Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts

Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts

Do you want practical learning with real-time examples? Then, refer to Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts. You can easily enhance your skills by practicing the problems from Big Ideas Math Answer Key, Grade 5 Chapter 1 Place Value Concepts. Be the first to access Big Ideas Grade 5 Chapter 1 Math Answers PDF to start your practice. Check out each topic available on the below math answers. Every topic is given individually along with answers and explanations. It is easy to become a topper in the exam by practicing with the Big Ideas Grade 5 Math Answers Chapter 1 Place Value Concepts.

Big Ideas 5th Grade Chapter 1 Place Value Concepts Math Book Answer Key

Students can learn the quick way to solve problems using Big Ideas Grade 5 Chapter 1 Math Answers. Download Grade 5 Big Ideas Math Answers Chapter 1 Place Value Concepts for free. We provided solutions in an easy manner so that students can solve the problems in less time. Click on the links provided below and find every topic individually. Get the free pdf offline and practice whenever you want it.

Lesson 1: Place Value Patterns

Lesson 2 Place Value with Whole Numbers

Lesson 3 Patterns and Powers of 10

Lesson 4 Decimals to Thousandths

Lesson 5 Place Value with Decimals

Lesson 6 Compare Decimals

Lesson 7 Round Decimals

Performance Task

Lesson 1.1 Place Value Patterns

Explore and Grow

Write the whole number represented by each base ten block. Then use the base ten blocks to complete the table.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 1
Answer:

Rod is 10 times as much as Unit

The flat is 10 times as much as Rod

Cube is 3 times as much as Flat

Unit is 1/10 of Rod

Rod is 1/10 of Flat

The flat is 3/10 of Cube

Reasoning
Describe the patterns you see in a number as you move from one place value position to another place value position.

Answer: As we move from one place value position to another value position, The place value of a digit increase by ten times as move from the left.

Think and Grow: Place Value Patterns

You can use a place value chart to help write numbers that are 10 times as great as a number or \(\frac{1}{10}\) of a number.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 2

Show and Grow

Question 1.
Complete the statements.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 3

Apply and Grow: Practice

Use a place value chart to answer the question.
Question 2.
What number is 10 times as great as 6,000?
Answer:60,000

If 6,000 is multiplied by 10 times it becomes 60,000

Question 3.
What number is \(\frac{1}{10}\) of 300?
Answer:30

300 x 1/ 10 is 30

Question 4.
80 is 10 times as great as what number?
Answer: 8

if 8 is multiplied by 10 times it becomes 80

Question 5.
40,000 is \(\frac{1}{10}\) of what number?
Answer:4,00,000

4,00,000 x 1/10 is 40,000

The number is 4,00,000

Complete the table.
Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 4
Answer:

1. 100 ,1 ,

10,  10 times as number is 10 x 10 = 100

1/10 of 10 is 1

2. 4000,40

400, 10 times as number is 400 x 10 = 4000

1/10 of 400 is 40

3. 70,000,700

7,000, 10 times as number is 7,000 x 10 =70,000

1/10 of 7,000 is 700

4. 500,000,    5,000

50,000,10 times as number is 50,000 x 10 = 5,00,000

1/10 of 50,000 is 5,000

Question 7.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 5
Answer:

1. 2,000, 20

200,  10 times as number is 200 x 10 = 2,000

1/10 of 200 is 20

2. 3,00,000,   3,000

30,000,  10 times as number is 30,000 x 10 = 3,00,000

1/10 of  30,000 is 3,000

3. 900, 9

90, 10 times as number is 90 x 10 = 900

1/10 of 90 is 9

4. 80,000, 800

8,000,  10 times as number is 8,0,0 x 10 = 80,000

1/10 of 8,000 is 800

Question 8.
Patterns
Describe the relationship between any place value position and the next greater place value position.
Answer:

The relation between any place value position and next greater place value position increases ten times as we move.

Number Sense
Write whether the statement is true or false. If false, explain why.
Question 9.
600 is 100 times as great as 60,000.
______
Answer: True

600 X 100 = 60,000

Question 10.
9,000 is 1,000 times as great as 9.

_____
Answer: True

9 X 1,000 = 9,000

Think and Grow: Modeling Real Life

Example
Which state is about 10 times larger than Georgia?
The approximate land area of Georgia is 60,000 square miles.
Use a place value chart to find the number that is 10 times as great as 60,000.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 6
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 7
6,000_____ is 10 times as great as 60,000.
The land area of _____Alaska__ is about 600,000 square miles.
So, __Alaska____ is about 10 times larger than Georgia.

Show and Grow

Use the table above.
Question 11.
Which state is about 10 times larger than Hawaii?
Answer: Georgia is 10 times larger than Hawaii

Question 12.
Which state is about \(\frac{1}{10}\) the size of Wyoming?
Answer: Maryland is 1/10 the size of Wyoming

Question 13.
DIG DEEPER!
Which state is about 100 times larger than the District of Columbia?
Answer: Hawaii

District of Columbia = 60

100 times larger than the District of Columbia 60 X 100 = 6,000 is Hawaii

Question 14.
DIG DEEPER!
A mother rhinoceros weighs 2 tons. Her baby weighs \(\frac{1}{10}\) as much as her. What is the weight of the baby rhinoceros, in pounds.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 8
Answer: The weight of the baby rhinoceros is 20 Pounds.

2 tons Mother

Baby = 200 x 1/10 = 20 Pounds

Place Value Patterns Homework & Practice 1.1

Complete the statements.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 9
Question 1.
______ is 10 times as great as 2,000.
Answer:   20,000

2,000 x 10 = 20,000

Question 2.
_____ is \(\frac{1}{10}\) of 2,000.
Answer:      200

2,000 x 1/ 10 = 200

Use a place value chart to answer the question.
Question 3.
What number is 10 times as great as 50?
Answer:  500

50 x 10 = 500

Question 4.
What number is \(\frac{1}{10}\) of 4,000?
Answer:  400

4,000 x 1/10 = 400

Question 5.
800 is \(\frac{1}{10}\) of what number?
Answer:8,000

8,000 x 1/10 = 800

Question 6.
60,000 is 10 times as great as what number?
Answer:6,000

6,000 is 10 times means 6,000 x 10 =60,000

Complete the table.
Question 7.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 10
Answer:

1. 400, 4

40, 10 times as number is 40 x 10 = 400

1/10 of 40 is 4

2. 5,000, 50

500, 10 times as number is 500 x 10 = 5,000

1/10 of 500 is 50

3. 10,000  ,  100

1,000, 10 times as number is 1,000 x 10 = 10,000

1/10 of 1,000 is 100

4.8,00,000   , 8,000

80,000, 10 times as number is 80,000 x 10 = 8,00,000

1/10 of 80,000 is 8,000

Question 8.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 11
Answer:

1. 7,00,000   ,   7,000

70,000, 10 times as number is 70,000 x 10 = 7,00,000

1/10 of 70,000 is 7,000

2. 200 ,  2

20, 10 times as number is 20 x 10 = 200

1/10 of 20 is 2

3. 30,000, 300

3,000, 10 times as number is 3,000 x 10 = 30,000

1/10 of 3000 is 300

4. 1,000 ,  10

100, 10 times as number is 100 x 10 = 1,000

1/10 of 100 is 10

Question 9.
Patterns
Describe the relationship between any place value position and the next lesser place value position.

Answer:

The relation between any place value position and next lesser place value position decreases ten times as we move.

Question 10.
YOU BE THE TEACHER
Your friend says 6,700 is \(\frac{1}{10}\) of 67,000. Is your friend correct? Explain.
Answer: Yes My friend is correct, because if we divide 67,000 by 10 we get the result as 6,700 only.

Use the table.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.1 12
Question 11.
Modeling Real Life
Which city’s population is about 10 times the population of Newark?
Answer: Population of  Oakland.

Newark=40,000

40,000 x 10 = 4,00,000 is Oakland

Question 12.
Modeling Real Life
Which city’s population is about \(\frac{1}{10}\) the population of Marina?
Answer: Population of Del Ray Oaks.

Marina = 20,000

20,000 x 1/10 = 2,000=Del Ray Oaks

Question 13.
DIG DEEPER!
An archaeologist finds a ceramic bowl that is about 400,000 years old. He finds different artifact that is \(\frac{1}{100}\) times as old as the 100ceramic bowl. How much older is the ceramic bowl than the other artifact?
Answer: The Ceramic bowl is 4000 years old.

4,00,000 x 1/100= 4,000

Review & Refresh

Find the factor pairs for the number.
Question 14.
9
Answer: 1,3,9 are factor pairs of 9

Question 15.
24
Answer:  (1,24), (2,12) (3,8) and (4,6) are factor pairs of 24

Question 16.
15
Answer:(1,3,5,15) are factor pairs of 15

Lesson 1.2 Place Value with Whole Numbers

Model the number. Draw your model.
Then write the value of each digit.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 1
Compare the values of the 4s.
Answer: Thousands-4, Hundreds -4 , Tens – 4 and Ones- 2

4 is in thousand, 4 is in Hundreds and 4 is at tens value.

Repeated Reasoning
Is the value of the 4 in the tens place 10 times as much as the value of the 2 in the ones place? Explain.
Answer: No, Why means 4 in tens place means its value is 40 and 2 in ones place means only 2so 4 in the tens place is not 10 times as the value of 2.

Think and Grow: Place Value with Whole Numbers

Key Idea
A place value chart shows the value of each digit in a number. It also shows how the place values are grouped. Each group of three digits is called a period. Ina number, periods are separated by commas.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 2
Example
Write the number in standard form, word form, and expanded form.
Standard form:4,66,900
Word form: Four hundreds sixty six thousand and nine hundred
Expanded form:
4 × 1,00,000 + __6___ × 10,000_____ + ___6__ × _1,000_____ + _9____ × __100____

Show and Grow

Write the number in two other forms.
Question 1.
Standard form: 78,300
Word form:
Expanded form:
Answer:

Word Form : Seventy -Eight Thousand, Three Hundred.

Expanded Form : 7 x 10,000 + 8 x 1000 + 3 x 100=78,300

Question 2.
Standard form:
Word form: three hundred fifty thousand, fifty-eight
Expanded form:
Answer:

Standard Form : 300,50,058

Expanded Form : 3 x 10,00,000+ 5 x 10,000 + 5 x 10 +8=300,50,058

Question 3.
Compare the values of the 6s in the number 466,900.
Answer:

the values of the 6 s are one is in 60 thousand’s place and other is at 6 thousands place.

Apply and Grow: Practice

Write the number in two other forms.
Question 4.
Standard form:
Word form:
Expanded form: 6 × 100,000 + 8 × 1,000 + 4 × 100 + 5 × 10 + 9 × 1
Answer:

Standard Form : 6,08,459

Word Form : 6 hundred / six lakh, eight thousand , four hundred  fifty nine.

Question 5.
Standard form: 45,006,702
Word form:
Expanded form:

Answer:

Word form: forty five lakh ,six thousand seven hundred and two.

Expanded form:4 x 10000000 + 5 x 1000000 + 6 x 1000 +

7 x 100 +2

Question 6.
Compare the values of the 7s in the number 4,877,034.
Answer:

The value’s of 7s is at seventy thousand,(70,000) and again at  seven thousand(7,000).

Question 7.
Compare the values of the 3s in the number 5,338.
Answer:

The values of 3s is at 3-Hundred(300) and at thirty (30)[3 tens)]

or

3 at hundreds and 3 at tens place.

Compare.
Question 8.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 3
Answer:

In 8,046 The value of 4 is in Ten’s place and 6 is in one’ s place

and in 8,460 the value of 4 is in hundreds place and 6 in ten’s place.

Question 9.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 4
Answer:

In 28,517 the value of 2 is at twenty thousand place, 8 at thousands and 5 at hundreds, 1 at tens and 7 at one’s place.

and 28,509 the value of 2 is at twenty thousand place, 8 at thousands and 5 at hundreds, zero at tens and 9 at one’s place.

Question 10.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 5

Answer:

In 5,854,331- Fifty Lakhs , 8 at lakhs, Five at ten thousand, 4 at thousand,3 at hundred, 3 at ten’s and 1 at one’s place.

and in 5,854,231 – Fifty Lakhs , 8 at lakhs, Five at ten thousand, 4 at thousand,2 at hundred, 3 at ten’s and 1 at one’s place.

Question 11.
The white truffle is the world’s most expensive edible fungus, which costs up to three thousand dollars per kilogram. Write this number in standard form.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 6
Answer:

Standard Form: 3000$ per Kg.

the white truffle is 3000$ per kg.

Question 12.
YOU BE THE TEACHER
Your friend says that in the number 45,951, one 5 is 10 times as great as the other 5. Is your friend correct? Explain.
Answer:

Yes,  because  at first the 5 is at tens place and in next time 5 is at thousands place so friend is right 5 is 10 times greater as the other 5 . As we move from right value to left twice tens value place becomes thousand value place.

Question 13.
Logic
Newton is thinking of a 6-digit number in which all of the digits are the same. The value of the digit in the thousands place is 8,000. What is Newton’s number?
Answer:

8,88,888

8 x 1,00,000 + 8 x 10,000+8 x 1,000+8 x 100 + 8 x 10 + 8 x 1

Think and Grow: Modeling Real Life

Example
Compare the values of the 3sin Jupiter’s average distance from the Sun.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 7
Use a place value chart to help you find the value of each 3.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 8
Each place value is 10 times as great as the place value to its right. The digits are two places apart. So, multiply 30,000 by 10 × 10 = 100.
So, the value of the 3 in the millions place is _______ times the value of the 3 in the ten thousands place.

Show and Grow

Use the table above.
Question 14.
Compare the values of the 7s in Mars’s average distance from the Sun.
Answer:

the value of 7 s in first is in thousands place and next 7 s is at hundreds place, the digits are one place apart, so the value of first 7 is in thousands place is 10 times the value of the next 7s in hundreds place.

Question 15.
Compare the values of the 4s in Saturn’s average distance from the Sun.
Answer:

the value of 4 s in first is in four hundred thousands place and next 4 s is at hundreds place, the digits are two places apart, so the value of first  4 is in hundred thousands place is 100 times the value of the next 4s in hundreds place.

Question 16.
DIG DEEPER!
An organization wants to donate all of the money raised through fund raisers and raffles to a children’s charity. Complete the donation check.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 9
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 10
Answer:

1.

$ 194,918 + $ 35,187 = $ 230,005

Word Form :  Two Hundred, Thirty thousand and Five dollars.

Standard Form : $ 230,005

Place Value with Whole Numbers Homework & Practice 1.2

Write the value of the underlined digit.
Question 1.
740,225
Answer:

4 is at Forty thousand place

Question 2.
604,197,872
Answer:

6 is at Six Hundred Lakh place

Question 3.
12,405,287
Answer:

2  is at  twenty Lakh or twenty hundred thousand place

Question 4.
392,183
Answer:

3 is at 3 lakhs or 3 hundred thousand place

9  is at ninety thousand place

2 is at 2 thousand place

1 is at one hundred place

8 is at eighty place

3 is at ones place place

Write the number in two other forms.
Question 5.
Standard form: 450,014
Word form:
Expanded form:
Answer:

Word Form : 4 lakhs or 4 hundred thousand ,  fifty thousand and fourteen

Expanded  Form : 4 x 1,00,000 +5 x 10000 + 1 x 10 + 4

Question 6.
Standard form:
Word form: fourteen thousand, two hundred one
Expanded form:
Answer:

Standard Form : 14,201

Expanded Form : 1 X 10000 + 4 X 1000 + 2 X 100 +1

Question 7.
Compare the values of the 9s in the number 537,499.
Answer:

First 9 is at ones place, and second 9 is at tens place.

Question 8.
Compare the values of the 5s in the number 78,550,634.
Answer:

First 5 is at Ten Thousands place,

Second 5 is at Five Hundred Thousands place.

Compare.
Question 9.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 11
Answer:

67,893 < 67,943

6 at sixty thousand place ,6 at sixty thousand place

7 at thousand place, 7 at thousand place

8 at eight hundred place,9 at nine hundred is great

9 at tens place and 4 at four at tens place is less

and 3 at ones place and 3 at ones place is same

Question 10.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 12
Answer:

450,823 > 405,823

4 at Four hundred thousand,

5 at fifty thousand and 0 is smaller at ten thousands place

0 at thousands place is smaller than 5 at thousands place

8  at hundreds place is same as 8 at hundreds place

2 at tens place is same as 2 at tens place

and 3 at ones place is same as 3 at ones place

Question 11.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 13
Answer:

176,994 = 176,994

1 at one hundred thousand, is same at 1 at One Hundred Thousand place

7 at Seventy thousand is same at 7 at Seventy Thousand place

6 at six thousand is same at 6 at six thousand place

9 at hundreds place is same at 9 at hundreds place

9 at ninety or 9 tens place is same as 9 at Ninety or 9 Tens place

and 4 at ones place is same as 4 at Ones place

Question 12.

Your body contains about 60,000 miles of blood vessels. Write this number in word form.
Answer:

60,000 miles of blood vessels in  Word Form : Sixty Thousand miles.

Question 13.
Which One Doesn’t Belong?
Which number does belong with the other three?
1 × 10,000 + 4 × 1,000 + 2 × 100 + 6 × 1 fourteen thousand, two hundred six
140,206
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 14
Answer:

No number is repeated, so no number belongs to other

one at tens place

4 at ones place

2 at hundredths place

and 6 at thousandths place.

Question 14.
DIG DEEPER!
Find the difference in the values of the underlined digits.
856,092 37,841
Answer:

8 is at 80 Million,

and

8 is at hundreds value

the difference is 8 X 1,00,00,000 times the other  8 value.

Use the table.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 15
Question 15.
Modeling Real Life
Compare the values of the 3s in the music cost.
Answer: 3 is at 3 millions place and other 3 is at 3 hundreds place.

Question 16.
Modeling Real Life
Compare the values of the 2s in the cost.
Answer:

Costs Places

In cast 2 is at two hundred thousands place and 2 at thousands place.

In Director 2 is at 2 millions place and 2 at thousands place.

In Editing 2 value is not there.

In Music also 2 is not there

In Producers 2 value is at 2 millions place

In Visual effects 2 place is first at thousands place and 2 is in ones place

Question 17.
DIG DEEPER!
What is the total cost for the director and producers? Write your answer in word form.
Answer: Total cost of director-$ 2,712,480+ and cost of roducer –

$ 2,759,981 is $ 5,472,461

Word Form : five million, four hundred thousand, seventy two thousand, four hundred sixty one dollars.

Review & Refresh

Compare
Question 18.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 16

Answer:

0.14 < 0.15

0.14-0+ 1 x 1/10+4/100

0 is at ones place ,1 is at 1/10 place, and 4 is at 1 /100 place is small

0.15-0+1×1/10+5/100

0 is at ones place ,1 is at 1/10 place, and 5 is at 1 / 100 place

Question 19.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 17

Answer:

2.2 = 2.20

2.2-2 is at ones place and .2 is at 2/10 place

2.20- 2 is at ones place ,.2 is at 2/10 place is same

Question 20.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.2 18

Answer:

5.8 > 5.08

5.8 – 5 is at ones place and .8 is at 1/10 place

5.08  – 5 is at ones place and .8 is at 1/100 place  is small

Lesson 1.3 Patterns and Powers of 10

Explore and Grow

Write a multiplication expression to answer each question.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 1
How many units are in 1 rod?

How many units are in 10 rods?

How many units are in 100 rods?

How many units are in 1,000 rods?
Answer:

In 1 rod its 1 x 100 =100

In 10 rods its 1 x 10 = 101

In 100 rods its 1 x 10 x 10 = 102

In 1,000 rods it is 1 x 10 x 10 x 10 =103

Repeated Reasoning
How many tens are in 100? in 1,000? in 10,000?
Answer: In 100 its 10 tens, in 1,000 its 100 tens and in 10,000 its 1000 tens

Think and Grow: Patterns and Powers of 10

Key Idea
A power is a product of repeated factors. The base of a power is the repeated factor. The exponent of a power gives the number of times the base is used as a factor.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 2

Example
Find the value of 4 × 103.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 3
Multiply 4 by powers of 10. Look for a pattern.
4 × 101 = 4 × 10 = _40____
4 × 102 = 4 × 10 × 10 = _400_____
4 × 103 = 4 × 10 × 10 × 10 = __4,000____
So, 4 × 103 = ____4,00,000_.
Notice the pattern: In each product, the number of zeros after 4 is equal to the exponent.

Show and Grow

Question 1.
Identify the base, exponent, and power for the expression 106.
Answer:

base : 10,

exponent: 6

power :106= 10 x 10 x 10 x 10 x 10 x 10

Question 2.
Write the product 10 × 10 × 10 × 10 as a power.

Answer:

10 x 10 x 10 x 10  as a power is 10 4

Question 3.
Find the value of 5 × 102.
Answer:

5 × 102 is 

5 x 10 x 10 = 500

Apply and Grow: Practice

Find each product. Use patterns to help.
Question 4.
2 × 10 = _____
2 × 100 = _____
2 × 1,000 = _____
2 × 10,000 = ____
Answer:

2 x 10 =20

2 x 100=200

2 x 1,000= 2,000

2 x 10,000 = 20,000

Question 5.
9 × 10 = _____
9 × 100 = _____
9 × 1,000 = _____
9 × 10,000 = ____
Answer:

9 x 10 = 90

9 x 100 = 900

9 x 1,000= 9,000

9 x 10,000 = 90,000

Question 6.
5 × 10 = _____
5 × 100 = _____
5 × 1,000 = _____
5 × 10,000 = ____
Answer:

5 x 10 = 50

5 x 100= 500

5 x 1,000 = 5,000

5 x 10,000 = 50,000

Find the value of the expression.
Question 7.
104
Answer:

104= 10 x 10 x 10 x 10 = 10,000

Question 8.
6 × 105
Answer:

6 × 105 = 6 x 10 x 10 x 10 x 10 x 10 = 6,00,000

Question 9.
7 × 102
Answer:

7 × 102 = 7 x 10 x 10 = 700

Question 10.
5 ×10 4
Answer:

5 ×10 4= 5 x 10 x 10 x 10 x 10 = 50,000

Rewrite the number as a whole number multiplied by a power of 10.
Question 11.
20,000
Answer:

20,000=2 x 104

Question 12.
500
Answer:

500=5 X 102

Question 13.
900,000
Answer:

9,00,000=9 x 105

Number Sense
Write the number in expanded form using exponents.
Question 14.
53,124
(5 × 104) + ______
Answer:

53,214=(5 × 104) +( 3 x 103)+(1 x 102) + (2 x101)+ (4 x100)

Question 15.
8624
(8 × 102) + _______
Answer:

8624=(8 x 103 )+(6 x 102) +(2 x 101)+(4×100)

Question 16.
DIG DEEPER!
Your friend writes (3 × 104) + (5 × 103) + (2 × 102) + 4 as the expanded form of thirty-five thousand, twenty-four. Explain what your friend did wrong.
Answer:

(3 × 104) is 3 x 10 x 10 x 10 x 10 = 30,000 is Thirty Thousand

(5 × 103) is 5 x 10 x 10 x 10 = 5,000 is five thousand

(2 × 102) is 2 x 10 x 10 = 200 is two hundred not twenty

and 4 is four,

it is  thirty-five thousand, two hundred and four= 35,204

not thirty-five thousand, twenty-four ≠ 35,024

35,204 is right

35,024 is wrong

Think and Grow: Modeling Real Life

Example
Newton and Descartes are running for mayor. How many people voted in the election?
Find the number of votes for each candidate.
Newton: 105 = ______
Descartes: 9 × 104 = ______
Add the votes for Newton and Descartes.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 4
______ people voted in the election.

Newton: 105 =  10 x 10 x 10 x 10 x 10 is 1,00,000

Descartes: 9 × 104 = 9 x 10 x 10 x 10 x 10 is 90,000

so total people voted is 1,00,00 + 90,000 = 1,90,000

 

 1,00,000
 + 90,000
=1,90,000

1,90,000 Voted in the election.

Show and Grow

Question 17.
A surf shop has been in business for two years. What are the total sales for Year 1 and Year 2 combined?
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 5
Answer:

Year 1 $ 104 = 10 x 10 x 10 x 10 =10,000

Year 2 $ 6 x 105= 6 x 10 x 10 x 10 x 10 x 10 = 6,00,000

       10,000
 + 6,00,000
=  6,10,000

there fore total sales for Year 1 and Year 2 combined is  is 6,10,000 .

Question 18.
Which migration is farther? About how much farther is it?
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 6
Answer:

Chinook Salmon about 4 x 103km

4 X 10 x 10 x 10 = 4000 km

and

Leatherback Turtle about 2 X 104 km

2 x 10 x 10 x 10 x 10 = 20,000 km

So Leatherback Turtle is  farther and

20,000 km – 4000 km = 16,000 km

it is 16,000 km farther

Question 9.
DIG DEEPER!
A human has about 104 taste buds. A cow has about 3 times as many taste buds as a human. About how many taste buds does a cow have? Write your answer as a whole number multiplied by a power of 10.
Answer:

(104) 3 time means (104)

as per the law powers are multiplied 4 x 3

(104×3) = (1012)

(104×3)= 1 X (104) x (104)  x (104)

(1012)= 1 x (10 x 10 x 10 x 10 ) x (10 x 10 x 10 x 10 ) x (10  x 10 x 10 x 10)

Patterns and Powers of 10 Homework & Practice 1.3

Question 1.
Identify the base, exponent, and power for the expression 103.
Answer:

base: 10

exponent:  3

power : 103= 10 x 10 x 10

Question 2.
Write 10 × 10 × 10 × 10 a power.
Answer:

10 x 10 x 10 x 10 = 104

Find each product. Use patterns to help.
Question 3.
6 × 10 = ______
6 × 100 = _____
6 × 1,000 = _____
6 × 10,000 = _____
Answer:

6 x 10=6 × 101 = 60

6 x 100 =6 × 102 = 600

6 X 1,000=6 × 103 =  6,000

6 x 10,000 = 6 × 104= 60,000

Question 4.
8 × 10 = ______
8 × 100 = _____
8 × 1,000 = _____
8 × 10,000 = _____
Answer:

8 x 10=8 × 101 = 80

8 x 100 =8 × 102 = 800

8 X 1,000=8 × 103 = 8,000

8 x 10,000 = 8 × 104= 80,000

Question 5.
4 × 10 = ______
4 × 100 = _____
4 × 1,000 = _____
4 × 10,000 = _____
Answer:

4 x 10=4 × 101 = 40

4 x 100 =4 × 102 = 400

4 X 1,000= 4× 103 = 4,000

4 x 10,000 = 4 × 104= 40,000

Find the value of the expression.
Question 6.
103
Answer:

103 = 10 x 10 x 10 = 1,000

Question 7.
2 × 104
Answer:

2 × 104 = 2 x 10 x 10 x 10 x 10 = 20,000

Question 8.
9 × 105
Answer:

9 × 105   =  9 x 10 x 10 x 10 x 10 x 10 = 9,00,000

Question 9.
3 × 102
Answer:

3 × 102  = 3 x 10 x 10 = 300

Rewrite the number as a whole number multiplied by a power of 10.
Question 10.
100,000
Answer:

100,000 = 1 x 10 x 10 x 10 x 10 x 10 = 1 × 105

Question 11.
70
Answer:

70 = 7 x 10 = 7 x 101

Question 12.
6,000
Answer:

6,000 = 6 x 10 x 10 x 10 = 6 x 10

Number Sense
Write the number in standard form.
Question 13.
(3 × 102) + (8 × 101)
Answer:

(3 × 102) + (8 × 101)  = 3 x 10 x 10 + 8 x 10

3 x 100 + 80

300+ 80 = 380

Question 14.
(2 × 103) + (5 × 102) + (4 × 101)
Answer:

(2 × 103) + (5 × 102) + (4 × 101) =  2 x 10 x 10 x 10 + 5 x 10 x 10 + 4 x 10

2,000 + 500 + 40

2,540

Question 15.
YOU BE THE TEACHER
Newton 6 says 106 = 10 × 6. Is he correct? Explain
Answer:

No he is wrong because  106  ≠  10 × 6 it is 10 is to be multiplied by  6 times ,

106 = 10 x 10 x 10 x 10 x 10 x 10=10,00,000 is correct

Question 16.
Which One Doesn’t Belong?
Which one does not belong with the other three?
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 7
Answer:

Both are different blocks , in first block it is  square and next it is cube.  

Question 17.
Modeling Real Life
Each student at an elementary school votes once on this year’s field trip. How many students voted in all?
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 8
Answer:

Aquarium = 7 x 102 =   7 x 10 x 10 = 700

Amusement Park = 103  = 10 x 10 x 10 =  1000

700 + 1000 = 1,700

So total number of students voted are =  1,700.

Question 18.
Modeling Real Life
On which day did more people attend the event? How many more people?
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 9
Answer:

Friday = 10 x 10 x 10 = 1,000

Saturday = 5 x 102 = 5 x 10 x 10 = 500

on Friday more people attend the event

and more are 1,000 – 500 = 500 , s0 500 more people attended.

Review & Refresh

Divide. Then check your answer
Question 19.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 10
Answer:

Step 1) Start by dividing 65 by 6 to get the decimal answer as illustrated below. Note that we round the answer if necessary, but don’t worry, the final answer will still be exact.

65 / 6 = 10.83

Step 2) Next we take the Whole part of the answer in Step 1 and multiply it by the Divisor. As you can see, it does not matter if we rounded in the previous step because the Decimal part is ignored. Furthermore, the Divisor in 65 divided by 6 is 6. Thus, the Whole multiplied by the Divisor is:

10 x 6 = 60

Step 3) Finally, we will subtract the answer in Step 2 from the Dividend to get the answer. The Dividend in 65 divided by 6 is 65. Thus, our final calculation to get the answer is:

65 – 60 = 5

the answer is 5.

Question 20.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 11
Answer:

Step 1) Start by dividing 50 by 4 to get the decimal answer as illustrated below. Note that we round the answer if necessary, but don’t worry, the final answer will still be exact.

50 / 4 = 12.50

Step 2) Next we take the Whole part of the answer in Step 1 and multiply it by the Divisor. As you can see, it does not matter if we rounded in the previous step because the Decimal part is ignored. Furthermore, the Divisor in 50 divided by 4 is 4. Thus, the Whole multiplied by the Divisor is:

12 x 4 = 48

Step 3) Finally, we will subtract the answer in Step 2 from the Dividend to get the answer. The Dividend in 50 divided by 4 is 50. Thus, our final calculation to get the answer is:

50 – 48 = 2

the answer is 2.

Question 21.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.3 12
Answer:

Step 1) Start by dividing 45 by 3 to get the decimal answer as illustrated below. Note that we round the answer if necessary, but don’t worry, the final answer will still be exact.

45 / 3 = 15

the answer is 0.

Lesson 1.4 Decimals to Thousandths

Explore and Grow

Divide the square into10 equal parts. Shade one part. What part of the whole is shaded?
Fraction: Decimal:

Divide each of the 10 parts into 10 equal parts. Shade one part using a different color. What part of the whole is shaded with the second color?
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 1
Fraction: Decimal:

If you divide each of the 100 equal parts into10 equal parts, how many parts will the model have?

If you shade one of those parts, what part of the whole is shaded?
Fraction: Decimal:
Answer:

the model will have 10 parts , only 1/ 10th part is shaded, 0.1

Structure
Compare the number of hundredths to the number of tenths. Compare the number of hundredths to the number of thousandths. What do you notice?
Answer:

number of hundredths to the number of tenths is 10 to 100 , 10/100=1/10 = 0.1

number of hundredths to the number of thousandths is

100 to 1000, 100/1000 =1/10= 0.1

both has equal values 1/10 = 0.1

the number of hundredths to the number of tenths is equal to the number of hundredths to the number of thousandths

Think and Grow: Thousandths

Key Idea
In a decimal, the third place to the right of the decimal point is the thousandths place. You can write thousandths as fractions or decimals.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 2
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 3

Show and Grow

Write the decimal as a fraction.
Question 1.
0.009

Answer:

0.009= 9 / 1,000 = 9 x 1/1,000

Question 2.
0.063
Answer:

0.063 = 63 / 1,000= 63 x 1/1,000

Question 3.
0.194
Answer:

0.194 = 194 / 1,000= 194 x 1/1,000

Write the fraction as a decimal
Question 4.
\(\frac{3}{1,000}\)
Answer:

\(\frac{3}{1,000}\)= 3 x 1/ 1,000=  0.003

Question 5.
\(\frac{91}{1,000}\)
Answer:

\(\frac{91}{1,000}\)= 91 x 1/ 1,000 = 0.091

Question 6.
\(\frac{607}{1,000}\)
Answer:

\(\frac{607}{1,000}\)= 607 x 1/1,000= 0.607

Apply and Grow: Practice

Write the decimal as a fraction.
Question 7.
0.645
Answer:

0.645= 645x 1/1,000= 645/1,000

Question 8.
0.002
Answer:

0.002=2 x 1/1,000=  2/1,000

Question 9.
0.98
Answer:

0.98= 98 x 1/ 1,000= 98/1,000

Question 10.
0.6
Answer:

0.6 = 6 x 1/10= 0.6/10

Write the fraction as a decimal.
Question 11.
\(\frac{884}{1,000}\)
Answer:

\(\frac{884}{1,000}\)= 884x 1/1,000= 0.884

Question 12.
\(\frac{8}{1,000}\)
Answer:

\(\frac{8}{1,000}\)= 8 x 1/1,000= 0.008

Question 13.
\(\frac{39}{100}\)
Answer:

\(\frac{39}{100}\)= 39 x 1/ 100= 0.39

Question 14.
\(\frac{1}{10}\)
Answer:

\(\frac{1}{10}\)= 1 x 1/10 = 0.1

Question 15.
0.4 is \(\frac{1}{10}\) of what number?
Answer:

\(\frac{4}{10}\) is 0.4

Question 16.
0.52 is 10 times as great as what number?
Answer:

0.52 is 10 times as great as 0.052

Question 17.
You use 47 of the cotton balls for an art project. What portion of the bag of cotton balls do you use? Write your answer as a decimal.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 4
Answer:

47/100= 47 x 1/100= 0.47

Question 18.
Which One Doesn’t Belong?
Which number does not belong with the other three?
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 5
Answer:

0.29 does not belongs to other three

Question 19.
YOU BE THE TEACHER
Your friend says the value of the 7 in the hundredths place of 0.877 is 10 times as great as the 7 in the thousandths place. Is your friend correct? Explain.
Answer:

Yes , because the value of 7 in the hundredths place as compared is 10 times as great as the 7 in the thousands place.

Question 20.
Write each fraction as a decimal. What do you notice?
\(\frac{4}{10}\), \(\frac{40}{100}\) and \(\frac{400}{1,000}\)
Answer:

\(\frac{4}{10}\)=0.4

\(\frac{40}{100}\) =0.4

\(\frac{400}{1,000}\)=0.4

number of tenths, tenths number of hundredths and hundredths number of thousandths are same.

Think and Grow: Modeling Real Life

Example
You put together 156 pieces of the puzzle before lunch and 148 pieces of the puzzle after lunch. What portion of the puzzle did you put together? Write your answer as a decimal.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 6
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 7
Write the fraction as a decimal.
You put together ____0.304__ of the puzzle.

Show and Grow

Question 21.
You make flash cards out of index cards. You use 50 index cards for social studies and 25 index cards for science. What portion of the pack of index cards do you use? Write your answer as a decimal.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 8
Answer:

50 for social studies, 25 for science total is 50 + 25 = 75, total number of flash cards is 1,000 ,

75 / 1,000 or 75 by 1,000 or 75 x 1/1,000= 0.075

Question 22.
There are 458 knock-knock jokes in the book. not What fraction of the jokes in the book are knock-knock jokes?
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 9
Answer:

458/1,000 or 458 x 1/1,000 of jokes books in knock- knock jokes.

Question 23.
DIG DEEPER!
A newly hatched caterpillar was 0.02 inches long. After 2 weeks, the caterpillar grew 10 times as long as its length when it hatched. After another 2 weeks, the caterpillar grew 10 times as long as its length after 2 weeks. How long is the caterpillar now?
Answer: First week it is 0.02 inches,

in two weeks -2 weeks-0.02 x 10 = 0.2 inches

again after 2 weeks -0.2 x 10 = 2 inches

Decimals to Thousandths Homework & Practice 1.4

Write the decimal as a fraction.
Question 1.
0.735
Answer:

0.735= 735 / 1,000= 735 x 1/1,000

Question 2.
0.051
Answer:

0.051= 51 / 1,000= 51 x 1/1,000

Question 3.
0.804
Answer:

0.804 = 804 / 1,000= 804 x 1/ 1,000

Question 4.
0.2
Answer:

0.2 = 2 / 10= 2 x 1/10

Write the fraction as a decimal.
Question 5.
\(\frac{98}{1,000}\)
Answer:

\(\frac{98}{1,000}\)=98 x 1/1,000= 0.098

Question 6.
\(\frac{67}{100}\)
Answer:

\(\frac{67}{100}\)= 67 x 1/100=0.67

Question 7.
\(\frac{4}{100}\)
Answer:

\(\frac{4}{100}\)= 4 x 1/100=0.04

Question 8.
\(\frac{9}{10}\)
Answer:

\(\frac{9}{10}\)= 9 x 1/10= 0.9

Question 9.
0.08 is 10 times as great as what number?
Answer:

0.008

if 0.008 is multiplied by 10 times it becomes 0.08

Question 10.
0.001 is \(\frac{1}{10}\) of what number?
Answer:

0.0001

if 0.0001 is \(\frac{1}{10}\) equals to 0.001

Question 11.
YOU BE THE TEACHER
Your friend says that \(\frac{16}{1,000}\) can be written as 0.16. Is your friend correct? Explain.
Answer: No, because if 16 / 1,000 or 16 x 1/1,000 is 0.016 not 0.16,

0.016 ≠ 0.16

so he is incorrect.

Question 12.
Precision
Thirteen unit cubes are taken from the thousand cube. Write a fraction and a decimal to represent how many unit cubes are left.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 10
Answer:

13 units cubes are taken

Total number of units are  1000

left are from 1,000-13/1,000=87/1000

87 / 1,000 and 0.087 cubes are left

Question 13.
DIG DEEPER!
Write the number represented by each point on the number line.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 11
Point X: _____
Point Y: ___
Answer:

on the number line Point X = 7.633 as we move from 7.63 to three places forward

so Point X is 7.633

on the number line Point Y = 7.638 as we move from 7.63 its eighth place forward

and Point Y is 7.638

Question 14.
Modeling Real Life
A restaurant owner has a 1,000-count box of napkins. She puts 125 of the napkins on tables. What portion of the box of napkins does she use for the tables? Write your answer as a decimal.
Answer:

Restaurant has 1,000 count box of napkins and keeps 125 on table so portion of the box she uses is  125/1,000= 125 x 1/ 1,000= 0.125

Question 15.
DIG DEEPER!
Your friend has a recipe book with 1,000 recipes. She wants to try two new recipes each week. What fraction of the recipes in the book will she have tried after 1 year?
Answer:

In a year there are almost 52 weeks. Each week 2 means 2 x 52 =approximately  104 recipes in a year.

so in a year she would have tired 104 / 1,000= 104 x 1/1,000 = 0.104 recipes.

Review & Refresh

Question 16.
Extend the pattern of shapes by repeating the rule “square, octagon, pentagon, octagon ”What is the 48th shape in the pattern?
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1.4 12
Answer:

” square, octagon, pentagon, octagon, hexagon, octagon , heptagon , octagon,

octagon, octagon, nonagon, octagon , decagon, octagon,

11 hendecagon Octagon
12 dodecagon Octagon
13 tridecagon Octagon
14 tetradecagon Octagon
15 pentadecagon Octagon
16 hexadecagon Octagon
17 heptadecagon Octagon
18 octadecagon Octagon
19 enneadecagon Octagon
20 icosagon Octagon
21 icosikaihenagon Octagon
22 icosikaidigon Octagon
23 icosikaitrigon Octagon
24 icosikaitetragon Octagon
25 icosikaipentagon Octagon
26 icosikaihexagon Octagon
27 icosikaiheptagon Octagon
28 icosikaioctagon Octagon
29 icosikaienneagon Octagon
30 triacontagon Octagon
31 triacontakaihenagon Octagon
32 triacontakaidigon Octagon
33 triacontakaitrigon Octagon
34 triacontakaitetragon Octagon
35 triacontakaipentagon Octagon
36 triacontakaihexagon Octagon
37 triacontakaiheptagon Octagon
38 triacontakaioctagon Octagon
39 triacontakaienneagon Octagon
40 tetracontagon Octagon
41 tetracontakaihenagon Octagon
42 tetracontakaidigon Octagon
43 tetracontakaitrigon Octagon
44 tetracontakaitetragon Octagon
45 tetracontakaipentagon Octagon
46 tetracontakaihexagon Octagon
47 tetracontakaiheptagon Octagon
48 tetracontakaioctagon Octagon

Lesson 1.5 Place Value with Decimals

Explore and Grow

Model the number. Draw your model.
Then write the value of each digit.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 1
Answer:

from the model 3.33 =

0nes digit is – 3

tenths digit is 3/10 =0.3

hundredths digit is 3/ 100 = 0.03

Repeated Reasoning
Compare the value of the ones digit to the value of the tenths digit. Then do the same with the tenths and the hundredths digits. Explain why you can use base ten blocks to model ones, tenths, and hundredths.
Answer:

a digit in one place represents one and 10 times more what it represents in the place to its right and that is tenths digit.
Similarly a digit in tenths place represents tenth and 10 times more what it represents in the place to its right and that is hundredths digit.

Each place to the left is 10 times the size of the place to the right, and base 10 blocks are the best way to model ones, tenths, and hundredths.

OPERATIONS WITH DECIMALS. Using Base Ten Blocks to Multiply Decimals Flat = one (1) Long = one tenth (0.1) rod = one hundredth (0.01) - ppt download

Think and Grow: Place Value with Decimals

Key Idea
In a place value chart, whole numbers are to the left of the decimal point. Decimals are to the right of the decimal point.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 2
Standard form:2.557
Word form:” Two and five hundred fifty-seven  thousandths”
Expanded form: 2 × 1 + __5__ ×\(\frac{1}{10}\)____ + 5 × \(\frac{1}{100}\) + _7_ × \(\frac{1}{1000}\)__

Show and Grow

Write the number in two other forms.
Question 1.
Standard form: 0.398
Word form:
Expanded form:

Answer:

Word form: “Three hundred ninety – eight thousandths”
Expanded form: 3 x (1/10) + 9 x (1/100) + 8 x (1/1,000)

Question 2.
Standard form:
Word form: eight and forty-six thousandths
Expanded form:
Answer:

Standard form:8.046

Expanded form: 8 x 1 + (4 /100) + (6 / 1,000)

Question 3.
Compare the values of the 5s in the number 2.557.
Answer: at first 5s place value is at tenths and next its place value is at hundredths .

Apply and Grow: Practice

Write the value of the underlined digit.
Question 4.
0.418
Answer:

4 is at tenths value place

Question 5.
5.296
Answer:

9 is at hundredths value place

Question 6.
3.806
Answer:

8 is at tenths value place

6 is at  thousandths value place

Question 7.
0.547
Answer:

7 is at thousandths value place

Write the number in two other forms.
Question 8.
Standard form:
Word form:
Expanded form: 4 × 1 + 9 × \(\frac{1}{10}\) + 8 × \(\frac{1}{1,000}\)
Answer:

Standard form: 4.908
Word form:” Four and nine hundredth – eight thousandths”

Question 9.
Standard form: 0.125
Word form:
Expanded form:
Answer:

Word form: “one hundred twenty – five thousandths”
Expanded form: 1 x (1/10) + 2 x (1 /100) +5 x (1 / 1,000)

Question 10.
Compare the values of the 4s in the number 0.844.
Answer:

at first 4s place value is at hundredths and next 4s place value is at thousandths

Question 11.
Compare the values of the 3s in the number 3.367.
Answer: at first 3s place value is at ones place and next 3s place value is at tenths place

Question 12.
A pygmy jerboa weighs one hundred thirty-two thousandths pound. Write this number in standard form.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 3
Answer:

The Standard Form of pygmy jerboa’s -one hundred thirty-two thousandths pound weighs is  0.132 pound

Question 13.
Reasoning
Is 9.540 equivalent to 9.54? Explain.
Answer:

Yes. 9.540 is equivalent to 9.54 because at thousandths value its 0, so zero multiplied by any number is zero.

Therefore both are equivalent.

Question 14.
DIG DEEPER!
Write three decimals that are equivalent to 6 × 1 + 4 × \(\frac{1}{10}\) .
Answer:

the three equivalent decimals are of 6 .04 are 6.040, 6.0400, 6.04000

Think and Grow: Modeling Real Life

Example
How do the values of the 3s in the masses of the fruits compare?
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 4
Use a place value chart to help you find the value of each 3.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 5
The value of the 3 in the mass of the tomato is _one_____ .
The value of the 3 in the mass of the chili pepper is __tenths___.
So, the value of the 3 in the mass of the tomato is __10_____ times the value of the 3 in the mass of the chili pepper. Also, the value of the 3 in the mass of the chili pepper is ___1/10__ the value of the 3 in the mass of the tomato.

Show and Grow

Question 15.
Two baseball players have batting averages of 0.358 and 0.345. How do the values of the 5s in the batting averages compare?
Answer:

In 0.358 the place value of 5s is at tenths place and in

0.345 the place value of 5s  is at thousandths place.

Question 16.
The stopwatch shows a runner’sritethe100-meter dash time. Write the time in words.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 6
Answer:

15.76 seconds time in words  is Fifteen and seven tenths and six hundredths

Question 17.
DIG DEEPER!
You exchange 1 U.S. dollar for Australian dollars and 1 U.S. dollar for Kuwaiti dinars. Do you have 10 times as many Australian dollars as Kuwaiti dinars? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 7
Answer:

Yes, Because 1 Australian dollars = 1.302 of 0.302 of Kuwaiti dinars,

if we multiply 0.302 by 10 it becomes 1.302 which is equal to 1 Australian dollars.

[0.302 x 10 = 1.302]

so 1 Australian dollars is 10 times more than the 1  Kuwaiti dinars.

Place Value with Decimals Homework & Practice 1.5

Write the value of the underlined digit.
Question 1.
5.437
Answer:

5 at ones place

4 at tenths place

3 at hundredths place

7 at thousandths place

Question 2.
0.852
Answer:

the underlined digit is 2 at thousandths place,

its value is 2 x 1/1000= 2/1000

Question 3.
0.962
Answer:

the underlined digit is 6 at hundredths place

its value is 6 x 1/100= 6 /100

Question 4.
4.165

the underlined digit is 1 at tenths place

its value is 1x 1/10 = 1/10
Answer:

Write the number in two other forms.
Question 5.
Standard form: 9.267
Word form:
Expanded form:
Answer:

Word form:  nine and two tenths six hundredths seven thousandths.

Expanded form : 9+ 2 x 1/10 + 6 x 1/100 + 7 x 1/1000

Question 6.
Standard form:
Word form: two and forty-three thousandths
Expanded form:
Answer:

Standard form : 0.243

Expanded form: 2 x 1/10+4 x 1/100+3 x 1/1000

Question 7.
Compare the values of the 6s in the number 1.668.
Answer:

first 6s at tenths place and next 6s at hundredths place

Question 8.
Compare the values of the 7s in the number 7.704.
Answer:

first 7s at ones place and next 7s at tenths place

Question 9.
A pygmy possum weighs 0.097 pound. Write this number in word form.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 8
Answer:

pygmy weighs 0.097 pound and its

word form is  nine hundredths and seven thousandths

Questio 10.
Which One Doesn’t Belong?
Which one does not belong with the other three?
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 9
Answer:

no number belongs to other three because 5 is at tenths place

one at hundredths place and 4 at thousandths all are at different places.

Question 11.
Reasoning
Which number cards are equal to the value of the underlined digit?
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 10
Answer:

2 x 1/1,000 , two thousandths and 0.002 are equal to the value of the underlined digit.

Question 12.
Modeling Real Life
How do the values of the 5s in the heights of the plants compare?
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 11
Answer:

In Peace lily the 5s place is at  thousandths

and in Venus flytrap the 5s place is at hundredths

Question 13.
Modeling Real Life
The world’s largest gold nugget is located in Las Vegas, Nevada. It has a mass of about 27.247 kilograms. Write how to say the nugget’s mass in words.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 12
Answer:

the mass of gold nugget in Las Vegas is 27.247 kilograms given

In Words form it is twenty seven and two tenths  four hundredths and  seven thousandths

Review & Refresh

Compare.
Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 13
Answer:

8/10 =0.8

80/100=0.8

both are equal 8/10 = 80/100

Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 14
Answer:

5/8 = 0.625

3/6= 0.5

0.625 > 0.5,

5/8 > 3/ 6, 5/8 is greater than 3/6

Question 16.
Big Ideas Math Answer Key Grade 5 Chapter 1 Place Value Concepts 1.5 15

Answer:

7/2= 3.5

10/8=1.25

so 7/2 is greater than 10/8.

7/2 > 10/8

Lesson 1.6 Compare Decimals

Explore and Grow

Use models to compare the decimals.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 1
Answer:

0.62 >0.26, 0.62 is greater than 0.26, 0.6  is greater than o.2

0.80= 0.8, 0.80 both are  equal

3.5 < 3.55, 3.55 is greater than 3.50

Reasoning
How can you use a place value chart to compare two decimals? Use a place value chart to check your answers above.
Answer:

we use another table to compare with the previous and write the answer.

Think and Grow: Compare Decimals

Example
Compare 3.769 and 3.749.
Use a place value chart. Start at the left. Compare the digits in each place until the digits differ.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 2
The digits in the ones place and the tenths place are the same. Compare the hundredths.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 3

Example
Compare 2.4 and 2.405.
Use place value. Line up the decimal points. Start at the left. Compare the digits in each place until the digits differ.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 4

2.4<2.405

Show and Grow

Compare
Question 1.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 5
Answer:

Ones . Tenths Hundredths Thousandths
9 . 0 6 3
9 . 0 6 7
Same . Same Same Greater

So 9.063 < 9.067

Question 2.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 6
Answer:

Ones . Tenths Hundredths Thousandths
0 . 8 9 0
0 . 8 0 9
Same . Same Greater Greater

So 0.89 > 0.809,

Apply and Grow: Practice
Compare.
Question 3.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 7
Answer:

Ones . Tenths Hundredths Thousandths
8 . 5 3 7
8 . 5 4 1
Same . Same Greater Greater

8.537 < 8.541

Question 4.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 8
Answer:

6.401 < 6.409, since  0.009 is greater than 0.001

Question 5.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 9
Answer:

7.409 > 7.049 since 7.4 is greater than 7.0

Question 6.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 10
Answer:

0.25 = 0.250

Both are equal

Question 7.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 11
Answer:

2.701 >2.700, since 0.001 is greater than 0.000

Question 8.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 12
Answer:

4.006 < 4.61, since 4.6 is greater than 4.0

Question 9.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 13
Answer:

0.041 < 41.6, 41 is greater than 0

Question 10.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 14
Answer:

0.007 < 0.7 as 0.7 is greater than 0.007

Order the decimals from least to greatest.
Question 11.
321.499, 325.499, 321.489
Answer:

321.499, 325.499, 321.489 from least to greatest

as 321.489 is smaller than 321.499 and 321.499 is smaller than 325.499

so 321.489 , 321.499 , 325.499

Question 12.
9.7, 9.64, 9.78
Answer:

9.7, 9.64, 9.78 from least to greatest

9.64 is smaller than 9.7 and  9.7 is smaller than 9.78

so 9.64, 9.7 , 9.78

Open-Ended
Complete the number to make the statement true.
Question 13.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 15
Answer:

10.321 > 10.311

as 10.311 is smaller than 10.321

Question 16.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 16
Answer:

28.60 = 28.600

Both are equal

Question 15.
Number Sense
Is 0.472 greater than or less than \(\frac{47}{1,000}\)? Explain.
Answer:

0.472, 0.047

0.475 is greater than 0.047,

as 4 in the tenths  place is greater than 0 in the others tenth place

Question 16.
YOU BE THE TEACHER
Your friend says that 45.6 is less than 45.57 because 6 is less than 57. Is your friend correct? Explain.
Answer:

Friend says 45.6 is less than 45.57

No ,he is wrong as the 6th in tenths place is greater than 5, in the tenths place

so he is wrong  45.60 > 45.57 not less

Think and Grow: Modeling Real Life

Example
You, your friend, and your cousin compete at a gymnastics competition. Your floor routine score is 15.633. Your friend’s score is 15.533, and your cousin’s score is 15.635. Order the scores from least to greatest.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 17
Use a place value chart. Compare the digits in each place until the digits differ.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 18
_15.533______ is the least
Write the remaining numbers in the place value chart. Compare the digits in each place until the digits differ.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 19

15.633 ,15.635

as 15.533 is less than 15.633 as  5 at tenths place is less than 6 at tenths place

and 15.633 is less than 15.653 as 3 at hundredths place is less than 5 at hundredths place

So, the scores from least to greatest are 15.533, 15.633 and 15.635

Show and Grow

Question 17.
You stand on one leg for 2.75 minutes, your friend stands on one leg for 2 minutes, and your cousin stands on one leg for 2.25 minutes. Order the amounts of time from least to greatest.
Answer:

You – 2.75 min , Friend – 2.00 min and cousin for 2.25 min

2.00< 2.25 ,2.00 is less than 2.25 as 0 at tenths place is less than 2 at tenths place

2.25 < 2.75, 2.25 is less than 2.75 as 2 at tenths place is less than 7 at tenths place

so From Least to Greatest : 2.00 min, 2.25 min, 2.75 min

Question 18.
DIG DEEPER!
You, Newton, Descartes, and your friend each have a tablet. The table shows the screen display sizes. Your friend’s tablet has the greatest display size. Your tablet’s display size is greater than Newton’s but less than Descartes’s. What is the display size of your tablet.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 20
Answer:

Y-Your, N-Newton’s ,F-Friend- 12.9, D-Descartes

Given Friend has greatest display size F= 12.9

Y – N ,given Your tablet’s display size is greater than Newton’s, N< Y and yours is less than Descartes’s Y < D

therefore  N < Y <D < F, Newton’s< Yours<Descartes’s<Friend

so N-Newton’s- 7.9, Y-Your-9.7,D-Descartes-10.5,F-Friend-12.9

Compare Decimals Homework & Practice 1.6

Write which place to use when comparing the numbers.
Question 1.
0.521
0.576
Answer:

2 at hundredths place  is smaller than 7 at hundredths place

so 0.521 < 0.576

Question 2.
17.422
17.946
Answer:

4 at tenths place  is small than 9 at tenths place

so 17.422 < 17.946

Question 3.
9.678
9.67
Answer:

8 at thousandths place is greater than 0 at thousandths place

9.678 > 9.670

Compare.
Question 4.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 21
Answer:

4 at hundredths  place is smaller than 7 at hundredths place

so 3.445 is smaller than 3.472

3.445 < 3.472

Question 5.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 22

Answer:

0 at tenths place is smaller than 4 at tenths place

so 23.049 < 23.409

Question 6.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 23
Answer:

4 at tenths place is greater than 3 at tenths place

75.4 > 75.391

Question 7.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 24
Answer:

All  given place values are same

so 14.100 or 14.10 = 14.100

Question 8.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 25
Answer:

the value of 5s at hundredths place is more than 0 in other hundredths place

4.05> 4.005

Question 9.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 26
Answer:

15.2, 15.002

2 at tenths place is greater than 0 at tenths place

15.2 > 15.002

Question 10.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 27
Answer:

0.021, 0.026

1 at thousandths is smaller than 6 at thousandths place

0.021 < 0.026

Order the decimals from least to greatest.
Question 11.
2.75, 0.2, 0.275
Answer:

0.2 < 0.275 as 0 at hundredths place is less than 7 at hundredths place

0.275<2.75 as 0 at ones place is less than 2 at at ones place

so from least to greatest 0.2, 0.275, 2.75

Question 12.
56.01, 56.1, 56.001
Answer:

56.001 < 56.01 as 0 at hundredths place is less than 1 at hundredths place

56.01 < 56.1 as o at tenths place is less than 1 at tenths place

56.001 , 56.01, 56.1

Open-Ended
Complete the number to make the statement true.
Question 13.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 28
Answer:

29.030 = 29.030

both the place values are same

Question 14.
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 29
Answer:

3.562 <3.562

as the value of  6 is at hundredths place so the other value at hundredths place is 6.

Question 15.
YOU BE THE TEACHER
Newton says 8.51 is less than 8.492 because 8.51 has fewer digits after the decimal point than 8.492. Is he correct? Explain.
Answer:

No, he is not correct its not the digits after the decimal point is fewer

but the 5s at tenths place is greater than 4 at the tenths place so

8.51 is greater than 8.492, 8.51 > 8.492

Question 16.
Open-Ended
Descartes is thinking of a number less than 46.922 and greater than 46.915. What could Descartes’s number be?
Answer:

The numbers can be 46.916,  4 6.917, 46.918, 46.919, 46.920 or 46.921 and

all these numbers are less than 46.922 and greater than 46.915.

Question 17.
Modeling Real Life
Player A’s batting average is 0.300, Player B’s batting average is 0.333, and Player C’s batting average is 0.313. Order the batting averages from greatest to least.
Answer:

B- 0.333, C-0.313 , A – 0.300

0.333 is great than 0.313 as 3 at hundredths place is great than 1 at hundredths place

so 0.333>0.313

0.313 is greater than 0.300 as 1 at hundredths place is great than 0 at hundredths place

0.313 > 0.300

the batting averages from greatest to least are B> C> A=0.333 > 0.313 >0.300

batting averages from greatest to least 0.333,0.313,0.300

Question 18.
Modeling Real Life
A gasoline station customer pumps more than 9.487 gallons of gasoline but less than 10 gallons. Which display could be his?
Big Ideas Math Answers 5th Grade Chapter 1 Place Value Concepts 1.6 30
Answer:

Given A gasoline station customer pumps more than 9.487 gallons but less than 10 gallons

as 9.003 is less than 9.487 so not 9.003

as 9.499 is greater than 9.487  and even less than 10.000 so it is 9.499

as 9.406 is less than 9.487 so it cannot be 9.406

as 9.872 is greater than 9.487  and even less than 10.000 so it can be 9.872

so the displays can be  9.499 or 9.82

as both are more than 9.487 gallons and less than 10 gallons

Review & Refresh

Round the number to the place of the underlined digit.
Question 19.
7,851
Answer:

the round place of 5 is 7,850

Question 20.
9,462
Answer:

the round place of 9 means 10,000 or 9,500

Question 21.
4,983
Answer:

the round place of 9 – 5,000

Question 22.
51,504
Answer:

the round place of 1- 52,504

Lesson 1.7 Round Decimals

Explore and Grow

Plot the numbers on the number line. Which numbers round to 3? Which numbers round to4? How do you know?
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.7 1
Answer:

3,  3.09, 3.5, 3.51, 3.6, 3.77, 3.9 , 4

The numbers round to 3 are 3.09

The numbers round to 4 are 3.51, 3.6, 3.77 , 3.9

Repeated Reasoning
Show how you can use a number line to round 3.09, 3.51, and 3.77 to the nearest tenth.
Answer:

3.09 to the nearest tenth is 3.10

3.51 to the nearest tenth is 3.50

3.77 to the nearest tenth is 3.80

Think and Grow: Round Decimal Number

Example
Use a number line to round 7.36 to the nearest tenth.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.7 2
7.36 is closer to 7.4 than it is to 7.3.
So, 7.36 rounded to the nearest tenth is __7.40____.

Example
Use place value to round 2.185 to the nearest whole number and to the nearest hundredth.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.7 3
So, 2.185 rounded to the nearest whole number is __2.2_____.
2.185 rounded to the nearest hundredth is ___2.200___.

Show and Grow

Round the number to the place of the underlined digit.
Question 1.
12.67
Answer:

the round place of  digit 6 is 12.70

Question 2.
0.439
Answer:

the round place of  digit 4 is 0.5

the round place of  digit 3 is 0.44

the round place of digit  9 is 0.440

Question 3.
2.555
Answer:

the round place of  digit 2 is 3.000

Question 4.
5.409

the round place of digit 4 is 5.400

Question 5.
Round 0.68 to the nearest tenth.
Answer:

0.68 to the nearest tenth  is 0.70, 6 at tenths place becomes 7

Question 6.
Round 1.715 to the nearest hundredth.
Answer:

1.715 to the nearest hundredth is  1.720, 1 at hundredths place becomes 2

Question 7.
Round 4.07 to the nearest whole number.
Answer:

4.07 to the nearest whole number becomes  4.00 or 4

Question 8.
Round 0.289 to the nearest tenth.
Answer:

0.289 to the nearest tenth is 0.300 as 2 becomes 3 at tenths place.

Apply and Grow: Practice

Round the number to the place of the underlined digit.
Question 9.
1.482
Answer:

the underlined digit is 8 ,so its value becomes 1.490

Question 10.
5.093
Answer:

the underlined digit is 0 so its value  becomes 5.100

Question 11.
8.502
Answer:

the underlined digit is 8 so its value becomes 9.000

Question 12.
34.748
Answer:

if it is underlined at 3 it becomes 35.000

if it is underlined at 4 it becomes 35.000

if it is underlined at 7 it becomes 35.000

if it is underlined at 4 it becomes 34.800

if it is underlined at 8 it becomes 34.750

Question 13.
Round 2.619 to the nearest whole number.
Answer:

the value of 2.619 becomes 3.000

Question 14.
Round 7.825 to the nearest tenth.
Answer:

the value of 7.825 to the nearest tenth is 7.900

Question 15.
Round 92.701 to the nearest ten.
Answer:

the value of 92.701 to the nearest ten 93.000

Question 16.
Round 4.263 to the nearest hundredth.
Answer:

the value of 4.263 to the nearest hundredth is 4.270

Question 17.
Round 0.829.
Nearest whole number:
Nearest tenth:
Nearest hundredth:
Answer:

Round 0.829
Nearest whole number:0.900
Nearest tenth:0.830
Nearest hundredth:0.830

Question 18.
Round 18.062.
Nearest whole number:18.100
Nearest tenth:18.070
Nearest hundredth:18.063
Answer:

Question 19.
A baby harp seal weighs 25.482 pounds. Round this weight to the nearest tenth of a pound.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.7 4
Answer:

Given a baby harp seal weighs 25.482 pounds and the nearest tenth is 25.000 pounds

So the weight of baby harp seal is 25.000 pounds.

Name the place value to which the number was rounded.
Question 20.
8.942 to 8.94
Answer:

It was rounded at hundredths value place

Question 21.
0.164 to 0.2
Answer:

It was rounded at tenths value place

Question 22.
15.826 to 16
Answer:

It was rounded at whole value place

Question 23.
Writing
Explain what happens when you round 2.999 to the nearest tenth.
Answer:

2.999 round value becomes 3.000 as all value places at tenths, hundredths, thousandths are 9 it becomes increased as we move so it ones value increases by 1 and becomes round 3.000.

Question 24.
DIG DEEPER!
To what place should you round 23.459 to get the greatest number? the least number? Explain.
Answer:

To make 23.459 to  greatest number the value of 4s at tenths  becomes 5,

23.500 and to make 23.500 round make 5 at tenths value increased and make ones value 3 as 4 so we get 24.000

and to make 23.459 to least number the value at hundredths 5 becomes 0 ,

23.400  and to make 23.400 round make 4 at tenths value as decreased to 0 and ones value same as 3 it becomes as 23.000

Think and Grow: Modeling Real Life

Example
Gasoline prices are listed to the nearest thousandth of a dollar. The final price is rounded to the nearest hundredth. About how much does a customer pay for 1 gallon of regular gasoline at the station shown?
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.7 5
Think: What do you know? What do you need to find? How will you solve?
Use place value to round the price of 1 gallon of regular gasoline, $2.799, to the nearest hundredth.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.7 6
So, a customer will pay about $ ___$ 2.800____ for 1 gallon of gasoline.

Show and Grow

Use the table.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.7 7
Question 25.
What is the length of the praying mantis rounded to the nearest hundredth?
Answer:

Given  praying mantis as 3.254 so the  nearest hundredth becomes 3.26

Praying mantis becomes 3.260

Question 26.
What is the length of the cicada rounded to the nearest tenth?
Answer:

Given cicada 1.48 so rounded to the nearest tenth becomes 1.50

Cicada becomes 1.50

Question 27.
What is the length of the hissing cockroach rounded to the nearest tenth?
Answer:

Given Hissing cockroach 2.682 so rounded to the nearest tenth is 2.700

Hissing cockroach becomes 2.700

Question 28.
DIG DEEPER!
You have about $3 in coins. Write one possible combination of coins that represents the least amount of money you could have. Write another combination of coins for the greatest amount of money you could have.
Answer:

$3 Least amount of money combinations – 1. $ 1.0, $1.0    2.$ 1.0, $1.5

3.$0.5, $ 2.0 all combinations becomes less than $3

$ 3 greatest amount of money combinations- 1. $1.0, $ 2.5   2.$1.5 ,$2.0

3. $2.0 , $ 2.0 all combinations becomes more than $3

Round Decimals Homework & Practice 1.7

Round the number to the place of the underlined digit.
Question 1.
49.012
Answer:

the underline digit is 4 its round number 4 becomes 5 so it is 50.000

Question 2.
2.308
Answer:

the underline digit is 2  its round number 2 becomes as 2.000

Question 3.
9.647
Answer:

the underline digit is 6 its round number 6 becomes 7 so it is 9.700

Question 4.
7.519
Answer:

the underline digit is 1 its round number 1 becomes  2 so it is 7.520

Question 5.
Round 8.436 to the nearest hundredth.
Answer:

8.436 to the nearest hundredth, 3 becomes 4 so it is 8.440

Question 6.
Round 15.159 to the nearest ten.
Answer:

15.159 to the nearest ten ,1 becomes 2 so it is 15.200

Question 7.
Round 1.602 to the nearest whole number.
Answer:

1.602 to the nearest whole number is 2.0

Question 8.
Round 3.619 to the nearest tenth.
Answer:

3.619 to the nearest tenth, so 6 becomes 7 it is 3.700

Question 9.
Round 4.183.
Nearest whole number:
Nearest tenth:
Nearest hundredth:
Answer:

Round 4.183.
Nearest whole number:4.000
Nearest tenth:4.200
Nearest hundredth:4.200

Question 10.
Round 9.076.
Nearest whole number:
Nearest tenth:
Nearest hundredth:
Answer:

Round 9.076.
Nearest whole number:9.000
Nearest tenth:9.100
Nearest hundredth:9.080

Name the place value to which each number was rounded.
Question 11.
16.932 to 20
Answer:

16.932 at Tens value it is rounded  so it becomes 20

Question 12.
0.581 to 0.58
Answer:

0.581 to 0.58

0.581 at Thousandths value has been rounded so 0.58

Question 13.
7.429 to 7.4
Answer:

7.429 to 7.4

7.429 at Hundredths value has been rounded so 7.4

Question 14.
Structure
Round ★ to the nearest tenth.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.7 8
Answer:

the nearest tenth * in the number line is showing at 5.64

Question 15.
Precision
The area of a campground is exactly halfway between 25.9 acres and 26 acres. What is the area of the campground?
Answer:

Halfway of 25.9 and 26.0 is 25.90+25.60=51.50/2 = 25.75 acres

So the area of the campground is 25.75 acres.

Question 16.
Open-Ended
Name two different numbers that round to 3.8 when rounded to the nearest tenth.
Answer:

The two numbers that are round to 3.8 when rounded to the nearest tenth place the value becomes 4.0 and  3.90

Question 17.
Open-Ended
Name two different numbers that round to 7.42 when rounded to the nearest hundredth.
Answer:

The two numbers that round to 7.42 when rounded to the nearest hundredth place the value becomes 7.50 and 7.40

Use the table.
Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts 1.7 9
Question 18.
Modeling Real Life
Your science class designs and tests four model boats to find out how much weight they can hold without sinking. What is the greatest weight rounded to the nearest tenth that a boat can hold?
Answer:

Given weights held without sinking in kilograms are

0.694,0.605,0.592,0.547 among all the weights the greatest weight rounded to the nearest tenth that a boat can hold is (0.694) i.e  0.7 kilograms

Question 19.
Modeling Real Life
What is the least weight rounded to the nearest hundredth that a boat can hold?
Answer:

The least weight rounded to the nearest hundredth is (0.547)- o.5 kilograms a boat can hold

Review & Refresh

Find the product.
Question 20.
7 × 40
Answer:

The product of 7 X 40 = 280

Question 21.
5,000 × 9
Answer:

The product of 5,000 x 9 = 45,000

Question 22.
8 × 200
Answer:

The product of 8 x 200=1,600

Place Value Concepts Performance Task

There are 18 species of penguins. Scientists have estimated the populations of 16 penguin species.
Question 1.
What fraction of penguin species have unknown populations?
Answer:

Given total is 18 species of penguins out of which Scientists have estimated the populations of 16 penguin species. so unknown is 18-16/18=2/18,

so 2/18=1/9 of penguin species have unknown populations

Question 2.
Several species of penguins and their estimated populations and locations are shown.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 1
a. Are there more emperor penguins or rockhopper penguins? Explain.
b. Which species of penguin has the greatest population? Explain.
c. About how many penguins live in Antarctica? Round your answer to the nearest hundred thousand.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 2
d. The Galápagos penguin is an endangered species. There are about 1,000 times as many macaroni penguins as Galápagos penguins. About how many Galápagos penguins are there?
Answer:

a. Rock hopper penguins are more than Emperor

2,460,000>595,000 as  2,460,000 is greater than 595,000 so Rock hopper penguins are more

B. Macaroni species 18 x 106 of penguin has the greatest population as compared to Emperor 595,000,Adelie 4,000,000+7,00,000+40,000=4,740,000 and

Rockhopper-2,460,000

among all the species Macaroni 18 x 106– species is more.

C. Macaroni + Adelie + Rockhopper +Emperor

18,000,000+4,740,000+ 2,460,000 + 595,000 = 25,795,000

25,785,000  to the nearest hundred thousand is 26,000,000

So there are almost 26,000,000 penguins live in Antarctica.

d. 1,000 times as many macaroni penguins as Galápagos penguins is

18 x 106  x 1 x 1000 = 18 x 10

So there are 18 x 10Galápagos penguins available

Place Value Concepts Activity

Place Value Plug In
Directions:
1. Players take turns.
2. On your turn, roll six dice. Arrange the dice into a six-digit number that matches one of the descriptions.
3. Write your number on the lines.
4. The first player to fill in all of the numbers wins!
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts 3
Answer:

Place Value Concepts Chapter Practice

1.1 Place Value Patterns

Use a place value chart to answer the question.
Question 1.
What number is 10 times as great as 4,000?
Answer:

10 times as great as 4,000 is 10 x 4,000= 40,000

Question 2.
What number is \(\frac{1}{10}\) of 8,000?
Answer:

\(\frac{1}{10}\) of 8,000 is 8,000/10 = 800

Question 3.
10,000 is 10 times as great as what number?
Answer:

10,000 is 10 times as great as 1,000

Question 4.
70 is \(\frac{1}{10}\) of what number?
Answer:

70 is \(\frac{1}{10}\) of 700

Question 5.
Complete the table.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts chp 5
Answer:

300 is 10 times as great as 30  and 30 x 1/10 is 3

6,000 is 10 times as great as 600 and 600 x 1/10  is 60

90,000 is 10 times as great as 9,000 and 9,000 x 1/10 is 900

2,00,000 is 10 times as great as 20,000 and 20,000 x 1/10=2,000

Question 6.
YOU BE THE TEACHER
Your friend says 500 is 10 times as great as 5,000. Is your friend correct? Explain.
Answer:

No, My friend is wrong because  500 is not 10 times great as 5,000,

500 < 5,000.

1.2 Place Value with Whole Numbers

Question 7.
Write the number in two other forms.
Standard form: 456,701
Word form:
Expanded form:
Answer:

Word form: Four hundred fifty six thousand, seven hundred and one
Expanded form:4 x 100000 + 5 x 10000 + 6 x 1000 + 7 x 100 + 1

Question 8.
Write the number in two other forms.
Standard form:
Word form: Eight million, sixty thousand, five hundred seventy-three
Expanded form:
Answer:

Standard form:8,060,573
Expanded form:8 x 1000000+6 x 10000 + 5 x 100 + 7 x 10 + 3

Question 9.
Compare the values of the 4s in the number 900,441,358.
Answer:

4s value is at lakh or hundred thousandths place and another 4s place is at ten thousandths place.

Question 10.
Write the values of the 6s in the number 96,672.
Answer:

The values of the 6s in the number 96,672 are

6s place is at thousand and another 6s is at hundreds place

Compare.
Question 11.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts chp 11
Answer:

83,802 > 83,082

The value at hundred 8 is more/great to 0 at hundreds place

Question 12.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts chp 12
Answer:

2,498,576 > 2,477,583

The value at 9 at ten thousands place is more/great than 7 at ten thousands place

1.3 Patterns and Powers of 10

Question 13.
Write 10 × 10 as a power.
Answer:

10 × 10 as a power is 102

Find each product. Use patterns to help.
Question 14.
4 × 10 = _____
4 × 100 = ______
4 × 1,000 = _____
4 × 10,000 = ______
Answer:

4 × 10 = 40
4 × 100 = 400
4 × 1,000 = 4,000
4 × 10,000 =40,000

Question 15.
3 × 10 = _____
3 × 100 = ______
3 × 1,000 = _____
3 × 10,000 = ______
Answer:

3 × 10 = 30
3 × 100 = 300
3 × 1,000 = 3,000
3 × 10,000 = 30,000

Question 16.
7 × 10 = _____
7 × 100 = ______
7 × 1,000 = _____
7 × 10,000 = ______
Answer:

7 × 10 = 70
7 × 100 = 700
7 × 1,000 = 7,000
7 × 10,000 = 70,000

Find the value of the expression.
Question 17.
105
Answer:

105=1,00,000

Question 18.
8 × 101
Answer:

8 × 101= 8 x 10 = 80

Question 19.
7 × 104
Answer:

7 × 104 = 7 x  10 x 10 x 10 x 10= 70,000

Question 20.
3 × 105
Answer:

3 × 105 = 3 X 10 X 10 X 10 X 10 X 10=3,00,000

Rewrite the number as a whole number multiplied by a power of 10.
Question 21.
5,000
Answer:

5,000= 5 X 103

Question 22.
600,000
Answer:

6,00,000= 6 X 105

Question 23.
90
Answer:

90= 90 X 101

1.4 Decimals to Thousandths

Write the decimal as a fraction.
Question 24.
0.062
Answer:

0.062= 62 X 1/1,000

Question 25.
0.008
Answer:

0.008= 8 X 1/1,000

Question 26.
0.195
Answer:

0.195= 195 X 1/1,000

Write the fraction as a decimal.
Question 27.
\(\frac{2}{1,000}\)
Answer:

\(\frac{2}{1,000}\) = 2 x 1/1,000= 0.002

Question 28.
\(\frac{37}{1,000}\)
Answer:

\(\frac{37}{1,000}\) = 37 x 1/1,000= 0.037

Question 29.
\(\frac{409}{1,000}\)
Answer:

\(\frac{409}{1,000}\)= 409/1,000=0.409

Question 30.
0.0.7 is 10 times as great as what number?
Answer:

0.07 is 10 times great as 0.007

Question 31.
0.04 is \(\frac{1}{10}\) of what number?
Answer:

0.04 is \(\frac{1}{10}\) of 0.4

1.5 Place Value with Decimals

Write the number in two other forms.
Question 32.
Standard form:
Word form:
Expanded form: 5 × 1 + 3 × \(\frac{1}{10}\) + 8 × \(\frac{1}{100}\) + 4 × \(\frac{1}{1,000}\)
Answer:

Standard form:5+0.3+0.08+0.004=5.384
Word form: five and three tenths, eight hundredths and 4 thousandths

Question 33.
Standard form: 2.059
Word form:
Expanded form:
Answer:

Word form: two and five hundredths and nine thousandths

Expanded form: 2 x 1 + 5 x 1/100 + 9 x 1/1000

Question 34.
Compare the values of the 5s in the number 1.055.
Answer:

5s value is at hundredths and another 5s value is at thousandths place

Question 35.
Compare the values of the 8s in the number 6.884.
Answer:

8s place value is at tenths value and other 8s place value is at hundredths place

1.6 Compare Decimals

Compare.
Question 36.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts chp 36
Answer:

15.891 > 15.791

the 8 value at tenths place is greater than 7 at tenths place

Question 37.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts chp 37
Answer:

8.205 < 8.250

the 0 at hundredths is less than 5 at hundredths

Question 38.
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts chp 38

Answer:

Both are same

3.600 = 3.6

Order the decimals from least to greatest.
Question 39.
7.008, 7.09, 7.180
Answer:

7.008<7.09

7.09 <7.18

so the decimals from least to greatest are 7.008 , 7.09, 7.180

Question 40.
50.426, 50.42, 50
Answer:

50 < 50.42

50.42 < 50.426

so the decimals from least to greatest are 50, 50.42, 5.0426

Question 41.
Modeling Real Life
Newton weighs a treat at the pet store. He says it weighs less than 0.519 ounce but more than 0.453 ounce. Which treats could he have weighed?
Big Ideas Math Solutions Grade 5 Chapter 1 Place Value Concepts chp 41
Answer:

Newton says it weighs less than 0.519 ounce  but more than 0.453 ounce,

so Blue berry waffle 0.512 which is in between 0.453<0.512<0.519

and We have more than 0.453 is 0.459 ounce which is Peanut butter and is less than 0.519 ounce

So Peanut butter 0.453 is in between 0.453<0.459<0.519

so newton would have weighed Peanut butter 0.459 ounce

and it can be even Blueberry  waffle 0.512  ounce

newton’s treats could be Peanut butter ,Blueberry waffle

1.7 Round Decimals

Round the number to the place of the underlined digit.
Question 42.
9.514
Answer:

if 9 is digit then it is rounded as 9.0

if 5 then  it is rounded as 9.500

if 1 then it is rounded as 9.520

if  4 then it is rounded as 9.520,9.510

Question 43.
1.027
Answer:

1.027 at 2 it is rounded as 1.03

Question 44.
8.469
Answer:

8.469, 8 is rounded as 8.500 or 8 or 8.000

Question 45.
32.501
Answer:

in 32.501

3 is rounded as  30.000

2- 33.000

5- 32.6 or 33

0-32.5

1 -32.500

Question 46.
Round 0.176 to the nearest
Answer:

0.200

Question 47.
Round 6.538 to the nearest tenth. hundredth.
Answer:

6.500 

Question 48.
Round 7.425.
Nearest whole number:
Nearest tenth:
Nearest hundredth:
Answer:

Nearest whole number:8.000, 8
Nearest tenth:7.500
Nearest hundredth:7.430

Question 49.
Round 2.108.
Nearest whole number:
Nearest tenth:
Nearest hundredth:
Answer:

Nearest whole number:2.000, 2
Nearest tenth:2.100
Nearest hundredth:2.100

Final Words:

It is very important for the students to understand and learn the fundamentals at the primary level itself. Here we have prepared the questions as per the latest edition 2019. Keep the textbook aside and try to solve the problems by referring to our Big Ideas Math Answers Grade 5 Chapter 1 Place Value Concepts. To make you comfortable we have provided the solution key for Big Ideas Math Grade 5 Chapter 1 Place Value Concepts in the pdf format. Stay tuned to our CCSS Math Answers to get the latest updates of BIM Grade 5 Chapters.

HDFCLIFE Pivot Calculator

Big Ideas Math Answers Grade 8 Chapter 7 Functions

Big Ideas Math Answers Grade 8 Chapter 7

Big Ideas Math Grade 8 Chapter 7 Functions is given by subject experts adhering to the latest syllabus guidelines. Become Proficient in the Concepts of Big Ideas Math Grade 8 Functions by consistently practicing from our BIM Book 8th Grade Ch 7 Functions Answer Key. To make it easy for you to understand the concepts we have provided step-by-step solutions in the Big Ideas Math Answers Grade 8 Chapter 7 Functions. You can download the quick resources without even paying a single penny.

Big Ideas Math Book 8th Grade Answer Key Chapter 7 Functions

Big Ideas Math Grade 8 Ch 7 Functions Solutions provided here cover the questions from Lessons 7.1 to 7.5, Practice Tests, Review Tests, Cumulative Practice, Assessment Tests, etc. You can have deeper insights into all the Topics of Functions such as the representation of functions, linear functions, analyzing and sketching graphs, etc. Use the below available quick links for Big Ideas Math 8th Grade Chapter 7 Functions Answer Key and clear all your ambiguities regarding the concerned topics.

STEAM Video/Performance Task

Getting Ready for Chapter 7

Lesson 1 Relations and Functions

Lesson 2 Representations of Functions 

Lesson 3 Linear Functions

Lesson 4 Comparing Linear and Non Linear Functions

Lesson 5 Analyzing and Sketching Graphs

Functions Connecting Concepts

Functions STEAM Video/Performance Task

STEAM Video

Apparent Temperature
Sometimes it feels hotter or colder outside than the actual apparent temperature. How hot or cold it feels is called the temperature. What weather factors might contribute to the apparent temperature?
Watch the STEAM Video “Apparent Temperature.” Then answer the following questions.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 1
1. Robert says that the Wet-Bulb Globe Temperature (WBGT)index is used as a measure of apparent temperature.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 2
In the formula, TW is the natural wet-bulb temperature, TG is the black-globe temperature, TD and is the dry-bulb temperature. Find WBGT when TW = 75ºF, TG = 100ºF, and TD = 84ºF.
2. Different categories of Wet-Bulb Globe Temperatures are shown in the chart. Each category can be represented by a different-colored flag. Which flag color is displayed when WGBT = 87.5ºF?

