 12

Question 2.
4 × 9
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 16
Answer:
Rewrite 4 as 2 + 2.
4 × 9 = (2 + 2) × 9
4 × 9 = (2 × 9) + (2 × 9)
4 × 9 = 18 + 18
4 × 9 = 36

Question 3.
8 × 4 = ___
Answer:

Rewrite 4 as 2 + 2.
8 × 4 = 8 × (2 + 2)
8 × 4 = (8 × 2) + (8 × 2)
8 × 4 = 16 + 16
8 × 4 = 32

Question 4.
4 × 4 = ____
Answer:

Rewrite 4 as 2 + 2.
4 × 4 = 4 × (2 + 2)
4 × 4 = (4 × 2) + (4 × 2)
4 × 4 = 8 + 8
4 × 4 = 16

Question 5.
10 × 4 = _____
Answer:
Rewrite 4 as 2 + 2.
10 × 4 = 10 × (2 + 2)
10 × 4 = (10 × 2) + (10 × 2)
10 × 4 = 20 + 20
10 × 4 = 40

Question 6.
4 × 6 = _____
Answer:
Rewrite 4 as 2 + 2.
4 × 6 = (2 + 2) × 6
4 × 6 = (2 × 6) + (2 × 6)
4 × 6 = 12 + 12
4 × 6 = 24

Question 7.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 17
Answer:
Rewrite 4 as 2 + 2.
4 × 5 = (2 + 2) × 5
4 × 5 = (2 × 5) + (2 × 5)
4 × 5 = 10 + 10
4 × 5 = 20

Question 8.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 18
Answer:
Any number multiplied by 0 is always 0.
0 × 4 = 0

Question 9.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 19
Answer:
Any number multiplied by 1 is always the same number.
1 × 4 = 4

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 20
Answer:
Rewrite 4 as 2 + 2.
2 × 4 = 2 × (2 + 2)
2 × 4 = (2 × 2) + (2 × 2)
2 × 4 = 4 + 4
2 × 4 = 8

Find the missing factor
Question 11.
10 × ____ = 40
Answer: 4

Explanation:
Let the missing factor be p.
10 × p = 40
p = 40/10 = 4
p = 4
Therefore the missing factor is 4.

Question 12.
____ × 1 = 4
Answer: 4

Explanation:
Let the missing factor be q.
q × 1 = 4
q = 4/1
q = 4
Therefore the missing factor is 4.

Question 13.
8 = ____ × 4
Answer: 2

Explanation:
Let the missing factor be r.
8 = r × 4
r = 8/4
r = 2
Therefore the missing factor is 2.

Question 14.
A string quartet has 4 musicians. Each musician’s instrument has 4 strings. How many total strings are there in a string quartet?
Answer: 16

Explanation:
Given that,
A string quartet has 4 musicians. Each musician’s instrument has 4 strings.
1 musician = 4 strings
4 musicians = 4 × 4 strings = 16 strings
Therefore there are 16 strings in a string quartet.

Question 15.
Which One Doesn’tBelong?
Which one does belong with the other three?
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 21
Answer:
(2 × 3) + (2 × 3) doesn’t belong to the other three expressions.
4 × 6, 6 × 4, (2 × 6) + (2 × 6) belong to other expressions. Because
4 × 6 = 24
6 × 4 = 24
(2 × 6) + (2 × 6) = 12 + 12 = 24

Question 16.
Modeling Real Life
The tally chart shows the number of tables a carpenter makes each day. Each table has 4 legs. How many legs does the carpenter make on the busiest day?
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3.2 22
Answer:
Given,
The tally chart shows the number of tables a carpenter makes each day. Each table has 4 legs.
Monday – 5 + 5 = 10
Tuesday – 5 + 3 = 8
Wednesday – 4
Monday – 10 × 4 = 40
Tuesday – 8  × 4 = 32
Wednesday – 4 × 4 = 16
Thus the carpenter makes 40 legs on the busiest day.

Question 17.
DIG DEEPER!
Newton is packing lunches. Each lunch needs 1 sandwich, 2 celery sticks, 3 carrot sticks, and 4 strawberries. How many of each item does Newton need to make 4 lunches?
Answer:
Given,
Newton is packing lunches. Each lunch needs 1 sandwich, 2 celery sticks, 3 carrot sticks, and 4 strawberries.
1 lunch – 1 sandwich, 2 celery sticks, 3 carrot sticks, 4 strawberries
4 lunches – 1 sandwich × 4, 2 celery sticks × 4, 3 carrot sticks × 4, 4 strawberries × 4 = 4 + 8 + 12 + 16 = 40
Thus newton needs 4 sandwiches, 8 celery sticks, 12 carrot sticks, and 16 strawberries to make 4 lunches.

Review & Refresh

Write the number in expanded form and word form
Question 18.
837
____ + ____ + _____
_______________

Answer:
The expanded form of 837 is 800 + 30 + 7
The word form of 837 is Eight Hundred and Third Seven.

Question 19.
954
____ + ____ + ____
_______________

Answer:
The expanded form of 954 is 900 + 50 + 4
The word form of 954 is Nine Hundred and Fifty-Four.

Lesson 3.3 Multiply by 6

Explore and Grow
Question 1.
Use equal groups to find the product. Draw your model.
4 × 6 = _____
Answer:
Big-Ideas-Math-Grade-6-Answers-Chapter-3-More-Multiplication-Facts-&-Strategies
6 + 6 + 6 + 6 = 24

Repeated Reasoning
Explain how you can use equal groups to multiply. Complete the table.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 1
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.3-1

Think and Grow: Multiply by 6

Example
Find 5 × 6.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 2
Answer:
Rewrite 6 as 5 + 1.
5 × 6 = 5 × (5 + 1)
5 × 6 = (5 × 5) + (5 × 1)
5 × 6 = 25 + 5
5 × 6 = 30

Show and Grow

Find the product
Question 1.
3 × 6
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 3
Answer:
Distribute 6 as 3 + 3.
3 × 6 = 3 × (3 + 3)
3 × 6 = (3 × 3) + (3 × 3)
3 × 6 = 9 + 9
3 × 6 = 18

Question 2.
6 × 5
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 4
Answer:
Distribute 6 as 3 + 3.
6 × 5 = (3 + 3) × 5
6 × 5 = (3 × 5) + (3 × 5)
6 × 5 = 15 + 15
6 × 5 = 30

Apply and Grow: Practice

Find the product
Question 3.
6 × 7
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 5
Answer:

Distribute 6 as 3 + 3.
6 × 7 = (3 + 3) × 7
6 × 7 = (3 × 7) + (3 × 7)
6 × 7 = 21 + 21
6 × 7 = 42

Question 4.
4 × 6
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 6
Answer:
Distribute 6 as 3 + 3.
4 × 6 = 4 × (3 + 3)
4 × 6 = (4 × 3) + (4 × 3)
4 × 6 = 12 + 12
4 × 6 = 24

Question 5.
8 × 6 = ___
Answer:
Distribute 6 as 3 + 3.
8 × 6 = 8 × (3 + 3)
8 × 6 = (8 × 3) + (8 × 3)
8 × 6 = 24 + 24
8 × 6 = 48

Question 6.
7 × 6 = ____
Answer:
Distribute 6 as 3 + 3.
7 × 6 = 7 × (3 + 3)
7 × 6 = (7 × 3) + (7 × 3)
7 × 6 = 21 + 21
7 × 6 = 42

Question 7.
6 × 0 = ___
Answer: 0
Any number multiplied by 0 will be always 0.
6 × 0 = 0

Question 8.
6 × 4 = ____
Answer:
Distribute 6 as 3 + 3.
6 × 4 = (3 + 3) × 4
6 × 4 = (3 × 4) + (3 × 4)
6 × 4 = 12 + 12
6 × 4 = 24

Question 9.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 7
Answer:
Distribute 6 as 3 + 3.
6 × 6 = 6 × (3 + 3)
6 × 6 = (6 × 3) + (6 × 3)
6 × 6 = 18 + 18
6 × 6 = 36

Question 10.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 8
Answer: 6
Distribute 6 as 3 + 3.
6 × 1 = (3 + 3) × 1
6 × 1 = (3 × 1) + (3 × 1)
6 × 1 = 3 + 3
6 × 1 = 6

Question 11.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 9
Answer: 12

Distribute 6 as 3 + 3.
2 × 6 = 2 × (3 + 3)
2 × 6 = (2 × 3) + (2 × 3)
2 × 6 = 6 + 6
2 × 6 = 12

Question 12.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 10
Answer: 54
Distribute 6 as 3 + 3.
9 × 6 = 9 × (3 + 3)
9 × 6 = (9 × 3) + (9 × 3)
9 × 6 = 27 + 27
9 × 6 = 54

Compare
Question 13.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 11
Answer: =

Explanation:
8 × 6 = 48
48 = 48

Question 14.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 12
Answer: <

Explanation:
6 × 0 = 0
10 × 6 = 60
6 × 0 < 10 × 6

Question 15.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 13
Answer: <

Explanation:
6 × 1 = 6
6 × 3 = 18
6 is less than 18.
6 × 1 < 6 × 3

Question 16.
There are 9 volleyball teams in a tournament. There are 6 players on each team. How many volleyball players are in the tournament?
Answer: 54

Explanation:
Given,
There are 9 volleyball teams in a tournament. There are 6 players on each team.
9 × 6 = 54
Thus there are 54 volleyball players in the tournament.

Question 17.
YOU BE THE TEACHER
Your friend says that all multiples of 3 are also multiples of 6. Is your friend correct? Explain.
Answer: No. Multiples of 3 are not the multiples of 6.

Think and Grow: Modeling Real Life

You have 5 apples. You cut each apple into8 slices. You have6 oranges and cut each orange into4 slices. Do you have more apple slices or orange slices?
Multiplication equations:
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 14
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 15
You have more _____ slices.

Answer:
Given,
You have 5 apples. You cut each apple into 8 slices. You have 6 oranges and cut each orange into 4 slices.
1 apple = 8 slices
5 apples = 5 × 8 slices = 40 slices
1 orange = 4 slices
6 oranges = 6 × 4 slices = 24 slices
40 slices is greater than 24 slices.
Therefore you have more apple slices.

Show and Grow

Question 18.
Your friend makes 6 bracelets each day for 7 days. Your cousin makes 10 bracelets each day for 3 days. Who makes fewer bracelets?
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 16
Answer:
Given,
Your friend makes 6 bracelets each day for 7 days.
Your cousin makes 10 bracelets each day for 3 days.
1 day = 6 bracelets
7 days = 7 × 6 bracelets = 42 bracelets
1 day = 10 bracelets
3 days = 3 × 10 bracelets = 30 bracelets
By this we can say that my cousin makes fewer bracelets.

Question 19.
You draw 6 octagons. How many sides do you draw?
Answer: A octagon contains 8 sides. Thus I need to draw 48 sides for all 6 octagon.

Question 20.
You draw 9 hexagons. How many sides do you draw?
Answer: A hexagon contains 6 sides. Thus I need to draw 54 sides for all 9 hexagons.

Multiply by 6 Homework & Practice 3.3

Find the product
Question 1.
2 × 6
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 17
Answer:
Distribute 6 as 3 + 3.
2 × 6 = 2 × (3 + 3)
2 × 6 = (2 × 3) + (2 × 3)
2 × 6 = 6 + 6
2 × 6 = 12

Question 2.
6 × 6
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 18
Answer:
Distribute 6 as 3 + 3.
6 × 6 = (3 + 3) × 6
6 × 6 = (3 × 6) + (3 × 6)
6 × 6 = 18 + 18
6 × 6 = 36

Question 3.
9 × 6 = ____
Answer:
Distribute 6 as 3 + 3.
9 × 6 = 9 × (3 + 3)
9 × 6 = (9 × 3) + (9 × 3)
9 × 6 = 27 + 27
9 × 6 = 54

Question 4.
5 × 6 = _____
Answer:
Distribute 6 as 3 + 3.
5 × 6 = 5 × (3 + 3)
5 × 6 = (5 × 3) + (5 × 3)
5 × 6 = 15 + 15
5 × 6 = 30

Question 5.
6 × 3 = ____
Answer:
Distribute 6 as 3 + 3.
6 × 3 = (3 + 3) × 3
6 × 3 = (3 × 3) + (3 × 3)
6 × 3 = 9 + 9
6 × 3 = 18

Question 6.
4 × 6 = ____
Answer:
Distribute 6 as 3 + 3.
4 × 6 = 4 × (3 + 3)
4 × 6 = (4 × 3) + (4 × 3)
4 × 6 = 12 + 12
4 × 6 = 24

Question 7.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 19
Answer:
Distribute 6 as 3 + 3.
8 × 6 = 8 × (3 + 3)
8 × 6 = (8 × 3) + (8 × 3)
8 × 6 = 24 + 24
8 × 6 = 48

Question 8.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 20
Answer:
Any number multiplied by 0 is always 0.
So, 6 × 0 = 0

Question 9.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 21
Answer:
Any number multiplied by 1 is always the same number.
6 × 1 = 6

Question 10.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 22
Answer:
Distribute 6 as 3 + 3.
10 × 6 = 10 × (3 + 3)
10 × 6 = (10 × 3) + (10 × 3)
10 × 6 = 30 + 30
10 × 6 = 60

Compare
Question 11.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 23
Answer: <

Explanation:
4 × 6 = 24
6 × 6 = 36
24 < 36
Thus 4 × 6 < 6 × 6

Question 12.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 24
Answer: >

Explanation:
6 × 5 = 30
4 × 5 = 20
30 > 20
Thus 6 × 5 > 4 × 5

Question 13.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 25
Answer: =

Explanation:
6 × 7 = 42
42 = 42
Thus 42 = 6 × 7

Question 14.
There are 6 faces on a standard die. How many faces are on 5 dice?
Answer: 30 faces

Explanation:
Given that,
There are 6 faces on a standard die.
1 dice – 6 faces
5 dice – 5 × 6 faces = 30 faces
Thus 5 dice has 30 faces.

Question 15.
DIG DEEPER!
You have a muffin tin with 6 cups. You want to bake 36 muffins. How many times must you use the muffin tin?
Answer:
Given,
You have a muffin tin with 6 cups. You want to bake 36 muffins.
1 muffin tin – 6 cups
x – 36 muffins
36/6 = 6
Thus I must use 6 muffin tins.

Question 16.
Modeling Real Life
You practice the saxophone 1 hour in the morning and 2 hours at night each day. How many hours do you practice in 6 days?
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 26
Answer:
Given that,
You practice the saxophone for 1 hour in the morning and 2 hours at night each day.
1 day = 1 + 2 = 3 hours
6 days = 6 × 3 = 18 hours.
Thus you practice 18 hours in 6 days.

Question 17.
Modeling Real Life You want to make 6 pentagons using toothpicks. How many toothpicks do you need?
Answer:
We know that a pentagon contains 5 sides.
So to make the pentagon you need 5 toothpicks
6 pentagons = 6 × 5 = 30 toothpicks.
Thus you need 30 toothpicks to make 6 pentagons.

Review & Refresh

Question 18.
Find the total value.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.3 27
Total value: _____
Answer:
2 dimes = $0.2
2 pennies = $0.02
1 Nickel = $0.05
1 Washington quarter = $0.25
1 cent = $0.01
Total Value = $0.2+ $0.02 + $0.05 + $0.25 + $0.01
= $0.53
Thus the total value is $0.53

Lesson 3.4 Multiply by 7

Explore and Grow
Question 1.
Use the number line to find the product
2 × 7 = ____
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 1
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.4-1
Multiply 2 and 7.
7 + 7 = 14

Repeated Reasoning
Complete the table. Explain how you found the missing products.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 2

Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.4-2

Think and Grow: Multiply by 7

Example
Find 3 × 7
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 3
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
3 × 7 = 3 × (5 + 2)
3 × 7 = (3 × 5) + (3 × 2)
3 × 7 = 15 + 6
3 × 7 = 21

Show and Grow

Find the product
Question 1.
4 × 7
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 4
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
4 × 7 = 4 × (5 + 2)
4 × 7 = (4 × 5) + (4 × 2)
4 × 7 = 20 + 8
4 × 7 = 28

Question 2.
7 × 9
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 5
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
7 × 9 = (5 + 2) × 9
7 × 9 = (5 × 9) + (2 × 9)
7 × 9 = 45 + 18
7 × 9 = 63

Apply and Grow: Practice

Find the product
Question 3.
7 × 7
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 6
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
7 × 7 = 7 × (5 + 2)
7 × 7 = (7 × 5) + (7 × 2)
7 × 7 = 35 + 14
7 × 7 = 49

Question 4.
7 × 6
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 7
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
7 × 6 = (5 + 2) × 6
7 × 6 = (5 × 6) + (2 × 6)
7 × 6 = 30 + 12
7 × 6 = 42

Question 5.
7 × 5 = ___
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
7 × 5 = (5 + 2) × 5
7 × 5 = (5 × 5) + (2 × 5)
7 × 5 = 25 + 10
7 × 5 = 35

Question 6.
7 × 1 = ___
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
7 × 1 = (5 + 2) × 1
7 × 1 = (5 × 2) + (2 × 1)
7 × 1 = 10 + 2
7 × 1 = 12

Question 7.
2 × 7 = ___
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
2 × 7 = 2 × (5 + 2)
2 × 7 = (2 × 5) + (2 × 2)
2 × 7 = 10 + 4
2 × 7 = 14

Question 8.
3 × 7 = ____
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
3 × 7 = 3 × (5 + 2)
3 × 7 = (3 × 5) + (3 × 2)
3 × 7 = 15 + 6
3 × 7 = 21

Question 9.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 8
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
10 × 7 = 10 × (5 + 2)
10 × 7 = (10 × 5) + (10 × 2)
10 × 7 = 50 + 20
10 × 7 = 70

Question 10.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 9
Answer: 0
Any number multiplied by 0 is always 0.

Question 11.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 10
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
8 × 7 = 8 × (5 + 2)
8 × 7 = (8 × 5) + (8 × 2)
8 × 7 = 40 + 16
8 × 7 = 56

Question 12.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 11
Answer:
Rewrite 7 as 5 + 2.
Use Distributive Property.
7 × 9 = (5 + 2) × 9
7 × 9 = (5 × 9) + (2 × 9)
7 × 9 = 45 + 18
7 × 9 = 63

Find the missing factor
Question 13.
7 × ___ = 0
Answer: 0

Explanation:
Let the missing factor be x.
7 × x = 0
x = 0/7
x = 0
Thus the missing factor is 0.

Question 14.
___ × 7 = 35
Answer: 5

Explanation:
Let the missing factor be y.
y × 7 = 35
y = 35/7
y = 5
Thus the missing factor is 5.

Question 15.
70 = ___ × 7
Answer: 10

Explanation:
Let the missing factor be z.
70 = z × 7
z = 70/7
z = 10
Thus the missing factor is 10.

Question 16.
How many days are in 4 weeks?
Answer: 28 days

Explanation:
Convert from weeks to days.
1 week – 7 days
4 weeks – 4 × 7 = 28 days
Therefore there are 28 days in 4 weeks.

Question 17.
Number Sense
How can you use the Commutative Property of Multiplication to find 7 × 3?
Answer:
The commutative property is a math rule that says that the order in which we multiply numbers does not change the product.
The Commutative Property of Multiplication of 7 × 3 is 3 × 7.

Think and Grow: Modeling Real Life

A child ticket costs $7. An adult ticket costs 4 times as much as the child ticket. Descartes has $40. Can he buy an adult ticket?
Multiplication equation:
Descartes _____ buy an adult ticket.

Answer:
Given,
A child ticket costs $7.
An adult ticket costs 4 times as much as the child ticket.
4 × $7 = $28
Descartes has $40.
$40 – $28 = $12
Therefore Descartes can buy an adult ticket.

Show and Grow

Question 18.
A small painting costs $5. A large painting costs 7 times as much as the small painting. Newton has $30.Can he buy a large painting?
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 12
Answer:
Given,
A small painting costs $5.
A large painting costs 7 times as much as a small painting.
7 × $5 = $35
Newton has $30.
$30 – $35 = -$5
Thus he cannot buy a large painting.

Question 19.
DIG DEEPER!
You study your spelling words for 5 minutes twice a day. How many minutes do you spend studying your spelling words in one week?
Answer:
Given,
You study your spelling words for 5 minutes twice a day.
1 day – 2 × 5 = 10 minutes
Convert from week to days.
1 week – 7 days
7 × 10 = 70 minutes

Question 20.
DIG DEEPER!
Your dentist tells you to brush your teeth for 3 minutes three times a day. How many minutes should you spend brushing your teeth in one week?
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 13
Answer:
Given,
Your dentist tells you to brush your teeth for 3 minutes three times a day.
3 minutes – 3 × 3 = 9 minutes a day.
Convert from week to day.
1 week – 7 days
7 × 9 minutes = 63 minutes a week.

Multiply by 7 Homework & Practice 3.4

Question 1.
5 × 7 =
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 14
Answer: 35

Explanation:
Rewrite 7 as 2 + 5.
5 × 7 = 5 × (2 + 5)
5 × 7 = (5 × 2) + (5 × 5)
5 × 7 = 10 + 25
5 × 7 = 35

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 15
Answer: 14

Explanation:
Rewrite 7 as 2 + 5.
7 × 2 = (2 + 5) × 2
7 × 2 = (2 × 2) + (5 × 2)
7 × 2 = 4 + 10
7 × 2 = 14

Question 3.
0 × 7 = ___
Answer: 0

Explanation: Any number multiplied by 0 will be always 0.

Question 4.
7 × 7 = ____
Answer: 49

Explanation:
Rewrite 7 as 2 + 5.
7 × 7 = (2 + 5) × 7
7 × 7 = (2 × 7) + (5 × 7)
7 × 7 = 14 + 35
7 × 7 = 49

Question 5.
10 × 7 = ____
Answer: 70

Explanation:
Rewrite 7 as 2 + 5.
10 × 7 = 10 × (2 + 5)
10 × 7 = (10 × 2) + (10 × 5)
10 × 7 = 20 + 50
10 × 7 = 70

Question 6.
8 × 7 = ____
Answer: 56

Explanation:
Rewrite 7 as 2 + 5.
8 × 7 = 8 × (2 + 5)
8 × 7 = (8 × 2) + (8 × 5)
8 × 7 = 16 + 40
8 × 7 = 56

Question 7.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 16
Answer: 7

Explanation:
Rewrite 7 as 2 + 5.
7 × 1 = (2 + 5) × 1
7 × 1 = (2 × 1) + (5 × 1)
7 × 1 = 2 + 5
7 × 1 = 7

Question 8.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 17
Answer: 63

Explanation:
Rewrite 7 as 2 + 5.
7 × 9 = (2 + 5) × 9
7 × 9 = (2 × 9) + (5 × 9)
7 × 9 = 18 + 45
7 × 9 = 63

Question 9.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 18
Answer: 28

Explanation:
Rewrite 7 as 2 + 5.
4 × 7 = 4 × (2 + 5)
4 × 7 = (4 × 2) + (4 × 5)
4 × 7 = 8 + 20
4 × 7 = 28

Question 10.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 19
Answer: 21

Explanation:
Rewrite 7 as 2 + 5.
3 × 7 = 3 × (2 + 5)
3 × 7 = (3 × 2) + (3 × 5)
3 × 7 = 6 + 15
3 × 7 = 21

Find the missing factor
Question 11.
7 × ____ = 7
Answer: 1

Explanation:
Let the missing factor be x.
7 × x = 7
x = 7/7
x = 1
Thus the missing factor is 1.

Question 12.
____ × 7 = 14
Answer: 2

Explanation:
Let the missing factor be y.
y × 7 = 14
y = 14/7
y = 2
Thus the missing factor is 2.

Question 13.
56 = ____ × 7
Answer: 8

Explanation:
Let the missing factor be t.
56 = t × 7
t = 56/7
t = 8
Thus the missing factor is 8.

Question 14.
You go to school for 5 days each week. You spend 7 hours at school each day. How many hours do you spend at school in one week?
Answer: 35

Explanation:
Given that,
You go to school for 5 days each week.
You spend 7 hours at school each day.
1 day – 7 hours
5 days – x
x = 5 × 7 hours
x = 35 hours
Thus you spend 35 hours at school in one week.

Question 15.
Number Sense
Circle the multiples of 7
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 20
Answer: Multiples of 7 are 21, 35, 56, 63
Big-Ideas-Math-Solutions-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.4-20

Question 16.
Structure
Find in two different ways 7 × 6
Answer: The two different ways to find 7 × 6 are
Multiply the two numbers 7 and 6
7 × 6 = 42
The other way to find the product 7 × 6 is by using the distributive property.
Rewrite 6 as 3 + 3.
7 × 6 = 7 × (3 + 3)
7 × 6 = (7 × 3) + (7 × 3)
7 × 6 = 21 + 21
7 × 6 = 42

Question 17.
Modeling Real Life
A pair of regular shoes costs $9. A pair of light-up shoes costs 7 times as much as the pair of regular shoes. Newton has $60. Can he buy a pair of light-up shoes?
Answer:
Given that,
A pair of regular shoes costs $9. A pair of light-up shoes costs 7 times as much as a pair of regular shoes.
9 × 7 = $63
Newton has $60.
$60 – $63 = -$3
By this, we can say that Newton cannot buy a pair of light-up shoes.

Question 18.
DIG DEEPER!
A veterinarian tells you to feed your dog 2 cups of food twice a day. How many cups of food should you feed your dog in one week?
Answer:
Given,
A veterinarian tells you to feed your dog 2 cups of food twice a day.
First, convert from week to day.
1 week = 7 days
7 × 2 cups = 14 cups
Thus you feed 14 cups in one week.

Review & Refresh

Question 19.
What is the best estimate of the length of a thumbtack?
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 21
Answer:
By seeing the figure we can estimate the length of a thumbtack as 2 centimeters.