Performance Task

Heat Index
After completing this chapter, you will be able to use the STEAM concepts you learned to answer the questions in the Video Performance Task. You will be given information about heat index.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 3
You will be asked to create a graph of the temperatures and heat indices. Why is it useful to know the heat index?

Functions Getting Ready for Chapter 7

Chapter Exploration

Work with a partner. Copy and complete the diagram.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 4

1.  Answer: ( 1 , 2 ) , ( 2 , 4 ) , ( 3 , 6 ) , ( 4 , 8 ) .

Explanation:
Given , Width w = 2 , Length l = x  where x = 1 , 2 , 3 , 4 .
To find Area A of a rectangle we have , A = w × l
for x = 1 , A = 2 × 1 = 2 ,
for x = 2 , A = 2 × 2 = 4 ,
for x = 3 , A = 2 × 3 = 6 ,
for x = 4 , A = 2 × 4 = 8 ,
So, for every value of Input x = 1 , 2 , 3 , 4 we have Output A = 2 , 4 , 6 , 8 , respectively .
That is ( 1 , 2 ) , ( 2 , 4 ) , ( 3 , 6 ) , ( 4 , 8 ) .

2. Answer: ( 1 , 6 ) , ( 2 , 8 ) , ( 3 , 10 ) , ( 4 , 12 ).

Explanation:
Given , Width w = 2 , Length l = x  where x = 1 , 2 , 3 , 4 .
To find Perimeter of a rectangle we have , P = 2( l + w )
for x = 1 ,P = 2( 1 + 2 ) = 2 × 3 = 6 ,
for x = 2 , P = 2( 2 + 2 ) = 2 × 4 = 8  ,
for x = 3 , P = 2( 3 + 2 ) = 2 × 5 = 10  ,
for x = 4 , P = 2( 4 + 2 ) = 2 × 6 = 12  ,
So, for every value of Input x = 1 , 2 , 3 , 4 we have Output P = 6 , 8 , 10 , 12 , respectively .
That is ( 1 , 6 ) , ( 2 , 8 ) , ( 3 , 10 ) , ( 4 , 12 ) .

3. Answer : ( 1 , 6 ) , ( 2 , 12 ) , ( 3 , 18 ) , ( 4 , 24 ) .

Explanation:
Given , Radius of a circle , where as  r = 1 , 2 , 3 , 4
To find the circumference of a circle , we have C = 2Òr , Ò = 3.14 , or we can write it as 3 .
for r = 1 , C = 2 × 3 × 1 = 6 ,
for r = 2 , C = 2 × 3 × 2 = 12 ,
for r = 3 , C = 2 × 3 × 3 = 18 ,
for r = 4 , C = 2 × 3 × 4 = 24 ,
So, for every value of Input r = 1 , 2 , 3 , 4 we have Output C = 6 , 12 , 18 , 24 , respectively .
That is ( 1 , 6 ) , ( 2 , 12 ) , ( 3 , 18 ) , ( 4 , 24 ) .

4. Answer: ( 1 , 9 ) , ( 2 , 18 ) , ( 3 , 27 ) , ( 4 , 36 )

Explanation:
Given , Two Edges of a cube = 3 , h = 1 , 2 , 3 , 4
To find the Volume of the cube we have , V = a³
for h = 1 , V = 3 × 3 × 1 = 9 ,
for h = 2 , V = 3 × 3 × 2 = 12 ,
for h = 3 , V = 3 × 3 × 3 = 27 ,
for h = 4 , V = 3 × 3 × 4 = 36 ,
So, for every value of Input h = 1 , 2 , 3 , 4 we have Output V = 9 , 18 , 27 , 36 , respectively .
That is ( 1 , 9 ) , ( 2 , 18 ) , ( 3 , 27 ) , ( 4 , 36 ) .

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
input
mapping diagram
nonlinear function
output
linear function

Answer : Input : The ordered pairs can be used to show inputs and outputs of a function ,The input is the number you feed into the expression, and the output is what you get after the calculations are finished.

mapping diagram : A relation pairs inputs with outputs , A relation can be represented by ordered pairs or a mapping diagram .

nonlinear function : nonlinear functions are functions which are not linear. Quadratic functions are one type of nonlinear function. It is a relation between two variables , function that does not form a line when graphed.

output ; The ordered pairs can be used to show inputs and outputs of a function ,The input is the number you feed into the expression, and the output is what you get after the calculations are finished.

linear function : A linear function is a relation between two variables that produces a straight line when graphed. And it has one dependent variable and one independent variable .

Lesson 7.1 Relations and Functions

EXPLORATION 1

Interpreting Diagrams
Work with a partner. Describe the relationship between the inputs and outputs in each diagram. Then complete each diagram. Is there more than one possible answer? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 1
Answer: a. The relation between the inputs and outputs is outputs are the result of twice as many times the inputs.
b. The relation between the inputs and outputs is outputs are the result of colors of inputs . In this case we can notice that , for any one input we can have more than one output .

Explanation:
a. As shown in the diagrams , The relation between the inputs and outputs is outputs are the result of twice as many times the inputs , so for input 1 = 1 × 1 = 1 as output ,
for input 2 = 2 × 2 = 4  ,
for input 3 = 3 × 3 = 9  ,
for input 5 = 5 × 5 = 25 ,
for input 8 = 8 × 8 = 64 ,
for input 9 = 9 × 9 = 81 ,

So, for every value of Input = 1 , 2 , 3 , 5 , 8 , 9 , we have Output = 1 , 4 , 9 , 25 , 64 , 81 , respectively .
That is ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) , ( 5 , 25 ) , ( 8 , 64 ) , ( 9 , 81 ) .

b. The relation between the inputs and outputs is outputs are the result of colors of inputs . 
for input Blueberry = color is blue as output
for  input lemon = color is yellow as output
for input Apple = color is yellow , red and green as output
for input Grape = color is green as output.

In this case we can notice that , for any one input we can have more than one output .

EXPLORATION 2

Describing Relationships Between Quantities
Work with a partner. The diagrams show the numbers of tickets bought by customers for two different plays and the total costs (in dollars).
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 2
a. For each diagram, how many outputs does each input have?
b. Describe the prices of tickets for each play.
c. A person buys 4 tickets for each play. Can you determine the total cost of all 8 tickets? Explain.
Answer:
a. For Play A ,The number of inputs are equal to number of outputs ,
For Play B , The number of inputs are not equal to number of outputs .
So, input 1 = 2 outputs , input 2 = 3 outputs , input 3 = 4 outputs .

b. For Play A , The price of the each ticket is $8 .
For Play B , The price of each ticket is $4 or $8 .

c. For Play A , each ticket is $8 , Then for 4 tickets = 4 × $8 = $32 .
For Play B , each ticket is $4 or $8 , Then for 4 tickets = 4 × $8 = $32 . or 4 × $4 = $16 .

Explanation:
a. For Play A ,
The number of inputs are equal to number of outputs , 4 inputs = 4 outputs
That is ( 1 , 8 ) , ( 2 , 16 ) , ( 3 , 24 ) , ( 4 , 32 ) .
For Play B ,
The number of inputs are not equal to number of outputs , 3 inputs are not equal to 7 outputs
That is , for input 1 = 4 , 8 as outputs ,
for input 2 = 8 , 12 , 16 as outputs ,
for input 3 = 12 , 16 , 20 , 24 as outputs ,
So, input 1 = 2 outputs , input 2 = 3 outputs , input 3 = 4 outputs .

b. For Play A ,
The price of the each ticket is $8 .
For Play B ,
The price of each ticket is $4 or $8 .

c. Given , A person buys 4 tickets for each play.
For Play A , each ticket is $8 , Then for 4 tickets = 4 × $8 = $32 .
And for 8 tickets = 8 × $8 = $64 .
For Play B , each ticket is $4 or $8 , Then for 4 tickets = 4 × $8 = $32 . or 4 × $4 = $16 .
And for 8 tickets =8 × $4 =$32  or 8 × $8 = $64 .

Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 3

Try It

List the ordered pairs shown in the mapping diagram.
Question 1.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 4
Answer: Ordered pairs are ( 0 , 12 ) , ( 2 , 10 ) , ( 4 , 8 ) , ( 6 , 6 ) .

Explanation:
As shown , Ordered pairs are the combinations of input and output
So , Ordered pairs are ( 0 , 12 ) , ( 2 , 10 ) , ( 4 , 8 ) , ( 6 , 6 ) .

Question 2.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 5
Answer: Ordered pairs are ( 1 , -1 ) , ( 1 , -2 ) , ( 2 , -3 ) , ( 2 , -4 ) .

Explanation:
As shown , Ordered pairs are the combinations of input and output
So , Ordered pairs are ( 1 , -1 ) , ( 1 , -2 ) , ( 2 , -3 ) , ( 2 , -4 ) .

Determine whether the relation is a function.
Question 3.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 6
Answer: The relation is not a function

Explanation:
The each input has  more than two outputs , Even one of those inputs are unclear of outputs
So , The relation is not a function .

Question 4.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 7
Answer: The relation is a function .

Explanation:
Each input has exactly one output ,
So , The relation is a function .

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
PRECISION
Describe how relations and functions are different.
Answer: Relations are nothing but the ordered pairs with Inputs and Outputs . On the other hand , Functions are The relation that pairs with one input with exactly one output  are called Functions.

IDENTIFYING FUNCTIONS List the ordered pairs shown in the mapping diagram. Then determine whether the relation is a function.
Question 6.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 8
Answer: The ordered pairs are ( 10 , 1 ) , ( 15 , 1 ) , ( 20 , 13 ) , ( 25 , 7 ) and The relation is a function .

Explanation:
As shown , The ordered pairs are ( 10 , 1 ) , ( 15 , 1 ) , ( 20 , 13 ) , ( 25 , 7 ) .
Each input has exactly one output ,
So, The relation is a function .

Question 7.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 9
Answer: The ordered pairs are ( 0 , -5 ) , ( 0 , -4 ) , ( 1 , -4 ) , ( 2 , -3 ) , ( 3 , -2 ) and relation is not a function .

Explanation:
As shown , The ordered pairs are ( 0 , -5 ) , ( 0 , -4 ) , ( 1 , -4 ) , ( 2 , -3 ) , ( 3 , -2 ) .
The input 0 has more than one output ,
So, The relation is not a function .

Question 8.
OPEN-ENDED
Copy and complete the mapping diagram at the left to represent a relation that is a function. Then describe how you can not modify the mapping diagram so that the relation is a function.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 10
Answer: ordered pairs are ( -8 , -4 ) , ( 0 , -2 ) , ( 8 , 0 ) , ( 16 , 2 ) . To have the relation as a function we must have only one output for one input.

Explanation:
The ordered pairs of the diagram are ( -8 , -4 ) , ( 0 , -2 ) , ( 8 , 0 ) , ( 16 , 2 ) .
Each Input must have only one output in order to be the relation is a function ,
If ,The mapping diagram has the right to left representation or each input has more than one output , then the relation is not a function .
So , To have the relation as a function we must have only one output for one input.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
The mapping diagram represents the costs of reserving a hotel room for different numbers of nights.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 11
a. Is the cost a function of the number of nights reserved?
b. Describe the relationship between the cost and the number of nights reserved.
Answer: a. Yes , The cost is a function of the number of nights reserved .
b. The relationship between the cost and the number of nights reserved is , For every night reservation of the room is increasing by $85 with increase in the next reservation ,

Explanation:
a. From the diagram we have ,
Ordered pairs are ( 1 , -$85 ) , ( 2 , $170 ) , ( 3 , $255 ) , ( 4 , $340 ) . each input has exactly one output ,
So , the relation is a function and ,
Yes , The cost is a function of the number of nights reserved .

b. The relationship between the cost and the number of nights reserved is ,
For every night reservation of the room is increasing by $85 with increase in the next reservation,
that is , input 1 = $85 as output
Input 2 = $85 + $85 = $170 as output
Input 3 = $170 + $85 = $255 as output
Input 2 = $255 + $85 = $340 as output

So, The relationship between the cost and the number of nights reserved is ,
For every night reservation of the room is increasing by $85 with increase in the next reservation.

Question 10.
DIG DEEPER!
The graph represents the number of contestants in each round of a talent competition.
a. Is the number of contestants a function of the round number?
b. Predict the number of contestants in the talent competition during Round 7. Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 12
Answer: a. The number of contestants is a function of the round number.
b. The number of contestants in the talent competition during Round 7  are 2.

Explanation:
a. From the given graph , The ordered pairs are ( 1 , 128 ) , ( 2 , 64 ) , ( 3 , 32 ) , ( 4 , 16 ) .
Each input has only one output , The relation is a function .
So , the number of contestants is a function of the round number.

b. Firstly , The relation between the input and output is,
With every increase in the round number the number of contestants are decreasing by half the number of the previous round , That is, for  input 1 = 128 as output
For input 2 = 128 – 64 = 64  as output
For input 3 = 64 – 32 = 32  as output
For input 4 = 32 – 16 = 16  as output
For input 5 = 16 – 8 = 8  as output
For input 6 = 8 – 4 = 4  as output
For input 7 = 4 – 2 = 2  as output,
So, The number of contestants in the talent competition during Round 7  are 2 .

Relations and Functions Homework & Practice 7.1

Review & Refresh

Choose an appropriate data display for the situation. Explain your reasoning.
Question 1.
the number of runners in each
age group at a marathon
Answer: In a marathon ,the people of all age group are participating for a promotion on healthy lifestyle, The number of runners in each  group has kids, adults and old people to spread the awareness of leading a healthy life by running daily in the morning . Running or jogging in the morning can help us to maintain our body mass index at an optimal level which is good for heart. The Marathon is conducted by the government of health ministry to be example for the future generations.

Question 2.
the high temperature and the
attendance at a water park each day
Answer:  Generally, The water park is normally crowded depending on the season and the temperature, In summer the attendance in the waterpark is at the utmost point because of the high temperature and the seasonal vacation. Going to the water park in summer is super fun due to the number of  water slides , water rides will be a nice place to the whole family trip and as well as friends . In order to be there at a less crowded time spring is also a nice time to visit the water park .

Graph the linear equation.
Question 3.
y = 2x – 3
Answer:
Explanation:
Given , y = 2x – 3 , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 1 , then y = 2(1) – 3 = 2 – 3 = -1 . co-ordinates are (1 , -1)
if x = 2 , then y = 2(2) – 3 = 4 – 3 = 1 , co-ordinates are (2 , 1)
The co-ordinates (1 , -1) , (2 , 1) form a straight line .
So, y = 2x – 3 is a linear equation.

Question 4.
y = – 0.5x
Answer:
Explanation:
Given , y = -0.5x , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = -0.5(0) = 0 . co-ordinates are (0 , 0)
if x = 2 , then y = -0.5(2) = -1 , co-ordinates are (2 , -1)
The co-ordinates (0 , 0) , (2 , -1) form a straight line .
So, y = -0.5x is a linear equation.

Question 5.
y = – 3x + 4
Answer:
Explanation:
Given , y = – 3x + 4 , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = – 3(0) + 4 = 4 . co-ordinates are (0 , 4)
if x = 1 , then y = – 3(1) + 4 = -3 + 4 = 1 , co-ordinates are (1 , 1)
if x = 2 , then y = – 3(2) + 4 = -6 + 4 = -2 , co-ordinates are (2 , -2)
The co-ordinates (0 , 4) , (1 , 1) , (2 , -2) form a straight line .
So, y = – 3x + 4 is a linear equation.

Question 6.
Which word best describes two figures that have the same size and the same shape?
A. congruent
B. adjacent
C. parallel
D. similar
Answer:  A. congruent

Explanation:
Two figures which have the same size and shape are congruent.

Concepts, Skills, &Problem Solving

INTERPRETING DIAGRAMS Describe the relationship between the inputs and outputs in the diagram. Then complete the diagram. Is there more than one possible answer? Explain your reasoning. (See Exploration 1, p. 275.)
Question 7.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 13
Answer: The relationship between the inputs and outputs in the diagram is ,
For every increase in number of input is having the output of adding -4 to the previous output.

Explanation:
The relationship between the inputs and outputs in the diagram is ,
For every increase in number of input is having the output of adding -4 to the previous output ,
for input 1 = -1 as output
for input 2 = -1 + (-4) = -5 as output
for input 3 = -5 + (-4) = -9 as output
for input 4 = -9 + (-4) = -13 as output
for input 5 = -13 + (-4) = -17 as output
for input 6 = -17 + (-4) = -21 as output.
So, The relationship between the inputs and outputs in the diagram is ,
For every increase in number of input is having the output of adding -4 to the previous output .

In this case , we are witnessing only one output for one input.

Question 8.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 14
Answer: The relationship between the inputs and outputs in the diagram is,
Each input has the sports name and the output has the Starting letter of sports name.

Explanation:
The relationship between the inputs and outputs in the diagram is,
Each input has the sports name and the output has the Starting letter of sports name.
For input basketball = b as output
For input baseball = b as output
For input football = f as output
For input soccer = s as output
For input swimming = s as output,
So, The relationship between the inputs and outputs in the diagram is,
Each input has the sports name and the output has the Starting letter of sports name.

In this case we have more than one output for input.

LISTING ORDERED PAIRS List the ordered pairs shown in the mapping diagram.
Question 9.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 15
Answer: Ordered pairs are ( 0 , 4 ) , ( 3 , 5 ) , ( 6 , 6 ) , ( 9 , 7 ) .

Explanation:
As shown , Ordered pairs are the combinations of input and output
So , Ordered pairs are ( 0 , 4 ) , ( 3 , 5 ) , ( 6 , 6 ) , ( 9 , 7 ) .

Question 10.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 16
Answer: Ordered pairs are ( 1 , 8 ) , ( 3 , 8 ) , ( 3 , 4 ) , ( 5 , 6 ) , ( 7 , 2 ).

Explanation:
As shown , Ordered pairs are the combinations of input and output
So , Ordered pairs are ( 1 , 8 ) , ( 3 , 8 ) , ( 3 , 4 ) , ( 5 , 6 ) , ( 7 , 2 ).

Question 11.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 17
Answer: Ordered pairs are ( 6 , -5 ) , ( 7 , -5 ) , ( 8 , -10 ) , ( 9 , -10 ).

Explanation:
As shown , Ordered pairs are the combinations of input and output
So , Ordered pairs are ( 6 , -5 ) , ( 7 , -5 ) , ( 8 , -10 ) , ( 9 , -10 ).

IDENTIFYING FUNCTIONS Determine whether the relation is a function.
Question 12.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 18
Answer: The relation is not a function .

Explanation:
The each input has  more than two outputs , That is one input has multiple number of outputs.
Here , input 0 has two outputs which are 10 and 20 .
So , The relation is not a function .

Question 13.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 19
Answer: The relation is a function .

Explanation:
Each input has exactly one output ,
So , The relation is a function .

Question 14.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 20
Answer: The relation is a function .

Explanation:
Each input has exactly one output ,
So , The relation is a function .

Question 15.
YOU BE THE TEACHER
Your friend determines whether the relation shown in the mapping diagram is a function. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 21
Answer: The relation is not a function .

Explanation:
The each input has  more than two outputs , That is one input has multiple number of outputs.
Here , input 4 has four outputs which are 5, 6 , 7 and 8.
So , The relation is not a function .

REASONING Draw a mapping diagram that represents the relation. Then determine whether the relation is a function. Explain.
Question 16.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 22
Answer:  The mapping diagram representing the relation is

Explanation:
From the given graph , co-ordinates of the ordering pairs are( 1 , 1 ), ( 3 , 3 ), ( -1 , -1 ), ( -3 , -3 ).
Each input has exactly one output ,
So , The relation is a function .

Question 17.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 23
Answer: The mapping diagram representing the relation is

Explanation:
From the given graph , co-ordinates of the ordering pairs are( 0 , 8 ),( 2 , 8 ),( 4 , 8 ),( 6 , 8 ),( 8 , 8 ),( -2 , 8 ),( -4 , 8 ). Each input has exactly one output ,
So , The relation is a function.

Question 18.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 24
Answer: The mapping diagram representing the relation is

Explanation:
From the given graph , co-ordinates of the ordering pairs are( -2 , 1 ),( -2 , 2 ),( -2 , 3 ),( -2 , 4 ),( -2 , 5 ),( -2 , 6 ).
Each input has more than one output ,
So , The relation is not a function.

Question 19.
MODELING REAL LIFE
The normal pressure at sea level is 1 atmosphere of pressure(1 ATM). As you dive below sea level, the pressure changes. The mapping diagram represents the pressures at different depths.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 25
a. Complete the mapping diagram.
b. Is pressure a function of depth?
c. Describe the relationship between pressure and depth.
d. List the ordered pairs. Then plot the ordered pairs in a coordinate plane. What do you notice about the points?
e. RESEARCH What are common depths for beginner scuba divers? What are common depths for experienced scuba divers?
Answer: The detailed explanation of  all the answers are given below .

Explanation:
a. The mapping diagram is
b. Yes , the pressure is a function of depth, Because depth is related to pressure in the given mapping diagram.

c. The relationship between pressure and depth is,
for every 10m increase in Depth of input there is an increase in 1 ATM pressure .

d. The ordered pairs are ( 0 , 1 ) , ( 10 , 2 ) , (20 , 3 ) , ( 30 , 4 ) , ( 40 , 5 ), ( 50 , 6 ).
The plot of the ordered pairs in a coordinate plane is

From the graph, we have seen that, if the depth of the diving of scuba drivers increases then the water pressure increases with increase in depth. So, the graph have straight line .

e. The common depths for beginner scuba divers is 30 feet to 60 feet or 9 to 18 meters ,
The common depths for experienced scuba divers is more than 60 feet or more than 18 meters .

Question 20.
DIG DEEPER!
The table shows the cost of purchasing 1, 2, 3, or 4 T-shirts from a souvenir shop.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 26
a. Is the cost a function of the number of T-shirts purchased?
b. Describe the relationship between the cost and the number cost per T-shirt of T-shirts purchased. How does the change as you purchase more T-shirts?
Answer: The detailed explanation of  all the answers are given below .

Explanation:
a. Yes , The cost is a function of the number of T-shirts purchased, Because the cost of the purchased T-shirts is varying with the number of T-shirts purchased.

b. The relationship between the cost and the number cost per T-shirt of T-shirts purchased is,
Input is the cost of 1 T-shirt is $10 as output , Then for 2 T-shirts cost will be $20
If 2 T-shirts will be purchased at same time, cost will be decreased by $2 so it will be $10 + 8 = $18 for 2 T-shirts.
As per the single T-shirt cost , For 3 T-shirts will be $30,
So in the table given that 3 T-shirts will cost $24 , because it cost $18 + 6 = $24 for 3 T-shirts.
It goes same for 4 T-shirts , For 4 T-shirts will be $40, because it cost $24 + 4 = $28 for 4 T-shirts.

The change as you purchase more T-shirts is For every increase in purchase of the number of T-shirts is decrease in the cost of total T-shirts purchased.

Question 21.
REPEATED REASONING
The table shows the outputs for several inputs. Use two methods to predict the output for an input of 200.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.1 27
Answer: The output for an input of 200 is 1025.

Explanation:
Method 1. The relation between inputs and outputs is as follows,
y = 25 + 5x
As input increases by 1 , output increases by 5 units,
To find output of 200 as input ,
put x = 200 in the equation,
y = 25 + 5(200)
= 25 + 1000
= 1025.
So , y = 1025.

Method 2.  As the table shown, for every increase in input there is an increase in 5 numbers in output,
So , For 1 input = 25 + 5 = 30 as output
For 2 input = 30 + 5 = 35 as out put
For 3 input = 35 + 5 = 40 as out put
For 4 input = 40 + 5 = 45 as out put
By doing this for number 200 as input we have , 1025 as output.

Lesson 7.2 Representations of Functions

EXPLORATION 1

Using a Table to Describe Relationships
Work with a partner. Make a table that shows the relationship  between the figure number x and the area A of each figure. Then use an equation to find which figure has an area of 81 square units when the pattern continues.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 1

Answer: a. The equation is y = 2x – 1, For figure has an area of 81 square units is 41.
b. The equation is y = x²,  For figure has an area of 81 square units is 9.

Explanation:
a. figure shows the 1 square unit of each box  for and it has a pattern of  2x – 1
figure 1 = 1 square unit
figure 2 =3 square units
figure 3 = 5 square unit and so on
So, the equation is y = 2x – 1 , it is in the form of y = mx + c,
Given to which figure has an area of 81 square units
substitute y  = 81, we have
y = 2x – 1
81 = 2x – 1
2x = 82
x = 41
So, For figure has an area of 81 square units is 41.

b. As shown above , we know that ,
figure 1 = 1 square unit
figure 2 =4 square units
figure 3 = 9 square unit and so on
Here we have a pattern of power of its own number,
So, the Equation  will be y = x²
Given to which figure has an area of 81 square units
substitute y  = 81, we have
x  = 9
So, For figure has an area of 81 square units is 9.

EXPLORATION 2

Using a Graph
Work with a partner. Use a graph to test the truth of each statement. If the statement is true, write an equation that shows how to obtain one measurement from the other.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 2
a. “You can find the horsepower of a race-car car engine if you know its volume in cubic inches”
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 3
b. “You can find the volume of a race-car engine in cubic centimeters if you know its volume in cubic inches.”
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 4

Answer: a. Given ordered pairs are (200 , 375) , (350 , 650) , (350 , 250) , (500 , 600)
We can not find the horsepower of a race-car car engine if you know its volume in cubic inches
b. Given ordered pairs are (100 , 1640) , (200 , 3280) , (300 , 4920) ,
Yes, You can find the volume of a race-car engine in cubic centimeters if you know its volume in cubic inches

Explanation:
a. Given ordered pairs are (200 , 375) , (350 , 650) , (350 , 250) , (500 , 600)
We can not find the horsepower of a race-car car engine if you know its volume in cubic inches

b. Given ordered pairs are (100 , 1640) , (200 , 3280) , (300 , 4920) ,
Yes, You can find the volume of a race-car engine in cubic centimeters if you know its volume in cubic inches

Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 5

Try It

Question 1.
Write a function rule for “The output is one-fourth of the input.”
Answer:  y = \(\frac{x}{4}\)

Explanation:
Let us say x is input and y is output , then
The output is one-fourth of the input, will be ,
y = \(\frac{x}{4}\).

Find the value of y when x = 5.
Question 2.
y = 4x – 1
Answer: y = 19.

Explanation:
Given, y = 4x – 1
substitute x = 5 , we get
y = 4(5) – 1
y = 20 – 1 = 19
So, y = 19.

Question 3.
y = 10x
Answer: y = 50

Explanation:
Given, y =10x
substitute x = 5 , we get
y = 10(5)
y = 50
So, y = 50.

Question 4.
y = 7 – 3x
Answer: y = -8.

Explanation:
Given, y = 7 – 3x
substitute x = 5 , we get
y = 7 – 3(5)
y = 7 – 15 = -8
So, y = -8.

Graph the function.
Question 5.
y = x + 1
Answer:

Explanation:
Given , y = x + 1  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 0 + 1 = 1 . co-ordinates are (0 , 1)
if x = 1 , then y = 1 + 1 = 2 . co-ordinates are (1 , 2)
if x = 2 , then y = 2 + 1 = 3 , co-ordinates are (2 , 3)
The co-ordinates (0 , 1) , (1 , 2) , (2 , 3) form a straight line .

Question 6.
y = – 3x
Answer:

Explanation:
Given , y = – 3x  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = -3(0) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = -3(1) = -3 . co-ordinates are (1 , -3)
if x = 2 , then y = -3(2) = -6 , co-ordinates are (2 , -6)
if x = 3 , then y = -3(3) = -9 , co-ordinates are (3 , -9)
The co-ordinates (0 , 0) , (1 , -3) , (2 , -6) ,(3 , -9) form a straight line .

Question 7.
y = 3x + 2
Answer:

Explanation:
Given , y = 3x + 2  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =3(0) + 2 = 2 . co-ordinates are (0 , 2)
if x = 1 , then y = 3(1) + 2= 5 . co-ordinates are (1 , 5)
if x = 2 , then y =3(2) + 2 = 7 , co-ordinates are (2 , 7)
The co-ordinates (0 , 2) , (1 , 5) , (2 , 7) form a straight line .

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

WRITING FUNCTION RULES Write a function rule for the statement.
Question 8.
The output is three times the input.
Answer: y = 3x

Explanation:
Let us say x is input and y is output , then
The output is three times the input. will be ,
So , y = 3x .

Question 9.
The output is eight more than one-seventh of the input.
Answer: y = 8 + \(\frac{x}{7}\) .

Explanation:
Let us say x is input and y is output , then
The output is eight more than one-seventh of the input., will be ,
So, y = 8 + \(\frac{x}{7}\) .

EVALUATING A FUNCTION Find the value of y when x = 5.
Question 10.
y = 6x
Answer: y = 30

Explanation:
Given, y = 6x
substitute x = 5 , we get
y = 6(5) =30
So, y = 30

Question 11.
y = 11 – x
Answer: y = 6

Explanation:
Given, y = 11 – x
substitute x = 5 , we get
y = 11 – 5 = 6
So, y = 6.

Question 12.
y = \(\frac{1}{5}\)x + 1
Answer:  y = 2.

Explanation:
Given, y = \(\frac{1}{5}\)x + 1
substitute x = 5 , we get
y = \(\frac{x}{5}\) + 1
y= \(\frac{5}{5}\) + 1
y = 1 + 1 = 2
So, y = 2 .

GRAPHING A FUNCTION Graph the function.
Question 13.
y = – 2x
Answer:

Explanation:
Given , y = – 2x  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = – 2(0) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = – 2(1)= -2 . co-ordinates are (1 , -2)
if x = 2 , then y =- 2(2) = -4 , co-ordinates are (2 , -4)
if x = 3 , then y =- 2(3) = -6 , co-ordinates are (3 , -6)
The co-ordinates (0 , 0) , (1 , -2) , (2 , -4) , (3 , -6) form a straight line .

Question 14.
y = x – 3
Answer:

Explanation:
Given , y = x – 3 , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 0 – 3 = -3 . co-ordinates are (0 , -3)
if x = 1 , then y = 1 – 3= -2 . co-ordinates are (1 , -2)
if x = 2 , then y = 2 – 3 = -1 , co-ordinates are (2 , -1)
if x = 3 , then y = 3 – 3 = 0 , co-ordinates are (3 , 0)
The co-ordinates (0 , -3) , (1 , -2) , (2 , -1) , (3 , 0) form a straight line .

Question 15.
y = 9 – 3x
Answer: 

Explanation:
Given , y = 9 – 3x , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 9 – 3(0) = 9 . co-ordinates are (0 , 9)
if x = 1 , then y = 9 – 3(1) = 6 . co-ordinates are (1 , 6)
if x = 2 , then y = 9 – 3(2) = 3 , co-ordinates are (2 , 3)
if x = 3 , then y = 9 – 3(3) = 0 , co-ordinates are (3 , 0)
The co-ordinates (0 , 9) , (1 , 6) , (2 , 3) , (3 , 0) form a straight line .

Question 16.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 6
Answer: As mentioned in the explanation below  a & d , b & c are different .

Explanation:
Given ,
a. what output is 4 more than twice the input 3?
Let us say that , y is output and x is input and given as 3 ,
then, we have y = 4 + 2(3) = 10.
b. What output is twice the sum of the input 3 and 4?
Let us say that , y is output and x is input and given as 3 ,
then, we have y = 2( 3 + 4 ) = 14.
c. what output is the sum of 2 times the input 3 and 4?
Let us say that , y is output and x is input and given as 3 ,
then, we have y = 2( 3 + 4 ) = 14.
d. what output is 4 increased by twice the input 3?
Let us say that , y is output and x is input and given as 3 ,
then, we have y = 4 + 2(3) = 10.

So, a & d , b & c are different .

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.
The World Health Organization(WHO) suggests having 23 health-care workers for every 10,000 people. How many health-care workers are needed to meet the WHO suggestion for a population of 250,000 people? Justify your answer using a graph.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 7
Answer: So, 575 health-care workers are needed to meet the WHO suggestion for a population of 250,000 people

Explanation:
Given, The World Health Organization(WHO) suggests having 23 health-care workers for every 10,000 people.
we need to find how many health-care workers are needed to meet the WHO suggestion for a population of 250,000 people,
For every 10,000 people we have 23 care takers
Then for 250,000 people we have
\(\frac{23 × 250,000}{10,000}\)
= 23 × 25
= 575
So, 575 health-care workers are needed to meet the WHO suggestion for a population of 250,000 people

Question 18.
DIG DEEPER!
A truck produces 22 pounds of carbon dioxide for every gallon of diesel fuel burned. The fuel economy of the truck is 18 miles per gallon. Write and graph a function that describes the relationship between carbon dioxide produced and distance traveled.
Answer: y = 22x + 18  is the linear equation

Explanation:
Given, A truck produces 22 pounds of carbon dioxide for every gallon of diesel fuel burned.
The fuel economy of the truck is 18 miles per gallon.
So, we have y = 22x + 18 is in the form of y = mx +c
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 22(0) + 18 = 18 . co-ordinates are (0 , 18)
if x = 1 , then y =22(1) + 18= 40  . co-ordinates are (1 , 40)
if x = 2 , then y =22(2) + 18 = 62 , co-ordinates are (2 , 62)
if x = 3 , then y =22(3) + 18 = 84  , co-ordinates are (3 , 84)
The co-ordinates (0 , 18) , (1 , 40) , (2 , 62) , (3 , 84) form a straight line .
The graph is

Representations of Functions Homework & Practice 7.2

Review & Refresh

Determine whether the relation is a function. 
Question 1.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 8
Answer:  The relation is a function .

Explanation:
Each input has exactly one output ,
So , The relation is a function .

Question 2.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 9
Answer:  The relation is a function .

Explanation:
Each input has exactly one output ,
So , The relation is a function .

Question 3.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 10
Answer: The relation is not a function .

Explanation:
The each input has  more than two outputs , That is one input has multiple number of outputs.
Here , input 2 has two outputs which are 0 and -4 .
So , The relation is not a function .

Find the slope of the line.
Question 4.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 11
Answer: slope = 1.

Explanation:
By using the slope equation , we know that
Slope = \(\frac{change in y}{change in x}\) or
slope = \(\frac{▲y}{▲x}\)
From the graph we know that change in y or ▲y is change from -2 to -4 =2
change in x or ▲x is change from 1 to 3 = 2 ,
So, slope = \(\frac{▲y}{▲x}\)
slope = \(\frac{2}{2}\)
slope = 1.

Question 5.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 12
Answer: slope = \(\frac{5}{2}\) .

Explanation:
By using the slope equation , we know that
Slope = \(\frac{change in y}{change in x}\) or
slope = \(\frac{▲y}{▲x}\)
From the graph we know that change in y or ▲y is change from -4 to 1 = 5
change in x or ▲x is change from -1 to -3 = 2 ,
So, slope = \(\frac{▲y}{▲x}\)
slope = \(\frac{5}{2}\) .

Question 6.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 13
Answer:  slope = \(\frac{1}{3}\) .

Explanation:
By using the slope equation , we know that
Slope = \(\frac{change in y}{change in x}\) or
slope = \(\frac{▲y}{▲x}\)
From the graph we know that change in y or ▲y is change from -4 to -3 = 1
change in x or ▲x is change from 1 to 4 = 3 ,
So, slope = \(\frac{▲y}{▲x}\)
slope = \(\frac{1}{3}\) .

Concepts, Skills, & Problem Solving

USING A GRAPH Use a graph to test the truth of the statement. If the statement is true, write an equation that shows how to obtain one measurement from the other measurement. (See Exploration 2, p. 281.)

Question 7.
“You can find the weight of a cell phone in ounces if you know its screen size in inches.”
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 14
Answer: we can does not find the weight of a cell phone in ounces if you know its screen size in inches.

From the given table , Ordered pairs are (4 , 4) , (4.7 , 4.8) , (5 , 4.8) , (5.5 , 6.4)
First find the slope m of the line containing the two given points (4, 4) and (4.7, 4.8)
m = (y2-y1) / (x2-x1)
m= (4.8 – 4) / (4.7 – 4)
m = 0.8/0.7 .
So, we can does not find the weight of a cell phone in ounces if you know its screen size in inches.

Question 8.
“You can find the age of a child in years if you know the age of the child in months.”
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 15
Answer: YES, y = 0.08x + 0.04 is a linear equations

Explanation:
From the given table , Ordered pairs are (9 , 0.75) , (12 , 1) , (15 , 1.25) , (24 , 2)
First find the slope m of the line containing the two given points (12 ,1) and (24, 2)
m = (y2-y1) / (x2-x1)
m= (2 – 1) / (24 – 12)
m = 1/12
m = 0.08.
substitute the slope in the (12 ,1) to get point slope to form a line.
y-y1 = m (x-x1)
y – 1 = 0.08(x – 12)
y –1 = 0.08x – 0.96
y = 0.08x –0.96 + 1
y =0.08 x + 0.04
So, y = 0.08x + 0.04 is a linear equation

WRITING FUNCTION RULES Write a function rule for the statement.
Question 9.
The output is half of the input.
Answer: y = \(\frac{x}{2}\).

Explanation:
Let us say x is input and y is output , then
The output is half of the input, will be ,
y = \(\frac{x}{2}\).

Question 10.
The output is eleven more than the input.
Answer: y = x + 11

Explanation:
Let us say x is input and y is output , then
The output is eleven more than the input, will be ,
y = x + 11

Question 11.
The output is three less than the input.
Answer: y = x – 3

Explanation:
Let us say x is input and y is output , then
The output is three less than the input, will be ,
y = x – 3

Question 12.
The output is the cube of the input.
Answer: y = x³

Explanation:
Let us say x is input and y is output , then
The output is the cube of the input, will be ,
y = x³

Question 13.
The output is six times the input.
Answer: y = 6x

Explanation:
Let us say x is input and y is output , then
The output is six times the input, will be ,
y = 6x

Question 14.
The output is one more than twice the input.
Answer: y = 2x + 1

Explanation:
Let us say x is input and y is output , then
The output is one more than twice the input, will be ,
y = 2x + 1

EVALUATING A FUNCTION Find the value of y for the given value of x.
Question 15.
y = x + 5; x = 3
Answer: y = 8

Explanation:
Given, y = x + 5
substitute x = 3 , we get
y = 3 + 5
So, y = 8.

Question 16.
y = 7x; x = – 5
Answer:  y = -35.

Explanation:
Given, y = 7x
substitute x = -5 , we get
y = 7(-5)
So, y = -35.

Question 17.
y = 1 – 2x; x = 9
Answer: y = -17

Explanation:
Given, y = 1 – 2x
substitute x = 9 , we get
y = 1 – 2(9)
y = 1 – 18
So, y = -17.

Question 18.
y = 3x + 2; x = 0.5
Answer: y = 5.5

Explanation:
Given, y = 3x + 2
substitute x = 0.5 , we get
y = 3(0.5) + 2
y = 3.5 + 2
So, y = 5.5 .

Question 19.
y = 2x3; x = 3
Answer: y = 54

Explanation:
Given, y = 2x3
substitute x = 3 , we get
y = 2(3)³
y = 2 × 27 = 54
So, y = 54.

Question 20.
y = \(\frac{x}{2}\) + 9; x = – 12
Answer: y = 3

Explanation:
Given, y = \(\frac{x}{2}\) + 9
substitute x = -12 , we get
y = \(\frac{-12}{2}\) + 9
y = -6 + 9
So, y = 3 .

GRAPHING A FUNCTION Graph the function.
Question 21.
y = x + 4
Answer: 

Explanation:
Given , y = x + 4  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 0 + 4 = 4 . co-ordinates are (0 , 4)
if x = 1 , then y = 1 + 4 = 5 . co-ordinates are (1 , 5)
if x = 2 , then y = 2 + 4 = 6 , co-ordinates are (2 , 6)
if x = 3 , then y = 3 + 4 = 7 , co-ordinates are (3 , 7)
The co-ordinates (0 , 4) , (1 , 5) , (2 , 6) , (3 , 7) form a straight line .

Question 22.
y = 2x
Answer:

Explanation:
Given , y = 2x  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 2(0) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = 2(1) = 2 . co-ordinates are (1 , 2)
if x = 2 , then y = 2(2) = 4 , co-ordinates are (2 , 4)
if x = 3 , then y = 2(3) = 6 , co-ordinates are (3 , 6)
The co-ordinates (0 , 0) , (1 , 2) , (2 , 4) , (3 , 6) form a straight line .

Question 23.
y = – 5x + 3
Answer:

Explanation:
Given , y = – 5x + 3  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =- 5(0) + 3 = 3 . co-ordinates are (0 , 3)
if x = 1 , then y = – 5(1) + 3 = -2 . co-ordinates are (1 , -2)
if x = 2 , then y = – 5(2) + 3 = -7 , co-ordinates are (2 , -7)
if x = 3 , then y = – 5(3) + 3 = -12 , co-ordinates are (3 , -12)
The co-ordinates (0 , 3) , (1 , -2) , (2 , -7) , (3 , -12) form a straight line .

Question 24.
y = \(\frac{x}{4}\)
Answer:

Explanation:
Given , y = \(\frac{x}{4}\) , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = \(\frac{0}{4}\) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = \(\frac{1}{4}\) = 0.25 . co-ordinates are (1 , 0.25)
if x = 2 , then y = \(\frac{2}{4}\) = 0.5 , co-ordinates are (2 , 0.5)
if x = 3 , then y = \(\frac{3}{4}\) = 0.75 , co-ordinates are (3 , 0.75)
The co-ordinates (0 , 0) , (1 , 0.25) , (2 , 0.5) , (3 , 0.75) form a straight line .

Question 25.
y = \(\frac{3}{2}\)x + 1
Answer:

Explanation:
Given , y = \(\frac{3}{2}\)x + 1  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =\(\frac{3}{2}\)(0) + 1 = 1 . co-ordinates are (0 , 1)
if x = 1 , then y = \(\frac{3}{2}\)(1) + 1= 2.5 . co-ordinates are (1 , 2.5)
if x = 2 , then y = \(\frac{3}{2}\)(2) + 1 = 4 , co-ordinates are (2 , 4)
if x = 3 , then y = \(\frac{3}{2}\)(3) + 1 = 5.5 , co-ordinates are (3 , 5.5)
The co-ordinates (0 , 1) , (1 , 2.5) , (2 , 4) , (3 , 5.5) form a straight line .

Question 26.
y = 1 + 0.5x
Answer:

Explanation:
Given , y = 1 + 0.5x  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =1 + 0.5(0) = 1 . co-ordinates are (0 , 1)
if x = 1 , then y = 1 + 0.5(1) = 1.5 . co-ordinates are (1 , 1.5)
if x = 2 , then y = 1 + 0.5(2) = 2 , co-ordinates are (2 , 2)
if x = 3 , then y = 1 + 0.5(3) = 2.5 , co-ordinates are (3 , 2.5)
The co-ordinates (0 , 1) , (1 , 1.5) , (2 , 2) , (3 , 2.5)  form a straight line .

MATCHING Match the graph with the function it represents.
A. y = \(\frac{x}{3}\)
B. y = x + 1
C. y = – 2x + 6
Question 27.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 16
Answer:  B. y = x + 1.

Explanation:
Given , y = x + 1  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 0 + 1 = 1 . co-ordinates are (0 , 1)
if x = 1 , then y = 1 + 1 = 2 . co-ordinates are (1 , 2)
if x = 2 , then y = 2 + 1 = 3 , co-ordinates are (2 , 3)
The co-ordinates (0 , 1) , (1 , 2) , (2 , 3) form a straight line .

Question 28.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 17
Answer: c. y = – 2x + 6

Explanation:
Given , y = – 2x + 6  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = – 2(0) + 6 = 6 . co-ordinates are (0 , 6)
if x = 1 , then y = – 2(1) + 6 = 4 . co-ordinates are (1 , 4)
if x = 2 , then y = – 2(2) + 6 = 2 , co-ordinates are (2 , 2)
if x = 3 , then y = – 2(3) + 6 = 0 , co-ordinates are (3 , 0)
The co-ordinates (0 , 6) , (1 , 4) , (2 , 2) , (3 , 0) form a straight line .

Question 29.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 18
Answer: A. y = \(\frac{x}{3}\)

Explanation:
Given , y =  \(\frac{x}{3}\) , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =\(\frac{0}{3}\) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = \(\frac{1}{3}\)= 0.3 . co-ordinates are (1 , 0.3)
if x = 2 , then y = \(\frac{2}{3}\)= 0.6 , co-ordinates are (2 , 0.6)
The co-ordinates (0 , 0) , (1 , 0.3) , (2 , 0.6) form a straight line .

Question 30.
YOU BE THE TEACHER
Your friend graphs the function represented by the input-output table. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 19
Answer: Yes , He is correct

Explanation:

Ordered pairs are (-1 , -4) , (1 , -2) , (3 ,0) , (5 , 2)
these points form a straight line when graphed.
Yes , He is correct

Question 31.
MODELING REAL LIFE
A dolphin eats 30 pounds of fish per day.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 20
a. Write and graph a function that relates the number p of pounds of fish that a dolphin eats in d days.
b. How many total pounds of fish does a dolphin eat in 30 days?
Answer:

Explanation:
a. Given , A dolphin eats 30 pounds of fish per day.
by each passing day eating fish is increased by the day passes .
So, y = 30x is the function,
The graph represents the function as

b. Given , A dolphin eats 30 pounds of fish per day.
then for 30 days ,
30 × 30 = 900 pounds
So, A dolphin eats 900 pounds of fish in 30 days

Question 32.
MODELING REAL LIFE
You fill a fish tank with 55 gallons of water on Saturday. The water evaporates at a rate of 1.5 gallons per day. You plan to add water when the tank reaches 49 gallons. When will you add water? Justify your answer.
Answer: As the action starts on Saturday , the tank will reach 49 gallons after 4 days , That is on Wednesday.

Explanation:
Given data ,, implies that slope of the function m = -1.5
The y intercept b= 55,
Then the equation  will be y = 55 – 1.5x
Given , You plan to add water when the tank reaches 49 gallons.
determine x for y = 49 ,
So, 49 = 55 – 1.5x ,
1.5x = 55 – 49
1.5x = 6
x = \(\frac{6}{1.5}\)
x = 4.

As the action starts on Saturday , the tank will reach 49 gallons after 4 days , That is on Wednesday.

USING AN EQUATION Find the value of x for the given value of y.
Question 33.
y = 5x – 7; y = – 22
Answer: x = -3

Explanation:
Given, y = 5x – 7
x = \(\frac{y + 7}{5}\)
substitute y = -22 , we get
x = \(\frac{-22 + 7}{5}\)
x = \(\frac{- 15}{5}\)
x = -3
So, x = -3 .

Question 34.
y = 9 – 7x; y = 37
Answer: x = -4

Explanation:
Given, y = 9 – 7x
x = \(\frac{9 – y}{7}\)
substitute y = 37 , we get
x = \(\frac{9 – 37}{7}\)
x = \(\frac{- 28}{7}\)
x = -4
So, x = -4 .

Question 35.
y = \(\frac{x}{4}\) – 7; y = 2
Answer: x = 36

Explanation:
Given, y = \(\frac{x}{4}\) – 7
x = 4( y + 7)
substitute y = 2 , we get
x = 4( 2 + 7)
x = 4(9)
x = 36
So, x = 36 .

Question 36.
PROBLEM SOLVING
You decide to make and sell bracelets. The cost of your materials is $84.00. You charge $3.50 for each bracelet.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 21
a. P Write a function that represents the profit for selling b bracelets.
b. Which variable is independent? dependent? Explain.
c. You will break even when the cost of your materials equals your income. How many bracelets must you sell to break even?
Answer: a. A function that represents the profit for selling b bracelets is p = 3.5b – 84.
b. Here , the profit depends on the number of bracelets sold , b is the independent variable and p is the dependent variable.
c. To break even you must sell 24  bracelets.

Explanation:
a. Given , The cost of your materials is $84.00. You charge $3.50 for each bracelet,
Let p be the profit , b be the number of bracelets sold,
So, profit = income – cost .
p = 3.5b – 84.
Thus , A function that represents the profit for selling b bracelets is p = 3.5b – 84.

b. Here , the profit depends on the number of bracelets sold , b is the independent variable and p is the dependent variable.

c. set the income expression from part a equal to the cost of 84 and solve for b ,
So, income = cost .
3.5b = 84 ,
b = \(\frac{84}{3.5}\)
b = 24.

To break even you must sell 24  bracelets.

Question 37.
MODELING REAL LIFE
A furniture store is having a sale where everything is 40% off.
a. Write and graph a function that represents the amount of discount on an item at regular price.
b. You buy a bookshelf that has a regular price of $85. What is the sale price of the bookshelf?
Answer: a. The function is y = 0.4x and the graph is given below.
b. The sale price of the bookshelf s $51.

Explanation:
a. A function that represents the amount of discount on an item at regular price is ,
Given , 40% = 0.4 ,
To find the percent of the number , we should multiply the number by the percent in the decimal form ,
so, the equation is d = 0.4p ,
let us convert it in to a function form , y = 0.4x
we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 0.4(0) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = 0.4(1)= 0.4 . co-ordinates are (1 , 0.4)
if x = 2 , then y =0.4(2) = 0.8 , co-ordinates are (2 , 0.8)
if x = 3 , then y = 0.4(3) = 1.2 , co-ordinates are (3 , 1.2)
The co-ordinates (0 , 0) , (1 , 0.4) , (2 , 0.8) , (3 , 1.2) form a straight line .
The graph is
b. Given , You buy a bookshelf that has a regular price of $85.
The sale price of the bookshelf is ,
substituting the given price in p = 85 ,
it will be the discount d = 0.4 (85) = 34
Then the sale price is $85 – $34 = $51.

So, The sale price of the bookshelf s $51.

Question 38.
REASONING
You want to take a two-hour air boat tour. Which is a better deal, Snake Tours or Gator Tours? Use functions to justify your answer.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 22
Answer: By using functions , $50 > $40 , So, Gator tours are cheaper than the snake tours .

Explanation:
Given , You want to take a two-hour air boat tour.
Let x be the hours of  air boat tour and y be the cost of air boat tour ,
Snake tours , y = 25x
putt x = 2 ,
So , y = 25 (2) = 50 .
y = 50.

Gator tour , y = 35 + \(\frac{5}{2}\)x
Put x = 2 ,
So, y = 35 + \(\frac{5}{2}\) x
y = 35 + 2.5x
y = 35 + 2.5 (2)
y = 35 + 5
y = 40 .

Finally $50 > $40 , So, Gator tours are cheaper than the snake tours

Question 39.
REASONING
The graph of a function is a line that passes through the points (3, 2), (5, 8), and (8, y). What is the value of y?
Answer: The value of y is 17 , so, The third given point is (8, 17)

Explanation:
First find the slope m of the line containing the two given points (3,2) and (5,8)
m = (y2-y1) / (x2-x1)
m= (8 – 2) / (5 – 3)
m = 6 / 2
m = 3
Then use the slope and one of the given points (3,2) to find the y-intercept
y = mx +
2 = 3(3) + b
2 = 9 + b
-7 = b
The equation is   y = 3x -7
Then find the third point (8, y) by replacing x by 8
y = 3x -7
y = 3(8) -7
y = 24 -7
y = 17

so the third given point is (8, 17)

Question 40.
CRITICAL THINKING
Make a table where the independent variable is the side length of a square and the dependent variable is the perimeter. Make a second table where the independent variable is the side length of a square and the dependent variable is the area. Graph both functions in the same coordinate plane. Compare the functions.
Answer: The graph for the perimeter is linear , The graph for the Area is Quadratic .

Explanation:
Let us say , s be the side length of the square ,
Then the perimeter is P = 4s ,
The function will be y= 4x,
we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 4(0) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = 4(1) = 4 . co-ordinates are (1 , 4)
if x = 2 , then y = 4(2) =8 , co-ordinates are (2 , 8)
if x = 3 , then y = 4(3) = 0 , co-ordinates are (3 , 12)
The co-ordinates (0 , 0) , (1 , 4) , (2 ,8) , (3 , 12) form a straight line .

Table will be ,

Let us say , s be the side length of the square ,
Then the Area is A = s² ,
The function will be y=x²,
we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 0² = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = 1² = 1 . co-ordinates are (1 , 1)
if x = 2 , then y = 2² =4 , co-ordinates are (2 , 4)
if x = 3 , then y = 3² = 9 , co-ordinates are (3 , 9)
The co-ordinates (0 , 0) , (1 , 1) , (2 ,4) , (3 , 9) form a straight line .

Second table is
Then the graph is 
The graph for the perimeter is linear , The graph for the Area is Quadratic .

Question 41.
PUZZLE
The blocks that form the diagonals of each square are shaded. Each block has an area of one square unit. Find the “green area” of Square 20. Find the “green area” of Square 21. Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 7 Functions 7.2 23
Answer:  The green area of the Square 20 is 46 square units and The green area of the Square 21 is 48 square units.

Explanation:
Given , Each block has an area of one square unit,
Square 1 has  the diagonals of each square are shaded. the “green area” is 3 + 3 = 6 square units ,
Square 2 has  the diagonals of each square are shaded. the “green area” is 4 + 4 = 8 square units ,
Square 3 has  the diagonals of each square are shaded. the “green area” is 5 + 5 = 10 square units ,
Square 4 has  the diagonals of each square are shaded. the “green area” is 6 + 6 = 12 square units,
Square 5 has  the diagonals of each square are shaded. the “green area” is 7 + 7 = 14 square units ,
Here , The number of squares are increasing by one block with the square numbers.
So for the , Square 20 has  the diagonals of each square are shaded. the “green area” is 23 + 23 = 46 square units,
And Square 21 has  the diagonals of each square are shaded. the “green area” is 24 + 24 = 48 square units.

Lesson 7.3 Linear Functions

EXPLORATION 1

Writing and Graphing Functions
Work with a partner. Each table shows a familiar pattern from geometry.

  • Determine what the variables x and y represent. Then write a function rule that relates y to x.
  • Is the function a linear function? Explain your reasoning.

Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 1
Answer: All of them are explained below

Explanation:
The variables x and y represents a rectangle
a. From the given table , Ordered pairs are (1 , 10) , (2 , 12) , (3 , 14) , (4 , 16)
First find the slope m of the line containing the two given points (1 ,-1) and (2, -2)
m = (y2-y1) / (x2-x1)
m= (-2 – (-1)) / (2 – 0)
m = -1 / 2 .
substitute the slope in the (1 , 10) to get point slope to form a line.
y-y1 = m (x-x1)
y – 10 = -1/2 ( x –1)
2(y – 10) = -x  + 1
2y – 20 = -x+ 1
2y = -x  + 21
y = \(\frac{-1}{2}\) (x – 21)
So ,  y = \(\frac{-1}{2}\) (x – 21) is linear function.

b. The variables x and y represent a circle
Ordered pairs are (1 , 3.14 ) , (2 , 6.28) , (3 , 9.42) , (4 , 12.5 )
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (1 , 3.14 ) , (2 , 6.28)
m = (y2-y1) / (x2-x1)
m= (6.28 –3.14) / (2– 1)
m = 3.14/1
m = 3.14
substitute the slope in the (8 , 4) to get point slope to form a line.
y-y1 = m (x-x1)
y – 3.14 =3.14 ( x –1)
y – 3.14 = 3.14x – 3.14
y = 3.14x – 3.14 + 3.14
y = 3.14x
So ,  y = 3.14x  is linear function.
Where x is the diameter of the circle.

c. The variables x and y represents a trapezoid
a. From the given table , Ordered pairs are (1 , 5) , (2 , 6) , (3 , 7) , (4 , 8)
First find the slope m of the line containing the two given points (1 ,5) and (2, 6)
m = (y2-y1) / (x2-x1)
m= (6 – 5) / (2 – 1)
m = 1 .
substitute the slope in the (1 ,5) to get point slope to form a line.
y-y1 = m (x-x1)
y – 5 = 1(x – 1)
y – 5 = x – 1
y = x – 1 + 5
y = x + 4
So, y = x + 4 is a linear equation

d. The variables x and y represents a cube
a. From the given table , Ordered pairs are (1 , 28) , (2 , 40) , (3 , 52) , (4 , 64)
First find the slope m of the line containing the two given points (1 ,28) and (2, 40)
m = (y2-y1) / (x2-x1)
m= (40 – 28) / (2 – 1)
m = 12 .
substitute the slope in the (1 ,28) to get point slope to form a line.
y-y1 = m (x-x1)
y – 28 = 12(x – 1)
y – 28 = 12x – 12
y = 12x – 12 + 28
y = 12x + 16
So, y = 12x + 16 is a linear equation.

Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 2

Try It

Question 1.
Use the graph to write a linear function that relates y to x.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 3
Answer: the linear function is y = \(\frac{-1}{2}\)x -1.

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (-4 , 1) , (-2 , 0 ) , (0 , -1) , ( 2, -2 )
First find the slope m of the line containing the two given points (0 ,-1) and (2, -2)
m = (y2-y1) / (x2-x1)
m= (-2 – (-1)) / (2 – 0)
m = -1 / 2 .
Because the line crosses the y axis at ( 0, -1 ) , The y intercept is -1.
So , the linear function is y = \(\frac{-1}{2}\)x -1.

Question 2.
Use the table to write a linear function that relates y to x.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 4
Answer: the linear function is y = (0)x + 2.

Explanation:
Ordered pairs are (-2 , 2) , (-1 , 2) , (0 , 2) , (1 , 2)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (0 ,2) and (1, 2)
m = (y2-y1) / (x2-x1)
m= (2 – 2) / (1 – 0)
m = 0
Because the line crosses the y axis at ( 0, 2 ) , The y intercept is 2.
So , the linear function is y = (0)x + 2.

Question 3.
WHAT IF?
The rate of descent doubles. Repeat parts (a) and (b).
Answer: a. the linear function is y = -1x + 65.
b. The slope indicates that the height decreases 1000 feet per minute.
The y intercept indicates that the descent begins at a cruising altitude of 65,000 feet.

Explanation:
a. From the Given table , The rate of descents is 5
If it doubles , then The rate of descents is 10.
The the ordered pairs will be (0 , 65) , (10 ,55) , (20 , 45) .
First find the slope m of the line containing the two given points (0 ,65) and (10, 55)
m = (y2-y1) / (x2-x1)
m= (55 – 65) / (10 – 0)
m = -10 / 10
m = -1
Because the line crosses the y axis at ( 0, 65 ) , The y intercept is 65.
So , the linear function is y = -1x + 65.

b. The slope indicates that the height decreases 1000 feet per minute.
The y intercept indicates that the descent begins at a cruising altitude of 65,000 feet.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
WRITING A LINEAR FUNCTION
Use the graph to write a linear function that relates y to x.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 5
Answer:  The linear function is y = -4x -2 .

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (-2 , 6) , (-1 , 2 ) , (0 , -2) , ( 1, -6 )
First find the slope m of the line containing the two given points (0 ,-2) and (1, -6)
m = (y2-y1) / (x2-x1)
m= (-6 – (-2)) / (1 – 0)
m = -4 .
Because the line crosses the y axis at ( 0, -2) , The y intercept is -2.
So , the linear function is y = -4x -2 .

Question 5.
INTERPRETING A LINEAR FUNCTION
The table shows the revenue R (in millions of dollars) of a company when it spends A (in millions of dollars) on advertising.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 6
a. Write and graph a linear function that relates R to A.
b. Interpret the slope and the y-intercept.
Answer:  a. The linear function is y = 2x + 2. and the graph is shown below
b. The slope indicates that the increasing in the amount of spending on advertising by 2 million dollars
The y intercept indicates that the Revenue begins to increasing from the 2 million dollars.

Explanation:
a. From the given table ,
The the ordered pairs will be (0 , 2) , (2 ,6) , (4 , 10) , (6 , 14) , (8 ,18) .
The graph is
First find the slope m of the line containing the two given points (0 ,2) and (2, 6)
m = (y2-y1) / (x2-x1)
m= (6 – 2) / (2 – 0)
m = 4 / 2
m = 2
Because the line crosses the y axis at ( 0, 2 ) , The y intercept is 2.
So , the linear function is y = 2x + 2.

b. The slope indicates that the increasing in the amount of spending on advertising by 2 million dollars
The y intercept indicates that the Revenue begins to increasing from the 2 million dollars.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
Manager A earns $15 per hour and receives a $50 bonus. The graph shows the earnings of Manager B. (a) Which manager has a greater hourly wage? (b) After how many hours does Manager B earn more money than Manager A?
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 7
Answer: a. Manager B has the greater hourly wage than Manager A .
b. As manager A receives a $50 bonus , Manager B has to work an hour extra to earn more money than Manager A .

Explanation:
a. Manager A earns $15 per hour and receives a $50 bonus.
The ordered pairs will be  (0 , 0) , (1 , 15) , (2 , 30) , (3 , 45)
The graph shows the earnings of Manager B.
Ordered pairs from the graph are  (0 , 0) , (1 , 25) , (2 , 50) , (3 , 75)
So, Manager B has the greater hourly wage than Manager A .

b. As manager A receives a $50 bonus , Manager B has to work an hour extra to earn more money than Manager A .

Question 7.
Each month, you start with 2 gigabytes of data and use 0.08 gigabyte per day. The table shows the amount (in gigabytes) of data that your friend has left days after the start of each month. Who runs out of data first? Justify your answer.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 8
Answer:  you will be run out of data first

Explanation:
a. Given , Each month, you start with 2 gigabytes of data and use 0.08 gigabyte per day.
Let x be the number of days and y be the total data in gigabytes.
So, y = -0.08x + 2 ,
You will be out of data if , -0.08x + 2 = 0 ,
-0.08x + 2 = 0
2 = 0.08x
x = \(\frac{2}{0.08}\)
x = 25.
Hence ,you will be run out of data in 25 days.
b. Daily data usage for the friend will be given by the slope of the graph.
The the ordered pairs will be (0 , 3) , (7 ,2.3) , (14 , 1.6) .
First find the slope m of the line containing the two given points (7 ,2.3) and (14, 1.6)
m = (y2-y1) / (x2-x1)
m= (1.6 – 2.3) / (14 – 7)
m = -0.7 / 7
m = -0.1
Because the line crosses the y axis at ( 0, 3 ) , The y intercept is 3.
So , the linear function is y = -0.1x + 3.
Your friend will be out of data if ,
-0.1x + 3 = 0
3 = 0.1x
x = \(\frac{3}{0.1}\)
x = 30 .
Hence ,Friend will be run out of data in 30 days

So , you will be run out of data first

Linear Functions Homework & Practice 7.3

Review & Refresh

Write a function rule for the statement. Then graph the function.
Question 1.
The output is ten less than the input.
Answer: y = x – 10.

Explanation:
Let us say x is input and y is output , then
The output is ten less than the input, will be ,
y = x – 10.

Question 2.
The output is one-third of the input.
Answer: y = \(\frac{x}{3}\)

Explanation:
Let us say x is input and y is output , then
The output is one-third of the input, will be ,
y = \(\frac{x}{3}\) .

Solve the system.
Question 3.
y = x + 5
y = – 3x + 1
Answer: X = 0 , Y = 5

Explanation:
Y=3X+5 ——————-(1)
Y=X+5 ——————(2)
Substitute Y=X+5 in equation (1)
X+5=3X+5
Solve it for X
X+3X=55
4X=0
X=0/4=0
X = 0
Substitute X=0 in equation (1)
Y=0+5
Y=5

Question 4.
x + y = – 4
6x + 2y = 4
Answer:  X = 3 , Y= -7 .

Explanation:
2Y=−6X+4 ——————-(1)
Y= –X-4 ——————(2)
Substitute Y= –X-4 in equation (1)
2Y = −6X+4
2 ( X – 4 ) = −6X + 4
-2X – 8 = -6X + 4
6X -2X = 8 + 4
4X = 12
X = 3
Substitute X=3 in equation (2)
Y=– 3 – 4
Y= -7 .

Question 5.
– 4x + 3y = 14
y = 2x + 8
Answer:  X = -5 , Y = -2 .

Explanation:
3Y = 4X+14 ——————-(1)
Y = 2X + 8 ——————(2)
Substitute Y= 2X + 8 in equation (1)
3Y = 4X+14
3(2X + 8) = 4X+14
6X + 24 = 4X + 14
6X – 4X = 14 – 24
2X = -10
X = -5
Substitute X= -5 in equation (2)
Y= 2(-5) + 8
Y= -10 + 8
Y = -2.

Concepts, Skills, &Problem Solving

WRITING AND GRAPHING FUNCTIONS The table shows a familiar pattern from geometry. (a) Determine what the variables x and y represent. Then write a function rule that relates y to x. (b) Is the function a linear function? Explain your reasoning. (See Exploration 1, p. 289.)
Question 6.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 9
Answer: a. The variables x and y represent a right angle triangle
b. y = 2x  is linear function.

Explanation:
In order to write the function we have to write the ordered pairs
Ordered pairs are  (1 , 2) , (2 , 4) ,  (3 , 6 ) , (4 , 8), (5 , 10 ) .
a. the variables x and y represent a right angle triangle
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (1 , 2) , (2 , 4)
m = (y2-y1) / (x2-x1)
m= (4 – 2) / (2– 1)
m = 2/1
m = 2
b. substitute the slope in the (2 , 4) to get point slope to form a line.
y-y1 = m (x-x1)
y – 4 = 2 ( x – 2)
y – 4 = 2x – 4
y = 2x – 4 + 4
y = 2x
So ,  y = 2x  is linear function.

Given side of triangle is 4 then x= 4/2 = 2
x = 2 and y = 4.

Question 7.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 10
Answer: y = 3.14x  is linear function. and The variables x and y represent a circle

Explanation:
Ordered pairs are (1 , 3.14 ) , (2 , 6.28) , (3 , 9.42) , (4 , 12.5 )
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (1 , 3.14 ) , (2 , 6.28)
m = (y2-y1) / (x2-x1)
m= (6.28 –3.14) / (2– 1)
m = 3.14/1
m = 3.14
substitute the slope in the (8 , 4) to get point slope to form a line.
y-y1 = m (x-x1)
y – 3.14 =3.14 ( x –1)
y – 3.14 = 3.14x – 3.14
y = 3.14x – 3.14 + 3.14
y = 3.14x
So ,  y = 3.14x  is linear function.
Where x is the diameter of the circle.

WRITING LINEAR FUNCTIONS Use the graph or table to write a linear function that relates y to x.
Question 8.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 11
Answer: The linear function is y = \(\frac{4}{3}\)x +2

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (-3 , -2) , (0 , 2 ) , (3 , 6) , ( 6, 10 )
First find the slope m of the line containing the two given points (3 ,6) and (6, 10)
m = (y2-y1) / (x2-x1)
m= (10 – 6) / (6 – 3)
m = 4/3 .
Because the line crosses the y axis at ( 0, 2) , The y intercept is 2.
So , the linear function is y = \(\frac{4}{3}\)x +2 .

Question 9.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 12
Answer: The linear function is y = (0)x +3 .

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (-2 , 3) , (-1 , 3 ) , (0 , 3) , ( 1, 3 ) , (2 , 3)
First find the slope m of the line containing the two given points (1 ,3) and (2, 3)
m = (y2-y1) / (x2-x1)
m= (3 – 3) / (2 – 1)
m = 0 .
Because the line crosses the y axis at ( 0, 3) , The y intercept is 3.
So , the linear function is y = (0)x +3 .

Question 10.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 13
Answer: The linear function is y = \(\frac{-1}{4}\)x + 0.

Explanation:
Ordered pairs are (-8 , 2) , (-4 , 1) , (0 , 0) , (4 , -1)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (-8 ,2) and (-4, 1)
m = (y2-y1) / (x2-x1)
m= (1 – 2) / (-4 – (-8))
m = -1/4
Because the line crosses the y axis at ( 0, 0 ) , The y intercept is 0.
So , the linear function is y = \(\frac{-1}{4}\)x + 0.

Question 11.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 14
Answer: The linear function is y = \(\frac{2}{3}\)x + 5.

Explanation:
Ordered pairs are (-3 , 3) , (0 , 5) , (3 , 7) , (6 , 9)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (3 ,7) and (6, 9)
m = (y2-y1) / (x2-x1)
m= (9 – 7) / (6 – 3)
m = 2/3
Because the line crosses the y axis at ( 0, 5 ) , The y intercept is 5.
So , the linear function is y = \(\frac{2}{3}\)x + 5.

Question 12.
INTERPRETING A LINEAR FUNCTION
The table shows the length y (in inches) of a person’s hair after x months.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 15
a. Write and graph a linear function that relates y to x.
b. Interpret the slope and the y-intercept.
Answer: a. The linear function is y = 0.5x + 11.
b. The slope indicates that the increasing in the hair length
The y intercept indicates that the increasing in hair length by time.

Explanation:
a. Given ,
The ordered pairs will be (0 , 11) , (3 ,12.5) , (6 , 14) .
The graph is
First find the slope m of the line containing the two given points (3 ,12.5) and (6 , 14)
m = (y2-y1) / (x2-x1)
m= (14 – 12.5) / (6 – 3)
m = 1.5 / 3
m = 0.5
Because the line crosses the y axis at ( 0, 11 ) , The y intercept is 11.
So , the linear function is y = 0.5x + 11.

b. The slope indicates that the increasing in the hair length
The y intercept indicates that the increasing in hair length by time.

Question 13.
INTERPRETING A LINEAR FUNCTION
The table shows the percent (in decimal form) of battery power remaining x hours after you turn on a laptop computer.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 16
a. Write and graph a linear function that relates y to x.
b. Interpret the slope, the x-intercept, and the y-intercept.
c. After how many hours is the battery power at75%?
Answer: a. The linear function is y = -0.2x + 1.
b. given below the explanation.
c. Battery will be 75% after 1.25 hours.

Explanation:
a. Given ,
The ordered pairs will be (0 , 1) , (2 ,0.6) , (4 , 0.2) .
The graph is
First find the slope m of the line containing the two given points (2 ,0.6) and (4 , 0.2)
m = (y2-y1) / (x2-x1)
m= (0.2 – 0.6) / (4 – 2)
m = -0.4 / 2
m = -0.2
Because the line crosses the y axis at ( 0, 1 ) , The y intercept is 1.
So , the linear function is y = -0.2x + 1.

b. Slope is -0.2 which means that as time increases by 1 hour, Battery power remaining decreases by 20% .
y intercept is 1, which means initially the battery power remaining before usage was 100%.
x intercept is 5 which means the battery remaining will be 0 after 5 hours.

c. battery percent will be 75% of 0.75 if ,
-0.2x + 1 = 0.75
0.2x = 1 – 0.75
x = 0.25/0.2
x = 1.25
Battery will be 75% after 1.25 hours.

Question 14.
MODELING REAL LIFE
The number y of calories burned after x minutes of kayaking is represented by the linear function y = 4.5x. The graph shows the number of calories burned by hiking.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 17
a. Which activity burns more calories per minute?
b. You perform each activity for 45 minutes. How many total calories do you burn? Justify your answer.
Answer: a. hiking burns more calories than kayaking .
b. In kayaking, 202.5 calories are burnt per minute. and In hiking , 225 calories are burnt per minute.

Explanation:
a. The number y of calories burned after x minutes of kayaking is represented by the linear function y = 4.5x.
So, The ordered pairs of the graph are (0 , 0) , (1 , 4.5) , (2 , 9) , (3, 13.5)
Here , In kayaking burns 4.5 calories per minute .
For hiking ,
The ordered pairs of the graph are (0 , 0) , (1 , 5) , (2 , 10) , (3, 15)
Here , In hiking burns 5 calories per minute.
Thus , hiking burns more calories than kayaking .

b. Given , perform each activity for 45 minutes.
Liner function of the kayaking is y = 4.5x
substitute x = 45 in equation
y = 4.5 (45)
y = 202.5
In kayaking, 202.5 calories are burnt per minute.
Linear function of the hiking is y = 5x
substitute x = 45 in equation
y = 5 (45)
y = 225
In hiking , 225 calories are burnt per minute.

Question 15.
DIG DEEPER!
You and a friend race each other. You give your friend a 50-foot head start. The distance y (in feet) your friend runs after x seconds is represented by the linear function y = 14x + 50. The table shows your distance at various times throughout the race. For what distances will you win the race? Explain.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 18
Answer: you will win the race for distances greater than 190 feet

Explanation:
The distance y (in feet) your friend runs after x seconds is represented by the linear function y = 14x + 50.
The slope of the line is 14 so , your friend runs at the rate of 14 ft per second
To find your rate , the ordered pairs are (2 , 38) , (4 , 76) , (6 , 114) , (8 , 152)
First find the slope m of the line containing the two given points (2 ,38) and (4 , 76)
m = (y2-y1) / (x2-x1)
m= (76 – 38) / (4 – 2)
m = 38 / 2
m = 19
You are running at the rate of 19 ft per second.
To get the linear equation , substitute the slope in the (2 , 38) to get point slope to form a line.
Then we have , y = 19x
Now if x = 10 , to run faster then ,
y = 19(10)
y = 190 .
Your friend linear equation is y = 14x + 50 .
if x = 10 ,then
y = 14(10) + 50
y = 140 + 50
y = 190.
So , for x > 10 , means you will run farther than your friend which means you would win the race .
Therefore, you will win the race for distances greater than 190 feet.

Question 16.
REASONING
You and your friend are saving money to buy bicycles that cost $175 each. You have $45 to start and save an additional $5 each week. The graph shows the amount y(in dollars) that your friend has after x weeks. Who can buy a bicycle first? Justify your answer.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 19
Answer:  your friend will but the bicycle first.

Explanation:
Given , your friend savings are
the ordered pairs are (0,15) and (3,39)
First find the slope m of the line containing the two given points (0,15) and (3,39)
m = (y2-y1) / (x2-x1)
m= (39 – 15) / (3 – 0)
m = 24 / 3
m = 8
Because the line crosses the y axis at ( 0, 15 ) , The y intercept is 15.
So , the linear function is y = 8x + 15.
to buy bicycles that cost $175 each
if y = 175 , then
175 = 8x + 15
8x = 175 – 15
x = 160/8
x = 20
So, your friend need 20 weeks to buy the bicycle
Given, You have $45 to start and save an additional $5 each week
So , the linear function will be y = 5x + 45.
to buy bicycles that cost $175 each
if y = 175 , then
175 = 5x + 45
5x = 175 – 45
x = 130/5
x = 26
So, you need 26 weeks to buy the bicycle.
Hence, your friend will but the bicycle first.

Question 17.
CRITICAL THINKING
Is every linear equation a linear function? Explain your reasoning.
Answer: All linear equations produce straight lines when graphed, not all linear equations produce linear functions. In order to be a linear function, a graph must be both linear (a straight line) and a function (matching each x-value to only one y-value).

Question 18.
PROBLEM SOLVING
The heat index is calculated using the relative humidity and the temperature. For every 1 degree increase in the temperature from 94°F to 97°F at 75% relative humidity, the heat index rises 4°F. On a summer day, the relative humidity is 75%, the temperature is 94°F, and the heat index is 124°F. Estimate the heat index when the relative humidity is 75% and the temperature is 100°F. Use a function to justify your answer.
Big Ideas Math Answers Grade 8 Chapter 7 Functions 7.3 20
Answer:  Heat index is 148°F

Explanation:
The form of linear equation is y = mx + c
and the slope of the function is given by m = (y2-y1) / (x2-x1)
Let y be the heat index and x be the temperature
Given , (94, 124)
For every 1 degree increase in the temperature from 94°F to 97°F at 75% relative humidity, the heat index rises 4°F
that is m = 4
Since the line passes through (94, 124) means
124 = 4x + c
124 = 4(94) + c
124 = 376 + c
c = 124 – 376
c = -252
Linear function for the heat index is y = 4x – 252
put x = 100
So, y = 4(100) – 252
y = 400 – 252
y = 148.
Finally, Heat index is 148°F.


Lesson 7.4 Comparing Linear and Non Linear Functions

EXPLORATION 1

Comparing Functions
Work with a partner. Each equation represents the height h (in feet) of a falling object after t seconds.

  • Graph each equation. Explain your method.
  • Decide whether each graph represents a or function.
  • Compare the falling objects.

Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 1
Answer: Explained below

Explanation:
a. Given, h = 300 – 15t can be written as y = 300 – 15x
h = 300 – 15t  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =300 – 15(0) = 300 . co-ordinates are (0 , 300)
if x = 1 , then y = 300 – 15(1) = 285 . co-ordinates are (1 , 285)
if x = 2 , then y = 300 – 15(2) = 270  , co-ordinates are (2 , 270)
if x = 3 , then y = 300 – 15(3) = 255  , co-ordinates are (3 , 255)
The co-ordinates (0 , 300) , (1 , 285) , (2 , 270) , (3 , 255) form a straight line .

The graph is

Given , h = 300- 16t2  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =300- 16(0)2  = 300 . co-ordinates are (0 , 300)
if x = 1 , then y =300- 16(1)2 = 284 . co-ordinates are (1 , 284)
if x = 2 , then y = 300- 16(2)2 = 236 , co-ordinates are (2 , 236)
if x = 3 , then y = 300- 16(3)2 = 252  , co-ordinates are (3 , 252)
The co-ordinates (0 , 300) , (1 , 284) , (2 , 236) , (3 , 252) does not form a straight line .

The graph is

b. For, h = 300 – 15t , The graph is linear so the so it is a function,
For h = 300- 16t2 , The graph is linear so the so it is a function.

c. Sky diver has the slow fall while compared to the bowling ball , because parachute can be controlled with the wind and can be divert the destination point, and bowling ball cannot be controlled while falling.

Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 2

Try It

Does the table represent a linear or nonlinear function? Explain.
Question 1.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 3
Answer: y = 2x – 12 is linear function.

Explanation:
Ordered pairs are (2 , -8) , (4 , -4) , (6 , 0) , (8 , 4)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (6 ,0) and (8, 4)
m = (y2-y1) / (x2-x1)
m= (4 – 0) / (8– 6)
m = 4/2
m = 2
substitute the slope in the (8 , 4) to get point slope to form a line.
y-y1 = m (x-x1)
y – 4 = 2 ( x – 8)
y – 4 = 2x – 16
y = 2x – 16 + 4
y = 2x – 12
So ,  y = 2x – 12 is linear function.

Question 2.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 4
Answer: y = –\(\frac{5}{3}\)x + 25 is linear function.

Explanation:
Ordered pairs are (0 , 25) , (3 , 20) , (7 , 15) , (12 , 10)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (0 ,25) and (3, 20)
m = (y2-y1) / (x2-x1)
m= (20 – 25) / (3– 0)
m = -5/3
Because the line crosses the y axis at ( 0, 25 ) , The y intercept is 25.
So , the linear function is y = –\(\frac{5}{3}\)x + 25.
So , y = –\(\frac{5}{3}\)x + 25 is linear function.

Does the equation represent a linear or nonlinear function? Explain.
Question 3.
y = x + 5
Answer: y = x + 5 is a linear function

Explanation:
Given , y = x + 5  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 0 + 5 = 5 . co-ordinates are (0 , 5)
if x = 1 , then y = 1 + 5 = 6 . co-ordinates are (1 , 6)
if x = 2 , then y = 2 + 5 = 7 , co-ordinates are (2 , 7)
The co-ordinates (0 , 5) , (1 , 6) , (2 , 7) form a straight line .
Each x input has only one y output so it is a function .
And it  forms a straight line when graphed .
So, y = x + 5 is a linear function.

Question 4.
y = \(\frac{4x}{3}\)
Answer: y = \(\frac{4x}{3}\) is a linear function.

Explanation:
Given , y = \(\frac{4x}{3}\) , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = \(\frac{4(0)}{3}\) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = \(\frac{4(1)}{3}\) = \(\frac{4}{3}\)  = 1.3. co-ordinates are (1 , 1.3)
if x = 2 , then y = \(\frac{4(2)}{3}\) = \(\frac{8}{3}\) = 2.6 , co-ordinates are (2 , 2.6)
The co-ordinates (0 , 0) , (1 ,1.3 ) , (2 , 2.6) form a straight line .
Each x input has only one y output so it is a function .
And it forms a straight line when graphed .
So, y = \(\frac{4x}{3}\) is a linear function.

Question 5.
y = 1 – x2
Answer: y = 1 – x2 is not a linear function.

Explanation:
Given , y = 1 – x2  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 1 – 02 = 1 . co-ordinates are (0 , 1)
if x = 1 , then y = 1 – 12 = 0 . co-ordinates are (1 , 0)
if x = 2 , then y = 1 – 22 = -3 , co-ordinates are (2 , -3)
The co-ordinates (0 , 5) , (1 , 6) , (2 , 7) form a straight line .
Each x input has only one y output so it is a function .
And it does not forms a straight line when graphed .
So, y = 1 – x2 is not a linear function.

Does the graph represent a linear or nonlinear function? Explain.
Question 6.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 5
Answer: The graph represents a nonlinear function.

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (0 , 2) , (-1 , 0) ,  (-2 , -2 ) , (-3 , -4), (0 , 1 ) , (2 , -2) , ( 3, -4 )
The inputs have more than one output ,
And points form a straight line
So , the graph is non linear function

Question 7.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 6
Answer: The graph is a linear function

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (0 , 0) , (-1 , -1) ,  (-2 , -2 ) , (-3 , -3), (1 , 1 ) , (2 , 2) , ( 3, 3 )
The inputs have exactly one output ,
And points form a straight line
So , the graph is a linear function.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

IDENTIFYING FUNCTIONS Does the table or graph represent a linear or nonlinear function? Explain.
Question 8.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 7
Answer: It is not a linear function

Explanation:
Ordered pairs are (3 , 0) , (-1 , 2) , (-5 , 4) , (-9 , 6)
Each input has exactly one output
and it does not form a straight line when graphed
So, it is not a linear function .

Question 9.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 8
Answer: The graph is non linear function

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (0 , -1) , (-1 , 0) ,  (-2 , 3 ) , (1 , 0 ) , (2 , 3) .
The inputs have exactly one output ,
And points does not form a straight line
So , the graph is non linear function

Question 10.
WHICH ONE DOESN’T BELONG?
Which equation does not belong with the other three? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 9
Answer: 5xy = -2 does not belong with the other three.

Explanation:
15y = 6x , y = \(\frac{2}{5}\)x , 10y = 4x .
These are evaluated as 5y = 2x
5xy = -2 , is different from 5y = 2x.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11.
The loudness of sound is measured in (dB). The graph shows the loudness y of a sound (in decibels) x meters from the source of the sound. Is the relationship between loudness and distance linear or nonlinear? Approximate the loudness of the sound 12 meters from the source.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 10
Answer: The relationship between loudness and distance  is nonlinear Function. And the loudness of the sound 12 meters from the source is approximately 85dB as shown in the graph.

Explanation:
As shown in the graph , the plot of the points does not form a straight line ,
Its a parabolic decay , The amount of loudness decreases with the increase in distance,
So, The relationship between loudness and distance  is nonlinear Function.

And the loudness of the sound 12 meters from the source is approximately 85dB as shown in the graph.

Question 12.
A video blogger is someone who records a video diary. A new website currently hosts 90 video bloggers and projects a gain of 10 video bloggers per month. The table below shows the actual numbers of video bloggers. How does the projection differ from the actual change?
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 11
Answer: Projections are more than the actual values

Explanation:

So, Projections are more than the actual values

Comparing Linear and Non Linear Functions Homework & Practice 7.4

Review & Refresh

Write a linear function that relates y to x.
Question 1.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 12
Answer: The linear function is y = x – 2

In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (0 , -2) , (1 , -1 ) , (-1 , -3) , ( 2, 0), (3 , 1) , (4 , 2) , ( 5, 3)
First find the slope m of the line containing the two given points (2 ,0) and (3, 1)
m = (y2-y1) / (x2-x1)
m= (1 – 0) / (3 – 2)
m = 1 .
Because the line crosses the y axis at ( 0, -2 ) , The y intercept is -2.
So , the linear function is y = x – 2 .

Question 2.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 13
Answer: The linear function is y =\(\frac{-1}{1.5}\)x + 5.

Explanation:
Ordered pairs are (0 , 5) , (1.5 , 4) , (3 , 3) , (4.5 , 2)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (1.5 ,4) and (3, 3)
m = (y2-y1) / (x2-x1)
m= (3 – 4) / (3 – 1.5)
m = -1 /1.5
Because the line crosses the y axis at ( 0, 5 ) , The y intercept is 5.
So , the linear function is y =\(\frac{-1}{1.5}\)x + 5.

The vertices of a figure are given. Draw the figure and its image after a dilation with the given scale factor. Identify the type of dilation.
Question 3.
A (- 3, 1), B (- 1, 3), C (- 1, 1); k = 3
Answer: The New right angle triangle is larger than the original one So , its a increase .

Explanation:
Given , (- 3, 1),  (- 1, 3),  (- 1, 1) these pairs form a right angle triangle
K = 3 , For the dilation figure multiply the 3 with the given ordered pairs , then
(- 3, 1) × 3 = ( -9 , 3)
(- 1, 3) × 3 = ( -3 , 9)
(- 1, 1) × 3 = (-3 , 3)
From these new ordered pairs we form a new  right angle triangle
The figure is
The New right angle triangle is larger than the original one So , its a increase .

Question 4.
J (2, 4), K (6, 10), L (8, 10), M (8, 4); k = \(\frac{1}{4}\)
Answer: It is a reduction

Explanation:
Given , (2, 4),  (6, 10),  (8, 10) ,(8,4) these pairs forms a figure
K = 0.25 , For the dilation figure multiply the 3 with the given ordered pairs , then
(2, 4) × 0.25 = (0.5, 1)
(6, 10) × 0.25 = (1.5, 2.5)
(8, 10) × 0.25 = (2, 2.5)
(8, 4) × 0.25 = (2 , 1)
From these new ordered pairs we form a new figure
The figure is
The New figure is smaller than the original , So, It is a reduction .

Concepts, Skills, & Problem Solving

COMPARING FUNCTIONS Graph each equation. Decide whether each graph represents a linear or nonlinear function. (See Exploration 1, p. 295.)
Question 5.
h = 5 + 6t Equation 1
h = 5 + 6t2 Equation 2
Answer: h = 5 + 6t Equation 1 is a linear function
h = 5 + 6t2 Equation 2 is a non linear function .

Explanation:
Given , h = 5 + 6t  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 5 + 6(0) = 5 . co-ordinates are (0 , 5)
if x = 1 , then y = 5 + 6(1) = 11 . co-ordinates are (1 , 11)
if x = 2 , then y = 5 + 6(2) = 17 , co-ordinates are (2 , 17)
if x = 3 , then y = 5 + 6(3) = 23 , co-ordinates are (3 , 23)
The co-ordinates (0 , 5) , (1 , 11) , (2 , 17) , (3 , 23) form a straight line .

Given , h = 5 + 6t2  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 5 + 6(0)² = 5 . co-ordinates are (0 , 5)
if x = 1 , then y = 5 + 6(1)² = 11 . co-ordinates are (1 , 11)
if x = 2 , then y = 5 + 6(2)² = 26 , co-ordinates are (2 , 26)
if x = 3 , then y = 5 + 6(3)² = 59 , co-ordinates are (3 , 59)
The co-ordinates (0 , 5) , (1 , 11) , (2 , 26) , (3 , 59) does not form a straight line .

The graph of both equations is
So, h = 5 + 6t Equation 1 is a linear function
h = 5 + 6t2 Equation 2 is a non linear function .

Question 6.
y = – \(\frac{x}{3}\) Equation 1
y = – \(\frac{3}{x}\) Equation 2
Answer:  y = – \(\frac{x}{3}\) Equation 1 is a linear function
y = – \(\frac{3}{x}\) Equation 2 is a non linear function.

Given , y =- \(\frac{x}{3}\) , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =- \(\frac{0}{3}\) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = – \(\frac{1}{3}\) = – 0.3 . co-ordinates are (1 , – 0.3 )
if x = 2 , then y = – \(\frac{2}{3}\) = – 0.6 , co-ordinates are (2 ,-0.6)
if x = 3 , then y = – \(\frac{3}{3}\) = -1 , co-ordinates are (3 , -1)
The co-ordinates (0 , 0) , (1 , -0.3) , (2 , -0.6) , (3 , -1) form a straight line .

Given , y =- \(\frac{3}{x}\) , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =- \(\frac{3}{0}\) = no number
if x = 1 , then y = – \(\frac{3}{1}\) = – 3 . co-ordinates are (1 , – 1 )
if x = 2 , then y = – \(\frac{3}{2}\) = – 1.5 , co-ordinates are (2 ,-1.5)
if x = 3 , then y = – \(\frac{3}{3}\) = -1 , co-ordinates are (3 , -1)
The co-ordinates (1 , -1) , (2 , -1.5) , (3 , -1) form a straight line .

The graph of both the equations is
So,  y = – \(\frac{x}{3}\) Equation 1 is a linear function
y = – \(\frac{3}{x}\) Equation 2 is a non linear function.

IDENTIFYING FUNCTIONS FROM TABLES Does the table represent a linear or nonlinear function? Explain.
Question 7.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 14
Answer:  linear function is y = 4x + 4.

Explanation:
Ordered pairs are (0 , 4) , (1 , 8) , (2 , 12) , (3 , 16)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (2 , 12) and (3 , 16)
m = (y2-y1) / (x2-x1)
m= (16 – 12) / (3– 2)
m = 4/1
m = 4
Because the line crosses the y axis at ( 0, 4 ) , The y intercept is 4.
So , the linear equation is y = 4x + 4.
And it is a linear function.

The graph is

Question 8.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 15
Answer: y = 4x – 6 is linear function.

Explanation:
Ordered pairs are (6 , 21) , (5 , 15) , (4 , 10) , (3 , 6)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (4 , 10) and (3 , 6)
m = (y2-y1) / (x2-x1)
m= (6 – 10) / (3– 4)
m = -4/-1
m = 4
substitute the slope in the(4 , 10) to get point slope to form a line.
y-y1 = m (x-x1)
y – 10 = 4 ( x – 4)
y – 10 = 4x – 16
y = 4x – 16 + 10
y = 4x – 6
So ,  y = 4x – 6 is linear function.

The graph is

IDENTIFYING FUNCTIONS FROM EQUATIONS Does the equation represent a linear or nonlinear function? Explain.
Question 9.
2x + 3y = 7
Answer: The function is linear when m = \(\frac{-2}{3}\) and c = \(\frac{7}{3}\)

Explanation:
Given ,2x + 3y = 7
3y = 7 – 2x
y = \(\frac{-2}{3}\)x+ \(\frac{7}{3}\)
So, The function is linear when m = \(\frac{-2}{3}\) and c = \(\frac{7}{3}\)

Question 10.
y + x = 4x + 5
Answer: The function is linear when m = 3 and c = 5 .

Explanation:
Given , y + x = 4x + 5
y = 4x – x + 5
y = 3x + 5
So, The function is linear when m = 3 and c = 5 .

Question 11.
y = \(\frac{8}{x^{2}}\)
Answer: The function is linear when m = 8 and c = 0 .

Explanation:
Given , y = \(\frac{8}{x^{2}}\)
slope m = 8
c = 0
So, The function is linear when m = 8 and c = 0 .

IDENTIFYING FUNCTIONS FROM GRAPHS Does the graph represent a linear or nonlinear function? Explain.
Question 12.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 16
Answer: The graph is linear function

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (0 , 1) , (2 , 0) ,  (4 , -1 ) , (-2 , 2), ( -4, 3 )
The inputs have exactly one output ,
And points form a straight line
So , the graph is linear function

Question 13.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 17
Answer: The graph is non linear function.

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (0 , 0) , (-1 , -1) ,  (-4 , -2 ) , (1 , 1), ( 4, 2 )
The inputs have exactly one output ,
And points does not form a straight line
So , the graph is non linear function

Question 14.
IDENTIFYING A FUNCTION
The graph shows the volume V (in cubic feet) of a cube with an edge length of x feet. Does linear nonlinear the graph represent a linear or nonlinear function? Explain.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 18
Answer: The graph is non linear function

n order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (1 , 1) , (2 , 8) ,  (3 , 27 ) , (4 , 64)
The inputs have exactly one output ,
And points does not form a straight line
So , the graph is non linear function

Question 15.
MODELING REAL LIFE
The frequency y (in terahertz) of a light wave is a function of its wavelength x (in nanometers). Is the function relating the wavelength of light to its frequency linear or nonlinear?
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 19
Answer: The function is a non linear function

Explanation:
table is as follows
change in x is constant but change in y is not constant , it is increasing
So, the function is a non linear function .

Question 16.
DIG DEEPER!
The table shows the cost (in dollars) of pounds of sun flower seeds.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 20
a. What is the missing -value that makes the table represent a linear function?
b. Write a linear function that represents the cost of x pounds of seeds. Interpret the slope.
c. Does the function have a maximum value? Explain your reasoning.
Answer:  a. 3 pounds = $4.2
b. y = 1.4x  is linear function.
c.  If y has maximum value then the x also has maximum value.

Explanation:
a. As per the table 1 pound = $1.4
2 pounds = $2.8
3pounds = $4.2
4 pounds = $5.6
So, the price is increasing with weight of the seeds.

b. Ordered pairs are (2 , 2.8) , (3 , 4.2) , (4 , 5.6)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (3 , 4.2) and (4 , 5.6)
m = (y2-y1) / (x2-x1)
m= (5.6 – 4.2) / (4 – 3)
m = 1.4
substitute the slope in the (3 , 4.2) to get point slope to form a line.
y-y1 = m (x-x1)
y – 4.2 = 1.4 ( x – 3)
y – 4.2 = 1.4x – 4.2
y = 1.4x – 4.2 + 4.2
y = 1.4x
So ,  y = 1.4x  is linear function.

c. As shown in the table , and the function if y increases then x also increases with respect to the y
So, if y has maximum value then the x also has maximum value.

Question 17.
MODELING REAL LIFE
A birch tree is 9 feet tall and grows at a rate of 2 feet per year. The table shows the height h (in feet) of a willow tree after x years.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions 7.4 21
a. Does the table represent a linear or nonlinear function? Explain.
b. Which tree is taller after 10 years? Explain.
Answer: There is no linear relationship between x and y .

Explanation:
Table is as follows
Change in y is constant but change in x is increasing , not a constant
Hence, there is no linear relationship between x and y .

Question 18.
CRITICAL THINKING
In their first year, Show A has 7 million viewers and Show B has 5 million viewers. Each year, Show A has 90% of the viewers it had in the previous year. Show B loses 200,000 viewers each year.
a. Determine whether the function relating the year to the number of viewers is linear or nonlinear for each show.
b. Which show has more viewers in its sixth year?
Answer: a. The function relating the year to the number of viewers is linear
b. Both shows  has same number of viewers in the sixth year .

Explanation:
a. Given, In their first year, Show A has 7 million viewers and Show B has 5 million viewers. Each year, Show A has 90% of the viewers it had in the previous year. Show B loses 200,000 viewers each year.
For show A
So , In first year = 7
2 year = 90% of 7 = 6.3
3 year = 90% of 6.3 = 5.6
4 year = 90% of 5.6 = 5.04
5 year = 90% of 5 = 4.5
6 year = 90% of 4.5 = 4.05
So the ordered pairs are (1 , 7) , (2 , 6.3) , (3 , 5.6) , (4 , 5), (5 , 4.5) , (6 , 4)

For show B
In first year = 5 , As the viewers reduces by 2,00,000 in 5M
2 year = 5 – 0.2 = 4.8
3 year = 4.8 – 0.2 = 4.6
4 year = 4.6 – 0.2 = 4.4
5 year = 4.4 – 0.2 = 4.2
6 year = 54.2 – 0.2 = 4
So the ordered pairs are (1 , 5) , (2 , 4.8) , (3 , 4.6) , (4 , 4.4), (5 , 4.2) , (6 , 4)
As the year increases the viewers are also decreasing constantly as per the individual shows
So, The function relating the year to the number of viewers is linear .

b. As shown in part a , the ordered pairs having (6,4) represents the number of viewers to the year
So, Both shows  has same number of viewers in the sixth year .

Question 19.
NUMBER SENSE
The ordered pairs represent a function. (0,- 1), (1, 0), (2, 3), (3, 8), and (4, 15)
a. Graph the ordered pairs and describe the pattern. Is the function linear or nonlinear?
b. Write an equation that represents the function.
Answer: a. The graph is shown below and function is linear
b. The linear equation is y = 7x – 1.

Explanation:
Given, ordered pairs represent a function. (0,- 1), (1, 0), (2, 3), (3, 8), and (4, 15)
a. the graph is 
Each input has exactly one output and it forms a straight line So, the graph is linear
b. First find the slope m of the line containing the two given points (3 ,8) and (4, 15)
m = (y2-y1) / (x2-x1)
m= (15 – 8) / (4– 3)
m = 7/1
m = 7
Because the line crosses the y axis at ( 0, -1 ) , The y intercept is -1.
So , the linear equation is y = 7x – 1.

Lesson 7.5 Analyzing and Sketching Graphs

EXPLORATION 1

Matching Situations to Graphs
Work with a partner. Each graph shows your speed during a bike ride. Match each situation with its graph. Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 1
a. You increase your speed, then ride at a constant speed along a bike path. You then slow down until you reach your friend’s house. Analyze Relationships
b. You increase your speed, then go down a hill. You then quickly come to a stop at an intersection.
c. You increase your speed, then stop at a store for a couple of minutes. You then continue to ride, increasing your speed.
d. You ride at a constant speed, then go up a hill. Once on top of the hill, you increase your speed.
Answer: a – C ,
b – A ,
c – D ,
d – B ,

Explanation:
a. You increase your speed, then ride at a constant speed along a bike path. You then slow down until you reach your friend’s house. The graph C has the perfect graph representing the situation of given question.
b. You increase your speed, then go down a hill. You then quickly come to a stop at an intersection.
Because The graph A has the bike speed representing the situation for the time .
c. You increase your speed, then stop at a store for a couple of minutes. You then continue to ride, increasing your speed. Thus, The graph D is the final answer for the question
d. You ride at a constant speed, then go up a hill. Once on top of the hill, you increase your speed.
Because of the speed with respect to time the graph B is the correct answer for the question.

EXPLORATION 2

Interpreting a Graph
Work with a partner. Write a short paragraph that describe show the height changes over time in the graph shown. What situation can this graph represent?
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 2
Answer: The Graph can be representing a situation for low and high tides of the Ocean

Explanation:
As shown in the figure, The graph is plotted between the height and time,
We can take an example of an Ocean for its waves , As the time passes at the morning of a normal day, The waves of the ocean start rising higher at a period of time, and for the time being maintaining a peak height then drops to a lower height at a particular intervals of time , this process takes place for a while and vise versa.
Thus, the Graph can be representing a situation for low and high tides of the Ocean

Try It

Question 1.
The graph shows the location of a pelican relative to your location.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 3
a. Describe the path of the pelican.
b. Write an explanation for the decrease in the vertical distance of the pelican.
Answer: Both of them are explained below.

Explanation:
a. The path of the pelican is flying in the air , As they always fly in line and the amazing thing is the deeper the prey the higher they dive.
The graph shows the relationship between the horizontal distance that is the height from the land, vertical distance is the point from where its destination point is located, so at the starting point of the flight it has more distance from the ground means flying at a higher level , as the time passes it reaches to the closer point of its destination point so the altitude of the flight decreases with the decrease in the vertical distance and at a particular distance reaches its point of destination.

b. The decrease in the vertical distance of the pelican. is due to its flight to the destination point as it requires to stop flying to reach it, so in order to have a smooth landing on the ground , the bird gradually decreases its speed by decreasing its altitude.

Question 2.
A fully-charged battery loses its charge at a constant rate until it has no charge left. You plug it in, and it fully recharges at a constant rate. Then it loses its charge at a constant rate until it has no charge left. Sketch a graph that represents this situation.
Answer:  The graph is

Explanation:
In the graph , let the x-axis be time and y-axis be the battery charge ,
A fully-charged battery loses its charge at a constant rate until it has no charge left. So, line segment starts from 100 and decreases until it touches the x-axis.
You plug it in, and it fully recharges at a constant rate. Thus, line segment increases at a constant rate until it reaches 100
Then it loses its charge at a constant rate until it has no charge left. line segment decreases again at a constant rate until it again touches the x-axis .

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
ANALYZING GRAPHS
The graph shows the growth rate of a plant over time.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 4
a. Describe the change in growth rate.
b. Write an explanation for the decrease in growth rate and the increase in growth rate.
Answer: the answers are given below

Explanation:
a. the change in growth rate of a plant over the time is given by its size and height , So as the time passes the growth rate is constant from the the start and from a particular time the growth rate has been dropping slightly due to external or internal reasons of a plant and again at some time the growth rate is increasing at a constant rate until it reaches to its perfect growth of a plant.

b. The decrease in growth rate of the plant is due to some external causes like weather, rain, sunlight , watering, and the soil may effect its growth rate and the increase in growth rate is probably due to its soil fertility and sufficient sunlight providing sufficient chlorophyll.

Question 4.
SKETCHING GRAPHS
As you snowboard down a hill, you gain speed at a constant rate. You come to a steep section of the hill and gain speed at a greater constant rate. You then slow down at a constant rate until you come to a stop. Sketch a graph that represents this situation.
Answer: The graph is

Explanation:
In the graph , let the x-axis be time and y-axis be the speed ,
As you snowboard down a hill, you gain speed at a constant rate, line segment decreases at a constant rate
You come to a steep section of the hill and gain speed at a greater constant rate, line segment becomes steeper i.e., the line segment decreases at a high constant rate.
You then slow down at a constant rate until you come to a stop, line segment becomes flatter i.e., the constant rate of decrease becomes less until it touches its x-axis

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
Two rowing teams are in a race. The graph shows their distances from the finish line over time. Describe the speed of each team throughout the race. Then determine which team finishes first.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 5
Answer: Team B will finishes race first.

Explanation:
Team A , The relationship between the time and distance from the finish line is given in the graph,
At starting point Team A has maintained a fair speed at the Beginning of the race and has been a little slow while reaching out to the destination point, and for a while they have been balancing the speed with the distance representing a curving point in the graph and directly dropping to the finish line drastically creating a slope, until it reaches in the x-axis line.
Team B , The relationship between the time and distance from the finish line is given in the graph,
As same as the Team A , Team B has a perfect start but it has been a way different them Team A because Team B has a game plan to win the race, as shown in the graph they have maintained a constant speed while reaching out to the destination and also having a smooth drift at a level of decreasing their distance from the finish line.

Team B will  finishes the race first because they are having a constant and smooth decreasing speed which comes to an end gradually at the finishing line.

Question 6.
DIG DEEPER!
The graphs show the movements of two airplanes over time. Describe the movement of each airplane.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 6
Answer: Detailed explanation is given below.

Explanation:
As shown in the graph , x-axis is time and y-axis be the height above ground
Airplane A, the line segment at a constant rate at the time of starting of the takeoff and then drops to a point while decreasing in the height to the ground for landing and for a constant time it is at the ground level until again it takes off  having an increase in the height from the ground level , at last it maintains a constant speed.

Airplane B, the line segment at a constant rate at the time of starting of the takeoff and then drops to a point while decreasing in the height to the ground for landing and for a constant time it is at the ground level until again it takes off  having an increase in the height from the ground level , at last it maintains a constant speed.
It is as same as the airplane A.

Analyzing and Sketching Graphs Homework & Practice 7.5

Review & Refresh

Does the table or equation represent a linear or nonlinear function? Explain.
Question 1.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 7
Answer: y = -0.5x + 11.5 is a linear function.

Explanation:
Ordered pairs are (-5 , 14) , (-1 , 12) , (3 , 10) , (7 , 8)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (3 , 10) and (7 , 8)
m = (y2-y1) / (x2-x1)
m= (8 – 10) / (7 – 3)
m = -2/4
m = -0.5
substitute the slope in the(3 , 10) to get point slope to form a line.
y-y1 = m (x-x1)
y – 10 = -0.5 ( x – 3)
y – 10 = -0.5x + 1.5
y = -0.5x + 1.5 + 10
y = -0.5x + 11.5
So , the linear equation is y = -0.5x + 11.5
And it is a linear function.

The graph is

Question 2.
y = x2 + 8
Answer: The function is linear  when m= 1 and c = 8.

Explanation:
Given , y = x2 + 8 ,
slope m = 1
c = 8
So, the function is linear when m = 1 and c= 8.

Graph the linear equation.
Question 3.
– 4x + y = – 1
Answer: The graph is

Explanation:
we can write – 4x + y = – 1 as y = 4x – 1
Given , y = 4x – 1  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 4(0) – 1 = -1 . co-ordinates are (0 , -1)
if x = 1 , then y = 4(1) – 1 = 3 . co-ordinates are (1 , 3)
if x = 2 , then y = 4(2) – 1 = 7 , co-ordinates are (2 , 7)
if x = 3 , then y = 4(3) – 1= 11 , co-ordinates are (3 , 11)
The co-ordinates (0 , -1) , (1 , 3) , (2 , 7) , (3 , 11) form a straight line .

Question 4.
2x – 3y = 12
Answer: The graph is

Explanation:
we can write  2x – 3y = 12 as y = \(\frac{2x-12}{3}\) or y = \(\frac{2}{3}\)x – 4
Given , y =\(\frac{2}{3}\)x – 4 , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = \(\frac{2}{3}\)0 – 4= – 4 . co-ordinates are (0 , -4)
if x = 1 , then y = \(\frac{2}{3}\)1 – 4 = 0.66 – 4 = -3.3 . co-ordinates are (1 , -3.3)
if x = 2 , then y = \(\frac{2}{3}\)2 – 4 =0.66(2) – 4 =1.3 – 4 = -2.6, co-ordinates are (2 , -2.6)
if x = 3 , then y = \(\frac{2}{3}\)3 – 4 = 0.66(3) – 4 = 1.98 – 4 = -2.0  , co-ordinates are (3 , -2.0)
The co-ordinates (0 , -4) , (1 , -3.3) , (2 , -2.6) , (3 , -2) form a straight line .

Question 5.
5x + 10y = 30
Answer: The graph is

Explanation:
5x + 10y = 30 can be written as y = -0.5x + 3
take 5 common on both sides we get
x + 2y = 6
y = \(\frac{-x + 6}{2}\)
y = \(\frac{-x}{2}\) + 6
y = -0.5x + 3
Given , y =-0.5x + 3 , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = -0.5(0) + 3 = 3 . co-ordinates are (0 , 3)
if x = 1 , then y = -0.5(1) + 3= 2.5 . co-ordinates are (1 , 2.5)
if x = 2 , then y = -0.5(2) + 3 = 4 , co-ordinates are (2 , 4)
if x = 3 , then y = -0.5(3) + 3 = 4.5  , co-ordinates are (3 , 4.5)
The co-ordinates (0 , 3) , (1 , 2.5) , (2 , 4) , (3 , 4.5)does not form a straight line .

Concepts, Skills, &Problem Solving

MATCHING DESCRIPTIONS WITH GRAPHS The graph shows your speed during a run. Match the verbal description with the part of the graph it describes. (See Exploration 1, p. 301.)
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 8
Question 6.
You run at a constant speed.
Answer: C

Explanation:
Because the line segment of the graph at point C show that the running speed is constant for a particular time ,
Thus forming a straight horizontal line.

Question 7.
You slow down at a constant rate.
Answer: D

Explanation:
Because the line segment of the graph at point D show that the running speed is decreasing at a constant rate for a particular time ,
Thus forming a straight steep line down the time axis.

Question 8.
You increase your speed at a constant rate.
Answer: A

Explanation:
Because the line segment of the graph at point A show that the running speed is increasing at a constant rate at a starting point of the race on time ,
Thus forming a slope in the graph.

Question 9.
You increase your speed at a faster and faster rate.
Answer: B

Explanation:
Because the line segment of the graph at point B show that the running speed is increasing at a faster rate after starting the race and maintaining a gradual growth of the speed and after reaching the next point speed is doubled from before ,
Thus forming a slope with a curve in the graph.

ANALYZING GRAPHS Describe the relationship between the two quantities.
Question 10.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 9
Answer: As the Time passes there will be increase in the volume.

Explanation:
The graph shows the relation between the volume and time of a Balloon , To fill up the balloon with air, with the inlet of air increases in volume and the time taken to fill the balloon with air is shown ,
The line segment starts at initial point stating the balloon at a no air state, then gradually increases with constant halts having constant volume at a particular time and vice versa.

So, As the Time passes there will be increase in the volume.

Question 11.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 10
Answer: As the times passes Dollars are maintaining imbalance.

Explanation:
The relationship between the time and dollars is given in the graph, As we all know money is never ever constant with time , As if it only increases or decreases or having both simultaneously , in this graph the line segment is having a steep and at some point of time it is maintaining a slight growth constantly with the time.

So, As the times passes Dollars are maintaining imbalance.

Question 12.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 11
Answer: An engine power is directly proportional to the engine speed and its horse power

Explanation:
The relationship between the engine speed and horse power is given in the graph, Generally every automobile is is defined as the best for its horse power which is the heart of the engine and it highlights the speed of the vehicle, Here engine power is defined by the horse power and the engine speed the line segment is having a curve increment in the horse power due to the increase in engine speed.

So, An engine power is directly proportional to the engine speed and its horse power

Question 13.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 12
Answer: As time increases the process of grams decaying will be faster.

Explanation:
The relationship between grams and time is given in the graph, its obvious that every product has its own expiry date, and if it crosses that its starts to decay, the graph implies that with the increase time the quality of the gram decreases or grams start to decay . The line segment in the graph shows that the gradually decrease indicating the spoiling rate of the grams  with rate of change of time.

So, As time increases the process of grams decaying will be faster.

Question 14.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 13
Answer: At the particular intervals of time hair growth has stopped and at regular intervals again starts growing with respect to the time.

Explanation:
The graph shows the relationship between the length of the hair and time taken to the growth of the hair, of course hair growth is not constant every time, here we have the graph with the line segment  not constant and having breaks at the times of interval.

So, At the particular intervals of time hair growth has stopped and at regular intervals again starts growing with respect to the time.

Question 15.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 14
Answer: In a period of time the balance will be reduced gradually at regular intervals maintaining a constant balance to clear the loan.

Explanation:
The relationship between the balance of the loan with the time period of the loan to be cleared, The loan should be cleared in the time limit and should maintain a neat balance, every increase in time period the balance is debited from the loan , there will be decrease in the balance and gaps are occurred in the graph.

so, In a period of time the balance will be reduced gradually at regular intervals maintaining a constant balance to clear the loan.

Question 16.
ANALYZING GRAPHS
Write an explanation for the relationship shown in the graph in Exercise 10.
Answer: The graph shows the relation between the volume and time of a Balloon , To fill up the balloon with air, with the inlet of air increases in volume and the time taken to fill the balloon with air is shown ,
The line segment starts at initial point stating the balloon at a no air state, then gradually increases with constant halts having constant volume at a particular time and vice versa.

Question 17.
MODELING REAL LIFE
The graph shows the natural gas usage for a house.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 15
a. Describe the change in usage from January to March.
b. Describe the change in usage from March to May.
Answer: a. The change in usage from January to March, As shown in the graph the usage is at peaks in the month of January and started to decrease after that month and continuing to decrease in the month of March.
b.  The change in usage from March to May, As shown in the graph the usage is continuing to decrease in the month of March and started to increase again in the month of May and maintaining to use highly from the month of May.

Explanation:
a. The change in usage from January to March, As shown in the graph the usage is at peaks in the month of January and started to decrease after that month and continuing to decrease in the month of March.

b.  The change in usage from March to May, As shown in the graph the usage is continuing to decrease in the month of March and started to increase again in the month of May and maintaining to use highly from the month of May.

SKETCHING GRAPHS Sketch a graph that represents the situation.
Question 18.
The value of a television decreases at a constant rate, and then remains constant.
Answer: The graph is

Explanation:
Draw the axis and label the x- axis as time and y- axis as value, then sketch the graph,
The value of the television decreases at a constant rate: line segment starts to decrease at a constant rate,
And then remains constant, after reaching a certain value : line segment becomes parallel to horizontal axis.

Question 19.
The distance from the ground changes as your friend swings on a swing.
Answer: The graph is

Explanation:
Your friend starts close to the ground and then swings up. Then she falls back down close to the ground again and swings back . When she swings back, she gets higher than when she was swinging forward, she then starts to swing forward again getting close to the ground and then going up even higher than when she was swinging backward, she continues to getting higher and higher every time she swings forwards and backwards,

Question 20.
The value of a rare coin increases at a faster and faster rate.
Answer: The graph is

Explanation:
Draw the Axis and label them as x-axis as time and y – axis as distance,
The value of a rare coin increases at a faster and faster rate , so the curve moves upwards at an increasing rate.

Question 21.
You are typing at a constant rate. You pause to think about your next paragraph and then you resume typing at the same constant rate.
Answer: The graph is

Explanation:
A constant rate means that portion of the graph is linear , pausing means the number of words stays constant, typing again at the same constant rate means the last piece of the graph is linear again with the same slope as the first portion of the graph.

Question 22.
CRITICAL THINKING
The graph shows the speed of an object over time.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 16
a. Sketch a graph that shows the distance traveled by the object over time.
b. Describe a possible situation represented by the graphs.
Answer: a. The distance and time are directly proportional to each other.
b. As time passes the speed and time are relatively balancing each other in the graph.

Explanation:
a. The graph is 
In this graph the relationship between distance and time is shown, for example , let the object be a bike, the time taken to reach the destination for the bike is directly proportional to the distance travelled , So as time passes the distance is gradually increasing from the starting point.

So, the distance and time are directly proportional to each other.

b. Th graph shown , is the relationship between the speed and the time , let the object moving be Train,
it is running between the station so it has to be halted in the stations to be listed in the stoppings , So the line segment is started with a constant speed with the time and again at the time interval dropping the speed with respect to time it has maintaining the same speed .

So, As time passes the speed and time are relatively balancing each other in the graph.

Question 23.
MODELING REAL LIFE
The graph shows the average scores of two bowlers from the start of a season to the end of the season.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 17
a. Describe each bowler’s performance.
b. Who had a greater average score most of the season? Who had a greater average score at the end of the season?
c. Write an explanation for the change in each bowler’s average score throughout the bowling season.
Answer: All the answers are explained below

Explanation:
a.  Bowler A : As the graph represent the relationship between the score and the week, bowler A has started with the good take off and having able to grasp the same energy his performance on the goals are increasing rapidly over the week and had the good score than Bowler B.

Bowler B : As the graph represent the relationship between the score and the week, bowler B has started with the good take off but as the time passes the performance of the  bowler is has been intended to decrease his scores gradually over the week.

b.  Bowler A and Bowler B had a greater average score most of the season, but Bowler A had a greater average score at the end of the season

c. Bowler A has the same energy his performance on the goals are increasing rapidly over the week and had the good score than Bowler B. so it has a smaller change in average’s score in the bowling season .
While Bowler B has good take off but as the time passes the performance of the  bowler is has been intended to decrease his scores gradually over the week. so he has a drastic change in average’s score in the bowling season .

Question 24.
DIG DEEPER!
You can use a supply and demand model to understand how the price of a product changes in a market. The supply curve of a particular product represents the quantity suppliers will produce at various prices. The demand curve for the product represents the quantity consumers are willing to buy at various prices.
Big Ideas Math Answer Key Grade 8 Chapter 7 Functions 7.5 18

a. Describe and interpret each curve.
b. Which part of the graph represents a surplus? Explain your reasoning.
c. The curves intersect at the equilibrium point, which is where the quantity produced equals the quantity demanded. Suppose that demand for a product suddenly increases, causing the entire demand curve to shift to the right. What happens to the equilibrium point?
Answer:  All of them are explained below .

Explanation:
a. The supply curve of a particular product represents the quantity suppliers will produce at various prices, As shown in the graph, the relationship between the price and quantity is given , so if prices increases gradually Quantity increases .
The demand curve for the product represents the quantity consumers are willing to buy at various prices, As shown in the graph, the relationship between the price and quantity is given , so if prices decreases with increase in Quantity .

b. The graph does not implies any surplus because each demand and supply is given by their respective curve over the prices and quantity

c. As shown in the graph, The curves intersect at the equilibrium point, which is where the quantity produced equals the quantity demanded. Given, that demand for a product suddenly increases, causing the entire demand curve to shift to the right. Then the equilibrium point will be pointed where the two curves meet after the change in the demand graph so change in the supply graph is also possible.

Functions Connecting Concepts

Using the Problem-Solving Plan
Question 1.
The table shows the lengths x (in inches) and weights y(in pounds) of several infants born at a hospital. Determine whether weight is a function of length. Then estimate the weight of an infant that is 20 inches long.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cc 1
Understand the problem.
You know the lengths and weights of several infants. You are asked to determine whether weight is a function of length and to estimate the weight of a 20-inch-long infant.

Make a plan.
Determine whether any of the lengths are paired with more than one weight. Then use a graphing calculator to find an equation that represents the data. Evaluate the equation when x = 20 to estimate the weight of a 20-inch-long infant.

Solve and check.
Use the plan to solve the problem. Then check your solution.
Answer: Weight is the function of the length

Explanation:
From the table we have , Each length has only one weight , so weight is a function of length.
First find the slope m of the line containing the two given points (19.3 , 7.3) and (18.9 , 6.5)
m = (y2-y1) / (x2-x1)
m= (6.5 – 7.3) / (18.9 – 19.3)
m = 0.2
substitute the slope in the (19.3 , 7.3) to get point slope to form a line.
y-y1 = m (x-x1)
y – 7.3 = 0.2 ( x – 19.3)
y – 7.3 = 0.2x – 3.86
y = 0.2x – 3.86 + 7.3
y = 0.2x + 3.4
So ,  y = 0.2x + 3.4 is linear function.

For x = 20 ,
y = 0.2 (20) + 3.4
y = 4 + 3.4
y = 7.4

So, The weight of an infant that is 20 inches long. is 7.4.

Question 2.
Each mapping diagram represents a linear function. At what point do the graphs of the functions intersect? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cc 2
Answer:  The point of intersection is (-1, -4)

Explanation:
Function 1 – Ordered pairs are ( -8 , 24 ) , ( -3 , 4 ) , ( -1 , -4 ) , ( 1 , -12) .
Function 2 – Ordered pairs are ( 6 , 17 ) , ( 10 , 29 ) , ( 13 , 38 ) , ( 15 , 44 ) .
Graph the points we get, So, The point of intersection is (-1,-4).

Performance Task

Heat Index
At the beginning of this chapter, you watched a STEAM Video called “Apparent Temperature.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cc 3
Answer:

Functions Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 1
Input: Ordered pairs can be used to show inputs and outputs , inputs are represented by x

Output: Ordered pairs can be used to show inputs and outputs , Outputs are represented by y

Relation: A relation pairs inputs with outputs

Mapping diagram: A relation can be represented by ordered pairs or mapping diagrams.

Function: The relation that pairs each input with exactly one output is a function.

Function rule: it is an equation, that describes the relationship between inputs(independent variables) and outputs(dependent variables).

Linear function: A linear function is a function whose graph is a straight line i.e., non vertical line . A linear can be written in the form y = mx + c , where m is the slope and c is the y intercept

Non linear function: The graph of a linear function shows a constant rate of change, A non linear function does not have a constant rate of change, So its graph is a not a line.

Graphic Organizers
You can use an Example and Non-Example Chart to list examples and non-examples of a concept. Here is an Example and Non-Example Chart for functions.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 2

Choose and complete a graphic organizer to help you study the concept.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 3
1. linear functions
2. nonlinear functions
3. linear functions with positive slope
4. linear functions with negative slope

Answer: 1. linear functions

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 4

7.1 Relations and Functions (pp. 275–280)
Learning Target: Understand the concept of a function.

List the ordered pairs shown in the mapping diagram. Then determine whether the relation is a function.
Question 1.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 5
Answer: The ordered pairs are ( 1 , -4 ) , ( 3 , 6 ) , ( 5 , 0 ) , ( 7 , 6 ) , ( 7 , 8 ) and The relation is not a function .

Explanation:
As shown , The ordered pairs are ( 1 , -4 ) , ( 3 , 6 ) , ( 5 , 0 ) , ( 7 , 6 ) , ( 7 , 8 ) .
The input 7 has more than one output,
So, The relation is not a function .

Question 2.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 6
Answer: ordered pairs are ( 0 , 0 ) , ( 1 , 10 ) , ( 2 , 5 ) , ( 3 , 15 ) and The relation is a function .

Explanation:
As shown , The ordered pairs are ( 0 , 0 ) , ( 1 , 10 ) , ( 2 , 5 ) , ( 3 , 15 ).
Each input has exactly one output ,
So, The relation is a function .

Question 3.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 7
Answer: The ordered pairs are ( -1 , 0 ) , ( -1 , 1 ) , ( 0 , 1 ) , ( 1 , 2 ), ( 3 ,3 ) and The relation is not a function

Explanation:
As shown , The ordered pairs are ( -1 , 0 ) , ( -1 , 1 ) , ( 0 , 1 ) , ( 1 , 2 ), ( 3 ,3 ) .
The input -1 has more than one output ,
So, The relation is not a function .

Question 4.
For ordered pairs that represent relations, which coordinate represents the input? the output?
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 8
Answer: x coordinate is the input and y coordinate is the output

Explanation:
Ordered pairs from the given graph are ( 2 , 7 ) , ( 3 , 7 ) , ( 4 , 5 ) , ( 5 , 5 ) , ( 6 , 3 ) .
So , x coordinate is the input and y coordinate is the output

Question 5.
Draw a mapping diagram that represents the relation shown in the graph. Then determine whether the relation is a function. Explain.
Answer:

Explanation:
The mapping diagram is
each input has more than one output
So, relation is not a function.

Question 6.
The mapping diagram represents the lengths (in centimeters) of a rubber band when different amounts of force (in Newtons) are applied.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 9
a. Is the length of a rubber band a function of the force applied to the rubber band?
b. Describe the relationship between the length of a rubber band and the force applied to the rubber band.
Answer:  a. Yes
b. For every increase in 0.7 in input there is an increment of 2 in output.

Explanation:
a. The ordered pairs are  ( 0 , 5 ) , ( 0.7 , 7 ) , ( 1.4 , 9 ) , ( 2.1 , 11 )
Each input has exactly one output
So, the length of a rubber band a function of the force applied to the rubber band.

b. For every increase in 0.7 in input there is an increment of 2 in output.

7.2 Representations of Functions (pp. 281–288)
Learning Target: Represent functions in a variety of ways.

Write a function rule for the statement.
Question 7.
The output is two less than the input.
Answer: y = x – 2

Explanation:
Let us say x is input and y is output , then
The output is two less than the input, will be
y = x – 2

Question 8.
The output is two more than one-fourth of the input.
Answer: y = \(\frac{x}{4}\) + 2

Explanation:
Let us say x is input and y is output , then
The output is two more than one-fourth of the input, will be
y = \(\frac{x}{4}\) + 2

Find the value of y for the given value of x.
Question 9.
y = 2x – 3; x = – 4
Answer: y = -8

Explanation:
Given, y = 2x
substitute x = -4 , we get
y = 2(-4)
y = -8.

Question 10.
y = 2 – 9x ; x = \(\frac{2}{3}\)
Answer: y = – 3.4

Explanation:
Given , y = 2 – 9x
substitute x = \(\frac{2}{3}\) , we get
y = 2 – 9 (0.6)
y = 2 – 5.4
y = – 3.4

Question 11.
y = \(\frac{x}{3}\) + 5; x = 6
Answer: y = 7.

Explanation:
Given, y = \(\frac{x}{3}\) + 5
substitute x = 6 , we get
y = \(\frac{6}{3}\) + 5
y = 2 + 5
y = 7.

Graph the function.
Question 12.
y = x + 3
Answer: The graph is

Explanation:
Given , y = x + 3  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 0 + 3 = 3 . co-ordinates are (0 , 4)
if x = 1 , then y = 1 + 3  = 4 . co-ordinates are (1 , 5)
if x = 2 , then y = 2 + 3 = 5 , co-ordinates are (2 , 6)
if x = 3 , then y = 3 + 3 = 6 , co-ordinates are (3 , 7)
The co-ordinates (0 , 3) , (1 , 4) , (2 , 5) , (3 , 6) form a straight line .

Question 13.
y = – 5x
Answer: The graph is

Explanation:
Given , y = – 5x  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =- 5(0) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = – 5(1)  = – 5 . co-ordinates are (1 , – 5)
if x = 2 , then y = – 5(2) = -10 , co-ordinates are (2 , -10)
if x = 3 , then y =- 5(3) = -15 , co-ordinates are (3 , -15)
The co-ordinates (0 , 0) , (1 , -5) , (2 , -10) , (3 , -15) form a straight line .

Question 14.
y = 3 – 3x
Answer: The graph is

Explanation:
Given , y =3 – 3x , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 3 – 3(0) = 3 . co-ordinates are (0 , 3)
if x = 1 , then y = 3 – 3(1)  = 0 . co-ordinates are (1 , 0)
if x = 2 , then y = 3 – 3(2) = – 3 , co-ordinates are (2 , – 3)
if x = 3 , then y =3 – 3(3) = – 6 , co-ordinates are (3 , – 6)
The co-ordinates (0 , 3) , (1 , 0) , (2 , – 3) , (3 , – 6) form a straight line .

Question 15.
An online music store sells songs for $0.90 each.
a. Write a function that you can use to find the cost of buying songs.
b. What is the cost of buying 5 songs?
Answer: a. C = 0.90s
b. $4.5

Explanation:
a. The total cost is equal to the cost of each song times the number of songs, if each song is $0.90,
Then the total cost C of s songs is C = 0.90s.

b. Substituting s= 5 in C = 0.90s we get,
C = 0.90(5) = 4.5.
So, cost of 5 songs is $4.5.

7.3 Linear Functions (pp. 289–294)
Learning Target: Use functions to model linear relationships.

Use the graph or table to write a linear function that relates y to x.
Question 16.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 16
Answer: The linear function is y = \(\frac{1}{3}\)x + 3.

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (3 , 4) , (0 , 3 ) , (-3 , 2) , ( -6, 1 )
First find the slope m of the line containing the two given points (0 ,3) and (-3, 2)
m = (y2-y1) / (x2-x1)
m= (2 – 3) / (-3 – 0)
m = -1 / -3 .
m = 1/3 .
Because the line crosses the y axis at ( 0, 3 ) , The y intercept is 3.
So , the linear function is y = \(\frac{1}{3}\)x + 3.

Question 17.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 17
Answer: The linear function is y = −(0)x -7.

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (-2 , -7) , (0 , -7 ) , (2 , -7) , ( 4 , -7 )
First find the slope m of the line containing the two given points (0 ,-7) and (2, -7)
m = (y2-y1) / (x2-x1)
m= (-7 – (-7)) / (2 – 0)
m = 0 .
Because the line crosses the y axis at ( 0, -7 ) , The y intercept is -7.
So , the linear function is y = −(0)x -7.

Question 18.
The table shows the age x (in weeks) of a puppy and its weight y (in pounds).
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 18
a. Write and graph a linear function that relates y to x.
b. Interpret the slope and the y-intercept.
c. After how many weeks will the puppy weigh 33 pounds?
Answer: a. y = \(\frac{3}{2}\)x + 3
b. 3 pounds
c. Age is 20 weeks

Explanation:
a. In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (6 , 12) , (8 , 15 ) , (10 , 18) , ( 12 , 21 )
First find the slope m of the line containing the two given points ((6 ,12) and (8 , 15)
m = (y2-y1) / (x2-x1)
m= (15 – 12) / (8 – 6)
m = 3/2 .
substitute the slope in the (6 ,12) to get point slope to form a line.
y-y1 = m (x-x1)
y – 12 = 3/2 ( x – 6)
2(y – 12) = 3(x – 6)
2y – 24 = 3x – 18
2y = 3x – 18 + 24
2y  = 3x + 6
So ,  2y  = 3x + 6 or y = \(\frac{3}{2}\)x + 3 is linear function.

b. The slope measures the rate of change of weight due to change in weeks, Here the slope of 3/2 means that as one week passes, weight of the puppy increases by 3/2 pounds.
y intercept measures the weight of the puppy, when it was born which is 3 pounds in this case measured by c.

c. put y = 33,
33 = \(\frac3}{2}\)x + 3
30 = \(\frac{3}{2}\)x
30 × 2 = 3x
x = 60/3
x = 20.
So, Age is 20 weeks.

7.4 Comparing Linear and Nonlinear Functions (pp. 295–300)
Learning Target: Understand differences between linear and nonlinear functions.

Does the table represent a linear or nonlinear function? Explain.
Question 19.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 19
Answer: y = 3x – 8 is linear function.

Explanation:
Ordered pairs are (3 , 1 ) , (6 , 10) , (9 , 19) , (12 , 28)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (3 , 1 ) and (6 , 10)
m = (y2-y1) / (x2-x1)
m= (10 – 1) / (6– 3)
m = 9/3
m = 3
substitute the slope in the (3 , 1) to get point slope to form a line.
y-y1 = m (x-x1)
y – 1 = 3 ( x – 3)
y – 1 = 3x – 9
y = 3x – 9 + 1
y = 3x – 8
So ,  y = 3x – 8 is linear function.

Question 20.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 20
Answer: y = -x + 4 is linear function.

Explanation:
Ordered pairs are (1 , 3 ) , (3 , 1) , (5 , 1) , (7 , 3)
Plot the points in the table , Draw a line through the points
First find the slope m of the line containing the two given points (1 , 3 ) and (3 , 1)
m = (y2-y1) / (x2-x1)
m= (1 – 3) / (3– 1)
m = -2/2
m = -1
substitute the slope in the (1 , 3) to get point slope to form a line.
y-y1 = m (x-x1)
y – 3 = -1 ( x – 1)
y – 3 = -x + 1
y = -x + 1 + 3
y = -x + 4
So ,  y = -x + 4 is linear function.

Question 21.
Does the graph represent a linear or nonlinear function? Explain.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 21
Answer: The graph represent a non linear function.

Explanation:
As shown in the graph linear function represents a  straight line to which not happened here,
So , the graph is non linear function

Question 22.
Does the equation y = 2.3x represent a linear or nonlinear function? Explain.
Answer: y = 2.3x is a linear function.

Explanation:
Given , y = 2.3x  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y = 2.3(0) = 0 . co-ordinates are (0 , 0)
if x = 1 , then y = 2.3(1) = 2.3 . co-ordinates are (1 , 2.3)
if x = 2 , then y = 2.3(2) = 4.6 , co-ordinates are (2 , 4.6)
The co-ordinates (0 , 0) , (1 , 2.3) , (2 , 4.6) form a straight line .
Each x input has only one y output so it is a function .
And it  forms a straight line when graphed .
So, y = 2.3x is a linear function.

7.5 Analyzing and Sketching Graphs (pp. 301–306)
Learning Target: Use graphs of functions to describe relationships between quantities.

Question 23.
Describe the relationship between the two quantities in the graph.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 23
Answer: At certain point of time, line segment is increased with respect to time and maintained a constant change and again dropped gradually with the increase in time.

Explanation:
The relationship between the graph is population and time ,
At certain point of time, line segment is increased with respect to time and maintained a constant change and again dropped gradually with the increase in time.
So, the city population is not constant at all the time.

Sketch a graph that represents the situation.
Question 24.
You climb a climbing wall. You climb halfway up the wall at a constant rate, then stop and take a break. You then climb to the top of the wall at a greater constant rate.
Answer: The graph is 

Explanation:
You start climbing a wall at a constant rate so the first portion of the graph needs to be linear with a positive slope, you then take a break which means your height is constant so the second part of the graph needs to be a horizontal line, you then start climbing again at a constant rate, so the last part of the graph needs to be linear with a positive slope.

Question 25.
The price of a stock increases at a constant rate for several months before the stock market crashes. The price then quickly decreases at a constant rate.
Answer: The graph is

Explanation:
The stock price is increasing at a constant rate so the first part of the graph needs to be linear with positive slope, Then price begins to drop quickly so the second part of the graph needs to be linear with a steep negative slope.

Question 26.
The graph shows the sales of two companies during a particular year.
Big Ideas Math Answers 8th Grade Chapter 7 Functions cr 26
a. Describe the sales of each company.
b. Which company has greater total sales for the year?
c. Give a possible explanation for the change in each company’s sales throughout the year.
Answer: All The explanation is given below

a. Company A – The sales of the company is increasing at a constant rate so the first part of the graph needs to be linear with positive slope, and decreasing with a slight negative steep and again increasing at a constant rate increasing the sales of the company

Company B – The sales of the company is increasing at a constant rate so the first part of the graph needs to be increase in curve with a slight decrease in the graph leading to decrease in sales and vise versa.

b. Company A has the greater total sales for the year compared to Company B, with maintaining the sales up to the mark without losses.

c. The change in each company’s sales throughout the year,  Company A – The sales of the company is increasing at a constant rate so the first part of the graph needs to be linear with positive slope,
Company B – The sales of the company is increasing at a constant rate so the first part of the graph needs to be increase in curve with a slight decrease in the graph leading to decrease in sales

Functions Practice Test

Question 1.
List the ordered pairs shown in the mapping diagram. Then determine whether the relation is a function.
Big Ideas Math Answers Grade 8 Chapter 7 Functions pt 1
Answer: The relation is a function

Explanation:
As shown , Ordered pairs are the combinations of input and output
So , Ordered pairs are ( 2 , 9 ) , ( 4 , 9 ) , ( 6 , 10 ) , ( 8 , 11 ) .
Each input has exactly one output ,
So , The relation is a function .

Question 2.
Draw a mapping diagram that represents the relation. Then determine whether the relation is a function. Explain.
Big Ideas Math Answers Grade 8 Chapter 7 Functions pt 2
Answer: The mapping diagram is

Explanation:
Ordered pairs from the given graph are ( -3 , 5 ) , ( -1 , 1 ) , ( -1 , 3 ) , ( 1 , 2 ) , ( 3 , 4 ) .
Each input has exactly one output ,
So , The relation is a function .

Question 3.
Write a function rule for “The output is twice the input.”
Answer: y = 2x

Explanation:
Let us say x is input and y is output , then
The output is twice the input. will be
y = 2x

Question 4.
Graph the function y = 1 – 3x.
Answer: The graph is

Explanation:
Given , y = 1 – 3x  , we know y = mx + c , where m = slope , c = constant
To obtain the graph , we should have ordered pairs ,
So , if x = 0 , then y =1 – 3(0) = 1 . co-ordinates are (0 , 1)
if x = 1 , then y = 1 – 3(1)  = -2 . co-ordinates are (1 , -2)
if x = 2 , then y = 1 – 3(2) = -5 , co-ordinates are (2 , -5)
if x = 3 , then y =1 – 3(3) = -8 , co-ordinates are (3 , -8)
The co-ordinates (0 , 1) , (1 , -2) , (2 , -5) , (3 , -8) form a straight line .

Question 5.
Use the graph to write a linear function that relates y to x.
Big Ideas Math Answers Grade 8 Chapter 7 Functions pt 5
Answer: The linear function is y = 0.5x – 1

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (-4 , -3) , (-2 , -2 ) , (0 , -1) , ( 2 , 0 )
First find the slope m of the line containing the two given points (0 , -1) and ( 2 , 0 )
m = (y2-y1) / (x2-x1)
m= (0 – (-1)) / (2 – 0)
m = 1 / 2 .
m = 0.5 .
Because the line crosses the y axis at ( 0, -1 ) , The y intercept is -1.
So , the linear function is y = 0.5x – 1 .

Question 6.
Does the table represent a linear or nonlinear function? Explain.
Big Ideas Math Answers Grade 8 Chapter 7 Functions pt 6
Answer: The linear function is y = −4x + 8

Explanation:
In order to write the function we have to write the ordered pairs of the graph ,
Ordered pairs are  (0 , 8) , (2 , 0 ) , (4 , -8) , ( 6 , -16 )
First find the slope m of the line containing the two given points (0 , 8) and (2 , 0 )
m = (y2-y1) / (x2-x1)
m= (0 – 8) / (2 – 0)
m = -4
Because the line crosses the y axis at ( 0, 8 ) , The y intercept is 8.
So , the linear function is y = −4x + 8.

Question 7.
The table shows the number of y meters a water-skier travels in x minutes.
Big Ideas Math Answers Grade 8 Chapter 7 Functions pt 7
a. Write a function that relates y to x.
b. Graph the linear function.
c. At this rate, how many kilometers will the water-skier travel in 12 minutes?
d. Another water-skier travels at the same rate but starts a minute after the first water-skier. Will this water-skier catch up to the first water-skier? Explain.
Answer: All the answers are given below

Explanation:
Ordered pairs are  (1 , 600) , (2 , 1200 ) , (3 , 1800) , ( 4 , 2400 ) , (5 , 3000)
First find the slope m of the line containing the two given points(1 , 600) and (2 , 1200 )
m = (y2-y1) / (x2-x1)
m= (1200 – 600) / (2 – 1)
m = 600
So, the line is of the form y = 600x + c
put x= 3 and y = 1800 in the above equation we get,
1800 = 600(3) + c
c = 1800 – 1800
c = 0.
So, The line is y = 600x.

b. The graph is

c. put x = 12 in y = 600x
y = 600(12)
y = 7200
7200 meters, i.e., 7.2km

d. Another water skier travels at the same rate but starts a minute after the first water skier, Since both are travelling at the same rate , the water skier who was late will always be behind the first water skier.

Question 8.
The graph shows the prices of two stocks during one day.
Big Ideas Math Answers Grade 8 Chapter 7 Functions pt 8
a. Describe the changes in the price of each stock.
b. Which stock has a greater price at the end of the day?
c. Give a possible explanation for the change in the price of Stock B throughout the day.
Answer: Detailed Explanation is given below.

Explanation:
a. The changes in the price of each stock is Stock A has the constant increase in stock for a particular time and maintains a constant price forming a straight line in the graph, and again decreasing with a negative slope and vise versa, while Stock B is having steep negative slope that is decreasing in the prices with the time and having a constant horizontal line. and having a positive increase in the slope and same repeats again.

b. stock B has a greater price at the end of the day, having a positive increase in the slope

c. The change in the price of Stock B throughout the day, is having steep negative slope that is decreasing in the prices with the time and having a constant horizontal line. and having a positive increase in the slope and same repeats again, compared to stock A .

Question 9.
You are competing in a footrace. You begin the race by increasing your speed at a constant rate. You then run at a constant speed until you get a cramp and have to stop. You wait until your cramp goes away before you start increasing your speed again at a constant rate. Sketch a graph that represents the situation.
Answer: The graph is

Explanation:
You begin the race by increasing your speed at a constant rate so the first portion of the graph needs to be linear with a positive slope , you then run at a constant speed so the next portion of the graph needs to be horizontal line , you then stop and take a break , so your speed is zero, which means the next portion of the line needs to be
horizontal line on the x axis , you then increase your speed again at a constant rate sop that the last portion of the graph needs to be linear with a positive slope

Functions Cumulative Practice

Big Ideas Math Solutions Grade 8 Chapter 7 Functions cp 1
Question 1.
What is the slope of the line?
Big Ideas Math Solutions Grade 8 Chapter 7 Functions cp 2
Answer: Not in the options but the answer is m = -4/3

Explanation:
Ordered pairs are  (-4 , 5) , (1 , -3 ),
First find the slope m of the line containing the two given points
m = (y2-y1) / (x2-x1)
m= (-3 – 5) / (2 – (-4))
m = -8/6
m = -4/3.

Question 2.
Which value of a makes the equation 24 = \(\frac{a}{3}\) – 9 true?
F. 5
G. 11
H. 45
I. 99
Answer: I. 99

Explanation:
Substitute a = 99 , in the given equation we get,
24 = \(\frac{a}{3}\) – 9
24 = \(\frac{99}{3}\) – 9
24 = 33 – 9
24 = 24.
So, last option is the correct answer.

Question 3.
A mapping diagram is shown.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions cp 3
What number belongs in the box so that the equation describes the function represented by the mapping diagram?
Big Ideas Math Solutions Grade 8 Chapter 7 Functions cp 5
Answer: m = 7 , y = 7x + 5

Explanation:
Ordered pairs are  (4 , 33) , (7 , 54 ), (10 , 75) , (13 , 96 ),
First find the slope m of the line containing the two given points (4 , 33) and (7 , 54 )
m = (y2-y1) / (x2-x1)
m= (54 – 33) / (7 – 4)
m = 21/3
m = 7.
So, y = 7x + 5

Question 4.
What is the solution of the system of linear equations?
3x + 2y = 5
x = y + 5
A. (3, – 2)
B. (- 2, 3)
C. (- 1, 4)
D. (1, – 4)
Answer: A. (3, – 2)

Explanation:
Given 3x + 2y = 5
Then substitute , x = y + 5 in the above equation
3( y + 5) + 2y = 5
3y + 15 + 2y = 5
5y + 15 = 5
5( y + 3) = 5
y + 3 = 1
y = 1 – 3
y = -2,
substitute y = -2 in x = y + 5 then
x = 3
So, (3 , -2)

Question 5.
The director of a research lab wants to present data to donors. The data show how the lab uses a large amount of donated money for research and only a small amount of money for other expenses. Which type of display best represents these data?
F. box-and-whisker plot
G. circle graph
H. line graph
I. scatter plot
Answer: I. scatter plot

Explanation:
Scatter plot is the best graph for this type of data where vertical axis will show the amount of money and Horizontal axis will show research and other expenses.

Question 6.
Which graph shows a nonlinear function?
Big Ideas Math Solutions Grade 8 Chapter 7 Functions cp 6
Answer: option B

Explanation:
As all the other options are representing the linear function that is forming a straight line  expect for option B , it is representing a non linear equation.

Question 7.
Which equation of a line passes through the point (—2, 3) and has a slope of \(\frac{3}{4}\)?
Big Ideas Math Solutions Grade 8 Chapter 7 Functions cp 7
Answer:  F. y – 3 = \(\frac{3}{4}\)(x + 2)

Explanation:
Given, y – 3 = \(\frac{3}{4}\)(x + 2)
it is in the form of y = mx + c
so, slope m = \(\frac{3}{4}\)
Substitute the given points in this equation that is x = -2 and y = 3
3 – 3 = \(\frac{3}{4}\)(-2 + 2)
0 = 0.
So, F is the correct option.

Question 8.
The tables show the sales (in millions of dollars) for two companies over a five-year period.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions cp 8
Part A Does the first table show a linear function? Explain your reasoning.
Part B Does the second table show a linear function? Explain your reasoning.
Answer: Part A, the first table shows a linear function,
Part B the second table shows a linear function.

Explanation:
Part A – ordered pairs are (1 , 2) , (2 , 4) , (3 , 6) , (4 , 8) , (5 , 10)
Each input has exactly one output and forms a straight line when graphed
So, it is a linear function.

Part B – ordered pairs are (1 , 1) , (2 , 1) , (3 , 2) , (4 , 3) , (5 , 5)
Each input has exactly one output and does not form a straight line when graphed
So, it is a linear function.

Question 9.
The equations y = – x + 4 and y = \(\frac{1}{2}\)x – 8 form a system of linear equations. The table shows the values of y for given values of x.
Big Ideas Math Solutions Grade 8 Chapter 7 Functions cp 9
What can you conclude from the table?
A. The system has one solution, when x = 0.
B. The system has one solution, when x = 4.
C. The system has one solution, when x = 8.
D. The system has no solution.
Answer: C. The system has one solution, when x = 8.

Explanation:
Given , y = – x + 4 and y = \(\frac{1}{2}\)x – 8
for x = 8 we have
y = -8 + 4 = -4
y = 0.5(8) – 8 = 4 – 8 = -4
Both the equations have one solution for x = 8
So, The system has one solution, when x = 8.

Question 10.
The vertices of a triangle are A (- 1, 3), B (1, 2), and C (- 1, – 1). Dilate the triangle using a scale factor of 2. What is the y-coordinate of the image of B?
Big Ideas Math Solutions Grade 8 Chapter 7 Functions cp 10
Answer: The New right angle triangle is larger than the original one So , its a increase .

Explanation:
Given , (- 1, 3),  ( 1, 2 ),  (- 1, -1) these pairs form a right angle triangle
K = 2 , For the dilation figure multiply the 3 with the given ordered pairs , then
(- 1, 3) × 2 = ( -2 , 6)
( 1, 2) × 2 = ( 2 , 4)
(- 1, -1) × 2 = (-2 , -2)
From these new ordered pairs we form a new  right angle triangle

The New right angle triangle is larger than the original one So, it’s an increase.

Conclusion

Enhance your problem-solving skills and overall math proficiency using the Big Ideas Math Grade 8 Chapter 7 Functions. Give your best in exams by solving the Problems Over here aligned as per the Latest BIM Textbooks. In case of any suggestions do leave us your queries via the comment box so that we can get back to you.

Must Try:

CHOLAFIN Pivot Calculator

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes

Big Ideas Math Book Answers has created a sequence of lessons in all the chapters. Get Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes on this. This pdf link will make understanding concepts of 3-Dimensional shapes so easy. The 3-D shapes are cone, cylinder, cube, cuboid, etc. So following the Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes is necessary to get notified of the topics. So it will be easy for you to understand the concepts behind each and every lesson.

Big Ideas Math Book Grade K Answer Key Chapter 12 Identify Three-Dimensional Shapes

The topics covered in this chapter are Vocabulary, Two- and Three-Dimensional Shapes, Cubes and Spheres etc. So, Download Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes PDF for free. For more practice questions simply go to the performance task and cumulative practice which is given at the end of the chapter. Just click on the below-attached links and start your preparation from now.

Vocabulary

Lesson: 1 Two- and Three-Dimensional Shapes

Lesson: 2 Describe Three-Dimensional Shapes

Lesson: 3 Cubes and Spheres

Lesson: 4 Cones and Cylinders

Lesson: 5 Build Three-Dimensional Shapes

Lesson: 6 Positions of Solid Shapes

Chapter 12: Identify Three-Dimensional Shapes

Identify Three-Dimensional Shapes Vocabulary

Directions:
Circle each can. Draw a square around each box. Count and write how many of each two-dimensional shape you draw.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes v 1
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Identify-Three-Dimensional-Shapes-Vocabulary

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes v 2

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes v 4
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Identify-Three-Dimensional-Shapes-Vocabulary-1

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes v 6

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes v 8
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Identify-Three-Dimensional-Shapes-Vocabulary-2

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes v 10

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes v 12

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Identify-Three-Dimensional-Shapes-Vocabulary-3

Lesson 12.1 Two- and Three-Dimensional Shapes

Explore and Grow

Directions:
Circle any triangles, rectangles, squares, hexagons, and circles you see in the picture. Use another color to circle any objects in the picture that match the blue shapes shown. Tell what you notice about each shape.

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 1
Answer:

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.1-Two-and-Three-Dimensional-Shapes-Explore-and-Grow
two-dimensional
rectangle
circle
triangle
hexagon
Total11 shapes
three-dimensional
cylinder
sphere
cube
cone
Total 8 shapes
Explanation:
A two-dimensional shape is a shape that has length and width but no depth. … A circle is one example of a two-dimensional shape. Example Two. A rectangle is another example of a two-dimensional shape.

Triangles, Rectangles, Squares, Hexagons, and Circles all these shapes are all 2-D shapes.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width, and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
All the objects in the picture represent the 3-D shapes so, they are circled with a different color.

Think and Grow

Directions:
Circle any three-dimensional shapes. Draw rectangles around any two-dimensional shapes. Tell why your answers are correct.

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 2

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 3
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.1-Two-and-Three-Dimensional-Shapes-Think-and-Grow
two-dimensional
rectangle
circle
three-dimensional
cylinder
sphere
cuboid
cone
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Apply and Grow: Practice

Directions:
1 – 4 Circle any three-dimensional shapes. Draw rectangles around any two-dimensional shapes. Tell why your answers are correct.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 4
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.1-Two-and-Three-Dimensional-Shapes-Apply-and-Grow-Practice-Directions-Question-1
two-dimensional
rectangle
circle
three-dimensional
cylinder
cube
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangles.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 5
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.1-Two-and-Three-Dimensional-Shapes-Apply-and-Grow-Practice-Directions-Question-2
two-dimensional
circle
square
hexagon
three-dimensional
sphere
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 6
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.1-Two-and-Three-Dimensional-Shapes-Apply-and-Grow-Practice-Directions-Question-3
two-dimensional
Triangle
three-dimensional
Cuboid
Triangle prism
Cone
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Question 4.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 7
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.1-Two-and-Three-Dimensional-Shapes-Apply-and-Grow-Practice-Directions-Question-4
two-dimensional
0
three-dimensional
Cylinder
Sphere
Cone

Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Think and Grow: Modeling Real Life

Directions:
Circle any shapes in the picture that are solids. Draw rectangles around any shapes in the picture that are flats. Count and write how many solids and flats you find.

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 8
three-Dimensional
________
– – – – – – – –
________

two-dimensional
________
– – – – – – – –
________

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Lesson-12.1-Two-and-Three-Dimensional-Shapes-Think-and-Grow-Modeling-Real-Life

two-dimensional
Rectangles
Total 7 flat surfaces
three-dimensional
Cubes
Sphere
Cone
Cylinder
Total 8 solids
Explanation:
Solid figures are three-dimensional. A face is a flat surface of a solid.

Two- and Three-Dimensional Shapes Homework & Practice 12.1

Directions:
1 – 3 Circle any three-dimensional shapes. Draw rectangles around any two-dimensional shapes. Tell why your answers are correct.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 9
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Two-and-Three-Dimensional-Shapes-Homework -Practice-12.1-Question-1
two-dimensional
Triangle
Square
three-dimensional
Cylinder
Sphere
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 10
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Two-and-Three-Dimensional-Shapes-Homework -Practice-12.1-Question-2
two-dimensional
Hexagon
Circle
three-dimensional
Cube
Cone
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 11
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Two-and-Three-Dimensional-Shapes-Homework -Practice-12.1-Question-3
two-dimensional
Rectangle
Square
three-dimensional
Cube
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Directions:
4 and 5 Circle any three-dimensional shapes. Draw rectangles around any two-dimensional shapes. Tell why your answers are correct. 6 Circle any three-dimensional shapes in the picture. Count and write the number. Draw rectangles around any two-dimensional shapes in the picture. Count and write the number.

Question 4.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 12
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Two-and-Three-Dimensional-Shapes-Homework -Practice-12.1-Question-4
two-dimensional
Circle
three-dimensional
Cylinder
Sphere
Cone
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Question 5.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 13
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Two-and-Three-Dimensional-Shapes-Homework -Practice-12.1-Question-5
two-dimensional
Rectangle
Circle
Triangle
three-dimensional
Cube
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Question 6.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.1 14
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Two-and-Three-Dimensional-Shapes-Homework -Practice-12.1-Question-6
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Lesson 12.2 Describe Three-Dimensional Shapes

Explore and Grow

Directions:
Cut out the Roll, Stack, Slide Sort Cards. Sort the cards into the categories shown.

rolls

stacks

slides

Answer:

Think and Grow

Directions:

  • Look at the solid shape on the left that rolls. Circle the other solid shapes that roll.
  • Look at the solid shapes on the left that stack. Circle the other solid shapes that stack.
  • Look at the solid shape on the left that slides. Circle the other solid shapes that slide.

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 1

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 2
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.2-Describe-Three-Dimensional-Shapes-Explore-and-Grow
Explanation:
Solid shapes that can roll are circled with Brown.
Solid shapes that can slide are circled with Yellow.
Solid shapes that can stack are circled with Blue.

The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.
Shapes with a curved face can roll. Example sphere , cylinder , cone shape.

Apply and Grow: Practice

Directions:
1 Look at the solid shape on the left that rolls. Circle the other solid shapes that roll. 2 Circle the solid shapes that roll and slide. 3 Circle the solid shapes that stack and slide. 4 Circle the solid shape that does not stack or slide.

Question 1.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 3
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.2-Describe-Three-Dimensional-Shapes-Apply-and-Grow-Practice-Question-1

Given:
The cylinder can roll,roll and slide, stack and slide
The cube can slide and stack.
The sphere can only roll.

Explanation:
The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.
Shapes with a curved face can roll. Example sphere, cylinder, cone shape.

Question 2.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 4
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.2-Describe-Three-Dimensional-Shapes-Apply-and-Grow-Practice-Question-2
Given:
The cylinder can roll, stack and slide, roll and slide
Cone can roll, roll and slide.
The cube can slide and stack.
Explanation:
Solid shapes that can roll are circled with Brown.
Solid shapes that can slide are circled with Yellow.
Solid shapes that can stack are circled with Blue.

The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.
Shapes with a curved face can roll. Example sphere, cylinder, cone shape.

Question 3.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 5
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.2-Describe-Three-Dimensional-Shapes-Apply-and-Grow-Practice-Question-3
Given:
A ball that represents a Sphere. The ball can only roll.
A wooden log which represent cylinder. Log can roll, roll and slide, stack and slide.
The wooden box which represents cube. The box can slide and stack.
Hat represent cone. hat can roll, roll and slide.

Explanation:
Solid shapes that can roll are circled with Brown.
Solid shapes that can slide are circled with Yellow.
Solid shapes that can stack are circled with Blue.
The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.
Shapes with a curved face can roll. Example sphere, cylinder, cone shape.

Question 4.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 6
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.2-Describe-Three-Dimensional-Shapes-Apply-and-Grow-Practice-Question-4
Given:
The ball which represents Sphere. The ball can only roll.
Glue stick which represents cylinder. Glue stick can roll,roll and slide, stack and slide.
The box which represents cube. Box can slide and stack.
Birthday Hat represents cone. Birthday hat can roll, roll and slide.

Explanation:
Solid shapes that can roll are circled with Brown.
Solid shapes that can slide are circled with Yellow.
Solid shapes that can stack are circled with Blue.
The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.
Shapes with a curved face can roll. Example sphere, cylinder, cone shape.

Think and Grow: Modeling Real Life

Directions:
You stack the 3 objects shown. Write 1 below the object you place at the bottom of the stack, write 2 below the object you stack next, and write 3 below the object you stack last. Tell why you chose this order.

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 7

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 8
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.2-Describe-Three-Dimensional-Shapes-Think-and-Grow-Modeling-Real-Life

Given:
cylinder shaped Oats box and piggy bank. Oats box and piggy bank can roll, roll and slide, stack and slide.
cube shaped Cardboard box and a Wooden box . A cardboard box and a Wooden Box can slide and stack.
cone shaped Party hats. Party hats can roll, roll and slide.

Explanation:
The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.
Shapes with a curved face can roll. Example sphere , cylinder , cone shape.

Describe Three-Dimensional Shapes Homework & Practice 12.2

Directions:
1 Look at the solid shapes on the left that stack. Circle the other solid shapes that stack. 2 Look at the solid shape on the left that rolls. Circle the other solid shapes that roll. 3 Look at the solid shape on the left that slides. Circle the other solid shapes that slide.

Question 1.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 9
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Describe-Three-Dimensional-Shapes-Homework-Practice-12.2-Question-1
Explanation:
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.

Question 2.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 10
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Describe-Three-Dimensional-Shapes-Homework-Practice-12.2-Question-2
Explanation:
Shapes with a curved face can roll. Example sphere, cylinder, cone shape.

Question 3.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 11
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Describe-Three-Dimensional-Shapes-Homework-Practice-12.2-Question-3
Explanation:
The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.

Directions:
4 Circle the solid shapes that roll and stack. 5 Circle the solid shapes that stack and slide. 6 Circle the solid shape that does not roll. 7 You stack the 3 objects shown. Write 1 below the object you place at the bottom of the stack, write 2 below the object you stack next, and write 3 below the object you stack last. Tell why you chose this order.

Question 4.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 12
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Describe-Three-Dimensional-Shapes-Homework-Practice-12.2-Question-4
Cylinder shaped objects can roll and slide.

Explanation:
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.
Shapes with a curved face can roll. Example sphere, cylinder, cone shape.

Question 5.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 13
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Describe-Three-Dimensional-Shapes-Homework-Practice-12.2-Question-5

Given:
Cube shaped objects can slide and stake.
Cylinder shaped objects can roll and slide.

Explanation:
The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.

Question 6.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 14
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Describe-Three-Dimensional-Shapes-Homework-Practice-12.2-Question-6
Given:
Cube shaped objects can slide and stake. Cube shaped solids that does not roll.

Explanation:
The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.

Question 7.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.2 15
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Describe-Three-Dimensional-Shapes-Homework-Practice-12.2-Question-7

Explanation:
In the figure given we have 2 cylinder shaped objects and 1 cone shaped object.
Cylinders can stack, slide and roll. So, I used both the cylinder shaped objects at the bottom.
Cone can roll and slide. As cones shaped figures can not be stacked I used at the top.

Lesson 12.3 Cubes and Spheres

Explore and Grow

Directions:
Cut out the Cube and Sphere Sort Cards. Sort the cards into the categories shown.

Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 1
Answer:

Think and Grow

Directions:
Circle the cube. Draw a rectangle around the sphere. Tell why your answers are correct.

Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 2

Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 3

Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 4
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.3-Cubes-and-Spheres-Think-and-Grow
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges.

Apply and Grow: Practice

Directions:
1 Circle the cube. Draw a rectangle around the sphere. Tell why your answers are correct. 2 – 4 Circle any object that looks like a cube. Draw a rectangle around any object that looks like a sphere. Tell why your answers are correct.

Question 1.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 5
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.3-Cubes-and-Spheres-Apply-and-Grow-Practice-Question-1

Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges.

Question 2.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 6
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.3-Cubes-and-Spheres-Apply-and-Grow-Practice-Question-2
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges.

Question 3.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 7
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.3-Cubes-and-Spheres-Apply-and-Grow-Practice-Question-3
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges.

Question 4.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 8
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.3-Cubes-and-Spheres-Apply-and-Grow-Practice-Question-4
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges.

Think and Grow: Modeling Real Life

Directions:
Use Make a Cube to build your own number cube. Draw the shape of the flat surfaces of your cube. Count and write the number of flat surfaces.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 9
________
– – – – – – – –
________ flat surfaces
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes- Lesson-12.3-Cubes-and-Spheres- Think-and-Grow-Modeling-Real-Life

Cubes and Spheres Homework & Practice 12.3

Directions:
1 – 3 Circle the cube. Draw a rectangle around the sphere. Tell why your answers are correct.

Question 1.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 10
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cubes-and-Spheres-Homework-&-Practice-12.3-Question-1
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges

Question 2.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 11
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cubes-and-Spheres-Homework-&-Practice-12.3-Question-2
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges

Question 3.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 12
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cubes-and-Spheres-Homework-&-Practice-12.3-Question-3
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges

Directions:
4 – 6 Circle any object that looks like a cube. Draw a rectangle around any object that looks like a sphere. Tell why your answers are correct. 7 Draw the shape of the flat surfaces of a die. Count and write the number of flat surfaces.

Question 4.

Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 13
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cubes-and-Spheres-Homework-&-Practice-12.3-Question-4
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges

Question 5.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 14
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cubes-and-Spheres-Homework-&-Practice-12.3-Question-5
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges

Question 6.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 15
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cubes-and-Spheres-Homework-&-Practice-12.3-Question-6
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges.

Question 7.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.3 16
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cubes-and-Spheres-Homework-&-Practice-12.3-Question-7
Explanation:
Dice is similar to a cube.
A cube is a region of space formed by six identical square faces joined along their edges.

Lesson 12.4 Cones and Cylinders

Explore and Grow

Directions:
Cut out the Cone and Cylinder Sort Cards. Sort the cards into the categories shown.

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 1
Answer:

Think and Grow

Directions:
Circle the cone. Draw a rectangle around the cylinder. Tell why your answers are correct.

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 2

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 3

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 4
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.4-Cones-and-Cylinders-Think-and-Grow
Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Apply and Grow: Practice

Directions:
1 Circle the cone. Draw a rectangle around the cylinder. Tell why your answers are correct. 2 – 4 Circle any object that looks like a cone. Draw a rectangle around any object that looks like a cylinder. Tell why your answers are correct.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 5
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.4-Cones-and-Cylinders-Apply-and-Grow-Practice-Question-1

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 6
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.4-Cones-and-Cylinders-Apply-and-Grow-Practice-Question-2

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 7
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.4-Cones-and-Cylinders-Apply-and-Grow-Practice-Question-3

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Question 4.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 8
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.4-Cones-and-Cylinders-Apply-and-Grow-Practice-Question-4

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Think and Grow: Modeling Real Life

Directions:
Use Make a Cylinder to build a can of vegetables. Draw the shape of the flat surfaces of your can. Count and write the number of flat surfaces.

Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 9
__________
– – – – – – – – – –
__________ flat surfaces
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Lesson-12.4-Cones-and-Cylinders-Think-and-Grow-Modeling-Real-Life

Cones and Cylinders Homework & Practice 12.4

Directions:
1 – 3 Circle the cone. Draw a rectangle around the cylinder. Tell why your answers are correct.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 10
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cones-and-Cylinders-Homework-&-Practice-12.4- Directions-Question-1

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 11
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cones-and-Cylinders-Homework-&-Practice-12.4- Directions-Question-2

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 12
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cones-and-Cylinders-Homework-&-Practice-12.4- Directions-Question-3

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Directions:
4 – 6 Circle any object that looks like a cone. Draw a rectangle around any object that looks like a cylinder. Tell why your answers are correct. 7 Draw the shape of the flat surface of a cone. Count and write the number of flat surfaces.

Question 4.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 13
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cones-and-Cylinders-Homework-&-Practice-12.4- Directions-Question-4

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Question 5.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 14
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cones-and-Cylinders-Homework-&-Practice-12.4- Directions-Question-5

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Question 6.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 15
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cones-and-Cylinders-Homework-&-Practice-12.4- Directions-Question-6

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Question 7.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 12.4 16
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional Shapes-Cones-and-Cylinders-Homework-&-Practice-12.4- Directions-Question-7
Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

Lesson 12.5 Build Three-Dimensional Shapes

Explore and Grow

Directions:
Use your materials to build one of the three-dimensional shapes shown. Circle the three-dimensional shape that you build.

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 1
Answer:

Think and Grow

Directions:

  • Use your materials to build the 2 shapes shown.
  • Connect the 2 shapes that you build, as shown.
  • Tell what solid shape you build.

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 2
Answer:
Cube

Apply and Grow: Practice

Directions:
1 – 3 Use your materials to build the solid shape shown. 4 Use your materials to build a solid shape that has 6 square, flat surfaces. Circle the shape you build.