Question 20.
What is the best estimate of the height of a trampoline?
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.4 22
Answer:
By seeing the figure we can estimate the height of a trampoline as 4 centimeters.

Lesson 3.5 Multiply by 8

Explore and Grow

Use the tape diagram to find the product
3 × 8 = ____
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 1
Answer: 24
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.5-1

Repeated Reasoning
Explain how you can use a different model to solve. Complete the table.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 2
Answer: The tables for multiples of 8 are given below,
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.5-2

Think and Grow: Multiply by 8

Example
Find 5 × 8
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 3

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
5 × 8 = 5 × (4 + 4)
5 × 8 = (5 × 4) + (5 × 4)
5 × 8 = 20 + 20
5 × 8 = 40

Show and Grow

Find the product
Question 1.
4 × 8
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 4
Answer: 32

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
4 × 8 = 4 × (4 + 4)
4 × 8 = (4 × 4) + (4 × 4)
4 × 8 = 16 + 16
4 × 8 = 32

Question 2.
8 × 7
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 5
Answer: 56

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 7 = (4 + 4) × 7
8 × 7 = (4 × 7) + (4 × 7)
8 × 7 = 28 + 28
8 × 7 = 56

Find the product
Question 3.
8 × 8
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 6
Answer: 64

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 8 = (4 + 4) × 8
8 × 8 = (4 × 8) + (4 × 8)
8 × 8 = 32 + 32
8 × 8 = 64

Question 4.
3 × 8
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 7
Answer: 24

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
3 × 8 = 3 × (4 + 4)
3 × 8 = (3 × 4) + (3 × 4)
3 × 8 = 12 + 12
3 × 8 = 24

Question 5.
8 × 2 = ___
Answer: 16

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 2 = (4 + 4) × 2
8 × 2 = (4 × 2) + (4 × 2)
8 × 2 = 8 + 8
8 × 2 = 16

Question 6.
7 × 8 = ___
Answer: 56

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
7 × 8 = 7 × (4 + 4)
7 × 8 = (7 × 4) + (7 × 4)
7 × 8 = 28 + 28
7 × 8 = 56

Question 7.
8 × 6 = ____
Answer: 48

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 6 = (4 + 4) × 6
8 × 6 = (4 × 6) + (4 × 6)
8 × 6 = 24 + 24
8 × 6 = 48

Question 8.
10 × 8 = ____
Answer: 80

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
10 × 8 = 10 × (4 + 4)
10 × 8 = (10 × 4) + (10 × 4)
10 × 8 = 40 + 40
10 × 8 = 80

Compare
Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 8
Answer: 0

Explanation:
Any number multiplied by 0 will be 0.
0 × 8 = 0 × (4 + 4)
0 × 8 = (0 × 4) + (0 × 4)
0 × 8 = 0 + 0
0 × 8 = 0

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 9
Answer: 72

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 9 = (4 + 4) × 9
8 × 9 = (4 × 9) + (4 × 9)
8 × 9 = 36 + 36
8 × 9 = 72

Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 10
Answer: 40

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 5 = (4 + 4) × 5
8 × 5 = (4 × 5) + (4 × 5)
8 × 5 = 20 + 20
8 × 5 = 40

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 11
Answer: 8

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
1 × 8 = 1 × (4 + 4)
1 × 8 = (1 × 4) + (1 × 4)
1 × 8 = 4 + 4
1 × 8 = 8

Compare
Question 13.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 12
Answer: =

Explanation:
8 × 8 = 64
64 = 64
So, 8 × 8 = 64

Question 14.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 13
Answer: <

Explanation:
Any number multiplied by 0 will be 0.
Any number multiplied by 1 will be the same number.
8 × 0 = 0
8 × 1 = 8
0 < 8
So, 8 × 0 < 8 × 1

Question 15.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 14
Answer: >

Explanation:
8 × 6 = 48
5 × 8 = 40
48 > 40
8 × 6 > 5 × 8

Question 16.
Which One Doesn’tBelong?
Which expression does not belong with the other three?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 15
Answer:
4 × (4 + 3) = 4 × 7 = 28
3 × (1 + 7) = 3 × 8 = 24
3 × (3 + 5) = 3 × 8 = 24
3 × (4 + 4) = 3 × 8 = 24
Thus the first doesn’t belong with the other three.

Think and Grow: Modeling Real Life

A marching band has 7 rows with 8 musicians in each row. There is also a row of 6 people who carry flags. How many people are in the marching band in all?
Multiplication equation:
Addition equation:
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 16
There are _______ people in the marching band in all.

Answer:
A marching band has 7 rows with 8 musicians in each row.
1 row – 8 musicians
7 rows – 7 × 8 musicians = 56 musicians
There is also a row of 6 people who carry flags.
56 + 6 = 62 people
Therefore there are 62 people in the marching band in all.

Show and Grow

Question 17.
A table has 3 rows with 8 prizes in each row. There is also a row of 4 prizes on the floor. How many prizes are there in all?
Answer: 28 prizes

Explanation:
A table has 3 rows with 8 prizes in each row.
1 row – 8 prizes
3 rows – 8 × 3 = 24 prizes
There is also a row of 4 prizes on the floor
24 + 4 = 28 prizes
Thus there are 28 prizes in all.

Question 18.
DIG DEEPER!
One section of a parking lot has 2 rows of 8 cars. Another section of the parking lot has 8 rows of 6 cars. How many cars are in the parking lot in all?
Answer: 64 cars

Explanation:
Given that,
One section of a parking lot has 2 rows of 8 cars.
2 × 8 = 16 cars
Another section of the parking lot has 8 rows of 6 cars.
8 × 6 = 48 cars
16 + 48 = 64 cars
Thus 64 cars are in the parking lot in all.

Question 19.
DIG DEEPER!
One building has 8 rows of 5 windows. Another building has 9 rows of 8 windows. How many windows are on the two buildings in all?
Answer: 112 windows

Explanation:
Given that,
One building has 8 rows of 5 windows.
8 × 5 = 40 windows
Another building has 9 rows of 8 windows.
9 × 8 = 72 windows
40 windows + 72 windows = 112 windows
Therefore 112 windows are on the two buildings in all.

Multiply by 8 Homework & Practice 3.5

Question 1.
8 × 6
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 17
Answer: 48

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 6 = (4 + 4) × 6
8 × 6 = (4 × 6) + (4 × 6)
8 × 6 = 24 + 24
8 × 6 = 48

Question 2.
2 × 8
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 18
Answer: 16

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
2 × 8 = 2 × (4 + 4)
2 × 8 = (2 × 4) + (2 × 4)
2 × 8 = 8 + 8
2 × 8 = 16

Question 3.
8 × 0 = ____
Answer: 0

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 0 = (4 + 4) × 0
8 × 0 = (4 × 0) + (4 × 0)
8 × 0 = 0 + 0
8 × 0 = 0

Question 4.
4 × 8 = ____
Answer: 32

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
4 × 8 = 4 × (4 + 4)
4 × 8 = (4 × 4) + (4 × 4)
4 × 8 = 16 + 16
4 × 8 = 32

Question 5.
8 × 10 = ____
Answer: 80

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 10 = (4 + 4) × 10
8 × 10 = (4 × 10) + (4 × 10)
8 × 10 = 40 + 40
8 × 10 = 80

Question 6.
1 × 8 = ____
Answer: 8

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
1 × 8 = 1 × (4 + 4)
1 × 8 = (1 × 4) + (1 × 4)
1 × 8 = 4+ 4
1 × 8 = 8

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 19
Answer: 24

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 3 = (4 + 4) × 3
8 × 3 = (4 × 3) + (4 × 3)
8 × 3 = 12 + 12
8 × 3 = 24

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 20
Answer: 64

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 8 = (4 + 4) × 8
8 × 8 = (4 × 8) + (4 × 8)
8 × 8 = 32 + 32
8 × 8 = 64

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 21
Answer: 56

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
7 × 8 = 7 × (4 + 4)
7 × 8 = (7 × 4) + (7 × 4)
7 × 8 = 28 + 28
7 × 8 = 56

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 22
Answer: 72

Explanation:
Rewrite 8 as 4 + 4.
We can find the product by using the Distributive Property.
8 × 9 = (4 + 4) × 9
8 × 9 = (4 × 9) + (4 × 9)
8 × 9 = 36 + 36
8 × 9 = 72

Compare
Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 23
Answer: >

Explanation:
8 × 3 = 24
8 × 2 = 16
24 > 16
So, 8 × 3 > 8 × 2

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 24
Answer: <

Explanation:
8 × 7 = 56
56 < 64
So, 8 × 7 < 64

Question 13.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 25
Answer: =

Explanation:
8 × 1 = 8
1 × 8 = 8
8 = 8
So, 8 × 1 = 1 × 8

Question 14.
There are 8 batteries in a package. You buy 7 packages. How many batteries do you have?
Answer: 56 batteries

Explanation:
Given that,
There are 8 batteries in a package.
You buy 7 packages.
1 pack – 8 batteries
7 pack – 7 × 8 batteries
7 packs = 56 batteries
Therefore there are 56 batteries in all.

Question 15.
Patterns
Tell whether each statement is true or false. If false, explain.
The ones digit in multiples of 8 follows the pattern 8, 6, 4, 2, 0. _____
The product of an odd factor and 8 is always odd. ____

Answer:
The ones digit in multiples of 8 follows the pattern 8, 6, 4, 2, 0 is true. Because the multiples of 8 are 8, 16, 24, 32, 40.
Thus the statement is true.
The product of an odd factor and 8 is always odd is false. Because any odd number multiplied by an even number will be always an even number. Thus the statement is false.

Question 16.
Modeling Real Life
A trophy case has 4 shelves with 8 trophies on each shelf and 1 shelf with 3 trophies. How many trophies are there in all?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 26
Answer: 35 trophies

Explanation:
Given,
A trophy case has 4 shelves with 8 trophies on each shelf and 1 shelf with 3 trophies.
1 shelf – 8 trophies
4 shelves – 4 × 8 trophies = 32 trophies
1 shelf – 3 trophies
32 + 3 = 35 trophies
Therefore there are 35 trophies in all.

Question 17.
DIG DEEPER!
In a science experiment, one group of students tests 2 rows of 8 magnets. A different group tests 1 row of 8 magnets. How many magnets do they test in all?
Answer: 24 magnets

Explanation:
Given that,
In a science experiment, one group of students tests 2 rows of 8 magnets.
2 × 8 = 16 magnets
A different group tests 1 row of 8 magnets.
1 × 8 = 8 magnets
16 + 8 = 24 magnets
Thus there are 24 magnets in all.

Review & Refresh

Show and write the time.

Question 18.
half past 3
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 27
Answer: 3:30

Explanation:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.5-27
Half is nothing but 30 minutes.
So, half past 30 is 3:30

Question 19.
quarter to 11
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.5 28
Answer: 10:45

Explanation:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.5-28
The quarter is nothing but 15. Quarter to 11 means 10:45.

Lesson 3.6 Multiply by 9

Explore and Grow

Write an equation to match the array
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 1
Cross out the last column. Write an equation to match the new array
____ × ____ = ____
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.6-1
Now there are 4 rows of 9 columns.
So the equation to match the new arrays is 4 × 9 = 36

Repeated Reasoning
Explain how multiplying by 10 can help you multiple by 9. Complete the table.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 2
Answer:
The multiples of 9 are given in the below table.
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.6-2

Think and Grow: Multiply by 9

Example
Find 6 × 9
One way: Use the Distributive Property. Rewrite 9 as 5 + 4
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 3

Distributive Property (with subtraction)
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 4

Answer: 54

Explanation:
Rewrite 9 as 5 + 4
6 × 9 = 6 × (5 + 4)
6 × 9 = (6 × 5) + (6 × 4)
6 × 9 = 30 + 24
6 × 9 = 54
Distribution property with subtraction
Rewrite 9 as 10 – 1
6 × 9 = 6 × (10 – 1)
6 × 9 = (6 × 10) – (6 × 1)
6 × 9 = 60 – 6
6 × 9 = 54

Show and Grow

Question 1.
Find 3 × 9
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 5
Answer: 27

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
3 × 9 = 3 × (10 – 1)
3 × 9 = (3 × 10) – (3 × 1)
3 × 9 = 30 – 3
3 × 9 = 27

Apply and Grow: Practice

Find the product
Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 6
Answer: 36

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
4 × 9 = 4 × (10 – 1)
4 × 9 = (4 × 10) – (4 × 1)
4 × 9 = 40 – 4
4 × 9 = 36

Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 7
Answer: 72

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
9 × 8 = (10 – 1) × 8
9 × 8 = (10 × 8) – (1 × 8)
9 × 8 = 80 – 8
9 × 8 = 72

Question 4.
9 × 7 = ___
Answer: 63

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
9 × 7 = (10 – 1) × 7
9 × 7 = (10 × 7) – (1 × 7)
9 × 7 = 70 – 7
9 × 7 = 63

Question 5.
9 × 1 = ___
Answer: 9

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
9 × 1 = (10 – 1) × 1
9 × 1 = (10 × 1) – (1 × 1)
9 × 1 = 10 – 1
9 × 1 = 9

Question 6.
2 × 9 = ____
Answer: 18

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
2 × 9 = 2 × (10 – 1)
2 × 9 = (2 × 10) – (2 × 1)
2 × 9 = 20 – 2
2 × 9 = 18

Question 7.
0 × 9 = ____
Answer: 0

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
0 × 9 = 0 × (10 – 1)
0 × 9 = (0 × 10) – (0 × 1)
0 × 9 = 0 – 0
0 × 9 = 0

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 8
Answer: 90

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
9 × 10 = (10 – 1) × 10
9 × 10 = (10 × 10) – (1 × 10)
9 × 10 = 100 – 10
9 × 10 = 90

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 9
Answer: 45

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
9 × 5 = (10 – 1) × 5
9 × 5 = (10 × 5) – (1 × 5)
9 × 5 = 50 – 5
9 × 5 = 45

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 10
Answer: 72

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
8 × 9 = 8 × (10 – 1)
8 × 9 = (8 × 10) – (8 × 1)
8 × 9 = 80 – 8
8 × 9 = 72

Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 11
Answer: 81

Explanation:
Distribution property with subtraction
Rewrite 9 as 10 – 1
9 × 9 = 9 × (10 – 1)
9 × 9 = (9 × 10) – (9 × 1)
9 × 9 = 90 – 9
9 × 9 = 81

Find the missing factor
Question 12.
9 × ____ = 0
Answer: 0

Explanation:
Let the missing factor be p.
9 × p = 0
p = 0/9
p = 0
Thus the missing factor is 0.

Question 13.
___ × 9 = 45
Answer: 5

Explanation:
Let the missing factor be q.
q × 9 = 45
q = 45/9
q = 5
Thus the missing factor is 5.

Question 14.
9 × ____ = 36
Answer: 4

Explanation:
Let the missing factor be s.
9 × s = 36
s = 36/9
s = 4
Thus the missing factor is 4.

Question 15.
A softball team has 9 positions and 2 players for each position. How many players are on the team?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 12
Answer: 18 players

Explanation:
Given that,
A softball team has 9 positions and 2 players for each position.
9 × 2 = 18 players
Thus there are 18 players on the team.

Question 16.
Number Sense
Which are not multiples of 9?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 13
Answer: The numbers that are not the multiples of 9 are 50, 42, 10.

Think and Grow: Modeling Real Life

In geocaching, people search for a cache, or collection of objects, using a GPS device. You go geocaching for 9 days. You find 2 caches each day. Your goal is to find 20 caches. Do you reach your goal?
Multiplication equation:
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 14
You ______ reach your goal.

Answer: Yes

Explanation:
Given,
In geocaching, people search for a cache, or collection of objects, using a GPS device.
You go geocaching for 9 days.
You find 2 caches each day.
1 day – 2 caches
9 days – 9 × 2 caches = 18 caches
Your goal is to find 20 caches.
20 – 18 = 2 caches
Yes, you can reach your goal.

Show and Grow

Question 17.
You have 9 math problems for homework. You spend 4 minutes on each problem. Your goal is to finish your math homework in 40 minutes. Do you reach your goal?
Answer: Yes

Explanation:
Given,
You have 9 math problems for homework. You spend 4 minutes on each problem.
1 problem – 4 minutes
9 problems – 9 × 4 minutes = 36 minutes
Your goal is to finish your math homework in 40 minutes.
40 – 36 = 4 minutes
Yes, you reach your goal.

Question 18.
DIG DEEPER!
You exercise for 6 days. You exercise for 10 minutes each day. Your friend exercises for 8 days. Your friend exercises for 9 minutes each day. Who exercises the most minutes?
Answer: Your friend

Explanation:
Given,
You exercise for 6 days. You exercise for 10 minutes each day.
1 day – 10 minutes
6 days – 6 × 10 minutes = 60 minutes
Your friend exercises for 8 days. Your friend exercises for 9 minutes each day.
1 day – 9 minutes
8 days – 8 × 9  minutes = 72 minutes
Thus your friend exercises the most minutes.

Question 19.
DIG DEEPER!
Your school is collecting cans for a food drive. Seven students from your class each collect 9 cans. Eight students from your friend’s class each collect 8 cans. Which class collects the most cans?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 15
Answer: Your friends class collects the most cans

Explanation:
Given,
Your school is collecting cans for a food drive. Seven students from your class each collect 9 cans.
1 student – 9 cans
7 students – 7 × 9 cans = 63 cans
Eight students from your friend’s class each collect 8 cans.
1 student – 8 cans
8 students – 8 × 8 cans = 64 cans
Thus Your friend’s class collects the most cans.

Multiply by 9 Homework & Practice 3.6

Find the product
Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 16
Answer: 63

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction
7 × 9 = 7 × (10 – 1)
7 × 9 = (7 × 10) – (7 × 1)
7 × 9 = 70 – 7
7 × 9 = 63

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 17
Answer: 18

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction.
9 × 2 = (10 – 1) × 2
9 × 2 = (10 × 2) – (1 × 2)
9 × 2 = 20 – 2
9 × 2 = 18

Question 3.
1 × 9 = ____
Answer: 9

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction.
1 × 9 = 1 × (10 – 1)
1 × 9 = (1 × 10) – (1 × 1)
1 × 9 = 10 – 1
1 × 9 = 9

Question 4.
9 × 9 = _____
Answer: 81

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction.
9 × 9 = 9 × (10 – 1)
9 × 9 = (9 × 10) – (9 × 1)
9 × 9 = 90 – 9
9 × 9 = 81

Question 5.
10 × 9 = ____
Answer: 90

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction.
10 × 9 = 10 × (10 – 1)
10 × 9 = (10 × 10) – (10 × 1)
10 × 9 = 100 – 10
10 × 9 = 90

Question 6.
3 × 9 = _____
Answer: 27

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction.
3 × 9 = 3 × (10 – 1)
3 × 9 = (3 × 10) – (3 × 1)
3 × 9 = 30 – 3
3 × 9 = 27

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 18
Answer: 72

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction.
9 × 8 = (10 – 1) × 8
9 × 8 = (10 × 8) – (1 × 8)
9 × 8 = 80 – 8
9 × 8 = 72

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 19
Answer: 54

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction.
9 × 6 = (10 – 1) × 6
9 × 6 = (10 × 6) – (1 × 6)
9 × 6 = 60 – 6
9 × 6 = 54

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 20
Answer: 36

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction.
4 × 9 = 4 × (10 – 1)
4 × 9 = (4 × 10) – (4 × 1)
4 × 9 = 40 – 4
4 × 9 = 36

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 21
Answer: 0

Explanation:
Rewrite 9 as 10 – 1
We can find the product by using the distributive property with subtraction.
0 × 9 = 0 × (10 – 1)
0 × 9 = (0 × 10) – (0 × 1)
0 × 9 = 0
0 × 9 = 0

Find the missing factor
Question 11.
9 × ___ = 9
Answer: 1

Explanation:
Let the missing factor be a.
9 × a = 9
a = 9/9
a = 1
Therefore the missing factor is 1.

Question 12.
___ × 5 = 45
Answer: 9

Explanation:
Let the missing factor be r.
r × 5 = 45
r = 45/5
r = 9
Therefore the missing factor is 9.

Question 13.
90 = ___ × 10
Answer: 9

Explanation:
Let the missing factor be p.
90 = p × 10
p = 90/10
p = 9
Therefore the missing factor is 9.

Question 14.
You see 9 chipmunks on your walk to school. You see twice as many pigeons. How many pigeons do you see?
Answer: 18

Explanation:
Given that,
You see 9 chipmunks on your walk to school. You see twice as many pigeons.
Let the number of pigeons is x.
twice = 2x
2 × 9 = 18
Thus there are 18 pigeons.

Question 15.
Patterns
Use the table
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 22
What pattern do you notice in the ones digits? the tens digits?
What do you notice about the sum of the digits for each multiple of 9?
Answer:
The pattern in ones digit is 9, 8, 7, 6, 5, 4, 3, 2, 1, 0.
The pattern in tens digit is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
I observe that the sum of digits for each multiple of 9 is 9.
18 = 1 + 8 = 9
27 = 2 + 7 = 9
36 = 3 + 6 = 9
45 = 4 + 5 = 9
54 = 5 + 4 = 9
63 = 6 + 3 = 9
72 = 7 + 2 = 9
81 = 8 + 1 = 9
90 = 9 + 0 = 9

Question 16.
YOU BE THE TEACHER
Your friend says the product of 7 × 9 is 69. Is your friend correct? Explain.
Answer: My friend is incorrect. Because the product of 7 × 9 is 63.

Question 17.
Modeling Real Life
You sell 8 orchids. You want to raise $70. Do you meet your goal?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 23

Answer:
Given,
You sell 8 orchids.
The cost of 1 orchid – $9
8 orchids – 8 × $9 = $72
You want to raise $70.
$70 – $72 = -$2
No, I didn’t reach my goal.

DIG DEEPER!
Newton sells 9 lilies. Descartes sells 5 orchids. Who raises the most money?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.6 24
Answer:
Given that,
Newton sells 9 lilies. Descartes sells 5 orchids.
Cost of lily – $6
9 × $6 = $54
Cost of Orchid – $9
5 × $9 = $45
Thus Newton raises more money.

Review & Refresh

Complete the equation.
Question 18.
4 × 5 = 5 × ____
Answer: 4

Explanation:
The commutative property is a math rule that says that the order in which we multiply numbers does not change the product.
So, 4 × 5 = 5 × 4

Question 19.
8 × ____ = 7 × 8
Answer: 7

Explanation:
The commutative property is a math rule that says that the order in which we multiply numbers does not change the product.
8 × 7 = 7 × 8

Lesson 3.7 Practice Multiplication Strategies

Explore and Grow

Show how to find the product
6 × 7 = ____
Answer: 42

Explanation:
Multiply the two numbers 6 and 7.
6 × 7 = 42

Reasoning
What other strategies can you use to solve?
Answer: The other strategy to find the product is by using the distributive property.
Rewrite 7 as 5 + 2
6 × 7 = 6 × (5 + 2)
6 × 7 = (6 × 5) + (6 × 2)
6 × 7 = 30 + 12
6 × 7 = 42

Think and Grow: Practice Multiplication Strategies

Example
Use any strategy to find 4 × 3.
One way: Use a number line. Skip count by 3s four times.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 1
Another way: Use the Distributive Property
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 2

Answer:
Using Distributive Property
Rewrite 4 as 2 + 2.
4 × 3 = (2 + 2) × 3
4 × 3 = (2 × 3) + (2 × 3)
4 × 3 = 6 + 6
4 × 3 = 12

Show and Grow

Use any strategy to find the product
Question 1.
5 × 6 = ____
Answer: 30

Explanation:
Using Distributive Property
Rewrite 6 as 3 + 3.
5 × 6 = 5 × (3 + 3)
5 × 6 = (5 × 3) + (5 × 3)
5 × 6 = 15 + 15
5 × 6 = 30

Question 2.
3 × 8 = ____
Answer: 24

Explanation:
Using Distributive Property.
Rewrite 8 as 4 + 4.
3 × 8 = 3 × (4 + 4)
3 × 8 = (3 × 4) + (3 × 4)
3 × 8 = 12 + 12
3 × 8 = 24

Question 3.
7 × 8 = ____
Answer: 56

Explanation:
Using Distributive Property
Rewrite 8 as 4 + 4.
7 × 8 = 7 × (4 + 4)
7 × 8 = (7 × 4) + (7 × 4)
7 × 8 = 28 + 28
7 × 8 = 56

Question 4.
9 × 6 = ____
Answer: 54

Explanation:
Using Distributive Property
Rewrite 6 as 3 + 3.
9 × 6 = 9 × (3 + 3)
9 × 6 = (9 × 3) + (9 × 3)
9 × 6 = 27 + 27
9 × 6 = 54

Apply and Grow: Practice

Use any strategy to find the product
Question 5.
5 × 9 = _____
Answer: 45

Explanation:
Using Distributive Property
Rewrite 9 as 5 + 4.
5 × 9 = 5 × (5 + 4)
5 × 9 = (5 × 5) + (5 × 4)
5 × 9 = 25 + 20
5 × 9 = 45

Question 6.
5 × 7 = ____
Answer: 35

Explanation:
Using Distributive Property
Rewrite 7 as 5 + 2.
5 × 7 = 5 × (5 + 2)
5 × 7 = (5 × 5) + (5 × 2)
5 × 7 = 25 + 10
5 × 7 = 35

Question 7.
10 × 3 = ____
Answer: 30

Explanation:
Using Distributive Property
Rewrite 10 as 5 + 5.
10 × 3 = (5 + 5) × 3
10 × 3 = (5 × 3) + (5 × 3)
10 × 3 = 15 + 15
10 × 3 = 30

Question 8.
7 × 1 = ____
Answer: 7

Explanation:
Using Distributive Property
Rewrite 7 as 5 + 2.
7 × 1 = (5 + 2) × 1
7 × 1 = (5 × 1) + (2 ×1)
7 × 1 = 5 + 2
7 × 1 = 7

Question 9.
5 × 5 = ____
Answer: 25

Explanation:
Using Distributive Property
Rewrite 5 as 2 + 3
5 × 5 = 5 × (2 + 3)
5 × 5 = (5 × 2) + (5 × 3)
5 × 5 = 10 + 15
5 × 5 = 25

Question 10.
4 × 9 = ____
Answer: 36

Explanation:
Using Distributive Property
Rewrite 4 as 2 + 2.
4 × 9 = (2 + 2) × 9
4 × 9 = (2 × 9) + (2 × 9)
4 × 9 = 18 + 18
4 × 9 = 36

Question 11.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 3
Answer: 48

Explanation:
Using Distributive Property.
Rewrite 8 as 4 + 4.
8 × 6 = (4 + 4) × 6
8 × 6 = (4 × 6) + (4 × 6)
8 × 6 = 24 + 24
8 × 6 = 48

Question 12.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 4
Answer: 28

Explanation:
Using Distributive Property
Rewrite 4 as 2 + 2.
4 × 7 = (2 + 2) × 7
4 × 7 = (2 × 7) + (2 × 7)
4 × 7 = 14 + 14
4 × 7 = 28

Question 13.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 5
Answer: 10

Explanation:
Using Distributive Property
Rewrite 5 as 2 + 3
5 × 2 = (2 + 3) × 2
5 × 2 = (2 × 2) + (2 × 3)
5 × 2 = 4 + 6
5 × 2 = 10

Name the strategy or property used to solve

Question 14.
9 × 9 = 9 × (10 – 1)
9 × 9 = (9 × 10) – (9 × 1)
9 × 9 = 90 – 9
9 × 9 = 81
Answer: Distributive property with subtraction
The distributive property of multiplication over subtraction is like the distributive property of multiplication over addition. You can subtract the numbers and then multiply, or you can multiply and then subtract.

Question 15.
2 × 4 = 8
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 6
Answer: Number line

Question 16.
DIG DEEPER!
Use known facts to find 6 × 12. Explain your strategy 6 × 12 = ____
Answer: 72

Explanation:
Rewrite 6 as 3 + 3.
6 × 12 = (3 + 3) × 12
6 × 12 = (3 × 12) + (3 × 12)
6 × 12 = 36 + 36
6 × 12 = 72

Think and Grow: Modeling Real Life

You want to make a dragon that is 25 feet long for a parade. You have 6 pieces of fabric that are each 5 feet long. Do you have enough fabric to make the dragon?
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 7
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 8
You _____ have enough fabric to make the dragon.

Answer:
Given,
You want to make a dragon that is 25 feet long for a parade.
You have 6 pieces of fabric that are each 5 feet long
6 × 5 = 30
Big-Ideas-Math-Answers-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.7-8
Yes, you have enough fabric to make the dragon.

Show and Grow

Question 17.
A book is 70 pages long. You read 9 pages each day for one week. Do you finish the book in one week?
Answer: No

Explanation:
Given,
A book is 70 pages long. You read 9 pages each day for one week.
Convert week to day.
1 day – 9 pages
7 days – 7 × 9 pages = 63 pages
63 – 70 = -7
Thus you did not finish the book in one week.

Question 18.
DIG DEEPER!
You buy 4 packs of juice boxes. How many juice boxes do you buy in all?
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 9
Explain how you solved the problem.
Answer: 40 juice boxes

Explanation:
You buy 4 packs of juice boxes.
1 pack contains 5 grape, 3 apple, 2 berry
4 packs = 5 grapes × 4, 3 apple × 4, 2 berry × 4
= 20 grape, 12 apple, 8 berry
= 20 + 12 + 8 = 40
Thus you buy 40 juice boxes in all.

Practice Multiplication Strategies Homework & Practice 1.1

Use any strategy to find the product
Question 1.
0 × 6 = ____
Answer: 0

Explanation:
Any number multiplied by 0 will be always 0.

Question 2.
2 × 10 = ____
Answer: 20

Explanation:
You can find the product by using the distributive property.
Rewrite 10 as 5 + 5.
2 × 10 = 2 × (5 + 5)
2 × 10 = (2 × 5) + (2 × 5)
2 × 10 = 10 + 10
2 × 10 = 20

Question 3.
1 × 9 = ____
Answer: 9

Explanation:
You can find the product by using the distributive property.
Rewrite 9 as 6 + 3
1 × 9 = 1 × (6 + 3)
1 × 9 = (1 × 6) + (1 × 3)
1 × 9 = 6 + 3
1 × 9 = 9

Question 4.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 10
Answer: 20

Explanation:
You can find the product by using the distributive property.
Rewrite 4 as 2 + 2.
4 × 5 = (2 + 2) × 5
4 × 5 = (2 × 5) + (2 × 5)
4 × 5 = 10 + 10
4 × 5 = 20

Question 5.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 11
Answer: 72

Explanation:
You can find the product by using the distributive property.
Rewrite 9 as 6 + 3
8 × 9 = 8 × (6 + 3)
8 × 9 = (8 × 6) + (8 × 3)
8 × 9 = 48 + 24
8 × 9 = 72

Question 6.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 12
Answer: 21

Explanation:
You can find the product by using the distributive property.
Rewrite 7 as 5 + 2.
7 × 3 = (5 + 2) × 3
7 × 3 = (5 × 3) + (2 × 3)
7 × 3 = 15 + 6
7 × 3 = 21

Name the strategy or property used to solve.
Question 7.
3 × 5 = ?
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 13

Answer: The strategy used for the above product is table diagram.

Question 8.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 14
Answer: Distributive Property of Multiplication.

Explanation:
According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together.

Question 9.
Logic
Without multiplying, how can you tell which product will be greater,6 × 3 or 6 × 4? Explain.
Answer: We can say the greater product by comparing the multiples 4 is greater than 3. So, 6 × 3 is less than 6 × 4.

Question 10.
YOU BE THE TEACHER
Descartes uses the Distributive Property to solve 5 × 8. Is he correct? Explain.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 15
Answer: No Descartes solution is incorrect.

Explanation:
Rewrite 8 as 10 – 2.
5 × 8 = 5 × (10 – 2)
5 × 8 = (5 × 10) – (5 × 2)
5 × 8 = 50 – 10
5 × 8 = 40

Question 11.
Modeling Real Life
You order 24 eggs from a farmer. The farmer has 8 chickens. Each chicken lays 3 eggs. Does the farmer have enough eggs for your order? Explain.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 16
Answer:
Given that,
You order 24 eggs from a farmer.
The farmer has 8 chickens.
Each chicken lays 3 eggs.
8 × 3 eggs = 24 eggs
Yes, the farmer have enough eggs for your order.

Question 12.
DIG DEEPER!
You have 3 piles of sports cards. There are 3 baseball cards, 2 basketball cards, and 4 football cards in each pile. How many sports cards do you have in all?
Answer:
Given that,
You have 3 piles of sports cards.
There are 3 baseball cards, 2 basketball cards, and 4 football cards in each pile.
1 pile – 3 basketball cards, 2 basketball cards, and 4 football cards
3 piles – 9 basketball cards, 6 basketball cards, and 12 football cards
= 9 + 6 + 12 = 27 cards
Thus you have 27 sports cards in all.

Review & Refresh

Find the sum
Question 13.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 17
Answer: 92

Explanation:
Add all the three numbers.
32
+13
+47
92

Question 14.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 18
Answer: 84

Explanation:
Add all the three numbers.
46
+14
+24
84

Question 15.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 19
Answer: 72

Explanation:
Add all the three numbers.
55
+10
+12
72

Question 16.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.7 20
Answer: 63

Explanation:
Add all the three numbers.
21
+13
+29
63

Lesson 3.8 Multiply Three Factors

Explore and Grow

Model 2 arrays that each have 4 rows and 3 columns. Draw your model. Complete the equation for the arrays.
2 × ____ ×____ = _____
Answer: 2 × 4 × 3 = 24

Model 3 arrays that each have 2 rows and 4 columns. Draw your model. Complete the equation for the arrays.
3 × ___ × ____ = ____
Answer: 3 × 2 × 4 = 24

Structure
Compare the equations. How are they the same? How are they different?
Answer: The solutions for both equations are the same. But the arrays and rows and columns are different.

Think and Grow: Associative Property of Multiplication

Associative Property of Multiplication: Changing the grouping of factors does not change the product
Example
Find (5 × 2) × 3
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 1
Answer:
Associative Property of Multiplication
First find the factors in the bracket.
5 × 2 = 10
10 × 3 = 30
Commutative Property of Multiplication
5 × (2 × 3)
2 × 3 = 6
5 × 6 = 30

Example
Find 4 × (7 × 2)
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 2

Answer:
4 × (7 × 2) = 4 × (2 × 7)
4 × (2 × 7) = (4 × 2) × 7
(4 × 2) × 7 = 8 × 7 = 56
4 × (2 × 7) = 4 × 14 =56

Show and Grow

Find the product
Question 1.
(3 × 2) × 2 = _____
Answer: 12

Explanation:
First find the factors in the bracket.
(3 × 2) × 2
3 × 2 = 6
6 × 2 = 12

Question 2.
3 × (4 × 3) = ____
Answer: 36

Explanation:
First find the factors in the bracket.
3 × (4 × 3)
(4 × 3) = 12
3 × 12 = 36

Apply and Grow: Practice

Find the product
Question 3.
(3 × 2) × 7 = ____
Answer: 42

Explanation:
First find the factors in the bracket.
3 × 2 = 6
6 × 7 = 42

Question 4.
5 × (4 × 2) = _____
Answer: 40

Explanation:
First find the factors in the bracket.
4 × 2 = 8
5 × 8 = 40

Question 5.
(3 × 6) × 2 = ____
Answer: 36

Explanation:
First find the factors in the bracket.
3 × 6 = 18
18 × 2 = 36

Question 6.
3 × (5 × 3) = ____
Answer: 45

Explanation:
First find the factors in the bracket.
5 × 3 = 15
3 × 15 = 45

Question 7.
(4 × 10) × 2 = ____
Answer: 80

Explanation:
First find the factors in the bracket.
4 × 10 = 40
40 × 2 = 80

Question 8.
6 × (0 × 7) = ____
Answer: 0

Explanation:
First find the factors in the bracket.
0 × 7 = 0
6 × 0 = 0

Tell whether the equation is true or false. Explain.

Question 9.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 3
Answer: True

Explanation:
First find the factors in the bracket.
5 × 3 = 15
15 × 4 = 60
5 × 4 = 20
3 × 20 = 60
Thus the above statement is true.

Question 10.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 4
Answer: False

Explanation:
First find the factors in the bracket.
2 × 6 = 12
4 × 12 = 48
8 × 4 = 32
48 ≠ 32
Thus the above statement is false.

Question 11.
A water ride has 4 log boats. Each boat has 2 sections with 2 seats in each section. How many seats does the water ride have?
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 5
Answer:
Given that,
A water ride has 4 log boats. Each boat has 2 sections with 2 seats in each section.
1 section – 2 seats
2 sections – 2 × 2 seats = 4 seats
4 × 4 seats = 16 seats
Therefore the water ride have 16 seats.

Question 12.
DIG DEEPER!
Complete the square so that the product of the numbers in each row and each column equals 24.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 6
Answer:
Big-Ideas-Math-Solutions-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.8-6

Think and Grow: Modeling Real Life

There are 26 students in your class. Your teacher brings in 4 boxes of muffins. Each box has 4 packages with 2 muffins in each package. Are there enough muffins for the class?
Multiplication equation:
There ______ enough muffins for the class.

Answer:
Given,
There are 26 students in your class.
Your teacher brings in 4 boxes of muffins.
Each box has 4 packages with 2 muffins in each package.
1 package – 2 muffins
4 packages – 2 × 4 packages = 8 packages
4 × 8 packages = 32 packages
32 – 26 = 6
Thus there are enough muffins for the class.

Show and Grow

Question 13.
Newton needs to send out 50 letters. He buys 4 sheets of stamps. Each sheet has 2 rows with 6 stamps in each row. Does Newton have enough stamps to send out all the letters?
Answer:
Given that,
Newton needs to send out 50 letters.
He buys 4 sheets of stamps.
Each sheet has 2 rows with 6 stamps in each row.
2 × 6 = 12 stamps
4 × 12 stamps = 48 stamps
48 – 50 = -2
Newton does not have enough stamps to send out all the letters.

Question 14.
DIG DEEPER!
There are 60 people in line to ride the tram at a zoo. There are 5 benches on each tram car. Two people fit on each bench. The tram is 5 cars long. How many people will have to wait for the next tram?
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 7
Answer:
Given that,
There are 60 people in line to ride the tram at a zoo.
There are 5 benches on each tram car.
Two people fit on each bench.
5 × 2 = 10 people fit on 5 benches
The tram is 5 cars long
5 × 10 = 50 people
50 – 60 = -10 people
Thus 10 people will have to wait for the next time.

Multiply Three Factors Homework & Practice 3.8

Find the product
Question 1.
(2 × 4) × 1 = _____
Answer: 8

Explanation:
First multiply the factors in the bracket.
2 × 4 = 8
8 × 1 = 8
Thus (2 × 4) × 1 = 8

Question 2.
2 × (3 × 3) = ____
Answer: 18

Explanation:
First multiply the factors in the bracket.
3 × 3 = 9
2 × 9 = 18
Thus 2 × (3 × 3) = 18

Question 3.
(4 × 2) × 9 = _____
Answer: 72

Explanation:
First multiply the factors in the bracket.
4 × 2 = 8
8 × 9 = 72
Thus (4 × 2) × 9 = 72

Question 4.
2 × (8 × 5) = _____
Answer: 80

Explanation:
First multiply the factors in the bracket.
8 × 5 = 40
2 × 40 = 80
So, 2 × (8 × 5) = 80

Tell whether the equation is true or false. Explain
Question 5.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 8
Answer: False

Explanation:
Any number multiplied by 0 will be always 0.
0 × 3 = 0
7 × 0 = 0
7 × 3 = 21
Thus the equation 7 × (0 × 3) = 7 ×3 is false.

Question 6.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 9
Answer: True

Explanation:
6 × 5 = 30
2 × 30 = 60
6 × 2 = 12
12 × 5 = 60
Thus the equation 2 × (6 × 5) = (6 × 2) × 5 is false.

Question 7.
A sudoku puzzle is made of 9 large squares. Each large square is made of an array with 3 rows and 3 columns of small squares. How many small squares are there?
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 10
Answer: 81 small squares

Explanation:
Given that,
A sudoku puzzle is made of 9 large squares.
Each large square is made of an array with 3 rows and 3 columns of small squares.
3 × 3 = 9 small squares in each large square
9 × 9 = 81 small squares in sudoko puzzle.

Question 8.
DIG DEEPER!
Use the number of cards to complete the equations.
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 11

Answer:
Solve the equations by using the number of cards.
Big-Ideas-Math-Solutions-Grade-3-Chapter-3-More-Multiplication-Facts-and-Strategies-3.8-11

Question 9.
YOU BE THE TEACHER
Your friend says that (2 × 1) × 7 is equal to 2 × (1 × 7). Is your friend correct? Explain.
Answer:
Given,
Your friend says that (2 × 1) × 7 is equal to 2 × (1 × 7).
(2 × 1) × 7 = 2 × 7 = 14
2 × (1 × 7) = 2 × 7 = 14
Thus your friend is correct.

Question 10.
Writing
How do you know 2 × 9 × 5 is the same as 10 × 9?
Answer:
2 × 9 × 5 = 18 × 5 = 90
10 × 9 = 90
Thus the expression is correct.

Question 11.
Modeling Real Life
There are 64 soccer players. Some coaches bring 7 boxes of protein bars. Each box has 5 packages with 2 protein bars in each package. Are there enough protein bars for each soccer player to get one?
Big Ideas Math Solutions Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.8 12

Answer:
Given that,
There are 64 soccer players.
Some coaches bring 7 boxes of protein bars.
Each box has 5 packages with 2 protein bars in each package.
1 package – 2 protein bars
5 packages – 2 × 5 = 10 protein bars
7 × 10 = 70 protein bars
70 – 64 = 6
Thus each soccer player get enough protein bars.

Review & Refresh

Question 12.
A training academy certifies 12 fire fighting Dalmations. They divide the Dalmations among different cities with 3 Dalmatians in each city. How many cities receive firefighting Dalmations?
____ ÷ _____ = _____
Answer:
Given that,
A training academy certifies 12 fire fighting Dalmations. They divide the Dalmations among different cities with 3 Dalmatians in each city.
12 ÷ 3 = 4
Therefore 4 cities receive firefighting Dalmations.

Lesson 3.9 More Problem Solving: Multiplication

Explore and Grow

Model the story. A baseball league gives 8 new baseballs to each team. There are 8 teams. How many baseballs does the league need?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 1
The baseball league needs ______ baseballs.
Answer:
Given,
A baseball league gives 8 new baseballs to each team.
There are 8 teams.
8 × 8 baseballs = 64 baseballs
Thus, the baseball league needs 64 baseballs.

Reasoning
Explain how you can use a different strategy to solve.
Answer:
We can solve by using the distributive property.
8 can be written as (10 – 1)
8 × 8 = 8 × (10 – 2)
8 × 8 = (8 × 10) – (8 × 2)
8 × 8 = 80 – 16
8 × 8 = 64

Think and Grow: Using the Problem-Solving Plan

Example
You want to make 8 dream catchers. You have 30 feathers. You tie 3 feathers to each dream catcher. How many feathers do you have left?

Understand the Problem

What do you know?
• You want to make ______ dream
• You have ______ feathers in all
• You tie _____ feathers to each dream catcher.

Answer:
Fill the blanks with the help of the above question.
• You want to make 8 dream
• You have 30 feathers in all
• You tie 3 feathers to each dream catcher.

What do you need to find?
• You need to find how many ______ are left after catchers. you make dream catchers.

Answer:
By seeing the above question we can fill the blanks.
• You need to find how many feathers are left after catchers. you make dream catchers.

Make a Plan

How will you solve?
• Multiply _____ by ______ to find how many ______ you used to make 8 dream catchers.
• Subtract the product from ______.

Answer:
Fill in the blanks by using the above question.
• Multiply 8 by 3 to find how many feathers you used to make 8 dream catchers.
• Subtract the product from total number of feathers.

Solve
8 × 3 = ______
30 – ___ = _____
You have _____ feathers left.

Answer:
8 × 3 = 24
30 – 24 = 6
Thus, you have 6 feathers left.

Show and Grow

Question 1.
You want to make 6 cheese sandwiches. You have 8 slices of cheese in all. You put 2 slices on each sandwich. How many more slices of cheese do you need?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 2
Answer:
Given that,
You want to make 6 cheese sandwiches.
Each sandwich needs 2 slices.
6 × 2 = 12 slices
You have 8 slices of cheese in all. You put 2 slices on each sandwich.
12 – 8 = 4 slices
Therefore you need 4 slices of cheese.

Apply and Grow: Practice

Question 2.
You have 60 fluid ounces of water and 6 water bottles. You pour 8 fluid ounces of water into each bottle. What information do you know that will help you find how much water you have left?
Answer:
Given,
You have 60 fluid ounces of water and 6 water bottles. You pour 8 fluid ounces of water into each bottle.
To find how much water you have left we need to subtract total fluid ounces and amount of fluid ounces in 6 water bottles.
6 × 8 fluid ounces = 48 fluid ounces
60 – 48 = 12 fluid ounces
Thus 12 fluid ounces of water is left.

Question 3.
A sheet of stamps has 2 rows with 5 stamps in each row. You buy 3 sheets of stamps. How many stamps do you buy?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 3
Answer:
Given that,
A sheet of stamps has 2 rows with 5 stamps in each row.
2 × 5 = 10 stamps in each row.
You buy 3 sheets of stamps.
10 × 3 = 30 stamps
Thus you buy 30 stamps.

Question 4.
Your cousin works two jobs. She walks dogs 4 days each week for 2 hours each day. She babysits 2 days each week for 5 hours each day. How many hours does your cousin work in one week?
Answer:
Given that,
Your cousin works two jobs. She walks dogs 4 days each week for 2 hours each day.
4 × 2 = 8 hours each week
She babysits 2 days each week for 5 hours each day.
2 × 5 hours = 10 hours
8 + 10 = 18 hours in one week
Thus your cousin 18 hours in one week.

Question 5.
Four arcade tokens cost $1. You buy $5 in tokens. You play 6 games that cost 3 tokens each. How many tokens do you have left?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 4
Answer:
Given that,
Four arcade tokens cost $1. You buy $5 in tokens.
1 × 4 = 4 tokens
5 × 4 = 20 tokens
You play 6 games that cost 3 tokens each.
6 × 3 = 18 tokens
20 – 18 tokens = 2 tokens
Thus 2 tokens are leftover.

Think and Grow: Modeling Real Life

A basketball team scores 6 three-pointers, 9 two-pointers, and 7 free throws in the first quarter. How many points does the team score in all?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 5
Understand the problem:
Make a plan:
Solve:
The team scores ____ points in all.

Answer:
Given that,
A basketball team scores 6 three-pointers, 9 two-pointers, and 7 free throws in the first quarter.
6 × 3 points = 18 points
9 × 2 points = 18 points
7 × 1 point = 7 point
18 + 18 + 7 = 43 points
Thus the team scores 43 points in all.

Show and Grow

Question 6.
Descartes buys 4 shirts, 3 pairs of shorts, and 2 pairs of pants. How much does he spend in all?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 6
Answer:
Given that,
Descartes buys 4 shirts, 3 pairs of shorts, and 2 pairs of pants.
4 shirts – 4 × $5 = $20
3 shorts – 3 × $7 = $21
2 pants – 2 × $10 = $20
20 + 21 + 20 = $61
Thus Descartes spend $61 in all.

Question 7.
DIG DEEPER!
City workers set up 100 chairs for a concert in a park. Section A has 4 rows with 9 chairs in each row. Section B has 3 rows with 8 chairs in each row. Section C has 8 equal rows. How many chairs are in each of section C’s rows?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 7
Answer: 40

Explanation:
Given that,
City workers set up 100 chairs for a concert in a park.
Section A has 4 rows with 9 chairs in each row.
4 × 9 = 36 chairs
Section B has 3 rows with 8 chairs in each row.
3 × 8 = 24 chairs
Section C has 8 equal rows.
100 – 36 – 24 = 40 chairs
Thus there are 40 chairs in Section C.

More Problem Solving: Multiplication Homework & Practice 3.9

Question 1.
You and 3 of your friends each spend $9 on a present. How much does the present cost?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 8
Answer: $36

Explanation:
Given,
You and 3 of your friends each spend $9 on a present.
1 – $9
4 – 4 × $9 = $36
Thus the cost of the present is $36.

Question 2.
A landlord is replacing windows in an apartment complex. There are 6 apartments. Each apartment has 8 windows. Seven windows do not need replacing. How many windows does the landlord need to replace?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 9
Answer: 41 windows

Explanation:
Given,
A landlord is replacing windows in an apartment complex.
There are 6 apartments. Each apartment has 8 windows.
6 × 8 = 48 windows
Seven windows do not need replacing.
48 – 7 = 41
Therefore the landlord needs to replace 41 windows.

Question 3.
Women’s boxing consists of four 2-minute rounds. There is a 1-minute rest interval between each round. What information do you know that will help you find how long a boxing match lasts?
Answer: I need that information about the time that last four 2-minute rounds and 1-minute rest interval.

Explanation:
Given,
Women’s boxing consists of four 2-minute rounds.
4 × 2 = 8 minutes
There is a 1-minute rest interval between each round.
4 × 1 = 4 minutes
8 + 4 = 12 minutes
Thus the boxing match lasts for 12 minutes.

Question 4.
Writing
Write and solve your own word problem that involves multiplication.
Answer:
Find the product of 3 × 5
We can find the product by using the distributive property.
Rewrite 5 as 2 + 3.
3 × 5 = 3 × (2 + 3)
3 × 5 = (3 × 2) + (3 × 3)
3 × 5 = 6 + 9
3 × 5 = 15

Question 5.
Modeling Real Life
You play a trivia game with your friends. Team A answers 3 hard questions and 6 easy questions. Team B answers 5 hard questions and 2 easy questions. Which team is winning? by how many points?
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 10
Answer:
Given,
You play a trivia game with your friends. Team A answers 3 hard questions and 6 easy questions.
Hard – 5 points
Easy – 2 points
3 hard question = 3 × 5 = 15 points
6 easy questions = 6 × 2 = 12 points
15 + 12 = 27 points
Team B answers 5 hard questions and 2 easy questions.
5 hard question = 5 × 5 = 25 points
2 easy questions = 2 × 2 = 4 points
25 + 4 = 29 points
Thus Team B is winning the points by 2 points.

Review & Refresh

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 11
Answer:
i. 5 sides
ii. 5 vertices
iii. pentagon

Explanation:
By seeing the above figure we can say that the shape of the image is a pentagon with 5 sides and 5 vertices.

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 3 More Multiplication Facts and Strategies 3.9 12
Answer:
i. 4 sides
ii. 4 vertices
iii. Quadrilateral

Explanation:
The figure has 4 sides and 4 vertices so the shape is a quadrilateral.

More Multiplication Facts and Strategies Performance Task

Question 1.
You and your friend make origami animals.
a.You have two packages of paper. One package has 8 colors with 6 sheets of each color. Another package has 3 white sheets and 9 times as many colored sheets as white sheets. How many sheets of paper do you have in all?
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 1

Answer:
Given,
You have two packages of paper. One package has 8 colors with 6 sheets of each color.
8 × 6 = 48 colors in 6 sheets
Another package has 3 white sheets and 9 times as many colored sheets as white sheets.
9 × 3 = 27 sheets
= 27 + 6 = 33 sheets

b.You and your friend each want to make 5 origami animals every day for one week. Do you have enough paper? Explain.