Question 1.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 3
Answer:

Question 2.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 4
Answer:

Question 3.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 5
Answer:

Question 4.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 6
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Lesson-12.5-Build-Three-Dimensional-Shapes-Apply-Grow-Question-1-4

Think and Grow: Modeling Real Life

Directions:

  • Use your materials to build the castle tower in the picture.
  • Circle the solid shapes that you use to build the tower.

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 7

Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 8
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Lesson-12.5-Build-Three-Dimensional-Shapes- Think-Grow-Modeling Real-Life
Explanation
The above figure represent a cylinder base with cone on the top.

Build Three-Dimensional Shapes Homework & Practice 12.5

Directions:
1 and 2 Use your materials to build the solid shape shown. 3 Use your materials to build the solid shape that has a curved surface and only 1 flat surface. Circle the shape you build. 4 Use your materials to build a solid shape that has no flat surfaces. Circle the shape you build.

Question 1.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 9
Answer:

Question 2.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 10
Answer:

Question 3.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 11
Answer:

Question 4.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 12
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Build-Three-Dimensional-Shapes-Homework-Practice-12.5-Question-1-4

Directions:
5 Use your materials to build the totem pole in the picture. Circle the solid shapes that you use to make the totem pole.

Question 5.
Big Ideas Math Answers Grade K Chapter 12 Identify Three-Dimensional Shapes 12.5 13

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Build-Three-Dimensional-Shapes-Homework-Practice-12.5-Question-5

Explanation:
The above totem is a stack of three shapes. The bottom is in the shape of a Cube. The middle is in the shape of a Cylinder. The top is in the shape of a Cone.

Lesson 12.6 Positions of Solid Shapes

Explore and Grow

Directions:
Place a counter beside the bench. Place a counter in front of the tree. Place a counter next to below the stairs. Place a counter the baby swing.

Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.6 1
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.6-Positions –of-Solid-Shapes

Think and Grow

Directions:

  • Circle the object that looks like a cylinder that is next to the table. Draw a line through the object that looks like a cone that is below the shelf. Draw a rectangle around the object that looks like a sphere that is above the table.
  • Circle the object that looks like a cube that is behind the shovel. Draw a line through the object that looks like a cylinder that is beside the tree. Draw a rectangle around the object that looks like a sphere that is in front of the tree.

Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.6 2
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.6-Positions –of-Solid-Shapes-Think-and-Grow

Apply and Grow: Practice

Directions:
1 Circle the object that looks like a cylinder that is behind a paper cup. Draw a line through the object that looks like a sphere that is above the napkin dispenser. Draw a rectangle around the object that looks like a cone that is below a glass cup. 2 Circle the object that looks like a cone that is beside the log. Draw a line through the object that looks like a sphere that is above the log. Draw a rectangle around the object that looks like a cone that is in front of the log.

Question 1.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.6 3
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.6-Positions –of-Solid-Shapes-Apply-and-Grow-Practice-Question-1

Question 2.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.6 4
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Lesson-12.6-Positions –of-Solid-Shapes-Apply-and-Grow-Practice-Question-2

Think and Grow: Modeling Real Life

Directions: Use the City Scene Cards to place the objects on the picture.

  • Place a dog in front of the boy crossing the street.
  • Place a tree beside the building that looks like a cube.
  • Place an object that looks like a sphere above the buildings. Place that object behind a cloud.
  • Place an object that looks like a cone below the traffic light.
  • Place a streetlight next to the girl on the sidewalk.

Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.6 5
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Lesson-12.6-Positions-of-Solid-Shapes-Think-and-Grow-Modeling-Real-Life

Positions of Solid Shapes Homework & Practice 12.6

Directions:
1 Circle the object that looks like a sphere that is beside the pool. Draw a line through the object that looks like a cone that is next to the ball. Draw a rectangle around the object that looks like a cylinder that is behind the block.

Question 1.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.6 6
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Positions-of-Solid-Shapes-Homework-&-Practice-12.6-Question-1

Directions:
2 Circle the object that looks like a cone that is above the stuffed animal. Draw a line through the object that looks like a cylinder that is in front of the stuffed animal. Draw a rectangle around the object that looks like a cube that is below the stuffed animal. 3 Use the Construction Scene Cards to place the objects on the picture. Place a building below the object that is shaped like a cube. Place a tree beside that building. Place a blimp the traffic cone. Place a truck in front of the traffic cone.

Question 2.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.6 7
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Positions-of-Solid-Shapes-Homework-&-Practice-12.6-Question-2

Question 3.
Big Ideas Math Solutions Grade K Chapter 12 Identify Three-Dimensional Shapes 12.6 8
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Positions-Solid-Shapes-Homework-Practice-12.6-Question-3

Identify Three-Dimensional Shapes Performance Task

Directions: 1 You pick up trash in the park. Draw lines to match each item with its correct recycling bin.

  • The object that rolls but does not stack that is in front of the lamppost goes in the yellow bin.
  • The object below the bench that does not roll goes in the blue bin.
  • The object that has 1 flat surface that is behind an object that looks like a cylinder goes in the green bin.
  • The object that stacks, slides, and rolls that are above an object that looks like a cube goes in the orange bin.
  • The object in front of the tree that rolls and has 2 flat surfaces goes in the green bin.
  • The object next to the tree that stacks and slides and has only flat surfaces goes in the green bin.
  • The object that has a curved surface that does not stack that is beside the tree goes in the blue bin.
  • The object that slides and rolls that is next to an object that has 6 flat surfaces goes in the blue bin.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 1
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes-Identify-Three-Dimensional-Shapes-Performance-Task

Identify Three-Dimensional Shapes Activity

Solid Shapes: Spin and Cover
Directions:
Take turns using the spinner to find which type of three-dimensional shape to cover. Use a counter to cover an object on the page. Repeat this process until you have covered all of the objects.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes 2
Answer:

Identify Three-Dimensional Shapes Chapter Practice

Directions:
1 and 2 Circle any three-dimensional shapes. Draw rectangles around any two-dimensional shapes. Tell why your answers are correct. 3 Look at the solid shape on the left that rolls. Circle the other solid shapes that roll. 4 Circle the solid shapes that stack and slide.

12.1 Two- and Three-Dimensional Shapes

Question 1.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 1
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-12.1-Two-and-Three-Dimensional-Shapes-Question 1
two-dimensional
rectangle
circle
three-dimensional
Sphere
cube
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 2
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-12.1-Two-and-Three-Dimensional-Shapes-Question-2
two-dimensional
Triangle
three-dimensional
Cylinder
Cone
Explanation:
A two-dimensional shape is a shape that has length and width but no depth.
Examples: Circle, Triangle, Rectangle, Squares, Hexagons.
2-D shapes have been shaped with rectangle.
A three-dimensional shape can be defined as a solid figure or an object or shape that has three dimensions – length, width and height. Unlike two-dimensional shapes, three-dimensional shapes have thickness or depth.
Examples: Sphere, Torus, Cylinder, Cone, Cube, Cuboid, Triangular Pyramid, Square Pyramid.
3-D shapes have been shaped with circle.

12.2 Describe Three-Dimensional Shapes

Question 3.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 3
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-12.2-Describe-Three-Dimensional-Shapes-Question-3
Explanation:
Shapes with a curved face can roll. Example sphere , cylinder , cone shape

Question 4.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 4
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-12.2-Describe-Three-Dimensional-Shapes-Question-4
Explanation:
The object which has a flat surface can slide. Example Rectangle, cube, cuboid, cylinder shapes.
Shapes with a flat face can stack. Example Cube, Rectangle, Cylinder shape.

Directions:
5 Circle the cube. Draw a rectangle around the sphere. Tell why your answers are correct. 6 Circle any object that looks like a cube. Draw a rectangle around any object that looks like a sphere. Tell why your answers are correct. 7 and 8 Circle any object that looks like a cone. Draw a rectangle around any object that looks like a cylinder. Tell why your answers are correct.

12.3 Cubes and Spheres

Question 5.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 5
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-12.3-Cubes-and-Spheres-Question-5
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges.

Question 6.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 6
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-12.3-Cubes-and-Spheres-Question-6
Explanation:
A sphere is a round, ball-shaped solid. It has one continuous surface with no edges or vertices.
A cube is a region of space formed by six identical square faces joined along their edges.

12.4 Cones and Cylinders

Question 7.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 7
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-12.4-Cones-and-Cylinders-Question-7

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Question 8.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 8
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-12.4-Cones-and-Cylinders-Question-8

Explanation:
A Cone is a distinctive three-dimensional geometric figure that has a flat surface and a curved surface, pointed towards the top. The pointed end of the cone is called the apex, whereas the flat surface is called the base.

A Cylinder is a three-dimensional solid that holds two parallel bases joined by a curved surface, at a fixed distance.

Directions:
9 Use your materials to build the solid shape shown. 10 Use your materials to build a shape that has a curved surface and 2 flat surfaces. Circle the shape you build. 11 Use your materials to build the elf in the picture. Circle the solid shapes that you use to make the elf.

12.5 Build Three-Dimensional Shapes

Question 9.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 9

Question 10.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 10
Answer 9 – 10 :
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Lesson-12.5-Build-Three-Dimensional-Shapes-Question-9-10

Question 11.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 11
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12-Identify-Three-Dimensional-Shapes-Lesson-12.5-Build-Three-Dimensional-Shapes-Question-11

Directions:
12 Circle the object that looks like a cylinder that is below the hat. Draw a line through the object that looks like a cone that is beside the cooler. Draw a rectangle around the object that looks like a cylinder that is in front of the hat. 13 Circle the object that looks like a sphere that is above the cone. Draw a line through the object that looks like a cylinder that is next to the cone. Draw a rectangle around the object that looks like a sphere that is behind the cone.

Question 12.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 12
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-Question-12

Question 13.
Big Ideas Math Answer Key Grade K Chapter 12 Identify Three-Dimensional Shapes chp 13
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-12- Identify-Three-Dimensional-Shapes- Identify Three-Dimensional Shapes Chapter Practice-Question-13

Final Words:

You can learn the difference between 2-D and 3-D shapes from here. Write your new ideas on your book and solve the problems in own way. Also create questions on your own and try to understand the concepts in depth. We hope the given info is helpful for all the students of Grade K. If you have any doubts regarding the concept you can post the comments in the below-mentioned comment box. Hope this Big Ideas Math Grade K Solution Key helps you to score good marks in the exams.

Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems

Big Ideas Math Answers Grade 2 Chapter 12

Are you feeling difficulty while preparing for the 12th Chapter Solve Length Problems? Then stay tuned to this page. Here we are giving the Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems. Students can download the Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems pdf for free of cost.

Big Ideas Math Book 2nd Grade Answer Key Chapter 12 Solve Length Problems

Students can check out the topic-wise questions and solutions of Big Ideas Math Book Grade 2 Chapter 12 Solve Length Problems. The different lessons of BIM Grade 2 Chapter 12 Solve Length Problems are Solve Length Problems Vocabulary, Use a Number Line to Add and Subtract Lengths, Problem Solving: Length, Problem Solving: Missing Measurement, and Practice Measurement Problems.

Make use of Big Ideas Math 2nd Grade 12th Chapter Solve Length Problems Answer Key during your practice sessions. Make the most out of them and score better grades in your exams. Students can access whichever topic they feel like preparing by click on the quick links listed below. You will be directed to the chosen link.

Vocabulary

Lesson: 1 Use a Number Line to Add and Subtract Lengths

Lesson: 2 Problem Solving: Length

Lesson: 3 Problem Solving: Missing Measurement

Lesson: 4 Practice Measurement Problems

Chapter: 12 – Solve Length Problems

Solve Length Problems Vocabulary

Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems v 1
Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems v 2
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Solve-Length-Problems-Vocabulary-Organize-It
Explanation :
Bar model can be defined as a pictorial representation of a number in the form of bars or boxes used to solve number problems.

Define It
Match the review word to its model.
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems v 3
Answer :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Solve-Length-Problems-Vocabulary-Define-It

Lesson 12.1 Use a Number Line to Add and Subtract Lengths

Explore and Grow

Your goldfish is 4 centimetres long. It grows 6 more centimetres. Use the number line and your ruler to show how long the goldfish is now.
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 1

Answer :
Length of the gold fish = 4 centimetres
Increased length of gold fish = 6 centimetres
Length of gold fish now = 4 + 6 = 10 centimetres  .

Explanation:
Draw an arrow from 0 to 4 to represent 4. Then draw an arrow 6 units to the right representing adding +6.
So, 4 + 6 =0

What is the same about your ruler and the number line? What is different?
_________________________
_________________________
_________________________
_________________________
Answer:
A number line is just that – a straight, horizontal line with numbers placed at even increments along the length. It’s not a ruler, so the space between each number doesn’t matter, but the numbers included on the line determine how it’s meant to be used.
Ruler a straight strip, typically marked at regular intervals and used to draw straight lines or measure distances.

Show and Grow

Question 1.
You swim 15 meters and take a break. Then you swim 10 meters. How many meters do you swim?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 2
Answer:
Distance covered while swimming = 15 meters
Distance covered while swimming after break = 10 metresters .
Therefore, 25 meters traveled in swimming
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.1-Use-a-Number-Line-to-Add-Subtract-Lengths-Show-Grow-Question-1
Explanation :
Draw an arrow from 0 to 10 to represent 10. Then draw an arrow 15 units to the right representing adding +15.
So, 10 + 15 = 25

Question 2.
A ribbon is 16 yards long. You cut off 7 yards. How long is the ribbon now?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 3
Answer:
Length of the ribbon = 16 yards
Decreased in the length of the ribbon = 7 yards
Length of the ribbon now = 16 – 7 = 9 yards .
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.1-Use-a-Number-Line-to-Add-Subtract-Lengths-Show-Grow-Question-2
Explanation :
Draw an arrow from 0 to 16 to represent 16. Then draw an arrow 7 units to the left representing Subtracting -7.
So, 16  – 7 = 9

Apply and Grow: Practice

Question 3.
A snake is 24 inches long. It sheds 14 inches of its skin. How much skin does it not shed?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 4

Answer:
Length of the snake = 24
Length of snake sheds = 14
Length of the snake didn’t shed = Total Length – shed length = 24 – 14 = 10 inches
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.1-Use-a-Number-Line-to-Add-Subtract-Lengths-Apply-Grow-Practice-Question-3
Explanation :
Draw an arrow from 0 to 24 to represent 24. Then draw an arrow 14 units to the left representing Subtracting 10
So, 24 – 14 = 10 inches .

Question 4.
A photo is 15 centimetres long. You cut off 3 centimetres from the left and 3 centimetres from the right. How long is the photo now?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 5
Answer:
Length of the photo = 15 centimetres
Length of the photo cut from left = 3 centimetres
Length of the photo cut from right = 3 centimetres
Length of the photo now = 15 – 3 – 3 = 15 – 6 = 9 centimetres
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.1-Use-a-Number-Line-to-Add-Subtract-Lengths-Apply-Grow-Practice-Question-4

Question 5.
Structure
Write an equation that matches the number line.
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 6
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.1-Use-a-Number-Line-to-Add-Subtract-Lengths-Question-5

Explanation :
Draw an arrow from 0 to 11 to represent 11. Then draw an arrow 6 units to the right representing adding 6 and draw an arrow 3 units to the left representing subtracting 3 .
So, 11 + 6 – 3 = 14 .

Think and Grow: Modeling Real Life

You want to make a bracelet that is 6 inches long. You make 4 inches before lunch. You make 2 inches after lunch. Did you finish the bracelet?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 7
Model:
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 8
Did you finish? Yes No
Answer:
Length of the bracelet = 6 inches
Length of the bracelet made before lunch = 4 inches.
Length of the bracelet made after lunch = 2 inches.
Total Length of the bracelet made = 4 + 2 = 6 inches .
Yes , it is finished
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.1-Use-a-Number-Line-to-Add-Subtract-Lengths-Think-Grow-Modeling-Real-Life

Explanation :
Draw an arrow from 0 to 4 to represent 4. Then draw an arrow 2 units to the right representing adding 2
So, 4 + 2 = 6 inches .

Show and Grow

Question 6.
You are painting a fence that is 24 feet long. You paint 10 feet on Saturday. You paint 13 feet on Sunday. Did you finish painting the fence?
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.1-Use-a-Number-Line-to-Add-Subtract-Lengths-Question-6
Answer:
Length of the total fencing = 24 feet
Length of the fence painted on Saturday = 10 feet
Length of the fence painted on Sunday = 13 feet
Total length of the fencing painted = 10 + 13 = 23 feet.
No painting of fencing is not finished as it is painted 23 feet . still 1 feet left to paint ( 24 – 23 = 1 feet )
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.1-Use-a-Number-Line-to-Add-Subtract-Lengths-Question-7

Explanation :
Draw an arrow from 0 to 10 to represent 10. Then draw an arrow 13 units to the right representing adding 13
So, 10 + 13  = 23 feet .

Question 7.
DIG DEEPER!
You throw a disc 9 meters. On your second throw, the disc travels 3 meters more than your first throw. How many meters did the disc travel in all?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 10
______ meters
Answer:
Length of the first disc thrown = 9 metres
Length of the disc travels in second throw = 3 meters more than your first throw = 9 + 3 = 12
Total length the disc travels in first and second throw = 9 + 12 = 21 meters .
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.1-Use-a-Number-Line-to-Add-Subtract-Lengths-Question-7
Explanation :
Draw an arrow from 0 to 9 to represent 9. Then draw an arrow 12 units to the right representing adding 12
So, 9 + 12  = 21 meters .

Use a Number Line to Add and Subtract Lengths Homework & Practice 12.1

Question 1.
You kick a ball 13 yards. Your friend kicks it back 9 yards. How far is the ball from you now?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 11
Answer:
Distance traveled by ball when i kicked the ball  = 13 yards
Distance traveled by ball when my friend kicked the ball = 9 yards back ward = -9 yards.
Distance of ball from me now = 13 yards – 9 yards = 4 yards.
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Use-a-Number-Line-to-Add-Subtract-Lengths-Homework-Practice-12.1-Question-1
Explanation :
Draw an arrow from 0 to 13 to represent 13. Then draw an arrow 9 units to the left representing Subtracting 9
So, 13- 9 = 4 yards .

Question 2.
Your shoelace is 20 inches long. Your friend’s is 4 inches longer than yours. How long is your friend’s shoelace?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 12
Answer:
Length of my shoelace = 20 inches
Length of my friend’s shoelace = 4 inches longer than me = 4 + 20 = 24 inches.

Explanation :
Draw an arrow from 0 to 20 to represent 20. Then draw an arrow 4 units to the right representing adding 4
So, 20 + 4 = 24 inches .

Question 3.
Structure
One power cord is 7 feet long. Another power cord is 5 feet long. Use the number line to find the combined length of the power cords.
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 13
Answer:
Length of power cord = 7 feet
Length of another power cord = 5 feet
Total length of both cords = 7 + 5 = 12 feet
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Use-a-Number-Line-to-Add-Subtract-Lengths-Homework-Practice-12.1-Question-3

Explanation :
Draw an arrow from 0 to 7 to represent 7. Then draw an arrow 5 units to the right representing adding 5
So, 7 + 5  =12 feets .

Question 4.
Modeling Real Life
A worker needs to pave a bike path that is 25 feet long. He completes 13 feet on Monday and 11 feet on Tuesday. Did he complete the paving?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 14
Answer:
Length of the bike path = 25 feet
length paved on Monday = 13 feet
Length paved on Tuesday = 11 feet
Total length paved on Monday and Tuesday = 13 + 11 = 24 feet
1 feet is less to complete the length of bike path . So, Paving is not completed .
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Use-a-Number-Line-to-Add-Subtract-Lengths-Homework-Practice-12.1-Question-4

Explanation :
Draw an arrow from 0 to 13 to represent 13. Then draw an arrow 11 units to the right representing adding 11
So, 13+ 11  = 24 feets .

Question 5.
DIG DEEPER!
You throw a baseball 5 yards. On your second throw, the baseball travels 2 yards more than your first throw. How many yards did the baseball travel in all?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 15
_______ yards
Answer:
Distance traveled by a base ball = 5 yards
Distance traveled by a base ball in second throw = 2 yards more than your first throw = 5 + 2 = 7 yards .
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Use-a-Number-Line-to-Add-Subtract-Lengths-Homework-Practice-12.1-Question-5
Explanation :
Draw an arrow from 0 to 5 to represent 5. Then draw an arrow +2 units to the right representing adding 2
So, 5 + 2  = 7 yards .

Review & Refresh

Compare
Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 16
Answer:
210 = 200 + 10

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 12.1 17
Answer:
532 = 500 + 20 + 3

Lesson 12.2 Problem Solving: Length

Explore and Grow

How much longer is the red ribbon than the blue ribbon?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 1
______ inches
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.2-Problem-Solving-Length-Explore-Grow

Show and Grow

Question 1.
An orange fish is 10 centimeters long. A yellow fish is 35 centimeters long. A red fish is 19 centimeters long. How much longer is the yellow fish than the red fish?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 2
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 3
Answer:
Length of Orange fish =  10 centimeters
Length of Yellow fish = 35 centimeters
Length of red fish  = 19 centimeters
length of yellow fish is how much longer than the red fish = 35 – 19 = 16 centimeters
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.2-Problem-Solving-Length-Show-Grow-Question-1

Apply and Grow: Practice

Question 2.
A green scarf is 60 inches long. An orange scarf is 45 inches long. A red scarf is 36 inches long. How much longer is the green scarf than the red scarf?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 4
______ inches
Answer:
Length of green scarf = 60 inches
Length of Orange scarf = 45 inches .
Length of Red scarf = 36 inches .
Length of green scarf is how much longer than red scarf = 60 – 36 = 24 inches .

Question 3.
DIG DEEPER!
A pink ribbon is 90 centimeters long. A purple ribbon is 35 centimeters long. A blue ribbon is46 centimeters long. How much longer is the pink ribbon than the total length of the purple and blue ribbons?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 5
_______ centimeters
Answer:
Length of pink ribbon = 90 centimeters
Length of purple ribbon = 35 centimeters
Length of blue ribbon = 46 centimeters
The total length of the purple and blue ribbons = 35 + 46 = 81  centimeters.
Length of pink ribbon is how much longer than the total length of the purple and blue ribbons = 90 – 81 = 9 centimeters

Question 4.
Structure
How much taller is Student 3 than the shortest student?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 6
________ inches
Answer:
Taller student is student 3 = 53 inches .
Shorter student is student 2 = 48 inches .
Taller Student is how much taller than shorter student = 53 – 48 = 5 inches .

Think and Grow: Modeling Real Life

You hop 27 inches and then 24 inches. Your friend hops 3 inches less than you. How far does your friend hop?
Think: What do you know? What do you need to find?
Model:
_______ inches
Answer:
Length hoped by me is 27 and 24 inches = 27 + 24 = 51 inches
Length hoped by my friend = 3 inches less than me = 51 – 3 = 48 inches .
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.2-Problem-Solving-Length-Question-4

Show and Grow

Question 5.
You throw a ball 36 feet and then 41 feet. Your friend throws a ball 5 feet farther than you. How far does your friend throw the ball?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 7
______ feet
Answer:
Distance traveled by ball in first throw = 36 feet
Distance traveled by ball in Second throw = 41 feet
Total Distance traveled by balls = 36 + 41 = 77 feet
Distance traveled by ball when my friend throws = 5 feet farther than me = 77 + 5 = 82 feet .

Question 6.
DIG DEEPER!
A black horse runs 53 meters and then 45 meters. A brown horse runs 62 meters and then 31 meters. Which horse ran the longer distance in all? How many more meters did the horse run?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 8
Black horse Brown horse
_________ more meters
Answer:
Distance traveled by black horse is 53 meters and then 45 meters. = 53 + 45 = 98 meters
Distance traveled by Brown horse is 62 meters and then 31 meters = 62 + 31 = 93 meters
Black Horse runs longer distance than brown horse
Black horse travels 98 – 93 = 5 meters more than brown horse .

Problem Solving: Length Homework & Practice 12.2

Question 1.
The distance to the principal’s office is 24 yards. The distance to the bathroom is 15 yards. The distance to your teacher’s desk is 2 yards. How much farther away is the principal’s office than the bathroom?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 9
_______ yards
Answer:
The distance to the principal’s office = 24 yards
The distance to the bathroom = 15 yards
The distance to your teacher’s desk = 2 yards
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Problem-Solving-Length-Homework-Practice-12.2-Question-1
The principal’s office is 9 yards farther away than the bathroom .

Question 2.
YOU BE THE TEACHER
You launch a rocket 63 meters. Your friend launches it 28 meters, and your cousin launches it 86meters. Your cousin says that he launches the rocket 58 meters farther than you. Is he correct? Explain.
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 10
Answer:
Distance traveled by my rocket = 63 meters
Distance traveled by my friend rocket = 28 meters
Distance traveled by my cousin’s rocket = 86 meters
Distance differences between my rocket and my cousins rocket = 86 – 63 = 23 meters .
My Rockets is 23 meters farther than my cousin rocket not 58 meters.
So above statement is wrong .
Explanation :
Distance differences between my rocket and my cousins rocket = 86 – 63 = 23 meters .
My Rockets is 23 meters farther than my cousin rocket not 58 meters.

Question 3.
Modeling Real Life
You create a drawing that is 15 centimeters long and then add on 7 more centimeters. Your friend creates a drawing that is 3 centimeters longer than yours. How long is your friend’s drawing?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems 12.2 11
_______ centimeters
Answer:
Length of my Drawing = 15 centimeters
Length of my drawing after adding 7 centimeters = 15 + 7 = 22 centimeters
Length of my Friends drawing = 3 centimeters longer than my drawing = 22 + 3 = 25 centimeters
Therefore length of my friend’s drawing = 25 centimeters .

Question 4.
DIG DEEPER!
A frog hops 36 inches and then 22 inches. A toad hops 14 inches and then 43 inches. Which animal hopped the longer distance in all? How many more inches did the animal hop?
Frog Toad
________ more inches
Answer:
Distance hopped by frog is 36 inches and then 22 inches. = 36 + 22 = 58 inches .
Distance hopped by toad is 14 inches and then 43 inches = 14 + 43 = 57 inches .
Frog hopped more distance than toad
Difference of distance hopped by frog and toad = 58 – 57 = 1 inch .

Review & Refresh 

Question 5.
635 + 10 = ______
635 + 100 = _____
Answer:
635 + 10 = 645
635 + 100 = 735
Explanation :
The ones digit remains the same when you add ten. The tens digit increases by 1 every time you add ten
The ones digit remains the same and the tens digit remains the same when you add hundred. The hundred digit increases by 1 every time you add hundred

Question 6.
824 + _____ = 924
824 + _____ = 834
Answer:
824 + 100 = 924
824 + 10 = 834
Explanation :
The ones digit remains the same when you add ten. The tens digit increases by 1 every time you add ten
The ones digit remains the same and the tens digit remains the same when you add hundred. The hundred digit increases by 1 every time you add hundred

Question 7.
309 + _____ = 409
309 + _____ = 319
Answer:
309 + 100 = 409
309 + 10 = 319
Explanation :
The ones digit remains the same when you add ten. The tens digit increases by 1 every time you add ten
The ones digit remains the same and the tens digit remains the same when you add hundred. The hundred digit increases by 1 every time you add hundred

Lesson 12.3 Problem Solving: Missing Measurement

Explore and Grow

You and your friend each have a piece of yarn. The total length of both pieces is 16 centimeters. Use a ruler to measure your yarn. Then draw your friend’s yarn.
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 1
Answer:

Explain how you found the length of your friend’s yarn.
_______________________
_______________________
_______________________
Answer:
Total Length of the yarn  = 16 centimeters
Length of my yarn = 7.8 centimeters .
Length of my friend’s yarn = 16 – 7.8 =  8.2 centimeters
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.3-Problem-Solving-Missing-Measurement-Explore-Grow

Show and Grow

Question 1.
A rope is 31 meters long. You cut a piece off. Now the rope is 14 meters long. How much rope did you cut off?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 2
So, ? = ______.
______ inches
Answer:
Length of the rope = 31 meters
Length of the rope after cut off = 14 meters.
Length of the cut off rope = 31 – 14 = 17 meters .
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.3-Problem-Solving-Missing-Measurement-Show-Grow-Question-1

Apply and Grow: Practice

Question 2.
A celery stalk is 20 centimeters long. You cut off the leaves. Now it is 13 centimeters long. How much did you cut off?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 3
______ centimeters
Answer:
Length of celery stalk = 20 centimeters
Length of celery stacking after chopping the leaves = 13 centimeters .
Length of the chopped leaves = 20 – 13 = 7 centimeters .

Question 3.
Descartes walked some and then ran 39 yards. He went a total of 75 yards. How far did he walk?
______ yards
Answer:
Total Distance traveled by Descartes = 75 yards
Distance traveled by running = 39 yards
Distance Traveled by walking = 75 – 39 = 36 yards

Question 4.
Your coat zipper is 18 inches long. The zipper gets stuck at 11 inches. How much of the zipper will not zip?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 4
_____ inches
Answer:
Length of coat zipper = 18 inches
Length of zipper got stuck = 11 inches
Length of zipper will not zip = 18 – 11 = 7 inches .

Question 5.
Number Sense
The path to school is 181 meters long in all. How long is the missing part of the path?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 5
_______ meters
Answer:
Total path to school = 181 meters
The path from the house to first turn = 74 meters
The path from the first turn to the second turn = 86 meters.
Path from second turn to school is missing = 181 – 74 – 86 = 181 – 160 = 21 meters

Think and Grow: Modeling Real Life

You make a paper chain that is 8 feet long. You add 7 feet of chain to the end. Then 6 feet of the chain breaks off. How long is the chain now?
Think: What do you know? What do you need to find?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 6
_____ feet
Answer:
Length of paper chain = 8 feet
Length of chain added in the end = 7 feet .
Length of chain broken = 6 feet
Total length of the chain now = 8 + 7 – 6 = 15 – 6 = 9 feet .
Explanation :
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.3-Problem-Solving-Missing-Measurement-Think-Grow-Modeling-Real-Life

Show and Grow

Question 6.
You build a tower that is 48 centimeters tall. You add 34 centimeters to the tower height. Your tower breaks and 29 centimeters fall off. How tall is your tower now?
______ centimeters
Answer:
Height of the tower = 48 centimeters
Height added to the tower = 34 centimeters
Total Height of the tower now = 48 + 34 = 82 centimeters .
Height at which tower broken = 29 centimeters
Height of the tower after broken = 82 – 29 = 53 centimeters

Question 7.
A football team is 78 yards away from scoring. They gain 15 yards on the first play and 21 yards on the second play. How far is the team from scoring now?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 7
_______ yards
Answer:
The distance of foot ball team from scoring = 75 yards
Distance gain in first play = 15 yards
Distance gain in second play = 21 yards.
Distance of foot ball team from scoring now = 75 – 15 – 21 = 39 yards .

Problem Solving: Missing Measurement Homework & Practice 12.3

Question 1.
A piece of fabric is 36 inches long. Another piece is 18 inches long. What is the total length of both pieces of fabric?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 8
_____ inches
Answer:
Length of first fabric = 36 inches
Length of second fabric = 18 inches
Total length of both fabrics = 36 + 18 = 54 inches.

Question 2.
A rose is 61 centimeters long. You cut off some of the stem. Now it is 48 centimeters long. How much did you cut off?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 9
______ Centimeters
Answer:
Length of rose = 61 centimeters
Length of rose after chopping stem = 48 centimeters .
Length of chopped stem = 61 – 48 = 13 centimeters .

Question 3.
Number Sense
Newton’s balloon is 18 inches long. Descartes’s balloon is 23 inches long. Your friend’s balloon is 12 inches long. Which sentences are true?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 10
Newton’s balloon is 6 inches longer than your friend’s.
Your friend’s balloon is 11 inches longer than Descartes’s.
Descartes’s balloon is 5 inches longer than Newton’s.
Answer:
Statement 1 is true and Statement 3 is true
Length of Newton’s balloon =18 inches
Length of Descartes’s balloon = 23 inches .
Length of  my Friend’s balloon = 12 inches .
Explanation :
Statement 1 :
Newton balloon and my friends balloon difference in length = 18 – 12 = 6 inches.
It means Newton balloon is 6 inches longer than my friend balloon .
Statement 2 :
Descartes balloon and my friends balloon difference in length= 23 – 12 = 11 inches .
Descartes balloon is 11 inches longer than my friends balloon .
so statement is wrong .
Statement 3 :
Descartes balloon and newtons balloon difference in length = 23 – 18 = 5 inches.
Descartes balloon is 5 inches longer than newton balloon .
Statement is true .

Question 4.
DIG DEEPER!
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 11
Answer:
Equation 3 is true
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Problem-Solving-Missing-Measurement-Homework-Practice-12.3-Question-4

Question 5.
Modeling Real Life
A piece of wood is 16 feet long. You cut off 6 feet, but it is still too long. You cut off 2 more feet. How long is the piece of wood now?
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 12
______ feet
Answer:
Length of wood = 16 feet
Length of wood chopped = 6 feet
Length of wood chopped again = 2 feet
Length of wood after chopping = 16 – 6 – 2 = 16 – 8 = 8 feet .

Review & Refresh

Question 6.
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 13
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Problem-Solving-Missing-Measurement-Homework-Practice-12.3-Question-6

Question 7.
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 14
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Problem-Solving-Missing-Measurement-Homework-Practice-12.3-Question-7
Question 8.
Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems 12.3 15
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Problem-Solving-Missing-Measurement-Homework-Practice-12.3-Question-8

Lesson 12.4 Practice Measurement Problems

Explore and Grow

Newton’s piece of string is 24 centimeters long. He gives Descartes 12 centimeters of the string. How long is the string that Newton has left? Draw a picture and write an equation to solve.
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 1
______ cm
Answer:
Length of Newton’s peice of string = 24 centimeters
Length of string given to Descartes = 12 centimeters
Length of string left over with Newton = 24 – 12 = 12 centimeters .
24 – 12 = 12 is the equation .

Compare the lengths of string. Is one longer, or are they the same length? Explain.

__________________________
__________________________
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Lesson-12.4-Practice-Measurement-Problems-Explore-Grow
Both the lengths are same
Newton’s length = Descarte’s length = 12 centimeters .

Show and Grow

Question 1.
Your blanket is 66 inches long. Your friend’s blanket is 9 inches longer than yours. How long is your friend’s blanket?
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 2
So, ? = ______
_____ inches
Answer:
Length of my blanket = 66 inches
Length of my friend’s blanket = 9 inches longer than yours. = 66 + 9 = 75 inches
My friends blanket is 9 inches longer than my blanket .

Apply and Grow: Practice

Question 2.
A blue whale is 31 meters long. A humpback whale is 16 meters long. How much longer is the blue whale than the humpback whale?
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 3
_____ meters
Answer:
Length of blue whale = 31 meters
Length of hump back whale = 16 meters
Differences in the length of blue whale and hump back whale = 31 – 16 = 15 meters.
Length of blue whale is 15 meters longer than hump back whale .

Question 3.
Newton runs 450 meters. Descartes runs 25 meters less than Newton. How far do they run in all?
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 4
______ meters
Answer:
Distance traveled by Newtons in running= 450 meters
Distance traveled by Descartes in running= 25 meters less than Newton = 450 – 25 = 425 meters .
Total Distance traveled by Newton and Descartes = 450 + 425 = 875 meters .

Question 4.
Reasoning
Solve the problem below two different ways.
You want to read 100 books during the school year. You read 25 books in the fall and 54 books in the winter. How many books do you still need to read?
______ books
Answer:
Total Number of books to read = 100 books
Number of books read in fall = 25 books
Number of books read in winter = 54 books
Number of books still needed to read = 100 – 25 – 54 = 100 – 79 = 21 books

Think and Grow: Modeling Real Life

A yellow subway train is 18 meters longer than a blue subway train. The yellow subway train is 92 meters long. How long is the blue subway train?
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 5
Equation:
______ meters
Answer :
Length of yellow subway tarin = 92 meters .
Length of yellow subway train = 18 meters longer than a blue subway train
Length of blue subway train = 92 – 18 = 74

Show and Grow

Question 5.
A brown rabbit hops 24 inches less than a white rabbit. The brown rabbit hops 48 inches. How many inches does the white rabbit hop?
_____ inches
Answer:
Distance traveled by brown rabbit by hopping = 24 inches less than a white rabbit
Distance traveled by brown rabbit = 48 inches .
Distance traveled by white rabbit = 48 + 24 = 72 inches

Question 6.
Your kite string is 47 yards long. You tie 6 yards of string to the end. Now your kite string is 21 yards longer than your friend’s kite string. How long is your friend’s kite string?
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 6
_______ yards
Answer:
Length of the kite  string = 47 yards
Length attached to kite in the end = 6 yards
Length of my kite string now =47 + 6 = 53 yards
Length of my kite string= 21 yards longer than your friend’s kite string
Length of my friend’s kite string = 53 – 21 = 32 yards .

Practice Measurement Problems Homework & Practice 12.4

Question 1.
A swimming pool is 28 feet long. The pool cover is 32 feet long. How much longer is the pool cover?
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 7
So, ? = ______
_____ feet
Answer:
Length of swimming pool = 28 feet
Length of pool cover = 32 feet
Difference of length in pool and pool cover = 32 – 28 = 4 feet
So, the pool cover is 4 feet longer than pool

Question 2.
Writing
Write and solve a word problem about the colored pencils.
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 8
Answer:
Which color pencil is longer and how much centimetres it is longer from shorter pencil ?
Explanation :
The length of green pencil = 8 cm
The length of red pencil = 11 cm
The length of blue pencil = 15 cm
green pencil is the shorter pencil
Blue pencil is the longer pencil
Differences in the lengths of blue and green pencils = 15 – 8 = 7 cms
The blue pencil is 7 cms longer than green pencil .

Question 3.
Modeling Real Life
You cast out your fishing line 14 yards less than your friend. Your friend casts out her line 33 yards. How many yards do you cast out your fishing line?
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 9
______ yards
Answer:
length of my friend’s fishing line = 33 yards
length of my fishing line= 14 yards less than my friend’s
Length of my fishing line= 33 – 14 = 19 yards

Question 4.
Modeling Real Life
Your nightstand is 24 inches tall. You put a 20-inch lamp on it. Now your nightstand and lamp are 19 inches taller than your bed. How tall is your bed?
Big Ideas Math Solutions Grade 2 Chapter 12 Solve Length Problems 12.4 10
______ inches
Answer:
Height of my nightstand = 24 inches
Height of the lamp = 20 inch
Height of nightstand and lamp = 24 + 20 = 42 inches.
Height of bed = nightstand and lamp are 19 inches taller than your bed.
Height of bed = 42 – 19 = 23 inches .

Review & Refresh

Question 5.
Write the number in expanded form and word form.
645
______ + ____ + ______ _________
Answer:
645 = 600 + 40 + 5
Explanation :
Six hundred and forty five represents six hundreds plus four tens and five ones
When numbers are separated into individual place values and decimal places is called expanded form

Question 6.
Write the number in standard form and word form.
800 + 60 + 2
_____ _______________________
Answer:
862
Explanation:
eight hundred and sixty two = 8 hundreds plus 6 tens and 2ones .
When numbers are separated into individual place values and decimal places is called expanded form

Solve Length Problems Performance Task

Question 1.
A recorder is 1 foot long. A clarinet is 24 inches long. Which instrument is longer? How much longer is the instrument?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 1
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 2
Recorder Clarinet
_____ inches
Answer:
Length of the Recorder = 1 foot = 12 inches
Length of the clarinet = 24 inches
Clarinet is longer
Difference in the length of clarinet and Recorder = 24 – 12 = 12 inches .
Clarinet is 12 inches longer than Recorder .

Question 2.
A piano has 27 more keys than a keyboard. There are 52 white keys and 36 black keys on a piano.
a. How many keys are on the keyboard?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 3
_____ keys
Answer :
Number of white keys on piano = 52
Number of black keys on piano = 36
Total Number of keys on piano = 52 + 36 = 88 keys
piano has 27 more keys than a keyboard
Number of keys on key board = 88 – 27 = 61 keys .

b. The number of black keys on the piano is equal to the number of white keys on the keyboard. How many black keys are on the keyboard?
______ black keys
Answer:
Number of black keys on piano = 36 = number of white keys on keyboard
Number of black keys on keyboard = 61 – 36 = 25 keys .

Question 3.
A drum set has drums and cymbals on stands.
a. The cymbals are 77 centimeters from the ground. You raise the stand 18 centimeters. The cymbals are now 23 centimeters higher than one of the drums. What is the height of the drum?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 4
_______ centimeters
Answer :
Length of cymbals from ground = 77 centimeters .
Length of the stand = 18 centimeters .
Length of cymbals now = 77 + 18 = 95 centimeters
cymbals are now 23 centimeters higher than one of the drums.
Height of the drums = 95 – 23 =72 centimeters .

b. Another drum is 60 centimeters from the ground. You raise it 12 centimeters. Are both drums the same height?
Yes No
Answer:
Height of the drum = 60 centimeters
Height of the drum raised = 12 centimeters.
Height of the drum now = 60 + 12 = 72 centimeters .
Both the drums are at equal heights of 72 centimeters
Yes, both drums have the same height .

Solve Length Problems Activity

Draw and Cover
To Play: Players take turns. On your turn, pick a Draw and Cover Card and solve. Then cover the sea turtle that has the answer. Continue playing until all sea turtles are covered.
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems 5

Solve Length Problems Chapter Practice

12.1 Use a Number Line to Add and Subtract Lengths

Question 1.
You throw a ball 12 yards. Your friend throws it back 8 yards. How far is the ball from you now?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems chp 1
Answer:
Distance traveled by ball when thrown by me= 12 yards
Distance traveled by my friend after throwing = – 8 yards. ( – represents back direction )
Distance of ball far from me  = 12 – 8 = 4 yards.
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems- Solve-Length-Problems-Chapter-Practice-12.1-Use-Number-Line-Add-Subtract-Lengths-Question-1
Explanation:
Draw an arrow from 0 to 12 to represent 12. Then draw an arrow 4 units to the left representing subtracting 4.
So, 12 – 8 = 4 yards .

12.2 Problem Solving: Length

Question 2.
Your cat’s first collar was 6 inches long. Now your cat has a collar that is 13 inches long. Your puppy’s collar is 11 inches long. How much longer is your cat’s collar now?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems chp 2
Answer:
Length of cat first collar = 6 inches
Length of cat collar now  = 13 inches .
Difference in cat collar now and first collar = 13 – 6 = 7 inches
Cat collar is 7 inches longer than first collar
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems- Solve-Length-Problems-Chapter-Practice-12.2-Problem-Solving-Length-Question-2

12.3 Problem Solving: Missing Measurement

Question 3.
You must be 54 inches tall to ride a roller coaster. At 8 years old, you were 48 inches tall. You grow 3 inches the next year. How much more do you still need to grow to be able to ride the roller coaster?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems chp 3
______ inches
Answer:
Required Height to ride a aroller coaster = 54
My Height at the age of 8 yaers = 48 inches.
Next year my height = 48 + 3 = 51 inches.
Height required more for me to ride a roller coaster = 51 – 48 = 3 inches

Question 4.
Number Sense
A car tire is 61 centimeters tall. A truck tire is 84 centimeters tall. A monster truck tire is167 centimeters tall. Which sentences are true?
The car tire is 23 centimeters taller than the truck tire.
The truck tire is 83 centimeters shorter than the monster truck tire.
The monster truck tire is 106 centimeters taller than the car tire.
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems chp 4
Answer:
Statements 2 and 3 are true .
Length of car tire = 61 centimeters
Length of truck tire = 84 centimeters
Length of monster truck = 167 centimeters
Explanation:
Statement 1:
Car tire and truck tire difference in lengths = 84 – 61 = 23 centimeters
Truck tire is 23 centimeters  longer than car tire
Statement is false
Statement 2 :
Truck tire and monster tire lengths differences = 167 – 84 = 83 centimeters
The truck tire is 83 centimeters shorter than the monster truck tire.
Statement is true .
Statement 3 :
Monster truck and car tire lengths differences = 167 – 61 = 106 centimeters
The monster truck tire is 106 centimeters taller than the car tire.
Statement is true .

12.4 Practice Measurement Problems

Question 5.
A kangaroo jumps 24 feet. A frog jumps 19 feet less than the kangaroo. How far does the frog jump?
Big Ideas Math Answer Key Grade 2 Chapter 12 Solve Length Problems chp 6
______ feet
Answer:
Height of kangaroo jump = 24 feet
Height of frog jump = 19 feet less than the kangaroo. = 24 – 19 = 5 feet

Question 6.
A store owner wants to add on to the parking lot to make it 38 meters long. It is currently 21 meters long. How many meters does the store owner want to add?
_____ meters
Answer:
Length of parking after adding parking = 38 meters
Current length = 21 meters
Increase in the length = 38 – 21 = 17 meters .

Solve Length Problems Cumulative Practice

Question 1.
Which expressions have a sum less than 12?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems cp 1
Answer:
Expressions : 5 + 3 , 1 +0 and 4 +6
Explanation :
5 + 3 = 8
4 + 6 = 10
1 + 0 = 1
7 + 8 = 15

Question 2.
Find each difference.
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems cp 2
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Solve-Length-Problems-Cumulative-Practice-Question-2

Question 3.
A blue sailboat is 44 feet long. A white sailboat is 36 feet long. A green sailboat is 22 feet long. Which sentences are true?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems cp 3
Answer:
Statement 2 and 3 are true .
Length of blue sail boat = 44 feet
Length of white sail boat = 36 feet
Length of green sail boat = 22 feet
Explanation:
Statement 1 :
Difference of blue and green sail boat = 44 – 22 = 22 feet
The blue sail boat is 22 feet longer than green boat .
Statement 1 is wrong .
Statement 2 :
Difference of white and green sail boat = 36 – 22 = 14 feet.
The green sail boat is 14 feet shorter than white sail boat .
Statement 2 is true .
Statement 3 :
Difference of blue and green sail boat = 44 – 22 = 22 feet
The green sail boat is 22 feet shorter than blue boat .
Statement 3 is true .
Statement 4 :
Difference of blue and white sail boat = 44 – 36 = 8 feet
The blue sail boat is 8 feet longer than white sail boat .
Statement 4 is wrong .

Question 4.
What is the value of the underlined digit?
739
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems cp 4
Answer:
3 tens

Question 5.
Use mental math to solve.
403 – 10 = ______
898 – 100 = _____
640 – 10 = ______
204 – 10 = ______
843 – _____ = 833
_______ – 100 = 731
Answer:
403 – 10 = 393
898 – 100 = 798
640 – 10 = 630
204 – 10 = 194
843 – 10 = 833
831- 100 = 731

Question 6.
The cracker is about 2 inches long. What is the best estimate of the length of the cracker box?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems cp 6
Answer:
Length of cracker = 2 inches
The best estimate of the length of the cracker box = 3 inches

Question 7.
You take 14 pictures on Friday. You take 20 more on Saturday. Your friend takes 34 pictures in all on Friday and Saturday. How many pictures did you and your friend take in all?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems cp 7
Answer:
Number of pictures taken on Friday by me= 14 pictures
Number of pictures taken on Saturday by me= 20 more on Saturday = 20 pictures
Number of pictures taken by friend on Friday and Saturday = 34 pictures
Total Number of pictures taken by me and my friend = 14 + 20 + 34 = 68 pictures

Question 8.
Which expressions are equal to 245 + 386?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems cp 8
Answer:
245 + 386 = 631
Expression  are 631 and 200 + 300 + 40 + 80+ 5 + 6
Explanation :
631
500 + 130 + 11 = 630 + 11 = 641
200 + 300 + 40 + 80+ 5 + 6 = 500+120+11 = 620 +11 = 631
500 + 120 + 5 = 625

Question 9.
Newton runs 7 yards, takes a break, and runs 3 more yards. Which number line shows how many yards Newton runs?
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems cp 9
Answer:
Distance traveled by Newton =7 yards
Distance traveled by newton after break = 3 yards
Total Distance Traveled by Newton = 7 + 3 = 10 yards
Picture 4 represents correct .
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Solve-Length-Problems-Cumulative-Practice-Question-9
Explanation :
in picture 4 the distance traveled by newton will be shown where the arrow ends at 10 .

Question 10
Find the sum.
Big Ideas Math Answers 2nd Grade Chapter 12 Solve Length Problems cp 10
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-12-Solve-Length-Problems-Solve-Length-Problems-Cumulative-Practice-Question-10

Question 11.
Find each difference.
80 – 53 = ?
79 – 13 = ?
90 – 32 = ?
64 – 40 = ?
Answer:
80 – 53 = 27
79 – 13 = 66
90 – 32 = 58
64 – 40 = 24

Question 12.
Complete the sentences using centimeters or meters.
A teacher’s desk is about 2 ________ long.
A paper clip is about 8 ________ long.
A carrot is about 12 _________ long.
A boat is about 20 _______ long.
Answer:
A teacher’s desk is about 2 meters long.
A paper clip is about 8 centimeters long.
A carrot is about 12 centimeters long.
A boat is about 20 meters long.

Final Words:

We wish that the data mentioned here about Big Ideas Math Answers Grade 2 Chapter 12 Solve Length Problems are helpful for the students. You can complete the homework and assignments easily by downloading Big Ideas Math Book Grade 2 Chapter 12 Solve Length Problems Answer Key PDF. If you have any doubts, then please leave a comment below. Bookmark our site to get the solutions for other chapters of grade 2.

Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers

Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers

Are you in need of a perfect guide for solving all the questions during homework or assignments? Then, Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers is the right one to practice and understand the concepts easily. Don’t worry about scoring marks in the examinations as this BIM Algebra 2 Ch 3 Solution Key helps students to excel in maths and improve math skills.

After practicing from the BigIdeas Math Algebra 2 Chapter 3 Quadratic Equations & Complex Numbers Answer Book, you can easily understand the concepts and score good grades in the exams. Check out the rest of the article and find more details about Ch 3 Quadratic Equations and Complex Numbers Big Ideas Math Algebra 2 Answers.  

Big Ideas Math Book Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers

Download the Topic-wise BigIdeas math book textbook solutions of algebra 2 ch 3 Quadratic Equations and Complex Numbers pdf for free of charge and practice well at any time and anywhere you wish. Once you start preparing the concepts of chapter 3 from BIM Textbook Solution Key, you can stand out from the crowd and become a topper in the class. Just tap on the available links below and Download Big Ideas Math Book Algebra 2 Ch 3 Answers Key freely. These solutions of BIM Algebra 2 Quadratic Equations and Complex Numbers are prepared by the subject experts according to the guidelines of the Common core standards. 

Quadratic Equations and Complex Numbers Maintaining Mathematical Proficiency

Simplify the expression.
Question 1.
\(\sqrt{27}\)
Answer:

Question 2.
–\(\sqrt{112}\)
Answer:

Question 3.
\(\sqrt{\frac{11}{64}}\)
Answer:

Question 4.
\(\sqrt{\frac{147}{100}}\)
Answer:

Question 5.
\(\sqrt{\frac{18}{49}}\)
Answer:

Question 6.
–\(\sqrt{\frac{65}{121}}\)
Answer:

Question 7.
–\(\sqrt{80}\)
Answer:

Question 8.
\(\sqrt{32}\)
Answer:

Factor the polynomial.
Question 9.
x2 − 36
Answer:

Question 10.
x2 − 9
Answer:

Question 11.
4x2 − 25
Answer:

Question 12.
x2 − 22x + 121
Answer:

Question 13.
x2 + 28x + 196
Answer:

Question 14.
49x2 + 210x + 225
Answer:

Question 15.
ABSTRACT REASONING
Determine the possible integer values of a and c for which the trinomial ax2+ 8x+c is factorable using the Perfect Square Trinomial Pattern. Explain your reasoning.
Answer:

Quadratic Equations and Complex Numbers Mathematical Practices

Mathematically proficient students recognize the limitations of technology

Monitoring Progress

Question 1.
Explain why the second viewing window in Example 1 shows gaps between the upper and lower semicircles, but the third viewing window does not show gaps.
Answer:

Use a graphing calculator to draw an accurate graph of the equation. Explain your choice of viewing window.
Question 2.
y = \(\sqrt{x^{2}-1.5}\)
Answer:

Question 3.
y = \(\sqrt{x-2.5}\)
Answer:

Question 4.
x2 + y2= 12.25
Answer:

Question 5.
x2 + y2 = 20.25
Answer:

Question 6.
x2 + 4y2 = 12.25
Answer:

Question 7.
4x2 + y2 = 20.25
Answer:

Lesson 3.1 Solving Quadratic Equations

Essential Question How can you use the graph of a quadratic equation to determine the number of real solutions of the equation?

EXPLORATION 1

Matching a Quadratic Function with Its Graph
Work with a partner. Match each quadratic function with its graph. Explain your reasoning. Determine the number of x-intercepts of the graph.
a. f(x) = x2 − 2x
b. f(x) = x2 − 2x + 1
c. f(x) = x2 − 2x + 2
d. f(x) = −x2 + 2x
e. f(x) = −x2 + 2x − 1
f. f(x) = −x2 + 2x − 2
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 1

EXPLORATION 2

Solving Quadratic Equations
Work with a partner. Use the results of Exploration 1 to find the real solutions (if any) of each quadratic equation.
a. x2 − 2x = 0
b. x2 − 2x + 1 = 0
c. x2 − 2x + 2 = 0
d. −x2 + 2x = 0
e. −x2 + 2x − 1 = 0
f. −x2 + 2x − 2 = 0

Communicate Your Answer

Question 3.
How can you use the graph of a quadratic equation to determine the number of real solutions of the equation?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 2.1
Answer:

Question 4.
How many real solutions does the quadratic equation x2 + 3x + 2 = 0 have? How do you know? What are the solutions?
Answer:

Monitoring Progress

Solve the equation by graphing.
Question 1.
x2 − 8x + 12 = 0
Answer:

Question 2.
4x2 − 12x + 9 = 0
Answer:

Question 3.
\(\frac{1}{2}\)x2 = 6x − 20
Answer:

Solve the equation using square roots.
Question 4.
\(\frac{2}{3}\)x2 + 14 = 20
Answer:

Question 5.
−2x2 + 1 = −6
Answer:

Question 6.
2(x − 4)2 = −5
Answer:

Solve the equation by factoring.
Question 7.
x2 + 12x + 35 = 0
Answer:

Question 8.
3x2 − 5x = 2
Answer:

Find the zero(s) of the function.
Question 9.
f(x) = x2 − 8x
Answer:

Question 10.
f(x) = 4x2 + 28x + 49
Answer:

Question 11.
WHAT IF?
The magazine initially charges $21 per annual subscription. How much should the magazine charge to maximize annual revenue? What is the maximum annual revenue?
Answer:

Question 12.
WHAT IF?
The egg container is dropped from a height of 80 feet. How does this change your answers in parts (a) and (b)?
Answer:

Solving Quadratic Equations 3.1 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Explain how to use graphing to find the roots of the equation ax2 + bx + c = 0.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 2

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation by graphing.
Question 3.
x2 + 3x + 2 = 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 3

Question 4.
−x2 + 2x + 3 = 0
Answer:

Question 5.
y = x2 − 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 5

Question 6.
−8 = −x2 − 4
Answer:

Question 7.
8x = −4 − 4x2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 7

Question 8.
3x2 = 6x − 3
Answer:

Question 9.
7 = −x2 − 4x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 9

Question 10.
2x = x2 + 2
Answer:

Question 11.
\(\frac{1}{5}\)x2 + 6 = 2x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 11

Question 12.
3x = \(\frac{1}{4}\)x2 + 5
Answer:

In Exercises 13–20, solve the equation using square roots.
Question 13.
s2 = 144
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 13

Question 14.
a2 = 81
Answer:

Question 15.
(z − 6)2 = 25
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 15

Question 16.
(p − 4)2 = 49
Answer:

Question 17.
4(x − 1)2 + 2 = 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 17

Question 18.
2(x + 2)2 − 5 = 8
Answer:

Question 19.
\(\frac{1}{2}\)r2 − 10 = \(\frac{3}{2}\)r2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 19

Question 20.
\(\frac{1}{5}\)x2 + 2 = \(\frac{3}{5}\)x2
Answer:

Question 21.
ANALYZING RELATIONSHIPS
Which equations have roots that are equivalent to the x-intercepts of the graph shown?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 3
A. −x2 − 6x − 8 = 0
B. 0 = (x + 2)(x + 4)
C. 0 = −(x + 2)2 + 4
D. 2x2 − 4x − 6 = 0
E. 4(x + 3)2 − 4 = 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 21

Question 22.
ANALYZING RELATIONSHIPS
Which graph has x-intercepts that are equivalent to the roots of the equation (x − \(\frac{3}{2}\))2 = \(\frac{25}{4}\)? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 4
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in solving the equation.
Question 23.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 23

Question 24.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 6
Answer:

Question 25.
OPEN-ENDED
Write an equation of the form x2 = d that has (a) two real solutions, (b) one real solution, and (c) no real solution.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 25

Question 26.
ANALYZING EQUATIONS
Which equation has one real solution? Explain.
A. 3x2 + 4 = −2(x2 + 8)
B. 5x2 − 4 = x2 − 4
C. 2(x + 3)2 = 18
D. \(\frac{3}{2}\)x2 − 5 = 19
Answer:

In Exercises 27–34, solve the equation by factoring.
Question 27.
0 = x2 + 6x + 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 27

Question 28.
0 = z2 − 10z + 25
Answer:

Question 29.
x2 − 8x = −12
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 29

Question 30.
x2 − 11x = −30
Answer:

Question 31.
n2 − 6n = 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 31

Question 32.
a2 − 49 = 0
Answer:

Question 33.
2w2 − 16w = 12w − 48
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 33

Question 34.
−y + 28 + y2 = 2y + 2y2
Answer:

MATHEMATICAL CONNECTIONS In Exercises 35–38, find the value of x.
Question 35.
Area of rectangle = 36
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 35

Question 36.
Area of circle = 25π
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 8
Answer:

Question 37.
Area of triangle = 42
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 37

Question 38.
Area of trapezoid = 32
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 10
Answer:

In Exercises 39–46, solve the equation using any method. Explain your reasoning.
Question 39.
u2 = −9u
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 39

Question 40.
\(\frac{t^{2}}{20}\) + 8 = 15
Answer:

Question 41.
−(x + 9)2 = 64
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 41

Question 42.
−2(x + 2)2 = 5
Answer:

Question 43.
7(x − 4)2 − 18 = 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 43

Question 44.
t2 + 8t + 16 = 0
Answer:

Question 45.
x2 + 3x + \(\frac{5}{4}\) = 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 45

Question 46.
x2 − 1.75 = 0.5
Answer:

In Exercises 47–54, find the zero(s) of the function.
Question 47.
g(x) = x2 + 6x + 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 47

Question 48.
f(x) = x2 − 8x + 16
Answer:

Question 49.
h(x) = x2 + 7x − 30
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 49

Question 50.
g(x) = x2 + 11x
Answer:

Question 51.
f(x) = 2x2 − 2x − 12
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 51

Question 52.
f(x) = 4x2 − 12x + 9
Answer:

Question 53.
g(x) = x2 + 22x + 121
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 53

Question 54.
h(x) = x2 + 19x + 84
Answer:

Question 55.
REASONING
Write a quadratic function in the form f(x) = x2 + bx + c that has zeros 8 and 11.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 55

Question 56.
NUMBER SENSE
Write a quadratic equation in standard form that has roots equidistant from 10 on the number line.
Answer:

Question 57.
PROBLEM SOLVING
A restaurant sells 330 sandwiches each day. For each $0.25 decrease in price, the restaurant sells about 15 more sandwiches. How much should the restaurant charge to maximize daily revenue? What is the maximum daily revenue?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 11
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 57

Question 58.
PROBLEM SOLVING
An athletic store sells about 200 pairs of basketball shoes per month when it charges $120 per pair. For each $2 increase in price, the store sells two fewer pairs of shoes. How much should the store charge to maximize monthly revenue? What is the maximum monthly revenue?
Answer:

Question 59.
MODELING WITH MATHEMATICS
Niagara Falls is made up of three waterfalls. The height of the Canadian Horseshoe Falls is about 188 feet above the lower Niagara River. A log falls from the top of Horseshoe Falls.
a. Write a function that gives the height h (in feet) of the log after t seconds. How long does the log take to reach the river?
b. Find and interpret h(2) − h(3).
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 59

Question 60.
MODELING WITH MATHEMATICS
According to legend, in 1589, the Italian scientist Galileo Galilei dropped rocks of different weights from the top of the Leaning Tower of Pisa to prove his conjecture that the rocks would hit the ground at the same time. The height h (in feet) of a rock after t seconds can be modeled by h(t) = 196 − 16t2.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 12
a. Find and interpret the zeros of the function. Then use the zeros to sketch the graph.
b. What do the domain and range of the function represent in this situation?
Answer:

Question 61.
PROBLEM SOLVING
You make a rectangular quilt that is 5 feet by 4 feet. You use the remaining 10 square feet of fabric to add a border of uniform width to the quilt. What is the width of the border?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 13
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 61

Question 62.
MODELING WITH MATHEMATICS
You drop a seashell into the ocean from a height of 40 feet. Write an equation that models the height h (in feet) of the seashell above the water after t seconds. How long is the seashell in the air?
Answer:

Question 63.
WRITING
The equation h = 0.019s2 models the height h (in feet) of the largest ocean waves when the wind speed is s knots. Compare the wind speeds required to generate 5-foot waves and 20-foot waves.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 14
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 63

Question 64.
CRITICAL THINKING
Write and solve an equation to find two consecutive odd integers whose product is 143.
Answer:

Question 65.
MATHEMATICAL CONNECTIONS
A quadrilateral is divided into two right triangles as shown in the figure. What is the length of each side of the quadrilateral?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 15
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 65

Question 66.
ABSTRACT REASONING
Suppose the equation ax2 + bx + c = 0 has no real solution and a graph of the related function has a vertex that lies in the second quadrant.
a. Is the value of a positive or negative? Explain your reasoning.
b. Suppose the graph is translated so the vertex is in the fourth quadrant. Does the graph have any x-intercepts? Explain.
Answer:

Question 67.
REASONING
When an object is dropped on any planet, its height h (in feet) after t seconds can be modeled by the function h = −\(\frac{g}{2}\)t2 + h0, where h0 is the object’s initial height and g is the planet’s acceleration due to gravity. Suppose a rock is dropped from the same initial height on the three planets shown. Make a conjecture about which rock will hit the ground first. Justify your answer.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 16
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 67

Question 68.
PROBLEM SOLVING
A café has an outdoor, rectangular patio. The owner wants to add 329 square feet to the area of the patio by expanding the existing patio as shown. Write and solve an equation to find the value of x. By what distance should the patio be extended?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 17
Answer:

Question 69.
PROBLEM SOLVING
A flea can jump very long distances. The path of the jump of a flea can be modeled by the graph of the function y = −0.189x2 + 2.462x, where x is the horizontal distance (in inches) and y is the vertical distance (in inches). Graph the function. Identify the vertex and zeros and interpret their meanings in this situation.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 69.1
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 69.2

Question 70.
HOW DO YOU SEE IT?
An artist is painting a mural and drops a paintbrush. The graph represents the height h (in feet) of the paintbrush after t seconds.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 18
a. What is the initial height of the paintbrush?
b. How long does it take the paintbrush to reach the ground? Explain.
Answer:

Question 71.
MAKING AN ARGUMENT
Your friend claims the equation x2 + 7x =−49 can be solved by factoring and has a solution of x = 7. You solve the equation by graphing the related function and claim there is no solution. Who is correct? Explain.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 71

Question 72.
ABSTRACT REASONING
Factor the expressions x2 − 4 and x2 − 9. Recall that an expression in this form is called a difference of two squares. Use your answers to factor the expression x2 − a2. Graph the related function y = x2 − a2. Label the vertex, x-intercepts, and axis of symmetry.
Answer:

Question 73.
DRAWING CONCLUSIONS
Consider the expression x2 + a2, where a > 0.
a. You want to rewrite the expression as (x + m)(x + n). Write two equations that m and n must satisfy.
b. Use the equations you wrote in part (a) to solve for m and n. What can you conclude?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 73

Question 74.
THOUGHT PROVOKING
You are redesigning a rectangular raft. The raft is 6 feet long and 4 feet wide. You want to double the area of the raft by adding to the existing design. Draw a diagram of the new raft. Write and solve an equation you can use to find the dimensions of the new raft.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 19
Answer:

Question 75.
MODELING WITH MATHEMATICS
A high school wants to double the size of its parking lot by expanding the existing lot as shown. By what distance x should the lot be expanded?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 20
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 75

Maintaining Mathematical Proficiency

Find the sum or difference.
Question 76.
(x2 + 2) + (2x2 − x)
Answer:

Question 77.
(x3 + x2 − 4) + (3x2 + 10)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 77

Question 78.
(−2x + 1) − (−3x2 + x)
Answer:

Question 79.
(−3x3 + x2 − 12x) − (−6x2 + 3x − 9)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 79

Find the product.
Question 80.
(x + 2)(x − 2)
Answer:

Question 81.
2x(3 − x + 5x2)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 81

Question 82.
(7 − x)(x − 1)
Answer:

Question 83.
11x(−4x2 + 3x + 8)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 83

Lesson 3.2 Complex Numbers

Essential Question What are the subsets of the set of complex numbers?
In your study of mathematics, you have probably worked with only real numbers, which can be represented graphically on the real number line. In this lesson, the system of numbers is expanded to include imaginary numbers. The real numbers and imaginary numbers compose the set of complex numbers.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 1

EXPLORATION 1

Classifying Numbers
Work with a partner. Determine which subsets of the set of complex numbers contain each number.
a. \(\sqrt{9}\)
b. \(\sqrt{0}\)
c. −\(\sqrt{4}\)
d. \(\sqrt{\frac{4}{9}}\)
e. \(\sqrt{2}\)
f. \(\sqrt{-1}\)

EXPLORATION 2

Complex Solutions of Quadratic Equations
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 2
Work with a partner. Use the definition of the imaginary unit i to match each quadratic equation with its complex solution. Justify your answers.
a. x2 − 4 = 0
b. x2 + 1 = 0
c. x2 − 1 = 0
d. x2 + 4 = 0
e. x2 − 9 = 0
f. x2 + 9 = 0
A. i
B. 3i
C. 3
D. 2i
E. 1
F. 2

Communicate Your Answer

Question 3.
What are the subsets of the set of complex numbers? Give an example of a number in each subset.
Answer:

Question 4.
Is it possible for a number to be both whole and natural? natural and rational? rational and irrational? real and imaginary? Explain your reasoning.
Answer:

Monitoring Progress

Find the square root of the number.
Question 1.
\(\sqrt{-4}\)
Answer:

Question 2.
\(\sqrt{-12}\)
Answer:

Question 3.
−\(\sqrt{-36}\)
Answer:

Question 4.
2\(\sqrt{-54}\)
Answer:

Find the values of x and y that satisfy the equation.
Question 5.
x + 3i = 9 − yi
Answer:

Question 6.
9 + 4yi = −2x + 3i
Answer:

Question 7.
WHAT IF?
In Example 4, what is the impedance of the circuit when the capacitor is replaced with one having a reactance of 7 ohms?
Answer:

Perform the operation. Write the answer in standard form.
Question 8.
(9 − i ) + (−6 + 7i )
Answer:

Question 9.
(3 + 7i ) − (8 − 2i )
Answer:

Question 10.
−4 − (1 + i) − (5 + 9i)
Answer:

Question 11.
(−3i)(10i)
Answer:

Question 12.
i(8 − i)
Answer:

Question 13.
(3 + i)(5 −i)
Answer:

Solve the equation.
Question 14.
x2 = −13
Answer:

Question 15.
x2= −38
Answer:

Question 16.
x2 + 11 = 3
Answer:

Question 17.
x2 − 8 = −36
Answer:

Question 18.
3x2 − 7 = −31
Answer:

Question 19.
5x2 + 33 = 3
Answer:

Find the zeros of the function.
Question 20.
f(x) = x2 + 7
Answer:

Question 21.
f(x) = −x2 − 4
Answer:

Question 22.
f(x) = 9x2 + 1
Answer:

Complex Numbers 3.2 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What is the imaginary unit i defined as and how can you use i?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 1

Question 2.
COMPLETE THE SENTENCE
For the complex number 5 + 2i, the imaginary part is ____ and the real part is ____.
Answer:

Question 3.
WRITING
Describe how to add complex numbers.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 3

Question 4.
WHICH ONE DOESN’T BELONG?
Which number does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 3
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, find the square root of the number.
Question 5.
\(\sqrt{-36}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 5

Question 6.
\(\sqrt{-64}\)
Answer:

Question 7.
\(\sqrt{-18}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 7

Question 8.
\(\sqrt{-24}\)
Answer:

Question 9.
2\(\sqrt{-16}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 9

Question 10.
−3\(\sqrt{-49}\)
Answer:

Question 11.
−4\(\sqrt{-32}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 11

Question 12.
6\(\sqrt{-63}\)
Answer:

In Exercises 13–20, find the values of x and y that satisfy the equation.
Question 13.
4x + 2i = 8 + yi
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 13

Question 14.
3x + 6i = 27 + yi
Answer:

Question 15.
−10x + 12i = 20 + 3yi
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 15

Question 16.
9x − 18i = −36 + 6yi
Answer:

Question 17.
2x − yi = 14 + 12i
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 17

Question 18.
−12x + yi = 60 − 13i
Answer:

Question 19.
54 − \(\frac{1}{7}\)yi = 9x− 4i
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 19

Question 20.
15 − 3yi = \(\frac{1}{2}\)x + 2i
Answer:

In Exercises 21–30, add or subtract. Write the answer in standard form.
Question 21.
(6 − i) + (7 + 3i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 21

Question 22.
(9 + 5i) + (11 + 2i )
Answer:

Question 23.
(12 + 4i) − (3 − 7i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 23

Question 24.
(2 − 15i) − (4 + 5i)
Answer:

Question 25.
(12 − 3i) + (7 + 3i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 25

Question 26.
(16 − 9i) − (2 − 9i)
Answer:

Question 27.
7 − (3 + 4i) + 6i
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 27

Question 28.
16 − (2 − 3i) − i
Answer:

Question 29.
−10 + (6 − 5i) − 9i
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 29

Question 30.
−3 + (8 + 2i) + 7i
Answer:

Question 31.
USING STRUCTURE
Write each expression as a complex number in standard form.
a. \(\sqrt{-9}+\sqrt{-4}-\sqrt{16}\)
b. \(\sqrt{-16}+\sqrt{8}+\sqrt{-36}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 31

Question 32.
REASONING
The additive inverse of a complex number z is a complex number za such that z + za = 0. Find the additive inverse of each complex number.
a. z = 1 + i
b. z = 3 − i
c. z = −2 + 8i
Answer:

In Exercises 33–36, find the impedance of the series circuit.
Question 33.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 33

Question 35.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 35

In Exercises 37–44, multiply. Write the answer in standard form.
Question 37.
3i(−5 + i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 37

Question 38.
2i(7 − i)
Answer:

Question 39.
(3 − 2i)(4 + i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 39

Question 40.
(7 + 5i)(8 − 6i)
Answer:

Question 41.
(4 − 2i)(4 + 2i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 41

Question 42.
(9 + 5i)(9 − 5i)
Answer:

Question 43.
(3 − 6i)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 43

Question 44.
(8 + 3i)2
Answer:

JUSTIFYING STEPS In Exercises 45 and 46, justify each step in performing the operation.
Question 45.
11 − (4 + 3i) + 5i
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 45

Question 46.
(3 + 2i)(7 − 4i)
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 7
Answer:

REASONING In Exercises 47 and 48, place the tiles in the expression to make a true statement.
Question 47.
(____ − ____i) – (____ − ____i ) = 2 − 4i
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 47

Question 48.
____i(____ + ____i ) = −18 − 10i
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 9
Answer:

In Exercises 49–54, solve the equation. Check your solution(s).
Question 49.
x2 + 9 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 49

Question 50.
x2 + 49 = 0
Answer:

Question 51.
x2 − 4 = −11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 51

Question 52.
x2 − 9 = −15
Answer:

Question 53.
2x2 + 6 = −34
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 53

Question 54.
x2 + 7 = −47
Answer:

In Exercises 55–62, find the zeros of the function.
Question 55.
f(x) = 3x2 + 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 55

Question 56.
g(x) = 7x2 + 21
Answer:

Question 57.
h(x) = 2x2 + 72
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 57

Question 58.
k(x) = −5x2 − 125
Answer:

Question 59.
m(x) = −x2 − 27
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 59

Question 60.
p(x) = x2 + 98
Answer:

Question 61.
r(x) = − \(\frac{1}{2}\)x2 − 24
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 61

Question 62.
f(x) = −\(\frac{1}{5}\)x2 − 10
Answer:

ERROR ANALYSIS In Exercises 63 and 64, describe and correct the error in performing the operation and writing the answer in standard form.
Question 63.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 63

Question 64.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 11
Answer:

Question 65.
NUMBER SENSE
Simplify each expression. Then classify your results in the table below.
a. (−4 + 7i) + (−4 − 7i)
b. (2 − 6i) − (−10 + 4i)
c. (25 + 15i) − (25 − 6i)
d. (5 + i)(8 − i)
e. (17 − 3i) + (−17 − 6i)
f. (−1 + 2i)(11 − i)
g. (7 + 5i) + (7 − 5i)
h. (−3 + 6i) − (−3 − 8i)
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 65

Question 66.
MAKING AN ARGUMENT
The Product Property ofSquare Roots states \(\sqrt{a}\) • \(\sqrt{b}\) = \(\sqrt{ab}\) . Your friend concludes \(\sqrt{-4}\) • \(\sqrt{-9}\) = \(\sqrt{36}\) = 6. Is your friend correct? Explain.
Answer:

Question 67.
FINDING A PATTERN
Make a table that shows the powers of i from i1 to i8 in the first row and the simplified forms of these powers in the second row. Describe the pattern you observe in the table. Verify the pattern continues by evaluating the next four powers of i.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 67

Question 68.
HOW DO YOU SEE IT?
The graphs of three functions are shown. Which function(s) has real zeros? imaginary zeros? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 13
Answer:

In Exercises 69–74, write the expression as a complex number in standard form.
Question 69.
(3 + 4i) − (7 − 5i) + 2i(9 + 12i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 69

Question 70.
3i(2 + 5i) + (6 − 7i) − (9 + i)
Answer:

Question 71.
(3 + 5i)(2 − 7i4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 71

Question 72.
2i3(5 − 12i )
Answer:

Question 73.
(2 + 4i5) + (1 − 9i6) − (3 +i7)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 73

Question 74.
(8 − 2i4) + (3 − 7i8) − (4 + i9)
Answer:

Question 75.
OPEN-ENDED
Find two imaginary numbers whose sum and product are real numbers. How are the imaginary numbers related?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 75

Question 76.
COMPARING METHODS
Describe the two different methods shown for writing the complex expression in standard form. Which method do you prefer? Explain.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 14
Answer:

Question 77.
CRITICAL THINKING
Determine whether each statement is true or false. If it is true, give an example. If it is false, give a counterexample.
a. The sum of two imaginary numbers is an imaginary number.
b. The product of two pure imaginary numbers is a real number.
c. A pure imaginary number is an imaginary number.
d. A complex number is a real number.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 77

Question 78.
THOUGHT PROVOKING
Create a circuit that has an impedance of 14 − 3i.
Answer:

Maintaining Mathematical Proficiency

Determine whether the given value of x is a solution to the equation.
Question 79.
3(x − 2) + 4x − 1 = x − 1; x = 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 79

Question 80.
x3 − 6 = 2x2 + 9 − 3x; x = −5
Answer:

Question 81.
−x2 + 4x = 19 — 3x2; x = −\(\frac{3}{4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 81

Write a quadratic function in vertex form whose graph is shown.
Question 82.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 15
Answer:

Question 83.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 83

Question 84.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 17
Answer:

Lesson 3.3 Completing the Square

Essential Question How can you complete the square for a quadratic expression?

EXPLORATION 1

Using Algebra Tiles to Complete the Square
Work with a partner. Use algebra tiles to complete the square for the expression x2 + 6x.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 1
a. You can model x2 + 6x using one x2-tile and six x-tiles. Arrange the tiles in a square. Your arrangement will be incomplete in one of the corners.
b. How many 1-tiles do you need to complete the square?
c. Find the value of c so that the expression x2 + 6x + c is a perfect square trinomial.
d. Write the expression in part (c) as the square of a binomial.

EXPLORATION 2

Drawing Conclusions
Work with a partner.
a. Use the method outlined in Exploration 1 to complete the table.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 2
b. Look for patterns in the last column of the table. Consider the general statement x2 + bx + c = (x + d)2. How are d and b related in each case? How are c and d related in each case?
c. How can you obtain the values in the second column directly from the coefficients of x in the first column?