Answer:
Given,
You and your friend each want to make 5 origami animals every day for one week.
5 + 5 = 10 origami animals every day for one week.
10 × 7 = 70 origami animals
Yes, you have enough paper.
c.You make a jumping frog. The first step is to fold the paper into fourths that are shaped like triangles. Draw lines to show how you would fold your paper.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 2
d. It takes 12 steps to make a crane. Your friend makes 8 paper cranes. How many steps does your friend do in all? Explain.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 3

Answer:
Given,
It takes 12 steps to make a crane. Your friend makes 8 paper cranes.
12 × 8 = 96 steps
Your friend does 96 steps in all.

More Multiplication Facts and Strategies Activity

Product Lineup
Directions:
1. Players take turns flipping two number cards.
2. On your turn, multiply the two numbers and place a counter on the product.
3. If you flip two of the same number, take another turn.
4. The first player to create two lines of 5 in a row, horizontally, vertically, or diagonally, wins!
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies 4

Answer:
Multiply 5 by row, horizontally and vertically.
5 × 27 = 135
5 × 64 = 320
5 × 3 = 15

More Multiplication Facts and Strategies Chapter Practice

3.1 Multiply by 3

Draw the model to find the product
Question 1.
3 × 8
Answer:
BIM Grade 3 Chapter 3 More Multiplication facts and strategies img_1

Question 2.
3 × 4
Answer:
Bigideas math answers grade 3 chapter 3 more multiplication facts and strategies img_2

Find the product
Question 3.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 3
Answer: 15

Explanation:
Multiply the two numbers 3 and 5.
3 × 5 = 15

Question 4.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 4
Answer: 0

Explanation:
Multiply the two numbers 0 and 3.
Any number multiplied by 0 will be always 0.
So, 0 × 3 = 0

Question 5.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 5
Answer: 3

Explanation:
Multiply the two numbers 1 and 3.
Any number multiplied by 1 will be always the same number.
1 × 3 = 3

Question 6.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 6
Answer: 18

Explanation:
Multiply the two numbers 3 and 6.
3 × 6 = 18

3.2 Multiply by 4

Find the product
Question 7.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 7
Answer: 32

Explanation:
Multiply the two numbers 4 and 8.
4 × 8 = 32

Question 8.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 8
Answer: 40

Explanation:
Multiply the two numbers 10 and 4.
10 × 4 = 40

Question 9.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 9
Answer: 28

Explanation:
Multiply the two numbers 7 and 4.
7 × 4 = 28

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 10
Answer: 16

Explanation:
Multiply the two numbers 4 and 4.
4 × 4 = 16

Find the missing factor
Question 11.
1 × ___ = 4
Answer: 4

Explanation:
Let the missing factor be a.
1 × a = 4
a = 4/1
a = 4
Thus the missing factor is 4.

Question 12.
____ × 4 = 0
Answer: 0

Explanation:
Let the missing factor be b.
b × 4 = 0
b = 0/4
b = 0
Thus the missing factor is 0.

Question 13.
20 = ____ × 4
Answer: 5

Explanation:
Let the missing factor be c.
20 = c × 4
c = 20/4
c = 5
Thus the missing factor is 5.

Question 14.
Number Sense
How can you use 2 × 5 to find 4 × 5?
Answer:
Rewrite 4 as 2 + 2.
4 × 5 = (2 + 2) × 5
4 × 5 = (2 × 5) + (2 × 5)
4 × 5 = 10 + 10
4 × 5 = 20

3.3 Multiply by 6

Find the product
Question 15.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 15
Answer: 30

Explanation:
Multiply the two numbers 6 and 5.
6 × 5 = 30

Question 16.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 16
Answer: 12

Explanation:
Multiply the two numbers 6 and 2.
6 × 2 = 12

Question 17.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 17
Answer: 54

Explanation:
Multiply the two numbers 9 and 6.
9 × 6 = 54

Question 18
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 18
Answer: 18

Explanation:
Multiply the two numbers 3 and 6.
3 × 6 = 18

Compare
Question 19.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 19
Answer: <

Explanation:
Compare both expressions
7 × 6 = 42
6 × 8 = 48
42 is less than 48.
7 × 6 < 6 × 8

Question 20.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 20
Answer: >

Explanation:
Compare both expressions
6 × 9 = 54
4 × 6 = 24
54 is greater than 24.
6 × 9 > 4 × 6

Question 21.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 21
Answer: =

Explanation:
Compare both expressions
6 × 6 = 36
36 = 6 × 6

3.4 Multiply by 7

Find the product
Question 22.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 22
Answer: 0

Explanation:
Multiply the two numbers 0 and 7.
Any number multiplied by 0 is always 0.
0 × 7 = 0

Question 23.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 23
Answer: 49

Explanation:
Multiply the two numbers 7 and 7.
7 × 7 = 49

Question 24.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 24
Answer: 70

Explanation:
Multiply the two numbers 10 and 7.
10 × 7 = 70

Question 25.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 25
Answer: 14

Explanation:
Multiply the two numbers 7 and 2.
7 × 2 = 14

Question 26.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 26
Answer: 7

Explanation:
Multiply the two numbers 7 and 1.
Any number multiplied by 1 is always the same number.
7 × 1 = 7

Question 27.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 27
Answer: 63

Explanation:
Multiply the two numbers 7 and 9.
7 × 9 = 63

Question 28.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 28
Answer: 28

Explanation:
Multiply the two numbers 4 and 7.
4 × 7 = 28

Question 29.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 29
Answer: 56

Explanation:
Multiply the two numbers 8 and 7.
8 × 7 = 56

3.5 Multiply by 8

Find the product
Question 30.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 30
Answer: 48

Explanation:
Multiply the two numbers 8 and 6.
8 × 6 = 48

Question 31.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 31
Answer: 72

Explanation:
Multiply the two numbers 9 and 8.
9 × 8 = 72

Question 32.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 32
Answer: 32

Explanation:
Multiply the two numbers 4 and 8.
4 × 8 = 32

Question 33.
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 33
Answer: 40

Explanation:
Multiply the two numbers 8 and 5.
8 × 5 = 40

Question 34.
Precision
There are 8 fluid ounces in 1 cup. How many fluid ounces are in 3 cups?
Big Ideas Math Answers 3rd Grade Chapter 3 More Multiplication Facts and Strategies chp 34
Answer: 24 fluid ounces

Explanation:
Given,
There are 8 fluid ounces in 1 cup.
1 cup – 8 fluid ounces
3 cups – 3 × 8 fluid ounces = 24 fluid ounces
Therefore there are 24 fluid ounces in 3 cups.

3.6 Multiply by 9

Find the product
Question 35.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies chp 35
Answer: 90

Explanation:
Multiply the two numbers 10 and 9.
10 × 9 = 90

Question 36.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies chp 36
Answer: 36

Explanation:
Multiply the two numbers 9 and 4.
9 × 4 = 36

Question 37.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies chp 37
Answer: 54

Explanation:
Multiply the two numbers 9 and 6.
9 × 6 = 54

Question 38.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies chp 38
Answer: 18

Explanation:
Multiply the two numbers 3 and 6.
3 × 6 = 18
Thus the product of 3 and 6 is 18.

Question 39.
Reasoning
An artist needs 80 flower petals for a craft. She picks 9 flowers that each have 8 petals. Does she have enough flower petals for her craft? Explain.
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies chp 39
Answer:
Given,
An artist needs 80 flower petals for a craft.
She picks 9 flowers that each have 8 petals.
9 × 8 = 72 flower petals.
80 – 72 = 8  flower petals
Therefore, She has enough flower petals for her craft.

3.7 Practice Multiplication Strategies

Use any strategy to find the product
Question 40.
2 × 7 = ____
Answer: 14

Explanation:
We can find the product by using the distributive property.
Rewrite 7 as 5 + 2.
2 × 7 = 2 × (5 + 2)
2 × 7 = (2 × 5) + (2 × 2)
2 × 7 = 10 + 4
So, 2 × 7 = 14

Question 41.
6 × 6 = ____
Answer: 36

Explanation:
We can find the product by using the distributive property.
Rewrite 6 as 3 + 3
6 × 6 = 6 × (3 + 3)
6 × 6 = (6 × 3) + (6 × 3)
6 × 6 = 18 + 18
Thus 6 × 6 = 36

Question 42.
5 × 1 = _____
Answer: 5

Explanation:
We can find the product by using the distributive property.
Rewrite 5 as 2 + 3
5 × 1 = (2 + 3) × 1
5 × 1 = (2 × 1) + (3 × 1)
5 × 1 = 2 + 3
Thus 5 × 1 = 5

Question 43.
8 × 10 = ____
Answer: 80

Explanation:
We can find the product by using the distributive property.
Rewrite 10 as 5 + 5
8 × 10 = 8 × (5 + 5)
8 × 10 = (8 ×5) + (8 × 5)
8 × 10 = 40 + 40
So, 8 × 10 = 80

Question 44.
4 × 0 = ____
Answer: 0

Explanation:
We can find the product by using the distributive property.
Any number multiplied by 0 will be always 0.
So, 4 × 0 = 0

Question 45.
10 × 10 = ____
Answer: 100

Explanation:
We can find the product by using the distributive property.
Rewrite 10 as 5 + 5
10 × 10 = 10 × (5 + 5)
10 × 10 = (10 ×5) + (10 × 5)
10 × 10 = 50 + 50
Thus 10 × 10 = 100

Question 46.
Modeling Real Life
Newton has 4 bundles of balloons. There is 1 blue balloon, 2 purple balloons, and 1 green balloon in each bundle. How many balloons does he have in all?
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies chp 46
Answer: 16 balloons

Explanation:
Given that,
Newton has 4 bundles of balloons.
There is 1 blue balloon, 2 purple balloons, and 1 green balloon in each bundle.
4 bundles = 1 blue balloon × 4 + 2 purple balloons × 4 + 1 green balloon × 4
= 4 blue ballon + 8 purple balloon + 4 green balloon
= 16 balloons
Thus there are 16 balloons in all.

3.8 Multiply Three Factors

Find the product
Question 47.
(4 × 2) × 2 = _____
Answer: 16

Explanation:
Given the expression as (4 × 2) × 2
First, multiply the factors in the bracket.
4 × 2 = 8
8 × 2 = 16
Thus (4 × 2) × 2 = 16

Question 48.
3 × (3 × 3) = ____
Answer: 27

Explanation:
Given the expression as 3 × (3 × 3)
First, multiply the factors in the bracket.
3 × 3 = 9
3 × 9 = 27
Thus 3 × (3 × 3) = 27

Question 49.
(3 × 3) × 6 = ____
Answer: 54

Explanation:
Given the expression as (3 × 3) × 6
First, multiply the factors in the bracket.
3 × 3 = 9
9 × 6 = 54
Thus (3 × 3) × 6 = 54

Question 50.
2 × (8 × 4) = _____
Answer: 64

Explanation:
Given the expression as 2 × (8 × 4)
First, multiply the factors in the bracket.
8 × 4 =32
2 × 32 = 64
Thus 2 × (8 × 4) = 64

Question 51.
(2 × 0) × 4 = _____
Answer: 0

Explanation:
Given the expression as (2 × 0) × 4
First, multiply the factors in the bracket.
2 × 0 = 0
0 × 4 = 0
Thus (2 × 0) × 4 = 0

Question 52.
0 × (8 × 5) = ____
Answer: 0

Explanation:
Given the expression as 0 × (8 × 5)
First, multiply the factors in the bracket.
8 × 5 = 40
0 × 40 = 0
Thus 0 × (8 × 5) = 0

3.9 More Problem Solving: Multiplication

Question 53.
You find 2 four-leaf clovers and 9 three-leaf clovers. What is the total number of leaves on the clovers you find?
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies chp 53
Answer:
Given that,
You find 2 four-leaf clovers and 9 three-leaf clovers
2 four-leaf clovers = 4 + 4 = 8 leaves
9 three-leaf clovers = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 =27 leaves
8 + 27 = 35
Thus the total number of leaves on the cloves are 35.

Question 54.
A bakery sells croissants for $2 each. A tray contains 2 rows of 4 croissants. You buy 2 trays. How many croissants do you buy?
Big Ideas Math Answers Grade 3 Chapter 3 More Multiplication Facts and Strategies chp 54
Answer: 16 croissants

Explanation:
Given that,
A bakery sells croissants for $2 each.
A tray contains 2 rows of 4 croissants.
2 × 4 = 8
You buy 2 trays.
8 × 2 = 16 croissants
Cost of croissants = 16 × $2 = $32
Thus I can buy 16 croissants.

Final Words:

I hope the answers provided in Big Ideas Math Grade 3 Chapter 3 More Multiplication Facts and Strategies Answer Key is helpful for all the students to become an expert in this concept. If you have any queries regarding Chapter 3 More Multiplication Facts and Strategies topic, you can leave a comment below. Stay in touch with us to get solutions for all questions of Big Ideas Math Book Grade 3 Chapter 3 Chapter 3 More Multiplication Facts and Strategies.

Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs

Big Ideas Math Geometry Answers Chapter 2

Reasoning and Proofs Chapter Answers provided are aligned as per the Big Ideas Math Geometry Textbooks. Have an overview of the concepts you need to learn in BIM Geometry Ch 2 Reasoning and Proofs and test your understanding. Master the topics of BIM Geometry Chapter 2 Reasoning and Proofs by practicing from the quick links available below. Identify the knowledge gap and concentrate on the areas you are lagging and improvise on them accordingly. You will find the Big Ideas Math Geometry Answers and attempt the exam with utmost confidence.

Big Ideas Math Book Geometry Answer Key Chapter 2 Reasoning and Proofs

Become proficient in the concepts of BIM Geometry Chapter 2 Reasoning and Proofs by referring to the quick links available. Solve the Questions available in BIM Book Geometry Chapter 2 Reasoning and Proofs Answer Key on a frequent basis and get a good hold of the concepts. Geometry Big Ideas Math Chapter 2 Reasoning and Proofs Answers here include questions from Lessons, Review Tests, Cumulative Practice, Assessment Tests, Practice Tests, etc. aligned as per the Textbooks. Download the Big Ideas Math Book Geometry Ch 2 Reasoning and Proofs Solution Key for free of cost and ace your preparation.

Reasoning and Proofs Maintaining Mathematical Proficiency

Write an equation for the nth term of the arithmetic sequence. Then find a50.

Question 1.
3, 9, 15, 21, ……..
Answer:

Question 2.
– 29, – 12, 5, 22, ……..
Answer:

Question 3.
2.8, 3.4, 4.0, 4.6, ………
Answer:

Question 4.
\(\frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{5}{6}\), ………
Answer:

Question 5.
26, 22, 18, 14, ………
Answer:

Question 6.
8, 2, – 4, – 10, ………
Answer:

Solve the literal equation for x.

Question 7.
2y – 2x = 10
Answer:

Question 8.
20y + 5x = 15
Answer:

Question 9.
4y – 5 = 4x + 7
Answer:

Question 10.
y = 8x – x
Answer:

Question 11.
y = 4x + zx + 6
Answer:

Question 12.
z = 2x + 6xy
Answer:

Question 13.
ABSTRACT REASONING
Can you use the equation for an arithmetic sequence to write an equation for the sequence 3, 9, 27, 81. . . . ? Explain our reasoning.
Answer:

Reasoning and Proofs Mathematical Practices

Monitoring Progress

Decide whether the syllogism represents correct or flawed reasoning, If flawed, explain why the conclusion Is not valid.

Question 1.
All triangles are polygons.
Figure ABC is a triangle.
Therefore, figure ABC is a polygon.
Answer:

Question 2.
No trapezoids are rectangles.
Some rectangles are not squares.
Therefore, some squares are not trapezoids.
Answer:

Question 3.
If polygon ABCD is a square. then ills a rectangle.
Polygon ABCD is a rectangle.
Therefore, polygon ABCD is a square.
Answer:

Question 4.
If polygon ABCD is a square, then it is a rectangle.
Polygon ABCD is not a square.
Therefore, polygon ABCD is not a rectangle.
Answer:

2.1 Conditional Statements

Exploration 1

Determining Whether a Statement is True or False

Work with a partner: A hypothesis can either be true or false. The same is true of a conclusion. For a conditional statement to be true, the hypothesis and conclusion do not necessarily both have to be true. Determine whether each conditional statement is true or false. Justify your answer.

a. If yesterday was Wednesday, then today is Thursday.
Answer:

b. If an angle is acute. then it has a measure of 30°.
Answer:

c. If a month has 30 days. then it is June.
Answer:

d. If an even number is not divisible by 2. then 9 is a perfect cube.
Answer:

Exploration 2

Determining Whether a Statement is True or False

Work with a partner: Use the points in the coordinate plane to determine whether each statement is true or false. Justify your answer.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 1
a. ∆ABC is a right triangle.
Answer:

b. ∆BDC is an equilateral triangle.
Answer:

c. ∆BDC is an isosceles triangle.
Answer:

d. Quadrilateral ABCD is a trapezoid.
Answer:

e. Quadrilateral ABCD is a parallelogram.
Answer:

Exploration 3

Determining Whether a Statement is True or False

Work with a partner: Determine whether each conditional statement is true or false. Justify your answer.

CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to distinguish correct logic or reasoning from that which is flawed.

a. If ∆ ADC is a right triangle, then the Pythagorean Theorem is valid for ∆ADC.
Answer:

b. If ∠A and ∠B are complementary, then the sum of their measures is 180°.
Answer:

c. If figure ABCD is a quadrilateral, then the sum of its angle measures is 180°.
Answer:

d. If points A, B, and C are collinear, then the lie on the same line.
Answer:

e. It Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2 and Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 3 intersect at a point, then they form two pairs of vertical angles.
Answer:

Communicate Your Answer

Question 4.
When is a conditional statement true or false?
Answer:

Question 5.
Write one true conditional statement and one false conditional statement that are different from those given in Exploration 3. Justify your answer.
Answer:

Lesson 2.1 Conditional Statements

Monitoring Progress

Use red to identify the hypothesis and blue to identify the conclusion. Then
rewrite the conditional statement in if-then form.

Question 1.
All 30° angles are acute angles.
Answer:

Question 2.
2x + 7 = 1. because x = – 3.
Answer:

In Exercises 3 and 4, write the negation of the statement.

Question 3.
The shirt is green.
Answer:

Question 4.
The Shoes are not red.
Answer:

Question 5.
Repeat Example 3. Let p be “the stars are visible” and let q be “it is night.”
Answer:

Use the diagram. Decide whether the statement is true. Explain your answer using the definitions you have learned.

Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 4

Question 6.
∠JMF and ∠FMG are supplementary.
Answer:

Question 7.
Point M is the midpoint of \(\overline{F H}\).
Answer:

Question 8.
∠JMF and ∠HMG arc vertical angles.
Answer:

Question 9.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 5
Answer:

Question 10.
Rewrite the definition of a right angle as a single biconditional statement.
Definition: If an angle is a right angle. then its measure is 90°.
Answer:

Question 11.
Rewrite the definition of congruent segments as a single biconditional statement.
Definition: If two line segments have the same length. then they are congruent segments.
Answer:

Question 12.
Rewrite the statements as a single biconditional statement.
If Mary is in theater class, then she will be in the fall play. If Mary is in the fall play. then she must be taking theater class.
Answer:

Question 13.
Rewrite the statements as a single biconditional statement.
If you can run for President. then you are at least 35 years old. If you are at least 35 years old. then you can run for President.
Answer:

Question 14.
Make a truth table for the conditional statement p → ~ q.
Answer:

Question 15.
Make a truth table for the conditional statement ~(p → q).
Answer:

Exercise 2.1 Conditional Statements

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What type of statements are either both true or both false?
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which statement does not belong with the other three? Explain your reasoning.
If today is Tuesday, then tomorrow is Wednesday
If it is Independence Day, then it is July.
If an angle is acute. then its measure is less than 90°.
If you are an athlete, then you play soccer.
Answer:

In Exercises 3 – 6. copy the conditional statement. Underline the hypothesis and circle the conclusion.

Question 3.
If a polygon is a pentagon, then it has five sides.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 3

Question 4.
If two lines form vertical angles, then they intersect.
Answer:

Question 5.
If you run, then you are fast.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 5

Question 6.
If you like math. then you like science.
Answer:

In Exercises 7 – 12. rewrite the conditional statement in if-then form.

Question 7.
9x + 5 = 23, because x = 2.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 7

Question 8.
Today is Friday, and tomorrow is the weekend.
Answer:

Question 9.
You are in a hand. and you play the drums.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 9

Question 10.
Two right angles are supplementary angles.
Answer:

Question 11.
Only people who are registered are allowed to vote.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 11

Question 12.
The measures complementary angles sum to 90°
Answer:

In Exercises 13 – 16. write the negation of the statement.

Question 13.
The sky is blue.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 13

Question 14.
The lake is cold.
Answer:

Question 15.
The ball is not pink.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 15

Question 16.
The dog is not a Lab.
Answer:

In Exercises 17 – 24. write the conditional statement p → q. the converse q → p, the inverse ~ p → ~ q, and the contrapositive ~ q → ~ p in words. Then decide whether each statement is true or false.

Question 17.
Let p be “two angles are supplementary” and let q be “the measures of the angles sum to 180°
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 17

Question 18.
Let p be “you are in math class” and let q be “you are in Geometry:”
Answer:

Question 19.
Let p be “you do your math homework” and let q be “you will do well on the test.”
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 19

Question 20.
Let p be “you are not an only child” and let q be “you have a sibling.
Answer:

Question 21.
Let p be “it does not snow” and let q be I will run outside.”
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 21

Question 22.
Let p be “the Sun is out” and let q be “it is day time”
Answer:

Question 23.
Let p be “3x – 7 = 20” and let q be “x = 9.”
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 23

Question 24.
Let p be “it is Valentine’s Day” and let q be “it is February.
Answer:

In Exercises 25 – 28, decide whether the statement about the diagram is true. Explain your answer using the definitions you have learned.

Question 25.
m∠ABC = 90°
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 6
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 25

Question 26.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 7
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 8
Answer:

Question 27.
m∠2 + m∠3 = 180°
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 9
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 27

Question 28.
M is the midpoint of \(\overline{A B}\).
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 10
Answer:

In Exercises 29 – 32. rewrite the definition of the term as a biconditional statement.

Question 29.
The midpoint of a segment is the point that divides the segment into two congruent segments.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 29

Question 30.
Two angles are vertical angles when their sides form two pairs of opposite rays.
Answer:

Question 31.
Adjacent angles are two angles that share a common vertex and side but have no common interior points.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 31

Question 32.
Two angles are supplementary angles when the sum of their measures 180°.
Answer:

In Exercises 33 – 36. rewrite the statements as a single biconditional statement.

Question 33.
If a polygon has three sides. then it is a triangle.
If a polygon is a triangle, then it has three sides.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 33

Question 34.
If a polygon has four sides, then it is a quadrilateral.
If a polygon is a quadrilateral, then it has four sides.
Answer:

Question 35.
If an angle is a right angle. then it measures 90°.
If an angle measures 90°. then it is a right angle.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 35

Question 36.
If an angle is obtuse, then ii has a measure between 90° and 180°.
If an angle has a measure between 90° and 180°. then it is obtuse.
Answer:

Question 37.
ERROR ANALYSIS
Describe and correct the error in rewriting the conditional statement in if – then form.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 11
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 37

Question 38.
ERROR ANALYSIS
Describe and correct the error in writing the converse of the conditional statement.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 12
Answer:

In Exercises 39 – 44. create a truth table for the logical statement.
Question 39.
~ p → q
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 39

Question 40.
~ q → p
Answer:

Question 41.
~(~ p → ~ q)
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 41

Question 42.
~ (p → ~ q)
Answer:

Question 43.
q → ~ p
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 43

Question 44.
~ (q → p)
Answer:

Question 45.
USING STRUCTURE
The statements below describe three ways that rocks are formed.