Communicate Your Answer

Question 3.
How can you complete the square for a quadratic expression?
Answer:

Question 4.
Describe how you can solve the quadratic equation x2 + 6x = 1 by completing the square.
Answer:

Monitoring Progress

Solve the equation using square roots. Check your solution(s).
Question 1.
x2 + 4x + 4 = 36
Answer:

Question 2.
x2 − 6x + 9 = 1
Answer:

Question 3.
x2 − 22x + 121 = 81
Answer:

Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.
Question 4.
x2 + 8x + c
Answer:

Question 5.
x2 − 2x + c
Answer:

Question 6.
x2 − 9x + c
Answer:

Solve the equation by completing the square.
Question 7.
x2 − 4x + 8 = 0
Answer:

Question 8.
x2 + 8x − 5 = 0
Answer:

Question 9.
−3x2 − 18x − 6 = 0
Answer:

Question 10.
4x2 + 32x = −68
Answer:

Question 11.
6x(x + 2) = −42
Answer:

Question 12.
2x(x − 2) = 200
Answer:

Write the quadratic function in vertex form. Then identify the vertex.
Question 13.
y = x2 − 8x + 18
Answer:

Question 14.
y = x2 + 6x + 4
Answer:

Question 15.
y = x2 − 2x − 6
Answer:

Question 16.
WHAT IF?
The height of the baseball can be modeled by y = −16t2 + 80t + 2. Find the maximum height of the baseball. How long does the ball take to hit the ground?
Answer:

Completing the Square 3.3 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What must you add to the expression x2 + bx to complete the square?
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 1

Question 2.
COMPLETE THE SENTENCE
The trinomial x2 − 6x + 9 is a ____ because it equals ____.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, solve the equation using square roots. Check your solution(s).
Question 3.
x2 − 8x + 16 = 25
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 3

Question 4.
r2 − 10r + 25 = 1
Answer:

Question 5.
x2 − 18x + 81 = 5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 5

Question 6.
m2 + 8m + 16 = 45
Answer:

Question 7.
y2 − 24y + 144 = −100
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 7

Question 8.
x2 − 26x + 169 = −13
Answer:

Question 9.
4w2 + 4w + 1 = 75
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 9

Question 10.
4x2 − 8x + 4 = 1
Answer:

In Exercises 11–20, find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.
Question 11.
x2 + 10x + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 11

Question 12.
x2 + 20x + c
Answer:

Question 13.
y2 − 12y + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 13

Question 14.
t2 − 22t + c
Answer:

Question 15.
x2 − 6x + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 15

Question 16.
x2 + 24x + c
Answer:

Question 17.
z2 − 5z + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 17

Question 18.
x2 + 9x + c
Answer:

Question 19.
w2 + 13w + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 19

Question 20.
s2 − 26s + c
Answer:

In Exercises 21–24, find the value of c. Then write an expression represented by the diagram.
Question 21.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 3
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 21

Question 22.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 4
Answer:

Question 23.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 23

Question 24.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 6
Answer:

In Exercises 25–36, solve the equation by completing the square.
Question 25.
x2 + 6x + 3 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 25

Question 26.
s2 + 2s − 6 = 0
Answer:

Question 27.
x2 + 4x − 2 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 27

Question 28.
t2 − 8t − 5 = 0
Answer:

Question 29.
z(z + 9) = 1
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 29

Question 30.
x(x + 8) = −20
Answer:

Question 31.
7t2 + 28t + 56 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 31

Question 32.
6r2 + 6r + 12 = 0
Answer:

Question 33.
5x(x + 6) = −50
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 33

Question 34.
4w(w − 3) = 24
Answer:

Question 35.
4x2 − 30x = 12 + 10x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 35

Question 36.
3s2 + 8s = 2s − 9
Answer:

Question 37.
ERROR ANALYSIS
Describe and correct the error in solving the equation.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 7
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 37

Question 38.
ERROR ANALYSIS
Describe and correct the error in finding the value of c that makes the expression a perfect square trinomial.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 8
Answer:

Question 39.
WRITING
Can you solve an equation by completing the square when the equation has two imaginary solutions? Explain.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 39

Question 40.
ABSTRACT REASONING
Which of the following are solutions of the equation x2 − 2ax + a2 = b2? Justify your answers.
A. ab
B. −a − b
C. b
D. a
E. a − b
F. a + b
Answer:

USING STRUCTURE In Exercises 41–50, determine whether you would use factoring, square roots, or completing the square to solve the equation. Explain your reasoning. Then solve the equation.
Question 41.
x2 − 4x − 21 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 41

Question 42.
x2 + 13x + 22 = 0
Answer:

Question 43.
(x + 4)2 = 16
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 43

Question 44.
(x − 7)2 = 9
Answer:

Question 45.
x2 + 12x + 36 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 45

Question 46.
x2 − 16x + 64 = 0
Answer:

Question 47.
2x2 + 4x − 3 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 47

Question 48.
3x2 + 12x + 1 = 0
Answer:

Question 49.
x2 − 100 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 49

Question 50.
4x2 − 20 = 0
Answer:

MATHEMATICAL CONNECTIONS In Exercises 51–54, find the value of x.
Question 51.
Area of rectangle = 50
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 9
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 51

Question 52.
Area of parallelogram = 48
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 10
Answer:

Question 53.
Area of triangle = 40
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 11
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 53

Question 54.
Area of trapezoid = 20
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 12
Answer:

In Exercises 55–62, write the quadratic function in vertex form. Then identify the vertex.
Question 55.
f(x) = x2 − 8x + 19
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 55

Question 56.
g(x) = x2 − 4x − 1
Answer:

Question 57.
g(x) = x2 + 12x + 37
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 57

Question 58.
h(x) = x2 + 20x + 90
Answer:

Question 59.
h(x) = x2 + 2x − 48
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 59

Question 60.
f(x) = x2 + 6x − 16
Answer:

Question 61.
f(x) = x2 − 3x + 4
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 61

Question 62.
g(x) = x2 + 7x + 2
Answer:

Question 63.
MODELING WITH MATHEMATICS
While marching, a drum major tosses a baton into the air and catches it. The height h (in feet) of the baton t seconds after it is thrown can be modeled by the function h = −16t2 + 32t + 6.
a. Find the maximum height of the baton.
b. The drum major catches the baton when it is 4 feet above the ground. How long is the baton in the air?
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 63

Question 64.
MODELING WITH MATHEMATICS
A firework explodes when it reaches its maximum height. The height h (in feet) of the firework t seconds after it is launched can be modeled by h = \(-\frac{500}{9} t^{2}+\frac{1000}{3} t\) + 10. What is the maximum height of the firework? How long is the firework in the air before it explodes?
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 13
Answer:

Question 65.
COMPARING METHODS
A skateboard shop sells about 50 skateboards per week when the advertised price is charged. For each $1 decrease in price, one additional skateboard per week is sold. The shop’s revenue can be modeled by y = (70 − x)(50 + x).
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 14
a. Use the intercept form of the function to find the maximum weekly revenue.
b. Write the function in vertex form to find the maximum weekly revenue.
c. Which way do you prefer? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 65

Question 66.
HOW DO YOU SEE IT?
The graph of the function f(x) = (x − h)2 is shown. What is the x-intercept? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 15
Answer:

Question 67.
WRITING
At Buckingham Fountain in Chicago, the height h (in feet) of the water above the main nozzle can be modeled by h = −162 + 89.6t, where t is the time (in seconds) since the water has left the nozzle. Describe three different ways you could find the maximum height the water reaches. Then choose a method and find the maximum height of the water.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 67

Question 68.
PROBLEM SOLVING
A farmer is building a rectangular pen along the side of a barn for animals. The barn will serve as one side of the pen. The farmer has 120 feet of fence to enclose an area of 1512 square feet and wants each side of the pen to be at least 20 feet long.
a. Write an equation that represents the area of the pen.
b. Solve the equation in part (a) to find the dimensions of the pen.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 16
Answer:

Question 69.
MAKING AN ARGUMENT
Your friend says the equation x2 + 10x = −20 can be solved by either completing the square or factoring. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 69

Question 70.
THOUGHT PROVOKING
Write a function g in standard form whose graph has the same x-intercepts as the graph of f(x) = 2x2 + 8x + 2. Find the zeros of each function by completing the square. Graph each function.
Answer:

Question 71.
CRITICAL THINKING
Solve x2 + bx + c = 0 by completing the square. Your answer will be an expression for x in terms of b and c.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 71

Question 72.
DRAWING CONCLUSIONS
In this exercise, you will investigate the graphical effect of completing the square.
a. Graph each pair of functions in the same coordinate plane.
y = x2 + 2x y = x2 − 6x
y = (x + 1)2 y = (x − 3)2
b. Compare the graphs of y = x2 + bx and y = (x + \(\frac{b}{2}\))2. Describe what happens to the graph of y = x2 + bx when you complete the square.
Answer:

Question 73.
MODELING WITH MATHEMATICS
In your pottery class, you are given a lump of clay with a volume of 200 cubic centimeters and are asked to make a cylindrical pencil holder. The pencil holder should be 9 centimeters high and have an inner radius of 3 centimeters. What thickness x should your pencil holder have if you want to use all of the clay?
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 17
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 73

Maintaining Mathematical Proficiency

Solve the inequality. Graph the solution.
Question 74.
2x − 3 < 5
Answer:

Question 75.
4 − 8y ≥ 12
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 75

Question 76.
\(\frac{n}{3}\) + 6 > 1
Answer:

Question 77.
−\(\frac{2s}{5}\) ≤ 8
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 77

Graph the function. Label the vertex, axis of symmetry, and x-intercepts.
Question 78.
g(x) = 6(x − 4)2
Answer:

Question 79.
h(x) = 2x(x − 3)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 79

Question 80.
f(x) = x2 + 2x + 5
Answer:

Question 81.
f(x) = 2(x + 10)(x − 12)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 81

Quadratic Equations and Complex Numbers Study Skills: Creating a Positive Study Environment

3.1–3.3 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 18

Core Concepts
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 19

Mathematical Practices
Question 1.
Analyze the givens, constraints, relationships, and goals in Exercise 61 on page 101.
Answer:

Question 2.
Determine whether it would be easier to find the zeros of the function in Exercise 63 on page 117 or Exercise 67 on page 118.
Answer:

Study Skills: Creating a Positive Study Environment

  • Set aside an appropriate amount of time for reviewing your notes and the textbook, reworking your notes, and completing homework.
  • Set up a place for studying at home that is comfortable, but not too comfortable. The place needs to be away from all potential distractions.
  • Form a study group. Choose students who study well together, help out when someone misses school, and encourage positive attitudes.
    Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 20

Quadratic Equations and Complex Numbers 3.1–3.3 Quiz

Solve the equation by using the graph. Check your solution(s).
Question 1.
x2 − 10x + 25 = 0
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 1
Answer:

Question 2.
2x2 + 16 = 12x
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 2
Answer:

Question 3.
x2 = −2x + 8
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 3
Answer:

Solve the equation using square roots or by factoring. Explain the reason for your choice.
Question 4.
2x2 − 15 = 0
Answer:

Question 5.
3x2 − x − 2 = 0
Answer:

Question 6.
(x + 3)2 = 8
ans;

Question 7.
Find the values of x and y that satisfy the equation 7x − 6i = 14 + yi.
Answer:

Perform the operation. Write your answer in standard form
Question 8.
(2 + 5i) + (−4 + 3i)
Answer:

Question 9.
(3 + 9i) − (1 − 7i)
Answer:

Question 10.
(2 + 4i)(−3 − 5i)
Answer:

Question 11.
Find the zeros of the function f(x) = 9x2 + 2. Does the graph of the function intersect the x-axis? Explain your reasoning.
Answer:

Solve the equation by completing the square.
Question 12.
x2 − 6x + 10 = 0
Answer:

Question 13.
x2 + 12x + 4 = 0
Answer:

Question 14.
4x(x + 6) = −40
Answer:

Question 15.
Write y = x2 − 10x + 4 in vertex form. Then identify the vertex.
Answer:

Question 16.
A museum has a café with a rectangular patio. The museum wants to add 464 square feet to the area of the patio by expanding the existing patio as shown.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 4
a. Find the area of the existing patio.
b. Write an equation to model the area of the new patio.
c. By what distance x should the length of the patio be expanded?
Answer:

Question 17.
Find the impedance of the series circuit.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 5
Answer:

Question 18.
The height h (in feet) of a badminton birdie t seconds after it is hit can be modeled by the function h = −16t2 + 32t + 4.
a. Find the maximum height of the birdie.
b. How long is the birdie in the air?
Answer:

Lesson 3.4 Using the Quadratic Formula

Essential Question How can you derive a general formula for solving a quadratic equation?

EXPLORATION 1

Deriving the Quadratic Formula
Work with a partner. Analyze and describe what is done in each step in the development of the Quadratic Formula.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 1

EXPLORATION 2

Using the Quadratic Formula
Work with a partner. Use the Quadratic Formula to solve each equation.
a. x2 − 4x + 3 = 0
b. x2 − 2x + 2 = 0
c. x2 + 2x − 3 = 0
d. x2 + 4x + 4 = 0
e. x2 − 6x + 10 = 0
f. x2 + 4x + 6 = 0

Communicate Your Answer

Question 3.
How can you derive a general formula for solving a quadratic equation?
Answer:

Question 4.
Summarize the following methods you have learned for solving quadratic equations: graphing, using square roots, factoring, completing the square, and using the Quadratic Formula.
Answer:

Monitoring Progress

Solve the equation using the Quadratic Formula.
Question 1.
x2 − 6x + 4 = 0
Answer:

Question 2.
2x2 + 4 = −7x
Answer:

Question 3.
5x2 = x + 8
Answer:

Solve the equation using the Quadratic Formula.
Question 4.
x2 + 41 = −8x
Answer:

Question 5.
−9x2 = 30x + 25
Answer:

Question 6.
5x − 7x2 = 3x + 4
Answer:

Find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
Question 7.
4x2 + 8x + 4 = 0
Answer:

qm 8.
\(\frac{1}{2}\)x2 + x − 1 = 0
Answer:

Question 9.
5x2 = 8x − 13
Answer:

Question 10.
7x2 − 3x = 6
Answer:

Question 11.
4x2 + 6x = −9
Answer:

Question 12.
−5x2 + 1 = 6 − 10x
Answer:

Question 13.
Find a possible pair of integer values for a and c so that the equation ax2 + 3x + c = 0 has two real solutions. Then write the equation.
Answer:

Question 14.
WHAT IF?
The ball leaves the juggler’s hand with an initial vertical velocity of 40 feet per second. How long is the ball in the air?
Answer:

Using the Quadratic Formula 3.4 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
When a, b, and c are real numbers such that a ≠ 0, the solutions of the quadratic equation ax2 + bx + c = 0 are x= ____________.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 81

Question 2.
COMPLETE THE SENTENCE
You can use the ____________ of a quadratic equation to determine the number and type of solutions of the equation.
Answer:

Question 3.
WRITING
Describe the number and type of solutions when the value of the discriminant is negative.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 3

Question 4.
WRITING
Which two methods can you use to solve any quadratic equation? Explain when you might prefer to use one method over the other.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–18, solve the equation using the Quadratic Formula. Use a graphing calculator to check your solution(s).
Question 5.
x2 − 4x + 3 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 5

Question 6.
3x2 + 6x + 3 = 0
Answer:

Question 7.
x2 + 6x + 15 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 7

Question 8.
6x2 − 2x + 1 = 0
Answer:

Question 9.
x2 − 14x = −49
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 9

Question 10.
2x2 + 4x = 30
Answer:

Question 11.
3x2 + 5 = −2x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 11

Question 12.
−3x = 2x2 − 4
Answer:

Question 13.
−10x = −25 − x2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 13

Question 14.
−5x2 − 6 = −4x
Answer:

Question 15.
−4x2 + 3x = −5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 15

Question 16.
x2 + 121 = −22x
Answer:

Question 17.
−z2 = −12z + 6
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 17

Question 18.
−7w + 6 = −4w2
Answer:

In Exercises 19–26, find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
Question 19.
x2 + 12x + 36 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 19

Question 20.
x2 − x + 6 = 0
Answer:

Question 21.
4n2 − 4n − 24 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 21

Question 22.
−x2 + 2x + 12 = 0
Answer:

Question 23.
4x2 = 5x − 10
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 23

Question 24.
−18p = p2 + 81
Answer:

Question 25.
24x = −48 − 3x2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 25

Question 26.
−2x2 − 6 = x2
Answer:

Question 27.
USING EQUATIONS
What are the complex solutions of the equation 2x2− 16x+ 50 = 0?
A. 4 + 3i, 4 − 3i
B. 4 + 12i, 4 − 12i
C. 16 + 3i, 16 − 3i
D. 16 + 12i, 16 − 12i
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 27

Question 28.
USING EQUATIONS
Determine the number and type of solutions to the equation x2 + 7x = −11.
A. two real solutions
B. one real solution
C. two imaginary solutions
D. one imaginary solution
Answer:

ANALYZING EQUATIONS In Exercises 29–32, use the discriminant to match each quadratic equation with the correct graph of the related function. Explain your reasoning.
Question 29.
x2 − 6x + 25 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 29

Question 30.
2x2 − 20x + 50 = 0
Answer:

Question 31.
3x2 + 6x − 9 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 31

Question 32.
5x2 − 10x − 35 = 0
Answer:

Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 2

ERROR ANALYSIS In Exercises 33 and 34, describe and correct the error in solving the equation.
Question 33.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 33

Question 34.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 4
Answer:

OPEN-ENDED In Exercises 35–40, find a possible pair of integer values for a and c so that the quadratic equation has the given solution(s). Then write the equation.
Question 35.
ax2 + 4x + c = 0; two imaginary solutions
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 35

Question 36.
ax2 + 6x + c = 0; two real solutions
Answer:

Question 37.
ax2 − 8x + c = 0; two real solutions
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 37

Question 38.
ax2 − 6x + c = 0; one real solution
Answer:

Question 39.
ax2 + 10x = c; one real solution
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 39

Question 40.
−4x + c = −ax2; two imaginary solutions
Answer:

USING STRUCTURE In Exercises 41–46, use the Quadratic Formula to write a quadratic equation that has the given solutions.
Question 41.
x = \(\frac{-8 \pm \sqrt{-176}}{-10}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 41

Question 42.
x = \(\frac{15 \pm \sqrt{-215}}{22}\)
Answer:

Question 43.
x = \(\frac{-4 \pm \sqrt{-124}}{-14}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 43

Question 44.
x = \(\frac{-9 \pm \sqrt{137}}{4}\)
Answer:

Question 45.
x = \(\frac{-4 \pm 2}{6}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 45

Question 46.
x = \(\frac{2 \pm 4}{-2}\)
Answer:

COMPARING METHODS In Exercises 47–58, solve the quadratic equation using the Quadratic Formula. Then solve the equation using another method. Which method do you prefer? Explain.
Question 47.
3x2 − 21 = 3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 47

Question 48.
5x2 + 38 = 3
Answer:

Question 49.
2x2 − 54 = 12x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 49

Question 50.
x2 = 3x + 15
Answer:

Question 51.
x2 − 7x + 12 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 51

Question 52.
x2 + 8x − 13 = 0
Answer:

Question 53.
5x2 − 50x = −135
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 53

Question 54.
8x2 + 4x + 5 = 0
Answer:

Question 55.
−3 = 4x2 + 9x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 55.1
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 55.2

Question 56.
−31x + 56 = −x2
Answer:

Question 57.
x2 = 1 − x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 57

Question 58.
9x2 + 36x + 72 = 0
Answer:

MATHEMATICAL CONNECTIONS In Exercises 59 and 60, find the value for x.
Question 59.
Area of the rectangle = 24 m2
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 59

Question 6.
Area of the triangle = 8ft2
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 6
Answer:

Question 61.
MODELING WITH MATHEMATICS
A lacrosse player throws a ball in the air from an initial height of 7 feet. The ball has an initial vertical velocity of 90 feet per second. Another player catches the ball when it is 3 feet above the ground. How long is the ball in the air?
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 7
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 61

Question 62.
NUMBER SENSE
Suppose the quadratic equation ax2 + 5x + c = 0 has one real solution. Is it possible for a and c to be integers? rational numbers? Explain your reasoning. Then describe the possible values of a and c.
Answer:

Question 63.
MODELING WITH MATHEMATICS
In a volleyball game, a player on one team spikes the ball over the net when the ball is 10 feet above the court. The spike drives the ball downward with an initial vertical velocity of 55 feet per second. How much time does the opposing team have to return the ball before it touches the court?
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 63

Question 64.
MODELING WITH MATHEMATICS
An archer is shooting at targets. The height of the arrow is 5 feet above the ground. Due to safety rules, the archer must aim the arrow parallel to the ground.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 8
a. How long does it take for the arrow to hit a target that is 3 feet above the ground?
b. What method did you use to solve the quadratic equation? Explain.
Answer:

Question 65.
PROBLEM SOLVING
A rocketry club is launching model rockets. The launching pad is 30 feet above the ground. Your model rocket has an initial vertical velocity of 105 feet per second. Your friend’s model rocket has an initial vertical velocity of 100 feet per second.
a. Use a graphing calculator to graph the equations of both model rockets. Compare the paths.
b. After how many seconds is your rocket 119 feet above the ground? Explain the reasonableness of your answer(s).
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 65

Question 66.
PROBLEM SOLVING
The number A of tablet computers sold (in millions) can be modeled by the function A = 4.5t2 + 43.5t + 17, where t represents the year after 2010.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 9
a. In what year did the tablet computer sales reach 65 million?
b. Find the average rate of change from 2010 to 2012 and interpret the meaning in the context of the situation.
c. Do you think this model will be accurate after a new, innovative computer is developed? Explain.
Answer:

Question 67.
MODELING WITH MATHEMATICS
A gannet is a bird that feeds on fish by diving into the water. A gannet spots a fish on the surface of the water and dives 100 feet to catch it. The bird plunges toward the water with an initial vertical velocity of −88 feet per second.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 10
a. How much time does the fish have to swim away?
b. Another gannet spots the same fish, and it is only 84 feet above the water and has an initial vertical velocity of −70 feet per second. Which bird will reach the fish first? Justify your answer.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 67.1
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 67.2

Question 68.
USING TOOLS
You are asked to find a possible pair of integer values for a and c so that the equation ax2 − 3x + c = 0 has two real solutions. When you solve the inequality for the discriminant, you obtain ac < 2.25. So, you choose the values a = 2 and c = 1. Your graphing calculator displays the graph of your equation in a standard viewing window. Is your solution correct? Explain.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 11
Answer:

Question 69.
PROBLEM SOLVING
Your family has a rectangular pool that measures 18 feet by 9 feet. Your family wants to put a deck around the pool but is not sure how wide to make the deck. Determine how wide the deck should be when the total area of the pool and deck is 400 square feet. What is the width of the deck?
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 12
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 69

Question 70.
HOW DO YOU SEE IT?
The graph of a quadratic function y = ax2 + bx + c is shown. Determine whether each discriminant of ax2 + bx + c = 0 is positive, negative, or zero. Then state the number and type of solutions for each graph. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 13
Answer:

Question 71.
CRITICAL THINKING
Solve each absolute value equation.
a. |x2 – 3x – 14| = 4
b. x2 = |x| + 6
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 71.1
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 71.2

Question 72.
MAKING AN ARGUMENT
The class is asked to solve the equation 4x2 + 14x + 11 = 0. You decide to solve the equation by completing the square. Your friend decides to use the Quadratic Formula. Whose method is more efficient? Explain your reasoning.
Answer:

Question 73.
ABSTRACT REASONING
For a quadratic equation ax2 + bx + c = 0 with two real solutions, show that the mean of the solutions is \(\frac{b}{2a}\). How is this fact related to the symmetry of the graph of y = ax2 + bx + c?
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 73

Question 74.
THOUGHT PROVOKING
Describe a real-life story that could be modeled by h = −16t2 + v0t + h0 . Write the height model for your story and determine how long your object is in the air.
Answer:

Question 75.
REASONING
Show there is no quadratic equation ax2+bx+c= 0 such that a, b, and c are real numbers and 3i and −2i are solutions.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 75

Question 76.
MODELING WITH MATHEMATICS
The Stratosphere Tower in Las Vegas is 921 feet tall and has a “needle” at its top that extends even higher into the air. A thrill ride called Big Shot catapults riders 160 feet up the needle and then lets them fall back to the launching pad.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 14
a. The height h (in feet) of a rider on the Big Shot can be modeled by h = −16t2 + v0 t + 921, where t is the elapsed time (in seconds) after launch and v0 is the initial vertical velocity (in feet per second). Find v0 using the fact that the maximum value of h is 921 + 160 = 1081 feet.
b. A brochure for the Big Shot states that the ride up the needle takes 2 seconds. Compare this time to the time given by the model h = −16t2 + v0 t + 921, where v0 is the value you found in part (a). Discuss the accuracy of the model.
Answer:

Maintaining Mathematical Proficiency

Solve the system of linear equations by graphing.
Question 77.
−x + 2y = 6
x + 4y = 24
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 77

Question 78.
y = 2x − 1
y = x + 1
Answer:

Question 79.
3x + y = 4
6x + 2y = −4
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 79

Question 80.
y = −x + 2
−5x + 5y = 10
Answer:

Graph the quadratic equation. Label the vertex and axis of symmetry.
Question 81.
y = −x2 + 2x + 1
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 81

Question 82.
y = 2x2 − x + 3
Answer:

Question 83.
y = 0.5x2 + 2x + 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 83

Question 84.
y = −3x2 − 2
Answer:

Lesson 3.5 Solving Nonlinear Systems

Essential Question How can you solve a nonlinear system of equations?

EXPLORATION 1

Solving Nonlinear Systems of Equations
Work with a partner. Match each system with its graph. Explain your reasoning. Then solve each system using the graph.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 1

EXPLORATION 2

Solving Nonlinear Systems of Equations
Work with a partner. Look back at the nonlinear system in Exploration 1(f). Suppose you want a more accurate way to solve the system than using a graphical approach.
a. Show how you could use a numerical approach by creating a table. For instance, you might use a spreadsheet to solve the system.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 2
b. Show how you could use an analytical approach. For instance, you might try solving the system by substitution or elimination.

Communicate Your Answer

Question 3.
How can you solve a nonlinear system of equations?
Answer:

Question 4.
Would you prefer to use a graphical, numerical, or analytical approach to solve the given nonlinear system of equations? Explain your reasoning.
Answer:

Solve the system using any method. Explain your choice of method.
Question 1.
y = −x2 + 4
y = −4x + 8
Answer:

Question 2.
x2 + 3x + y = 0
2x + y = 5
Answer:

Question 3.
2x2 + 4x − y =−2
x2 + y = 2
Answer:

Solve the system.
Question 4.
x2 + y2 = 16
y = −x + 4
Answer:

Question 5.
x2 + y2 = 4
y = x + 4
Answer:

Question 6.
x2 + y2 = 1
y = \(\frac{1}{2}\)x + \(\frac{1}{2}\)
Answer:

Solve the equation by graphing.
Question 7.
x2 − 6x + 15 = −(x − 3)2 + 6
Answer:

Question 8.
(x + 4)(x − 1) = −x2 + 3x + 4
Answer:

Solving Nonlinear Systems 3.5 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Describe the possible solutions of a system consisting of two quadratic equations.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which system does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 3
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, solve the system by graphing. Check your solution(s).
Question 3.
y = x + 2
y = 0.5(x + 2)2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 3

Question 4.
y = (x − 3)2 + 5
y = 5
Answer:

Question 5.
y = \(\frac{1}{3}\)x + 2
y = −3x2 − 5x − 4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 5

Question 6.
y = −3x2 − 30x − 71
y = −3x − 17
Answer:

Question 7.
y = x2 + 8x + 18
y = −2x2 − 16x − 30
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 7

Question 8.
y = −2x2 − 9
y = −4x − 1
Answer:

Question 9.
y = (x − 2)2
y = −x2 + 4x − 2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 9

Question 10.
y = \(\frac{1}{2}\)(x + 2)2
y = −\(\frac{1}{2}\)x2 + 2
Answer:

In Exercises 11–14, solve the system of nonlinear equations using the graph.
Question 11.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 11

Question 12.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 5
Answer:

Question 13.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 6
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 13

Question 14.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 7
Answer:

In Exercises 15–24, solve the system by substitution.
Question 15.
y = x + 5
y = x2 − x + 2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 15

Question 16.
x2 + y2 = 49
y = 7 − x
Answer:

Question 17.
x2 + y2 = 64
y = −8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 17

Question 18.
x = 3
−3x2 + 4x − y = 8
Answer:

Question 19.
2x2 + 4x − y = −3
−2x + y = −4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 19

Question 20.
2x − 3 = y + 5x2
y = −3x − 3
Answer:

Question 21.
y = x2 − 1
−7 = −x2 − y
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 21

Question 22.
y + 16x − 22 = 4x2
4x2 − 24x + 26 + y = 0
Answer:

Question 23.
x2 + y2 = 7
x + 3y = 21
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 23

Question 24.
x2 + y2 = 5
−x + y = −1
Answer:

Question 25.
USING EQUATIONS
Which ordered pairs are solutions of the nonlinear system?
y = \(\frac{1}{2}\)x2 − 5x + \(\frac{21}{2}\)
y = −\(\frac{1}{2}\)x + \(\frac{13}{2}\)
A. (1, 6)
B. (3, 0)
C. (8, 2.5)
D. (7, 0)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 25

Question 26.
USING EQUATIONS
How many solutions does the system have? Explain your reasoning.
y = 7x2 − 11x + 9
y = −7x2 + 5x − 3
A. 0
B. 1
C. 2
D. 4
Answer:

In Exercises 27–34, solve the system by elimination.
Question 27.
2x2 − 3x −y =−5
−x + y = 5
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 27

Question 28.
−3x2 + 2x − 5 = y
−x + 2 = −y
Answer:

Question 29.
−3x2 + y = −18x + 29
−3x2 − y = 18x − 25
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 29

Question 30.
y = −x2 − 6x 10
y = 3x2 + 18x + 22
Answer:

Question 31.
y + 2x = −14
−x2 − y − 6x = 11
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 31

Question 32.
y = x2 + 4x + 7
−y = 4x + 7
Answer:

Question 33.
y = −3x2 − 30x − 76
y = 2x2 + 20x + 44
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 33

Question 34.
−10x2 + y = −80x + 155
5x2 + y = 40x − 85
Answer:

Question 35.
ERROR ANALYSIS
Describe and correct the error in using elimination to solve a system.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 35

Question 36.
NUMBER SENSE
The table shows the inputs and outputs of two quadratic equations. Identify the solution(s) of the system. Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 9
Answer:

In Exercises 37–42, solve the system using any method. Explain your choice of method.
Question 37.
y = x2 − 1
−y = 2x2 + 1
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 37

Question 38.
y = −4x2 − 16x − 13
−3x2 + y + 12x = 17
Answer:

Question 39.
−2x + 10 + y = \(\frac{1}{3}\)x2
y = 10
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 39

Question 40.
y = 0.5x2 − 10
y = −x2 + 14
Answer:

Question 41.
y = −3(x − 4)2 + 6
(x − 4)2 + 2 − y = 0
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 41

Question 42.
−x2 + y2 = 100
y = −x + 14
Answer:

USING TOOLS In Exercises 43–48, solve the equation by graphing.
Question 43.
x2 + 2x = −\(\frac{1}{2}\)x2 + 2x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 43

Question 44.
2x2 − 12x − 16 = −6x2 + 60x − 144
Answer:

Question 45.
(x + 2)(x − 2) = −x2 + 6x − 7
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 45

Question 46.
−2x2 − 16x − 25 = 6x2 + 48x + 95
Answer:

Question 47.
(x − 2)2 − 3 = (x + 3)(−x + 9) − 38
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 47

Question 48.
(−x + 4)(x + 8) − 42 = (x + 3)(x + 1) − 1
Answer:

Question 49.
REASONING
A nonlinear system contains the equations of a constant function and a quadratic function. The system has one solution. Describe the relationship between the graphs.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 49

Question 50.
PROBLEM SOLVING
The range (in miles) of a broadcast signal from a radio tower is bounded by a circle given by the equation x2 + y2= 1620.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 10
A straight highway can be modeled by the equation y = −\(\frac{1}{3}\)x + 30.
For what lengths of the highway are cars able to receive the broadcast signal?
Answer:

Question 51.
PROBLEM SOLVING
A car passes a parked police car and continues at a constant speed r. The police car begins accelerating at a constant rate when it is passed. The diagram indicates the distance d (in miles) the police car travels as a function of time t (in minutes) after being passed. Write and solve a system of equations to find how long it takes the police car to catch up to the other car.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 11
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 51

Question 52.
THOUGHT PROVOKING
Write a nonlinear system that has two different solutions with the same y-coordinate. Sketch a graph of your system. Then solve the system.
Answer:

Question 53.
OPEN-ENDED
Find three values for m so the system has no solution, one solution, and two solutions. Justify your answer using a graph.
3y = −x2 + 8x − 7
y = mx + 3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 53

Question 54.
MAKING AN ARGUMENT
You and a friend solve the system shown and determine that x = 3 and x = −3. You use Equation 1 to obtain the solutions (3, -3), (3, −3), (−3, 3), and (−3, −3). Your friend uses Equation 2 to obtain the solutions (3, 3) and (−3, −3). Who is correct? Explain your reasoning.
x2 + y2 = 18 Equation 1
x − y = 0 Equation 2
Answer:

Question 55.
COMPARING METHODS
Describe two different ways you could solve the quadratic equation. Which way do you prefer? Explain your reasoning.
−2x2 + 12x − 17 = 2x2 − 16x + 31
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 55

Question 56.
ANALYZING RELATIONSHIPS
Suppose the graph of a line that passes through the origin intersects the graph of a circle with its center at the origin. When you know one of the points of intersection, explain how you can find the other point of intersection without performing any calculations.
Answer:

Question 57.
WRITING
Describe the possible solutions of a system that contains (a) one quadratic equation and one equation of a circle, and (b) two equations of circles. Sketch graphs to justify your answers.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 57.1
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 57.2
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 57.3

Question 58.
HOW DO YOU SEE IT?
The graph of a nonlinear system is shown. Estimate the solution(s). Then describe the transformation of the graph of the linear function that results in a system with no solution.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 12
Answer:

Question 59.
MODELING WITH MATHEMATICS
To be eligible for a parking pass on a college campus, a student must live at least 1 mile from the campus center.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 13
a. Write equations that represent the circle and Oak Lane.
b. Solve the system that consists of the equations in part (a).
c. For what length of Oak Lane are students not eligible for a parking pass?
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 59.1
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 59.2

Question 60.
CRITICAL THINKING
Solve the system of three equations shown.
x2 + y2 = 4
2y = x2 − 2x + 4
y = −x + 2
Answer:

Maintaining Mathematical Proficiency

Solve the inequality. Graph the solution on a number line.
Question 61.
4x − 4 > 8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 61

Question 62.
−x + 7 ≤ 4 − 2x
Answer:

Question 63.
−3(x − 4) ≥ 24
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 63

Write an inequality that represents the graph.
Question 64.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 14
Answer:

Question 65.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 15
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 65

Question 66.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 16
Answer:

Lesson 3.6 Quadratic Inequalities

Essential Question How can you solve a quadratic inequality?

EXPLORATION 1

Solving a Quadratic Inequality
Work with a partner. The graphing calculator screen shows the graph of f(x) = x2 + 2x − 3.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 1
Explain how you can use the graph to solve the inequality x2 + 2x − 3 ≤ 0.
Then solve the inequality.

EXPLORATION 2

Solving Quadratic Inequalities
Work with a partner. Match each inequality with the graph of its related quadratic function. Then use the graph to solve the inequality.
a. x2 − 3x + 2 > 0
b. x2 − 4x + 3 ≤ 0
c. x2 − 2x − 3 < 0
d. x2 + x − 2 ≥ 0
e. x2 − x − 2 < 0
f. x2 − 4 > 0
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 2

Communicate Your Answer

Question 3.
How can you solve a quadratic inequality?
Answer:

Question 4.
Explain how you can use the graph in Exploration 1 to solve each inequality. Then solve each inequality.
Answer:

Monitoring Progress

Graph the inequality.
Question 1.
y ≥ x2 + 2x − 8
Answer:

Question 2.
y ≤ 2x2 −x − 1
Answer:

Question 3.
y > −x2 + 2x + 4
Answer:

Question 4.
Graph the system of inequalities consisting of y ≤ −x2 and y > x2 − 3.
Answer:

Solve the inequality.
Question 5.
2x2 + 3x ≤ 2
Answer:

Question 6.
−3x2 − 4x + 1 < 0
Answer:

Question 7.
2x2 + 2 > −5x
Answer:

Question 8.
WHAT IF?
In Example 6, the area must be at least 8500 square feet. Describe the possible lengths of the parking lot.
Answer:

Quadratic Inequalities 3.6 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Compare the graph of a quadratic inequality in one variable to the graph of a quadratic inequality in two variables.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 1

Question 2.
WRITING
Explain how to solve x2 + 6x − 8 < 0 using algebraic methods and using graphs.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, match the inequality with its graph. Explain your reasoning.
Question 3.
y ≤ x2 + 4x + 3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 3

Question 4.
y > −x2 + 4x − 3
Answer:

Question 5.
y < x2 − 4x + 3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 5

Question 6.
y ≥ x2 + 4x + 3
Answer:

Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 3

In Exercises 7–14, graph the inequality.
Question 7.
y < −x2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 7

Question 8.
y ≥ 4x2
Answer:

Question 9.
y > x2 − 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 9

Question 10.
y < x2 + 5
Answer:

Question 11.
y ≤ x2 + 5x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 11

Question 12.
y ≥ −2x2 + 9x − 4
Answer:

Question 13.
y > 2(x + 3)2 − 1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 13

Question 14.
y ≤ (x − \(\frac{1}{2}\))2 + \(\frac{5}{2}\)
Answer:

ANALYZING RELATIONSHIPS In Exercises 15 and 16, use the graph to write an inequality in terms of f (x) so point P is a solution.
Question 15.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 15

Question 16.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 5
Answer:

ERROR ANALYSIS In Exercises 17 and 18, describe and correct the error in graphing y ≥ x2 + 2.
Question 17.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 6
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 17

Question 18.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 7
Answer:

Question 19.
MODELING WITH MATHEMATICS
A hardwood shelf in a wooden bookcase can safely support a weight W (in pounds) provided W ≤ 115x2, where x is the thickness (in inches) of the shelf. Graph the inequality and interpret the solution.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 19

Question 20.
MODELING WITH MATHEMATICS
A wire rope can safely support a weight W (in pounds) provided W ≤ 8000d2, where d is the diameter (in inches) of the rope. Graph the inequality and interpret the solution.
Answer:

In Exercises 21–26, graph the system of quadratic inequalities.
Question 21.
y ≥ 2x2
y < −x2 + 1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 21

Question 22.
y > −5x2
y > 3x2 − 2
Answer:

Question 23.
y ≤ −x2 + 4x − 4
y < x2 + 2x − 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 23

Question 24.
y ≥ x2 − 4
y ≤ −2x2 + 7x + 4
Answer:

Question 25.
y ≥ 2x2 + x − 5
y < −x2 + 5x + 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 25

Question 26.
y ≥ x2 − 3x − 6
y ≥ x2 + 7x + 6
Answer:

In Exercises 27–34, solve the inequality algebraically.
Question 27.
4x2 < 25
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 27

Question 28.
x2 + 10x + 9 < 0
Answer:

Question 29.
x2 − 11x ≥ −28
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 29

Question 30.
3x2 − 13x > −10
Answer:

Question 31.
2x2 − 5x − 3 ≤ 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 31

Question 32.
4x2 + 8x − 21 ≥ 0
Answer:

Question 33.
\(\frac{1}{2}\)x2 − x > 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 33.1
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 33.2

Question 34.
−\(\frac{1}{2}\)x2 + 4x ≤ 1
Answer:

In Exercises 35–42, solve the inequality by graphing.
Question 35.
x2 − 3x + 1 < 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 35

Question 36.
x2 − 4x + 2 > 0
Answer:

Question 37.
x2 + 8x > −7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 37

Question 38.
x2 + 6x < −3
Answer:

Question 39.
3x2 − 8 ≤ − 2x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 39

Question 40.
3x2 + 5x − 3 < 1
Answer:

Question 41.
\(\frac{1}{3}\)x2 + 2x ≥ 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 41

Question 42.
\(\frac{3}{4}\)x2 + 4x ≥ 3
Answer:

Question 43.
DRAWING CONCLUSIONS
Consider the graph of the function f(x) = ax2 + bx + c.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 8
a. What are the solutions of ax2 + bx + c < 0?
b. What are the solutions of ax2 + bx + c > 0?
c. The graph of g represents a reflection in the x-axis of the graph of f. For which values of x is g(x) positive?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 43

Question 44.
MODELING WITH MATHEMATICS
A rectangular fountain display has a perimeter of 400 feet and an area of at least 9100 feet. Describe the possible widths of the fountain.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 9
Answer:

Question 45.
MODELING WITH MATHEMATICS
The arch of the Sydney Harbor Bridge in Sydney, Australia, can be modeled by y = −0.00211x2 + 1.06x, where x is the distance (in meters) from the left pylons and y is the height (in meters) of the arch above the water. For what distances x is the arch above the road?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 45

Question 46.
PROBLEM SOLVING
The number T of teams that have participated in a robot-building competition for high-school students over a recent period of time x(in years) can be modeled by T(x) = 17.155x2 + 193.68x + 235.81, 0 ≤ x ≤ 6.
After how many years is the number of teams greater than 1000? Justify your answer.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 11
Answer:

Question 47.
PROBLEM SOLVING
A study found that a driver’s reaction time A(x) to audio stimuli and his or her reaction time V(x) to visual stimuli (both in milliseconds) can be modeled by
A(x) = 0.0051x2 − 0.319x + 15, 16 ≤ x ≤ 70
V(x) = 0.005x2 − 0.23x + 22, 16 ≤ x ≤ 70
where x is the age (in years) of the driver.
a. Write an inequality that you can use to find the x-values for which A(x) is less than V(x).
b. Use a graphing calculator to solve the inequality A(x) < V(x). Describe how you used the domain 16 ≤ x ≤ 70 to determine a reasonable solution. c. Based on your results from parts (a) and (b), do you think a driver would react more quickly to a traffic light changing from green to yellow or to the siren of an approaching ambulance? Explain.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 47

Question 48.
HOW DO YOU SEE IT?
The graph shows a system of quadratic inequalities.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 12
a. Identify two solutions of the system.
b. Are the points (1, −2) and (5, 6) solutions of the system? Explain.
c. Is it possible to change the inequality symbol(s) so that one, but not both of the points in part (b), is a solution of the system? Explain.
Answer:

Question 49.
MODELING WITH MATHEMATICS
The length L (in millimeters) of the larvae of the black porgy fish can be modeled by L(x) = 0.00170x2 + 0.145x + 2.35, 0 ≤ x ≤ 40 where x is the age (in days) of the larvae. Write and solve an inequality to find at what ages a larva’s length tends to be greater than 10 millimeters. Explain how the given domain affects the solution.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 13
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 49

Question 50.
MAKING AN ARGUMENT
You claim the system of inequalities below, where a and b are real numbers, has no solution. Your friend claims the system will always have at least one solution. Who is correct? Explain.
y < (x + a)2
y < (x + b)2
Answer:

Question 51.
MATHEMATICAL CONNECTIONS
The area A of the region bounded by a parabola and a horizontal line can be modeled by A= \(\frac{2}{3}\)bh, where b and h are as defined in the diagram. Find the area of the region determined by each pair of inequalities.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 14
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 51.1
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 51.2

Question 52.
THOUGHT PROVOKING
Draw a company logo that is created by the intersection of two quadratic inequalities. Justify your answer.
Answer:

Question 53.
REASONING
A truck that is 11 feet tall and 7 feet wide is traveling under an arch. The arch can be modeled by y = −0.0625x2 + 1.25x + 5.75, where x and y are measured in feet.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 15
a. Will the truck fit under the arch? Explain.
b. What is the maximum width that a truck 11 feet tall can have and still make it under the arch?
c. What is the maximum height that a truck 7 feet wide can have and still make it under the arch?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 53.1
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 53.2

Maintaining Mathematical Proficiency

Graph the function. Label the x-intercept(s) and the y-intercept.
Question 54.
f(x) = (x + 7)(x − 9)
Answer:

Question 55.
g(x) = (x − 2)2 − 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 55

Question 56.
h(x) = −x2 + 5x − 6
Answer:

Find the minimum value or maximum value of the function. Then describe where the function is increasing and decreasing.
Question 57.
f(x) = −x2 − 6x − 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 57

Question 58.
h(x) = \(\frac{1}{2}\)(x + 2)2 − 1
Answer:

Question 59.
f(x) = −(x − 3)(x + 7)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 59

Question 60.
h(x) = x2 + 3x − 18
Answer:

Quadratic Equations and Complex Numbers Performance Task: Algebra in Genetics: The Hardy-Weinberg Law

3.4–3.6 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 16

Core Concepts
Section 3.4
Solving Equations Using the Quadratic Formula, p. 122
Analyzing the Discriminant of ax2+bx+c= 0, p. 124
Methods for Solving Quadratic Equations, p. 125
Modeling Launched Objects, p. 126

Section 3.5
Solving Systems of Nonlinear Equations, p. 132
Solving Equations by Graphing, p. 135

Section 3.6
Graphing a Quadratic Inequality in Two Variables, p. 140
Solving Quadratic Inequalities in One Variable, p. 142

Mathematical Practices
Question 1.
How can you use technology to determine whose rocket lands first in part (b) of Exercise 65 on page 129?
Answer:

Question 2.
What question can you ask to help the person avoid making the error in Exercise 54 on page 138?
Answer:

Question 3.
Explain your plan to find the possible widths of the fountain in Exercise 44 on page 145.
Answer:

Performance Task: Algebra in Genetics: The Hardy-Weinberg Law
Some people have attached earlobes, the recessive trait. Some people have free earlobes, the dominant trait. What percent of people carry both traits?
To explore the answers to this question and more, go to BigIdeasMath.com.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 17

Quadratic Equations and Complex Numbers Chapter Review

3.1 Solving Quadratic Equations (pp. 93–102)

Question 1.
Solve x2 − 2x − 8 = 0 by graphing.
Answer:

Solve the equation using square roots or by factoring.
Question 2.
3x2 − 4 = 8
Answer:

Question 3.
x2 + 6x − 16 = 0
Answer:

Question 4.
2x2 − 17x = −30
Answer:

Question 5.
A rectangular enclosure at the zoo is 35 feet long by 18 feet wide. The zoo wants to double the area of the enclosure by adding the same distance x to the length and width. Write and solve an equation to find the value of x. What are the dimensions of the enclosure?
Answer:

3.2 Complex Numbers (pp. 103–110)

Question 6.
Find the values x and y that satisfy the equation 36 − yi = 4x + 3i.
Answer:

Perform the operation. Write the answer in standard form.
Question 7.
(−2 + 3i ) + (7 − 6i )
Answer:

Question 8.
(9 + 3i ) − (−2 − 7i )
Answer:

Question 9.
(5 + 6i )(−4 + 7i )
Answer:

Question 10.
Solve 7x2 + 21 = 0.
Answer:

Question 11.
Find the zeros of f(x) = 2x2 + 32.
Answer:

3.3 Completing the Square (pp. 111–118)

Question 12.
An employee at a local stadium is launching T-shirts from a T-shirt cannon into the crowd during an intermission of a football game. The height h (in feet) of the T-shirt after t seconds can be modeled by h = −16t2 + 96t + 4. Find the maximum height of the T-shirt.
Answer:

Solve the equation by completing the square.
Question 13.
x2 + 16x + 17 = 0
Answer:

Question 14.
4x2 + 16x + 25 = 0
Answer:

Question 15.
9x(x − 6) = 81
Answer:

Question 16.
Write y = x2 − 2x + 20 in vertex form. Then identify the vertex.
Answer:

3.4 Using the Quadratic Formula (pp. 121–130)

Solve the equation using the Quadratic Formula.
Question 17.
−x2 + 5x = 2
Answer:

Question 18.
2x2 + 5x = 3
Answer:

Question 19.
3x2 − 12x + 13 = 0
Answer:

Find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
Question 20.
−x2 − 6x − 9 = 0
Answer:

Question 21.
x2 − 2x − 9 = 0
Answer:

Question 22.
x2 + 6x + 5 = 0
Answer:

3.5 Solving Nonlinear Systems (pp. 131–138)

Solve the system by any method. Explain your choice of method.
Question 23.
2x2 − 2 = y
−2x + 2 = y
Answer:

Question 24.
x2 − 6x + 13 = y
−y = −2x + 3
Answer:

Question 25.
x2 + y2 = 4
−15x + 5 = 5y
Answer:

Question 26.
Solve −3x2 + 5x − 1 = 5x2 − 8x − 3 by graphing.
Answer:

3.6 Quadratic Inequalities (pp. 139–146)

Graph the inequality.
Question 27.
y > x2 + 8x + 16
Answer:

Question 28.
y ≥ x2 + 6x + 8
Answer:

Question 29.
x2 + y ≤ 7x − 12
Answer:

Graph the system of quadratic inequalities.
Question 30.
x2 − 4x + 8 > y
−x2 + 4x + 2 ≤ y
Answer:

Question 31.
2x2 − x ≥ y − 5
0.5x2> y − 2x− 1
Answer:

Question 32.
−3x2 − 2x ≤ y + 1
−2x2 + x − 5 > −y
Answer:

Solve the inequality.
Question 33.
3x2 + 3x − 60 ≥0
Answer:

Question 34.
−x2 − 10x < 21
Answer:

Question 35.
3x2 + 2 ≤ 5x
Answer:

Quadratic Equations and Complex Numbers Chapter Test

Solve the equation using any method. Provide a reason for your choice.
Question 1.
0 = x2 + 2x + 3
Answer:

Question 2.
6x = x2 + 7
Answer:

Question 3.
x2 + 49 = 85
Answer:

Question 4.
(x + 4)(x − 1) = −x2 + 3x + 4
Answer:

Explain how to use the graph to find the number and type of solutions of the quadratic equation. Justify your answer by using the discriminant.
Question 5.
\(\frac{1}{2}\)x2 + 3x + \(\frac{9}{2}\) = 0
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 5
Answer:

Question 6.
4x2 + 16x + 18 = 0
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 6
Answer:

Question 7.
−x2 + \(\frac{1}{2}\)x + \(\frac{3}{2}\) = 0
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 7
Answer:

Solve the system of equations or inequalities.
Question 8.
x2 + 66 = 16x − y
2x − y = 18
Answer:

Question 9.
y ≥ \(\frac{1}{4}\)x2 − 2
y < −(x + 3)2x − y = 18 + 4
Answer:

Question 10.
0 = x2 + y2 − 40
y = x + 4
Answer:

Question 11.
Write (3 + 4i )(4 − 6i ) as a complex number in standard form.
Answer:

Question 12.
The aspect ratio of a widescreen TV is the ratio of the screen’s width to its height, or 16 : 9. What are the width and the height of a 32-inch widescreen TV? Justify your answer. (Hint: Use the Pythagorean Theorem and the fact that TV sizes refer to the diagonal length of the screen.)
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 12
Answer:

Question 13.
The shape of the Gateway Arch in St. Louis, Missouri, can be modeled by y = −0.0063x2 + 4x, where x is the distance (in feet) from the left foot of the arch and y is the height (in feet) of the arch above the ground. For what distances x is the arch more than 200 feet above the ground? Justify your answer.
Answer:

Question 14.
You are playing a game of horseshoes. One of your tosses is modeled in the diagram, where x is the horseshoe’s horizontal position (in feet) and y is the corresponding height (in feet). Find the maximum height of the horseshoe. Then find the distance the horseshoe travels. Justify your answer.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 14
Answer:

Quadratic Equations and Complex Numbers Cumulative Assessment

Question 1.
The graph of which inequality is shown?
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 1
A. y2 > x2 + x – 6
B. y ≥ x2 + x – 6
C. y > x2 – x – 6
D. y ≥ x2 – x – 6
Answer:

Question 2.
Classify each function by its function family. Then describe the transformation of the parent function.
a. g(x) = x + 4
b. h(x) = 5
c. h(x) = x2 − 7
d. g(x) = −∣x + 3∣− 9
e. g(x) = \(\frac{1}{4}\)(x − 2)2 − 1
f. h(x) = 6x+ 11
Answer:

Question 3.
Two baseball players hit back-to-back home runs. The path of each home run is modeled by the parabolas below, where x is the horizontal distance (in feet) from home plate and y is the height (in feet) above the ground. Choose the correct symbol for each inequality to model the possible locations of the top of the outfield wall.(HSA-CED.A.3)
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 3
Answer:

Question 4.
You claim it is possible to make a function from the given values that has an axis of symmetry at x = 2. Your friend claims it is possible to make a function that has an axis of symmetry at x = −2. What values can you use to support your claim? What values support your friend’s claim?(HSF-IF.B.4)
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 4
Answer:

Question 5.
Which of the following values are x-coordinates of the solutions of the system?
y = x2 – 6x + 14
y = 2x + 7
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 5
Answer:

Question 6.
The table shows the altitudes of a hang glider that descends at a constant rate. How long will it take for the hang glider to descend to an altitude of 100 feet? Justify your answer.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 6
A. 25 seconds
B. 35 seconds
C. 45 seconds
D. 55 seconds
Answer:

Question 7.
Use the numbers and symbols to write the expression x2 + 16 as the product of two binomials. Some may be used more than once. Justify your answer.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 7
Answer:

Question 8.
Choose values for the constants h and k in the equation x = \(\frac{1}{4}\)( y − k)2 + h so that each statement is true.(HSA-CED.A.2)
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 8
Answer:

Question 9.
Which of the following graphs represents a perfect square trinomial? Write each function in vertex form by completing the square.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 9
Answer:

Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations

Big Ideas Math Answers Grade 1 Chapter 3

The download pdf link of Big Ideas Math Grade 1 Answer Key Chapter 3 More Addition and Subtraction Situations is available on this page. This pdf includes the fundamentals of addition and subtraction in multiple methods. Thus, the students who want to improve their math skills must solve the questions given in the Big Ideas Math Answers Grade 1 Ch 3 More Addition and Subtraction Situations.

Big Ideas Math Book 1st Grade Answer Key Chapter 3 More Addition and Subtraction Situations

As per the student’s understanding level, only these Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations are designed and helping them to learn all primary mathematical concepts in a simple way. Learn the basic concepts of addition and subtraction from our Big Ideas Math Book 1st Grade Answers Chapter 3 More Addition and Subtraction Situations. Check out the topics covered in this chapter from the below section.

Lesson 1: Solve Add To Problems with Start Unknown

Lesson 2: Solve Take From Problems with Change Unknown

Lesson 3: Solve Take From Problems with Start Unknown

Lesson 4: Compare Problems: Bigger Unknown

Lesson 5: Compare Problems: Smaller Unknown

Lesson 6: True or False Equations

Lesson 7: Find Numbers That Make 10

Lesson 8: Fact Families

Performance Task

More Addition and Subtraction Situations Vocabulary

Organize It

Review Words:
count on
number line

Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 1
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and -Subtraction-Situations-More-Addition-and-Subtraction-Situations- Vocubulary

Define It

Use your vocabulary cards to identify the words.

Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 2
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and -Subtraction-Situations-More-Addition-and-Subtraction-Situations- Vocubulary-Define-It

Lesson 3.1 Solve Add To Problems with Start Unknown

Explore and Grow

Use counters to model each problem.

Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 3
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and -Subtraction-Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Explore-and-Grow
Explaination:
Given Sum is 5 and a addend of 2. So, to get the sum 5 we need to count 3 more from 2.
2 + 3 = 5.

Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 4
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and -Subtraction-Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Explore-and-Grow-2
Explaination:
Given 5 as sum and an addend 2. To get the sum 5 we need to count 3 more from 2.
3 + 2 = 5

Show and Grow

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 5
_______ + 5 = 8
Answer:
3 + 5 = 8
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and -Subtraction-Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Show-and-Grow
Explaination:
Given: 8 as sum and an addend 5. To get the sum 8 we need to add 3 more to 5.
3 + 5 = 8

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 6
_______ + 2 = 8
Answer:
1 + 2 = 3
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and -Subtraction-Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Show-and-Grow-2
Explaination:
Given: 3 as sum and an addend 2. To get the sum 3 we need to add 1 more to 2.
1 + 2 = 3

Apply and Grow: practice

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 7
_________ + 3 = 7
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Apply-and-Grow-practice-question-3

Explaination:
Given: 7 as sum and an addend 3. To get the sum 7 we need to add 4 more to 3.
4 + 3 = 7

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 8
_______ + 8 = 10
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Apply-and-Grow-practice-question-4

Explaination:
Given: 10 as sum and an addend 8.  We need to add 2 more to 8 to get 10 as sum.
2 + 8 = 10.

Question 5.
_______ + 4 = 9
Answer:
5 + 4 = 9
Explaination:
Given: 9 as sum and an addend 4. To get the sum 9 we need to add 5 more to 4.
5 + 4 = 9

Question 6.
_________ + 0 = 5
Answer:
5 + 0 = 5
Explaination:

Adding 0 to any number gives the sum as the number itself.
Given: 5 as sum and an addend 0.
so we add 5 to get the sum 5.

Question 7.
MP Structure
Circle the equation that matches the model.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 9
8 + 2 = 10            2 + 6 = 8
2 + 2 = 4             4 + 6 = 10
Answer:
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 9
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Apply-and-Grow-practice-question-7
Explaination :
There are 2 counters in 1st  box and 6 counters in 2nd box.
Total number of boxes = 2
Add the number of boxes=
2 + 6 = 8.

Think and Grow: Modeling Real Life

There are some ladybugs. 2 more join them. Now there are 9. How many ladybugs were there to start?
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 10
Model:
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 11
Addition equation:
______ ladybugs.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Think-and-Grow-Modeling-Real-Life
Addition equation: 7 + 2 = 9

There were 7 ladybugs at the start.
Explaination:

given

Show and Grow

Question 8.
You have some books. You get 4 more books. Now you have 10. How many books did you have to start?
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 12
Model:
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 13
Addition equation:

_________ books
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Apply-and-Grow-practice-question-8

Addition equation: 6 + 4 = 10

I have to start with 6 books.

Solve Add To Problems with Start Unknown Practice 3.1

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 14
________ + 5 = 10
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Practice-3.1-question-1
5 + 5 = 10

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 15
_________ + 1 = 4
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Practice-3.1-question-2
3 + 1 = 4

Question 3.
_________ + 2 = 7
Answer:
5 + 2 = 7

Question 4.
_________ + 3 = 9
Answer:
6 + 3 = 9

Question 5.
MP Structure
Circle the equations that match the model.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 16
6 + 1 = 7                      4 + 2 = 6
3 + 4 = 7                      1 + 5 = 6
Answer:
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 16
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Practice-3.1-question-5-MP-Structure

Question 6.
Modeling Real Life
There are some hippos. 6 more join them. Now there are 9. How many hippos were there to start?
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 17

___________ hippos
Answer:
Total number of hippos = 9
Number of hippos joined = 6
Number of hippos at the start are 3
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.1-Solve-Add-To-Problems-with-Start-Unknown-Practice-3.1-question-6-Modeling-Real-Life

Review & Refresh

Question 7.
There are 8 Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 18 on a free.
5 Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 18 fall off.
How many Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 18 are left?

________ – ________ = _________ Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 18
Answer:
Number of leaf on the trees = 8
Number of fallen leafs = 5
Total number of leafs left on the tree=
8 – 5 = 3 leafs left.

Question 8.
You have 3 Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 19.
You lose 1 Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 19.
How many im – 19 do you have left?
________ – ________ = _________ Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 19
Answer:
Number of balls i have = 3
Number of balls i lose = 1
Total number of left with me =
3 – 1 = 2 balls left.

Lesson 3.2 Solve Take From Problems with Change Unknown

Explore and Grow

Use counters to model each problem.

Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 20
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.2-Solve-Take-From-Problems-with-Change- Unknown-Explore-and-Grow-1

Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 21
Answer:

Show and Grow

Question 1.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 22
7 – _____ = 2
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.2-Solve-Take-From-Problems-with-Change- Unknown-Show-and-Grow-question-1
7 – 5 = 2

Question 2.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 23
5 – ______ = .3
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.2-Solve-Take-From-Problems-with-Change- Unknown-Show-and-Grow-question-2
5 – 2 = 3

Apply and Grow: Practice

Question 3.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 24
8 – ______ = 2
Answer:

8 – 6 = 2

Question 4.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 25
9 – ______ = 6
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.2-Solve-Take-From-Problems-with-Change- Unknown-Apply-and-Grow-Practice-question-4
9 – 3 = 6

Question 5.
3 – _______ = 0
Answer:
3 – 3 = 0

Question 6.
10 – ________ = 5
Answer:
10 – 5 = 5

Question 7.
MP Repeated Reasoning
Match each model with its correct equation.
Answer:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 26
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.2-Solve-Take-From-Problems-with-Change- Unknown-Apply-and-Grow-Practice-question-7

Think and Grow: Modeling Real Life

You have 9 coins. You toss some of them into a fountain. You have 5 left. How many coins did you toss?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 27
Model:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 28
Subtraction equation

________ coins.
Answer:
Total number of coins i have = 9 coins
Number of coins left with me = 5 coins
Number of coins i tossed in fountain = 9 – 5 = 4 coins
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.2-Solve-Take-From-Problems-with-Change- Unknown-Think-and-Grow-Modeling-Real-Life

Show and Grow

Question 8.
You have 8 crayons. You lose some of them. You have 2 left. How many crayons did you lose?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 29
Model:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 30
Subtraction equation:

__________ crayons
Answer:
Total number of crayons = 8
Number of crayons left after loosing = 2
Number of crayons left = 8 – 2 = 6
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.2-Solve-Take-From-Problems-with-Change- Unknown-question-8

Solve Take From Problems with Change Unknown Practice 3.2

Question 1.
8 – ? = 4
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 318 – ______ = 4
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-Solve-Take-From-Problems-with-Change-Unknown-Practice-3.2-question-1

Question 2.
7 – ? = 3
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 32
7 – ______ = 3
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-Solve-Take-From-Problems-with-Change-Unknown-Practice-3.2-question-2

Question 3.
5 – _______ = 4
Answer:
5 – 1 = 4

Question 4.
6 – ______ = 6
Answer:
6 – 0 = 6

Question 5.
MP Repeated Reasoning
Match each model with its correct equation.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 33
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-Solve-Take-From-Problems-with-Change-Unknown-Practice-3.2-question-5

Question 6.
Modeling Real Life
You have 10 toys. Your friend borrows some of them. You have 7 left. How many toys did your friend borrow?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 34

________ toys
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-Solve-Take-From-Problems-with-Change-Unknown-Practice-3.2-question-6
Total number of toys = 10
Number of toys left after borrowing = 7
Number of toys my friend borrowed = 10 – 7 = 3 toys.

Review & Refresh

Question 7.
You have 6 Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 35
You buy 3 more Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 35.
How many Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 35 do you have now?
_______ + _______ = ______ Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 35
Answer:
Number of banana’s i have = 6
Number of banana’s bought = 3
Total number of banana’s= 6 + 3 = 9 banana’s.

Question 8.
You have Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 36.
You buy 1 more Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 36.
How many Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 36 do you have now?
_______ + _______ = ______ Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 36
Answer:
number of bat i have =1
number of more bat i bought = 1
total number of bat i have = 1 + 1 = 2 bats.

Lesson 3.3 Solve Take From Problems with Start Unknown

Explore and Grow

Use counters to model each problem.

3 + 4 = _______

_____ – 3 = 4

Answer:

3 + 4 = 7
7 – 3 = 4

Show and Grow

Question 1.
? – 2 = 4
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 37
Think 2 + 4 = _______.
So, ______ – 2 = 4
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.3-Solve-Take-From-Problems-with-Start- Unknown- Show-and-Grow-question-1
Think 2 + 4 = 6.
So, 6 – 2 = 4

Question 2.
? – 1 = 1
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 38
Think 1 + 1 = _______.
So, ______ – 1 = 1
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.3-Solve-Take-From-Problems-with-Start- Unknown- Show-and-Grow-question-2
Think 1 + 1 = 2.
So, 2 – 1 = 1.

Apply and Grow: Practice

Question 3.
? – 2 = 5
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 39
Think 2 + 5 = _______.
So, ______ – 2 = 5
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.3-Solve-Take-From-Problems-with-Start- Unknown-Apply-and-Grow-Practice-question-3
Think 2 + 5 = 7.
So, 7 – 2 = 5

Question 4.
? – 3 = 6
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 40
Think 3 + 6 = _______.
So, ______ – 3 = 6.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.3-Solve-Take-From-Problems-with-Start- Unknown-Apply-and-Grow-Practice-question-4
Think 3 + 6 = 9.
So, 9 – 3 = 6.

Question 5.
? – 4 = 1
Think 4 + 1 = _______.
So, ______ – 4 = 1.
Answer:
Think 4 + 1 = 5.
So, 5 – 4 = 1.

Question 6.
? – 6 = 4
Think 6 + 4 = _______.
So, ______ – 6 = 4.
Answer:
Think 6 + 4 = 10.
So, 10 – 6 = 4.

Question 7.
MP Structure
Circle the equations that match the model.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 41
8 – 2 = 6                  8 + 2 = 10
6 – 2 = 4                  2 + 6 = 8
Answer:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 41
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.3-Solve-Take-From-Problems-with-Start- Unknown-Apply-and-Grow-Practice-question-7

Think and Grow: Modeling Real Life

A group of students are at a playground. 2 of them leave. There are 8 left. How many students were there to start?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 42
Model:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 43
Subtraction equation:

__________ students
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.3-Solve-Take-From-Problems-with-Start- Unknown-Think-and-Grow-Modeling-Real-Life
Number of students leave = 2
Number of students remaining = 8
Total number of students = 8 + 2 = 10 students.
subtraction equation: 10 – 2 = 8.

Show and Grow

Question 8.
You have some strawberries. You eat 9 of them. You have 0 left. How many strawberries did you have to start?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 44
Model:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 45
Subtraction equation:

_________ strawberries
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations-Lesson-3.3-Solve-Take-From-Problems-with-Start- Unknown-Show-and-Grow-question-8
Number of strawberries i ate = 9
Number of strawberries left after eating = 0
Total number of strawberries i had at the start = 9 + 0 = 9 strawberries.
Subtraction equation: 9 – 9 = 0.

Solve Take From Problems with Start Unknown Practice 3.3

Question 1.
? – 7 = 2
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 46
Think 7 + 2 = _______.
So, ______ – 7 = 2.
Answer:

Think 7 + 2 = 9.
So, 9 – 7 = 2.

Question 2.
? – 2 = 8
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 47
Think 2 + 8 = _______.
So, ______ – 2 = 8.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations- Solve-Take-From-Problems-with-Start-Unknown-Practice-3.3-question-2
Think 2 + 8 = 10.
So, 10 – 2 = 8.

Question 3.
? – 3 = 0
Think 3 + 0 = _______.
So, ______ – 3 = 0.
Answer:
Think 3 + 0 = 3.
So, 3 – 3 = 0.

Question 4.
? – 2 = 0
Think 2 + 6 = _______.
So, ______ – 2 = 6.
Answer:
Think 2 + 6 = 8.
So, 8 – 2 = 6.

Question 5.
MP Structure
Circle the equations that match the model.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 48
8 – 1 = 7 9 + 1 = 10
9 – 8 = 1 8 + 1 = 9
Answer:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 48
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction -Situations- Solve-Take-From-Problems-with-Start-Unknown-Practice-3.3-MP-Structure-question-5

Question 6.
Modeling Real Life
There are some people on a trolley. 4 of them exit. There are 4 people left. How many people were on the trolley to start?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 49

__________ people
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Solve-Take-From-Problems-with-Start-Unknown- Practice-3.3-Modeling-Real-Life-question-6
Number of people who exit the trolley = 4
Number of people left on trolley = 4
Total number of people on the trolley = 4 + 4 = 8

Review & Refresh

Question 7.
There are 5 blue balloons and 3 red balloons. How many more blue balloons are there?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 50
________ – _________ = ________ more blue balloons
Answer:
Given
Blue Balloons = 5
Red Balloons = 3
5 – 3 = 2 more blue balloons are there.

Lesson 3.4 Compare Problems: Bigger Unknown

Explore and Grow

Use counters to model the story.

Newton has 5 balls. Descartes has 2 more balls than Newton. How many balls does Descartes have?

Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 51
Answer:
Number of balls with newton = 5
Number of balls with Descartes = 5 + 2 = 7.

Show and Grow

Question 1.
Your friend has 7 trading cards. You have 3 more than your friend. How many trading cards do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 52
7 + _______ = _______ trading cards
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.4-Compare-Problems-Bigger-Unknown-Show-and-Grow-question-1
7 + 3 = 10 trading cards.

Apply and Grow: Practice

Question 2.
Your friend has I soccer ball. You have 2 more than your friend. How many soccer balls do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 53
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 54
1 + _______ = _______ soccer balls
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.4-Compare-Problems-Bigger-Unknown-Apply-and-Grow-Practice-question-2
1 + 2 = 3 soccer balls

Question 3.
Your friend swims 4 more laps than you. You swim 3 laps. How many laps does your friend swim?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 55
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 56
_______ + _______ = _______ laps
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.4-Compare-Problems-Bigger-Unknown-Apply-and-Grow-Practice-question-3

Question 4.
MP Precision
Your friend catches 5 more fish than you. You catch 2 fish. How many fish does your friend catch? Circle the bar model that matches the problem.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 57
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.4-Compare-Problems-Bigger-Unknown-Apply-and-Grow-Practice-MP-Precision-question-4

Think and Grow: Modeling Real Life

Your friend has 1 yellow flower and 2 red flowers. You have 3 more flowers than your friend. How many flowers do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 58
Model:
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 59
Addition equation:

________ flowers
Answer:
Model:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.4-Compare-Problems-Bigger-Unknown-Think-and-Grow- Modeling-Real-Life
Addition equation: 3 + 3 = 6 flowers.

Show and Grow

Question 5.
Your friend has 6 gray shirts and 2 blue shirts. You have 2 more shirts than your friend. How many shirts do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 60
Model:
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 61
Addition equation:

_________ shirts
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.4-Compare-Problems-Bigger-Unknown-Show-and-Grow-question-5
Addition equation: 8 + 2 = 10 shirts

Compare Problems: Bigger Unknown Practice 3.4

Question 1.
You have 3 key chains. Your friend has 5 more than you. How many key chains does your friend have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 62
3 + ______ = ______ key chains
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Compare-Problems-Bigger-Unknown-Practice-3.4-question-1
3 + 5 = 8 key chains

Question 2.
You have 8 more bracelets than your friend. Your friend has 2 bracelets. How many bracelets do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 63
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 64
_______ + ______ = ______ bracelets
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Compare-Problems-Bigger-Unknown-Practice-3.4-question-2
2 + 8 = 10 bracelets.

Question 3.
MP Precision
You have 1 seashell. Your friend has 8 more than you. How many seashells does your friend have? Circle the bar model that matches the problem.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 65
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Compare-Problems-Bigger-Unknown-Practice-3.4-MP-Precision-question-3

Question 4.
Modeling Real Life
Your friend has 2 comic books and 2 mystery books. You have 3 more books than your friend. How many books do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 66

_________ books
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Compare-Problems-Bigger-Unknown-Practice-3.4-Modeling-Real-Life-question-4
4 + 3 = 7 books.

Review & Refresh

Question 5.
There are 2 blue crayons and 6 red crayons. How many fewer blue crayons are there?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 67
_______ – ______ = _______ fewer blue crayons
Answer:
Number of red crayons = 6
Number of blue crayons = 2
Number of fewer blue crayons = 6 – 2 = 4 blue crayons.

Lesson 3.5 Compare Problems: Smaller Unknown

Explore and Grow

Use counters to model the story.

Newton has 5 treats. Descartes has 2 fewer treats than Newton. How many treats does Descartes have?

Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 68
Answer:
Number of treats with newton = 5
Number of treats with Descartes = 5 – 2 = 3.

Show and Grow

Question 1.
Your friend has 8 stones. You have I fewer than your friend. How many stones do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 69
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 70
_______ – _______ = ________ stones
_______ + _______ = ________ stones
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.5-Compare-Problems-Smaller-Unknown-Show-and-Grow-question-1
8 – 1 = 7 stones
7 + 1 = 8 stones

Apply and Grow Practice

Question 2.
You blow 5 bubbles. Your friend blows 2 fewer than you. How many bubbles does your friend blow?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 71
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 72
_______ – _______ = ________ bubbles
_______ + _______ = ________ bubbles
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.5-Compare-Problems-Smaller-Unknown-Apply-and-Grow- Practice-question-2
5 – 2 = 3 bubbles
3 + 2 = 5 bubbles.

Question 3.
You have 3 fewer oranges than your friend. Your friend has 9 oranges. How many oranges do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 73
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 74
_______ ○ ______ = _______ oranges
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.5-Compare-Problems-Smaller-Unknown-Apply-and-Grow- Practice-question-3
9 – 3 = 6 oranges

Question 4.
DIG DEEPER!
Complete the bar model. Do both equations match the bar model?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 75
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.5-Compare-Problems-Smaller-Unknown-Apply-and-Grow- Practice-DIG-DEEPER-question-4
Both the equations matches.

Think and Grow: Modeling Real Life

Your friend has 2 blue markers and 7 yellow I markers. You have 5 fewer markers than your friend. How many markers do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 76
Model:
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 77
Equation:

_________ markers
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.5-Compare-Problems-Smaller-Unknown-Think-and-Grow- Modeling-Real-Life

Equation: 9 – 5 = 4
I have 4 markers

Show and Grow

Question 5.
Your friend has 6 tennis balls and I baseball. You have 2 fewer balls than your friend. How many balls do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 78
Model:
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 79
Equation:

_________ balls
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.5-Compare-Problems-Smaller-Unknown-Show-and-Grow-question-5

Equation: 7 – 2 = 5

I have 5 balls

Compare Problems: Smaller Unknown Practice 3.5

Question 1.
Your friend has 9 awards. You have 5 fewer than your friend. How many awards do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 80
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 81
______ – _____ = _____ awards
______ + _____ = _____ awards
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.5-Compare-Problems-Smaller-Unknown-Think-and-Grow- Modeling-Real-Life
9 – 5 = 4 awards
4 + 5 = 9 awards

Question 2.
Your friend finds 2 fewer bugs than you. You find 4 bugs. How many bugs does your friend find?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 82
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 83
_______ ○ ______ = _______ bugs
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Compare-Problems-Smaller-Unknown-Practice-3.5-question-2
4 – 2 = 2 bugs
My friend found 2 bugs.

Question 3.
MP Reasoning
Complete the bar model. Circle the equation that matches the bar model.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 84
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Compare-Problems-Smaller-Unknown-Practice-3.5-MP-Reasoning-question-3

Question 4.
Modeling Real Life
Your friend has 8 black cats and 2 orange cats. You have 7 fewer cats than your friend. How many cats do you have?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 85

_______ cats
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Compare-Problems-Smaller-Unknown-Practice-3.5-Modeling Real Life-question-4
Number of cats my friend have= 8 + 2 = 10 cats
Number of cats i have = 10 – 7 = 3

Review & Refresh

Question 5.
Write the numbers of shirts and shorts. Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no.

Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Compare-Problems-Smaller-Unknown-Practice-3.5-Review-&-Refresh-question-5

Lesson 3.6 True or False Equations

Explore and Grow

Color the flowers that have a sum or difference of 6.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 87
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Explor-and-Grow

Show and Grow

Is the equation true or false?

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 88
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Show-and-Grow-question-1

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 89
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Show-and-Grow-question-2

Apply and Grow: Practice

Is the equation true or false?

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 90
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Apply-and-Grow-Practice-question-3

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 91
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Apply-and-Grow-Practice-question-4

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 92
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Apply-and-Grow-Practice-question-5
10 – 4 = 6
6 – 0 = 6

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 93
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Apply-and-Grow-Practice-question-6
5- 2 = 3
7 – 4 = 3

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 94
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Apply-and-Grow-Practice-question-7

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 95
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Apply-and-Grow-Practice-question-8

Question 9.
MP Number Sense
Circle all of the equations that are true.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 96
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Apply-and-Grow-Practice- MP-Number-Sense-question-9

Think and Grow: Modeling Real Life

You have 7 marbles. You lose 2 of them. Your friend has 4 marbles and finds 3 more. Do you and your friend have the same number of marbles?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 97
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 98
Answer:
Number of marbles I have = 7 – 2 = 5
Number of marbles my friend have = 4 + 3 = 7
Equation:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Think-and-Grow-Modeling-Real-Life

Show and Grow

Question 10.
You have I balloon. You blow up 3 more. Your friend has 5 balloons. I of your friend’s balloons pops. Do you and your friend have the same number of balloons?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 99
Equation:
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 98
Answer:
Number of balloons i have= 1 + 3 = 4
Number of balloons my friend have = 5 – 1 = 4
Equation:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction- Situations-Lesson-3.6-True-or-False-Equations-Show-and-Grow-question-10

True or False Equations Practice 3.6

Is the equation true or false?

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 100
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Equations-Practice-3.6-question-1

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 101
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Equations-Practice-3.6-question-2

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 102
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Equations-Practice-3.6-question-3

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 103
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Equations-Practice-3.6-question-4

Question 5.
MP Number Sense
Circle all of the equations that are false.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 104
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Equations-Practice-3.6-MP-Number-Sense-question-5

Question 6.
Modeling Real Life
You have 5 crayons. You find 3 more. Your friend has 7 crayons and finds I more. Do you and your friend have the same number of crayons?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 105
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 106
Answer:
Number of crayons with me = 5 + 3 = 8
Number of crayons with my friend = 7 + 1 = 8
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Equations-Practice-3.6-Modeling-Real-Life-question-6

Review & Refresh

Question 7.
Circle the triangles.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 107
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Equations-Practice-3.6-Review-&-Refresh-question-7

Lesson 3.7 Find Numbers That Make 10

Explore and Grow

Place some red counters on the ten frame. Add yellow counters to fill the frame. Write an equation to match.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 108
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.7-Find-Numbers-That-Make-10-Explore-and- Grow

Show and Grow

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 109
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.7-Find-Numbers-That-Make-10-Show-and-Grow-question-1

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 110
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.7-Find-Numbers-That-Make-10-Show-and-Grow-question-2

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 111
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.7-Find-Numbers-That-Make-10-Show-and-Grow-question-3

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 112
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.7-Find-Numbers-That-Make-10-Show-and-Grow-question-4

Apply and Grow Practice

Question 5.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 113
4 + _____ = 10
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.7-Find-Numbers-That-Make-10-Apply-and-Grow-Practice-question-5

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 114
7 + _____ = 10
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.7-Find-Numbers-That-Make-10-Apply-and-Grow-Practice-question-6

Question 7.
1 + _____ = 10
Answer:
1 + 9 = 10

Question 8.
8 + _____ = 10
Answer:
8 + 2 = 10

Question 9.
_____ + 2 = 10
Answer:
8 + 2 = 10

Question 10.
_____ + 5 = 10
Answer:
5 + 5 = 10

Question 11.
_____ + 3 = 10
Answer:
7 + 3 = 10

Question 12.
_____ + 0 = 10
Answer:
10 + 0 = 10

Question 13.
DIG DEEPER!
Match the numbers that have a sum of 10.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 115
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.7-Find-Numbers-That-Make-10-DIG-DEEPER-Practice-question-13

Think and Grow: Modeling Real Life

There are 7 jump ropes. Your teacher buys some more. Now there are 10. How many jump ropes did your teacher buy?
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 116
Model:

Addition equation:

________ jump ropes
Answer:
Number of jump ropes= 7
Teacher bought some ropes and total = 10
Number of ropes teacher brought = 10 – 7 = 3

Addition equation: 7 + 3 = 10

3 jump ropes.

Show and Grow

Question 14.
There are 2 penguins. Some more join them. Now there are 10. How many more penguins joined them?
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 117
Model:

Addition equation:

________ penguins
Answer:
Number of penguins at present= 2
Total number of penguins = 10
Number of penguins joined = 10 – 2 = 8

Addition equation: 2 + 8 = 10

8 penguins joined.

Find Numbers That Make 10 Practice 3.7

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 118
8 + _______ = 10
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Find-Numbers-That-Make-10-Practice-3.7-question-1

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 119
1 + _______ = 10
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Find-Numbers-That-Make-10-Practice-3.7-question-2

Question 3.
10 + _______ = 10
Answer:
10 + 0 = 10

Question 4.
5 + _______ = 10
Answer:
5 + 5 = 10

Question 5.
_______ + 4 = 10
Answer:
6 + 4 = 10

Question 6.
_______ + 0 = 10
Answer:
10 + 0 = 10

Question 7.
DIG DEEPER!
Match the numbers that have a sum of 10.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 120
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Find-Numbers-That-Make-10-Practice-3.7-DIG-DEEPER-question-7

Question 8.
Modeling Real Life
You have 3 baseball cards. Your friend gives you some more. Now you have 10. How many baseball cards did your friend give you?
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 121

________ baseball cards
Answer:
Number of baseball cards with me= 3
Total number of baseball cards i have = 10
Number of baseball cards given by my friend = 10 -3 = 7 cards.

Review & Refresh

Find the sum. Then change the order of the addends. Write the new addition problem.

Question 9.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 122
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Find-Numbers-That-Make-10-Practice-3.7-Review-&-Refresh-question-9

Question 10.
Big Ideas Math Answers 1st Grade 1 Chapter 3 More Addition and Subtraction Situations 123
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Find-Numbers-That-Make-10-Practice-3.7-Review-&-Refresh-question-10

Lesson 3.8 Fact Families

Explore and Grow

Use linking cubes to model the equations.

Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 124

Show and Grow

Question 1.
Complete the fact family.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 125
1 + 8 = ______                          9 – ______ = ______

______ + ______ = ______            9 – ______ = ______
Answer:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 125

1 + 8 = 9                         9 – 8 = 1

8 + 1 = 9                         9 – 1 = 8

Apply and Grow: Practice

Complete the fact family.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 126
4 + 2 = ______                              6 – ______ = ______

______ + ______ = ______                6 – ______ = ______
Answer:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 126

4 + 2 = 6                             6 – 2 = 4

2 + 4 = 6                             6 – 4 = 2

Question 3.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 127
3 + 6 = ______                                 ______ – 6 = 3

______ + ______ = ______                   ______ – ______ = ______
Answer:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 127

3 + 6 = 9                                9 – 6 = 3

6 + 3 = 9                                9 – 3 = 6

Question 4.
7 + 1 = ______                                   ______ – 7 = _______

______ + ______ = ______                    ______ – 1 = ______
Answer:

7 + 1 = 8                                   8 – 7 = 1

1 + 7 = 8                                   8 – 1 = 7

Question 5.
DIG DEEPER!
Cross out the equation that does not belong in the fact family.
5 + 3 = 8             5 – 3 = 2
3 + 5 = 8             8 – 5 = 3
Answer:
5 + 3 = 8
3 + 5 = 8             8 – 5 = 3

Think and Grow: Modeling Real Life

You have 3 puppets. Your friend has 7 puppets. How many fewer puppets do you have?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 128
Model:

Equation:

________ fewer puppets
Answer:
Number of puppets i have = 3
Number of puppets my friend have = 7
Difference between puppets between me and my friend:

Equation: 7 – 3 = 4

I have 4 fewer puppets.

Show and Grow

Question 6.
There are 2 spoons and 8 forks. How many more forks are there?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 129
Model:

Equation:

________ more forks
Answer:
Number of spoons = 2
Number of forks = 8
difference between spoons and folks:
Equation: 8 – 2 = 6

There are 6 more forks.

Fact Families Practice 3.8

Question 1.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 130
1 + 5 = ______                          6 – ______ = _______

______ + ______ = ______            6 – ______ = ______
Answer:
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 130

1 + 5 = 6                         6 – 5 = 1

5 + 1 = 6                         6 – 1 = 5

Question 2.
3 + 7 = ______                          ______ – 7 = 3

______ + ______ = ______            ______ – _______ = ______
Answer:

3 + 7 = 10                          10 – 7 = 3

7 + 3 = 10                          10 – 3 = 7

Question 3.
Complete the fact family.
3 + 0 = ______                           ______ – 3 = _______

______ + ______ = ______             ______ – 0 = ______
Answer:

3 + 0 = 3                           3 – 3 = 0

0 + 3 = 3                           3 – 0 = 3

Question 4.
DIG DEEPER!
Cross out the equation that does not belong in the fact family.
6 + 4 = 10               10 – 6 = 4
4 + 6 = 10                6 – 4 = 2
Answer:
6 + 4 = 10               10 – 6 = 4
4 + 6 = 10

Question 5.
Modeling Real Life
There are 7 fish and 2 frogs. How many fewer frogs are there?

________ fewer frogs
Answer:
Number of fish = 7
Number of frogs = 2
Difference between frogs and fish = 7 – 2 = 5.
There are 5 fewer frogs.

Review & Refresh

Circle the objects that holds more.

Question 6.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 131
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.8-Fact-Families-Review-&-Refresh-question-6

Question 7.
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 132
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Lesson-3.8-Fact-Families-Review-&-Refresh-question-7

More Addition and Subtraction Situations Performance Task

Question 1.
You and your friend bake banana bread and raisin bread. You have 8 loaves of bread. 3 of them are raisin bread. Your friend has 10 loaves of bread. 6 of them are banana bread. How many more loaves of banana bread does your friend have than you?

_______ loaf
Answer:
Total number of bread loaves i have = 8.
Number of raisin breads i have = 3.
Number of banana breads i have = 8 – 3 = 5.
Total number of bread loaves my friend have = 10.
Number of banana bread my friend have = 6.
Difference between the banana breads is : 6 – 5 = 1.
My friend have 1 more loaves of banana bread than me.