Igneous rock is formed from the cooling of Molten rock.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 13

Sedimentary rock is formed from pieces of other rocks.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 14

Metamorphic rock is formed by changing, temperature, pressure, or chemistry.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 15

a. Write each sLaternenl in if-then form.
b. Write the converse of each of the statements in part (a). Is the converse of each statement true? Explain your reasoning.
c. Write a true if-then statement about rocks that is different from the ones in parts (a) and (b). Is the converse of our statement true or false? Explain your reasoning
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 45.1
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 45.2

Question 46.
MAKING AN ARGUMENT
Your friend claims the statement “If I bought a shirt, then I went to the mall’ can he written as a true biconditional statement. Your sister says you cannot write it as a biconditional. Who is correct? Explain your reasoning.
Answer:

Question 47.
REASONING
You are told that the contrapositive of a statement is true. Will that help you determine whether the statement can be written as a true biconditional statement’? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 47

Question 48.
PROBLEM SOLVING
Use the conditional statement to identify the if-then statement as the converse. inverse. or contrapositive of the conditional statement. Then use the symbols to represent both statements.
Conditional statement: It I rode my bike to school, then I did not walk to school.
If-then statement: If did not ride my bike to school, then I walked to school.
p      q     ~    →    ↔
Answer:

USING STRUCTURE
In Exercises 49 – 52. rewrite the conditional statement in if-then form. Then underline the hypothesis and circle the conclusion.
Question 49.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 16
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 49

Question 50.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 17
Answer:

Question 51.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 18
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 51

Question 52.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 19
Answer:

Question 53.
MATHEMATICAL CONNECTIONS
Can the statement “If x2 – 10 = x + 2. then x = 4″ be combined with its converse to form a true biconditional statement?
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 53

Question 54.
CRITICAL THINKING
The largest natural arch in the United States is Landscape Arch. located in Thompson, Utah. h spans 290 feet.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 20
a. Use the information to write at least two true conditional statements.
Answer:

b. Which type of related conditional statement must also be true? Write the related conditional statements.
Answer:

C. What are the other two types of related conditional statements? Write the related conditional statements. Then determine their truth values. Explain your reasoning.
Answer:

Question 55.
REASONING
Which statement has the same meaning as the given statement?
Given statement:
You can watch a movie after you do your homework.
(A) If you do your homework, then you can watch a movie afterward.
(B) If you do not do your homework, then you can watch a movie afterward.
(C) If you cannot watch a movie afterward. then do your homework.
(D) If you can watch a movie afterward, then do not do your homework.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 55

Question 56.
THOUGHT PROVOKING
Write three conditional statements. where one is always true, one is always false, and one depends on the person interpreting the statement.
Answer:

Question 57.
CRITICAL THINKING
One example of a conditional statement involving dates is “If today is August 31, then tomorrow is September 1 Write a conditional statement using dates from two different months so that the truth value depends on when the statement is read.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 57

Question 58.
HOW DO YOU SEE IT?
The Venn diagram represents all the musicians at a high school. Write three conditional statements in if-then form describing the relationships between the various groups of musicians.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 21
Answer:

Question 59.
MULTIPLE REPRESENTATIONS
Create a Venn diagram representing each conditional statement. Write the converse of each conditional statement. Then determine whether each conditional statement and its converse are true or false. Explain your reasoning.
a. If you go to the zoo to see a lion, then you will see a Cat.
b. If you play a sport. then you wear a helmet.
c. If this month has 31 days. then it is not February.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 59.1
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 59.2

Question 60.
DRAWING CONCLUSIONS
You measure the heights of your classmates to gel a data set.
a. Tell whether this statement is true: If s and y are the least and greatest values in your data set, then the mean of the data is between x and y.
Answer:

b. Write the converse of the statement in part (a). Is the converse true? Explain your reasoning.
Answer:

c. Copy and complete the statement below using mean, median, or mode to make a conditional statement that is true for an data set. Explain your reasoning.
If a data set has a mean. median, and a mode. then the _____________ of the data set will always be a data value.
Answer:

Question 61.
WRITING
Write a conditional statement that is true, but its converse is false.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 61

Question 62.
CRITICAL THINKING
write a series of if-then statements that allow you to find the measure of each angle, given that m∠1 = 90° Use the definition of linear pairs.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 22
Answer:

Question 63.
WRITING
Advertising slogans such as “Buy these shoes! They will make you a better athlete!” often imply conditional statements. Find an advertisement or write your own slogan. Then write it as a conditional statement.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 63

Maintaining Mathematical Proficiency

Find the pattern. Then draw the next two figures in the sequence.

Question 64.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 23
Answer:

Question 65.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 24
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 65

Find the pattern. Then write the next two numbers.

Question 66.
1, 3, 5, 7 ……..
Answer:

Question 67.
12, 23, 34, 45 ……..
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 67

Question 68.
2, \(\frac{4}{3}, \frac{8}{9}, \frac{16}{27}\), ……..
Answer:

Question 69.
1, 4, 9, 16, ……..
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.1 Question 69

2.2 Inductive and Deductive Reasoning

Exploration 1

Writing a Conjecture

Work with a partner: Write a conjecture about the pattern. Then use your conjecture to draw the 10th object in the pattern.
a.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 25
Answer:

b.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 26
Answer:

c.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 27
Answer:

Exploration 2

Using a Venn Diagram

Work with a partner: Use the Venn diagram to determine whether the statement is true or false. Justify your answer. Assume that no region of the Venn diagram is empty.

CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to justify your conclusions and communicate them to others.

Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 28

a. If an item has Property B. then it has Property A.
Answer:

b. If an item has Property A. then it has Property B.
Answer:

c. If an item has Property A, then it has Property C.
Answer:

d. Some items that have Property A do not have Property B.
Answer:

e. If an item has Property C. then it does not have Property B.
Answer:

f. Sonic items have both Properties A and C.
Answer:

g. Some items have both Properties B and C.
Answer:

Exploration 3

Reasoning and Venn Diagrams

Work with a partner: Draw a Venn diagram that shows the relationship between different types of quadrilateral: squares. rectangles. parallelograms. trapezoids. rhombuses, and kites. Then write several conditional statements that are shown in your diagram. such as “If a quadrilateral is a square. then it is a rectangle.”
Answer:

Communicate Your Answer

Question 4.
How can you use reasoning to solve problems?
Answer:

Question 5.
Give an example of how you used reasoning to solve a real-life problem.
Answer:

Lesson 2.2 Inductive and Deductive Reasoning

Monitoring Progress

Question 1.
Sketch the fifth figure in the pattern in Example 1.
Answer:

Question 2.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 29
Answer:

Question 3.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 30
Answer:

Question 4.
Make and test a conjecture about the sign o1 the product of any three negative integers.
Answer:

Question 5.
Make and test a conjecture about the sum of any five consecutive integers.
Answer:

Find a counterexample to show that the conjecture is false.

Question 6.
The value of x2 is always greater than the value of x.
Answer:

Question 7.
The sum of two numbers is always greater than their difference.
Answer:

Question 8.
If 90° ∠ m ∠ R ∠ 180°, then ∠R is obtuse. The measure of ∠R is 155°. Using the Law of Detachment. what statement can you make?
Answer:

Question 9.
Use the Law of Syllogism to write a new conditional statement that follows from the pair of true statements.
If you get an A on your math test. then you can go to the movies.
If you go to the movies, then you can watch your favorite actor.
Answer:

Question 10.
Use inductive reasoning to make a conjecture about the sum of a number and itself. Then use deductive reasoning to show that the conjecture is true.
Answer:

Question 11.
Decide whether inductive reasoning or deductive reasoning is used to reach the
conclusion. Explain your reasoning.
All multiples of 8 are divisible by 4.
64 is a multiple of 8.
So, 64 is divisible by 4.
Answer:

Exercise 2.2 Inductive and Deductive Reasoning

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
How does the prefix “counter” help you understand the term counterexample?
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 1

Question 2.
WRITING
Explain the difference between inductive reasoning and deductive reasoning.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, describe the pattern. Then write or draw the next two numbers, letters, or figures.

Question 3.
1, – 2, 3, – 4, 5, ……..
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 3

Question 4.
0, 2, 6, 12, 20, ……..
Answer:

Question 5.
Z, Y, X, W, V, ……..
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 5

Question 6.
J, F, M, A, M, ……..
Answer:

Question 7.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 31
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 7

Question 8.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 32
Answer:

In Exercises 9 – 12. make and test a conjecture about the given quantity.

Question 9.
the product of any two even integers
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 9

Question 10.
the sum of an even integer and an odd integer
Answer:

Question 11.
the quotient of a number and its reciprocal
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 11

Question 12.
the quotient of two negative integers
Answer:

In Exercises 13 – 16, find a counter example to show that the conjecture is false.

Question 13.
The product of two positive numbers is always greater than either number,
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 13

Question 14.
If n is a nonzero integer, then \(\frac{n+1}{n}\) is always greater than 1.
Answer:

Question 15.
If two angles are supplements of each other. then one of the angles must be acute.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 15

Question 16.
A line s divides \(\overline{M N}\) into two line segments. So, the lines is a segment bisector of \(\overline{M N}\)
Answer:

In Exercises 17 – 20. use the Law of Detachment to determine what you can conclude from the given information, if possible.

Question 17.
If you pass the final, then you pass the class. You passed the final.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 17

Question 18.
If your parents let you borrow the ear, then you will go to the movies with your friend. you will go to the movies with your friend.
Answer:

Question 19.
If a quadrilateral is a square. then it has four right angles. Quadrilateral QRST has four right angles.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 19

Question 20.
If a point divides a line segment into two congruent line segments. then the point is a midpoint. Point P divides \(\overline{L H}\) into two congruent line segments.
Answer:

In Exercises 21 – 24, use the Law of Syllogism to write a new conditional statement that follows from the pair of true statements, if possible.

Question 21.
If x < – 2, then |x| > 2. If x > 2. then |x| > 2.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 21

Question 22.
If a = 3. then 5a = 15. If \(\frac{1}{2}\)a = 1\(\frac{1}{2}\), then a = 3.
Answer:

Question 23.
If a figure is a rhombus then the figure is a parallelogram. If a figure is a parallelogram, then the figure has two pairs of opposite sides that are parallel.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 23

Question 24.
If a figure is a square, then the figure has four congruent sides. If a figure is a square, then the figure has tour right angles.
Answer:

In Exercises 25 – 28. state the law of logic that is illustrated.

Question 25.
If you do your homework, then you can watch TV If you watch TV, then you can watch your favorite show.
If you do your homework. then you can watch your favorite show.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 25

Question 26.
If you miss practice the day before a game. then you will not be a starting player in the game.
You miss practice on Tuesday. You will not start the game Wednesday.
Answer:

Question 27.
If x > 12, then x + 9 > 20. The value of x is 14. So, x + 9 > 20.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 27

Question 28.
If ∠1 and ∠2 are vertical angles. then ∠1 ≅∠2.
If ∠1 ≅∠2 then m∠1 ≅ m∠2.
If ∠1 and ∠2 are vertical angles. then m∠1 = m∠2.
Answer:

In Exercises 29 and 30, use inductive reasoning to make a conjecture about the given quantity. Then use deductive reasoning to show that the conjecture is true.

Question 29.
the sum of two odd integers
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 29

Question 30.
the product of two odd integers
Answer:

In Exercises 31 – 34. decide whether inductive reasoning or deductive reasoning is used to reach the conclusion. Explain your reasoning.

Question 31.
Each time your mom goes to the store. she buy s milk. So. the next time your mom goes to the store. she will buy milk.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 31

Question 32.
Rational numbers can be written as fractions. Irrational numbers cannot be written as tractions. So. \(\frac{1}{2}\) is a rational number
Answer:

Question 33.
All men are mortal. Mozart is a man. so Mozart is mortal.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 33

Question 34.
Each time you clean your room. you are allowed to go out with your friends. So, the next time you clean your room. you will be allowed to go out with your friends.
Answer:

ERROR ANALYSIS
In Exercises 35 and 36, describe and correct the error in interpreting the statement.

Question 35.
If a figure is a rectangle. then the figure has four sides.
A trapezoid has four sides.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 33
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 35

Question 36.
Each day, you get to school before your friend.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 34
Answer:

Question 37.
REASONING
The table Shows the average weights of several subspecies of tigers. What conjecture can you make about the relation between the weights of female tigers and the weights of male tigers? Explain our reasoning.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 35
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 37

Question 38.
HOW DO YOU SEE IT?
Determine whether you can make each conjecture from the graph. Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 36
a. More girls will participate in high school lacrosse in Year 8 than those who participated in Year 7.
Answer:

b. The number of girls participating in high school lacrosse will exceed the number of boys participating in high school lacrosse in Year 9.
Answer:

Question 39.
MATHEMATICAL CONNECTIONS
Use inductive reasoning to write a formula for the sum of the first n positive even integers.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 39

Question 40.
FINDING A PATTERN
The following are the first nine Fibonacci numbers.
1, 1, 2, 3, 5, 8, 13, 21, 34, …….
a. Make a conjecture about each of the Fibonacci numbers after the first two.
Answer:

b. Write the next three numbers in the pattern.
Answer:

c. Research to find a real-world example of this pattern.
Answer:

Question 41.
MAKING AN ARGUMENT
Which argument is correct? Explain your reasoning.
Argument 1: If two angles measure 30° and 60° then the angles are complementary. ∠1 and ∠2 are complementary. So. m∠1 = 30° and m∠2 = 60°

Argument 2: If two angles measure 30° and 60°. then the angles are complementary. The measure of ∠1 is 30° and the measure of ∠2 is 60°. So, ∠1 and ∠2 are complementary.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 41

Question 42.
THOUGHT PROVOKING
The first two terms of a sequence are \(\frac{1}{4}\) and \(\frac{1}{2}\) Describe three different possible Patterns for the sequence. List the first five terms for each sequence.
Answer:

Question 43.
MATHEMATICAL CONNECTIONS
Use the table to make a conjecture about the relationship between x and y. Then write an equation for y in terms of x. Use the equation to test your conjecture for other values of x.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 37
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 43

Question 44.
REASONING
Use the pattern below. Each figure is made of squares that are 1 unit by 1 unit.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 38
a. Find the perimeter of each figure. Describe the pattern of the perimeters.
Answer:

b. Predict the perimeter of the 20th figure.
Answer:

Question 45.
DRAWING CONCLUSIONS
Decide whether each conclusion is valid. Explain your reasoning.

  • Yellowstone is a national park in Wyoming.
  • You and your Friend went camping at Yellowstone National Park.
  • When you go camping. you go canoeing.
  • If you go on a hike, your Friend goes with you.
  • You go on a hike.
  • There is a 3-mile-long trail near your campsite.

a. You went camping in Wyoming.
b. Your Friend went canoeing.
c. Your friend went on a hike.
d. You and your Friend went on a hike on a 3-mile-long trail.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 45

Question 46.
CRITICAL THINKING
Geologists use the Mohs’ scale to determine a mineral’s hardness. Using the scale. a mineral with a higher rating will leave a scratch on a mineral with a lower rating. Testing a mineral’s hardness can help identify the mineral.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 39
a. The four minerals are randomly labeled A, B, C, and D. Mineral A is scratched by Mineral B. Mineral C is scratched by all three of the other minerals. What can you conclude? Explain your reasoning.
Answer:

b. What additional test(s) can you use to identify all the minerals in part (a)?
Answer:

Maintaining Mathematical Proficiency

Determine which postulate is illustrated by the statement.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 40

Question 47.
AB + BC = AC
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 47

Question 48.
m∠DAC = m∠DAE + m∠EAB
Answer:

Question 49.
AD is the absolute value of the difference of the coordinates of A and D.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.2 Question 49

Question 50.
m∠DAC is equal to the absolute value of the difference between the real numbers matched with \(\vec{A}\)D and \(\vec{A}\)C on a protractor.
Answer:

2.3 Postulates and Diagrams

Exploration 1

Looking at a Diagram

Work with a partner. On a piece of paper. draw two perpendicular lines. Label them Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 41 and Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 42. Look at the diagram from different angles. Do the lines appear perpendicular regardless of the angle at which you look at them? Describe all the angles at which you can l00k at the lines and have them appear perpendicular.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 43
Answer:

Exploration 2

Interpreting a Diagram

Work with a partner: When you draw a diagram, you are communicating with others. It is important that you include sufficient information in the diagram. Use the diagram to determine which of the following statements you can assume to be true. Explain your reasoning.

ATTENDING TO PRECISION
To be proficient in math, you need to state the meanings of the symbols you choose.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 44
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 45
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 46
Answer:

Communicate Your Answer

Question 3.
In a diagram, what can be assumed and what needs to be labeled?
Answer:

Question 4.
Use the diagram in Exploration 2 to write two statements you can assume to be true and two statements you cannot assume to be true. Your statements should be different from those given in Exploration 2. Explain our reasoning.
Answer:

Lesson 2.3 Postulates and Diagrams

Monitoring progress

Question 1.
Use the diagram in Example 2. Which postulate allows you to say that the intersection of plane P and plane Q is a line?
Answer:

Question 2.
Use the diagram in Example 2 to write an example of the postulate.
a. Two Point Postulate
Answer:

b. Line-Point Postulate
Answer:

c. Line Intersection Postulate
Answer:

Refer back to Example 3.

Question 3.
If the given information states that \(\overline{P W}\) and \(\overline{Q W}\) arc congruent. how can you indicate that in the diagram?
Answer:

Question 4.
Name a pair of supplementary angles in the diagram. Explain.
Answer:

Use the diagram in Example 4.

Question 5.
Can you assume that plane S intersects plane T at Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 47?
Answer:

Question 6.
Explain how you know Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 48
Answer:

Exercise 2.3 Postulates and Diagrams

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Through any ___________ non collinear points. there exists exactly one plane.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 1

Question 2.
WRITING
Explain why you need at least three noncollinear points to determine a plane.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4. state the postulate illustrated by the diagram.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 49
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 50
Answer:

In Exercises 5 – 8, use the diagram to write an example of the postulate.

Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 51

Question 5.
Line-Point Postulate (Postulate 2.2)
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 5

Question 6.
Line Intersection Postulate (Postulate 2.3)
Answer:

Question 7.
Three Point Postulate (Postulate 2.4)
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 7

Question 8.
Plane-Line Postulate (Postulate 2.6)
Answer:

In Exercises 9 – 12. sketch a diagram of the description.

Question 9.
plane P and line m intersection plane P at a 90° angle
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 9

Question 10.
\(\overline{X Y}\) in plane P, \(\overline{X Y}\) bisected by point A. and point C not
on \(\overline{X Y}\)
Answer:

Question 11.
\(\overline{X Y}\) intersecting \(\overline{W V}\) at point A. so that XA = VA
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 11

Question 12.
\(\overline{A B}\), \(\overline{C D}\), and \(\overline{E F}\) are all in plane P. and point x is the midpoint of all three segments.
Answer:

In Exercises 13 – 20, use the diagram to determine whether you can assume the statement.

Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 52

Question 13.
Planes Wand X intersect at Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 53.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 13

Question 14.
Points K, L, M, and N are coplanar.
Answer:

Question 15.
Points Q, J, and M are collinear.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 15

Question 16.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 54
Answer:

Question 17.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 55 lies in plane X.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 17

Question 18.
∠PLK is a right angle.
Answer:

Question 19.
∠NKL and ∠JKM are vertical angles.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 19

Question 20.
∠NKI and ∠JKM are supplementary angles.
Answer:

ERROR ANALYSIS
In Exercises 21 and 22. describe and correct the error in the statement made about the diagram.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 56

Question 21.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 57
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 21

Question 22.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 58
Answer:

Question 23.
ATTENDING TO PRECISION
Select all the statements about the diagram that you cannot conclude.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 59
(A) A, B, and C are coplanar.
(B) Plane T intersects plane S in Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 60.
(C) Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 61 intersects Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 62.
(D) H, F, and D are coplanar.
(E) Plane T ⊥ plane S.
(F) Point B bisects \(\overline{H C}\).
(G) ∠ABH and ∠HBF are a linear pair.
(H) Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 63
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 23

Question 24.
HOW DO YOU SEE IT?
Use the diagram of line m and point C. Make a conjecture about how many planes can be drawn so that line m and point C lie in the same plane. Use postulates too justify your conjecture.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 64
Answer:

Question 25.
MATHEMATICAL CONNECTIONS
One way to graph a linear equation is to plot two points whose coordinates satisfy the equation and then connect them with a line. Which postulate guarantees this process works for any linear equation?
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 25

Question 26.
MATHEMATICAL CONNECTIONS
A way to solve a system of two linear equations that intersect is to graph the lines and find the coordinates of their intersection. Which postulate guarantees this process works for an two linear equations?
Answer:

In Exercises 27 and 28, (a) rewrite the postulate in if-then form. Then (b) write the converse, inverse, and contrapositive and state which ones are true.

Question 27.
Two Point Postulate (Postulate 2.1)
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 27

Question 28.
Plane-Point Postulate (Postulate 2.5)
Answer:

Question 29.
REASONING
Choose the correct symbol to go between the statements.
65
<         ≤         =        ≥          >
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 29

Question 30.
CRITICAL THINKING
If two lines intersect, then they intersect in exactly one point by the Line Intersection Postulate (Postulate 2.3). Do the two lines have to be in the same plane ? Draw a picture to support your answer. Then explain your reasoning.
Answer:

Question 31.
MAKING AN ARGUMENT
Your friend claims that even though two planes intersect in a line, it is possible for three planes to intersect in a point. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 31

Question 32.
MAKING AN ARGUMENT
Your friend claims that by the Plane Intersection Postulate (Post. 2.7), any two planes intersect in a line. Is your friend’s interpretation 0f the Plane Intersection Postulate (Post. 2.7) correct? Explain your reasoning.
Answer:

Question 33.
ABSTRACT REASONING
Points E, F, and G all lie in plane P and in plane Q. What must be true about points E, F. and G so that planes P and Q are different planes? What must be true about points E, F, and G to force planes P and Q to be the same plane? Make sketches to support your answers.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 33

Question 34.
THOUGHT PROVOKING
The postulates in this book represent Euclidean geometry. In spherical geometry. all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. A plane is the surface of the sphere. Find a postulate on page 84 that is not true in spherical geometry. Explain your reasoning.
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Tell which algebraic property of equality you used.

Question 35.
t – 6 = – 4
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 35

Question 36.
3x = 21
Answer:

Question 37.
9 + x = 13
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.3 Question 37

Question 38.
\(\frac{x}{7}\) = 5
Answer:

2.1 – 2.3 Study Skills: Using the Features of Your Textbook to Prepare for Quizzes and Tests

Mathematical Practices

Question 1.
Provide a counter example for each false conditional statement in Exercises 17 – 24 on page 71.
(You do not need to consider the converse. inverse, and contrapositive statements.)

Question 2.
Create a truth table for each of your answers to Exercise 59 on page 74.
Answer:

Question 3.
For Exercise 32 on page 88. write a question you would ask your friend about his or her interpretation.
Answer:

2.1 – 2.3 Quiz

Rewrite the conditional statement in if-then form. Then write the converse, inverse, and contrapositive of the conditional statement. Decide whether each statement is true or false.

Question 1.
An angle measure of 167° is an obtuse angle.
Answer:

Question 2.
You are in a physics class, so you always have homework.
Answer:

Question 3.
I will take my driving test, So I will get my driver’s license.
Answer:

Find a countereample to show that the conjecture is false.

Question 4.
The sum of a positive number and a negative number is always positive.
Answer:

Question 5.
If a figure has four sides, then it is a rectangle.
Answer:

Use inductive reasoning to make a conjecture about the given quantity. Then use deductive reasoning to show that the conjecture is true.

Question 6.
the sum of two negative integers
Answer:

Question 7.
the difference of two even integers
Answer:

Use the diagram to determine whether you can assume the statement.

Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 66

Question 8.
Points D, B, and C are coplanar.
Answer:

Question 9.
Plane EAF is parallel to plane DBC.
Answer:

Question 10.
Line m intersects line Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 67 at point it.
Answer:

Question 11.
Line Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 68 lies in plane DBC.
Answer:

Question 12.
m∠DBG = 90°
Answer:

Question 13.
You and your friend are bowling. Your friend claims that the statement “If I got a strike, then I used thegreen ball” can be written as a true biconditional statement. Is your friend correct? Explain your reasoning. (Section 2.1)
Answer:

Question 14.
The table shows the 1 – mile running times of the members of a high school track team.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 69
a. What conjecture can you make about the running times of females and males?
Answer:

b. What type of reasoning did you use? Explain.
Answer:

Question 15.
List five of the seven Point, Line, and Plane Postulates on page 84 that the diagram of the house demonstrates. Explain how the postulate is demonstrated in the diagram.
Big Ideas Math Geometry Answer Key Chapter 2 Reasoning and Proofs 70
Answer:

2.4 Algebraic Reasoning

Exploration 1

Justifying steps in a solution

Work with a partner: In previous courses. you studied diíIrcnt properties. such as the properties of equality and the Distributive, Commutative, and Associative Properties. Write the property that justifies each of the following solution steps.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 69
Answer:

Exploration 2

Stating Algebraic Properties

Work with a partner: The symbols Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 70 and Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 71 represent addition and multiplication (not necessarily in that order). Determine which symbol represents which operation. Justify your answer. Then state each algebraic property being illustrated.

LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 72
Answer:

Communicate Your Answer

Question 3.
How can algebraic properties help you solve an equation?
Answer:

Question 4.
Solve 3(x + 1) – 1 = – 13. Justify each step.
Answer:

Lesson 2.4 Algebraic Reasoning

Monitoring Progress

Solve the equation. Justify each step.

Question 1.
6x – 11 = – 35
Answer:

Question 2.
– 2p – 9 = 10p – 17
Answer:

Question 3.
39 – 5z = -1 + 5z
Answer:

Question 4.
3(3x + 14) = – 3
Answer:

Question 5.
4 = – 10b + 6(2 – b)
Answer:

Question 6.
Solve the formula A = \(\frac{1}{2}\)bh for b. Justify each step. Then find the base of a
triangle whose area is 952 square feet and whose height is 56 feet.
Answer:

Name the property of equality that the statement illustrates.

Question 7.
If m∠6 = m∠7, then m∠7 = m∠6.
Answer:

Question 8.
34° = 34°
Answer:

Question 9.
m∠1 = m∠2 and m∠2 = m∠5. So, m∠1 = m∠5.
Answer:

Question 10.
If JK = KL and KL = 16, then JK = 16.
Answer:

Question 11.
PQ = ST, so ST = PQ.
Answer:

Question 12.
ZY = ZY
Answer:

Question 13.
In Example 5. a hot dog stand is located halfway between the shoe store and the pizza shop. at point H. Show that PH = HM.
Answer:

Exercise 2.4 Algebraic Reasoning

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The statement “The measure of an angle is equal to itself” is true because of what property?
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find both answers.
What property justifies the following statement?
If c = d, then d = c.
If JK = LW. then LM = JK.
If e = f and f = g, then e = g.
If m∠R = m∠S, then m∠S = m∠R.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, write the property that justifies each step.

Question 3.
3x – 12 = 7x + 8             Given
– 4x – 12 = 8                   ___________
– 4x = 20                        ___________
x = – 5                            ___________
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 3

Question 4.
5(x – 1) = 4x + 13                Given
5x – 5 = 4x + 13                  ___________
x – 5 = 13                            ___________
x = 18                                 ___________
Answer:

In Exercises 5 – 14. solve the equation. Justify each step.

Question 5.
5x – 10 = – 40
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 5

Question 6.
6x + 17 = – 7
Answer:

Question 7.
2x – 8 = 6x – 20
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 7

Question 8.
4x + 9 = 16 – 3x
Answer:

Question 9.
5(3x – 20) = – 10
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 9

Question 10.
3(2x + 11) = 9
Answer:

Question 11.
2(- x – 5) = 12
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 11

Question 12.
44 – 2(3x + 4) = – 18x
Answer:

Question 13.
4(5x – 9) = – 2(x + 7)
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 13

Question 14.
3(4x + 7) = 5(3x + 3)
Answer:

In Exercises 15 – 20, solve the equation for y. Justify each step.