Question 2.
You give away 3 loaves of banana bread and 3 loaves of raisin bread. Your friend gives away 1 more loaf of bread than you. How many loaves of bread does your friend give away?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 133

_______ loaves
Answer:
Number of banana bread loaves i gave away = 3
Number of raisin bread loaves i gave away = 3
Total number of bread loaves i gave away = 3 + 3 = 6.
The number of bread loaf my friend gave away is 1 more then = 6 + 1 = 7.
My friend gave away 7 loaves of bread.

Question 3.
You and your friend make boxes of muffins. Does each box have the same number of muffins?
Big Ideas Math Answers Grade 1 Chapter 3 More Addition and Subtraction Situations 134
Yes              No
Answer:
Number of muffins in the boxes i make = 3 + 7 = 10.
Number of muffins in the boxes my friend make = 5 + 4 = 9.
Difference between the muffins = 10 – 9 = 1
NO  the boxes does not have same number of muffins.

More Addition and Subtraction Situations Chapter Practice

Solve Add To Problems with Start Unknown Homework & Practice 3.1

Question 1.
? + 4 = 6
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 135
_______ + 4 = 6
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Chapter-Practice-Solve-Add-To-Problems-with-Start-Unknown-Homework-&-Practice-3.1-Question-1
2 + 4 = 6

Question 2.
? + 2 = 8
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 136
_______ + 2 = 8
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Chapter-Practice-Solve-Add-To-Problems-with-Start-Unknown-Homework-&-Practice-3.1-Question-2
6 + 2 = 8

Solve Take From Problems with Change Unknown Homework & Practice 3.2

Question 3.
5 – ? = 4
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 137
5 – ______ = 4
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations- Solve-Take-From-Problems-with-Change-Unknown-Homework-&-Practice-3.2-question-3
5 – 1 = 4

Question 4.
7 – ? = 7
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 138
7 – ______ = 7
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations- Solve-Take-From-Problems-with-Change-Unknown-Homework-&-Practice-3.2-question-4
7 – 0 = 7

Solve Take From Problems with Start Unknown Homework & Practice 3.3

Question 5.
? – 6 = 3
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 139
Think 6 + 3 = ______.
So, ______ – 6 = 3.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations- Solve-Take-From-Problems-with-Change-Unknown-Homework-&-Practice-3.3-question-5

Think 6 + 3 = 9.
So, 9 – 6 = 3.

Question 6.
? – 3 = 1
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 140
Think 3 + 1 = ______.
So, ______ – 3 = 1.
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations- Solve-Take-From-Problems-with-Change-Unknown-Homework-&-Practice-3.3-question-6

Think 3 + 1 = 4.
So, 4 – 3 = 1.

Question 7.
MP Structure
Circle the equation that matches the model.
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 141
6 – 6 = 0 4 – 2 = 2 6 – 2 = 4
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations- Solve-Take-From-Problems-with-Change-Unknown-Homework-&-Practice-3.3-MP-Structure-question-7

Compare Problems: Bigger Unknown Homework & Practice 3.4

Question 8.
Your friend has 3 stickers. You have 4 more than your friend. How many stickers do you have?
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 142
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Compare-Problems-Bigger-Unknown-Homework-&-Practice-3.4-Question-8

Compare Problems: Smaller Unknown Homework & Practice 3.5

Question 9.
Your friend has 5 stuffed animals. You have 2 fewer than your friend. How many stuffed animals do you have?
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 143
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Compare-Problems-Smaller-Unknown-Homework-&-Practice-3.5-Question-9

Question 10.
Modeling Real Life Your friend has 4 dogs and 2 cots. You have I fewer pet than your friend. How many pets do you have?
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 144
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Compare-Problems-Smaller-Unknown-Homework-&-Practice-3.5-Modeling-Real-Life-Question-10

True or False Homework & Practice 3.6

Is the equation true or false?

Question 11.
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 145
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Homework-&-Practice-3.6-question-11

Question 12.
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 146
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Homework-&-Practice-3.6-question-12

Question 13.
MP Number Sense
Circle all of the equations that are true.
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 147
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-True-or-False-Homework-&-Practice-3.6-MP-Number-Sense-question-13

Find the number That Make 10 Homework & Practice 3.7

Question 14.
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 148
3 + _______ = 10
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Find-the-number-That-Make-10-Homework-&-Practice-3.7-question-14

Question 15.
Big Ideas Math Solutions Grade 1 Chapter 3 More Addition and Subtraction Situations 149
6 + _______ = 10
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-Find-the-number-That-Make-10-Homework-&-Practice-3.7-question-15

Fact Families Homework & Practice 3.8

Question 16.
Complete the fact family.
8 + 1 = _______                         ________ – 8 = 1
________ + _______ = _______      _______ – _______ = ______
Answer:
8 + 1 = 9                         9 – 8 = 1
1 + 8 = 9                         9 – 1 = 8

Question 17.
Modeling Real Life
There are 2 slides and 6 swings on a playground. How many more swings are there?

__________ more swings
Answer:
Number of slides = 2
Number of swings = 6
Number of more swings = 6 – 2 = 4 more swings.

More Addition and Subtraction Situations Cumulative Practice 1 – 3

Question 1.
Shade the circle next to the answer.
4 + 4 = ______
○ 4
○ 6
○ 9
○ 8
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-More-Addition-and-Subtraction-Situations-Cumulative-Practice-1-3-Question-1

Question 2.
Shade the circle next to the addition equation that you can use to solve 8 – 3.
○ 8 + 3 = 11
○ 3 + 5 = 8
○ 1 + 8 = 9
○ 5 + 2 = 7
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-More-Addition-and-Subtraction-Situations-Cumulative-Practice-1-3-Question-2

Question 3.
Circle the equation that matches the bar model.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 150
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-More-Addition-and-Subtraction-Situations-Cumulative-Practice-1-3-question-3

Question 4.
You take 10 pictures. Your friend takes 3 pictures. Shade the circle next to the equation that shows how many more pictures you take.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 151
○ 3 + 3 = 6
○ 10 – 3 = 7
○ 10 + 3 = 13
○ 3 – 1 = 2
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-More-Addition-and-Subtraction-Situations-Cumulative-Practice-1-3-Question-4

Question 5.
Shade the circle next to the number that completes the addition equation.
________ + 7 = 9
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 152
○ 1
○ 2
○ 3
○ 4
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-More-Addition-and-Subtraction-Situations-Cumulative-Practice-1-3-Question-5

Question 6.
There are 6 Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 153.
3 more Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 153 join them.
How many Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 153 are there now?
_________ + __________ = _________ Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 153
Answer:
Number of Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 153 = 6
Total number of Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 153 = 6 + 3 = 9 Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 153

Question 7.
Is each equation true or false?
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 154
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-More-Addition-and-Subtraction-Situations-Cumulative-Practice-1-3-question-7

Question 8.
Use the picture to write a subtraction equation.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 155
________ – 0 = _______
Answer:
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 155
5 – 0 = 5

Question 9.
You have 8 beads. 5 are orange. The rest are blue. Shade the circles next to the equations that describe the beads.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 156
○ 3 + 5 = 8
○ 8 – 2 = 6
○ 4 + 4 = 8
○ 8 – 5 = 3
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-More-Addition-and-Subtraction-Situations-Cumulative-Practice-1-3-Question-9

Question 10.
Use the numbers shown to write two equations.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 157
Answer:
Given   3     8     5
3 + 5 = 8      5 + 3 = 8

Question 11.
Shade the circle next to the equation that does not belong in the fact family.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 158
○ 3 + 1 = 4
○ 4 – 3 = 1
○ 1 + 3 = 4
○ 3 – 1= 2
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-More-Addition-and-Subtraction-Situations-Cumulative-Practice-1-3-Question-11

Question 12.
There are 3 rabbits. 3 more join them. Shade the circle next to the equation that shows how many rabbits there are in all.
Big Ideas Math Answer Key Grade 1 Chapter 3 More Addition and Subtraction Situations 159
○ 3 + 2 = 5
○ 3 + 4 = 7
○ 4 + 1 = 5
○ 3 + 3 = 6
Answer:
Big-Ideas-Math-Book-1st-Grade-Answer-Key-Chapter-3-More-Addition-and-Subtraction-Situations-More-Addition-and-Subtraction-Situations-Cumulative-Practice-1-3-Question-12

Conclusion:

Hope you are all satisfied with the answers provided in the Bigideas Math Grade 1 Chapter 3 More Addition and Subtraction Situations. Get different methods to solve the problems in our Big Ideas Math Solution Key 1st Grade Chapter 3 More Addition & Subtraction Situations. Thus, the students who are interested to learn the basics quickly can follow the methods given here.

Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120

Big Ideas Math Answers Grade 1 Chapter 6

Download Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 pdf for free from here. The solutions for all the questions are given in a simple manner by subject experts. The Chapter Count and Write Numbers to 120 includes Count to 120 by Ones, Count to 120 by Tens, Count and Write Numbers to 120, etc. This chapter discusses the methods to count the numbers to 120. Go through the below sections to find topic-wise links.

Big Ideas Math Book 1st Grade Answer Key Chapter 6 Count and Write Numbers to 120

To make your preparation perfect you have to follow the best guide. Big Ideas Math Book 1st Grade Answer Key Chapter 6 Count and Write Numbers to 120 is the best study material which consists of clear-cut explanations for all the problems. So, access the links and start your preparation for the exams.

Vocabulary

Lesson: 1 Count to 120 by Ones

Lesson: 2 Count to 120 by Tens

Lesson: 3 Compose Numbers 11 to 19

Lesson: 4 Tens

Lesson: 5 Tens and Ones

Lesson: 6 Make Quick Sketches

Lesson: 7 Understand Place Value

Lesson: 8 Write Numbers in Different Ways

Lesson: 9 Count and Write Numbers to 120

Chapter: 6 – Count and Write Numbers to 120

Count and Write Numbers to 120 Vocabulary

Organize It

Review Words:
column
decade numbers
hundred chart
row

Use the review words to complete the graphic organizer.

Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 1
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-1

Explanation:
The given graphic organizer title is called as hundred chart,
and numbers 41 to 50 are called as row, numbers 4,14,24
to 84, 94 are called as column and numbers 10,20,30 to 90,100
are called decade numbers.

Define It

Use your vocabulary cards to identify the words.

Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 2

Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-2

Explanation:
Given in the picture to identify words 2 – Two and 3 – Three.

Lesson 6.1 Count to 120 by Ones

Explore and Grow

Point to each number as you count to 120. Color the first two rows and the last two rows. How are the rows the same? How are they different?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 3
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-3

Explanation:
As we move from 1,2,3 count to 120  have pointed with an arrow for each number. Colored with green first two rows from 1 to 10, 11 to 20 and the last two rows from 101 to 110, 111 to 120.
The rows are same – as we move from number 1 to 120 from first number to second number the number is increased by 1 as 1, 1 + 1 = 2, 2 + 1 = 3 and so on 119 + 1 = 120 means the next number is addition of 1 plus the previous number.
The rows are different – as we see the numbers are not the same they differ its 1,2 ,3 to 119, 120 from 1 to 120.

Show and Grow

Count by ones to write the missing numbers.

Question 1.
82, _____83_____, ____84____, _____85_____, ____86______, ____87______
Answer:
The missing numbers after 82 are  83, 84, 85, 86 ,87

Explanation:
Given the starting number as 82 we count by ones and
write the next missing numbers as  82 + 1 = 83,
83 + 1 = 84, 84 + 1 = 85 , 85 +1 = 86, 86 + 1 = 87,
now the missing numbers after 82  are 83, 84, 85, 86 ,87.

Question 2.
103, ____104______, ____105______, ___106_______, ____107______, ____108______
Answer:
The missing numbers after 103 are 104,105,106,107,108.

Explanation:
Given the starting number as 103 we count by ones and
write the next missing numbers as 103 + 1  = 104,
104 + 1 =105, 105 + 1 = 106, 106 + 1 = 107,107 + 1 = 108,
now the missing numbers after 103 are 104,105,106,107,108.

Apply and Grow: Practice

Count by ones to write the missing numbers.

Question 3.
56, ____57______, ____58____, ____59______, ____60______, ____61______
Answer:
The missing numbers after 56 are 57,58,59,60,61

Explanation:
Given the starting number as 56 we count by ones and
write the next missing numbers as 56 + 1 = 57, 57 + 1 = 58,
58 + 1 = 59, 59 + 1 = 60, 60 + 1 = 61,
now the missing numbers after 56 are 57,58,59,60,61.

Question 4.
98, _____99_____, ____100______, ____101______, ____102______, ____103______
Answer:
The missing numbers after 98 are 99,100,101,102,103

Explanation:
Given the starting number as 98 we count by ones and
write the next missing numbers as 98 + 1 = 99,99 + 1 = 100,
100 + 1 = 101, 101 + 1 = 102, 102 + 1 = 103,
So the missing numbers after 98 are 98,99,100,101,102,103.

Question 5.
115, ____116______, ____117______, ____118______, ____119______, ___120_______
Answer:
The missing numbers after 115 are 116,117,118,119,120

Explanation:
Given the starting number as 115 we count by ones and
write the next missing numbers as 115 + 1 = 116,
116 + 1 = 117, 117 + 1 = 118, 118 + 1 = 119, 119 + 1 = 120,
So the missing numbers after 115 are 116,117,118,119,120.

Question 6.
___40_______, ____41______, 42, ____43______, ____44______, ____45______
Answer:
The missing numbers before and after 42 are 40,41,43,44,45

Explanation:
Given number is 42 and 2 numbers before 42 are
count 1 before 42 is 42- 1= 41 and count 1 before 41 is
41 – 1 = 40 and count 1 after 42 is 42 + 1 = 43, 43 + 1 = 44,
44 + 1 = 45, So the missing numbers before and after 42 are 40,41,43,44,45.

Write the missing numbers in the chart.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 4
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-4
The missing numbers in the chart are 33,35,42,44.

Explanation:
The missing number after 32 is 32 + 1 = 33,
the number after 34 is 34 + 1 = 35,
the number before 43 is 43 – 1 = 42 and
number after 43 is 43 + 1 = 44, So the missing
numbers in the chart are 33,35,42,44.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 5
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-5
The missing numbers in the chart are 81,83,84,92,94.

Explanation:
The missing number before 82 is 82 – 1 = 81,
the number after 82 is 82 + 1 = 83,now number
after 83 is 83 + 1 = 84.
The number after 91 is 91 + 1 = 92,
the number after 93 is 93 + 1 = 94, therefore the
missing numbers in the chart are 81,83,84,92,94.

Question 9.
MP Structure
Structure Write a number between 95 and 105. Then count by ones to write the next 7 numbers.
____96___, _____97____, _____98_____, ____99____, ___ 100___, ____101___, ___102__, ___103___
Answer:
Let the number be 96 which is between 95 and 105 and
the next 7 numbers after 96 are 97,98,99,100,101,102,103.

Explanation:
Given to write a number between 95 and 105,
Let us take the number as 96, now we will write the
next 7 numbers are 96 + 1 = 97,97+ 1 = 98, 98 + 1 =99,
99 + 1 = 100,100 + 1 = 101, 101 + 1 = 102, 102 + 1 = 103,
therefore the next 7 numbers after 96 are 97,98,99,100,101,102,103.

Think and Grow: Modeling Real Life

You hove 108 bouncy balls. You want 112. How many more bouncy bails do you need?
Draw more bails to show 112:
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 6

______4______ more bouncy balls
Answer:
4 more bouncy  balls are needed,
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-5

Explanation:
Given I have 108 bouncy balls and I want 112,
So I need 108 + 1 = 109, 109 + 1 = 110, 110 + 1 = 111,
111+ 1 = 112 or 112 – 108 = 4, so we need 4 more bouncy  balls.

Show and Grow

Question 10.
You have 66 rocks. You want 75. How many more rocks do you need?
Draw more rocks to show 75:
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 7

______9_______ more rocks
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-6
I need 9 more rocks.

Explanation:
Given I have 66 rocks and I want 75,
So more rocks I needed are 75 – 66 = 9.

Count to 120 by Ones Practice 6.1

Count by ones to write the missing numbers.

Question 1.
57, _____58_____, _____59_____, ____60______, ____61______, ____62______
Answer:
The missing numbers after 57 are 58,59,60,61,62.

Explanation:
Given the starting number as 57 we count by ones and
write the next missing numbers as 57 + 1 = 58,
58 + 1 = 59, 59 + 1 = 60, 60 + 1 = 61, 61 + 1 = 62,
So the missing numbers after 57 are 58,59,60,61,62.

Question 2.
109, ____110___, ____111___, ____112___, ____113___, __114_____
Answer:
The missing numbers after 109 are 110,111,112,113,114

Explanation:
Given the starting number as 109 we count by ones and
write the next missing numbers as 109 + 1 = 110,
110 + 1 =111,111 + 1 = 112, 112 + 1 = 113, 113 + 1 = 114,
So the missing numbers after 109 are 110,111,112,113,114.

Question 3.
40, ____41___, ____42___, ___43___, ___44___, ___45___
Answer:
The missing numbers after 40 are 41,42,43,44,45

Explanation:
Given the starting number as 40 we count by ones and
write the next missing numbers as 40 + 1 = 41,
41 + 1 = 42,42 + 1 = 43, 43 + 1 = 44, 44 + 1 = 45,
So the missing numbers after 40 are 40,41,42,43,44,45.

Question 4.
_____97_____, ____98______, ____99______, 100, ____101______, ____102______
Answer:
The missing numbers before and after 100 are 97,98,99,101,102

Explanation:
Given number is 100 and 3 numbers before 100 are
count 1 before 100 is 100- 1= 99, count 1
before 99 is 99 – 1 = 98 and count 1 before 98 is
98 – 1 = 97, numbers after 100 are 100 + 1 = 101 and
101 + 1 = 102,therefore the missing numbers
before and after 100 are 97,98,99,101,102.

Write the missing numbers in the chart.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 8
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-7

The missing numbers in the chart are 21,23,30,32.

Explanation:
The missing number after 20 is 20 + 1 = 21,
the number after 22 is 22 + 1 = 23,
the number before 31 is 31 – 1 = 30 and
number before 33 is 33 – 1 = 32, So the missing
numbers in the chart are 21,23,30,32.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 9
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-8

The missing numbers in the chart are 102,104,105,113,115.

Explanation:
The missing number before 103 is 103 – 1 = 102,
the number after 103 is 103 + 1 = 104,
the number after 104 is 104 + 1 = 105,
the number after 112 is 112 + 1 = 113 and
the number after 114 is 114 + 1 = 115
So the missing numbers in the chart are 102,104,105,113,115.

Question 7.
MP Structure
Write a number between 85 and 95. Then count by ones to write the next 7 numbers.

__87__, _88___, _89___, __90___, _91__, _92_, _93_, _94__
Answer:
Let the number be 87 which is between 85 and 95 and
the next 7 numbers after 87 are 88,89,90,91,92,93,94.

Explanation:
Given to write a number between 85 and 95,
Let us take the number as 87, now we will write the
next 7 numbers are 87 + 1 = 88,88+ 1 = 89, 89 + 1 =90,
90 + 1 = 91,91 + 1 = 92, 92 + 1 = 93, 93 + 1 = 94,
therefore the next 7 numbers after 87 are 88,89,90,91,92,93,94.

Question 8.
Modeling Real Life
There are 110 tokens. You want 119. How many more tokens do you need?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 10

____9_____ more tokens
Answer:
I need 9 more tokens.

Explanation :
Given there are 110 tokens. and I want 119,
so I needed 119 – 110 = 9 more tokens.

Review & Refresh

Question 9.
? – 6 = 4
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 11
Think 6 + 4 = ___10_____ .
So, ______10__ – 6 = 4.
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-9
6 + 4 = 10
so, 10 – 6 = 4
Explanation:
Given ___ – 6 = 4, Let us take the missing number as X,
therefore X – 6 = 4, X = 4 + 6 =10,the equation is 10 – 6 = 4.

Lesson 6.2 Count to 120 by Tens

Explore and Grow

Count to 10. Circle the number. Count 10 more. Circle the number. Continue until you reach 120.

Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 12
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-10

Explanation:

Start counting from 1 to 10 circled, Next from 11 to 20 counted
circled, counted till 120 and circled.

Show and Grow

Count by tens to write the missing numbers.

Question 1.
70, __80___, __90__, __100__, _110__, __120__
Answer:
The missing numbers after 70 are 80,90,100,110,120

Explanation:
Started from 70 counting by 10’s so 70 + 10 = 80,
80 + 10 = 90, 90 + 10 = 100, 100 + 10 = 110, 110 + 10 = 120,
So the missing numbers after 70 are 80,90,100,110,120

Question 2.
31, __41___, __51__, _61__, __71__, _81__
Answer:
The missing numbers after 31 are 41,51,61,71,81

Explanation:
Started from 31 counting by 10’s so 31 + 10 = 41,
41 + 10 = 51, 51 + 10 =61, 61 + 10 = 71 ,71 +10 = 81,
So the missing numbers after 31 are 41,51,61,71,81.

Apply and Grow: Practice

Count by tens to write the missing numbers

Question 3.
62, __72__, __82__, _92__, _102__, _112___
Answer:
The missing numbers after 62 are 72,82,92,102,112

Explanation:
Started from 62 counting by 10’s are 62 + 10 = 72,
72 + 10 = 82, 82 + 10 =92, 92 + 10 = 102 ,102 +10 = 112,
So the missing numbers after 62 are 72,82,92,102,112

Question 4.
43, __53__, _63__, __73__, __83__, _93__
Answer:
The missing numbers after 43 are 53,63,73,83,93

Explanation:
Started from 43 counting by 10’s are 43 + 10 = 53,
53 + 10 = 63, 63 + 10 =73, 73 + 10 = 83 ,83 +10 = 93,
So the missing numbers after 43 are 53,63,73,83,93.

Question 5.
___10__, __20___, 30, __40__, __50___, __60__
Answer:
The missing numbers before and after 30 are 10,20,40,50,60

Explanation:
Started from 30 counting by 10’s numbers before and after 30 are
before 30 – 10 = 20, 20 – 10 = 10 and after 30 are 30 + 10 = 40,
40 + 10 = 50, 50 + 10 = 60, therefore numbers before and after 30
are 10,20,40,50,60.

Question 6.
Write the missing numbers from the chart. Then count on by tens to write the next three numbers.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 13
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-11
The missing numbers in the chart are 71,81 and
count on 10’s the next three numbers after 81 are 91,101,111.

Explanation:
Given numbers in the chart, the number missing before 72
is 72 – 1 = 71 and number missing before 82 in the chart is
82 – 1 = 81,now counting by 10’s next three numbers after 81 are
81 + 10 = 91,91 + 10 = 101,101 + 10 = 111.
Therefore the missing numbers in the chart are 71,81 and
count on 10’s the next three numbers after 81 are 91,101,111.

Question 7.
YOU BE THE TEACHER
Your friend counts by tens starting with 27. Is your friend correct? Show how you know.
27, 37, 47, 67, 77, 87
Answer:
No, Friend is incorrect. As 57 is missing.

Explanation:
Given friend counts by tens starting with 27 and
gets 27, 37, 47, 67, 77, 87 but after 47 we should
get 47 + 10 =  57 as 57 is missing friend is incorrect.

Think and Grow: Modeling Real Life

You have 50 points. On your next turn, you knock over 6 cans. How many points do you have now?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 14
Write the numbers:

____110______ points
Answer:
I have 110 points now.

Explanation:
Given I have 50 points on my next turn I knock
over 6 cans and 10 points for each knocked over can
means I got 6 X 10 = 60 points, So in total I have
50 + 60 = 110 points now.

Show and Grow

Question 8.
You have 21 points. On your next turn, 3 beanbags land in the circle. How many points do you have now?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 15
Write the numbers:

_____51_____ points
Answer:
I have 51 points now.

Explanation:
Given I have 21 points on my next turn 3 beanbags
land in the circle 10 points for each beanbag
means I got 3 X 10 = 30 points, So in total I have
21 + 30 = 51 points now.

Count to 120 by Tens Practice 6.2

Count by tens to write the missing numbers.

Question 1.
69, __79_, _89__, _99_, _109_, _119__
Answer:
The missing numbers after 69 are 79,89,99,109,119

Explanation:
Started from 69 counting by 10’s numbers after 69 are
69 + 10 = 79, 79 + 10 = 89, 89 + 10 = 99, 99 + 10 = 109,
109 + 10 = 119, therefore the missing numbers after 69
are 79,89,99,109,119.

Question 2.
41, __51__, _61___, _71__, _81__, _91__
Answer:
The missing numbers after 41 are 51,61,71,81,91

Explanation:
Started from 41 counting by 10’s numbers after 41 are
41 + 10 = 51,51 + 10 = 61, 61 + 10 = 71, 71 + 10 = 81,
81 + 10 =91, So the missing numbers after 41 are 51,61,71,81,91.

Question 3.
16, __26__, _36__, _46__, _56__, __66__
Answer:
The missing numbers after 16 are 26,36,46,56,66

Explanation:
Started from 16 counting by 10’s numbers after 16 are
16 + 10 = 26, 26 + 10 = 36,36 + 10 = 46, 46 + 10 = 56,
56 + 10 = 66, So the missing numbers after 16 are 26,36,46,56,66.

Question 4.
_64__, __74__, __84__, 94, __104_, __114__
Answer:
The missing numbers before 94 are 84,74 ,64 and
after 94 are 104, 114.

Explanation:
Started from 94 counting by 10’s numbers before and after 94 are
before 94 – 10 = 84, 84 – 10 = 74 , 74 – 10 = 64 and after are
94 + 10 = 104,104 + 10 = 114, So the missing numbers before
94 are 84,74 ,64 and after 94 are 104, 114.

Question 5.
Write the missing numbers from the chart. Then count on by tens to write the next three numbers.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 16
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-12

The missing numbers in the chart are 6,16 and
count on 10’s the next three numbers after 16 are 26,36,46.

Explanation:
Given numbers in the chart, the number missing after 5
is 5 + 1  = 6 and number missing after 15 in the chart is
15 + 1 = 16,now counting by 10’s next three numbers after 16 are
16 + 10 = 26,26 + 10 = 36,36 + 10 = 46.
Therefore the missing numbers in the chart are 6,16 and
count on 10’s the next three numbers after 16 are 26,36,46.

Question 6.
DIG DEEPER!
You count to 50. You only count 5 numbers. Did you count by ones or by tens? Show how you know.
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-13
I count by tens.

Explanation:
Given I count to 50 and count only 5 numbers , If I count by ones
I will get 50 numbers like 1,1+1=2, 2+ 1=3,…till I reach 49+ 1 = 50
as shown in the above picture, If I count by tens I get 5 numbers,
Like 10,10 + 10 = 20,20 + 30 = 40, 40 + 10 = 50, as I count 5 numbers
So I count by tens.

Question 7.
Modeling Real Life
You have 30 points. On your next turn, 4 balls stick to the wall. How many points do you have now?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 17
____70_____ points
Answer:
I have 70 points now.

Explanation:
Given I have 30 points. On my next turn,
4 balls stick to the wall and each ball we have 10 points,
So 4 X 10 = 40 points means in total I have 30 + 40 = 70 points now.

Review & Refresh

Question 8.
3 + 1 = ___4_____
Answer:
3 + 1 = 4

Explanation:
We are going to add 1 to 3 we get 4.

Question 9.
5 – 1 = ____4_____
Answer:
5 – 1 = 4

Explanation:
We will subtract 1 from 5 we get 4.

Lesson 6.3 Compose Numbers 11 to 19

Explore and Grow

Color to show 13 and 17. What is the same about the numbers? What is different?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 18
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-14
Both are same- 13, 17 are prime numbers.
Both 13 and 17 are different as
both are not same numbers, they both are not equal.

Explanation:
Taken numbers from 1 to 20 in the chart,
colored numbers 13 and 17 with green.
Both are the same- 13 and 17 are prime numbers-
these are the numbers, which are only divisible by 1 or
the number itself as 13 and 17 are divisible by 1 and only
by them they both are prime.
Both are different-As 13 and 17 are equal they differ.

Show and Grow

Question 1.
Circle 10 feathers. Complete the sentence.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 19
____One____ ten and ___six____ ones is __sixteen____ .
Answer:

One ten and six ones is sixteen.
1 ten and 6 ones is 16.

Explanation:
Circled 10 feathers, after counting we have total
16 feathers we write the sentence as
one ten and six ones is sixteen means 10 + 6 = 16.
1 ten and 6 ones is 16.

Apply and Grow: Practice

Circle 10 feathers. Complete the sentence.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 20
___One_____ ten and __five___ ones is ____fifteen_____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-16
One ten and five ones is fifteen.
1 ten and 5 ones is 15.

Explanation:
Circled 10 fishes, after counting we have total
15 fishes we write the sentence as
one ten and five ones is fifteen means 10 + 5 = 15.
1 ten and 5 ones is 15.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 21
____One____ ten and ___three__ ones is ____thirteen_____ .

Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-17
One ten and three ones is thirteen.
1 ten and 3 ones is 13.

Explanation:
Circled 10 sun flowers, after counting we have total
13 sunflowers we write the sentence as
one ten and three ones is thirteen means 10 + 3 = 13.
1 ten and 3 ones is 13.

Question 4.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 22
___One_ ten and __seven__ ones is ___seventeen___ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-18

One ten and seven ones is seventeen.
1 ten and 7 ones is 17.

Explanation:
Circled 10 bees, after counting we have total
17 bees we write the sentence as
one ten and seven ones is seventeen means 10 + 7 = 17.
1 ten and 7 ones is 17.

Question 5.
MP Number Sense
Color to show the number. Complete the sentence.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 23
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 24
___one___ ten and __six___ ones is __sixteen____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-19
One ten and six ones is sixteen.
1 ten and 6 ones is 16.

Explanation:
Colored the shown number 16 with blue,
we write the sentence as
one ten and six ones is sixteen means 10 + 6 = 16.
1 ten and 6 ones is 16.

Think and Grow: Modeling Real Life

You have 15 footballs. A bag can hold 10. You fill a bag. How many footballs are not in the bag?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 25
Draw a picture:

Write the missing numbers: __one_ ten and __five_ ones is __fifteen__ footballs
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-19
One ten and five ones is fifteen footballs.
1 ten and 5 ones is 15.
5 footballs are not there in the bag,
Zero ten and five ones or 0 ten and 5 ones is five or 5 footballs are missing.

Explanation:
Given I have 15 footballs means one ten and five ones is
fifteen or 1 ten and 5 ones is 15 and a bag can hold 10 footballs,
In the picture there are total 15 footballs in that I
have taken 10 in the bag so footballs which are not there
in the bag are 15 – 10 = 5, So missing number of footballs
in the bag are zero ten and five ones is five or 0 ten and 5 ones is 5.

Show and Grow

Question 6.
Your teacher has 18 calculators. A case can hold 10. Your teacher fills a case. How many calculators are not in the case?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 26
Draw a picture:

Write the missing numbers: __one__ ten and __eights ones _ is eighteen_ calculators
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-20
One ten and eight ones is eighteen calculators.
1 ten and 8 ones is 18.
8 calculators are not there in the case,
Zero ten and eight ones is eight or
0 ten and 8 ones is 8 calculators are missing,

Explanation:
Given teacher ha 18 calculators means one ten and eight ones is
eighteen or 1 ten and 8 ones is 18 and a case can hold 10,
In the picture there are total 18 calculators  in that I
have taken 10 in the case so number of calculators which
are not there in the case are 18 – 10 = 8, So missing calculators
in the case are zero ten and eight ones is eight or 0 ten and 8 ones is 8.

Compose Numbers 11 to 19 Practice 6.3

Circle 10 objects. Complete the sentence.

Question 1.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 27
__one___ ten and ___one___ ones is eleven_ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-21

One ten and one ones is eleven.
1 ten and 1 ones is 11.

Explanation:
Circled 10 balls, after counting we have total
11 balls we write the sentence as
one ten and one ones is eleven means 10 + 1 = 11.
1 ten and 1 ones is 11.

Question 2.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 28
__one____ ten and __nine____ ones is _nineteen_____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-22

One ten and nine ones is nineteen.
1 ten and 9 ones is 19.

Explanation:
Circled 10 fishes, after counting we have total
19 fishes we write the sentence as
one ten and nine ones is nineteen means 10 + 9 = 19.
1 ten and 9 ones is 19.

Question 3.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 29
___one___ ten and __three____ ones __thirteen____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-23

One ten and three ones is thirteen.
1 ten and 3 ones is 13.

Explanation:
Circled 10 honeybees, after counting we have total
13 we write the sentence as
one ten and three ones is thirteen means 10 + 3 = 13.
1 ten and 3 ones is 13.

Question 4.
MP Number Sense
Color to show the number. Complete the sentence.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 30
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 31
__one____ ten and _eight___ ones is eighteen__ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-24

One ten and eight ones is eighteen.
1 ten and 8 ones is 18.

Explanation:
Colored the shown number 18 with yellow,
we write the sentence as
one ten and eight ones is eighteen means 10 + 8 = 18.
1 ten and 8 ones is 18.

Question 5.
MP Number Sense Match.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 32
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-25

Explanation:
Matched 1 ten and 3 ones with 13, 1 ten and 8 ones with 12
and 12 ones with 12 in the picture above.

Question 6.
Modeling Real Life
You have 16 books. A backpack can hold 10. You fill a backpack. How many books are not in the backpack?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 33

_____6_____ books
Answer:
There are 6 books which are not in the backpack.

Explanation:
Given I have 16 books means and a backpack
can hold 10 books, out of 16 books I
have taken 10 in the backpack so books which are not there
in the backpack are 16 – 10 = 6 books.

Review & Refresh

Question 7.
10 + 10 = ___20_____
Answer:
10 + 10 = 20

Explanation:
We are going to add 10 with 10 we get 20.

Question 8.
10 + 10 = ___20______
Answer:
10 + 10 = 20

Explanation:
We are going to add 10 with 10 we get 20.

Lesson 6.4 Tens

Explore and Grow

Circle groups of 10. Write the number of groups.

Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 34
Answer:

How many counters are there in all?
____20______ counters
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-26
There are 20 counters in all.

Explanation:
In the given picture circled groups of 10, There are 2 groups,
and counted there are total 20 counters in all.

Show and Grow

Circle groups of 10. Complete the sentence.

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 35

____6__ ten and _0__ ones is 60______ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-27
6 groups, 6 ten and 0 ones is 60.

Explanation:
In the given picture circled groups of 10, There are 6 groups,
and counted there are total 60 in all. 6 ten and 0 ones is 60.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 36

__5____ ten and ___0___ ones is 50_ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-28
5 groups, 5 ten and 0 ones is 50.

Explanation:
In the given picture circled groups of 10, There are 5 groups,
and counted there are total 50 in all. 5 ten and 0 ones is 50.

Apply and Grow: Practice

Circle groups of 10. Complete the sentence.

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 37

___7___ ten and ___0___ ones is 70______ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-29
7 groups, 7 ten and 0 ones is 70.

Explanation:
In the given picture circled groups of 10, There are 7 groups,
and counted there are total 70 in all. 7 ten and 0 ones is 70.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 38

___9___ ten and ___0___ ones is _90__ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-30
9 groups, 9 ten and 0 ones is 90.

Explanation:
In the given picture circled groups of 10, There are 9 groups,
and counted there are total 90 in all. 9 ten and 0 ones is 90.

Question 5.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 39

___1___ ten and ___0___ ones is __10____ .
Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-31
1 group, 1 ten and 0 ones is 10.

Explanation:
In the given picture circled groups of 10, There are 1 group,
and counted there are total 10 in all. 1 ten and 0 ones is 10.

Question 6.
MP Number Sense
You have 4 groups of 10 linking cubes. How many linking cubes do you have?

_____40_______ linking cubes
Answer:
I have 40 linking cubes.

Explanation:
Given I have 4 groups of 10 linking cubes, So I have
4 X 10 = 40 linking cubes.

Think and Grow: Modeling Real Life

You read 10 books every month. You 40 books. How many months does it take?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 40
Draw a picture:
Write the missing numbers: ___4___ tens and ___0___ ones __4____ months

Answer:
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120-32

4 tens and 0 ones is 40 books.
It will take 4 months.

Explanation:
Given I read 10 books every month and read 40 books,
4 tens and 0 ones is 40 books.
So we will divide 40 by 10 we get 4, therefore it will take 4 months,
Show and Grow

Question 7.
There are 10 dog bones in each box. You need 20 bones. How many boxes do you need?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 41
Draw a picture:

Write the missing numbers: ___2___ tens and ___0___ ones __2____ boxes
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-33

2 tens and 0 ones is 20 boxes.
We need 2 boxes.

Explanation:
Given there are 10 dog bones in each box and need 20 bones.
2 tens and 0 ones is 20 bones.
So we will divide 20 by 10 we get 2, therefore we need 2 boxes.

Tens Practice 6.4

Circle groups of 10. Complete the sentence.

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 42

___8___ tens and __0____ ones is 80__ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-34

8 groups, 1 ten and 0 ones is 10.

Explanation:
In the given picture circled groups of 10, There are 8 groups,
and counted there are total 80 in all. 8 ten and 0 ones is 80.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 43

___6___ tens and __0____ ones is __60____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-35

6 groups, 6 ten and 0 ones is 60.

Explanation:
In the given picture circled groups of 10, There are 6 groups,
and counted there are total 60 in all. 6 ten and 0 ones is 60.

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 44

___3___ tens and __0____ ones _is 30_____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-36

3 groups, 3 ten and 0 ones is 30.

Explanation:
In the given picture circled groups of 10, There are 3 groups,
and counted there are total 30 in all. 3 ten and 0 ones is 30.

Circle groups of 10. Complete the sentence.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 45

___2___ tens and __0____ ones is __20____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-37

2 groups, 2 ten and 0 ones is 20.

Explanation:
In the given picture circled groups of 10, There are 2 groups,
and counted there are total 20 in all. 2 ten and 0 ones is 20.

Question 5.
MP Number Sense
You have 7 groups of 10 linking cubes. How many linking cubes do you have?

_____70______ linking cubes
Answer:

I have 70 linking cubes.

Explanation:
Given I have 7 groups of 10 linking cubes, So I have
7 X 10 = 70 linking cubes.

Question 6.
Modeling Real Life
You swim 10 laps at every practice. You want to swim 50 laps. How many practices will it take?

_____5______ Practices
Answer:

5 tens and 0 ones is 50 laps
It will take 5 practices.

Explanation:
Given I swim 10 laps at every practice, I want to swim 50 laps
5 tens and 0 ones is 50 laps.
So I will divide 50 by 10 I get 5, therefore I will take 5 practices.

Review & Refresh

Question 7.
4 + 3 + 4 = ____11________
Answer:
4 + 3 + 4 = 11

Explanation:
First we add 4 and 3 then add 4 we get 11.

Question 8.
1 + 5 + 9 = _____15______
Answer:
1 + 5 + 9 = 15

Explanation:
First we add 1 and 5 then add 9 we get 15.

Question 9.
2 + 2 + 1 = _____5______
Answer:
2 + 2 + 1 = 5

Explanation:
First we add 2 and 2 then add 1 we get 5.

Question 10.
7 + 3 + 6 = _____16______
Answer:
7 + 3 + 6 = 16

Explanation:
First we add 7 and 3 then add 6 we get 16.

Lesson 6.5 Tens and Ones

Explore and Grow

Model 2 tens and 3 ones. Write the number.

Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 46
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-38
2 tens and 3 ones is 23.

Explanation:
Model shown in the picture 2 tens and 3 ones,
So placed 2 ten blocks in tens place and 3 blocks in ones place,
making 2 tens and 3 ones ,therefore the number is 23.

Show and Grow

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 47

__2___ tens and ___1___ ones is ___21___ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-39

2 tens and 1 ones is 21

Explanation:
Given in the picture there are 2 tens block in tens place
and 1 block in ones make 2 tens and 1 ones is 21.

Apply and Grow: Practice

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 48

___3___ tens and __5____ ones is ___35___ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-40
3 tens and 5 ones is 35

Explanation:
Given in the picture there are 3 ten blocks in tens place
and 5 blocks in ones make 3 tens and 5 ones is 35.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 49

___6___ tens and ___6___ ones is ___6___ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-41
6 tens and 6 ones is 66

Explanation:
Given in the picture there are 6 ten blocks in tens place
and 6 blocks in ones make 6 tens and 6 ones is 66.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 50

__8____ tens and __9____ ones is __89____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-42
8 tens and 9 ones is 89

Explanation:
Given in the picture there are 8 ten blocks in tens place
and 9 blocks in ones make 8 tens and 9 ones is 89.

Question 5.
YOU BE THE TEACHER
You have 92 linking cubes. Your friend says that there are 2 tens and 9 ones.
Is your friend correct? Show how you know.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-43
No, friend is incorrect.

Explanation:
Given I have 92 linking cubes, friend says that there are 2 tens and 9 ones,
but 92 means as shown in picture we get 9 tens and 2 ones,
therefore friend is in correct.

Think and Grow: Modeling Real Life

Your teacher has 2 packages of dice and 3 extra dice. Each package has 10 dice. How many dice are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 51
Draw a picture:

Write the missing numbers: ___2___ tens and __3____ ones _is__23___ dice
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-44
2 tens and 3 ones is 23 dice are there in all.

Explanation:
Given my teacher has 2 packages of dice and 3 extra dice.
Each package has 10 dice, 2 packages means 2 X 10 = 20 dice
and 3 extra dice making in all 2 tens and 3 ones is 23 dice as shown in picture.

Show and Grow

Question 6.
You have 3 boxes of colored pencils and 4 extra colored pencils. Each box has 10 pencils. How many colored pencils are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 52
Draw a picture:

Write the missing numbers: ___3___ tens and __4____ ones _is_34__ colored pencils
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-45
3 tens and 4 ones is 34 colored pencils.

Explanation:
Given I have 3 boxes of colored pencils and 4 extra colored pencils.
Each box has 10 pencils, So 3 boxes have 3 X 10 = 30 pencils and 4 extra
makes 30 + 4 = 34, 3 tens and 4 ones is 34 colored pencils are there in all.

Tens and Ones Practice 6.5

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 53

__7____ tens and ___9___ ones is __79____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-46
7 tens and 9 ones is 79.

Explanation:
Given in the picture there are 7 ten blocks in tens place
and 9 blocks in ones make 7 tens and 9 ones is 79.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 54

__8____ tens and ___1___ ones _is 81_____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-47
8 tens and 1 ones is 81.

Explanation:
Given in the picture there are 8 ten blocks in tens place
and 1 blocks in ones make 8 tens and 1 ones is 81.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 55

___2___ tens and __4___ ones is 24_ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-48
2 tens and 4 ones is 24.

Explanation:
Given in the picture there are 2 ten blocks in tens place
and 4 blocks in ones make 2 tens and 4 ones is 24.

Question 4.
YOU BE THE TEACHER
You have 17 linking cubes. Your friend says that there is 1 ten and 7 ones. Is your friend correct? Show how you know.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-49

Yes, friend is correct.

Explanation:
Given I have 17 linking cubes, friend says that there is 1 ten and 7 ones,
means as shown in picture we get 1 ten and 7 ones, similar to what friend says,
therefore friend is  correct.

Question 5.
Modeling Real Life
You have 5 bags of apples and 1 extra apple. Each bag has 10 apples. How many apples are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 56

_____51______ apples
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-50
5 tens and 1 ones is 51 apples.

Explanation:
Given I have 5 bags of apples and 1 extra apple.
Each bag has 10 apples, So 5 bags have 5 X 10 = 50 apples and 1 extra apple
makes 50 + 1 = 51, 5 tens and 1 ones is 51 apples are there in all.

Review & Refresh

Question 6.
_____4_____ + 6 = 10
Answer:
4 + 6 = 10

Explanation:
Given ____ + 6 = 10, let us take the missing
number as X , X + 6 = 10, X = 10 – 6 = 4 making
the equation as 4 + 6 = 10.

Question 7.
_____6______ + 2 = 8
Answer:
6 + 2 = 8

Explanation:
Given ____ + 2 = 8, let us take the missing
number as X , X + 2 = 8, X = 8 – 2 = 6 making
the equation as 6 + 2 = 8.

Lesson 6.6 Make Quick Sketches

Explore and Grow

Model the number 2.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 57
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-51

2 ones is 2.

Explanation:
Model shown in the picture are 2 ones,
So 2 blocks in ones place, as shown in picture
making 2 ones ,therefore the number is 2.

Show and Grow

Make a quick sketch. Complete the sentence.

Question 1.
72
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 58
__72____ is ____7____ tens and __2____ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-52

72 is 7 tens and 2 ones.

Explanation:
The picture has 7 ten blocks in tens place
and 2 blocks in ones makes 72 as 7 tens and 2 ones.

Question 2.
36
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 59
__36____ is ___3__ tens and __6_ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-53

36 is 3 tens and 6 ones.

Explanation:
The picture has 3 ten blocks in tens place
and 6 blocks in ones makes 36 as 3 tens and 6 ones.

Apply and Grow: Practice

Make a quick sketch. Complete the sentence.

Question 3.
45
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 60
___45___ is ___4_____ tens and ___5___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-54

45 is 4 tens and 5 ones.

Explanation:
The picture has 4 ten blocks in tens place
and 5 blocks in ones makes 45 as 4 tens and 5 ones.

Question 4.
87
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 60
___87___ is ___8___ tens and ___7___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-55

87 is 8 tens and 7 ones.

Explanation:
The picture has 8 ten blocks in tens place
and 7 blocks in ones makes 87 as 8 tens and 7 ones.

Question 5.
64
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 60
__64____ is ___6____ tens and ___4___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-56

64 is 6 tens and 4 ones.

Explanation:
The picture has 6 ten blocks in tens place
and 4 blocks in ones makes 64 as 6 tens and 4 ones.

Question 6.
DIG DEEPER!
Which sketch shows 54?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 61
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-57
First sketch shows 54.

Explanation:
Given 2 sketches in that first counts 54 and second counts
45, So first sketch shows 54, So selected first one for 54.

Think and Grow: Modeling Real Life

You need 58 plates for a party. You have 51. How many more plates do you need?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 62
Complete the model:
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 63

____7_______ more plates
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-58
I need 7 more plates.

Explanation:
Given I need 58 plates for a party and I have 51,
therefore more plates needed are 58 – 51 = 7 plates
as shown in the picture above.

Show and Grow

Question 7.
You need 80 tickets for a prize. You have 73. How many more tickets do you need?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 64
Complete the model:
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 65

_____7______ more tickets
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-59
I need 7 more tickets.

Explanation:
Given I need 80 tickets for a prize and I have 73,
therefore more tickets needed are 80 – 73 = 7 plates
as shown in the picture above.

Make Quick Sketches Practice 6.6

Make a quick sketch. Complete the sentence.

Question 1.
27
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 66
__27____ is ____2____ tens and ___7___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-60

27 is 2 tens and 7 ones.

Explanation:
The picture has 2 ten blocks in tens place
and 7 blocks in ones makes 27 as 2 tens and 7 ones.

Question 2.
61
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 66
__61__ is ___6__ tens and ___1___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-61

61 is 6 tens and 1 one.

Explanation:
The picture has 6 ten blocks in tens place
and 1 blocks in one makes 61 as 6 tens and 1 one.

Question 3.
92
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 66
___92___ is ___9_____ tens and ___2___ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-62

92 is 9 tens and 2 ones.

Explanation:
The picture has 9 ten blocks in tens place
and 2 blocks in ones makes 92 as 9 tens and 2 ones.

Question 4.
DIG DEEPER
Which sketch shows 87?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 67
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-63

First sketch shows 87.

Explanation:
Given 2 sketches in that first counts 87 and second counts
78, So first sketch shows 87, So selected first one for 87.

Question 5.
Modeling Real Life
You need 55 beads to make a necklace. You have 48. How many more beads do you need?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 68

____7______ more beads
Answer:
I need 7 more beads.

Explanation:
Given I need 55 beads to make necklace and I have 48,
therefore more beads needed are 55 – 48 = 7 beads.

Review & Refresh

Question 6.
Color the shapes that have only 4 sides.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 69
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-64

Explanation:
Given some shapes and to color that have 4 sides so selected shapes
that has 4 sides and  have colored as shown above in the picture.

Lesson 6.7 Understand Place Value

Explore and Grow

Newton has 2 rods. Make a quick sketch.
Write the number.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 70
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-65
Newton has 2 rods or 20 cubes

Explanation:
Given Newton has 2 rods as shown in the picture means 20 cubes.

Descartes has 2 cubes. Make a quick sketch.
Write the number.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 71
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-66
Descartes has 2 cubes

Explanation:
Given Descartes has 2 cubes as shown in picture 2 cubes.

How are the models alike? How are they different?
Answer:
Both use cubes so alike,
Both differ in number so different.

Explanation:
Models are alike means both use cubes.
They are different Newton has 20 cubes and Descartes have 2 cubes,
20 and 2 are not the same and are not equal so they are different.

Show and Grow

Question 1.
Make a quick sketch. Complete the sentences.
64
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 72
_____6_____ tens is ____60______ .
____4______ ones is ____4______ .
_____6_____ tens and ____4______ ones is ___64_______.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-67
6 tens is 60,
4 ones is 4, 6 tens and 4 ones is 64.

Explanation:
As shown in the picture above first we take 6 tens cubes as 60
and 4 ones cubes as 4, making 6 tens and 4 ones is 64.

Apply and Grow: Practice

Make a quick sketch. Complete the sentences.

Question 2.
72
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 72
_____7_____ tens is ____70______ .
_____2_____ ones is ____2______ .
______7____ tens and ____2______ ones is ___72__.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-68
7 tens is 70,
2 ones is 2, 7 tens and 2 ones is 72.

Explanation:
As shown in the picture above first we take 7 tens cubes as 70
and 2 ones cubes as 2, making 7 tens and 2 ones is 72.

Question 3.
98
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 72
__9____ tens is ___90____ .
___8____ ones is ___8_____ .
___9____ tens and ____8____ ones is ___98___.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-69

9 tens is 90,
8 ones is 8, 9 tens and 8 ones is 98.

Explanation:
As shown in the picture above first we take 9 tens cubes as 90
and 8 ones cubes as 8, making 9 tens and 8 ones is 98.

Question 4.
57
____5______ tens is ___50_______ .
_____7_____ ones is ____7______ .
______5____ tens and _____7_____ ones is ____57______.
Answer:
5 tens is 50,
7 ones is 7, 5 tens and 7 ones is 57.

Explanation:
First we take 5 tens cubes as 50
and 7 ones cubes as 7, making 5 tens and 7 ones is 57.

Think and Grow: Modeling Real Life

You have 94 charms to make bracelets. There are lo charms on each bracelet. How many bracelets can you make?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 73
Model:

Write the missing numbers: ___9___ tens and ___4___ ones ___9___ bracelets
Answer:
9 tens and 4 ones is 94,
I make 9 bracelets.

Explanation:
Given I have 94 charms to make bracelets. There are lo charms
on each bracelet. So I divide 94 with 10 I get 9 as whole
and 4 as remainder so I cannot make with 4 charms,
I consider only whole so I take 9,therefore I can make 9 bracelets
or as I have 9 tens and 4 ones I take only tens so 9 bracelets I can make.

Show and Grow

Question 5.
You have 67 seeds. You plant 10 seeds in a row. How many rows can you plant?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 74
Model:

Write the missing numbers: ___6___ tens and __7____ ones __6____ rows
Answer:
6 tens and 7 ones is 67,
I can plant 6 rows.

Explanation:
Given I have 67 seeds. I can plant 10 seeds in a row
So I divide 67 with 10 I get 6 as whole
and 7 as remainder so I cannot make with 7 seeds a row ,
I consider only whole so I take 6,therefore I can make 6 rows only,
or as I have 6 tens and 7 ones I take only tens so 6 rows I can make.

Understand Place Value Practice 6.7

Make a quick sketch. Complete the sentences.

Question 1.
81
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 75
_____8_____ tens is ___80_______ .
_____1_____ ones is ____1______ .
_____8_____ tens and ____1______ ones is ____81______.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-70

8 tens is 80,
1 one is 1, 8 tens and 1 one is 81.

Explanation:
As shown in the picture above first we take 8 tens cubes as 80
and 1 one cubes as 1, making 8 tens and 1 one is 81.

Question 2.
53
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 75
_____5_____ tens is _____50_____ .
______3____ ones is ____3______ .
______5____ tens and _____3____ ones is ___53_____.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-71

5 tens is 50,
3 ones is 3, 5 tens and 3 ones is 53.

Explanation:
As shown in the picture above first we take 5 tens cubes as 50
and 3 ones cubes as 3, making 5 tens and 3 ones is 53.

Question 3.
49
___4_______ tens is ____40______ .
_____9_____ ones is ____9______ .
______4____ tens and ____9__ ones is __49__.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-72

4 tens is 40,
9 ones is 9, 4 tens and 9 ones is 49.

Explanation:
As shown in the picture above first we take 4 tens cubes as 40
and 9 ones cubes as 9, making 4 tens and 9 ones is 49.

Question 4.
76
____7______ tens is ____70______ .
______6____ ones is ____6______ .
____7______ tens and ____6______ ones is ____76______.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-73

7 tens is 70,
6 ones is 6, 7 tens and 6 ones is 76.

Explanation:
As shown in the picture above first we take 7 tens cubes as 70
and 6 ones cubes as 6, making 7 tens and 6 ones is 76.

Question 5.
Modeling Real Life
You have 77 crayons. A box can hold 10 crayons. How many boxes can you fill?
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 76

_____7____ boxes
Answer:
7 tens and 7 ones is 77,
I can fill 7 boxes.

Explanation:
Given I have 77 crayons, A box can hold 10 crayons,
So I divide 77 with 10 I get 7 as whole
and 7 as remainder I consider only whole so I take 7,
therefore I need 7 boxes to fill or as I have 7 tens and 7 ones to
fill I take only tens so 7 boxes I can fill.

Review & Refresh

Is the equation true or false?

Question 6.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 77
Answer:
4 + 9 ≠  2 + 3 + 5
Not equal, So false

Equation:
First we see 4 and 9 as 4 + 9 is 13 and
if we add 2,3 and 5 we get 10 ,
therefore 4 + 9 ≠ 2 + 3 +5, So false.

Question 7.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 78
Answer:
5 + 3 = 4 + 4
Is equal, So true

Explanation:
First we see 5 + 3 as we add 5 and 3 we get 8 and
4 + 4 is 8 both sides are equal, So equations are true.

Lesson 6.8 Write Numbers in Different Ways

Explore and Grow

Model 27 two ways.
Big Ideas Math Answers Grade 1 Chapter 6 Count and Write Numbers to 120 79
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 80
_____2___ tens and __7_____ ones is ___27_____ .
_____1___ ten and ___17____ ones is ___27_____ .
Answer:
one way is
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-74
2 tens and 7 ones is 27.
Other way is
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-75
1 ten and 17 ones is 27.

Explanation:
One way is we add 2 tens and 7 ones equal to 27,
20 + 7 = 27 and other way is 1 ten and 17 ones is 27,
10 + 17 = 27.

Show and Grow

Question 1.
Model 25 two ways.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 81
____2____ tens and ____5___ ones is 25.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-76
2 tens and 5 ones is 25.

Explanation:
we add 20 and 5 we get 25, So 2 tens and 5 ones is 25.

Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
____1____ tens and __15_____ ones is 25.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-77
1 ten and 15 ones is 25

Explanation:
we add 10 and 25 we get 25, So 1 ten and 15 ones is 25.

Apply and Grow: Practice

Question 2.
Model 52 two ways.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
____5____ tens and ___2____ ones is 52.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-78
5 tens and 2 ones is 52.

Explanation:
We add 50 and 2 we get 52, So 5 tens and 2 ones is 52.

Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
______4__ tens and ____12___ ones is 52.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-79
4 tens and 12 ones is 52.

Explanation:
First we take 40 and add to 12 we get 52,
40 + 12 = 52, therefore 4 tens and 12 ones is 52.

Question 3.
Model 14 two ways.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
___1_____ tens and ___4____ ones is 14.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-80
1 ten and 4 ones is 14

Explanation:
we add 10 + 4 = 14,
So 1 ten and 4 ones is 14.

Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 82
___0_____ tens and ___14____ ones is 14.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-81
14 ones is 14

Explanation:
we add 0 + 14 = 14,
0 tens and 14 ones is 14.

Question 4.
DIG DEEPER!
Circle all of the ways that show 39.
3 + 9          2 tens and 9 ones      9 tens and 3 ones
10 + 29      3 tens and 19 ones    39 ones
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-82
Circled the ways that show 39 are 10 + 29 and 39 ones

Explanation:
As shown in picture circled all of the ways that show 39,
3 + 9 = 12 ≠ 39 so not circled,
2 tens and 9 ones is 29 ≠ 39 so not circled,
9 tens and 3 ones is 93 ≠ 39 so not circled,
10 + 29 = 39 = 39 so circled it,
3 tens and 19 ones is 30 + 19 is 49 ≠ 39 so not circled,
39 ones is 39 = 39 so circled it.

Think and Grow: Modeling Real Life

The models show how many seashells you and your friend have. Does your friend have the same number of seashells as you?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 83
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 84
Circle: Yes         No

Show how you know:
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-83
No friend does not have the same number of
seashells as much as I have.

Explanation:
After counting I have total 25 seashells and friend have 30,
Both do not have same or equal number of seashells ,
So circled No as shown in the picture above.

Show and Grow

Question 5.
The models show how many erasers you and your friend have. Does your friend have the same number of erasers as you?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 85
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 86
Circle: Yes          No

Show how you know:
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-84
Yes, I and my friend have the same number of erasers.

Explanation:
After counting I have total 40 erasers and friend also have 40,
Both have the same or equal number of erasers,
So circled Yes as shown in the picture above.

Write Numbers in Different Ways Practice 6.8

Question 1.
Model 49 two ways.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 87
__4______ tens and
___9____ ones
is 49.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-84
4 tens and 9 ones is 49.

Explanation:
Adding 40 to 9 is 40 + 9 = 49,
so 4 tens and 9 ones is 49.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 87
___30_____ tens and
____19____ ones
is 49.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-85

30 tens and 19 ones is 49.

Explanation:
We add 30 and 19 we get 49,
So 30 tens and 19 ones make 49.

Question 2.
DIG DEEPER!
Circle all of the ways that show 45.
40 + 5              45 tens and 0 ones       4 tens and 5 ones
20 + 15            2 tens and 25 ones       54 ones
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-86

Circled the ways that show 45 are 40 + 5,
4 tens and 5 ones and 2 tens and 25 ones.

Explanation:
As shown in picture circled all of the ways that show 45,
40 + 5 = 45 = 45 so circled it,
45 tens and 0 ones is 450 ≠ 45 so not circled,
4 tens and 5 ones is 45 = 45 so circled it,
20 + 15 = 35 ≠ 45 so not circled,
2 tens and 25 ones is 20 + 25 is 45 = 45  so circled it,
54 ones is 54 ≠ 45 so not circled.

Question 3.
Modeling Real Life
The models show the number of toy cars you and your friend have. Does your friend have the same number of toy cars as you?
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 88
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 89
Circle: Yes        No

Show how you know:
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-87
Yes, friend have the same number of toy cars as I have.

Explanation:
After counting I have total 34 toy cars and friend also have 34 ,
Both have same or equal number of toy cars,
So circled Yes as shown in the picture above.

Review & Refresh

Circle the heavier object.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 90
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-88
Circled ball.

Explanation:
Given ball and balloon as ball is
heavier object than balloon I circled it.

Question 5.
Big Ideas Math Solutions Grade 1 Chapter 6 Count and Write Numbers to 120 91
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-89
Circled 4 bananas

Explanation:
Given 4 bananas and 1 banana as 4 bananas are
heavier object than 1 banana I circled it.

Lesson 6.9 Count and Write Numbers to 120

Explore and Grow

How many balls are there? How did you count?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 92
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-89
Number of balls are 106. Counted one after another.

Explanation:
There are total 106 balls, we start pointing and counting
from first ball as one add next ball and so on counting
till we reach the last ball ,we get total 106 balls.

Show and Grow

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 93
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-90
10 tens and 7 ones is 107.

Explanation:
Given in the picture if we count there are
100 + 7 = 107 , 10 tens and 7 ones is 107.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 94
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-91
11 tens and 5 ones is 115.

Explanation:
Given in the picture if we count there are
110 + 5 = 107 , 11 tens and 5 ones is 115.

Apply and Grow: Practice

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 95
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-92
6 tens and 4 ones is 64.

Explanation:
Given in the picture if we count there are
60 + 4 = 64 , 6 tens and 4 ones is 64.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 96
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-93
9 tens and 8 ones is 98.

Explanation:
Given in the picture if we count there are
90 + 8 = 98 , 9 tens and 8 ones is 98.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 97
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-94
10 tens and 1 one is 101.

Explanation:
Given in the picture if we count there are
100 + 1 = 101 , 10 tens and 1 one is 101.

Question 6.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 98
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-95
10 tens and 3 ones is 103.

Explanation:
Given in the picture if we count there are
100 + 3 = 103 , 10 tens and 3 ones is 103.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 99
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-96
11 tens and 0 ones is 110.

Explanation:
Given in the picture if we count there are
110 + 0 = 110 , 11 tens and 0 ones is 110.

Question 8
DIG DEEPER!
What number is equal to lo tens and 8 ones? Show how you know.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-97
The number equal to 10 tens and 8 ones is 108

Explanation:
Given to find number equal to 10 tens and 8 ones,
as shown in figure taken 10 tens and 8 ones ,
the number is equal to 100 + 8 = 108.

Think and Grow: Modeling Real Life

Your teacher has 12 bags of balloons. Each bag has 10 balloons. How many balloons are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 100
Model:

____120_______ balloons
Answer:
There are 120 balloons are there in all.

Explanation:
Given my teacher has 12 bags of balloons and
each bag has 10 balloons so in all there are
12 X 10 = 120 balloons.

Show and Grow

Question 9.
A dentist has 10 boxes of toothbrushes and 9 extra toothbrushes. Each box has 10 toothbrushes. How many toothbrushes are there in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 101
Model:

___109________ toothbrushes
Answer:

10 tens and 9 ones is 109.
There are 109 tooth brushes are there in all.

Explanation:
Given a dentist has 10 boxes of toothbrushes and
9 extra toothbrushes, each box has 10 toothbrushes
means 10 x 10 = 100 and 9 extra makes 100 + 9 = 109 or
10 tens and 9 ones is 109 tooth brushes are there in all.

Count and Write Numbers to 120 Practice 6.9

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 102
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-98

7 tens and 9 ones is 79.

Explanation:
Given in the picture if we count there are
70 + 9 = 79 , 7 tens and 9 ones is 79.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 103
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-99

11 tens and 5 ones is 115.

Explanation:
Given in the picture if we count there are
110 + 5 = 115 , 11 tens and 5 ones is 115.

Question 3.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 104
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-100

12 tens and 0 ones is 120.

Explanation:
Given in the picture if we count there are
120 + 0 = 120 , 12 tens and 0 ones is 120.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 105
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-101

9 tens and 2 ones is 92.

Explanation:
Given in the picture if we count there are
90 + 2 = 92 , 9 tens and 2 ones is 92.

Question 5.
DIG DEEPER!
What number is equal to 11 tens and 2 ones? Show how you know.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-102
The number equal to 11 tens and 2 ones is 112.

Explanation:
We have 11 tens and 2 ones as shown in picture
11 tens is 110 and 2 ones is 2 so 110 + 2= 112.

Question 6.
Modeling Real Life
You have 4 packs of baseball cards and 6 packs of football cards. Each pack has 10 cards. How many cards do you have in all?
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 106
_____100_______ cards
Answer:
I have in total 100 cards in all.

Explanation:
Given I have 4 packs of baseball cards and
6 packs of football cards, Each pack has 10 cards,
therefore 4 packs of baseball means 4 X 10 = 40 cards,
6 packs of football cards have 6 X 10 = 60 cards, In total
I have 40 + 60 =100 cards in all.

Review & Refresh

Make a 10 to add.

Question 7.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 107
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-103

Sum is 6 + 4 + 1 = 11  or  10 + 1 = 11,
So 6 + 5 = 11

Explanation:
First we make sum 10 , For 6 to make 10 we need 10-6=4,
as we have to  add 4 to 6 now  we will minus 4 from 5 we get 5 – 4 = 1,
Now we write the total sum as 6 + 4 + 1 = 11  or
10 + 1 = 11 or 6 + 5 = 11.

Question 8.
Big Ideas Math Answer Key Grade 1 Chapter 6 Count and Write Numbers to 120 108
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-104

Sum is 7 + 3 + 5 = 15  or  10 + 5 = 15,
So 7 + 8 = 15

Explanation:
First we make sum 10 , For 7 to make 10 we need 10-7=3,
as we have to  add 3 to 7 now  we will minus 3 from 8 we get 8 – 3 = 5,
Now we write the total sum as 7 + 3 + 5 = 15  or
10 + 5 = 15 or 7 + 8 = 15.

Count and Write Numbers to 120 Performance Task

Question 1.
Your class sells candles for a fundraiser. You earn 10 dollars for every large candle you sell and 1 dollar for every small candle.

a. You sell 6 large candles. How much money do you raise?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 109
Answer:
I raised 60 dollars.

Explanation:
Given my class sells candles for a fundraiser.
I earn 10 dollars for every large candle I sell ,
I sold 6 large candles means 6 X 10 = 60 dollars,
So I raised 60 dollars.

b. You want to raise 72 dollars. How much more money do you need to raise?

_____12_____ more dollars
Answer:
I need 12 more dollars to raise 72 dollars.

Explanation:
I know I raised 60 dollars to raise 72 dollars
I need more is 72 – 60 = 12 dollars.

c. You also sell 12 small candles. Do you reach your goal?
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 110
Yes          No
Answer:
Yes, I reached my goal.

Explanation:
Given I earn1 dollar for every small candle I sell,
I sold 12 small candles means 12 X 1= 12 dollars,
In total I sold 6 large candles and 12 small candles
I raised total 60 + 12 = 72 dollars which is similar to
which I want to raise, So yes I reached my goal.

Question 2.
Your friend wants to raise 54 dollars. What are two ways your friend can sell large and small candles to reach her goal?
_____5____ large candles and ____4______ small candles
_____4_____ large candles and ___14_______ small candles
Answer:
5 large candles and 4 small candles,
4  large candles and 14 small candles.

Explanation:
I know if I sell 1 large candle I earn 10 dollars
and if I sell 1 small candles I earn 1 dollar to raise
54 dollars my friend can sell 5 tens and 4 ones is 54,
5 X 10 + 4 X 1 = 50 + 4 = 54,
So 5 large candles and 4 small candles makes 54 dollars
She can even sell 4 tens and 14 ones is 54,
4 X 10 + 14 X 1 = 40 + 14 = 54,
So 4  large candles and 14 small candles makes 54 dollars.

Count and Write Numbers to 120 Chapter Practice

Count to 120 by Ones Homework & Practice 6.1

Question 1.
Count by ones to write the missing numbers.
99, __100___, ___101___, __102___, __103___, ___104___
Answer:
The missing numbers after 99 are 100,101,102,103,104

Explanation:
Started from 99 counting by 1’s numbers after 99 are
99 + 1 = 100, 100 + 1 = 101,101 + 1 = 102, 102 + 1 = 103,
104 + 1 = 105, So the missing numbers after 99 are 100,101,102,103,104.

write the missing numbers.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 111
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-105

The missing numbers in the chart are 69,71,78,80.

Explanation:
The missing number after 68 is 68 + 1 = 69,
the number after 70 is 70 + 1 = 71,
the number before 79 is 79 – 1 = 78 and
number before 81 is 81 – 1 = 80, So the missing
numbers in the chart are 69,71,78 and 80.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 112
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-106

The missing numbers in the chart are 101,103,110,112,113

Explanation:
The missing number after 100 is 100 + 1 = 101,
the number after 102 is 102 + 1 = 103,
the number before 111 is 111 – 1 = 110,
the number after 111 is 111 + 1 = 112,
the number after 112 is 112 + 1 = 113,
So the missing numbers in the chart are 101,103,110,112,113.

Count to 120 by Tens Homework & Practice 6.2

Question 4.
Write the missing numbers from the chart. Then count on by tens to write the next two numbers.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 113
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-107

The missing numbers in the chart are 84,94 and
count on 10’s the next two numbers after 94 are 104,114.

Explanation:
Given numbers in the chart, the number missing before 85
is 85 – 1 = 84 and number missing after 93 and before 94 in the chart is
93 + 1= 94 or 95 – 1 = 94,now counting by 10’s next
two numbers after 94 are 94 + 10 = 104,104 + 10 = 114,
Therefore the missing numbers in the chart are 84 and 94
count on 10’s the next two numbers after 94 are 104,114.

Question 5.
YOU BE THE TEACHER
Your friend counts by tens starting with 53. Is your friend correct? Show how you know.
53, 63, 73, 83, 103
Answer:
No friend is Incorrect.

Explanation:
Given friend counts by tens starting with 53 after 53 its is
53 + 10 = 63 then 63 + 10 = 73, 73 + 10 = 83 and after 83
it is 83 + 10 = 93 not 103 therefore friend is incorrect.

Compose Numbers 11 to 19 Homework & Practice 6.3

Question 6.
Circle 10 ducks. Complete the sentence.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 114
____1______ ten and ____2______ ones is ___12____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-108

1 ten and 2 ones is 12.

Explanation:
Circled 10 ducks, after counting we have total
12 ducks we write the sentence as
one ten and 2 ones is twelve means 10 + 2 = 12.
1 ten and 2 ones is 12.

Question 7.
Modeling Real Life
You have 19 tennis balls. A bag can hold 10. You fill a bag. How many tennis balls are not in the bag?

_____9_______ tennis balls
Answer:

9 tennis balls are not there in the bag.

Explanation:
Given I have 19 tennis balls and a bag can hold 10 and
I have filled a bag so balls that are not there in the bag are
19 – 10 = 9 , So 9 tennis balls are not there in the bag.

Tens Homework & Practice 6.4

Question 8.
Circle groups of 10. Complete the sentence.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 115

______5____ tens and ___0___ ones is ___50______ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-109

fife tens and zero ones is fifty.
5 tens and 0 ones is 50.

Explanation:
Circled 5 tens ,five tens and zero ones is fifty or
5 tens and 0 ones is 50.

Tens and Ones Homework & Practice 6.5

Question 9.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 116
____9_____ tens and ____9_____ ones is ___99_____ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-110

9 tens and 9 ones is 99

Explanation:
Given in the picture  there are 9 ten blocks in tens place
and 9 blocks in ones make 9 tens and 9 ones is 99.

Question 10.
Modeling Real Life
You have 6 boxes of plastic cups and 3 extra cups. Each box has 10 cups. How many cups are there in all?

_____63_____ plastic cups
Answer:
In all there are 63 plastic cups.

Explanation:
Given I have 6 boxes of plastic cups and 3 extra cups and
each box has 10 cups means 6 X 10 = 60 and 3 extra cups make
60 + 3 = 63 or 6 tens and 3 ones is  63 Plastic cups.

Make Quick Sketches Homework & Practice 6.6

Make a quick sketch. Complete the sentence.

Question 11.
17

____17____ is ____1___ ten and ____7____ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-111

17 is 1 tens and 7 ones.

Explanation:
Given in the picture  there are 1 ten blocks in tens place
and 7 blocks in ones make 17 as 1 tens and 7 ones.

Question 12.
84

___84____ is ____8___ tens and ___4__ ones.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-112

84 is 8 tens and 4 ones is 84

Explanation:
Given in the picture  there are 8 ten blocks in tens place
and 4 blocks in ones make 84 as 8  tens and 4 ones.

Understand Place Value Homework & Practice 6.7

Question 13.
39
______3____ tens is ____30______ .
______9____ ones is ____9______ .
_____3____ tens and ____9______ ones is ____39______ .
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-113

3 tens is 30,
9 ones is 9,
So 3 tens and 9 ones is 39.

Explanation:
As shown in the picture above first we take 3 tens cubes as 30
and 9 ones cubes as 9, making 3 tens and 9 ones is 39.

Write Numbers in Different Ways Homework & Practice 6.8

Question 14.
Model 59 two ways.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 117
_____5_____ tens and _____9_____ ones is 59.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-114

5 tens and 9 ones is 59

Explanation:
we add 5 X 10 = 50 and 9 X 1 = 9
we get 50 + 9 = 59,
So 5 tens and 9 ones is 59.

Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 117
_____2_____ tens and ____39______ ones is 59.
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-115
2 tens and 39 ones is 59

Explanation:
we add 2  X 10 = 20 and 39 X 1 = 39
we get  20 + 39 = 59,
So 2 tens and 39 ones is 59.

Count and Write Numbers to 120 Homework & Practice 6.9

Question 15.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 118
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-116

11 tens and 4 ones is 114

Explanation:
As shown in the picture there are 11 ten blocks in tens place
and 4 blocks in ones make 11 tens and 4 ones is 114.

Question 16.
Big Ideas Math Answers 1st Grade 1 Chapter 6 Count and Write Numbers to 120 119
Answer:
Big-Ideas-Math-Answers-Grade-1-Chapter-6-Count-and-Write-Numbers-to-120-117

10 tens and 8 ones is 108

Explanation:
As shown in the picture there are 10 ten blocks in tens place
and 8 blocks in ones make 10 tens and 8 ones is 108.

Conclusion:

Start solving the problems in the performance task and chapter test which are at the end of the chapter and check the solutions from here. Hope the information about Big Ideas Math Book Grade 1 Answer Key Chapter 6 Count and Write Numbers to 120 is beneficial for you to overcome the difficulties in maths. Check out the links and start solving all the questions.

Big Ideas Math Answers Grade 8 Chapter 2 Transformations

Big Ideas Math Answers Grade 8 Chapter 2

Big Ideas Math Answers Grade 8 Chapter 2 Transformations provided helps students to learn the associated lessons easily. All the Solutions provided in BIM 8th Grade Chapter 2 Transformations are aligned according to the Transformation Textbook Chapter. Students can identify the knowledge gap accordingly and improvise on them and get a good hold of the concepts. Be on the right track and achieve success in your journey of learning math. Download the Big Ideas Math Grade 8 Answers Chapter 2 Transformations for free of cost and clear your queries.

Big Ideas Math Book 8th Grade Answer Key Chapter 2 Transformations

Avail the handy Big Ideas Math Grade 8 Chapter 2 Transformations Solution Key over here and solve any kind of problem easily. 8th Grade Big Ideas Math Ch 2 Transformations Answers are given by experts after doing extensive research. Simply access them by clicking on the direct links available and prepare accordingly. Enhance your Problem Solving Ability and math skills with regular practice and attempt the exams with confidence.

Performance

Lesson: 1 Translations

Lesson: 2 Reflections

Lesson: 3 Rotations

Lesson: 4 Congruent Figures

Lesson: 5 Dilations

Lesson: 6 Similar Figures

Lesson: 7 Perimeters and Areas of Similar Figures

Chapter 2: Transformations 

Transformations STEAM Video/Performance

STEAM Video

Shadow Puppets

Some puppets are controlled using strings or wires. How else can a puppet be controlled?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 1.1

Watch the STEAM Video “Shadow Puppets.” Then answer the following questions.

Question 1.
Tory and Robert are using a light source to display puppets on a screen. Tory wants to show the pig jumping from the floor to the window. Should she use a translation, reflection, rotation, or dilation? Explain.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 1.2

Answer:
In the situation given, if we translate first, we move the pre-image closer to the center of dilation than if we translate second. That will result in a different image.

Question 2.
How can Tory show the pig getting smaller as it jumps out the window?

Performance Task

Master Puppeteer

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given the coordinates of a kite being used bya puppeteer.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 1.3
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 1.4
You will be asked to identify transformations for given movements of the kite. When might a puppeteer want to use a reflection?

Transformations Getting Ready for Chapter 2

Getting Ready for Chapter 2

Chapter Exploration
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 2

Question 1.
Work with a partner. Form each triangle on a geoboard.

  • Which of the triangles are congruent to the triangle at the right?
  • Measure the sides of each triangle with a ruler. Record your results in a table.
  • Write a conclusion about the side lengths of triangles that are congruent.

Big Ideas Math Answers Grade 8 Chapter 2 Transformations 3
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 4

Answer: option d

When two triangles are congruent they will have exactly the same three sides and exactly the same three angles. The equal sides and angles may not be in the same position

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what the terms might mean and record your thoughts.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 5

Lesson 2.1 Translations

EXPLORATION 1
Work with a partner.
a. For each figure below, draw the figurein a coordinate plane. Then copy the figureonto a piece of transparent paper and slide the copy to a new location in the coordinate plane. Describe the location of the copy compared to the location of the original.

  • point
  • triangle
  • line segment
  • rectangle
  • line

Big Ideas Math Answers Grade 8 Chapter 2 Transformations 6
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 6.1
b. When you slide figures, what do you notice about sides, angles, and parallel lines?
c. Describe the location of each point below compared to the point A(x, y).
B(x + 1, y + 2)
C(x – 3, y + 4)
D(x – 2, y + 3)
E(x + 4, y – 1)
d. You copy a point with coordinates (x, y) and slide it horizontally a units and vertically b units. What are the coordinates of the copy?

2.1 Lesson

Try It

Tell whether the blue figure is a translation of the red figure.

Question 1.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 7

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same but the size is different. The red figure slide to form a blue figure but it is not the same size. So blue figure is not the translation of red figure.

Question 2.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 8

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same and also the size is the same. The red figure slide to form exactly blue figure. So blue figure is the translation of red figure.

Try It

Question 3.
WHAT IF?
The red triangle is translated 4 units left and 2 units up. What are the coordinates of the image?

Answer:
Big Ideas Math Grade 8 Chapter 2 Answer Key img_6
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(-2,1) B(2,5), C(1,2) and a = -4, b = 2
A'(-2+a, -2+b) = A'(-2-4, 1+2) = A'(-6,3)
B'(2+a, 5+b) = B'(2-4, 5+2) = B'(-2,7)
A'(1+a, 2+b) = C'(1-4, 2+2) = C'(-3,4)
Hence the coordinate of image are A'(-6,3), B'(-2,7), C'(-3,4).

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

IDENTIFYING A TRANSLATION
Tell whether the blue figure is a translation of the red figure.

Question 4.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 9

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same but the size is different. The red figure slide to form a blue figure but it is not the same size. So blue figure is not the translation of red figure.

Question 5.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 10

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same, and also the size is the same. The red figure slide to form the exactly blue figure. So blue figure is the translation of red figure.

Question 6.
The vertices of a triangle are A(2, 2), B (0, 2), and C (3, 0). Translate the triangle 1 unit left and 2 units up. What are the coordinates of the image?

Answer:
Bigideas Math Answers Grade 8 Chapter 2 img_7
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(2,2) B(0,2), C(3,0) and a = -1, b = 2
A'(2+a, 2+b) = A'(2-1, 2+2) = A'(1,4)
B'(0+a, 2+b) = B'(0-1, 2+2) = B'(-1,4)
A'(3+a, 0+b) = C'(3-1, 0+2) = C'(2,2)
Hence the coordinate of image are A'(1,4), B'(-1,4), C'(2,2).

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
A neighborhood planner uses a coordinate plane to design a new neighborhood. The coordinates A(1, -1), B(1, -2), and C (2, -1) represent House A, House B, and House C. The planner decides to place a playground centered at the origin, and moves the houses to make space. House A is now located at A'(3, -4). What are the new coordinates of House B and House C when each house is moved using the same translation? Justify your answer.

Answer:
Big Ideas Math Answers 8th Grade Chapter 2 img_8
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(1,-1) B(1,-2), C(2,1) and A'(3,-4)
A'(1+a, -1+b) = A'(3,-4) So, a = 2, b = -3
New coordinates of the houses are
B'(1+a, -2+b) = B'(1+2, -2-3) = B'(3,-5)
C'(2+a, -1+b) = C'(2+2, -1-3) = C'(4,-4)
Hence the coordinate of image are B'(3,-5),C'(4,-4)

Question 8.
The locations of a quarterback and a wide receiver on a football field are represented in a coordinate plane. The quarterback throws the football to the point (6, -2). Use a translation to describe a path the wide receiver can take to catch the pass.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 11

Answer:
Coordinate of Receiver: (1,3) and football point:(-6,-2)
Horizontal shift: a = x2-x1 = 6 – 1 = 5
Verrical shift: b = y2 – y1 = -2 – 3 = -5
Hence the path which receiver will take 5 unit right and 5 unit down.

Translations Homework & Practice 2.1

Review & Refresh

Solve the equation for y.

Question 1.
6x + y = 12

Answer:
Given
6x + y = 12
Subtract 6x from both sides
y = 12 – 6x
Now arranging the terms
y = -6x + 12
y = 6 (-x + 2)
Thus y = 6(-x + 2)

Question 2.
9 = x + 3y

Answer:
Given,
9 = x + 3y
3y = 9 – x
Dividing by 3 on both sides
y = (9 – x)/3
y =  \(\frac{9}{3}\) – \(\frac{x}{3}\)
Now arranging the terms
y = – \(\frac{x}{3}\) + 3
Thus y = – \(\frac{x}{3}\) + 3

Question 3.
\(\frac{1}{3}\)x + 2y = 8

Answer:
Given,
\(\frac{1}{3}\)x + 2y = 8
Subtracting x/3 from both sides
2y = 8 – \(\frac{x}{3}\)
Now arranging the terms
2y = – \(\frac{x}{3}\) + 8
y = – \(\frac{x}{6}\) + 4

Question 4.
You put $550 in an account that earns 4.4% simple interest per year. How much interest do you earn in 6 months?
A. $1.21
B. $12.10
C. $121.00
D. $145.20

Answer: $12.10

Explanation:
Given:
You put $550 in an account that earns 4.4% simple interest per year.
Principal amount: P = $550
Rate of Interest: r = 4.4%
Time: t = 6 months = 0.5 year
We know that formula for Simple Interest is SI = prt/100
SI = (550 × 4.4 × 0.5)/100
SI = 1210/100
SI = 12.10
Hence the simple interest is $12.10
Thus the correct answer is option B.