Question 15.
5x + y = 18
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 15

Question 16.
– 4x + 2y = 8
Answer:

Question 17.
2y – 0.5x = 16
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 17

Question 18.
\(\frac{1}{2} x-\frac{3}{4} y\) = – 2
Answer:

Question 19.
12 – 3y = 30x + 6
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 19

Question 20.
3x + 7 = – 7 + 9y
Answer:

In Exercises 21 – 24. solve the equation for the given variable. Justify each step

Question 21.
C = 2πr; r
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 21

Question 22.
I = Prt;P
Answer:

Question 23.
S = 180(n – 2); n
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 23

Question 24.
5 = 2πr2 + 2πrh; h
Answer:

In Exercises 25 – 32, name the property of equality that the statement illustrates.

Question 25.
If x = y, then 3x = 3y.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 25

Question 26.
If AM = MB. then AM + 5 = MB + 5.
Answer:

Question 27.
x = x
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 27

Question 28.
If x = y, then y = x.
Answer:

Question 29.
m∠Z = m∠Z
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 29

Question 30.
If m∠Z = 29° and m∠B = 29°, then m∠A = m∠B
Answer:

Question 31.
If AB = LM, then LM = AB.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 31

Question 32.
If BC = XY and XY = 8, then BC = 8.
Answer:

In Exercises 33 – 40. use the property to copy and complete the statement.

Question 33.
Substitution Property of Equality:
If AB = 20. then AB + CD = ________ .
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 33

Question 34.
Symmetric Property oÌ Equality:
If m∠1 = m∠2. then ________ .
Answer:

Question 35.
Addition Property of Equality:
If AB = CD. then AB + EF = ________ .
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 35

Question 36.
Multiplication Property of Equality:
If AB = CD, then 5 • AB = ________ .
Answer:

Question 37.
Subtraction Property of Equality:
If LM = XY, then LM – GH = ________ .
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 37

Question 38.
Distributive Property:
If 5(x + 8) = 2, then ___ + ___ = 2.
Answer:

Question 39.
Transitive Property of Equality:
If m∠1 = m∠2 and m∠2 = m∠3, then ________ .
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 39

Question 40.
Reflexive Properly of Equality:
m∠ABC = ________ .
Answer:

ERROR ANALYSIS
In Exercises 41 and 42, describe and correct the error in solving the equation.

Question 41.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 73
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 41

Question 42.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 74
Answer:

Question 43.
REWRITING A FORMULA
The formula for the perimeter P of a rectangle is P = 2l + 2w, where l is the length and w is the width. Solve the formula for l. Justify each step. Then find the length of a rectangular lawn with a perimeter of 32 meters and a width of 5 meters.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 43

Question 44.
REWRITING A FORMULA
The formula for the area
A of a trapezoid is A = \(\frac{1}{2}\)h (b1 + b2), where h is the
height and b1 and b2 are the lengths of the two bases. Solve the formula for b1. Justify each step. Then find the length of one of the bases of the trapezoid when the area of the trapezoid is 91 square meters. the height is 7 meters. and the length of the other base is 20 meters.
Answer:

Question 45.
ANALYZING RELATIONSHIPS
In the diagram,
m∠ABD = m∠CBE. Show that m∠1 = m∠3.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 75
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 45

Question 46.
ANALYZING RELATIONSHIPS
In the diagram,
AC = BD. Show that AB = CD.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 76
Answer:

Question 47.
ANALYZING RELATIONSHIPS
Copy and complete the table to show that m∠2 = m∠3.

EquationReason
m∠1 = m∠4, m∠EFH = 90°, m∠GHF = 90°Given
m∠EHF = m∠GHF
m∠EHF = m∠1 + m∠2
m∠GHF = m∠3 + m∠4
m∠1 + m∠2 = m∠3 + m∠4
Substitution Property of Equality
m∠2 = m∠3

Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 47

Question 48.
WRITING
Compare the Reflexive Property of Equality with the Symmetric Property of Equality. How are the properties similar? How are they different?
Answer:

REASONING
In Exercises 49 and 50. show that the perimeter of ∆ABC is equal to the perimeter of ∆ADC.

Question 49.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 78
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 49

Question 50.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 79
Answer:

Question 51.
MATHEMATICAL CONNECTIONS
In the figure, \(\overline{Z Y}\) ≅ \(\overline{X W}\), ZX = 5x + 17, YW = 10 – 2x, and YX = 3. Find ZY and XW
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 80
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 51

Question 52.
HOW DO YOU SEE IT?
The bar graph shows the number of hours each employee works at a grocery store. Give an example of the Reflexive, Symmetric, and Transitive Properties of Equality.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 81
Answer:

Question 53.
ATTENDING TO PRECISION
Which of the following statements illustrate the Symmetric Property of Equality? Select all that apply.
(A) If AC = RS, then RS = AC.
(B) If x = 9 then 9 = x.
(C) If AD = BC, then DA = CB.
(D) AB = BA
(E) If AB = LW and LM = RT, then AB = RT.
(F) If XY = EF, then FE = XY.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 53

Question 54.
THOUGHT PROVOKING
Write examples from your everyday life lo help you remember the Reflexive, Symmetric, and Transitive Properties of Equality. Justify your answers.
Answer:

Question 55.
MULTIPLE REPRESENTATIONS
The formula to convert
a temperature in degrees Fahrenheit (°F) to degrees
Celsius (°C) is C = \(\frac{5}{9}\)(F – 32).
a. Solve the formula for F. Justify each step.
b. Make a table that shows the conversion to Fahrenheit for each temperature: 0°C. 20°C, 32°C. and 41°C.
c. Use your table to graph the temperature in degrees Fahrenheit as a function of the temperature in degrees Celsius. Is this a linear function?

Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 55.1
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 55.2
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 55.3

Question 56.
REASONING
Select all the properties that would also apply to inequalities. Explain your reasoning.
(A) Addition Property
Answer:

(B) Subtraction Properly
Answer:

(C) Substitution Property
Answer:

(D) Reflexive Property
Answer:

(E) Symmetric Property
Answer:

(F) Transitive Property
Answer:

Maintaining Mathematical Proficiency

Name the definition property, or postulate that is represented by each diagram.

Question 57.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 82
XY + YZ = XZ
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 57

Question 58.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 83
Answer:

Question 59.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 84
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.4 Question 59

Question 60.
Big Ideas Math Geometry Solutions Chapter 2 Reasoning and Proofs 85
m∠ABD + m∠DBC = m∠ABC
Answer:

2.5 Proving Statements about Segments and Angles

Exploration 1

Writing Reasons in a proof

Work with a partner: Four steps of a proof are shown. Write the reasons for each statement
Given AC = AB + AB
Prove AB = BC
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 86

REASONING ABSTRACTLY
To be proficient in math, you need to know and be able to use algebraic properties.

StatementsReasons
1. AC = AB + AB1. Given
2. AB = BC  = AC2. ______________________
3. AB + AB = AB + BC3. ______________________
4. AB = BC4. ______________________

Answer:

Exploration 2

Writing Steps in a Proof

Work with a partner: Six steps of a proof are shown. Complete the statements that correspond to each reason.
Given m∠1 = m∠3
Prove m∠EBA = m∠CBD
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 87

StatementsReasons
1. ___________________________1. Given
2. m∠EBA = m∠2 + m∠32. Angle Addition Postulate (Post.1.4)
3. m∠EBA = m∠2 + m∠13. Substitution Property of Equality
4. m∠EBA = ___________________________4. Commutative Property of Addition
5. m∠1 + m∠2 = ______________________5. Angle Addition Postulate (Post. 1.4)
6. ________________________________________6. Transitive Property of Equality

Answer:

Communicate Your Answer

Question 3.
How can you prove a mathematical statement?
Answer:

Question 4.
Use the given information and the figure to write a proof for the statement.
Given B is the midpoint of \(\overline{A C}\).
C is the midpoint of \(\overline{B D}\).
Prove AB = CD
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 88
Answer:

Lesson 2.5 Proving Statements about Segments and Angles

Monitoring Progress

Question 1.
Six Steps of a two-column proof are shown. Copy and complete the proof.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 89
Given T is the midpoint of \(\overline{S U}\).
Prove x = 5

StatementReason
1. T is the midpoint of \(\overline{S U}\).1. ________________________________________
2. \( \overline{S T} \cong \overline{T U} \)2. Definition of midpoint
3. ST = TU3. Definition of congruent segments
4. 7x = 3x + 204. ________________________________________
5. ________________________________________5. Subtraction Property of Equality
6. x = 56. ________________________________________

Answer:

Exercise 2.5 Proving Statements about Segments and Angles

Vocabulary and Core Concept Check

Question 1.
WRITING
How is a theorem different from a postulate?
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 1

Question 2.
COMPLETE THE SENTENCE
In a two-column proof, each __________ is on the left and each __________ is on the right.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4. copy and complete the proof.

Question 3.
Given PQ = RS
Prove PR = QS
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 90

StatementsReasons
1. PQ = RS1. ________________________________________
2. PQ + QR = RS + QR2. ________________________________________
3. ________________________________________3. Segment Addition Postulate (Post. 1.2)
4. RS + QR = QS4. Segment Addition Postulate (Post. 1.2)
5. PR = QS5. ________________________________________

Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 3

Question 4.
Given ∠1 is a complement of ∠2.
∠2 ≅ ∠3
Prove ∠1 is a complement of ∠3.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 91

StatementsReasons
1. ∠1 is a complement of ∠2.1. Given
2. ∠2 ≅ ∠32. _____________________________
3. m∠1 + m∠2 = 90°3. _____________________________
4. m∠2 = m∠34.  Definition of congruent angles
5. _____________________________5. Substitution Property Of Equality
6. ∠1 is a complement of ∠3.6. _____________________________

Answer:

In Exercises 5-10, name the property that the statement illustrates.

Question 5.
If \(\overline{P Q} \cong \overline{S T}\) and \(\overline{S T} \cong \overline{U V}\), then \(\overline{P Q} \cong \overline{U V}\).\
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 5

Question 6.
∠F ≅ ∠F
Answer:

Question 7.
If ∠G ≅∠H. then ∠H ≅ ∠G.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 7

Question 8.
\(\overline{D E} \cong \overline{D E}\)
Answer:

Question 9.
If \(\overline{X Y} \cong \overline{U V}\), then \(\overline{U V} \cong \overline{X Y}\).
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 9

Question 10.
If ∠L ≅∠M and ∠M ≅∠N, then ∠L ≅∠N.
Answer:

PROOF
In Exercises 11 and 12, write a two-column proof for the property.

Question 11.
Reflexive Property of Segment Congruence (Thm. 2.1)
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 11

Question 12.
Transitive Property of Angle Congruence (Thm. 2.2)
Answer:

PROOF
Exercises 13 and 14. write a two-column proof.

Question 13.
Given ∠GFH ≅ ∠GHF
Prove ∠EFG and ∠GHF are supplementary
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 92
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 13

Question 14.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 93
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 94
Answer:

Question 15.
ERROR ANALYSIS
In the diagram \(\overline{M N} \cong \overline{L Q}\) and \(\overrightarrow{L Q} \cong \overrightarrow{P N}\). Describe and correct the error in the reasoning.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 95
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 15

Question 16.
MODELING WITH MATHEMATICS
The distance fr the restaurant to the shoe store is the same as the distance from the cafe to the florist. The distance from the shoe store to the movie theater is the same as the distance from the movie theater to the cafe, and from the florist to the dry cleaners.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 96
Use the steps below to prove that the distance from the restaurant to the movie theater is the same as the distance from the cafe to the dry cleaners.
a. State what is given and what is to be proven for the situation.
Answer:

b. Write a two-column proof.
Answer:

Question 17.
REASONING
In the sculpture shown, \(\angle 1 \cong \angle 2\) and \(\angle 2 \cong \angle 3\) classify the triangle and justify your answer.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 97
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 17

Question 18.
MAKING AN ARGUMENT
In the figure, \(\overline{S R} \cong \overline{C B}\) and \(\overline{A C} \cong \overline{Q R}\) Your friend claims that, because of this. \( \overline{C B} \cong \overline{A C}\) by the Transitive Property of Segment Congruence (Thin. 2. 1). Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 98
Answer:

Question 19.
WRITING
Explain why you do not use inductive reasoning when writing a proof.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 19

Question 20.
HOW DO YOU SEE IT?
Use the figure to write Given and Prove statements for each conclusion.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 99
a. The acute angles of a right triangle are complementary.
b. A segment connecting the midpoints of two sides of a triangle is half as long as the third side.
Answer:

Question 21.
REASONING
Fold two corners of a piece of paper So their edges match. as shown.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 100
a. What do you notice about the angle formed at the top of the page by the folds?
b. Write a two-column proof to show that the angle measure is always the same no matter how you make the folds.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 21

Question 22.
THOUGHT PROVOKING
The distance from Springfield
to Lakewood City is equal to the distance from Springfield Lo BettsilIe. Janisburg is 50 miles farther from Springfield titan Bettsville. Moon Valley is 50 miles Farther from Springfield than Lakewood City is. Use line segments to draw a diagram that represents this situation.
Answer:

Question 23.
MATHEMATICAL CONNECTIONS
Solve for x using the given information. Justify each step.
Given \(\overline{Q R} \cong \overline{P Q}, \overline{R S} \cong \overline{P Q}\)
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 101
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 23

Maintaining Mathematical Proficiency

Use the figure

Question 24.
∠ 1 is a complement of ∠4. and m∠1 33°. Find in m∠4.
Answer:

Question 25.
∠3 is a supplement of ∠2, and m∠2 = 147°. Find m∠3.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.5 Question 25

Question 26.
Name a pair of vertical angles.
Big Ideas Math Answer Key Geometry Chapter 2 Reasoning and Proofs 102
Answer:

2.6 Proving Geometric Relationships

Exploration 1

Matching Reasons in a Flowchart Proof

work with a partner: Match each reason with the correct step in the flowchart.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 103
Given AC = AB + AB
Prove AB = BC
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 104

MODELING WITH MATHEMATICS
To be proficient in math, you need to map relationships using such tools as diagrams, two-way tables, graphs, flowcharts, and formulas.

A. Segment Addition Postulate (Post. 1.2)
Answer:

B. Given
Answer:

C. Transitive Property of Equality
Answer:

D. Subtraction Property of Equality
Answer:

Exploration 2

Matching Reasons in a Flowchart Proof

Work with a partner. Match each reason with the Correct step in the flowchart.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 105
Given m∠1 = m∠3
Prove m∠EBA = m∠CBD
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 106
A. Angle Addition Postulate (Post. 1.4)
Answer:

B. Transitive Property of Equality
Answer:

C. Substitution Property of Equality
Answer:

D. Angle Addition Postulate (Post. 1.4)
Answer:

E. Given
Answer:

F. Commutative Property of Addition
Answer:

Communicate Your Answer

Question 3.
How can you use a flowchart to prove a mathematical statement?
Answer:

Question 4.
Compare the flowchart proofs above with the two-column proofs in the Section 2.5 Explorations. Explain the advantages and disadvantages of each.
Answer:

Lessson 2.6 Proving Geometric Relationships

Monitoring Progress

Question 1.
Copy and complete the flowchart proof. Then write a two-column proof.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 107
Given \(\overline{A B}\) ⊥ \(\overline{B C}\), \(\overline{D C}\) ⊥ \(\overline{B C}\)
Prove ∠B ≅∠C
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 108
Answer:

Question 2.
Copy and complete the two-column proof. Then write a flowchart proof.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 109
Given AB = DE, BC = CD
Prove \(\overline{A C} \cong \overline{C E}\)

StatementsReasons
1. AB = DE, BC = CD1. Given
2. AB + BC = BC + DE2. Addition Property of Equality
3. _____________________________3. Substitution Property of Equality
4. AB + BC = AC, CD + DE = CE4. _____________________________
5. _____________________________5. Substitution Property of Equality
6. \( \overline{A C} \cong \overline{C E} \)6. _____________________________

Answer:

Question 3.
Rewrite the two-column proof in Example 3 without using the Congruent Supplements Theorem. How many steps do you save by using the theorem?
Answer:

Use the diagram and the given angle measure to find the other three angle measures.

Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 110

Question 4.
m∠1 = 117°
Answer:

Question 5.
m∠2 = 59°
Answer:

Question 6.
m∠4 = 88°
Answer:

Question 7.
Find the value of w.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 111
Answer:

Question 8.
write a paragraph proof.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 112
Given ∠1 is a right angle.
Prove ∠2 is a right angle.
Answer:

Exercise 2.6 Proving Geometric Relationships

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain why all right angles are congruent.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 1

Question 2.
VOCABULARY
What are the two types of angles that are formed by intersecting lines?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6. identify the pairs) of congruent angles in the figures. Explain how you know they are congruent.

Question 3.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 113
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 3

Question 4.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 114
Answer:

Question 5.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 115
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 5

Question 6.
∠ABC is supplementary to ∠CBD
∠CBD is supplementary to ∠DEF
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 116
Answer:

In Exercises 7 – 10. use the diagram and the given angle measure to find the other three measures.

Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 117

Question 7.
m∠1 = 143°
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 7

Question 8.
m∠3 = 159°
Answer:

Question 9.
m∠2 = 34°
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 9

Question 10.
m∠4 = 29°
Answer:

In Exercises 11 – 14, find the values of x and y.

Question 11.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 118
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 11

Question 12.

Answer:

Question 13.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 120
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 13

Question 14.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 121
Answer:

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in using the diagram to find the value of x.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 122

Question 15.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 123
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 15

Question 16.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 124
Answer:

Question 17.
PROOF
Copy and complete the flowchart proof. Then write a two-column proof.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 125
Given ∠1 ≅ ∠3
Prove ∠2 ≅ ∠4
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 126
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 17.1
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 17.2

Question 18.
PROOF
Copy and complete the two-column proof. Then write a flowchart proof.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 127
Given ∠ABD is a right angle
∠CBE is a right angle
Prove ∠ABC ≅ ∠DBE

StatementsReasons
1. ∠ABD is a right angle.

∠CBE is a right angle.

1. _____________________________
2. ∠ABC and ∠CBD are complementary.2. Definition of complementary
3. ∠DBE and ∠CBD are complementary3. _____________________________
4. ∠ABE ≅ ∠DBE4. _____________________________

Question 19.
PROVING A THEOREM
Copy and complete the paragraph proof be the Congruent Complements Theorem (Theorem 2.5). Then write a two-column proof.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 128
Given ∠1 and ∠2 are complementary
∠1 and ∠3 are complementary
Prove ∠2 ≅ ∠3
∠1 and ∠2 are complementary, and ∠1 and ∠3 are complementary. By the definition
of ____________ angles. m∠1 + m∠2 = 90° and ____________ = 90°. By the ____________ m∠1 + m∠2 = m∠1 + m∠3. By the Subtraction ____________
Property of Equality, ____________ . So. ∠2 ≅∠3 by the definition of ____________ .
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 19.1
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 19.2

Question 20.
PROVING A THEOREM
Copy and complete the two – column proof for the Congruent Supplement Theorem (Theorem 2.4). Then write a paragraph proof. (See Example 5.)
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 129
Given ∠1 and ∠2 are supplementary
∠3 and ∠4 are supplementary
∠1 and ∠4
Prove ∠2 ≅∠3

StatementsReasons
1. ∠1 and ∠2 are supplementary
∠3 and ∠4 are supplementary
∠1 ≅ ∠4
1. Given
2. m∠1 + m∠2 = 180
m∠3 + m∠4 = 180
2. _____________________________
3. ______________ = m∠3 + m∠43. Transitive Property of Equality
4. m∠1 = m∠44. Definition of Congruent angles
5. m∠1 + m∠2 = ___________________5. Substitution property of Equality
6. m∠2 = m∠36. _____________________________
7. __________________________7. _____________________________

Answer:

PROOF
In Exercises 21 – 24. write a proof using any format.

Question 21.
Given ∠QRS and ∠PSR are supplementary
Prove ∠QRL ≅ ∠PSR
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 130
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 21

Question 22.
Given ∠1 and ∠3 are complementary.
∠2 and ∠4 are complementary.
Prove ∠1 ≅ ∠4
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 131
Answer:

Question 23.
Given ∠AEB ≅ ∠DEC
Prove ∠AEC ≅ ∠DEB

Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 23.1
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 23.2

Question 24.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 133

Answer:

Question 25.
MAKING AN ARGUMENT
You overhear your friend discussing the diagram shown with a classmate. Your classmate claims ∠1 ≅∠4 because they are vertical angles Your friend claims they are not congruent because he can tell by looking at the diagram. Who is correct? Support your answer with definitions or theorems.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 135
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 25

Question 26.
THOUGHT PROVOKING
Draw three lines all intersecting at the same point. Explain how you can give two of the angle measures so that you can find the remaining four angle measures.
Answer:

Question 27.
CRITICAL THINKING
Is the converse of the Linear Pair Postulate (Postulate 2.8) true? If so, write a biconditional statement. Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 27

Question 28.
WRITING
How can you save time writing proofs?
Answer:

Question 29.
MATHEMATICAL CONNECTIONS
Find the measure of each angle in the diagram.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 136
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 29

Question 30.
HOW DO YOU SEE IT?
Use the student’s two-column proof.
Given ∠1 ≅ ∠2
∠1 and ∠2 are supplementary.
Prove ___________

StatementsReasons
1. ∠1 ≅ ∠2
∠1 and ∠2 are supplementary
1. Given
2. m∠1 = m∠22. Definitions of congruent angles
3. m∠1 + m∠2 = 180°3. Definition of supplementary angles
4. m∠1 + m∠1 = 180°4. substitution property of Equality
5. 2m∠1 = 180°5. Simplify
6. m∠1 = 90°6. Division Property of Equality
7. m∠2 = 90°7. Transitive Property of Equality
8. __________________________8. ________________________________________

a What is the student trying to prove?
Answer:

b. Your friend claims that the last line of the proof should be ∠1 ≅ ∠2. because the measures of the angles are both 90°. Is your friend correct? Explain.
Answer:

Maintaining Mathematical Proficiency

Use the cube

Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 137

Question 31.
Name three collinear points.
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 31

Question 32.
Name the intersection of plane ABF and plane EHG.
Answer:

Question 33.
Name two planes containing \(\overline{B C}\).
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 33

Question 34.
Name three planes containing point D.
Answer:

Question 35.
Name three points that are not collinear,
Answer:
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 2.6 Question 35

Question 36.
Name two planes containing point J.
Answer:

2.4 – 2.6 Performance Task: Induction and the Next Dimension

Mathematical Practices

Question 1.
Explain the purpose of justifying each step in Exercises 5-14 on page 96.
Answer:

Question 2.
Create a diagram to model each statement in Exercises 5-10 on page 103.
Answer:

Question 3.
Explain why you would not be able to prove the statement in Exercise 21 on page 113 if you were provided with the given information or able to use an postulates or theorems.
Answer:

Reasoning and Proofs Chapter Review

2.1 Conditional Statements

Write the if-then form, the converse, the inverse, the contrapositive. and the biconditional of the conditional statement.

Question 1.
Two lines intersect in a Point.

Question 2.
4x + 9 = 21 because x = 3.
Answer:

Question 3.
Supplementary angles sum to 180°.
Answer:

Question 4.
Right angles are 90°.
Answer:

2.2 Inductive and Deductive Reasoning

Question 5
conclusion can you make about the difference of any two odd integers?
Answer:

Question 6.
What conclusion can you make about the product of an even and an odd integer?
Answer:

Question 7.
Use the Law of Detachment to make a valid conclusion.
If an angle is a right angle, then the angle measures 90°. ∠B is a right angle.
Answer:

Question 8.
Use the Law of Syllogism to write a new conditional statement that follows from the pair of true statements: If x = 3, then 2x = 6. If 4x = 12. then x = 3.
Answer:

2.3 Postulates and Diagrams

Use the diagram at the right to determine whether you can assume the statement.

Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 138

Question 9.
Points A, B, C, and E are coplanar.
Answer:

Question 10.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 139
Answer:

Question 11.
Points F, B, and G are collinear.
Answer:

Question 12.
Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 140
Answer:

sketch a diagram of the description.

Question 13.
∠ABC, an acute angle, is bisected by Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 141.
Answer:

Question 14.
∠CDE, a straight angle, is bisected by Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 142.
Answer:

Question 15.
Plane P and plane R intersect perpendicularly in Big Ideas Math Geometry Answers Chapter 2 Reasoning and Proofs 143. \(\overline{Z W}\) lies in plane P
Answer:

2.4 Algebraic Reasoning

Solve the equation. Justify each step.

Question 16.
– 9x – 21 = – 20x – 87
Answer:

Question 17.
15x + 22 = 7x + 62
Answer:

Question 18.
3(2x + 9) = 30
Answer:

Question 19.
5x + 2(2x – 23) = – 154
Answer:

Name the property of equality that the statement illustrates.

Question 20.
If LM = RS and RS = 25, then LM = 25.
Answer:

Question 21.
AM = AM
Answer:

2.5 Proving Statements about Segments and Angles

Name the property that the statement illustrates.

Question 22.
If ∠DEF ≅∠JKL, then ∠JKL ≅ ∠DEF
Answer:

Question 23.
∠C ≅ ∠C
Answer:

Question 24.
If MN = PQ and PQ = RS. then MN = RS.
Answer:

Question 25.
Write a two-column proof be the Reflexive Property of Angle Congruence (Thm. 2.2).
Answer:

2.6 Proving Geometric Relationships

Question 26.
Write a proof using any format
Given ∠3 and ∠2 are complementary.
m∠1 + m∠2 = 90°
Prove ∠3 ≅∠1
Answer:

Reasoning and Proofs Test

Use the diagram to determine whether you can assume the statement.
Explain your reasoning.

Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 144

Question 1.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 145 ⊥ plane M
Answer:

Question 2.
Points F, G, and A are coplanar.
Answer:

Question 3.
Points E, C, and G are collinear.
Answer:

Question 4.
Planes M and P intersect at Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 146.
Answer:

Question 5.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 147 lies in plane P.
Answer:

Question 6.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 148
Answer:

Solve the equation. Justify each step.

Question 7.
9x + 31 = – 23 + 3x
Answer:

Question 8.
26 + 2(3x + 11) = – 18
Answer:

Question 9.
3(7x – 9) – 19x – 15
Answer:

Write the if-then form, the converse, the inverse, the contrapositive. and the biconditional of the conditional statement.

Question 10.
Two planes intersect at a line.
Answer:

Question 11.
A relation that pairs each input with exactly one output is a function.
Answer:

Use inductive reasoning to make a conjecture about the given quantity. Then use deductive reasoning to sIm that the conjecture is true.

Question 12.
the sum of three odd integers
Answer:

Question 13.
the product of three even integers
Answer:

Question 14.
Give an example of two statements for which the Law of Detachment does not apply.
Answer:

Question 15.
The formula for the area A of a triangle is A = \(\frac{1}{2}\)bh, where b is the base and h is the height. Solve the formula for h and justify each step. Then find the height of a standard yield sign when the area is 558 square inches and each side is 36 inches.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 149
Answer:

Question 16.
You visit the zoo and notice the following

  • The elephants, giraffes, lions, tigers, and zebras are located along a straight walkway.
  • The giraffes are halfway between the elephants and the lions.
  • The tigers are halfway between the lions and the zebras.
  • The lions are hallway between the giraffes and the tigers.

Draw and label a diagram that represents this information. Then prove that the distance between the elephants and the giraffes is equal to the distance between the tigers and the zebras. Use any proof format.
Answer:

Question 17.
Write a proof using an format.
Given ∠2 ≅∠3
\(\vec{T}\)V bisects ∠UTW.
Prove ∠1 ≅ ∠3
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 150
Answer:

Reasoning and Proofs Cumulative Assessment

Question 1.
Use the diagram to write an example of each postulate.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 151
a. Two Point Postulate (Postulate 2.1): Through any two points, there exists exactly one line.
Answer:

b. Line Intersection Postulate (Postulate 2.3): If two lines intersect, then their intersection is exactly one point.
Answer:

c. Three Point Postulate (Postulate 2.4): Through any three noncollinear points, there exists exactly one plane.
Answer:

d. Plane-Line Postulate (Postulate 2.6): If two points lie in a plane, then the line containing them lies in the plane.
Answer:

e. Plane Intersection Postulate (Postulate 2.7): If two planes intersect, then their intersection is a line
Answer:

Question 2.
Enter the reasons in the correct positions to complete the two-column proof.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 152
Answer:

Question 3.
Classify each related conditional statement. based on the conditional statement
“If I study, then I will pass the final exam.”
a. I will pass the final exam if and only if I study.
Answer:

b. If I do not study, then I will not pass the final exam.
Answer:

c. If I pass the final exam, then I studied.
Answer:

d. If I do not pass the final exam, then I did not study.
Answer:

Question 4.
List all segment bisectors given x = 3.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 153
Answer:

Question 5.
You are given m∠FHE = m∠BHG = m∠AHF = 90°. Choose the symbol that makes each statement true. State which theorem or postulate. if any, supports your answer.
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 154

=       ≅         ≠
a. ∠3 _____ ∠6
Answer:

b. m∠4 ______ m∠7
Answer:

c. m∠FHE _______ m∠AHG
Answer:

d. m∠AHG + m∠GHE _______180°
Answer:

Question 6.
Find the distance between each pair of points. Then order each line
segment from longest to shortest.
a. A(- 6, 1), B(- 1, 6)
Answer:

b. C(- 5, 8), D(5, 8)
Answer:

c. E(2, 7), F(4, – 2)
Answer:

d. G(7, 3), H(7, – 1)
Answer:

e. J(- 4, – 2), K(1, – 5)
Answer:

f. L(3, – 8), M(7, – 5)
Answer:

Question 7.
The proof shows that ∠MRL is congruent to ∠NSR. Select all other angles that are also congruent to ∠NSR.
Given ∠MRS and ∠NSR are supplementary.
Prove ∠MRL ≅ ∠VSR
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 155

StatementsReasons
1. ∠MRS and ∠NSR are supplementary1. Given
2. ∠MRL and ∠MRS are a linear pair.2. Definition of linear pair, as shown in the diagram
3. ∠MRL and ∠MRS are supplementary.3. Linear Pair Postulate (Postulate 2.8)
4. ∠MRL ≅ ∠NSR4. Congruent Supplements Theorem (Theorem 2.4)

∠PSK         ∠KSN        ∠PSR       ∠QRS          ∠QRL
Answer:

Question 8.
Your teacher assigns your class a homework problem that asks you to prove the Vertical Angles Congruence Theorem (Theorem 2.6) using the picture and information given at the right. Your friend claims that this can be proved without using the Linear Pair Postulate (Postulate 2.8). Is our friend correct? Explain your reasoning.
Given ∠1 and ∠3 arc vertical angles.
Prove ∠1 ≅ ∠3
Big Ideas Math Answers Geometry Chapter 2 Reasoning and Proofs 156
Answer:

Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays

Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis & Displays

Need an opportunity to become math proficient? Then, this article will help you to turn like that. As this page holds benefitted Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays Questions to solve and improve subject knowledge. Students who want to practice all topics from ch 11 Data Analysis and Displays can explore BIM algebra 1 ch 11 solution key pdf. Enhance your math skills with the help of 11.1 to 11.5 Exercises Questions, Practice Tests, Cumulative Assessment, Review Tests, Quiz, etc. involve in the BigIdeas Math Book Algebra 1 Chapter 11 Data Analysis and Displays Solution Key.

Big Ideas Math Algebra 1 Textbook Solutions for Chapter 11 Data Analysis and Displays

Make a great move at the time of your practice sessions or exam preparation by using the ultimate guide of Big Ideas Math Book Algebra 1 Answers Chapter 11 Data Analysis and Displays. This guide can help each and every learner by providing a deep level & step by step explanation about the concepts of Algebra 1 Ch 11. You can also make use of this BIM Algebra 1 Chapter 11 Data Analysis and Displays as quick revision material during exams.

These solutions are prepared by the subject experts as per the Common Core Curriculum. Click on the direct links available here and start practicing the respective Big Ideas Math Algebra 1 Textbook Answers of Chapter 11 concepts for better knowledge and performance in the annual exams.

Data Analysis and Displays Maintaining Mathematical Proficiency

The table shows the results of a survey. Display the data in a histogram.
Question 1.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 1
Answer:

Question 2.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 2
Answer:

The table shows the results of a survey. Display the data in a circle graph.
Question 3.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 3
Answer:

Question 4.
ABSTRACT REASONING
Twenty people respond “yes” or “no” to a survey question. Let a and b represent the frequencies of the responses. What must be true about the sum of a and b? What must be true about the sum when “maybe” is an option for the response?
Answer:

Data Analysis and Displays Mathematical Practices

Mathematically proficient students use diagrams and graphs to show relationships between data. They also analyze data to draw conclusions.
Using Data Displays
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 4

Monitoring Progress
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 5
Question 1.
The table shows the estimated populations of males and females by age in the United States in 2012. Use a spreadsheet, graphing calculator, or some other form of technology to make two different displays for the data.
Answer:

Question 2.
Explain why you chose each type of data display in Monitoring Progress Question 1. What conclusions can you draw from your data displays?
Answer:

Lesson 11.1 Measures of Center and Variation

Essential Question
How can you describe the variation of a data set?

EXPLORATION 1
Describing the Variation of Data
Work with a partner. The graphs show the weights of the players on a professional football team and a professional baseball team.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 1
a. Describe the data in each graph in terms of how much the weights vary from the mean. Explain your reasoning.
b. Compare how much the weights of the players on the football team vary from the mean to how much the weights of the players on the baseball team vary from the mean.
c. Does there appear to be a correlation between the body weights and the positions of players in professional football? in professional baseball? Explain.
Answer:

EXPLORATION 2

Describing the Variation of Data
Work with a partner. The weights (in pounds) of the players on a professional basketball team by position are as follows.
Power forwards: 235, 255, 295, 245; small forwards: 235, 235;
centers: 255, 245, 325; point guards: 205, 185, 205; shooting guards: 205, 215, 185
Make a graph that represents the weights and positions of the players. Does there appear to be a correlation between the body weights and the positions of players in professional basketball? Explain your reasoning.
Answer:

Communicate Your Answer
Question 3.
How can you describe the variation of a data set?
Answer:

Monitoring Progress

Question 1.
WHAT IF?
The park hires another student at an hourly wage of $8.45. (a) How does this additional value affect the mean, median, and mode? Explain. (b) Which measure of center best represents the data? Explain.
Answer:

Question 2.
The table shows the annual salaries of the employees of an auto repair service. (a) Identify the outlier. How does the outlier affect the mean, median, and mode? (b) Describe one possible explanation for the outlier.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 2
Answer:

Question 3.
After the first week, the 25-year-old is voted off Show A and the 48-year-old is voted off Show B. How does this affect the range of the ages of the remaining contestants on each show in Example 3? Explain.
Answer:

Question 4.
Find the standard deviation of the ages for Show B in Example 3. Interpret your result.
Answer:

Question 5.
Compare the standard deviations for Show A and Show B. What can you conclude?
Answer:

Question 6.
Find the mean, median, mode, range, and standard deviation of the altitudes of the airplanes when each altitude increases by \(\frac{1}{2}\) miles.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 3
Answer:

Measures of Center and Variation 11.1 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
In a data set, what does a measure of center represent? What does a measure of variation describe?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a1

Question 2.
WRITING
Describe how removing an outlier from a data set affects the mean of the data set.
Answer:

Question 3.
OPEN-ENDED
Create a data set that has more than one mode.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a3

Question 4.
REASONING
What is an advantage of using the range to describe a data set? Why do you think the standard deviation is considered a more reliable measure of variation than the range?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, (a) Find the mean, median, and mode of the data set and (b) determine which measure of center best represents the data. Explain. (See Example 1.)
Question 5.
3, 5, 1, 5, 1, 1, 2, 3, 15
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a5
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a13

Question 6.
12, 9, 17, 15, 10
Answer:

Question 7.
13, 30, 16, 19, 20, 22, 25, 31
Answer:

Question 8.
14, 15, 3, 15, 14, 14, 18, 15, 8, 16
Answer:

Question 9.
ANALYZING DATA
The table shows the lengths of nine movies.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 4
a. Find the mean, median, and mode of the lengths.
b. Which measure of center best represents the data? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a9

Question 10.
ANALYZING DATA
The table shows the daily changes in the value of a stock over 12 days.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 5
a. Find the mean, median, and mode of the changes in stock value.
b. Which measure of center best represents the data? Explain.
c. On the 13th day, the value of the stock increases by $4.28. How does this additional value affect the mean, median, and mode? Explain.
Answer:

In Exercises 11–14, find the value of x.
Question 11.
2, 8, 9, 7, 6, x; The mean is 6.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a11

Question 12.
12.5, -10, -7.5, x; The mean is 11.5.
Answer:

Question 13.
9, 10, 12, x, 20, 25; The median is 14.
Answer:

Question 14.
30, 45, x, 100; The median is 51.
Answer:

Question 15.
ANALYZING DATA
The table shows the masses of eight polar bears. (See Example 2.)
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 6
a. Identify the outlier. How does the outlier affect the mean, median, and mode?
b. Describe one possible explanation for the outlier.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a15.1
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a15.2

Question 16.
ANALYZING DATA
The sizes of emails (in kilobytes) in your inbox are 2, 3, 5, 2, 1, 46, 3, 7, 2, and 1.
a. Identify the outlier. How does the outlier affect the mean, median, and mode?
b. Describe one possible explanation for the outlier.
Answer:

Question 17.
ANALYZING DATA
The scores of two golfers are shown. Find the range of the scores for each golfer. Compare your results. (See Example 3.)
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a17

Question 18.
ANALYZING DATA
The graph shows a player’s monthly home run totals in two seasons. Find the range of the number of home runs for each season. Compare your results.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 8
Answer:

In Exercises 19–22, find (a) the range and (b) the standard deviation of the data set.
Question 19.
40, 35, 45, 55, 60
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a19

Question 20.
141, 116, 117, 135, 126, 121
Answer:

Question 21.
0.5, 2.0, 2.5, 1.5, 1.0, 1.5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a21

Question 22.
8.2, 10.1, 2.6, 4.8, 2.4, 5.6, 7.0, 3.3
Answer:

Question 23.
ANALYZING DATA
Consider the data in Exercise 17.
a. Find the standard deviation of the scores of Golfer A. Interpret your result.
b. Find the standard deviation of the scores of Golfer B. Interpret your result.
c. Compare the standard deviations for Golfer A and Golfer B. What can you conclude?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a23.1
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a23.2

Question 24.
ANALYZING DATA
Consider the data in Exercise 18.
a. Find the standard deviation of the monthly home run totals in the player’s rookie season. Interpret your result.
b. Find the standard deviation of the monthly home run totals in this season. Interpret your result.
c. Compare the standard deviations for the rookie season and this season. What can you conclude?
Answer:

In Exercises 25 and 26, find the mean, median, and mode of the data set after the given transformation.
Question 25.
In Exercise 5, each data value increases by 4.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a25

Question 26.
In Exercise 6, each data value increases by 20%.
Answer:

Question 27.
TRANSFORMING DATA
Find the values of the measures shown when each value in the data set increases by 14.
Mean: 62
Median: 55
Mode: 49
Range: 46
Standard deviation: 15.5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a27

Question 28.
TRANSFORMING DATA
Find the values of the measures shown when each value in the data set is multiplied by 0.5.
Mean: 320
Median: 300
Mode: none
Range: 210
Standard deviation: 70.6
Answer:

Question 29.
ERROR ANALYSIS
Describe and correct the error in finding the median of the data set.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 9
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a29

Question 30.
ERROR ANALYSIS
Describe and correct the error in finding the range of the data set after the given transformation.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 10
Answer:

Question 31.
PROBLEM SOLVING
In a bowling match, the team with the greater mean score wins. The scores of the members of two bowling teams are shown.
Team A: 172, 130, 173, 212
Team B: 136, 184, 168, 192
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 11
a. Which team wins the match? If the team with the greater median score wins, is the result the same? Explain.
b. Which team is more consistent? Explain.
c. In another match between the two teams, all the members of Team A increase their scores by 15 and all the members of Team B increase their scores by 12.5%. Which team wins this match? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a31.1
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a31.2

Question 32.
MAKING AN ARGUMENT
Your friend says that when two data sets have the same range, you can assume the data sets have the same standard deviation, because both range and standard deviation are measures of variation. Is your friend correct? Explain.
Answer:

Question 33.
ANALYZING DATA
The table shows the results of a survey that asked 12 students about their favorite meal. Which measure of center (mean, median, or mode) can be used to describe the data? Explain.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 12
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a33

Question 34.
HOW DO YOU SEE IT?
The dot plots show the ages of the members of three different adventure clubs. Without performing calculations, which data set has the greatest standard deviation? Which has the least standard deviation? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 13
Answer:

Question 35.
REASONING
A data set is described by the measures shown.
Mean: 27
Median: 32
Mode: 18
Range: 41
Standard deviation: 9
Find the mean, median, mode, range, and standard deviation of the data set when each data value is multiplied by 3 and then increased by 8.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a35

Question 36.
CRITICAL THINKING
Can the standard deviation of a data set be 0? Can it be negative? Explain.
Answer:

Question 37.
USING TOOLS
Measure the heights (in inches) of the students in your class.
a. Find the mean, median, mode, range, and standard deviation of the heights.
b. A new student who is 7 feet tall joins your class. How would you expect this student’s height to affect the measures in part (a)? Verify your answer.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a37

Question 38.
THOUGHT PROVOKING
To find the arithmetic mean of n numbers, divide the sum of the numbers by n. To find the geometric mean of n numbers a1, a2, a3, . . . , an, take the nth root of the product of the numbers.
geometric mean = \(\sqrt[n]{a_{1} \cdot a_{2} \cdot a_{3} \cdot \ldots \cdot a_{n}}\)
Compare the arithmetic mean to the geometric mean of n numbers.
Answer:

Question 39.
PROBLEM SOLVING
The circle graph shows the distribution of the ages of 200 students in a college Psychology I class.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.1 14
a. Find the mean, median, and mode of the students’ ages.
b. Identify the outliers. How do the outliers affect the mean, median, and mode?
c. Suppose all 200 students take the same Psychology II class exactly 1 year later. Draw a new circle graph that shows the distribution of the ages of this class and find the mean, median, and mode of the students’ ages.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a39.1
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a39.2

Maintaining Mathematical Proficiency

Solve the inequality.(Section 2.4)
Question 40.
6x + 1 ≤ 4x – 9
Answer:

Question 41.
-3(3y – 2) < 1 – 9y
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a41

Question 42.
2(5c – 4) ≥ 5(2c + 8)
Answer:

Question 43.
4(3 – w) > 3(4w – 4)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a43

Evaluate the function for the given value of x.(Section 6.3)
Question 44.
f(x) – 4x; x = 3
Answer:

Question 45.
f(x) = 7x; x = -2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a45

Question 46.
f(x) = 5(2)x; x = 6
Answer:

Question 47.
f(x) = -2(3)x; x = 4
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.1 a47

Lesson 11.2 Box-and-Whisker Plots

Essential Question How can you use a box-and-whisker plot to describe a data set?

EXPLORATION 1

Drawing a Box-and-Whisker Plot
Work with a partner. The numbers of first cousins of the students in a ninth-grade class are shown. A box-and-whisker plot is one way to represent the data visually.
a. Order the data on a strip of grid paper with 24 equally spaced boxes. Fold the paper in half to find the median.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 1
b. Fold the paper in half again to divide the data into four groups. Because there are 24 numbers in the data set, each group should have 6 numbers. Find the least value, the greatest value, the first quartile, and the third quartile.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 2
c. Explain how the box-and-whisker plot shown represents the data set.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 3
Answer:

CommunicateYour Answer

Question 2.
How can you use a box-and-whisker plot to describe a data set?
Answer:

Question 3.
Interpret each box-and-whisker plot.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 4
a. body mass indices (BMI) of students in a ninth-grade class
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 5
b. heights of roller coasters at an amusement park
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 6
Answer:

Monitoring Progress

Question 1.
A basketball player scores 14, 16, 20, 5, 22, 30, 16, and 28 points during a tournament. Make a box-and-whisker plot that represents the data.
Answer:

Use the box-and-whisker plot in Example 1.
Question 2.
Find and interpret the range and interquartile range of the data.
Answer:

Question 3.
Describe the distribution of the data.
Answer:

Question 4.
The double box-and-whisker plot represents the surfboard prices at Shop A and Shop B. Identify the shape of each distribution. Which shop’s prices are more spread out? Explain.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 7
Answer:

Box-and-Whisker Plots 11.2 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe how to find the first quartile of a data set.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Consider the box-and-whisker plot shown. Which is different? Find “both” answers.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 8
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, use the box-and-whisker plot to find the given measure.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 9
Question 3.
least value
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 3

Question 4.
greatest value
Answer:

Question 5.
third quartile
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 5

Question 6.
first quartile
Answer:

Question 7.
median
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 7

Question 8.
range
Answer:

In Exercises 9–12, make a box-and-whisker plot that represents the data.
Question 9.
Hours of television watched: 0, 3, 4, 5, 2, 4, 6, 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 9

Question 10.
Cat lengths (in inches): 16, 18, 20, 25, 17, 22, 23, 21
Answer:

Question 11.
Elevations (in feet): -2, 0, 5, -4, 1, -3, 2, 0, 2, -3, 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 11

Question 12.
MP3 player prices (in dollars): 124, 95, 105, 110, 95, 124, 300, 190, 114
Answer:

Question 13.
ANALYZING DATA
The dot plot represents the numbers of hours students spent studying for an exam. Make a box-and-whisker plot that represents the data.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 13

Question 14.
ANALYZING DATA
The stem-and-leaf plot represents the lengths (in inches) of the fish caught on a fishing trip. Make a box-and-whisker plot that represents the data.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 11
Answer:

Question 15.
ANALYZING DATA
The box-and-whisker plot represents the prices (in dollars) of the entrées at a restaurant.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 12
a. Find and interpret the range of the data.
b. Describe the distribution of the data.
c. Find and interpret the interquartile range of the data.
d. Are the data more spread out below Q1 or above or Q3? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 15

Question 16.
ANALYZING DATA
A baseball player scores 101 runs in a season. The box-and-whisker plot represents the numbers of runs the player scores against different opposing teams.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 13
a. Find and interpret the range and interquartile range of the data.
b. Describe the distribution of the data. c. Are the data more spread out between Q1 and Q2 or between Q2 and Q3? Explain.
Answer:

Question 17.
ANALYZING DATA
The double box-and-whisker plot represents the monthly car sales for a year for two sales representatives.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 14
a. Identify the shape of each distribution.
b. Which representative’s sales are more spread out? Explain.
c. Which representative had the single worst sales month during the year? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 17

Question 18.
ERROR ANALYSIS
Describe and correct the error in describing the box-and-whisker plot.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 15
Answer:

Question 19.
WRITING
Given the numbers 36 and 12, identify which number is the range and which number is the interquartile range of a data set. Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 19

Question 20.
HOW DO YOU SEE IT?
The box-and-whisker plot represents a data set. Determine whether each statement is always true. Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 16
a. The data set contains the value 11.
b. The data set contains the value 6.
c. The distribution is skewed right.
d. The mean of the data is 5.
Answer:

Question 21.
ANALYZING DATA
The double box-and-whisker plot represents the battery lives (in hours) of two brands of cell phones.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 17
a. Identify the shape of each distribution.
b. What is the range of the upper 75% of each brand?
c. Compare the interquartile ranges of the two data sets.
d. Which brand do you think has a greater standard deviation? Explain.
e. You need a cell phone that has a battery life of more than 3.5 hours most of the time. Which brand should you buy? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 21

Question 22.
THOUGHT PROVOKING
Create a data set that can be represented by the box-and-whisker plot shown. Justify your answer.
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.2 18
Answer:

Question 23.
CRITICAL THINKING
Two data sets have the same median, the same interquartile range, and the same range. Is it possible for the box-and-whisker plots of the data sets to be different? Justify your answer.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 23

Maintaining Mathematical Proficiency

Use zeros to graph the function. (Section 8.5)
Question 24.
f(x) = -2(x + 9)(x – 3)
Answer:

Question 25.
y = 3(x – 5)(x + 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 25

Question 26.
y = 4x2 – 16x = 48
Answer:

Question 27.
h(x) = -x2 + 5x + 14
Answer:
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 27.1
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays 11.2 a 27.2

Lesson 11.3 Shapes of Distributions

Essential Question How can you use a histogram to characterize the basic shape of a distribution?