Concepts, Skills, & Problem Solving

DESCRIBING RELATIONSHIPS
For each figure, describe the location of the blue figurerelative to the location of the red figure. (See Exploration 1, p. 43.)

Question 5.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 12

Answer: The path which the receiver will take is 6 units right and 3 units down.

Explanation:
Coordinate of Point A: (-3,2) and Point A’: (3,-5)
Horizontal shift: a = x2 – x1 = 3 – (-3) = 3 + 3 = 6
Vertical shift: b = y2 – y1 = -5 – (-2) = -5 + 2 = -3
Hence, The path which the receiver will take is 6 units right and 3 units down.

Question 6.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 13

Answer: The path which receive will take is 5 units left and 2 units down.

Explanation:
Coordinate of point A: (3,-2) and point A’: (-2,-4)
Horizontal shift: a = x2 – x1 = -2 – (3) = -2 – 3 = -5
Vertical shift: b = y2 – y1 = -4 – (-2) = -4 + 2 = -2
Hence, The path which receive will take is 5 unit left and 2 unit down.

IDENTIFYING A TRANSLATION
Tell whether the blue figure is a translation of the red figure.

Question 7.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 14

Answer:
Blue figure is the translation of red figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figure are the same. Also, the orientation of the blue figure is the same as the red figure. This means that the red figure sides to form the blue figure. So, the blue figure slides to form the red figure.

Question 8.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 15

Answer:
Blue figure is not the translation of the red figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figure are the same. But the orientation of the blue figure is different from the red figure. This means that the blue figure is not the translation of the red figure.

Question 9.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 16

Answer:
Blue figure is not the translation of the red figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figure are the same. But the orientation of the blue figure is different from the red figure. This means that the blue figure is not the translation of the red figure. The blue figure is the mirror image of the red figure.

Question 10.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 17

Answer:
Blue figure is the translation of red-figure.

Here in the given figure, we can see that the shape of both blue and red-figure are the same, and also the size of both the figure are the same. Also, the orientation of the blue figure is the same as the red figure. This means that the red figure slides to form the blue figure. so the figure slides to form the red figure.

Question 11.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 18

Answer:
Blue figure is the translation of red-figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figures are the same. Also, the orientation of the both figure is the same as the red figure. This means that the red figure slides to form the blue figure. So blue figure slides to form the red figure.

Question 12.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 19

Answer: Blue figure is not the translation of the red-figure.

Explanation:
Here in the given figure we can see that the shape of both blue and red figure are same and the size of both the figure are not same. The red figure are smaller as compared to the blue figure. This means that blue figure is not the translation of red figure.

TRANSLATE A FIGURE
The vertices of a triangle are L(0, 1), M(1, -2), and N(-2, 1). Draw the figure and its image after the translation.

Question 13.
1 unit left and 6 units up

Answer:
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_1

explanation:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to the x-coordinate and ‘b’ is added to the y-coordinates of the vertices.
A(x,y) – A'(x+a, y+b)
The value ‘a’ and ‘b’ will be positive if the shift is Right and Vertical up and the value of ‘a’ and ‘b’ will be negative if the shift is left and vertical down.
Given:L(0,1),M(1,-2),N(-2,1)anda=-1,b=6
L'(0+a,1+b)=l'(0-1,1+6)=L'(-1,7)
M'(1+a,-2+b)=M'(1-1,1+6)=M'(0,4)
N'(-2+a,1+b)=N'(-3-1,1+6)=N'(-4,7)
Hence,the coordinate of image are L'(-1,7),M'(0,4),N'(-4,7)

Question 14.
5 units right

Answer:
We know that to translate a figure ‘a’ units horizontal and ‘b’ units vertically in the coordinate plane, ‘a’ is added to X-coordinate and ‘b’ is added to Y-coordinate of the vertices.
a(x,y),=A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if the shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given:L(0,1),M(1,-2),N(-2,1)and a=5,b=0
L(0+a,1+b)=L'(0+5,1+0)=L'(5,1)
M'(1+a,-2=b)=M'(1+5,-2+0)=M'(6,-2)
N'(-2+a,1+b)=N'(-2+5,1+0)=N'(3,1)
Hence the coordinate of image are L'(5,1),M'(6,-2),N'(3,1).

Question 15.
(x + 2, y + 3)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: L(0,1), M(1,-2), N(-2,1) and (x+2,y+3)
So the value of: a = 2 and b = 3
L'(o + a, 1 + b) = L'(0 + 2,1 + 3) = L'(2,4)
M'(1+a, -2 + b) = M'(1 + 2, -2 + 3) = M'(3, 1)
N'(-2 + a, 1 + b) = N'(-2 + 2, 1 + 3) = N'(0, 4)
Hence the coordinate of the image is L'(2,4), M'(3,1), N'(0,4)

Question 16.
(x – 3, y – 4)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: L(0,1), M(1,-2), N(-2,1) and (x-3, y-4)
So the value of a = -3 and b = -4
L'(0 + a, 1 + b) = L'(0-3,1-4) = L'(-3,-3)
M'(1 + a, -2 + b) = M'(1 – 3, -2-4) = M'(-2,-6)
N'(-2 + a, 1 + b) = N'(-2 – 3, 1 – 4) = N'(-5, -3)
Hence the coordinate of the image are L'(-3,-3), M'(-2,-6), N'(-5, -3)

Question 17.
YOU BE THE TEACHER
Your friend translates point A 2 units down and 1 unit right. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 20

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given points,
A(3, 1) and a = 1, b = -2
A'(3+a, 1+b) = A'(3+1, 1-2) = A'(4, -1)
So, the point A’ translated by my friend is wrong. He has reversed the x and y coordinate for translation.
Hence the correct translate point is A'(4,-1)

Question 18.
TRANSLATING A FIGURE
Translate the triangle 4 units right and 3 units down. What are the coordinates of the image?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 21

Answer:
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations img_3
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given points from graph: J(-1,3), K(-1,1), L(-4,1)
and a = 4, b = -3
J'(-1 + a, 3 + b) = J'(-1+4,3-3) = J'(3,0)
K'(-1 + a, 1 + b) = K'(-1 + 4, 1 – 3) = K'(3,-2)
L'(-4 + a, 1 + b) = L'(-4 + 4, 1 – 3) = L'(0,-2)
Hence the coordinate of image is J'(3,0), K'(3,-2), L'(0,-2)

Question 19.
TRANSLATING A FIGURE
Translate the figure 2 units left and 4 units down. What are the coordinates of the image?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 22

Answer:
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations img_4
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: A(-1,4), B(2,3), C(3,0), D(-1,-1)
and a = -2, b = -4
A'(-1+a, 4+b) = A'(-1-2, 4-4) = A'(-3,0)
B'(2+a, 3+b) = B'(2-2, 3-4) = B'(0,-1)
C'(3+a, 0+b) = C'(3-2, 0-4) = C'(1,-4)
D'(-1+a, -1+b) = D'(-1-2, -1-4) = D'(-3,-5)
The coordinate of image are A'(-3,0), B'(0,-1), C'(1,-4), D'(-3,-5)

DESCRIBING A TRANSLATION
Describe the translation of the point to its image.

Question 20.
(3, 2) → (1,0)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Coordinate of the point A:(3,-2) and the image point A’:(1,0)
Horizontal shift: a = x2 – x1 = 1 – 3 = -2
Vertical shift: b = y2 – y1 = 0 – (-2) = 0 + 2 = 2
Hence the translation path will be 2 units left and 2 units up.

Question 21.
(-8, -4) → (-3, 5)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Coordinate of the point A:(-8,-4) and the image point A’:(-3,5)
Horizontal shift: a = x2 – x1 = -3 – (-8) = 5
Vertical shift: b = y2 – y1 = 5 – (-4) = 9
Hence the translation path will be 5 units left and 9 units up.

Question 22.
REASONING
You can click and drag an icon on a computer’s desktop. Is this an example of a translation? Explain.

Answer:
Yes, the dragging of an icon on a computer’s desktop is an example of translation.
Because when dragging an icon on desktop the icon directly slides and is stored in its new position. While dragging the icon there is no change in shape and size of the icon, thus fulfilling the criteria of translation.

Question 23.
MODELING REAL LIFE
The proposed location for a new oil platform is represented in a coordinate plane by a rectangle with vertices A(1, 3), B(1, 4), C(4, 4), and D(4, -3). An inspector recommends moving the oil platform 4 units right and 2 units down. Find the coordinates of the image. Then draw the original figureand the image in the coordinate plane.

Answer:
BIM 8th Grade Answers Chapter 2 Transformations img_5
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: A(1,-3), B(1,4), C(4,4), D(4,-3) and a = 4, b = -2
A'(1+a, -3+b) = A'(1+4, -3-2) = A'(5,-5)
B'(1+a, 4+b) = B'(1+4, 4-2) = B'(5,2)
C'(4+a, 4+b) = C'(4+4, 4-2) = C'(8,2)
D'(4+a, -3+b) = D'(4+4, -3-2) = D'(8,-5)
Hence the coordinate of image are A'(5,-5), B'(5,2), C'(8,2), D'(8,-5)

Question 24.
PROBLEM SOLVING
A school of fish translates from point F to point D.
a. Describe the translation of the school of fish.
b. Can the fishing boat make the same translation? Explain.
c. Describe a translation the fishing boat could make to get to point D.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 22.1

Answer:
a. Coordinate of the point F: (-3,2) and the point D: (2,3)
Horizontal shift: a = x2 – x1 = 2 – (-3) = 2 + 5 = 7
Vertical shift: b = y2 – y1 = 3 – 2 = 1
Hence the path of translation is 5 unit Right and 1 unit Up.
b. No, the fishing boat (point B) cannot make the same translation as by fish (point F). Because in path between from point B to point D the is an island which will interrupt the translation of fishing boat.
c. Coordinate of the point B:(-2,-1) and the point D: (2,3)
Horizontal shift: a = x2 – x1 = 2 – (-2) = 2 + 2 = 4
Verical shift: b = y2 – y1 = 3 -(-1) = 3 + 1 = 4
Hence the path of translation is 4 unit Right and 4 unit Up.

Question 25.
REASONING
The vertices of a triangle are A(0, -3), B(2, -1), and C(3, -3). You translate the triangle 5 units right and 2 units down. Then you translate the image 3 units left and 8 units down. Is the original triangle identical to the final image? Explain your reasoning.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: A(0,-3), B(2,-1), C(3,-3) and a1 = 5, b1 = -2
A'(0+a, -3+b) = A'(0+5, -3-2) = A'(5,-5)
B'(2+a, -1+b) = B'(2+5, -1-2) = B'(7,-3)
C'(3+a, -3+b) = C'(3+5,-3-2) = C'(8,-5)
Hence the coordinate of the first image are A'(5,-5), B'(7,-3), C'(8,-5)
Given: A'(5,-5), B'(7,-3), C'(8,-5) and a2 = -3, b2 = -8
A”(5+a, -5+b) = A”(5-3, -5-8) = A”(2,-13)
B”(7+a, -3+b) = B”(7-3, -3-8) = B”(4,-11)
C”(8+a, -5+b) = C”(8-3,-5-8) = C”(5,-13)
Hence the coordinate of the first image are A”(2,-13), B”(4,-11), C”(5,-13)
a = a1+a2 = 5 – 3 = 2, and b = b1 + b2 = -2 – 8 = -10
A'(0+a, -3+b) = A'(0+2, -3-10) = A'(2,-13)
B'(2+a, -1+b) = B'(2+2, -1-10) = B'(4,-11)
C'(3+a, -3+b) = C'(3+2,-3-10) = C'(5,-13)
Hence the original triangle is identical to the final image. This is because we can use both the translation by finding the resultant translation.
For final translation we can use:(x+2, y-10)

Question 26.
DIG DEEPER!
In chess, a knight can move only in an L-shaped pattern:

  • two vertical squares, then one horizontal square;
  • two horizontal squares, then one vertical square;
  • one vertical square, then two horizontal squares; or
  • one horizontal square, then two vertical squares.

Write a series of translations to move the knight from g8 to g5.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 23

Answer:
The series of translation to move knight from g8 to g5
1. Move 1 units Right to h8 and then 2 units Down to h6
2. Move 2 units Left to f6 and then 1 unit Up to f7
3. Move 2 units Down to f5 and then 1 unit Right to g5

Lesson 2.2 Reflections

Reflecting Figures

Work with a partner.
a. For each figure below, draw the figure in the coordinate plane. Then copy the axes and the figure onto a piece of transparent paper. Flip the transparent paper and align the origin and the axes with the coordinate plane. For each pair of figures, describe the line of symmetry.

  • point
  • triangle
  • line segment
  • rectangle
  • line

Big Ideas Math Answers Grade 8 Chapter 2 Transformations 24
b. When you reflect figures, what do you notice about sides, angles, and parallel lines?
c. Describe the relationship between each point below and the point A(4, 7) in terms of reflections.
d. A point with coordinates (x, y) is reflected in the x-axis. What are the coordinates of the image?
e. Repeat part(d) when the point is reflected in the y-axis

2.2 Lesson

Try It

Tell whether the blue figure is a reflection of the red figure.

Question 1.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 25

Answer: Blue figure is not the reflection of the red figure

Explanation:
By seeing the above figure we can say that the blue figure is not the mirror image of the red figure. Thus Blue figure is not the reflection of the red figure.

Question 2.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 26

Answer: Blue figure is the reflection of the red figure

Explanation:
By seeing the above figure we can say that the blue figure is the mirror image of the red figure. If the red figure is flipped it would form the shape of the blue figure. Thus Blue figure is the reflection of the red figure

Try It

Question 3.
The vertices of a rectangle are A(-4, -3), B(-4, -1), C(-1, -1), and D(-1, -3). Draw the figure and its reflection in (a) the x-axis and (b) the y-axis.

Answer:
Given,
The vertices of a rectangle are A(-4, -3), B(-4, -1), C(-1, -1), and D(-1, -3).
Reflection about the x-axis:
A(x,y) = A'(x,-y)
A(-4, -3) = A'(-4,3)
B(-4, -1) = B'(-4,1)
C(-1, -1) = C'(-1,1)
D(-1, -3) = D'(-1,3)
Reflection through x-axis:
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations img_9(i)
Reflection through y-axis:
A(x,y) = A'(-x,y)
A(-4, -3) = A'(4,-3)
B(-4, -1) = B'(4,-1)
C(-1, -1) = C'(1,-1)
D(-1, -3) = D'(1,-3)
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations img_9(ii)

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
REFLECTING A FIGURE
The vertices of a triangle are J(-3, -5), K(-2, 2), and L(1, -4). Draw the figure and its reflection in
(a) the x-axis and
(b) the y-axis.

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
The vertices of a triangle are J(-3, -5), K(-2, 2), and L(1, -4).
Reflection about the x-axis:
J(-3, -5) = J'(-3,5)
K(-2, 2) = K'(-2,-2)
L(1, -4) = L'(1,4)
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_10(i)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Reflection about the y-axis:
J(-3, -5) = J'(3,-5)
K(-2, 2) = K'(2,2)
L(1, -4) = L'(-1,-4)
BIM Grade 8 Answers Chapter 2 Transformations img_10(ii)

Question 5.
WHICH ONE DOESN’T BELONG?
Which transformation does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 27

Answer: 3rd figure is different from other figures. Because all the other three pictures are reflections of each other except the third one. The third picture is pointed in the same direction but all the other three figures are in opposite direction.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
You design a logo using the figure shown at the left. You want both the x-axis and the y-axis to be lines of reflection. Describe how to use reflections to complete the design. Then draw the logo in the coordinate plane.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 28

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
A(-4,2), B(-2,2), C(0,0), D(-2,0)
Reflection about the x-axis:
A(-4,2) = A'(-4,-2)
B(-2,2) = B'(-2,-2)
C(0,0) = C'(0,0)
D(-2,0) = D'(-2,0)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Reflection about the y-axis:
A(-4,2) = A”(4,2)
B(-2,2) = B”(2,2)
C(0,0) = C”(0,0)
D(-2,0) = D”(2,0)
Now to complete the Logo again we have to take a reflection of the image figure about the y-axis. In this way, the logo will be symmetric about both axis.
A”(4,2) = A”‘(4,-2)
B”(2,2) = B”‘(2,-2)
C”(0,0) = C”‘(0,0)
D”(2,0) = D”‘(-2,0)
Bigideas Math Answer Key Grade 8 Chapter 2 img_11

Question 7.
DIG DEEPER!
You hit the golf ball along the path shown, so that its final location is a reflection in the y-axis of its starting location.
a. Does the golf ball land in the hole? Explain.
b. Your friend tries the shot from the same starting location. He bounces the ball of the wall at the point (-0.5, 7) so that its path is a reflection. Does the golf ball land in the hole?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 29

Answer:
a. Coordinates of the location of golf ball = (2,4)
Coordinates of location of hole = (-3,4)
Location of ball after reflection through y-axis = (2,4) = (-2,4)
But the location of the hole is (-3,4)
So the ball will not go into a hole and it will miss the hole by 1 unit.
Hence the ball will not go into the hole.
b. Yes, when the ball bounces at the point (-0.5,7) then it will land in the hole.

Reflections Homework & Practice 2.2

Review & Refresh

The vertices of a quadrilateral are P(-1, -1), Q(0, 4), R(3, 1), and S(1, -2). Draw the figure and its image after the translation.

Question 1.
7 units down

Answer:
Big Ideas Math 8th Grade Answer Key for Chapter 2 img_12
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = 0, b = -7
P'(-1+a,-1+b) = P'(-1+0,-1-7) = P'(-1,-8)
Q'(0+a,4+b) = Q'(0+4,4-7) = Q'(4,-3)
R'(3+a, 1+b) = R'(3+0,1-7) = R'(3,-6)
S'(1+a,-2+b) = S'(1+0,-2-7) = S'(1,-9)
Thus the coordinate of the image is P'(-1,-8), Q'(4,-3), R'(3,-6), and S'(1,-9)

Question 2.
3 units left and 2 units up

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = -3, b = 2
P'(-1+a,-1+b) = P'(-1-3,-1+2) = P'(-4,1)
Q'(0+a,4+b) = Q'(0-3,4+2) = Q'(-3,6)
R'(3+a, 1+b) = R'(3-2,1+2) = R'(0,3)
S'(1+a,-2+b) = S'(1-3,-2+2) = S'(-2,0)
Thus the coordinate of the image are P'(-4,1), Q'(-3,6), R'(0,3) and S'(-2,0)
Big Ideas Math Answers Grade 8 Ch 2 Transformations img_11

Question 3.
(x + 4, y – 1)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = 4, b = -1
P'(-1+a,-1+b) = P'(-1+4,-1-1) = P'(3,-2)
Q'(0+a,4+b) = Q'(0+4,4-1) = Q'(4,3)
R'(3+a, 1+b) = R'(3+4,1-1) = R'(7,0)
S'(1+a,-2+b) = S'(1+4,-2-1) = S'(5,-3)
Thus the coordinate of the image are P'(3,-2), Q'(4,3), R'(7,0) and S'(5,-3)
Big ideas math answers grade 8 chapter 2 transformations img_12

Question 4.
(x – 5, y – 6)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = -5, b = -6
P'(-1+a,-1+b) = P'(-1-5,-1-6) = P'(-6,-7)
Q'(0+a,4+b) = Q'(0-5,4-6) = Q'(-5,-2)
R'(3+a, 1+b) = R'(3-5,1-6) = R'(-2,-5)
S'(1+a,-2+b) = S'(1-5,-2-6) = S'(-4,-8)
Thus the coordinate of the image are P'(-6,-7), Q'(-5,-2), R'(-2,-5) and S'(-4,-8)
BIM 8th Grade Answer Key Chapter 2 Transformations img_13

Tell whether the angles are complementary, supplementary or neither.

Question 5.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 30

Answer:
108° + 82° = 190°
Thus the angle is neither supplementary nor complementary.

Question 6.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 31

Answer: Complementary

Explanation:
43° + 47° = 90°
Two angles are called complementary when their measures add to 90 degrees.

Question 7.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 32

Answer:
38° + 62° = 100°
Hence the given angle is neither supplementary nor complementary.

Question 8.
36 is 75% of what number?
A. 27
B. 48
C. 54
D. 63

Answer: B. 48

Explanation:
Let x be the unknown value.
75% of x = 36
75% × x = 36
75/100 × x = 36
3/4x × x = 36
3x = 36 × 4
3x = 144
x = 144/3
x = 48
Thus the correct answer is option B.

Concepts, Skills, &Problem Solving
DESCRIBING RELATIONSHIPS
Describe the relationship between the given point and the point A(5, 3) in terms of reflections. (See Exploration 1, p. 49.)

Answer:
We know that when a point is reflected about x-axis then y-coordinate becomes the opposite.
P(x,y) = P'(x,-y)
We know that when a point is reflected about y-axis then x-coordinate becomes opposite.
P(x,y) = P'(-x,y)
Given: A(5,3), B(5,-3)
Hence the point A is reflected about the x-axis to get point B.

IDENTIFYING A REFLECTION
Tell whether the blue figure is a reflection of the red figure.

Question 12.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 33

Answer: No

Explanation:
The blue figure is not the mirror image of the red figure. If the red figure were flipped then the right of the blue and red figure should be facing each other. So, the blue figure is not a reflection of red figure.

Question 13.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 34

Answer: Yes

Explanation:
The blue figure is the mirror image of the red figure. If the red figure were flipped it will result in the blue figure. So, the blue figure is a reflection of red figure.

Question 14.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 35

Answer: Yes

Explanation:
The blue figure is the mirror image of the red figure. If the red figure were flipped it will result in the blue figure. So, the blue figure is a reflection of red figure.

Question 15.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 36

Answer: No

Explanation:
The blue figure is not the mirror image of the red figure. If the red figure were flipped then the right of the blue and red figure should be facing each other. So, the blue figure is not a reflection of red figure.

Question 16.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 37

Answer: Yes

Explanation:
The blue figure is the mirror image of the red figure. If the red figure were flipped it will result in the blue figure. So, the blue figure is a reflection of red figure.

Question 17.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 38

Answer: No

Explanation:
The blue figure is not the mirror image of the red figure. If the red figure were flipped then the right of the blue and red figure should be facing each other. So, the blue figure is not a reflection of red figure.

REFLECTING FIGURES
Draw the figure and its reflection in the x-axis. Identify the coordinates of the image.

Question 18.
A(3, 2), B(4, 4), C(1, 3)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
A(3, 2), B(4, 4), C(1, 3)
Reflection about the x-axis:
A(3, 2) = A'(3,-2)
B(4, 4) = B'(4,-4)
C(1, 3) = C'(1,-3)
Thus the coordinate of the image are A'(3,-2), B'(4,-4), C'(1,-3)
Bigideas Math Answers Grade 8 Chapter 2 img_14

Question 19.
M(-2, 1), N(0, 3), P(2, 2)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
M(-2, 1), N(0, 3), P(2, 2)
Reflection about the x-axis:
M(-2, 1) = M'(-2,-1)
N(0, 3) = N'(0,-3)
P(2, 2) = P'(2,-2)
Thus the coordinate of the image are M'(-2,-1), N'(0,-3), P'(2,-2)
Big Ideas Math Grade 8 Ch 2 Answer Key img_15

Question 20.
H(2, -2), J(4, -1), K(6, -3), L(5, -4)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
H(2, -2), J(4, -1), K(6, -3), L(5, -4)
Reflection about the x-axis:
H(2, -2) = H'(-2,-1)
J(4, -1) = J'(4,1)
K(6, -3) = K'(6,3)
L(5, -4) = L'(5,4)
Thus the coordinate of the image are H'(-2,-1), J'(4,1), K'(6,3) and L'(5,4)
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_16

Question 21.
D(-2, -5), E(0, -1), F(2, -1), G(0, -5)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
D(-2, -5), E(0, -1), F(2, -1), G(0, -5)
Reflection about the x-axis:
D(-2, -5) = D'(-2,5)
E(0, -1) = E'(0,1)
F(2, -1) = F'(2,1)
G(0, -5) = G'(0,5)
Thus the coordinate of the image are D'(-2,5), E'(0,1), F'(2,1), G'(0,5)
Big ideas Math Answers Grade 8 Chapter 2 Transformations img_17

REFLECTING FIGURES
Draw the figure and its reflection in the y-axis. Identify the coordinates of the image.

Question 22.
Q(-4, 2), R(-2, 4), S(-1, 1)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
Q(-4, 2), R(-2, 4), S(-1, 1)
Reflection about the x-axis:
Q(-4, 2) = Q'(4,2)
R(-2, 4) = R'(2,4)
S(-1, 1)= S'(1,1)
Thus the coordinate of the image is Q'(4,2), R'(2,4), S'(1,1)
Big Ideas Math Grade 8 Chapter 2 solution Key img_18

Question 23.
T(4, -2), U(4, 2), V(6, -2)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
T(4, -2), U(4, 2), V(6, -2)
Reflection about the y-axis
T(4,-2) = T'(-4,-2)
Y(4,2) = U'(-4,2)
V(6,-2) = V'(-6,-2)
Thus the coordinates of the figure are T'(-4,-2), U'(-4,2), V'(-6,-2)
Big Ideas Math Grade 8 Chapter 2 transformations answer key img_19

Question 24.
W(2, -1), X(5, -2), Y(5, -5), Z(2, -4)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
W(2, -1), X(5, -2), Y(5, -5), Z(2, -4)
Reflection about the y-axis:
W(2,-1) = W'(-2,-1)
X(5,-2) = X'(-5,-2)
Y(5,-5) = Y'(-5,-5)
Z(2,-4) = Z'(-2,-4)
Thus the coordinates of the figure are W'(-2,-1), X'(-5,-2), Y'(-5,-5), Z'(-2,-4)
Big Ideas Math Grade 8 2nd Chapter Answer Key for Transformations img_20

Question 25.
J(2, 2), K(7, 4), L(9, -2), M(3, -1)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
J(2, 2), K(7, 4), L(9, -2), M(3, -1)
Reflection about the y-axis
J(2, 2) = J(-2,2)
K(7, 4) = K'(-7,4)
L(9, -2) = L'(-9,-2)
M(3, -1) = M'(-3,-1)
Thus the coordinates of the figure are J(-2,2), K'(-7,4), L'(-9,-2), M'(-3,-1)
BIM Grade 8 Solution Key Chapter 2 Transformations img_21

Question 26.
REASONING
Which letters look the same when reflected in the line?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 39

Answer:
The letters which will look the same after being reflected through horizontal line are
B, C, D, E, H, I, K, O, X

STRUCTURE
The coordinates of a point and its image after a reflection are given. Identify the line of reflection.

Question 27.
(2, -2) → (2, 2)

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given A(2, -2) → A'(2, 2)
Here we can see that x-coordinate of both A & A’ is the same but the y-coordinate of A’ is just the opposite of A. This means that A’ is the reflection of A about the x-axis.
Hence the point A is reflected about the x-axis to get point A’.

Question 28.
(-4, 1) → (4, 1)

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given A(-4, 1) → A'(4, 1)
Here we can see that y-coordinate of both A & A’ is the same but the x-coordinate of A’ is just the opposite of A. This means that A’ is the reflection of A about y-axis.
Hence the point A is reflected about the y-axis to get point A’.

Question 29.
(-2, -5) → (4, -5)

Answer:
Given,
A(-2, -5) → A'(4, -5)
We observe that y-coordinate of both A and A’ is same but the x-coordinate of A’ is not opposite of A. This means that A’ is the reflection of A about a line x = a.
a = (x2+x1)/2 = (4-2)/2 = 2/2 = 1
Hence the point (-2,-5) is reflected about the line x = 1 to get point (4,-5)

Question 30.
(-3, -4) → (-3, 0)

Answer:
Given,
B(-3, -4) → B'(-3, 0)
We observe that x-coordinate of both B and B’ is the same but the y-coordinate of B’ is not the opposite of B. This means that A’ is the reflection of A about a line x = a.
b = (y2+y1)/2 = (0-4)/2 = -4/2 = -2
Hence the point (-3,-4) is reflected about the line y = -2 to get point (-3,0)

TRANSFORMING FIGURES
Find the coordinates of the figure after the transformations.

Question 31.
Translate the triangle 1 unit right and 5 units down. Then reflect the image in the y-axis.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 40

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given:
R(-4,1)
S(-4,4)
T(-2,1)
a = 1 and b = -5
R(-4,1) = R'(-4+a, 1+b) = R'(-4+1, 1-5) = R'(-3, -4)
S(-4,4) = S'(-4+a, 4+b) = S'(-4+1, 4-5) = S'(-3, -1)
T(-2,1) = T'(-2+a, 1+b) = T'(-2+1, 1-5) = T'(-1, -4)
Thus the coordinates of the image are R'(-3, -4), S'(-3, -1), T'(-1, -4)

Question 32.
Reflect the trapezoid in the x-axis. Then translate the image 2 units left and 3 units up.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 41

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
Given, W(-2,-2), X(-2,1), Y(2,1), and Z(4,-2)
Now reflection about the x-axis:
W(-2,-2) = W'(-2,2)
X(-2,1) = X'(-2,-1)
Y(2,1) = Y'(2,-1)
Z(4,-2) = Z'(4,2)
Thus the coordinates of the image: W'(-2,2), X'(-2,-1), Y'(2,-1), Z'(4,2)
Now translating the above image point:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given:
W(-2,-2), X(-2,1), Y(2,1), and Z(4,-2)
a = -2 and b = 3
W(-2,2) = W”(-2+a, 2+b) = W”(-2-2, 2+3) = W”(-4, 5)
X(-2,-1) = X”(-2+a, -1+b) = X”(-2-2, -1+3) = X”(-4, 2)
Y(2,-1) = Y”(2+a, -1+b) = Y”(2-2, -1+3) = Y”(0, 2)
Z(4,2) = Z”(4+a, 2+b) = Z”(4-2, 2+3) = Z”(2, 5)
Thus the coordinates of the image: W”(-4, 5), X”(-4, 2), Y”(0, 2), Z”(2, 5)

Question 33.
REASONING
In Exercises 31 and 32, is the original figure identical to the final image? Explain.

Answer: Yes, in exercises 31 and 32 the original figure is identical to the final image. Because the type of transformation used is reflection and translation. The shape and size of the image figure do not change when there is reflection or translation. The only position of the image changes in both cases when compared to the position of the original figure.

Question 34.
CRITICAL THINKING
Hold a mirror to the left side of the photo of the vehicle.
a. What word do you see in the mirror?

Answer: The word which we will see in the mirror will be AMBULANCE. Because the word is written in mirror image form on the vehicle.

b. Why do you think it is written that way on the front of the vehicle?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 42

Answer: Ambulance

Explanation:
It is written in that way because the ambulance will be behind any vehicle then the word “AMBULANCE” will correctly appear in the Rear-view mirror of the front vehicle.

Question 35.
DIG DEEPER!
Reflect the triangle in the line y = x. How are the x- and y-coordinates of the image related to the x- and y-coordinates of the original triangle?
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 43

Answer:
When a point is reflected about the line y = x then both x and y-coordinate become opposite.
A(x, y) = A'(-x, -y)
Given,
D(-1,-3)
E(-1,1)
F(-3,1)
Reflection about the line y = x
D(-1,-3) = D'(1,-3)
E(-1,1) = E'(1,-1)
F(-3,1) = F'(3,-1)
Hence the coordinates of the image: D'(1,-3), E'(1,-1), F'(3,-1)
Big Ideas Math Grade 8 Chapter 2 Solution Key img_21

Lesson 2.3 Rotations

EXPLORATION 1
Work with a partner.
a. For each figurebelow, draw the figure in the coordinate plane. Then copy the axes and the figure onto a piece of transparent paper. Turn the transparent paper and align the origin and the axes with the coordinate plane. For each pair of figures, describe the angle of rotation.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 44

  • point
  • triangle
  • line segment
  • rectangle

Big Ideas Math Answers Grade 8 Chapter 2 Transformations 45
b. When you rotate figures, what do you notice about sides, angles, and parallel lines?
c. Describe the relationship between each point below and the point A(3, 6) in terms of rotations.
d. What are the coordinates of a point P(x, y) after a rotation 90° counterclockwise about the origin? 180°? 270°?

2.3 Lesson

Try It
Tell whether the blue figure is a rotation of the red figure about the origin. If so, give the angle and direction of rotation.

Question 1.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 46

Answer: Yes blue figure is the rotation of red figure about the origin.

Explanation:
When we rotate the red figure 180 degrees clockwise or anti-clockwise about the origin we will get the same figure as the blue figure.

Question 2.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 47

Answer: Blue figure is not the rotation of the red figure.

Explanation:
When the red figure is rotated about the origin in any direction the distance of the center point of both the red figure and the blue figure will be the same from the origin point (0,0). The distance between the center of the object and the center of rotation always remains the same.
Thus Blue figure is not the rotation of the red figure.

Try It

The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 3.
J(-4, -4), K(-4, 2), L(-1, 0), M(-2, -3); 180° about the origin

Answer:
When a point is rotated 180 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-x, -y)
Given, J(-4, -4), K(-4, 2), L(-1, 0), M(-2, -3)
Rotation about the origin
J(-4, -4) = J'(4,4)
K(-4, 2) = K'(4,-2)
L(-1, 0) = L'(1,0)
M(-2, -3) = M'(2,3)
Hence the coordinate of the image are J'(4,4), K'(4,-2), L'(1,0), M'(2,3)

Question 4.
P(-3, 2), Q(6, 1), R(-1, -5); 90° counterclockwise about the origin

Answer:
When a point is rotated 90 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-y, x)
Given,
P(-3, 2), Q(6, 1), R(-1, -5)
Rotation about the origin
P(-3, 2) = P'(-2,-3)
Q(6, 1) = Q'(-1,6)
R(-1, -5) = R'(5,-1)
Hence the coordinate of the image is P'(-2,-3), Q'(-1,6), R'(5,-1)

Question 5.
A(5, 3), B(4, -1), C(1, -1); 90° clockwise about the origin

Answer:
When a point is rotated 270 degrees counterclockwise about the origin then both x and y-coordinates gets interchanged and the x-coordinate becomes the opposite.
A(x, y) = A'(y, -x)
Given,
A(5, 3), B(4, -1), C(1, -1)
Rotation about the origin
A(5, 3) = A'(3,-5)
B(4, -1) = B'(-1,-4)
C(1, -1) = C'(-1,-1)
Hence the coordinate of the image are A'(3,-5), B'(-1,-4), C'(-1,-1)

Try It

Question 6.
The vertices of a triangle are P(-1, 2), Q(-1, 0), and R(2, 0). Rotate the triangle 180° about the origin, and then reflect it in the x-axis. What are the coordinates of the image?

Answer:
When a point is rotated 180 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-x, -y)
Given,
P(-1, 2), Q(-1, 0), and R(2, 0)
Rotation about the origin
P(-1, 2) = P'(1,-2)
Q(-1, 0) = Q'(1,0)
R(2, 0) = R'(-2,0)
Hence the coordinate of the image is P'(1,-2), Q'(1,0), R'(-2,0)
Now reflecting above image point about x-axis:
When a point is reflected about the x-axis then the y-coordinate becomes opposite.
A(x, y) = A'(x, -y)
Given,
P'(1,-2), Q'(1,0), R'(-2,0)
Rotation about the origin
P'(1,-2) = P”(1,2)
Q'(1,0) = Q”(1,0)
R'(-2,0) = R”(-2,0)
Hence the coordinate of the image are P'(1,2), Q'(1,0), R'(-2,0)

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
IDENTIFYING A ROTATION
Tell whether the blue figure is a rotation of the red figure about point P. If so, give the angle and direction of rotation.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 48

Answer:
Yes, the blue figure is the rotation of the red figure about the origin.

Explanation:
Because when we will rotate the red figure 90 degrees anti-clockwise about the origin we will get the same figure as the blue figure.
By this, we can say that the blue figure is the result of the rotation of red figure by 90 degrees in the clock or anti-clockwise direction.

Question 8.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 49
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 50

Answer:
The statement which different from all other 3 statement is:
What are the coordinates of the image after a 270 degrees clockwise rotation about the origin?
Now coordinate of both the image are:
The rotation of an object 90 degrees clockwise is equal to the rotation of 270 degrees counterclockwise.
we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanges and x-coordinate becomes opposite.
A(x, y) = A'(y, -x)
Given,
A(2, 4)
B(4, 4)
C(4, 1)
Rotating 90 degrees clockwise about the origin
A(2,4) = A'(4,-2)
B(4, 4) = B'(4,-4)
C(4, 1) = C'(1,-4)
Hence the coordinate of the image are: A'(4,-2), B'(4,-4), C'(1,-4)
Image of statement which different from all 3 statement
The rotation of an object 270 degrees clockwise is equal to the rotation of 90 degrees counterclockwise.
we know that when a point is rotated 90 degrees counterclockwise about origin then both coordinate gets interchanges and x-coordinate becomes opposite
P(x, y) = P'(-y, x)
Given,
A(2, 4)
B(4, 4)
C(4, 1)
Rotating 90 degrees clockwise about the origin
A(2, 4) = A'(-4, 2)
B(4, 4) = B'(-4, 4)
C(4, 1) = C'(-1, 4)
Hence the coordinate of the image are: A'(4,-2), B'(-4,4), C'(-1,4)
Third statement “what are the coordinates of the image after a 270 degrees clockwise rotation about origin?” is different.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
You move the red game piece to the indicated location using a rotation about the origin, followed by a translation. What are the coordinates of the vertices of the game piece after the rotation? Justify your answer.
Big Ideas Math Answers Grade 8 Chapter 2 Transformations 51

Answer:
To move the red game piece in the indicated location the game piee should be rotated 90 degrees in clockwise direction. The rotation of red game piece 90 degrees clockwise is same as the rotation of 270 degrees counterclockwise.
we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanges and x-coordinate becomes opposite.
A(x,y) = A'(y, -x)
Conner point of red game piece:
A(0,-1), B(0,0), C(1,0), D(1,1), E(-2,1), F(-2,0), G(-1,0), H(-1,-1)
Rotating 90 degrees clockwise about the origin:
A(0,-1) = A'(-1,0)
B(0,0) = B'(0,0)
C(1,0) = C'(0,-1)
D(1,1) = D'(1,-1)
E(-2,1) = E'(1,2)
F(-2,0) = F'(0,2)
G(-1,0) = G'(0,1)
H(-1,-1) = H'(-1,1)
Hence the coordinate of corner of red game piece are A'(-1,0), B'(0,0), C'(0,-1), D'(1,-1), E'(1,2), F'(0,2), G'(0,1), H'(-1,1)

Question 10.
DIG DEEPER!
Skytypingis a technique that airplanes use to write messages in the sky. The coordinate plane shows a message typed in the sky over a city, where the positive y-axis represents north. What does the message say? How can you transform the message so that it is read from north to south?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 52

Answer: HELLO

Explanation:
The message above on the coordinate plane can be transformed from north to south by rotating the image 90 degrees anticlockwise.

Rotations Homework & Practice 2.3

Review & Refresh

Tell whether the blue figure is a reflection of the red figure.

Question 1.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 53

Answer: Yes, the blue figure is the reflection of red figure.

Explanation:
Because the blue figure is the exact mirror image of the red figure. If the red figure will be flipped it will result in the blue figure. So, the blue figure is the reflection of red figure.

Question 2.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 54

Answer: No, the blue figure is not the reflection of red figure.

Explanation:
Because the blue figure is not the mirror image of the red figure. If the red figure will be flipped it will not result in the blue figure. So blue figure is not reflection of red figure.

Find the circumference of the object. Use 3.14 or \(\frac{22}{7}\) for π.

Question 3.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 55

Answer:
Given diameter of disk D = 28 cm
Circumference of the circular disk is π × D
C = π × 28
C = 22/7 × 28
C = 22 × 4
C = 88 cm
Thus the circumference is 88 cm.

Question 4.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 56

Answer:
Given the diameter of disk D = 11.4 in
Circumference of the circular disk is π × D
C = π × 11.4
C = 22/7 × 11.4
C = 3.14 × 11.4
C = 35.796 in
Thus the circumference is 35.796 in

Question 5.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 57

Answer:
Given diameter of disk r = 0.5 ft
Circumference of the circular disk is 2π × r
C = 2π × 0.5
C = 6.28 × 0.5
C = 3.14 ft
Thus the circumference is 3.14 ft

Concepts, Skills, &Problem Solving

DESCRIBING RELATIONSHIPS
Describe the relationship between the given point and the point (2, 7) in terms of rotations. (See Exploration 1, p. 55.)

Question 6.
B(7, -2)

Answer:
Given,
A(7, 2) = B(7, -2)
Here we can see that after rotation x and y coordinate are interchanged and the y-coordinate is opposite. And we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinates get interchanged and the x-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Hence the above rotation is 270 degrees counterclockwise about the origin.

Question 7.
C(-7, 2)

Answer:
Given,
A(7, 2) = C(-7, 2)
Here we can see that after rotation x and y coordinate are interchanged and the y-coordinate is opposite. And we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanged and x-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Hence the above rotation is 90 degrees counterclockwise about the origin.

Question 8.
D(-2, -7)

Answer:
Given,
A(2, 7) = C(-2, -7)
Here we can see that after rotation x and y coordinate are interchanged and the y-coordinate is opposite. And we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanged and x-coordinate becomes opposite.
P(x, y) = P'(-x, -y)
Hence the above rotation is 180 degrees counterclockwise about the origin.

IDENTIFYING A ROTATION
Tell whether the blue figure is a rotation of the red figure about the origin. If so, give the angle and direction of rotation.

Question 9.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 58

Answer: No the blue figure is not the rotation of the red figure.

Explanation:
Because if the blue triangle were the result of the rotation of the red triangle then the hypotenuse of the blue triangle should have been parallel to the x-axis. so, it is not the case of rotation.

Question 10.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 59

Answer: Yes, the blue figure is the result of the rotation of the red figure.

Explanation:
Because if the red figure is rotated 90 degrees in a counterclockwise direction it will result in a blue figure.

Question 11.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 60

Answer: Yes, the blue figure is the result of the rotation of the red figure.

Explanation:
If the red figure is rotated 180 degrees in counterclockwise or clockwise direction it will result in blue figure.

Question 12.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 61

Answer: Yes, the blue figure is the result of the rotation of the red figure.

Explanation:
If the red figure is rotated 90 degrees in a clockwise direction it will result in blue figure.

ROTATING A FIGURE
The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 13.
A(2, -2), B(4, -1), C(4, -3), D(2, -4)
90° counterclockwise about the origin

Answer:
We know that when a point is rotated 90 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given,
A(2, -2), B(4, -1), C(4, -3), D(2, -4)
Rotating 90 degrees counterclockwise about the origin
A(2, -2) = A'(2,2)
B(4, -1) = B'(1,4)
C(4, -3) = C'(3,4)
D(2, -4) = D'(4,2)
Hence the coordinates of the image are A'(2,2), B'(1,4), C'(3,4), D'(4,2)

Question 14.
F(1, 2), G(3, 5), H(3, 2) 180° about the origin

Answer:
We know that when a point is rotated 180 degrees counterclockwise or clockwise direction about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-x, -y)
Given,
F(1, 2), G(3, 5), H(3, 2)
Rotating 180 degrees about the origin
F(1, 2) = F'(-1,-2)
G(3, 5) = G'(-3,-5)
H(3, 2) = H'(-3,-2)
Hence the coordinates of the image are F'(-1,-2), G'(-3,-5), H'(-3,-2)

Question 15.
J(-4, 1), K(-2, 1), L(-4, -3)
90° clockwise about the origin

Answer:
The rotation of an object 90 degrees clockwise is equal to the rotation of 270 degrees counterclockwise.
We know that when a point is rotated 270 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Given,
J(-4, 1), K(-2, 1), L(-4, -3)
Rotating 90 degrees clockwise about the origin
J(-4, 1) = J'(1,4)
K(-2, 1) = K'(1,2)
L(-4, -3) = L'(-3,4)
Hence the coordinates of the image are J'(1,4), K'(1,2), L'(-3,4)

Question 16.
P(-3, 4), Q(-1, 4), R(-2, 1), S(-4, 1)
270° clockwise about the origin

Answer:
The rotation of an object 270 degrees clockwise is equal to the rotation of 90 degrees counterclockwise.
We know that when a point is rotated 90 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given,
P(-3, 4), Q(-1, 4), R(-2, 1), S(-4, 1)
Rotating 90 degrees clockwise about the origin
P(-3, 4) = P'(-4,-3)
Q(-1, 4) = Q'(-4,-1)
R(-2, 1) = R'(-1,-2)
S(-4, 1) = S'(-1,-4)
Hence the coordinates of the image are P'(-4,-3), Q'(-4,-1), R'(-1,-2), S'(-1,-4)

Question 17.
W(-6, -2), X(-2, -2), Y(-2, -6), Z(-5, -6)
270° counterclockwise about the origin

Answer:
We know that when a point is rotated 270 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Given,
W(-6, -2), X(-2, -2), Y(-2, -6), Z(-5, -6)
Rotating 90 degrees clockwise about the origin
W(-6, -2) = W'(-2,6)
X(-2, -2) = X'(-2,2)
Y(-2, -6) = Y'(-6,2)
Z(-5, -6) = Z'(-6,5)
Hence the coordinates of the image are W'(-2,6), X'(-2,2), Y'(-6,2), Z'(-6,5)

Question 18.
A(1, -1), B(5, -6), C(1, -6)
90° counterclockwise about the origin

Answer:
We know that when a point is rotated 90 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given,
A(1, -1), B(5, -6), C(1, -6)
Rotating 90 degrees clockwise about the origin
A(1, -1) = A'(1,1)
B(5, -6) = B'(6,5)
C(1, -6) = C'(6,1)
Hence the coordinates of the image are A'(1,1), B'(6,5), C'(6,1)

Question 19.
YOU BE THE TEACHER
The vertices of a triangle are A(4, 4), B(1, -2), and C(-3, 0). Your friend finds the coordinates of the image after a rotation 90° clockwise about the origin. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 62

Answer:
We know that when a point is rotated 270 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Given,
A(4, 4), B(1, -2), and C(-3, 0).
Rotating 90 degrees clockwise about the origin
A(4, 4) = A'(4,-4)
B(1, -2) = B'(-2,-1)
C(-3,0) = C'(0,3)
Hence the coordinates of the image are A'(4,-4), B'(-2,-1), C'(0,3)
By this I can say that my friend is not correct.

Question 20.
PROBLEM SOLVING
A game show contestant spins the prize wheel shown. The arrow remains in a fixed position while the wheel rotates. The wheel stops spinning, resulting in an image that is a rotation 270° clockwise about the center of the wheel. What is the result?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 62.1

Answer: Free spin

Explanation:
The arrow is located at 90 degrees in the counterclockwise direction of free spin. So when the wheel is rotated 270 degrees in a clockwise direction the arrow will be on the free spin column.

PATTERN
A figure has rotational symmetry if a rotation of 180° or less produces an image that fits exactly on the original figure. Determine whether the figure has rotational symmetry. Explain your reasoning.

Question 21.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 63

Answer: Yes the given figure has rotational symmetry.

Explanation:
The given figure in the problem is rotated 120 degrees in any direction clockwise or counterclockwise then it will produce the same identical image. Since 120 degrees is less than 180 degrees so it will have rotational symmetry.

Question 22.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 64

Answer: No the given figure does not have rotational symmetry.

Explanation:
The given figure in the problem will produce the same identical image only when it is rotated 360 degrees. Since 360 degrees is greater than 180 degrees so it will not have rotational symmetry.

Question 23.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 65

Answer: Yes the given figure has rotational symmetry.

Explanation:
The given figure in the problem will produce the same identical image only when it is rotated 180 degrees. Since the maximum angle for rotational symmetry is 180 degrees so it will have rotational symmetry.

USING MORE THAN ONE TRANSFORMATION
The vertices of a figure are given. Find the coordinates of the image after the transformations given.

Question 24.
R(-7, -5), S(-1, -2), T(-1, -5)
Rotate 90° counterclockwise about the origin. Then translate 3 units left and 8 units up.

Answer:
We know that when a point is rotated 90 degrees counterclockwise about origin then both coordinates gets interchanges and y-coordinate becomes opposite.
P(x,y) = P'(-y,x)
Given, R(-7, -5), S(-1, -2), T(-1, -5)
Rotating 90 degrees counterclockwise about the origin
R(-7,-5) = R'(5,-7)
S(-1, -2) = S'(2,-1)
T(-1, -5) = T'(5,-1)
The coordinate of the image are R'(5,-7), S'(2,-1), T'(5,-1)
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given, R'(5,-7), S'(2,-1), T'(5,-1) and a = -3, b = 8
R'(5+a, -7+b) = R”(5-3, -7+8) = R”(2,1)
S'(2+a, -1+b) = R”(2-3, -1+8) = S”(-1,7)
R'(5+a, -1+b) = R”(5-3, -1+8) = T”(2,7)
The coordinate of the image are R”(2,1), S”(-1,7), T”(2,7)

Question 25.
J(-4, 4), K(-3, 4), L(-1, 1), M(-4, 1) Reflect in the x-axis, and then rotate 180° about the origin.

Answer:
We know that when a point is reflected about x-axis then y-coordinate becomes opposite.
A(x, y) = A'(x, -y)
Given J(-4, 4), K(-3, 4), L(-1, 1), M(-4, 1)
Reflection about the x-axis:
J(-4, 4) = J'(-4,-4)
K(-3, 4) = K'(-3,-4)
L(-1, 1) = L'(-1,-1)
M(-4, 1) = M'(-4,-1)
The coordinate of the image are J'(-4,-4), K'(-3,-4), L'(-1,-1), M'(-4,-1)
Now rotating the above image 180 degrees about the origin.
We know that when a point is reflected about x-axis then y-coordinate becomes opposite.
A(x, y) = A'(-x, -y)
J'(-4,-4), K'(-3,-4), L'(-1,-1), M'(-4,-1)
Rotating 180 degrees about the origin:
J'(-4,-4) = J”(4,4)
K'(-3,-4) = K”(3,4)
L'(-1,-1) = L”(1,1)
M'(-4,-1) = M”(4,1)
The coordinate of the image are J”(4,4), K”(3,4), L”(1,1), M”(4,1)

CRITICAL THINKING
Describe two different sequences of transformations in which the blue figure is the image of the red figure.

Question 26.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 66

Answer:
Two different ways of translating a red figure in to blue figure:
1. First rotate the red figure 90 degrees in the counterclockwise direction and then translate that image 5 units towards the left to get the blue figure.
2. First rotate the red figure 90 degrees in a clockwise direction and then translate that image 1 unit towards the Right and 5 units Up to get the blue figure.

Question 27.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 67

Answer:
Two different ways of translating a red figure in to blue figure:
1. First rotate the red figure 90 degrees in the counterclockwise direction and then translate that image 1 unit towards the left and 1 Down to get the blue figure.
2. First rotate the image in the x-axis and then translate that image 4 units towards the left and 2 units Up to get the blue figure.

Question 28.
REASONING
A trapezoid has vertices A(-6, -2), B(-3, -2), C(-1, -4), and D(-6, -4).
a. Rotate the trapezoid 180° about the origin. What are the coordinates of the image?

Answer:
A(x, y) = A'(-x, -y)
Given,
A(-6, -2), B(-3, -2), C(-1, -4), and D(-6, -4).
Rotating 180 degrees about the origin:
A(-6, -2) = A'(6,2)
B(-3, -2) = B'(3,2)
C(-1, -4) = C'(1,4)
D(-6, -4) = D'(6,4)
The coordinates of the image of trapezoid vertices are A'(6,2), B'(3,2), C'(1,4), D'(6,4)

b. Describe a way to obtain the same image without using rotations.

Answer:
In the above question, we can see that the coordinates of all the vertices of the trapezoid are negative and all the coordinates of the image vertices are positive. So there is another way to get the vertices of the image.
First, reflect the trapezoid in the x-axis and then in the y-axis or first reflect the trapezoid in the y-axis and then in the x-axis.

ROTATING A FIGURE
The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 29.
D(2, 1), E(2, -2), F(-1, 4)
90° counterclockwise about vertex D

Answer:
P(x, y) = P'(-(y – b) + a, (x -a) + b)
Given,
D(2, 1), E(2, -2), F(-1, 4)
(a, b) = (2, 1)
Rotation about the point D(2,1)
D(2, 1) = D'(2,1)
E(2, -2) = E'(-(-2-1) + 2, (2 – 2) + 1) = E'(5, 1)
F(-1, 4) = F'(-(4 – 1) + 2, (-1 – 2) + 1) = F'(-1, -2)
Hence the coordinate of the image: D'(2,1), E'(5, 1), F'(-1, -2)

Question 30.
L(-4, -3), M(-1, -1), N(2, -2)
180° about vertex M

Answer:
When a point is rotated 180 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite and ‘b’ and ‘a’ are subtracted from x and y coordinate respectively.
P(x, y) = P'(-(x – a) + b, -(y – b) + a)
Given,
L(-4, -3), M(-1, -1), N(2, -2)
Rotation about the point M(-1, -1):
L(-4, -3) = L'(-(-4 + 1) – 1, -(-3 + 1) – 1) = L'(2, 1)
M(-1, -1) = M'(-1, -1)
N(2, -2) = N'(-(2 + 1) – 1, -(-2 + 1) – 1) = N'(-4, 0)
Hence the coordinate of the image are L'(2, 1), M'(-1, -1), N'(-4, 0)

Question 31.
W(-5, 0), X(-1, 4), Y(3, -1), Z(0, -4)
270° counterclockwise about vertex W

Answer:
When a point is rotated 270 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite and ‘b’ and ‘a’ are subtracted from x and y coordinate respectively.
P(x, y) = P'(-(x – a) + b, -(y – b) + a)
Given,
W(-5, 0), X(-1, 4), Y(3, -1), Z(0, -4)
Rotation about the point W(-5, 0):
W(-5, 0) = W'(-5, 0)
X(-1, 4) = X'((4 – 0) – 5, -(-1 + 5) + 0) = X'(-1, -4)
Y(3, -1) = Y'((-1 – 0) – 5, -(3 + 5) + 0) = Y'(-6, -8)
Z(0, -4) = Z'((-4 – 0) – 5, -(0 + 5) + 0) = Z'(-9, -5)
Hence the coordinate of the image are W'(-5, 0), X'(-1, -4), Y'(-6, -8), Z'(-9, -5)

Question 32.
D(-3, -4), E(-5, 2), F(1, -1), G(3, -7)
270° clockwise about vertex E.

Answer:
When a point is rotated 90 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite and ‘b’ and ‘a’ are subtracted from x and y coordinate respectively.
P(x, y) = P'(-(x – a) + b, -(y – b) + a)
Given,
D(-3, -4), E(-5, 2), F(1, -1), G(3, -7)
Rotation about the point E(-5, 2):
D(-3, -4) = D'(-(-4 – 2) – 5, -(-3 + 5) + 2) = D'(1, 4)
E(-5, 2) = E'(-5, 2)
F(1, -1) = F'(-(-1 – 2) – 5, (1 + 5) + 2) = F'(-2, 8)
G(3, -7) = G'(-(-7 – 2) – 5, (3 + 5) + 2) = G'(4, 10)
Hence the coordinate of the image are D'(1, 4), E'(-5, 2), F'(-2, 8), G'(4, 10)

Question 33.
LOGIC
You want to find the treasure located on the map at Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 68. You are located at Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 69. The following transformations will lead you to the treasure, but they are not in the correct order. Find the correct order. Use each transformation exactly once.

  • Rotate 180° about the origin.
  • Reflect in the y-axis.
  • Rotate 90° counterclockwise about the origin.
  • Translate 1 unit right and 1 unit up.

Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 68.1

Answer:
The correct order of transformation to get the treasure are:

  • Rotate 180° about the origin.
  • Rotate 90° counterclockwise about the origin.
  • Reflect in the y-axis.
  • Translate 1 unit right and 1 unit up.

Question 34.
DIG DEEPER!
You rotate a triangle 90° counterclockwise about the origin. Then you translate its image 1 unit left and 2 units down. The vertices of the final image are (-5, 0), (-2, 2), and (-2, -1). What are the vertices of the original triangle?

Answer:
When a point is rotated 90 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite.
Let the three vertices of the triangle be: (x1, y1), (x2, y2), (x3, y3)
P(x, y) = P'(-y, x)
Rotating 90 degrees counterclockwise about the origin:
A(x1, y1) = A'(-y1, x1)
B(x2, y2) = B'(-y2, x2)
C(x3, y3) = C'(-y3, x3)
Now translating the image of the vertex for the final image
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x, y) = A'(x + a, y + b)
Given,
A'(-y1, x1), B'(-y2, x2), C'(-y3, x3) and a = -1, b = -2
A'(-y1 + a, x1 + b) = A”(-y1 – 1, x1 – 2)
B'(-y2 + a, x2 + b) = B”(-y2 – 1, x2 – 2)
C'(-y3 + a, x3 + b) = C”(-y3 – 1, x3 – 2)
The given coordinate of vertex point of final image are: (-5, 0), (-2, 2) and (-2, -1)
Now comparing the coordinate of the final image
(-y1 – 1, x1 – 2) = (-5, 0) so y1 = 4 and x1 = 2
(-y2 – 1, x2 – 2) = (-2, 2) so y2 = 1 and x2 = 4
(-y3 – 1, x3 – 2) = (-2, 1) so y3 = 1 and x3 = 1
Hence the vertices of original triangle are (2, 4), (4, 1) and (1, 1)

Lesson 2.4 Congruent Figures

EXPLORATION 1

Work with a partner.
a. For each pair of figures whose vertices are given below, draw the figures in a coordinate plane. Then copy one of the figures onto a piece of transparent paper. Use transformations to try to obtain one of the figures from the other figure.

  • A(-5, 1), B(-5, -4), C(-2, -4) and D(1, 4), E(1, -1), F(-2, -1)
  • G(1, 2), H(2, -6), J(5, 0) and L(-1, -2), M(-2, 6), N(-5, 0)
  • P(0, 0), Q(2, 2), R(4, -2) and X(0, 0), Y(3, 3), Z(6, -3)
  • A(0, 4), B(3, 8), C(6, 4), D(3, 0) and
    F(-4, -3), G(-8, 0), H(-4, 3), J(0, 0)
  • P(-2, 1), Q(-1, -2), R(1, -2), S(1, 1) and
    W(7, 1), X(5, -2), Y(3, -2), Z(3, 1)

Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 69.1
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 69.2
b. Which pairs of figures in part(a) are identical? Explain your reasoning.
c. FigureA and FigureB are identical. Do you think there must be a sequence of transformations that obtains Figure A from Figure B? Explain your reasoning.

2.4 Lesson

Try It

Question 1.
A triangle has vertices X(0, 4), Y(4, 4), and Z(4, 2). Is △XYZ congruent to any of the triangles in Example 1? Explain.

Answer:
Big Ideas Math Grade 8 Chapter 2 transformation key img_22
After plotting the triangle XYZ on the coordinate plane we can say that the triangle XYZ is congruent to triangle PQR among all the given triangle in the figure. In fact, if triangle PQR is rotated 90 degrees clockwise of 270 degrees counterclockwise it will result in the triangle XYZ.

Try It

Question 2.
Describing a different Sequence of rigid motions between the figures.

Answer:
Different sequence of rigid motion to get the blue figure from the red figure are:
1. First rotate the red figure 90 degrees clockwise and the origin.
2. Then translate the image 4 units Right and 1 unit Up.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
IDENTIFYING CONGRUENT FIGURES
Use the coordinate plane shown.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 70
a. Identify any congruent figures.

Answer: a. After seeing the figure we can say that triangle ABCD is congruent to triangle JKLM.

b. A rectangle has vertices W(4, 1), X(4, 2), Y(1, 2), and Z(-1, -1). Is Rectangle WXYZ congruent to any of the rectangles in the coordinate plane? Explain.

Answer:
Big ideas math Grade 8 ch 2 solution key img_23
Rectangle WXYZ is not congruent to any of the rectangles in the given figure because rectangle WXYZ is square of 3 units sides and the other rectangle in the figure does not have all the sides of 3 units.

RIGID MOTIONS
The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 4.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 71

Answer:
The sequence of the rigid motions from the red-figure to the blue figure:
1. First we will rotate the red figure 180 degrees clockwise or anticlockwise about the origin because the given red figure in the 4th quadrant and the blue figure is in the 2nd quadrant.
2. Then we will translate the image 1 unit left because one vertex (-1,-4) of red figure is on the negative side of the x-axis.

Question 5.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 72

Answer:
The sequence of the rigid motions from red figure to blue figure:
1. First we will rotate the red figure 90 degrees clockwise about the origin because the given red figure in the 1st quadrant.
2. Then we will translate the image 3 units right and 1 unit down.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
In the coordinate plane at the left, each grid line represents 50 feet. Each figure represents a pasture.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 73
a. Are the figures congruent? Use rigid motions to justify your answer.

Answer: No the blue figure and red figure are not congruent.

Explanation:
By reflecting the red figure about the y-axis and translating the image 4 units Up we will not the same blue figure. So both figure are not congruent to each other.

b. How many feet of fencing do you need to enclose each pasture?

Answer:
Given the length of each grid line = 50 feet
Total feet of fencing = 50 × total number of grid line along the boundary
For red figure fencing: 50 × 12 = 600 feet
For blue figure fencing: 50 × 12 = 600 feet

Question 7.
A home decorator uses a computer to design a floor tile. How can the decorator transform the tile as shown?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 74

Answer:
First, rotate the given tiles about 90 degrees in the clockwise direction and then take the mirror image about the vertical axis.

Congruent Figures Homework & Practice 2.4

Review & Refresh

The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 1.
A(1, 3), B(2, 5), C(3, 5), D(2, 3)
90° counterclockwise about the origin

Answer:
When a point is rotated 90 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-y, x)
Given,
A(1, 3), B(2, 5), C(3, 5), D(2, 3)
Rotating 90 degrees counterclockwise about the origin:
A(1, 3) = A'(-3, 1)
B(2, 5) = B'(-5, 2)
C(3, 5) = C'(-5, 3)
D(2, 3) = D'(-3, 2)
Hence the coordinate of the image are A'(-3, 1), B'(-5, 2), C'(-5, 3), D'(-3, 2)

Question 2.
F(-2, 1), G(-1, 3), H(3, 1)
180° about the origin

Answer:
When a point is rotated 180 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-x, -y)
Given,
F(-2, 1), G(-1, 3), H(3, 1)
Rotating 90 degrees counterclockwise about the origin:
F(-2, 1) = F'(2,-1)
G(-1, 3) = G'(1,-3)
H(3, 1) = H'(-3,-1)
Hence the coordinate of the image are F'(2,-1), G'(1,-3), H'(-3,-1)

Factor the expression using the greatest common factor.

Question 3.
4n – 32

Answer:
4n – 32
Take 4 as a common factor.
4(n – 8)
Thus the greatest common factor is 4(n – 8)

Question 4.
3w + 66

Answer:
3w + 66
Take 3 as a common factor.
3(w + 22)
Thus the greatest common factor is 3(w + 22)

Question 5.
2y – 18

Answer:
2y – 18
Take 2 as a common factor.
2(y – 9)
Thus the greatest common factor is 2(y – 9).

Concepts, Skills, & Problem Solving
TRANSFORMING FIGURES
The vertices of a pair of figures are given. Determine whether the figures are identical. (See Exploration 1, p. 63.)

Question 6.
G(0, 0), H(3, 2), J(1, -2) and L(-1, 0), M(2, 2), N(0, -3)

Answer:
Big Ideas Math 8th Grade Solution Key Chapter 2 img_24
After plotting the triangles GHJ and LMN we can say that the triangle LMN are bigger compared to the other triangle. Thus both the triangles are not identical.

Question 7.
A(-2, -1), B(-2, 2), C(-1, 1), D(-1, -2) and F(-2, 0), G(-1, 1), H(2, 1), J(1, 0)

Answer:
Answer Key for BIM Grade 8 Chapter 2 tranformations img_25
By seeing both the quadrilaterals ABCD and FGHJ we can say that they are identical.

IDENTIFYING CONGRUENT FIGURES
Identify any congruent figures in the coordinate plane.

Question 8.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 75

Answer:
On observing the diagram in the given figure we can see that the shape and size of pentagon ABCDE and pentagon FKJHG are the same. The length of each side of both the pentagon is the same. Thus they are congruent.

Question 9.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 76

Answer:
By seeing the above figure we can say that the shape and size of parallelogram EFGH and parallelogram BCDA are the same. The length of each side of both the parallelogram are same. Parallelogram BCDA can be obtained by rotating parallelogram EFGH 90 degrees clockwise and translating its image. Hence the parallelogram, EFGH is congruent to BCDA.

DESCRIBING A SEQUENCE OF RIGID MOTIONS
The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 10.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 77

Answer:
The sequence of rigid motions between the red and blue figures are:
1. First we will rotate the red figure 90 degrees clockwise about the origin because the given red figure is in the 2nd quadrant and the blue figure is in the 1st quadrant.
2. Then we will translate the image 1 unit left and 1 unit Down because one vertex of the red figure is at (-1, 1)

Question 11.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 78

Answer:
The sequence of rigid motions between the red and blue figures are:
1. First we will rotate the red figure 180 degrees clockwise or anticlockwise about the origin because the given red figure is in 4th quadrant and the blue figure is in the 2nd quadrant.
2. Then we will translate the image 1 unit Right and 1 unit Down because one vertex of red figure is at (2, -2)

Question 12.
YOU BE THE TEACHER
Your friend describes a sequence of rigid motions between the figures. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 79

Answer:
When a point is reflected about x-axis then the y-coordinate becomes opposite.
A(x, y) = A'(x, -y)
Coordinates of red figure are A(1, -1), B(3, -1), C(4, -3), D(2, -3)
Reflection about the x-axis:
A(1, -1) = A'(1, 1)
B(3, -1) = B'(3, 1)
C(4, -3) = C'(4, 3)
D(2, -3) = D'(2, 3)
Now translating the above image point 5 unit left.
Coordinate of the vertex of blue figure are: A”(-4, 1), B”(3,1), C”(-1, 3), D”(-3,3)
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given:
A(1, 1), B(3, 1), C(4, 3), D(2, 3) and a = -5, b = 0
A”(1+a, 1+b) = A”(1-5, 1+0) = A”(-4, 1)
B”(3+a, 1+b) = B”(3-5, 1+0) = B”(-2,1)
C”(4+a,3+b) = C”(4-5, 3+0) = C”(-1, 3)
D”(2+a,3+b) = D”(2-5, 3+0) = D”(-3,3)
Hence the coordinate of image are A”(-4, 1), B”(-2,1), C”(-1, 3), D”(-3,3)
Since the coordinate of the vertex of the blue is the same in both ways.
We can say that the blue figure is obtained by the rigid motion of the red figure.

NAMING CORRESPONDING PARTS
The figures are congruent. Name the corresponding angles and the corresponding sides.

Question 13.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 80

Answer:
Corresponding sides of the congruent figure are
AD = EH
AB = EF
BC = FG
CD = GH
Corresponding angles of the congruent figure are
∠A = ∠E
∠B = ∠F
∠C = ∠G
∠D = ∠H

Question 14.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 81

Answer:
Corresponding sides of the congruent figure are
PQ = WV
QR = VZ
RS = ZY
ST = YX
TP = XW
Corresponding angles of the congruent figure are
∠P = ∠W
∠Q = ∠V
∠R = ∠Z
∠S = ∠Y
∠T = ∠X

Question 15.
MODELING REAL LIFE
You use a computer program to transform an emoji. How can you transform the emoji as shown?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 82

Answer:
First, take the reflection of that emoji about vertical line and then rotate that image 90 degrees clockwise to get that given emoji.

Question 16.
CRITICAL THINKING
Two figures are congruent. Are the areas of the two figures the same? the perimeters? Explain your reasoning.

Answer:

  • The size of both figures should be the same.
  • The shape of both the figures should be the same.
  • All the corresponding angles should be the same.
  • Both the area and perimeter of two congruent figures are the same.

Question 17.
DIG DEEPER!
The houses are identical.
a. What is the length of side LM?

Answer:
Length of LM = length of CD
length of CD = 32 feet
So, the length of LM is 32 feet

b.Which angle of JKLMN corresponds to ∠D?
Answer:
∠D = ∠M
Thus ∠M corresponds to ∠D

c. Side AB is congruent to side AE. What is the length of side AB? What is the perimeter of ABCDE?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 83.

Answer:
AE = JN
The length of JN is 20 ft
So, the length of AE = 20 ft
Perimeter of ABCDE = AB + BC + CD + DE + EA
Perimeter of ABCDE = 20 + 12 + 32 + 12 + 20 = 96 feet
Thus the Perimeter of ABCDE  is 96 feet

Question 18.
REASONING
Two constellations are represented by the figures in the coordinate plane shown. Are the figures congruent? Justify your answer.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 84

Answer:
The above figure can be tranformed into below figure by rotating the figure 180 degrees clockwise or counterclockwise about the origin and translating the image 8 units Right and 8 units Up to get the above figure.

Lesson 2.5 Dilations

Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 85

EXPLORATION 1

Work with a partner. Use geometry software.
a. Draw a polygon in the coordinate plane. Then dilate the polygon with respect to the origin. Describe the scale factor of the image.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 86
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 86.1
b. Compare the image and the original polygon in part(a). What do you notice about the sides? the angles?
c. Describe the relationship between each point below and the point A(x, y) in terms of dilations.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 87
d. What are the coordinates of a point P(x, y) after a dilation with respect to the origin by a scale factor of k?

2.5 Lesson

Try It

Tell whether the blue figure is a dilation of the red figure.

Question 1.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 88

Answer:
No the blue figure is not the dilation of the red figure.

Explanation:
Blue and red figure has same size and same shape but the blue figure is reflection about vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

Question 2.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 89

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding verrtices meeting at a point. This means that blue figure is dilation of the red figure.

Try It

Question 3.
WHAT IF?
Triangle ABC is dilated by a scale factor of 2. What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: A (1, 3), B (2, 3), C (2, 1) and scale factor = 2
Dilating the figure by scale factor of 2
A (1, 3) = A'(1 . 2, 3 . 2) = A'(2, 6)
B (2, 3) = B'(2 . 2, 3 . 2) = B'(4, 6)
C (2, 1) = C'(2 . 2, 1 . 2) =  C'(4, 2)
Hence the coordinates of the image are A'(2, 6), B'(4, 6),  C'(4, 2)

Try It

Question 4.
WHAT IF?
Rectangle WXYZ is dilated by a scale factor of \(\frac{1}{4}\). What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the rectangle: W(-4, -6), X(-4, 8), Y(4, 8), Z(4, -6)
scale factor = 0.25
W(-4, -6) = W'(-4 × 0.25, -6 × 0.25) = W'(-1, -1.5)
X(-4, 8) = X'(-4 × 0.25, 8 × 0.25) = X'(-1, 2)
Y(4, 8) = Y'(4 × 0.25, 8 × 0.25) = Y'(1, 2)
Z(4, -6) = Z'(4 × 0.25, -6 × 0.25) = Z'(1, -1.5)
Hence the coordinates of the image are W'(-1, -1.5), X'(-1, 2), Y'(1, 2), Z'(1, -1.5)

Try It

Question 5.
WHAT IF?
Trapezoid ABCDis dilated using a scale factor of 3, and then rotated 180° about the origin. What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
a is the scaling factor
Given points of trapezoid: A(-2, -1), B(-1,1), C(0,1), D(0,-1) scale factor = 3
Dilating the figure by scale factor of 3
A(-2, 1) = A'(-2 . 3, -1 . 3) = A'(-6, -3)
B(-1, 1) = B'(-1 . 3, 1 . 3) = B'(-3, 3)
C(0, 1) = C'(0 . 3, 1 . 3) = C'(0, 3)
D(0, -1) = D'(0 . 3, -1 . 3) = D'(0, -3)
Thus the coodrinate of the image are A'(-6, -3), B'(-3, 3), C'(0, 3), D'(0, -3)
when a point is rotated 180 degrees about the origin then both x and y coordinate becomes opposite.
P(x, y) = P'(-x, -y)
Image points: A'(-6, -3), B'(-3, 3), C'(0, 3), D'(0, -3)
Rotating 180 degrees about the origin:
A'(-6, -3) = A”(6, 3)
B'(-3, 3) = B”(3, -3)
C'(0, 3) = C”(0, -3)
D'(0, -3) = D”(0, 3)
Thus the coodrinate of the image are A”(6, 3), B”(3, -3), C”(0, -3), D”(0, 3)

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

IDENTIFYING A DILATION
Tell whether the blue figure is a dilation of the red figure.

Question 6.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 90

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is dilation of the red figure.

Question 7.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 91

Answer: No the blue figure is not the dilation of the red figure.

Explanation:
Blue and red figure has same size and same shape but the blue figure is reflection about vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

Question 8.
DILATING A FIGURE
The vertices of a rectangle are J(4, 8), K(12, 8), L(12, 4), and M(4, 4). Draw the image after a dilation with a scale factor of \(\frac{1}{4}\). Identify the type of dilation.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
a is scale factor
Given, vertices of a rectangle are J(4, 8), K(12, 8), L(12, 4), and M(4, 4), scale factor = 0.25
J(4, 8) = J'(4 × 0.25, 8 × 0.25) = J'(1, 2)
K(12, 8) = K'(12 × 0.25, 8 × 0.25) = K'(3, 2)
L(12, 4) = L'(12 × 0.25, 4 × 0.25) = L'(3, 1)
M(4, 4) = M'(4 × 0.25, 4 × 0.25) = M'(1, 1)
Hence the coordinates of the image are J'(1, 2), K'(3, 2), L'(3, 1), M'(1, 1)
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_26

Question 9.
VOCABULARY
How is a dilation different from other transformations?

Answer:
The difference between dilation and other transformations are

  • In the case of dilate the size of the figure after the dilation either decrease or increase but the shape of the figure before and after dilation remains same. Also after the dilation the corresponding angles will be congruent.
  • In case of other transformations such as Rotation, translation, reflection the shape and size of figure before and after transformation remains the same.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
A photograph is dilated to fit in a frame, so that its area after the dilation is 9 times greater than the area of the original photograph. What is the scale factor of the dilation? Explain.

Answer: The scale factor of length and breadth will be 3.

Explanation:
Given,
The area after the dilation is 9 times greater than the area of the original photograph.
Area = length × breadth
p = 3 × 3
Hence the scale factor of length and breadth will be 3.

Question 11.
DIG DEEPER!
The location of a water treatment plant is mapped using a coordinate plane, where each unit represents 1 foot. The plant has vertices (0, 0), (0, 180), (240, 180), and (240, 0). You dilate the figure with a scale factor of \(\frac{1}{3}\). What are the coordinates of the image? What do you need to change so that the image accurately represents the location of the plant? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 92

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
a is the scaling factor
Location of water treatment plant: A(0, 0), B(0, 180), C(240, 180), D(240, 0)
scale factor = \(\frac{1}{3}\)
Dilating the figure by scale factor of \(\frac{1}{3}\)
A(0, 0) = A'(0 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = A'(0, 0)
B(0, 180) = B'(0 × \(\frac{1}{3}\), 180 × \(\frac{1}{3}\)) = B'(0, 60)
C(240, 180) = C'(240 × \(\frac{1}{3}\), 180 × \(\frac{1}{3}\)) = C'(80, 60)
D(240, 0) = D'(240 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = D'(80, 0)
Hence the coordinates of the image are A'(0, 0), B'(0, 60), C'(80, 60), D'(80, 0)

Dilations Homework & Practice 2.5

Review & Refresh

The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 1.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 93

Answer:
Sequence of rigid motion between the red and blue figure are
1. First rotate the blue figure 90 degrees in counterclockwise direction about the orgin because blue figure in 1st quadrant and red figure is in 3rd quadrant.
2. Then translate the image 1 unit left and 4 units down.

Question 2.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 94

Answer:
Sequence of rigid motion between the red and blue figure are
1. First reflect the blue fiure about x-axis. The image after the reflection will lies in the 3rd quadrant with same orientation.
2. Then translate the image 5 units Right.

Tell whether the ratios form a proportion.

Question 3.
3 : 5 and 15 : 20

Answer:
When two ratios are equal then it is called as proportion.
Given,
Given 3 : 5 and 15 : 20
3/5 and 15/20
3/5 and 3/4
Since the above two ratio are not equal hence they are not proportion

Question 4.
2 to 3 and 12 to 18

Answer:
When two ratios are equal then it is called as proportion.
Given,
2 to 3 and 12 to 18
2/3 and 12/18
2/3 and 2/3
Since the above two ratio are equal hence they are proportion.

Question 5.
7 : 28 and 12 : 48

Answer:
When two ratios are equal then it is called as proportion.
Given,
7 : 28 and 12 : 48
7/28 and 12/48
1/4 and 1/4
Since the above two ratio are equal hence they are proportion.

Concepts, Skills, &Problem Solving

DESCRIBING RELATIONSHIPS
Describe the relationship between the given point and the point A(8, 12) in terms of dilations. (See Exploration 1, p. 69.)

Question 6.
B(16, 24)

Answer:
Given a point and its image: A(8,12), B(16, 24)
Here we can see that both x-coordinate and y-coordinate of image point have increased to double.
This means that in this case, the image figure has become larger by the scale factor of 2 with respect to the origin.
Hence the dilation scale factor is 2.

Question 7.
C(2, 3)

Answer:
Given a point and its image: A(8,12), C(2, 3)
Here we can see that both x-coordinate and y-coordinate of image point has decreased to one-fourth.
This means that in this case, the image figure has become smaller by the scale factor of 0.25 with respect to the origin.
Thus the dilation scale factor is 1/4.

Question 8.
D(6, 9)

Answer:
Given a point and its image: A(8,12), D(6, 9)
Here we can see that both x-coordinate and y-coordinate of image point has decreased to three-fourth.
This means that in this case the image figure has become smaller by the scale factor of 0.75 with respect to the origin.
Thus the dilation scale factor is 3/4.

IDENTIFYING A DILATION
Tell whether the blue figure is a dilation of the red figure.

Question 9.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 95

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of the red figure are congruent to the angles of the red figure. Also there will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is a dilation of the red figure.

Question 10.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 96

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of the red figure are congruent to the angles of the red figure. Also there will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is a dilation of the red figure.

Question 11.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 97

Answer: No the blue figure is not the dilation of the red figure.

Explanation:
The Blue and red figure has same size and same shape but the blue figure is a reflection of vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

Question 12.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 98

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is dilation of the red figure.

Question 13.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 99

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is dilation of the red figure.

Question 14.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 100

Answer: No the blue figure is not the dilation of the red figure.

Explanation:
Blue and red figure has same size and same shape but the blue figure is reflection about vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

DILATING A FIGURE
The vertices of a figure are given. Draw the figure and its image after a dilation with the given scale factor. Identify the type of dilation.