EXPLORATION 1

Analyzing a Famous Symmetric Distribution
Work with a partner. A famous data set was collected in Scotland in the mid-1800s. It contains the chest sizes, measured in inches, of 5738 men in the Scottish Militia. Estimate the percent of the chest sizes that lie within (a) 1 standard deviation of the mean, (b) 2 standard deviations of the mean, and (c) 3 standard deviations of the mean. Explain your reasoning.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 1
Answer:

EXPLORATION 2

Comparing Two Symmetric Distributions
Work with a partner. The graphs show the distributions of the heights of 250 adult American males and 250 adult American females.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 2
Answer:
a. Which data set has a smaller standard deviation? Explain what this means in the context of the problem.
b. Estimate the percent of male heights between 67 inches and 73 inches.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 3
Answer:

CommunicateYour Answer
Question 3.
How can you use a histogram to characterize the basic shape of a distribution?
Answer:

Question 4.
All three distributions in Explorations 1 and 2 are roughly symmetric. The histograms are called “bell-shaped.”
a. What are the characteristics of a symmetric distribution?
b.Why is a symmetric distribution called “bell-shaped?”
c. Give two other real-life examples of symmetric distributions.Shapes of Distributions.
Answer:

Monitoring Progress

Question 1.
The frequency table shows the numbers of pounds of aluminum cans collected by classes for a fundraiser. Display the data in a histogram. Describe the shape of the distribution.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 4
Answer:

Question 2.
You record the numbers of email attachments sent by 30 employees of a company in 1 week. Your results are shown in the table. (a) Display the data in a histogram using six intervals beginning with 1–20. (b) Which measures of center and variation best represent the data? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 5
Answer:

Question 3.
Compare the distributions using their shapes and appropriate measures of center and variation.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 6
Answer:

Question 4.
Why is the mean greater than the median for the men?
Answer:

Question 5.
If 50 more women are surveyed, about how many more would you expect to own between 10 and 18 pairs of shoes?
Answer:

Shapes of Distributions 11.3 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Describe how data are distributed in a symmetric distribution, a distribution that is skewed left, and a distribution that is skewed right.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 1

Question 2.
WRITING
How does the shape of a distribution help you decide which measures of center and variation best describe the data?
Answer:

Monitoring Progress and Modeling with Mathematics

Question 3.
DESCRIBING DISTRIBUTIONS
The frequency table shows the numbers of hours that students volunteer per month. Display the data in a histogram. Describe the shape of the distribution.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 7
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 1

Question 4.
DESCRIBING DISTRIBUTIONS
The frequency table shows the results of a survey that asked people how many hours they spend online per week. Display the data in a histogram. Describe the shape of the distribution.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 8
Answer:

In Exercises 5 and 6, describe the shape of the distribution of the data. Explain your reasoning.
Question 5.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 9
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 5

In Exercises 7 and 8, determine which measures of center and variation best represent the data. Explain your reasoning.
Question 7.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 10
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 7

Question 8.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 12
Answer:

Question 9.
ANALYZING DATA
The table shows the last 24 ATM withdrawals at a bank.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 13
a. Display the data in a histogram using seven intervals beginning with 26–50.
b. Which measures of center and variation best represent the data? Explain.
c. The bank charges a fee for any ATM withdrawal less than $150. How would you interpret the data?
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 9.1
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 9.2

Question 10.
ANALYZING DATA
Measuring an IQ is an inexact science. However, IQ scores have been around for years in an attempt to measure human intelligence. The table shows some of the greatest known IQ scores.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 14
a. Display the data in a histogram using five intervals beginning with 151–166.
b. Which measures of center and variation best represent the data? Explain.
c. The distribution of IQ scores for the human population is symmetric. What happens to the shape of the distribution in part (a) as you include more and more IQ scores from the human population in the data set?
Answer:

ERROR ANALYSIS In Exercises 11 and 12, describe and correct the error in the statements about the data displayed in the histogram.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 15
Question 11.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 16
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 11

Question 12.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 17
Answer:

Question 13.
USING TOOLS
For a large data set, would you use a stem-and-leaf plot or a histogram to show the distribution of the data? Explain.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 13

Question 14.
REASONING
For a symmetric distribution, why is the mean used to describe the center and the standard deviation used to describe the variation? For a skewed distribution, why is the median used to describe the center and the  ve-number summary used to describe the variation?
Answer:

Question 15.
COMPARING DATA SETS
The double histogram shows the distributions of daily high temperatures for two towns over a 50-day period. Compare the distributions using their shapes and appropriate measures of center and variation.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 18
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 15

Question 16.
COMPARING DATA SETS
The frequency tables show the numbers of entrées in certain price ranges (in dollars) at two different restaurants. Display the data in a double histogram. Compare the distributions using their shapes and appropriate measures of center and variation.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 19
Answer:

Question 17.
OPEN-ENDED
Describe a real-life data set that has a distribution that is skewed right.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 17

Question 18.
OPEN-ENDED
Describe a real-life data set that has a distribution that is skewed left.
Answer:

Question 19.
COMPARING DATA SETS
The table shows the results of a survey that asked freshmen and sophomores how many songs they have downloaded on their MP3 players.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 20
a. Make a double box-and-whisker plot that represents the data. Describe the shape of each distribution.
b. Compare the number of songs downloaded by freshmen to the number of songs downloaded by sophomores.
c. About how many of the freshmen surveyed would you expect to have between 730 and 1570 songs downloaded on their MP3 players?
d. If you survey100 more freshmen, about how many would you expect to have downloaded between 310 and 1990 songs on their MP3 players?
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 19.1
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 19.2

Question 20.
COMPARING DATA SETS
You conduct the same survey as in Exercise 19 but use a different group of freshmen. The results are as follows.Survey size: 60; minimum: 200; maximum: 2400; 1st quartile: 640; median: 1670; 3rd quartile: 2150; mean: 1480; standard deviation: 500
a. Compare the number of songs downloaded by this group of freshmen to the number of songs downloaded by sophomores.
b. Why is the median greater than the mean for this group of freshmen?
Answer:

Question 21.
REASONING
A data set has a symmetric distribution. Every value in the data set is doubled. Describe the shape of the new distribution. Are the measures of center and variation affected? Explain.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 21

Question 22.
HOW DO YOU SEE IT?
Match the distribution with the corresponding box-and-whisker plot.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 21
Answer:

Question 23.
REASONING
You record the following waiting times at a restaurant.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 22
a. Display the data in a histogram using five intervals beginning with 0–9. Describe the shape of the distribution.
b. Display the data in a histogram using 10 intervals beginning with 0–4. What happens when the number of intervals is increased?
c. Which histogram best represents the data? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 23.1
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 23.2

Question 24.
THOUGHT PROVOKING
The shape of a bimodal distribution is shown. Describe a real-life example of a bimodal distribution.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 23
Answer:

Maintaining Mathematical Proficiency

Find the domain of the function.(Section 10.1)
Question 25.
f(x) = \(\sqrt{x+6}\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 25 width=

Question 26.
f(x) = \(\sqrt{2x}\)
Answer:

Question 27.
f(x) = \(\frac{1}{4} \sqrt{x-7}\)
Answer:
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.3 a 27

Data Analysis and Displays Study Skills: Studying for Finals

11.1–11.3What Did YouLearn?
Core Vocabulary
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 24

Core Concepts
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 25

Mathematical Practices
Question 1.
Exercises 15 and 16 on page 590 are similar. For each data set, is the outlier much greater than or much less than the rest of the data values? Compare how the outliers affect the means. Explain why this makes sense.
Answer:

Question 2.
In Exercise 18 on page 605, provide a possible reason for why the distribution is skewed left.
Answer:

Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays 11.3 26

Data Analysis and Displays 1.1–11.3 Quiz

Find the mean, median, and mode of the data set. Which measure of center best represents the data? Explain.(Section 11.1)
Question 1.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays q 1
Answer:

Question 2.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays q 2
Answer:

Find the range and standard deviation of each data set. Then compare your results.(Section 11.1)
Question 3.
Absent students during a week of school
Female: 6, 2, 4, 3, 4
Male: 5, 3, 6, 6, 9
Answer:

Question 4.
Numbers of points scored
Juniors: 19, 15, 20, 10, 14, 21, 18, 15
Seniors: 22, 19, 29, 32, 15, 26, 30, 19
Answer:

Make a box-and-whisker plot that represents the data.(Section 11.2)
Question 5.
Ages of family members:
60, 15, 25, 20, 55, 70, 40, 30
Answer:

Question 6.
Minutes of violin practice:
20, 50, 60, 40, 40, 30, 60, 40, 50, 20, 20, 35
Answer:

Question 7.
Display the data in a histogram. Describe the shape of the distribution. (Section 11.3)
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays q 7

Question 8.
The table shows the prices of eight mountain bikes in a sporting goods store. (Section 11.1 and Section 11.2)
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays q 8.1
a. Find the mean, median, mode, range, and standard deviation of the prices.
b. Identify the outlier. How does the outlier affect the mean, median, and mode?
c. Make a box-and-whisker plot that represents the data. Find and interpret the interquartile range of the data. Identify the shape of the distribution.
d. Find the mean, median, mode, range, and standard deviation of the prices when the store offers a 5% discount on all mountain bikes.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays q 8
Answer:

Question 9.
The table shows the times of 20 presentations. (Section 11.3)
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays q 9
a. Display the data in a histogram using five intervals beginning with 3–5.
b. Which measures of center and variation best represent the data? Explain.
c. The presentations are supposed to be 10 minutes long. How would you interpret these results?
Answer:

Lesson 11.4 Two-Way Tables

Essential Question How can you read and make a two-way table?

EXPLORATION 1

Reading a Two-Way Table
Work with a partner. You are the manager of a sports shop. The two-way tables show the numbers of soccer T-shirts in stock at your shop at the beginning and end of the selling season. (a) Complete the totals for the rows and columns in each table. (b) How would you alter the number of T-shirts you order for next season? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 1
Answer:

EXPLORATION 2

Making a Two-Way Table
Work with a partner. The three-dimensional bar graph shows the numbers of hours students work at part-time jobs.
a. Make a two-way table showing the data. Use estimation to find the entries in your table.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 2
b. Write two observations that summarize the data in your table.
Answer:

Communicate Your Answer
Question 3.
How can you read and make a two-way table?
Answer:

Monitoring Progress

Question 1.
You conduct a technology survey to publish on your school’s website. You survey students in the school cafeteria about the technological devices they own. The results are shown in the two-way table. Find and interpret the marginal frequencies.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 3
Answer:

Question 2.
You survey students about whether they are getting a summer job. Seventy-five males respond, with 18 of them responding “no.” Fifty-seven females respond, with 45 of them responding “yes.” Organize the results in a two-way table. Include the marginal frequencies.
Answer:

Question 3.
Use the survey results in Monitoring Progress Question 2 to make a two-way table that shows the joint and marginal relative frequencies. What percent of students are not getting a summer job?
Answer:

Question 4.
Use the survey results in Example 3 to make a two-way table that shows the conditional relative frequencies based on the row totals. Given that a student is a senior, what is the conditional relative frequency that he or she is planning to major in a medical field?
Answer:

Question 5.
Using the results of the survey in Monitoring Progress Question 1, is there an association between owning a tablet computer and owning a cell phone? Explain your reasoning.
Answer:

Two-Way Tables 11.4 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Each entry in a two-way table is called a(n) __________.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 1

Question 2.
WRITING
When is it appropriate to use a two-way table to organize data?
Answer:

Question 3.
VOCABULARY
Explain the relationship between joint relative frequencies, marginal relative frequencies, and conditional relative frequencies.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 3

Question 4.
WRITING
Describe two ways you can find conditional relative frequencies.
Answer:

Monitoring Progress and Modeling with Mathematics

You conduct a survey that asks 346 students whether they buy lunch at school. In Exercises 5–8, use the results of the survey shown in the two-way table.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 3
Question 5.
How many freshmen were surveyed?
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 5

Question 6.
How many sophomores were surveyed?
Answer:

Question 7.
How many students buy lunch at school?
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 7

Question 8.
How many students do not buy lunch at school?
Answer:

In Exercises 9 and 10, find and interpret the marginal frequencies.
Question 9.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 5
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 9

Question 10.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 6
Answer:

Question 11.
USING TWO-WAY TABLES
You conduct a survey that asks students whether they plan to participate in school spirit week. The results are shown in the two-way table. Find and interpret the marginal frequencies.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 7
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 11

Question 12.
USING TWO-WAY TABLES
You conduct a survey that asks college-bound high school seniors about the type of degree they plan to receive. The results are shown in the two-way table. Find and interpret the marginal frequencies.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 8
Answer:

USING STRUCTURE In Exercises 13 and 14, complete the two-way table..
Question 13.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 9
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 13

Question 14.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 10
Answer:

Question 15.
MAKING TWO-WAY TABLES
You conduct a survey that asks 245 students in your school whether they have taken a Spanish or a French class. One hundred nine of the students have taken a Spanish class, and 45 of those students have taken a French class. Eighty-two of the students have not taken a Spanish or a French class. Organize the results in a two-way table. Include the marginal frequencies.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 15

Question 16.
MAKING TWO-WAY TABLES
A car dealership has 98 cars on its lot. Fifty-five of the cars are new. Of the new cars, 36 are domestic cars. There are 15 used foreign cars on the lot. Organize this information in a two-way table. Include the marginal frequencies. In Exercises 17 and 18, make a two-way table that shows the joint and marginal relative frequencies.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 11.1
Answer:

In Exercises 17 and 18, make a two-way table that shows the joint and marginal relative frequencies.
Question 17.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 11
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 17

Question 18.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 12
Answer:

Question 19.
USING TWO-WAY TABLES
Refer to Exercise 17. What percent of students prefer aerobic exercise? What percent of students are males who prefer anaerobic exercise?
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 19

Question 20.
USING TWO-WAY TABLES
Refer to Exercise 18. What percent of the sandwiches are on wheat bread? What percent of the sandwiches are turkey on white bread?
Answer:

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in using the two-way table.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 13
Question 21.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 14
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 21

Question 22.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 15
Answer:

Question 23.
USING TWO-WAY TABLES
A company is hosting an event for its employees to celebrate the end of the year. It asks the employees whether they prefer a lunch event or a dinner event. It also asks whether they prefer a catered event or a potluck. The results are shown in the two-way table. Make a two-way table that shows the conditional relative frequencies based on the row totals. Given that an employee prefers a lunch event, what is the conditional relative frequency that he or she prefers a catered event?
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 16
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 23

Question 24.
USING TWO-WAY TABLES
The two-way table shows the results of a survey that asked students about their preference for a new school mascot. Make a two-way table that shows the conditional relative frequencies based on the column totals. Given that a student prefers a hawk as a mascot, what is the conditional relative frequency that he or she prefers a cartoon mascot?
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 17
Answer:

Question 25.
ANALYZING TWO-WAYTABLES
You survey college-bound seniors and find that 85% plan to live on campus, 35% plan to have a car while at college, and 5% plan to live off campus and not have a car. Is there an association between living on campus and having a car at college? Explain.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 25.1
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 25.2

Question 26.
ANALYZING TWO-WAYTABLES
You survey students and find that 70% watch sports on TV, 48% participate in a sport, and 16% do not watch sports on TV or participate in a sport. Is there an association between participating in a sport and watching sports on TV? Explain.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 18
Answer:

Question 27.
ANALYZING TWO-WAY TABLES
The two-way table shows the results of a survey that asked adults whether they participate in recreational skiing. Is there an association between age and recreational skiing?
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 19
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 27

Question 28.
ANALYZING TWO-WAY TABLES
Refer to Exercise 12. Is there an association between gender and type of degree? Explain.
Answer:

Question 29.
WRITING
Compare Venn diagrams and two-way tables.
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 29

Question 30.
HOW DO YOU SEE IT?
The graph shows the results of a survey that asked students about their favorite movie genre.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 20
a. Display the given information in a two-way table.
b. Which of the data displays do you prefer? Explain.
Answer:

Question 31.
PROBLEM SOLVING
A box office sells 1809 tickets to a play, 800 of which are for the main floor. The tickets consist of 2x + y adult tickets on the main floor, x – 40 child tickets on the main floor, x + 2y adult tickets in the balcony, and 3x – y – 80 child tickets in the balcony.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 21
a. Organize this information in a two-way table.
b. Find the values of x and y.
c. What percent of tickets are adult tickets?
d. What percent of child tickets are balcony tickets?
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 31.1
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 31.2

Question 32.
THOUGHT PROVOKING
Compare “one-way tables” and “two-way tables.” Is it possible to have a “three-way table?” If so, give an example of a three-way table.
Answer:

Maintaining Mathematical Proficiency

Tell whether the table of values represents a linear, an exponential, or a quadratic function. (Section 8.6)
Question 33.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 22
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.4 a 33

Question 34.
Big Ideas Math Algebra 1 Solutions Chapter 11 Data Analysis and Displays 11.4 23
Answer:

Lesson 11.5 Choosing a Data Display

Essential Question How can you display data in a way that helps you make decisions?

EXPLORATION 1

Displaying Data
Work with a partner. Analyze the data and then create a display that best represents the data. Explain your choice of data display.
a. A group of schools in New England participated in a 2-month study and reported 3962 animals found dead along roads.
birds: 307
mammals: 2746
amphibiAnswer: 145
reptiles: 75
unknown: 689
b. The data below show the numbers of black bears killed on a state’s roads from 1993 to 2012.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 1
c. A 1-week study along a 4-mile section of road found the following weights (in pounds) of raccoons that had been killed by vehicles.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 2
d. A yearlong study by volunteers in California reported the following numbers of animals killed by motor vehicles.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 3
Answer:

Communicate Your Answer
Question 2.
How can you display data in a way that helps you make decisions?
Answer:

Question 3.
Use the Internet or some other reference to find examples of the following types of data displays.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 4
bar graph
circle graph
scatter plot
stem-and-leaf plot
pictograph
line graph
box-and-whisker plot
histogram
dot plot

Monitoring Progress

Tell whether the data are qualitative or quantitative. Explain your reasoning.
Question 1.
telephone numbers in a directory
Answer:

Question 2.
ages of patients at a hospital
Answer:

Question 3.
lengths of videos on a website
Answer:

Question 4.
types of flowers at a florist
Answer:

Question 5.
Display the data in Example 2(a) in another way.
Answer:

Question 6.
Display the data in Example 2(b) in another way.
Answer:

Question 7.
Redraw the graphs in Example 3 so they are not misleading.
Answer:

Choosing a Data Display 11.5 Exercises

Vocabulary and Core Concept Check

Question 1.
OPEN-ENDED
Describe two ways that a line graph can be misleading.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which data set does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 5
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, tell whether the data are qualitative or quantitative. Explain your reasoning.
Question 3.
brands of cars in a parking lot
Answer:

Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 3

Question 4.
weights of bears at a zoo
Answer:

Question 5.
budgets of feature films
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 5

Question 6.
file formats of documents on a computer
Answer:

Question 7.
shoe sizes of students in your class
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 7

Question 8.
street addresses in a phone book
Answer:

In Exercises 9–12, choose an appropriate data display for the situation. Explain your reasoning.
Question 9.
the number of students in a marching band each year
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 9

Question 10.
a comparison of students’ grades (out of 100) in two different classes
Answer:

Question 11.
the favorite sports of students in your class
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 11

Question 12.
the distribution of teachers by age
Answer:

In Exercises 13–16, analyze the data and then create a display that best represents the data. Explain your reasoning.
Question 13.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 6
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 13.1
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 13.2
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 13.3

Question 14.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 7
Answer:

Question 15.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 8
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 15

Question 16.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 9
Answer:

Question 17.
DISPLAYING DATA
Display the data in Exercise 13 in another way.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 17.1
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 17.2

Question 18.
DISPLAYING DATA
Display the data in Exercise 14 in another way.
Answer:

Question 19.
DISPLAYING DATA
Display the data in Exercise 15 in another way.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 19

Question 20.
DISPLAYING DATA
Display the data in Exercise 16 in another way.
Answer:

In Exercises 21–24, describe how the graph is misleading. Then explain how someone might misinterpret the graph.
Question 21.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 10
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 21

Question 22.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 11
Answer:

Question 23.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 12
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 23

Question 24.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 13
Answer:

Question 25.
DISPLAYING DATA
Redraw the graph in Exercise 21 so it is not misleading.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 25

Question 26.
DISPLAYING DATA
Redraw the graph in Exercise 22 so it is not misleading.
Answer:

Question 27.
MAKING AN ARGUMENT
A data set gives the ages of voters for a city election. Classmate A says the data should be displayed in a bar graph, while Classmate B says the data would be better displayed in a histogram. Who is correct? Explain.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 27

Question 28.
HOW DO YOU SEE IT?
The manager of a company sees the graph shown and concludes that the company is experiencing a decline. What is missing from the graph? Explain why the manager may be mistaken.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 14
Answer:

Question 29.
REASONING
A survey asked 100 students about the sports they play. The results are shown in the circle graph.
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 15
a. Explain why the graph is misleading.
b. What type of data display would be more appropriate for the data? Explain.
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 29

Question 30.
THOUGHT PROVOKING
Use a spreadsheet program to create a type of data display that is not used in this section.
Answer:

Question 31.
REASONING
What type of data display shows the mode of a data set?
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 31

Maintaining Mathematical Proficiency

Determine whether the relation is a function. Explain.(Section 3.1)
Question 32.
(-5, -1), (-6, 0), (-5, 1), (-2, 2), (3 , 3)
Answer:

Question 33.
(0, 1), (4, 0), (8, 1), (12, 2), (16, 3)
Answer:
Big Ideas Math Answers Algebra 1 Chapter 11 Data Analysis and Displays 11.5 a 33

Data Analysis and Displays Performance Task: College Students Study Time

11.4–11.5 What Did You Learn?

Core Vocabulary
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 16

Core Concepts
Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 17

Mathematical Practices
Question 1.
Consider the data given in the two-way table for Exercises 5–8 on page 614. Your sophomore friend responded to the survey. Is your friend more likely to have responded “yes” or “no” to buying a lunch? Explain.
Answer:

Question 2.
Use your answer to Exercise 28 on page 622 to explain why it is important for a company manager to see accurate graphs.
Answer:

Big Ideas Math Answer Key Algebra 1 Chapter 11 Data Analysis and Displays 11.5 18

Data Analysis and Displays Chapter Review

11.1 Measures of Center and Variation (pp. 585–592)

Question 1.
Use the data in the example above. You run 4.0 miles on Day 11. How does this additional value affect the mean, median, and mode? Explain.
Answer:

Question 2.
Use the data in the example above. You run 10.0 miles on Day 11. How does this additional value affect the mean, median, and mode? Explain.
Answer:

Find the mean, median, and mode of the data.
Question 3.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays cr 1
Answer:

Question 4.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays cr 2
Answer:

Find the range and standard deviation of each data set. Then compare your results.
Question 5.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays cr 5
Answer:

Question 6.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays cr 6
Answer:

Find the values of the measures shown after the given transformation.
Mean: 109 Median: 104 Mode: 96 Range: 45 Standard deviation: 3.6
Question 7.
Each value in the data set increases by 25.
Answer:

Question 8.
Each value in the data set is multiplied by 0.6
Answer:

11.2 Box-and-Whisker Plots (pp. 593–598)

Make a box-and-whisker plot that represents the data. Identify the shape of the distribution.
Question 9.
Ages of volunteers at a hospital:
14, 17, 20, 16, 17, 14, 21, 18, 22
Answer:

Question 10.
Masses (in kilograms) of lions:
120, 230, 180, 210, 200, 200, 230, 160
Answer:

11.3 Shapes of Distributions (pp. 599–606)

Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays cr 7
Question 11.
The frequency table shows the amounts (in dollars) of money the students in a class have in their pockets.
a. Display the data in a histogram. Describe the shape of the distribution.
b. Which measures of center and variation best represent the data?
c. Compare this distribution with the distribution shown above using their shapes and appropriate measures of center and variation.
Answer:

11.4 Two-Way Tables (pp. 609–616)

Question 12.
The two-way table shows the results of a survey that asked shoppers at a mall about whether they like the new food court.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays cr 12
a. Make a two-way table that shows the joint and marginal relative frequencies.
b. Make a two-way table that shows the conditional relative frequencies based on the column totals.
Answer:

11.5 Choosing a Data Display (pp. 617–622)

Question 13.
Analyze the data in the table at the right and then create a display that best represents the data. Explain your reasoning.
Big Ideas Math Algebra 1 Answer Key Chapter 11 Data Analysis and Displays cr 13
Answer:

Tell whether the data are qualitative or quantitative. Explain.
Question 14.
heights of the members of a basketball team
Answer:

Question 15.
grade level of students in an elementary school
Answer:

Data Analysis and Displays Chapter Test

Describe the shape of the data distribution. Then determine which measures of center and variation best represent the data.
Question 1.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays c 1
Answer:

Question 2.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays c 2
Answer:

Question 3.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays c 3
Answer:

Determine whether each statement is always, sometimes, or never true. Explain your reasoning.
a. The sum of the marginal relative frequencies in the “total” row and the “total” column of a two-way table should each be equal to 1.
b. In a box-and-whisker plot, the length of the box to the left of the median and the length of the box to the right of the median are equal.
c. Qualitative data are numerical.
Answer:

Question 5.
Find the mean, median, mode, range, and standard deviation of the prices.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays c 5
Answer:

Question 6.
Repeat Exercise 5 when all the shirts in the clothing store are 20% off.
Answer:

Question 7.
Which data display best represents the data, a histogram or a stem-and-leaf plot? Explain.
15, 21, 18, 10, 12, 11, 17, 18, 16, 12, 20, 12, 17, 16
Answer:

Question 8.
The tables show the battery lives (in hours) of two brands of laptops.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays c 8
a. Make a double box-and-whisker plot that represents the data.
b. Identify the shape of each distribution.
c. Which brand’s battery lives are more spread out? Explain.
d. Compare the distributions using their shapes and appropriate measures of center and variation.
Answer:

Question 9.
The table shows the results of a survey that asked students their preferred method of exercise. Analyze the data and then create a display that best represents the data. Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays c 9
Answer:

Question 10.
You conduct a survey that asks 271 students in your class whether they are attending the class field trip. One hundred twenty-one males respond 92 of which are attending the field trip. Thirty-one females are not attending the field trip.
a. Organize the results in a two-way table. Find and interpret the marginal frequencies.
b. What percent of females are attending the class field trip?
Answer:

Data Analysis and Displays Cumulative Assessment

Question 1.
You ask all the students in your grade whether they have a cell phone. The results are shown in the two-way table. Your friend claims that a greater percent of males in your grade have cell phones than females. Do you support your friend’s claim? Justify your answer.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays ca 1
Answer:

Question 2.
Use the graphs of the functions to answer each question.
a. Are there any values of x greater than 0 where f (x) > h(x)? Explain.
b. Are there any values of x greater than 1 where g(x) > f(x)? Explain.
c. Are there any values of x greater than 0 where g(x) > h(x)? Explain.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays ca 2
Answer:

Question 3.
Classify the shape of each distribution as symmetric, skewed left, or skewed right.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays ca 3
Answer:

Question 4.
Complete the equation so that the solutions of the system of equations are (-2, 4) and (1, -5).
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays ca 4
Answer:

Question 5.
Pair each function with its inverse.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays ca 5
Answer:

Question 6.
The box-and-whisker plot represents the lengths (in minutes) of project presentations at a science fair. Find the interquartile range of the data. What does this represent in the context of the situation?
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays ca 6
A. 7; The middle half of the presentation lengths vary by no more than 7 minutes.
B. 3; The presentation lengths vary by no more than 3 minutes.
C. 3; The middle half of the presentation lengths vary by no more than 3 minutes.
D. 7; The presentation lengths vary by no more than 7 minutes.
Answer:

Question 7.
Scores in a video game can be between 0 and 100. Use the data set shown to fill in a value for x so that each statement is true.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays ca 7
a. When x = ____, the mean of the scores is 45.5.
b. When x = ____, the median of the scores is 47.
c. When x = ____, the mode of the scores is 63.
d. When x = ____, the range of the scores is 71.
Answer:

Question 8.
Select all the numbers that are in the range of the function shown.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays ca 8
Answer:

Question 9.
A traveler walks and takes a shuttle bus to get to a terminal of an airport. The function y = D(x) represents the traveler’s distance (in feet) after x minutes.
Big Ideas Math Algebra 1 Answers Chapter 11 Data Analysis and Displays ca 9
a. Estimate and interpret D(2).
b. Use the graph to find the solution of the equation D(x) = 3500. Explain the meaning of the solution.
c. How long does the traveler wait for the shuttle bus?
d. How far does the traveler ride on the shuttle bus?
e. What is the total distance that the traveler walks before and after riding the shuttle bus?
Answer:

Big Ideas Math Algebra 1 Answers Free PDF | Download BIM Algebra 1 Solution Key in PDF format

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