Question 15.
A(1, 1), B(1, 4), C(3, 1); k = 4

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: A(1, 1), B(1, 4), C(3, 1) and scale factor = 4
Dilating the figure by scale factor by 4
A(1, 1) = A'(1 × 4, 1 × 4) = A'(4, 4)
B(1, 4) = B'(1 × 4, 4 × 4) = B'(4, 16)
C(3, 1) = C'(3 × 4, 1 × 4) = C'(12, 4)
Hence the coordinate of the image are A'(4, 4), B'(4, 16), C'(12, 4)
Big Ideas Math 8th Grade Solution Key Chapter 2 Transformations img_27

Question 16.
D(0, 2), E(6, 2), F(6, 4); k = 0.5

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: D(0, 2), E(6, 2), F(6, 4) and scale factor = 0.5
Dilating the figure by scale factor by 0.5
D(0, 2) = D'(0 × 0.25, 2 × 0.25) = D'(0, 1)
E(6, 2) = E'(6 × 0.25, 2 × 0.25) = E'(3, 1)
F(6, 4) = F'(6 × 0.25, 4 × 0.25) = F'(3, 2)
Hence the coordinate of the image are D'(0, 1), E'(3, 1), F'(3, 2)
BIM 8th grade solution key for chapter 2 transformations img_28

Question 17.
G(-2, -2), H(-2, 6), J(2, 6); k = 0.25

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: G(-2, -2), H(-2, 6), J(2, 6) and scale factor = 0.25
G(-2, -2) = G'(-2 × 0.25, -2 × 0.25) = G'(-0.5, -0.5)
H(-2, 6) = H'(-2 × 0.25,6 × 0.25) = H'(-0.5, 1.5)
J(2, 6) = G'(2 × 0.25, 6 × 0.25) = J'(0.5, 1.5)
Hence the coordinate of the image are G'(-0.5, -0.5), H'(-0.5, 1.5), J'(0.5, 1.5)
Big Ideas Math Book Answers Grade 8 Chapter 2 Transformations img_29

Question 18.
M(2, 3), N(5, 3), P(5, 1); k = 3

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: M(2, 3), N(5, 3), P(5, 1) and scale factor = 3
M(2, 3) = M'(2 × 3, 3 × 3) = M'(6, 9)
N(5, 3) = N'(5 × 3, 3 × 3) = N'(15, 9)
P(5, 1) = P'(5 × 3, 1 × 3) = P'(15, 3)
Hence the coordinate of the image are M'(6, 9), N'(15, 9), P'(15, 3)
Big Ideas Math Grade 8 Chapter 2 Solution Key img_30

Question 19.
Q(-3, 0), R(-3, 6), T(4, 6), U(4, 0); k = \(\frac{1}{3}\)

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
Q(-3, 0), R(-3, 6), T(4, 6), U(4, 0) and scale factor = \(\frac{1}{3}\)
Q(-3, 0) = Q'(-3 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = Q'(-1, 0)
R(-3, 6) = R'(-3 × \(\frac{1}{3}\), 6 × \(\frac{1}{3}\)) = R'(-1, 2)
T(4, 6) = T'(4 × \(\frac{1}{3}\), 6 × \(\frac{1}{3}\)) = T'(4/3, 2)
U(4, 0) = U'(4 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = U'(4/3, 0)
Hence the coordinate of the image are Q'(-1, 0), R'(-1, 2), T'(4/3, 2), U'(4/3, 0)
Big Ideas Math Grade 8 ch 2 transformations answer key img_31

Question 20.
V(-2, -2), W(-2, 3), X(5, 3), Y(5, -2); k = 5

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
V(-2, -2), W(-2, 3), X(5, 3), Y(5, -2), scaling factor = 5
Dilating the figure by scale factor of 5
V(-2, -2) = V'(-2 × 5, -2 × 5) = V'(-10, -10)
W(-2, 3) = W'(-2 × 5, 3 × 5) = W'(-10, 15)
X(5, 3) = X'(5 × 5, 3 × 5) = X'(25, 15)
Y(5, -2) = Y'(5 × 5, -2 × 5) = Y'(25, -10)
Hence the coordinate of the image are V'(-10, -10),W'(-10, 15), X'(25, 15), Y'(25, -10)
Big Ideas Math Grade 8 ch 2 answer key img_32

Question 21.
YOU BE THE TEACHER
Your friend finds the coordinates of the image of △ABC after a dilation with a scale factor of 2. Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 101

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
The points of triangle: A(2, 5), B(2, 0), C(4, 0)
scale factor = 2
Dilating the figure by scale factor of 2
A(2, 5) = A'(2 × 2, 5 × 2) = A'(4, 10)
B(2, 0) = B'(2 × 2, 0 × 2) = B'(4, 0)
C(4, 0) = C'(4 × 2, 0 × 2) = C'(8, 0)
Hence the coordinate of the image are A'(4, 10), B'(4, 0), C'(8, 0)
By this, we can say that my friend is correct.

FINDING A SCALE FACTOR
The blue figure is a dilation of the red figure. Identify the type of dilation and find the scale factor.

Question 22.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 102

Answer:
Scale factor = side length of image/side length of original figure
Scale factor = A’B’/AB = 6/3 = 2
Scale factor = 2
Hence, type of dilation is enlargement with scale factor of 2.

Question 23.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 103

Answer:
Scale factor = side length of image/side length of original figure
Scale factor = X’Y’/XY= 2/8 = 1/4
Scale factor = 1/4
Hence, the type of dilation is reduction with the scale factor of 1/4

Question 24.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 104

Answer:
Scale factor = side length of image/side length of the original figure
Scale factor = J’K’/JK = 15/10 = 3/2
Scale factor = 3/2
Hence, the type of dilation is reduction with the scale factor of 3/2

Question 25.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 105

Answer:
Scale factor = side length of image/side length of original figure
Scale factor = Q’R’/QR = 4/8 = 1/2
Scale factor = 1/2
Hence, type of dilation is reduction with scale factor of 1/2

USING MORE THAN ONE TRANSFORMATION
The vertices of a figure are given. Find the coordinates of the image after the transformations given.

Question 26.
A(-5, 3), B(-2, 3), C(-2, 1), D(-5, 1)
Reflect in the y-axis. Then dilate using a scale factor of 2.

Answer:
We know that when a point is reflected about y-axis then is x-coordinate becomes opposite.
A(-5, 3), B(-2, 3), C(-2, 1), D(-5, 1)
A(x, y) = A'(-x, y)
A(-5, 3) = A'(5, 3)
B(-2, 3) = B'(2, 3)
C(-2, 1) = C'(2, 1)
D(-5, 1) = D'(5, 1)
Coordinate of the image are A'(5, 3), B'(2, 3), C'(2, 1), D'(5, 1)
A(-5, 3), B(-2, 3), C(-2, 1), D(-5, 1)
Reflect in the y-axis. Then dilate using a scale factor of 2
A'(5, 3) = A”(5 × 2, 3 × 2) = A”(10, 6)
B'(2, 3) = B”(2 × 2, 3 × 2) = B”(4, 6)
C'(2, 1) = C”(2 × 2, 1 × 2) = C”(4, 2)
D'(5, 1) = D”(5 × 2, 1 × 2) = D”(10, 2)
Coordinate of the image are A”(10, 6), B”(4, 6), C”(4, 2), D”(10, 2)
Big ideas math grade 8 chapter 2 solution key img_33

Question 27.
F(-9, -9), G(-3, -6), H(-3, -9).
Dilate using a scale factor of \(\frac{2}{3}\). Then translate 6 units up.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
F(-9, -9), G(-3, -6), H(-3, -9) and scale factor of \(\frac{2}{3}\)
Dilating the figure by scale factor of \(\frac{2}{3}\)
F(-9, -9) = F'(-9 × 2/3, -9 × 2/3) = F'(-6, -6)
G(-3, -6) = G'(-3 × 2/3, -6 × 2/3) = G'(-2, -4)
H(-3, -9) = H'(-3 × 2/3, -9 × 2/3) = H'(-2, -6)
Coordinate of the image are F'(-6, -6), G'(-2, -4), H'(-2, -6)
Now translating above image 6 units up
F'(-6, -6), G'(-2, -4), H'(-2, -6) and a = 0, b = 6
F”(-6 + a, -6 + b) = F”(-6 + 0, -6 + 6) = F”(-6, 0)
G”(-2 + a, -4 + b) = G”(-2 + 0, -4 + 6) = G”(-2, 2)
H”(-2 + a, -6 + b) = H”(-2 + 0, -6 + 6) = H”(-2, 0)
Coordinate of the image are F”(-6, 0), G”(-2, 2), H”(-2, 0)
BIM Grade 8 Answers Chapter 2 img_34

Question 28.
J(1, 1), K(3, 4), L(5, 1)
Rotate 90° clockwise about the origin. Then dilate using a scale factor of 3.

Answer:
The rotation of an object 90 degrees clockwise is equal to the rotation of 270 degrees counterclockwise.
When a point is rotated 270 degrees counterclockwise about the origin then both x and y-coordinates gets interchanged and the x-coordinate becomes the opposite.
A(x, y) = A'(y, -x)
J(1, 1), K(3, 4), L(5, 1)
Rotate 90° clockwise about the origin.
J(1, 1) = J'(1, -1)
K(3, 4) = K'(4, -3)
L(5, 1) = L'(1, -5)
Coordinate of the image are J'(1, -1), K'(4, -3), L'(1, -5)
Now dilate using a scale factor of 3.
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
J'(1, -1) = J”(1 . 3, -1 . 3) = J”(3, -3)
K'(4, -3) = K”(4 . 3, -3 . 3) = K”(12, -9)
L'(1, -5) = L”(1 . 3, -5 . 3) = L”(3, -15)
Coordinate of the image are J”(3, -3), K”(12, -9), L”(3, -15)
BIM 8th Grade Answers Ch 2 transformations img_35

Question 29.
LOGIC
You can use a flashlight and a shadow puppet (your hands) to project shadows on the wall.
a. Identify the type of dilation.p

Answer: The type of dilation is an enlargement

b. What does the flashlight represent?

Answer: Flashlight represents center of dilation because all the line connecting shadow and hand meet at the flashlight.
c. The length of the ears on the shadow puppet is 3 inches. The length of the ears on the shadow is 4 inches. What is the scale factor?

Answer: Scale factor = length of ears on shadow/length of ears on puppet
Scale factor = 4/3
d. Describe what happens as the shadow puppet moves closer to the flashlight. How does this affect the scale factor?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 106

Answer:
As the flashlight will move closer the shadow will become larger. Also, the scale factor will increase.

Question 30.
REASONING
A triangle is dilated using a scale factor of 3. The image is then dilated using a scale factor of \(\frac{1}{2}\). What scale factor can you use to dilate the original triangle to obtain the final image? Explain.

Answer:
Given the first scale factor of triangle S1 = 3
Given second scale factor of triangle S2 = 1/2
We know that the final scale factor S = S1 × S2
Final Scale factor S = 3 × 1/2 = 3/2
Hence, the scale factor of the final image will be the multiplication of the first and second dilation scale factor and the final scale factor will be 3/2.

CRITICAL THINKING
The coordinate notation shows how the coordinates of a figure are related to the coordinates of its image after transformations. What are the transformations? Are the figure and its image congruent? Explain.

Question 31.
(x, y) → (2x + 4, 2y – 3)

Answer:
Given, (x, y) → (2x + 4, 2y – 3)
We can see that both x-coordinate and y-coordinate has been multiplied by 2 this means that the point has been dilated by the scale factor of 2.
Also, 4 has been added to x-coordinate while 3 is added to y-coordinate which means that obtained after the dilation has been translated 4 unit Right and 3 units Down.
The final image will not be congruent because after the dilation the size of the image either increases or decreases that depend on the type of dilation.

Question 32.
(x, y) → (-x – 1, y – 2)

Answer:
Given, (x, y) → (-x – 1, y – 2)
We can see that 1 has been subtracted from x-coordinate while 2 is subtracted from y-coordinate which means that will image has translated 1 unit left and 2 units down. And also x-coordinate is opposite which means the image has been reflected about the y-axis.
Hence, transforms translation of 1 unit left and 2 units down followed by reflection about y-axis.

Question 33.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 107

Answer:
Given, (x, y) → (1/3x, -1/3y)
We can see that both x-coordinate and y-coordinate has been multiplied by 2 this means that the point has been dilated by the scale factor of 1/3. Also, y-coordinate is opposite which means that image obtained after the dilation has been reflected about the x-axis.
Thus transforms are dilation with the scale factor of 1/3 followed by reflection about the x-axis.

STRUCTURE
The blue figure is a transformation of the red figure. Use coordinate notation to describe the transformation. Explain your reasoning.

Question 34.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 108

Answer:
Coordinates of original figure A(1, 1) B(1, 2), C(2, 1)
Coordinates of red figure A'(2, 3) B'(2, 6), C'(4, 3)
Scale factor of x-coordinate = x-coordinate of image/x-coordinate of image = 2/1 = 2
Scale factor of y-coordinate = y-coordinate of image/y-coordinate of image = 3/1 = 3
Thus to transfer the red-figure into the blue figure x-coordinate of all the points has been multiplied by 2 and the y-coordinate of all the points has been multiplied by 3.

Question 35.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 109

Answer:
Coordinates of original figure A(4, 4) B(4, 8), C(8, 8), D(8, 4)
Coordinates of red figure A'(1, 2) B'(1, 4), C'(2, 4), D'(2. 2)
Scale factor of x-coordinate = x-coordinate of image/x-coordinate of image = 1/4 = 0.25
Scale factor of y-coordinate = y-coordinate of image/y-coordinate of image = 2/4 = 1/2 = 0.5
Thus to transfer the red figure into the blue figure x-coordinate of all the points has been multiplied by 0.25 and the y-coordinate of all the points has been multiplied by 0.50

Question 36.
NUMBER SENSE
You dilate a figure using a scale factor of 2, and then translate it 3 units right. Your friend translates the same figure 3 units right and then dilates it using a scale factor of 2. Are the images congruent? Explain.

Answer:
Blue the final image in both the case will be of the same shape and size.
Yes, the image in both cases will be the same.

Question 37.
PROBLEM SOLVING
The vertices of a trapezoid are A(-2, 3), B(2, 3), C(5, -2), and D(-2, -2). Dilate the trapezoid with respect to vertex A using a scale factor of 2. What are the coordinates of the image? Explain the method you used.

Answer:
When the points of a given figure are dilated about a point we simply multiply the distance of each side by the given scale factor. The coordinate of one point remains the same about which dilation occurs.
The vertices of a trapezoid are A(-2, 3), B(2, 3), C(5, -2), and D(-2, -2).
Scale factor = 2
So, here the coordinate of point A(-2, 3) will remains the same but all the other coordinates of points B’, C’, D’ will change according to the distance between each side of the trapezoid.
Big Ideas Math Key Grade 8 Chapter 2 transformations img_36
Image of the figure after dilating by a scale factor of 2
A(-2, 3) = A'(-2, 3)
B(2, 3) = B'(6, 3)
C(5, -2) = C'(12, -7)
D(-2, -2) = D'(-2, -7)
Thus the coordinate of the image are A'(-2, 3), B'(6, 3), C'(12, -7), D'(-2, -7)

Question 38.
DIG DEEPER!
A figure is dilated using a scale factor of -1. How can you obtain the image without using a dilation? Explain your reasoning.

Answer:
When a figure is dilated using a scale factor of -1 then both the x and y-coordinate of the image will become opposite.
Example:
A'(x . -1, y . -1) = A'(-x, -y)
But there are two ways to get the same image:
1. By rotating the figure 180 degrees clockwise or anticlockwise
A(x, y) rotating 180 degrees about the origin = A'(-x, -y)
2. By rotating the figure about the x-axis and y axis
A(x, y) reflecting about the origin = A'(-x, -y)

Lesson 2.6 Similar Figures

EXPLORATION 1

Work with a partner. Use geometry software.
a. For each pair of figures whose vertices are given below, draw the figures in a coordinate plane. Use dilations and rigid motions to try to obtain one of the figures from the other figure.

  • A(-3, 6), B(0, -3), C(3, 6) and G(-1, 2), H(0, 1), J(1, 2)
  • D(0, 0), E(3, 0), F(3, 3) and L(0, 0), M(0, 6), N(-6, 6)
  • P(1, 0), Q(4, 2), R(7, 0) and X(-1, 0), Y(-4, 6), Z(-7, 0)
  • A(-3, 2), B(-1, 2), C(-1, -1), D(-3, -1) and F(6, 4), G(2, 4), H(2, -2), J(6, -2)
  • P(-2, 2), Q(-1, -1), R(1, -1), S(2, 2) and W(2, 8), X(3, 3), Y(7, 3), Z(8, 8)

Big Ideas Math Answers 8th Grade Chapter 2 Transformations 110
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 111
b. Is a scale drawing represented by any of the pairs of figures in part(a)? Explain your reasoning.
c. Figure A is a scale drawing of Figure B. Do you think there must be a sequence of transformations that obtains Figure A from Figure B? Explain your reasoning.

2.6 Lesson

Try It

Question 1.
A triangle has vertices D(0, 4), E(5, 4), and F(5, 0). Is △DEF similar to △ABC and △JKL in Example 1? Explain.

Answer:
Given coordinate of the triangle ABC: A(0, 3), B(3, 3), C(3, 0)
Given coordinate of the triangle DEF: D(0, 4), E(5, 4), F(5, 0)
Given coordinate of the triangle JKL: J(0, 6), K(6, 6), L(6, 0)
Here we can see that there is no fixed relation between the coordinate between triangle ABC and DEF or triangle ABC and JKL. So no triangle is dilation with the triangle ABC.
Hence, triangle ABC is not similar △DEF and △JKL.

Try It

Question 2.
Can you reflect the red figure first, and then perform the dilation to obtain the blue figure? Explain.

Answer:
Because the final image will not depend on the order of transformation. When we will first reflect red figure then the image will be of the same size and after the dilation of the image obtained after reflection we will get the same image so we can use any two method but the final image will be the same

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
IDENTIFYING SIMILAR FIGURES
In the coordinate plane at the left, determine whether Rectangle ABCD is similar to Rectangle EFGH. Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 112

Answer: No rectangle ABCD is not similar to rectangle EFGH

Explanation:
Because the orientation of rectangle ABCD is not the same as the rectangle EFGH. Also, rectangle ABCD is not dilated with rectangle EFGH. So there is no similarity transformation between rectangle ABCD and rectangle EFGH.

Question 4.
SIMILARITY TRANSFORMATION
The red triangle is similar to the blue triangle. Describe a similarity transformation between the figures.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 113

Answer:
Coordinate of left vertex of the red triangle: A(5, 5)
Coordinate of the same vertex after dilation: A'(10, 10)
Now coordinate left vertex of the blue triangle: A”(0, 2)
So, the value of a = 0 – 10 = -10 and b = 2 – 10 = -8
It is given that the red triangle is similar to the blue triangle so the steps of transformation:
First, dilate the figure by the scale factor of 2 and then translate the image 10 units left and 8 units down.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
A medical supplier sells gauze in large and small rectangular sheets. A large sheet has a length of 9 inches and an area of 45 square inches. A small sheet has a length of 4 inches and a width of 3 inches. Are the sheets similar? Justify your answer.

Answer:
Condition for the rectangular sheets to be similar is that all the corresponding sides of bigger and smaller rectangular sheets should be in proportional.
Area of larger rectangular sheets a = 45 sq in
length of larger rectangular sheets l1 = 9 in
Width = a/l = 45/9 = 5 in
length of smaller rectangular sheets l2 = 4 in
width of smaller rectangular sheets b2 = 3 in
condition for similarity l1/l2 = b1/b2
9/4 ≠ 5/3
These sheets are not similar

Question 6.
The sail on a souvenir boat is similar in shape to the sail on a sailboat. The sail on the sailboat is in the shape of a right triangle with a base of 9 feet and a height of 24 feet. The height of the souvenir’s sail is 3 inches. What is the base of the souvenir’s sail?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 114

Answer:
Given,
The base of sail on sailboat b1 = 9 ft
height of sail on sailboat h1 = 24 ft
Given height of sail on souvenir boat: h2 = 3in = 0.25 ft
h1/h2 = b1/b2
24/0.25 = 9/h2
h2 = (9 × 0.25)/24 = 0.9375 ft = 1.125 in
Thus the height of sail of a souvenir boat is 1.125 in

Question 7.
DIG DEEPER!
A coordinate plane is used to represent a cheerleading formation. The vertices of the formation are A(4, 4), B(0, 8), C(4, 4), and D(0, 6). A choreographer creates a new formation similar to the original formation. Three vertices of the new formation are J(-2, -2), K(0, -4), and L(2, -2). What is the location of the fourth vertex? Explain.

Answer:
The vertices of the formation are A(4, 4), B(0, 8), C(4, 4), and D(0, 6). A choreographer creates a new formation similar to the original formation.
We observe the image point carefully that both the x and y coordinate of the image point is just half of the original point and each y-coordinate is opposite.
A(4, 4) = J(-2, -2)
B(0, 8) = K(0, -4)
C(4, 4) = L(2, -2)
D(0, 6) = M(x, y)
This means that the point A, B, C are dilated by using a scale factor of 0.5 and the image obtained from the dilation is reflected about the x-axis.
So, the image point is D(0, 6) = M(0, -3)
Big Ideas Math Answers Grade 8 Chapter 2 Transformations img_37

Similar Figures Homework & Practice 2.6

Review & Refresh

Tell whether the blue figure is a dilation of the red figure.

Question 1.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 115

Answer: No

Explanation:
Because the shape and size of both red and blue figure are the same which is not the property of dilation. The blue figure is the result of the reflection of the red figure 180 degrees in the clockwise or counterclockwise direction.

Question 2.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 116

Answer: Yes

Explanation:
When we see both red and blue figures closely we observe that all the angles of the red figure are congruent to the blue figure. Also, there will be the lines connecting corresponding vertices meeting at a point. This means that the blue figure is a dilation of red figure.

Question 3.
You solve the equation S = lw + 2wh for w. Which equation is correct?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 117

Answer: Option C

Explanation:
S = lw + 2wh
lw + 2wh = S
Taking w as a common factor
w(l + 2h) = s
w = s/(l + 2h)
Thus the correct answer is option C.

Concepts, Skills, &Problem Solving
TRANSFORMING FIGURES
The vertices of a pair of figures are given. Determine whether a scale drawing is represented by the pair of figures. (See Exploration 1, p. 77.)

Question 4.
A(-8, -2), B(-4, 2), C(-4, -2) and G(2, -1), H(4, -1), J(2, -3)

Answer: Yes

Explanation:
BIM Grade 8 Answer Key Chapter 2 Transformations img_38
After plotting both triangles we see that the original figure is exactly double of the image. Each side of the original triangle is double the length of the image triangle. So given vertices pair represent a scale drawing.
Scale factor = 2

Question 5.
A(0, 3), B(3, 4), C(5, 3), D(3, 2) and F(-4, 4), G(-1, 5), H(5, 3), J(3, 2)

Answer: No

Explanation:
BIM Grade 8 Answers Ch 2 img_39
After plotting both given figures we can see that there is no transformation relation between the original figure and the image figure. So, the given vertices pair does not represent a scale drawing.

IDENTIFYING SIMILAR FIGURES
Determine whether the figures are similar. Explain your reasoning.(See Exploration 1, p. 77.)

Question 6.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 118

Answer: Rectangle ABCD is similar to rectangle EFGH.

Explanation:
Because when you see both the given figure we can see that all the corresponding angles of rectangle ABCD and rectangle EFGH are equal. And also the corresponding sides of both rectangle are in proportional.
∠A = ∠E, ∠B = ∠F, ∠C = ∠G, ∠D = ∠H
AB/EF = BC/FG = GH/CD = DA/HE = 1/2
Hence rectangle ABCD is similar to rectangle EFGH.

Question 7.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 119

Answer: Both the triangle are not similar

Explanation:
Because when you see both the given figure we can see that all the corresponding angles of triangle ABC and triangle JKL are equal. And also the corresponding sides of both triangles are not proportional.
AB/JK ≠ KL/BC ≠ CA/LJ
Hence triangle ABC is not similar to triangle JKL

IDENTIFYING SIMILAR FIGURES
Draw the figures with the given vertices in a coordinate plane. Which figures are similar? Explain your reasoning.

Question 8.
Rectangle A: (0, 0), (4, 0), (4, 2), (0, 2)
Rectangle B: (0, 0), (6, 0), (6, 3), (0, 3)
Rectangle C: (0, 0), (4, 0), (4, 2), (0, 2)

Answer: Rectangle A and B are similar

Explanation:
Rectangle A: (0, 0), (4, 0), (4, 2), (0, 2)
Big Ideas Math Grade 8 Chapter 2 Transformations img_39(i)
Rectangle B: (0, 0), (6, 0), (6, 3), (0, 3)
Big Ideas Math Grade 8 Chapter 2 Transformations img_39(ii)
Rectangle C: (0, 0), (4, 0), (4, 2), (0, 2)
Big Ideas Math Grade 8 Chapter 2 Transformations img_39(iii)
By seeing the above figure we can say that rectangle A and rectangle B are similar and rectangle A and Rectangle C are congruent.

Question 9.
FigureA: (4, 2), (2, 2), (2, 0), (4, 0)
Figure B: (1, 4), (4, 4), (4, 1), (1, 1)
Figure C: (2, 1), (5, 1), (5, 3), (2, 3)

Answer: Rectangle A and B are similar

Explanation:
FigureA: (4, 2), (2, 2), (2, 0), (4, 0)
Big Ideas Math Grade 8 Chapter 2 Transformations img_40(i)
Figure B: (1, 4), (4, 4), (4, 1), (1, 1)
Big Ideas Math Grade 8 Chapter 2 Transformations img_40(ii)
Figure C: (2, 1), (5, 1), (5, 3), (2, 3)
Big Ideas Math Grade 8 Chapter 2 Transformations img_40(iii)
Rectangles A and B are similar because in rectangle A and B all the corresponding angles are equal and also all the corresponding sides are equal.

DESCRIBING A SIMILARITY TRANSFORMATION
The red figure is similar to the blue figure. Describe a similarity transformation between the figures.

Question 10.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 120

Answer:
1. First rotate the red figure 90 degrees anticlockwise because the red figure is in the first quadrant and the blue figure is in the second quadrant.
2. Then dilate the image obtained after the rotation by the scale factor of 2 because the blue figure is double the size of the red figure.
Scale factor = side of the image/side of the original figure = 4/2 = 2

Question 11.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 121

Answer:
First, dilate the red figure by the scale figure of 3 because the blue figure is triple the size of red figure.
scale factor = side of image/side of original figure = 6/2 = 3

Question 12.
MODELING REAL LIFE
A barrier in the shape of a rectangle is used to retain oil spills. On a blueprint, a similar barrier is 9 inches long and 2 inches wide. The width of the actual barrier is 1.2 miles. What is the length of the actual barrier?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 122

Answer:
Given,
Width of the actual barrier = 1.2 miles
Width of the barrier in the blueprint = 2 inches
2 inch dimension of blueprint = 1.2 miles of original
So, 1 inch dimension of blueprint = 1.2/2 = 0.6 miles.
Since the length of the barrier in the blueprint = 9 inches,
Thus the length of the actual barrier = 9(0.6) = 5.4 miles.

Question 13.
LOGIC
Are the following figures always, sometimes, or never similar? Explain.
a. two triangles
b. two squares
c. two rectangles

Answer:
a. Two triangles sometimes two triangles are similar when all the corresponding angles are equal and all the corresponding sides lengths are in proportion.
b. Two squares always two square are similar only when all the sides are proportional and all the angles are equal.
c. Two rectangles are similar when all the corresponding angles are equal but the lengths of the corresponding sides are not always in proportion.

Question 14.
CRITICAL THINKING
Can you draw two quadrilaterals each having two 130° angles and two 50° angles that are not similar? Justify your answer.

Answer:
Quadrilateral 1: 50°, 50°, 130°, 130° (trapezoid)
Quadrilateral 2: 50°, 130°, 50°, 130° (parallelogram)
Big Ideas Math Grade 8 ch 2 transformations answer key img_41

Question 15.
REASONING
The sign is rectangular.
a. You increase each side length by 20%. Is the new sign similar to the original? Explain your reasoning.

Answer:
Given,
You increase each side length by 20%
Scale factor = 1 + percentage increase/100 = 1 + 20/100 = 1.2

b. You increase each side length by 6 inches. Is the new sign similar to the original? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 123

Answer:
No, because when length and width are of a different size then adding 6 inches on each side will not increase the figure in a fixed proportion. So the corresponding length will be not proportional. This means that both the figure will be not in dilation, hence not similar.

Question 16.
DIG DEEPER!
A person standing 20 feet from a streetlight casts a shadow as shown. How many times taller is the streetlight than the person? Assume the triangles are similar.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 124

Answer:
Length of shadow l1 = 10 ft
height of man b1 = 6 ft
Total length of bigger triangle l2 = 20 + 10 = 30 ft
l1/l2 = b1/b2
10/30 = 6/b2
b2 = 180/10
b2 = 18 ft
The ratio of the height of streetlight and man is: 18/6 = 3
Hence, the streetlight is 3 times taller than that person.

Question 17.
GEOMETRY
Use a ruler to draw two different isosceles triangles similar to the one shown. Measure the heights of each triangle.
a. Are the ratios of the corresponding heights equivalent to the ratios of the corresponding side lengths?

Answer:
b1/b2 = h1/h2
6/3 = 4/2 = 2
Hence the ratio of corresponding heights is equivalent to the ratio of corresponding side lengths.

b. Do you think this is true for all similar triangles? Explain.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 125

Answer:
Yes, this will be true for all the similar triangles because the heights of the two similar triangles are multiplied by the same amount as the sides.

Question 18.
CRITICAL THINKING
Given △ABC ∼ △DEF and △DEF ∼ △JKL, is △ABC ∼ △JKL? Justify your answer.

Answer:
When △ABC is similar to △DEF and △DEF is similar to △JKL, then △ABC is similar to △JKL.

Lesson 2.7 Perimeters and Areas of Similar Figures

EXPLORATION 1

Work with a partner. Draw a rectangle in the coordinate plane.
a. Dilate your rectangle using each indicated scale factor k. Then complete the table for the perimeter P of each rectangle. Describe the pattern.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 126
b. Compare the ratios of the perimeters to the ratios of the corresponding side lengths. What do you notice?
c. Repeat part(a) to complete the table for the area A of each rectangle. Describe the pattern.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 127
d. Compare the ratios of the areas to the ratios of the corresponding side lengths. What do you notice?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 128
e. The rectangles shown are similar. You know the perimeter and the area of the red rectangle and a pair of corresponding side lengths. How can you find the perimeter of the blue rectangle? the area of the blue rectangle?
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 500

2.7 Lesson

Try It

Question 1.
The height of Figure A is 9 feet. The height of a similar Figure B is 15 feet. What is the value of the ratio of the perimeter of A to the perimeter of B?

Answer: The ratio of the perimeter of A to B is 3/5

Explanation:
We know that when two figures are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = Height of figure A/Height of figure B
Perimeter of figure A/Perimeter of figure B = 9/15 = 3/5
Thus the ratio of the perimeter of A to B is 3/5

Try It

Question 2.
The base of Triangle P is 8 meters. The base of a similar Triangle Q is 7 meters. What is the value of the ratio of the area of P to the area of Q?

Answer:
We know that when two figures are similar then the value of the ratio of their area is equal to the square of the value of the ratio of their corresponding side lengths.
The base of Triangle P is 8 meters. The base of a similar Triangle Q is 7 meters.
b1 = 8 m
b2 = 7 m
Area of triangle P/Area of triangle Q = base of triangle P/base of triangle Q

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

COMPARING PERIMETERS OF SIMILAR FIGURES
Find the value of the ratio (red to blue) of the perimeters of the similar figures.

Question 3.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 130

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
l1 = 9
l2 = 7
Perimeter of red figure/Perimeter of blue figure = Side length of red figure/Side length of blue figure
Perimeter of red figure/Perimeter of blue figure = 9/7
Thus the ratio of the perimeter of red to blue figure is 9/7

Question 4.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 131

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
b1 = 8
b2 = 10
Perimeter of red figure/Perimeter of blue figure = base length of red figure/base length of blue figure
Perimeter of red figure/Perimeter of blue figure = 8/10 = 4/5
Thus the ratio of the perimeter of red to blue triangle is 4/5

COMPARING AREAS OF SIMILAR FIGURES
Find the value of the ratio (red to blue) of the areas of the similar figures.

Question 5.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 132

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = (Side length of figure A/Side length of figure B)²
l1 = 12
l2 = 8
Perimeter of red figure/Perimeter of blue figure = side length of red figure/side length of blue figure
Perimeter of red figure/Perimeter of blue figure = (12/8)² = (3/2)² = 9/4
Thus the ratio of the perimeter of red to blue figure is 9/4

Question 6.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 133

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Area of figure A/Area of figure B = (Side length of A/Side length of B)²
l1 = 12
l2 = 8
Area of red figure/Area of blue figure = side length of red figure/side length of blue figure
Area of red figure/Area of blue figure = (4/5)² = 16/225
Thus the ratio of the perimeter of red to blue triangle is 16/225

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
Two similar triangular regions are prepared for development.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 134
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 136
a. It costs $6 per foot to install fencing. How much does it cost to surround the forest with a fence?

Answer:
Given,
It costs $6 per foot to install fencing.
Perimeter of grassland/perimeter of forest = Height of grassland/Height of forest
h1 = 60 yards
The perimeter of grassland = 240 yards
Height of forest h2 = 45 yards
Perimeter of grassland/perimeter of forest = 60/45
240/ perimeter of forest = 60/45
the perimeter of forest = 180 yards
Convert from yards to feet
180 yards = 540 feet
Thus the cost of fencing forest = 6 × 540 = $3,240

b. The cost to prepare 1 square yard of grassland is $15 and the cost to prepare 1 square yard of forest is $25. Which region costs more to prepare? Justify your answer.

Answer:
Perimeter of grassland/perimeter of forest = (Height of grassland/Height of forest)²
Height of grassland h1 = 60 yard
Height of forest h2 = 45 yards
Area of grassland = 2400 yd²
Cost to prepare 1 sq yd of grassland = $15
Cost to prepare 1 sq yd of forest = $25
Area of forest = (2400 × 9)/16
Thus the area of forest is 1350 yd²
Cost to prepare grassland = $15 × 2400 = $36,000
Cost to prepare of forest = $25 × 1350 = $33,750
Thus the grassland will cost more to prepare.

Question 8.
DIG DEEPER!
You buy a new television with a screen similar in shape to your old television screen, but with an area four times greater. The size of a television screen is often described using the distance between opposite corners of the screen. Your old television has a 30-inch screen. What is the size of your new television screen? Explain.

Answer:
Area of ΔABC/Area of ΔDEF = (Side length of AB/Side length of DE)²
Let the area of the screen of old television be x
Let the area of the screen of new television be 4x
l1 = 30 in
Area of the screen of new television/Area of the screen of new television= (distance of the screen of new television/distance of the screen of old television)²
4x/x = (distance of the screen of new television/30)²
distance of the screen of new television = 30 × 2 = 60 inch
Hence the distance of the screen of the new television is 60 inches.

Perimeters and Areas of Similar Figures Homework & Practice 2.7

Review & Refresh

The red figure is similar to the blue figure. Describe a similarity transformation between the figures.

Question 1.
Big Ideas Math Solutions Grade 8 Chapter 2 Transformations 501

Answer:
First, dilate the red figure using the scale factor of 3 because the side lengths of the blue figure are 3 times the side length of the red figure.
Scale factor = 6/2 = 3
Now reflect the image obtained after a dilation about the y-axis because both red and blue triangle is facing each other.

Question 2.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 137

Answer:
First, dilate the red figure using the scale factor of 0.5 because the side lengths of the blue figure are 3 times the side length of the red figure.
Scale factor = 2/4 = 0.5
Then rotate the image obtained after dilation in direction 90 degrees clockwise about the origin.

Find the area of the figure.

Question 3.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 138

Answer:
We know that the formula for the area of trapezoid = Base × height
h = 16 cm
b = 9 cm
Area of figure = 16 × 9 = 144 sq. cm
Hence the area of the given figure is 144 sq. cm

Question 4.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 139

Answer:
h = 5 in
b = 3 in
We know that,
A = 1/2 × b × h
A = 1/2 × 5 × 3
A = 7.5 sq. cm
Thus the area of the given figure is 7.5 sq. cm

Question 5.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 140

Answer:
h = 5 km
b1 = 6 km
b2 = 8 km
We know that,
Area of trapezoid = 1/2 × h × (b1 + b2)
A = 1/2 × 5 × 14 = 35 sq. km
Hence the area of the trapezoid is 35 sq. km

Concepts, Skills, &Problem Solving
COMPARING SIMILAR FIGURES
Dilate the figure using the indicated scale factor k. What is the value of the ratio (new to original) of the perimeters? the areas? (See Exploration 1, p. 83.)

Question 6.
a triangle with vertices (0, 0), (0, 2), and (2, 0); k = 3

Answer:
When the points of a given figure are dilated we simply multiply each x-coordinate and y-coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
where a is the scale factor
Given a triangle with vertices (0, 0), (0, 2), and (2, 0); k = 3
A(0, 0) = A'(0 . 3, 0 . 3) = A'(0, 0)
B(0, 2) = B'(0 . 3, 2 . 3) = B'(0, 6)
C(2, 0) = C'(2 . 3, 0 . 3) = C'(6, 0)
The coordinates of the image are A'(0, 0), B'(0, 6), C'(6, 0)
AB = √(2 – 0)² – (0 – 0)² = 2
A’B’ = √(6 – 0)² – (0 – 0)² = 6
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of new triangle/Perimeter of the original triangle = Distance of A’B’/Distance of AB = 6/2 = 3
Area of new triangle/Area of the original triangle = (Distance of A’B’/Distance of AB)² = (6/2)² = 3² = 9

Question 7.
a square with vertices (0, 0), (0, 4), (4, 4), and (4, 0); k = 0.5

Answer:
When the points of a given figure are dilated we simply multiply each x-coordinate and y-coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
where a is the scale factor
a square with vertices (0, 0), (0, 4), (4, 4), and (4, 0); k = 0.5
A(0, 0) = A'(0 . 0.5, 0 . 0.5) = A'(0, 0)
B(0, 4) = B'(0 . 0.5, 4 . 0.5) = B'(0, 2)
C(4, 4) = C'(4 . 0.5, 4 . 0.5) = C'(2, 2)
D(4, 0) = D'(4 . 0.5, 0 . 0.5) = D'(2, 0)
Coordinates of the image are A'(0, 0), B'(0, 2), C'(2, 2), D'(2, 0)
AB = √(4 – 0)² + (0 – 0)² = 4
A’B’ = √(2 – 0)² + (0 – 0)² = 2
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of new square/Perimeter of the original square = Distance of A’B’/Distance of AB = 2/4 = 1/2
Area of new square /Area of the original square = (Distance of A’B’/Distance of AB)² = (2/4)² = 1/4

PERIMETERS AND AREAS OF SIMILAR FIGURES
Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.

Question 8.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 141

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure= 11/6
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (11/6)² = 121/36

Question 9.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 142

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure= 5/8
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (5/8)² = 25/64

Question 10.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 143

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure= 4/7
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (4/7)² = 16/49

Question 11.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 144

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure = 14/9
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (14/9)² = 196/81

USING SIMILAR FIGURES
The figures are similar. Find x.

Question 12.
The ratio of the perimeters is 7 : 10.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 145

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of the figure B = Distance of figure A/Distance of figure B
7/10 = x/12
x = 84/10
x = 8.4
Thus the value of x is 8.4

Question 13.
The ratio of the perimeters is 8 : 5.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 146

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of the figure B = Distance of figure A/Distance of figure B
8/5 = x/16
x = 25.6
Thus the value of x is 25.6

Question 14.
COMPARING AREAS
The playing surfaces of two foosball tables are similar. The ratio of the corresponding side lengths is 10:7. What is the ratio of the areas?

Answer:
Area of figure A /Area of figure B = (Distance of figure A/Distance of figure B)²
Area of figure A /Area of figure B = (10/7)²
Area of figure A /Area of figure B = 100/49
Hence, the ratio of their areas is 100/49

Question 15.
CRITICAL THINKING
The ratio of the side length of Square A to the side length of Square B is 4:9. The side length of Square A is 12 yards. What is the perimeter of Square B?

Answer:
Given,
The ratio of the side length of Square A to the side length of Square B is 4/9.
The side length of Square A is 12 yards.
side length of Square A/side length of Square B = 4/9
12 /side length of Square B = 4/9
side length of Square B = 27 yards
We know that,
The perimeter of the square is = 4s
The perimeter of the square B = 4 × 27 = 108 yards

Question 16.
MODELING REAL LIFE
The cost of the piece of fabric shown is $1.31. What would you expect to pay for a similar piece of fabric that is 18 inches by 42 inches?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 147

Answer:
Given,
l = 21 in
w = 9 in
Area of the rectangle = l × w
A = 21 × 9 = 189 sq. in
The cost of the piece of fabric shown is $1.31
The cost of 1 sq. in of fabric = 1.31/189
l = 18 in
b = 42 in
Area of new fabric = 18 × 42 = 756 sq. in
Given the cost of new fabric = 1.31/189 × 756 = $5.24
Hence the cost of the new fabric is $5.24

Question 17.
PROBLEM SOLVING
A scale model of a merry-go-round and the actual merry-go-round are similar.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 148
a. How many times greater is the base area of the actual merry-go-round than the base area of the scale model? Explain.

Answer:
Radius of model merry go round = 6 in
Radius of actual merry go round = 10 ft = 120 in
Area of base of actual merry/Area of base of model merry = (Radius of actual merry/Radius of model merry)²
Area of base of actual merry/Area of base of model merry = (120/6)² = 400

b. What is the base area of the actual merry-go-round in square feet?

Answer:
The radius of model merry go round = 6 in
Radius of actual merry go round = 10 ft = 120 in
Area of base of actual merry = 450 sq. in
Area of base of actual merry/Area of base of model merry = (Radius of actual merry/Radius of model merry)²
Area of base of actual merry/450 = (120/6)² = 400
Area of base of actual merry = 400 × 450 = 180000 sq. in = 1250 ft²

Question 18.
STRUCTURE
The circumference of Circle K is π. The circumference of Circle L is 4π. What is the value of the ratio of their circumferences? of their radii? of their areas?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 149

Answer:
Given,
The circumference of Circle K is π
The circumference of Circle L is 4π.
circumference of Circle = 2πr
2πr = π
The radius of circle K r1 = 1/2
2πr = 4π
The radius of circle K r2 = 2
The ratio of their circumference = π/4π = 1/4
The ratio of radius of both circle = 1/4
The ratio of their area = π(r1)²/π(r2)² = 1/16

Question 19.
GEOMETRY
A triangle with an area of 10 square meters has a base of 4 meters. A similar triangle has an area of 90 square meters. What is the height of the larger triangle?

Answer:
Given,
A triangle with an area of 10 square meters has a base of 4 meters.
A similar triangle has an area of 90 square meters.
Area of the triangle = bh/2
h = 2a/b
h = (2 × 10)/4
h = 5 meters
Area of larger triangle/Area of smaller triangle = (height of larger triangle/height of smaller triangle)²
90/10 = (height of larger triangle/5)²
3 = (height of larger triangle/5)
Thus the height of larger triangle = 3 × 5 = 15 meters

Question 20.
PROBLEM SOLVING
You need two bottles of fertilizer to treat the flower garden shown. How many bottles do you need to treat a similar garden with a perimeter of 105 feet?
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 150

Answer:
The sides of the above figure are 4ft, 15 ft, 18 ft, 5 ft
Perimeter = 4ft + 15 ft + 18 ft + 5 ft = 42 ft
Number of bottle of fertilizer used in the above garden = 2
Number of bottle of fertilizer used in 1 ft = 2/42
Fertilizer used for 105 ft = 2/42 × 105 = 5
Thus 5 bottles will be used for 105 feet

Question 21.
REPEATED REASONING
Three square mirrors are used for a light reflection experiment. The ratio of the side length of Mirror A to the side length of Mirror B is 5 : 6. The ratio of the area of Mirror B to the area of Mirror C is 16 : 25. The perimeter of Mirror C is 280 centimeters. What is the area of Mirror A? Justify your answer.

Answer:
Given,
Three square mirrors are used for a light reflection experiment.
The ratio of the side length of Mirror A to the side length of Mirror B is 5 : 6 = 5/6
The ratio of the area of Mirror B to the area of Mirror C is 16 : 25 = 16/25
The perimeter of Mirror C is 280 centimeters
Side length of mirror C = Perimeter/4 = 280/4 = 70 cm
(Side length of mirror B/Side length of mirror C)² = Area of mirror A/Area of mirror B
(Side length of mirror B/Side length of mirror C)² = 16/25
(Side length of mirror B/Side length of mirror C) = 4/5
The side length of mirror B = 4/5 × Side length of mirror C
Side length of mirror B = 4/5 × 70 = 56 cm
A = s × s
A = 56 cm × 56 cm = 3136 sq. cm
Area of mirror A/Area of mirror B = (Side length of mirror A/Side length of mirror B)²
Area of mirror A/3136 = (5/6)²
Area of mirror A = 25/36 × 3136
Area of mirror A = 2177.7 sq. cm

Transformations Connecting Concepts

2 Connecting Concepts

Using the Problem-Solving Plan

Question 1.
A scale drawing of a helipad uses a scale of 1 ft : 20 ft. The scale drawing has an area of 6.25 square feet. What is the area of the actual helipad?
Understand the problem.
You know the scale of the drawing and the area of the helipad in the drawing. You are asked to find the area of the actual helipad.
Make a plan.
A scale drawing is similar to the actual object. So, use the scale 1 ft : 20 ft and the ratio 6.25 ft2 : A ft2 to write and solve a proportion that represents the area A of the actual helipad.
Solve and Check.
Use the plan to solve the problem. Then check your solution.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 150.1

Answer: 125 sq. ft

Question 2.
The locations of three cargo ships are shown in the coordinate plane. Each ship travels at the same speed in the same direction. After 1 hour, the x- and y-coordinates of Ship A increase 80%. Use a translation to describe the change in the locations of the ships. Then find the new coordinates of each ship.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 151
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 152

Question 3.
All circles are similar. A circle with a radius of 2 inches is dilated, resulting in a circle with a circumference of 22π inches. What is the scale factor? Justify your answer.

Answer:
Given,
A circle with a radius of 2 inches is dilated, resulting in a circle with a circumference of 22π inches.
C = 2π . r
22π = 2π . 2
p = 2π . 2
Thus the scale factor is 2.

Performance Task

Master Puppeteer

At the beginning of this chapter, you watched a STEAM Video called “Shadow Puppets.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 153

Transformations Connecting Concepts

2 Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 154

Graphic Organizers
You can use a Summary Triangle to explain a concept. Here is an example of Summary Triangle for translating a figure.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 155

Choose and complete a graphic organizer to help you study the concept.

  1. reflecting a figure
  2. rotating a figure
  3. congruent figures
  4. dilating a figure
  5. similar figures
  6. perimeters of similar figures
  7. areas of similar figures

Big Ideas Math Answers 8th Grade Chapter 2 Transformations 156

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 157

2.1 Translations (pp. 43–48)

Tell whether the blue figure is a translation of the red figure.

Question 1.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 158

Answer: No

Explanation:
The answer is no because in the case of translation the size of the figure does not change, only the position of the figure changes. But here the size of the blue figure is larger as compared to the red figure so this is not the translation. Here the blue figure is the result of the dilation of red figure.

Question 2.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 159

Answer: Yes

Explanation:
Yes, because in the case of translation the size of the figure does not change, only the position of the figure changes. Here the size of both blue figure and the red figure is the same but there is only a change in the position of the red figure to get blue figure. Here the blue figure is the result of the translation of the red figure.

Question 3.
The vertices of a quadrilateral are W(1, 2), X(1, 4), Y(4, 4), and Z(4, 2). Draw the figure and its image after a translation 3 units left and 2 units down.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: W(1, 2), X(1, 4), Y(4, 4), and Z(4, 2) and a = -3, b = -2
W'(1+a, 2+b) = W'(1-3, 2-2) = W'(-2,0)
X'(1+a, 4+b) = X'(1-3, 4-2) = X'(-2,2)
Y'(4+a, 4+b) = Y'(4-3, 4-2) = Y'(1, 2)
Z'(4+a, 2+b) = C'(4-3, 2-2) = Z'(1,0)
Hence the coordinate of image are W'(-2,0), X'(-2,2), Y'(1, 2), Z'(1,0)
big ideas math answers grade 8 chapter 2 img_41

Question 4.
The vertices of a triangle are A(-1, -2), B(-2, 2), and C(-3, 0). Draw the figure and its image after a translation 5 units right and 1 unit up.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(-1, -2), B(-2, 2), and C(-3, 0) and a = 5, b = 1
A'(-1+a, -2+b) = A'(-1-5, -2+1) = A'(4,-1)
B'(-2+a, 2+b) = B'(-2+5, 2+1) = B'(3,3)
C'(-3+a, 0+b) = C'(-3+5, 0+1) = C'(2, 1)
Hence the coordinate of image are A'(4,-1), B'(3,3), C'(2, 1)
BIM Grade 8 Chapter 2 Answer Key img_42

Question 5.
Your locker number is 20 and your friend’s locker number is 33. Describe the location of your friend’s locker relative to the location of your locker.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 159.1

Answer:
The location of my friend’s locker is first 1 locker Down and then 3 locker Right.

Question 6.
Translate the triangle 4 units left and 1 unit down. What are the coordinates of the image?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 160

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(3, 5), B(6, 3), and C(4, 1) and a = -4, b = -1
A'(3+a, 5+b) = A'(3-4, 5-1) = A'(-1,4)
B'(6+a, 3+b) = B'(6-4, 3-1) = B'(2,2)
C'(4+a, 1+b) = C'(4-4, 1-1) = C'(0, 0)
Hence the coordinate of image are A'(-1,4), B'(2,2), C'(0, 0)
BIM 8th Grade Answer Key Chapter 2 img_42

Question 7.
Describe a translation of the airplane from point A to point B.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 161

Answer:
First, move the aeroplane 6 units right from point A and then 4 units down.

2.2 Reflections (pp. 49 – 54)

Tell whether the blue figure is a reflection of the red figure.

Question 8.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 162

Answer:
No, because in the above figure the plane of reflection is inclined at 45 degrees with the horizontal line. So, the reflected figure will be perpendicular to the original figure. But in the given figure both are facing each other. This means that the blue figure is not the reflection of red figure.

Question 9.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 163

Answer:
The answer is no because the blue figure is not the mirror image of the red figure. The side of the red figure is not facing the side of the blue figure this means that the blue figure is not the reflection of red figure.

Question 10.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 163.1

Answer:
The answer is yes because the blue figure is the mirror image of the red figure. The side of the red figure is facing the side of the blue figure this means that the blue figure is the reflection of red figure.

Draw the figure and its reflection in (a) the x-axis and (b) the y-axis. Identify the coordinates of the image.

Question 11.
A(2, 0), B(1, 5), C(4, 3)

Answer:
A(x, y) = A'(x, -y)
Given: A(2, 0), B(1, 5), C(4, 3)
Reflection about x-axis:
A(2, 0) = A'(2, 0)
B(1, 5) = B'(1, -5)
C(4, 3) = C'(4, -3)
BIM Grade 8 Answers Chapter 2 img_43
Reflection about y-axis:
A(x, y) = A'(-x, y)
A(2, 0) = A'(-2, 0)
B(1, 5) = B'(-1, 5)
C(4, 3) = C'(-4, 3)
BIM Grade 8 Answers Chapter 2 img_44

Question 12.
D(-5, -5), E(-5, 0), F(-2, -2), G(-2, -5)

Answer:
Given, D(-5, -5), E(-5, 0), F(-2, -2), G(-2, -5)
Reflection about x-axis:
A(x, y) = A'(x, -y)
D(-5, -5) = D'(-5, 5)
E(-5, 0) = E'(-5, 0)
F(-2, -2) = F'(-2, 2)
G(-2, -5) = G'(-2, 5)
BIM Grade 8 Answers Chapter 2 img_45
Reflection about y-axis:
A(x, y) = A'(-x, y)
D(-5, -5) = D(5, -5)
E(-5, 0) = E'(5, 0)
F(-2, -2) = F'(2, -2)
G(-2, -5) = G'(2, -5)
BIM Grade 8 Answers Chapter 2 img_46

Question 13.
The vertices of a rectangle are E(-1, 1), F(-1, 3), G(-5, 3), and H(-5, 1). Find the coordinates of the figure after reflecting in the x-axis, and then translating 3 units right.

Answer:
We know that when a point is reflected about x-axis then y-coordinate becomes the opposite.
A(x, y) = A'(x, -y)
The vertices of a rectangle are E(-1, 1), F(-1, 3), G(-5, 3), and H(-5, 1).
Reflection about x-axis:
E(-1, 1) = E'(-1, -1)
F(-1, 3) = F'(-1, -3)
G(-5, 3) = G'(-5, 3)
H(-5, 1) = H'(-5, 1)
Thus the coordinates of the image are E'(-1, -1), F'(-1, -3), G'(-5, 3), H'(-5, 1)
Now translating the image 3 units Right.
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: E'(-1, -1), F'(-1, -3), G'(-5, 3), H'(-5, 1) a = 3, b = 0
E”(-1 + a, -1 + b) = E”(-1 + 3, -1 + 0) = E”(2, -1)
F”(-1 + a, -3 + b) = F”(-1 + 3, -3 + 0) = F”(2, -3)
G”(-5 + a, 3 + b) = G”(-5 + 3, 3 + 0) = G”(-2, 3)
H”(-5 + a, 1 + b) = H”(-5 + 3, 1 + 0) = H”(-2, 1)
Thus the coordinates of the image are E”(2, -1), F”(2, -3), G”(-2, 3), H”(-2, 1)

The coordinates of a point and its image after a reflection are given. Identify the line of reflection.

Question 14.
(-1, -3) → (1, -3)

Answer:
Given,
(-1, -3) → (1, -3)
We can see that the y-coordinate of both points and its image are the same but the x-coordinate of the image is the opposite of its points.
Hence, Y-axis is the line of reflection.

Question 15.
(2, 1) → (2, -1)

Answer:
Given,
(2, 1) → (2, -1)
We can see that the x-coordinate of both points and its image are the same but the y-coordinate of the image is the opposite of its points.
Hence, X-axis is the line of reflection.

Question 16.
You perform an experiment involving angles of refraction with a laser pen. You point a laser pen from point L at a mirror along the red path and the image is a reflection in the y-axis.
a. Does the light reach a cat at point C? Explain.

Answer:
Yes, the light will reach at point C.
Because the coordinate of point L is (4, 3) and the coordinate of point C is (-4, 3) and it is given problems that laser is reflected about the y-axis.
So when point L(4, 3) is reflected about the y-axis its x-coordinates become opposite and y-coordinates remain the same.
So when point L(4, 3) is reflected about the y-axis its image will be point C(-4, -3)

b. You bounce the light off the mirror so its path is a reflection. What line of reflection is needed for the light to reach the cat?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 164

Answer: The line of reflection will be y-axis.

2.3 Rotations (pp. 55–62)

Tell whether the blue figure is a rotation of the red figure about the origin. If so, give the angle and the direction of rotation.

Question 17.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 165

Answer:
The answer is no because the blue figure is the mirror image of the red figure. The blue figure is the result of the reflection of red figure about the y-axis. Also, both red and blue figure are facing each other with the y-axis in the center of both which remains that it is not the case of rotation.

Question 18.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 166

Answer:
The answer is yes, because the red figure is in the 1st quadrant and the blue figure in third quadrant. Also, both blue and red figure are facing each other in opposite directions which blue figure is the result of the rotation of red figure.
When red figure is rotated 180 degrees counterclockwise it will result in a blue figure.

The vertices of a triangle are A(-4, 2), B(-2, 2), and (-3, 4). Rotate the triangle as described. Find the coordinates of the image.

Question 19.
180° about the origin

Answer:
We know that when a point is rotated 180 degrees about origin then both x coordinate and y coordinate becomes opposite.
A(x, y) = A'(-x, -y)
Given points: A(-4, 2), B(-2, 2), and C(-3, 4)
Rotated 180 degrees about origin:
A(-4, 2) = A'(4, -2)
B(-2, 2) = B'(2, -2)
C(-3, 4) = C'(3, -4)
The coordinate of the image are A'(4, -2), B'(2, -2), C'(3, -4)

Question 20.
270° clockwise about the origin

Answer:
We know that when a point is rotated 90 degrees counterclockwise about origin then both x coordinate and y coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given points: A(-4, 2), B(-2, 2), and C(-3, 4)
Rotating 270 degrees clockwise about the origin:
A(-4, 2) = A'(-2, -4)
B(-2, 2) = B'(-2, -2)
C(-3, 4) = C'(-4, -3)
The coordinate of the image are A'(-2, -4), B'(-2, -2), C'(-4, -3)

Question 21.
A bicycle wheel is represented in a coordinate plane with the center of the wheel at the origin. Reflectors are placed on the bicycle wheel at points (7, 4) and (-5, -6). After a bike ride, the reflectors have rotated 90° counterclockwise about the origin. What are the locations of the reflectors at the end of the bike ride?

Answer:
We know that when a point is rotated 90 degrees counterclockwise about origin then both x coordinate and y coordinate becomes opposite.
P(x, y) = P'(-y, x)
Reflectors are placed on the bicycle wheel at points (7, 4) and (-5, -6)
A(7, 4) = A'(-4, 7)
B(-5, -6) = B'(6, -5)
Hence the new coordinate of the reflector are A'(-4, 7), B'(6, -5)

2.4 Congruent Figures (pp. 63–68)

Identify any congruent figures in the coordinate plane.

Question 22.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 167

Answer:
AB = ED
BC = DC
CA = CE
∠A = ∠E
∠B = ∠D
∠C = ∠C
When we see both triangles ABC and EDC closely we observe that both the triangles are the mirror images of each other with the y-axis as the line of reflection. So all the corresponding sides are equal and also all the corresponding angles, this means that both the triangle are congruent.
Hence ΔABC is congruent to ΔEDC
GH = JK
HF = KI
FG = IJ
∠G = ∠J
∠H = ∠K
∠F = ∠I
So we can see that all the corresponding sides are equal and also all the corresponding angles. this means that both the triangles are congruent.
Hence ΔGHF is congruent to ΔJKI

Question 23.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 168

Answer:
When we observe square ABCD and square EFGH we can see that
AB = EF
BC = FG
CD = GH
DA = HE
∠A = ∠E
∠B = ∠F
∠C = ∠G
∠D = ∠H
We can see that all the corresponding sides are equal and also all the corresponding angles, this means that both are congruent.
RS = IJ
ST = JK
TU = KL
UR = LI
∠R = ∠I
∠S = ∠J
∠T = ∠K
∠L = ∠U
Hence rectangle RSTU is congruent to rectangle IJKL

The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 24.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 169

Answer:
First, rotate the blue figure 90° clockwise because the blue figure is in a vertical position but the red figure is in the horizontal position.
Translate the image 5 units right because the first image formed after the rotation will in the second quadrant but the red figure in the first quadrant.

Question 25.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 170

Answer:
First, reflect the blue figure about the y-axis because both red and blue figure is facing each other and they are the mirror image of each other.
Then translate the image 2 units up because the first image formed after reflection will be at the same distance from the x-axis but the red figure is touchung the x-axis.

Question 26.
The figures are congruent. Name the corresponding angles and the corresponding sides.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 171

Answer:
Corresponding sides are
AB = KL
BC = LM
CA = MK
Corresponding angles
∠A = ∠K
∠B = ∠L
∠C = ∠M

Question 27.
Trapezoids EFGH and QRST are congruent.
a. What is the length of side QR ?

Answer:
Length of side:
QR = EF = 3 feet

b. Which angle in QRST corresponds to ∠H?
Answer:
The angle that corresponds to ∠H is ∠T

c. What is the perimeter of QRST ?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 172

Answer:
Perimeter of QSRT = QR + RS + ST + TQ
= EF + FG + GH + HE
= 3 + 5 + 4 + 8
= 20 ft

2.5 Dilations (pp. 69–76)

Tell whether the blue figure is a dilation of the red figure.

Question 28.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 173

Answer:
The answer is no because dilation the size of the image either increases or decreases that depend on the type of dilation. We can see that both the red and blue figure are of the same size which means that the blue figure is not dilation of the red figure.

Question 29.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 174

Answer:
The answer is yes because both red and blue figures are of the same shape and all the corresponding angles are equal but the blue figure is larger than the red figure. And in dilation, the size of the image is either increases or decreases but the shape always remains the same. So blue figure is the result of dilation of red figure.

The vertices of a figure are given. Draw the figure and its image after a dilation with the given scale factor. Identify the type of dilation.

Question 30.
P(-3, -2), Q(-3, 0), R(0, 0); k = 4

Answer:
A(x, y) = A'(x . a, y . a)
Where a is the scaling factor
Given,
P(-3, -2), Q(-3, 0), R(0, 0); k = 4
P(-3, -2) = P'(-3 . 4, -2 . 4) = P'(-12, -8)
Q(-3, 0) = Q'(-3 . 4, 0 . 4) = Q'(-12, 0)
R(0, 0) = R'(0 . 4, 0 . 4) = R'(0, 0)
Thus the coordinates of the image: P'(-12, -8), Q'(-12, 0), R'(0, 0)
Bigideas Math Answer Key for Grade 8 Chapter 2 img_45

Question 31.
B(3, 3), C(3, 6), D(6, 6), E(6, 3); k = \(\frac{1}{3}\)

Answer:
A(x, y) = A'(x . a, y . a)
Where a is the scaling factor
Given,
B(3, 3), C(3, 6), D(6, 6), E(6, 3); k = \(\frac{1}{3}\)
B(3, 3) = B'(3 . \(\frac{1}{3}\), 3 . \(\frac{1}{3}\)) = B'(1, 1)
C(3, 6) = C'(3 . \(\frac{1}{3}\), 6 . \(\frac{1}{3}\)) = C'(1, 2)
D(6, 6) = D'(6 . \(\frac{1}{3}\), 6 . \(\frac{1}{3}\)) = D'(2, 2)
E(6, 3) = E'(6 . \(\frac{1}{3}\), 3 . \(\frac{1}{3}\)) = E'(2, 1)
Thus the coordinates of the image: B'(1, 1), C'(1, 2), D'(2, 2), E'(2, 1)
BIm Grade 8 Chapter 2 Answers img_46

Question 32.
The blue figure is a dilation of the red figure. Identify the type of dilation and find the scale factor.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 175

Answer:
AB = 1 unit
A’B’ = 2 units
scale factor = size of image figure/size of actual figure
scale factor = 2/1 = 2
We can see from the above figure that the size of the image figure are larger as compared to the size of the original image so it is the Enlargement dilation.

Question 33.
The vertices of a rectangle are Q(-6, 2), R(6, 2), S(6, -4), and T(-6, -4). Dilate the rectangle with respect to the origin using a scale factor of \(\frac{3}{2}\). Then translate it 5 units right and 1 unit down. What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the rectangle: Q(-6, 2), R(6, 2), S(6, -4), and T(-6, -4), scale factor = \(\frac{3}{2}\)
Q(-6, 2) = Q'(-6 . \(\frac{3}{2}\), 2 . \(\frac{3}{2}\)) = Q'(-9, 3)
R(6, 2) = R'(6 . \(\frac{3}{2}\), 2 . \(\frac{3}{2}\)) = R'(9, 3)
S(6, -4) = S'(6 . \(\frac{3}{2}\), -4 . \(\frac{3}{2}\)) = S'(9, 6)
T(-6, -4) = T'(-6 . \(\frac{3}{2}\), -4 . \(\frac{3}{2}\)) = T'(-9, -6)
Thus the coordinates of the image: Q'(-9, 3), R'(9, 3), S'(9, 6), T'(-9, -6)
Q'(-9, 3), R'(9, 3), S'(9, 6), T'(-9, -6) a = 5, b = -1
Q”(-9 + a, 3 + b) = Q”(-9 + 5, 3 – 1) = Q”(-4, 2)
R”(9 + a, 3 + b) = R”(9 + 5, 3 – 1) = R”(14, 2)
S”(9 + a, -6 + b) = S”(9 + 5, -6 – 1) = S”(14, -7)
T”(-9 + a, -6 + b) = T”(-9 + 5, -6 – 1) = T”(-4, -7)
Thus the coordinates of the image: Q”(-4, 2), R”(14, 2), S”(14, -7), T”(-4, -7)

2.6 Similar Figures

Question 34.
Determine whether the two figures are similar. Explain your reasoning.

Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 175.1

Answer:
No, the above two figures are not similar.

Question 35.
Draw figures with the given vertices in a coordinate plane. Which figures are similar? Explain your reasoning.
Triangle A: (-4, 4), (-2, 4), (-2, 0)
Triangle B: (-2, 2), (-1, 2), (-1, 0)
Triangle C: (6, 6), (3, 6), (3, 0)

Answer:
Triangle A: (-4, 4), (-2, 4), (-2, 0)
BIM Grade 8 Solution Key Ch 2 img_48
Triangle B: (-2, 2), (-1, 2), (-1, 0)
Bigideas Math Answers Grade 8 Ch 2 img_49
Triangle C: (6, 6), (3, 6), (3, 0)
BIM Answers for Grade 8 Chapter 2 img_50

The figures are similar. Find x.

Question 36.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 177

Answer:
Ratio of sides of larger triangle = Ratio of sides of smaller triangle
20/14 = x/7
x = 10
Thus the value of x is 7 inches.

Question 37.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 178

Answer:
Ratio of sides of larger parallelogram= Ratio of sides of smaller parallelogram
x/6 = 6/4
x = 9
Thus the value of x is 9 cm

2.7 Perimeters and Areas of Similar Figures (pp. 83-88)

Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.

Question 38.
Big Ideas Math Answers 8th Grade Chapter 2 Transformations 179

Answer:
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
Perimeter of red figure/Perimeter of blue figure = 6/8 = 3/4
Hence the ratio of perimeter of red triangle to blue is 3/4
Area of figure A/Area of figure B = (Side length of A/Side length of B)²
Area of red figure/Area of blue figure = (6/8)² = 9/16
Thus the ratio of the perimeter of the red to the blue triangle is 9/16

Question 39.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 180

Answer:
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
Perimeter of red figure/Perimeter of blue figure = 28/16 = 7/4
Hence the ratio of the perimeter of red rectangle to blue is 7/4
Area of figure A/Area of figure B = (Side length of A/Side length of B)²
Area of red figure/Area of blue figure = (28/16)² = 49/16
Thus the ratio of the perimeter of the red to the blue rectangle is 49/16

The figures are similar. Find x.

Question 40.
The ratio of the perimeters is 5 :7.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 181

Answer:
5/7 = 12/x
x =(12 × 7)/5
x = 16.8 cm
Thus the value of x is 16.8 cm

Question 41.
The ratio of the perimeters is 6 : 5.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 182

Answer:
6/5 = x/6
x = 36/5
x = 7.2
Thus the value of x is 7.2

Question 42.
Two photos are similar. The ratio of the corresponding side lengths is 3 : 4. What is the ratio of the areas?

Answer:
Area of photo A/Area of photo B = (Side length of photo A/Side length of photo B)²
Area of photo A/Area of photo B = (3/4)² = 9/16
Thus the ratio of the area of two photos is 9/16

Question 43.
The ratio of side lengths of Square A to Square B is 2 : 3. The perimeter of Square A is 16 inches. What is the area of Square B?

Answer:
Given,
The ratio of side lengths of Square A to Square B is 2 : 3.
The perimeter of Square A is 16 inches.
Perimeter of Square A/Perimeter of Square B = Side length of Square A/Side length of Square B
The perimeter of Square B = (16 × 3)/2
The perimeter of Square B = 24 inches
Side length of square B = Perimeter/4 = 24/4 = 6 inch
Area of square B = s × s = 6 × 6 = 36 sq. in

Question 44.
The TV screen is similar to the computer screen. What is the area of the TV screen?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 183

Answer:
Area of computer screen = 108 sq. in
The side length of the computer screen = 12 in
The side length of the TV screen = 20 in
Area of TV screen = (25 × 108)/9
Area of TV screen = 300 sq. in
Hence the area of the TV screen is 300 sq. in

Transformations Practice Test

2 Practice Test

Triangles ABC and DEF are congruent.

Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 183.1

Question 1.
Which angle of DEF corresponds to ∠C ?

Answer:
Corresponding sides
Side AB = Side DE
Side BC = Side EF
Side CA = Side ED
Corresponding angles
∠A = ∠D
∠B = ∠E
∠C = ∠F
Thus the angle correspond to ∠C is ∠F

Question 2.
What is the perimeter of DEF ?

Answer:
Corresponding sides
Side AB = Side DE
Side BC = Side EF
Side CA = Side ED
Corresponding angles
∠A = ∠D
∠B = ∠E
∠C = ∠F
Perimeter of DEF = DE + EF + FD
= AB + BC + CA
= 5 + 4 + 6
= 15 cm
Thus the perimeter of the ΔDEF is 15 cm

Tell whether the blue figure is a translation, reflection, rotation, or dilation of the red figure.

Question 3.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 184

Answer: The scale factor of a dilation is greater than 1 because the shape of the blue figure is larger than the red figure.

Question 4.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 185

Answer:
The blue figure is the reflection of the red figure. Because the blue figure is the mirror image of red figure. Also, the shape and size of both red and blue figures are the same.

Question 5.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 186

Answer:
The blue figure is the translation of the red figure. Because the shape and size of both red and blue figures are the same. And both the figure is not facing to each other. This means that the blue figure is the result of the translation of the red figure.

Question 6.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 187

Answer:
The blue figure is the result of the rotation of the red figure. Because the shape and size of both the red and blue figure are the same but the figure is horizontal and the blue figure is vertical. This means that the blue figure is the result of the rotation of the term figure.

The vertices of a triangle are A(2, 5), B(1, 2), and C(3, 1). Find the coordinates of the image after the transformations given.

Question 7.
Reflect in the y-axis.

Answer:
A(x, y) = A'(-x, y)
A(2, 5), B(1, 2), C(3, 1)
Reflection about the y-axis:
A(2, 5) = A'(-2, 5)
B(1, 2) = B'(-1, 2)
C(3, 1) = C'(-3, 1)
Thus the coordinates of the image: A'(-2, 5), B'(-1, 2), C'(-3, 1)

Question 8.
Rotate 90° clockwise about the origin.

Answer:
A(x, y) = A'(x, -y)
Given, A(2, 5), B(1, 2), C(3, 1)
A(2, 5) = A'(2, -5)
B(1, 2) = B'(1, -2)
C(3, 1) = C'(3, -1)
Thus the coordinates of the image: A'(2, -5), B'(1, -2), C'(3, -1)

Question 9.
Reflect in the x-axis, and then rotate 90° counterclockwise about the origin.

Answer:
A(x, y) = A'(x, -y)
Given, A(2, 5), B(1, 2), C(3, 1)
A(2, 5) = A'(2, -5)
B(1, 2) = B'(1, -2)
C(3, 1) = C'(3, -1)
Thus the coordinates of the image: A'(2, -5), B'(1, -2), C'(3, -1)
A'(2, -5) = A”(2, 5)
B'(1, -2) = B”(1, 2)
C'(3, -1) = C”(3, 1)
Thus the coordinates of the image: A”(2, 5), B”(1, 2), C”(3, 1)

Question 10.
Dilate with respect to the origin using a scale factor of 2. Then translate 2 units left and 1 unit up.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: A(2, 5), B(1, 2), C(3, 1) and scale factor = 2
Dilating the figure by scale factor of 2
A (2, 5) = A'(2 . 2, 5 . 2) = A'(4, 10)
B (1, 2) = B'(1 . 2, 2 . 2) = B'(2, 4)
C (3, 1) = C'(3 . 2, 1 . 2) =  C'(6, 2)
Hence the coordinates of the image are A'(4, 10), B'(2, 4),  C'(6, 2)
Now translating image 2 unit left and 1 unit up.
Given: A'(4, 10), B'(2, 4),  C'(6, 2) a = -2, b = 1
A”(4 + a, 10 + b) = A”(4 – 2, 10 + 1) = A”(2, 11)
B”(2 + a, 4 + b) = B”(2 – 2, 4 + 1) = B”(0, 5)
C”(6 + a, 2 + b) = C”(6 – 2, 2 + 1) = C”(4, 3)
Hence the coordinates of the image are A”(2, 11), B”(0, 5), C”(4, 3)

Question 11.
In a coordinate plane, draw Rectangle A: (-4, 4), (0, 4), (0, 2), (-4, 2); Rectangle B: (-2, 2), (0, 2), (0, 1), (-2, 1); and Rectangle C:(-6, 6), (0, 6), (0, 3), (-6, 3). Which figures are similar? Explain your reasoning.

Answer:
Rectangle A: (-4, 4), (0, 4), (0, 2), (-4, 2)
BIM Grade 8 Chapter 2 Solutions img_51
Rectangle B: (-2, 2), (0, 2), (0, 1), (-2, 1)
BIM Grade 8 Chapter 2 Solutions img_52
Rectangle C:(-6, 6), (0, 6), (0, 3), (-6, 3)
BIM Grade 8 Chapter 2 Solutions img_53

Question 12.
Translate a point (x, y) 3 units left and 5 units up. Then translate the image 5 units right and 2 units up. What are the coordinates of the point after the translations?

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given: A(x,y) and a = -3, b = 5
A'(x+a, y+b) = A'(x – 3, y + 5)
Now translating image 5 units right and 2 units up.
Image after first translation: A'(x – 3, y + 5) and a = 5, b = 2
A”(x – 3 + a, y + 5 + b) = A”(x – 3 + 5, y + 5 + 2) = A”(x + 2, y + 7)
Thus the final image will be A”(x + 2, y + 7)

Question 13.
The two figures are similar.
(a) Find the value of x.

Answer:
Ratio of sides of red figure = Ratio of sides of blue figure
x/14 = 10/8
x = (10 × 14)/8
x = 17.5

(b) Find the values of the ratios (red to blue) of the perimeters and of the areas.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 188

Answer:
Perimeter of red figure/Perimeter of blue figure = Side length of red figure/Side length of blue figure
Perimeter of red figure/Perimeter of blue figure = 14/8 = 7/4
Thus the ratio of the perimeter of red to blue figure is 7/4
Area of red figure/Area of blue figure = (side length of red figure/side length of blue figure)²
Area of red figure/Area of blue figure = (14/8)² = 49/16
Thus the ratio of the perimeter of red to blue triangle is 49/16

Question 14.
A wide-screen television measures 36 inches by 54 inches. A movie theater screen measures 42 feet by 63 feet. Are the screens similar? Explain.

Answer:
Given,
A wide-screen television measures 36/54 = 2/3
A movie theater screen measures 42/63 = 2/3
We can see that the ratio of corresponding sides of the television screen is equal to the ratio of corresponding sides of the movie theatre. So television screens and movie theatres are similar.

Question 15.
You want to use the rectangular piece of fabric shown to make a pair of curtains for your window. Name the types of congruent shapes you can make with one straight cut. Draw an example of each type.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 189

Answer:
The types of congruent shapes that can be made with one straight cut
2 right triangles
2 rectangles
2 right trapezoid

Transformations Cumulative Practice

Cumulative Practice

Question 1.
A clockwise rotation of 90° is equivalent to a counterclockwise rotation of how many degrees?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 190

Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 191

Answer:
90° of clockwise rotation = (360 – 90)° of counterclockwise rotation
= 270° of counterclockwise rotation

Question 2.
The formula K = C + 273.15 converts temperatures from degrees Celsius C to Kelvin K. Which of the following formulas is not correct?
A. K – C = 273.
B. C = K – 273.15
C. C – K = -273.15
D. C = K + 273.15

Answer: C = K + 273.15

Question 3.
You want to solve the equation -3(x + 2) = 12x. What should you do first?
F. Subtract 2 from each side.
G. Add 3 to each side.
H. Multiply each side by -3.
I. Divide each side by -3.

Answer: I. Divide each side by -3.

Explanation:
-3(x + 2) = 12x
x + 2 = -4x
x = -4x – 2
x + 4x = -2
5x = -2
x = –\(\frac{2}{5}\)
Thus the correct answer is option I.

Question 4.
Which value of x makes the equation \(\frac{3}{4} x\) = 12 true?
A. 9
B. 11\(\frac{1}{4}\)
C. 16
D. 48

Answer: C. 16

Explanation:
\(\frac{3}{4} x\) = 12
3x = 12 × 4
3x = 48
x = \(\frac{48}{3}\)
x = 16
Thus the correct answer is option C.

Question 5.
A triangle is graphed in the coordinate plane. What are the coordinates of the image after a translation 3 units right and 2 units down?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 191.1
F. A'(1, 4), B'(1, 1), C'(3, 1)
G. A'(1, 2), B'(1, -1), C'(3, -1)
H. A'(-2, 2), B'(-2, -1), C'(0, -1)
I. A'(0, 1), B'(0, -2), C'(2, -2)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
A(-2, 4), B(-2, 1), C(0, 1) and a = 3, b = -2
A'(-2+a, 4+b) = A'(-2 + 3, 4 – 2) = A'(1, 2)
B'(-2+a, 1+b) = B'(-2 + 3, 1 – 2) = B'(1,-1)
C'(0+a, 1+b) = C'(0 + 3, 1 – 2) = C'(3, -1)
Coordinate of the image are: A'(1, 2), B'(1,-1), C'(3, -1)
Thus the correct answer is option G.

Question 6.
Your friend solved the equation in the box shown. What should your friend do to correct the error that he made?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 192
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 193

Answer:
–\(\frac{x}{3}\) + \(\frac{2}{5}\) = –\(\frac{7}{15}\)
–\(\frac{x}{3}\) = –\(\frac{13}{15}\)
x = 2\(\frac{3}{5}\)
Thus the correct answer is option C.

Question 7.
Your teacher dilates the rectangle using a scale factor of \(\frac{1}{2}\).
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 194
What is the area of the dilated rectangle in square inches?

Answer:
l = 10 in
b = 6 in
scale factor = \(\frac{1}{2}\)
New length after dilation = 10 × \(\frac{1}{2}\) = 5
New breadth after dilation = 6 × \(\frac{1}{2}\) = 3
Area of rectangle = l × b
A = 5 × 3 = 15 sq. in
The area of the dilated rectangle will be 5 in²

Question 8.
Your cousin earns $9.25 an hour at work. Last week she earned $222.00 How many hours did she work last week?
F. \(\frac{1}{24}\)
G. 22 hours
H. 24 hours
I. 212.75 hours

Answer: H. 24 hours

Explanation:
Given,
Your cousin earns $9.25 an hour at work.
Last week she earned $222.00
Total no. of working hour = total earning of week/rate of one hour
= \(\frac{222}{9.25}\)
= 24 hours
Thus the correct answer is option H.

Question 9.
Triangle EFG is a dilation of Triangle HIJ. Which proportion is not true for Triangle EFG and Triangle HIJ ?
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 195
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 196

Answer: \(\frac{EG}{HI}\) = \(\frac{FG}{IJ}\)
The correct answer is option B.

Question 10.
The red figure is congruent to the blue figure. Which of the following is a sequence of rigid motions between the figures?
F. Reflect the red triangle in the x-axis, and then translate it 3 units left.
G. Reflect the red triangle in the x-axis, and then translate it 3 units right.
H. Reflect the red triangle in the y-axis, and then translate it 3 units left.
I. Rotate the red triangle 90° clockwise about the origin.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 197

Answer:
1. First red triangle is reflected about the x-axis because both red and blue triangles are the mirror image of each other and also the red triangle is in the 1st quadrant and the blue triangle is in 4th quadrant.
2. Then translate the image 3 unit left because the base of both red and blue triangles is not opposite to each other.
Thus the correct answer is option F.

Question 11.
Several transformations are used to create the pattern.
Part A
Describe the transformation of Triangle GLM to Triangle DGH

Answer:
Both ΔGLM and Δ DGH are of the same shape and size but their position are different so the transformation will be translated.

Part B
Describe the transformation of Triangle ALQ to Triangle GLM.

Answer:
The size of the triangle ALQ is four times the size of triangle GLM and the shape of both triangles is the same so the transformation will be dilation.

Part C
Triangle DFN is a dilation of Triangle GHM. Find the scale factor.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 198
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 198.1

Answer:
The size of the triangle DFN is double the size of triangle GHM. So the scale factor of dilation will be 2.

Question 12.
A rectangle is graphed in the coordinate plane.
Big Ideas Math Answer Key Grade 8 Chapter 2 Transformations 199
What are the coordinates of the image after a reflection in the y-axis?
A. J'(4, 1), K'(4, 3), L'( 1, 3), M'(-1, 1)
B. J'(-4, 1), K(-4, -3), L'(1, -3), M'(1, 1)
C. J'(1, 4), K'(3, 4), L'(3, -1), M'(1, -1)
D. J'(-4, 1), K'(-4, 3), L'(1, 3), M'(1, 1)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x, y) = A'(-x, y)
J(-4, 1), K(-4, 3), L(1, 3), M(1, 1)
Reflection about the y-axis:
J(-4, 1) = J'(4, 1)
K(-4, 3) = K'(4, 3)
L(1, 3) = L'(-1, 3)
M(1, 1) = M'(-1, 1)
Coordinate of image are: J'(4, 1), K'(4, 3), L'(-1, 3), M'(-1, 1)
Thus the correct answer is option A.

Conclusion:

I wish the information provided in the above article regarding the Big Ideas Math Grade 8 Chapter 2 Transformations Answer Key is beneficial for all middle school students. Make use of the given links and score good marks in the exams. Share this pdf with your friends and help them to overcome the difficulties in maths. And also bookmark our page to get the latest updates of all Big Ideas Math Grade 8 Answers.

Big Ideas Math Answers Grade 7 Advanced | Big Ideas Math Modeling Real Life Grade 7 Answer Key

Big Ideas Math Answers Grade 7 Advanced

Big Ideas Math Book Grade 7 Advanced Answers provided helps students position for success and keeps them on right track. Practicing from the BIM Math Grade 7 Advanced Solutions helps you improve your learning experience at Chapter Level and Lesson Level. Gain a deeper understanding of concepts using Grade 7 Advanced Big Ideas Math Answers and attempt the exam with confidence and score better grades in the exams. Download the BIM Grade 7 Advanced Textbook Solutions PDF for free and take your preparation to the next level.

Big Ideas Math Book 7th Grade Advanced Answer Key | BIM Math 7th Grade Advanced Answers

Below is a comprehensive collection of Chapterwise Big Ideas Math Book Grade 7 Advanced Answers. The BIM Math Book 7th Grade Advanced Solutions provided here covers the questions from the BIM Textbooks and are as per the latest Common Core State Standards. BIM Math 7th Grade Advanced Answer Key provided includes questions belonging to Chapter Tests, Review Tests, Cumulative Practice, Assessment Tests, etc.

Benefits of Learning from Big Ideas Math Answer Key Grade 7 Advanced

Follow the below sections to know the importance of Big Ideas Math Answers and how it can help you in your preparation. They are in the below fashion

  • BIM 7th Grade Advanced Answer Key is provided by people of high subject expertise.
  • All the Grade 7 Advanced Big Ideas Math Solutions provided meets the Common Core State Standards Curriculum.
  • Big Ideas Math Grade 7 Advanced Answers are provided in PDF format and you can download them for free of cost.
  • Big Ideas Math 7th Grade Advanced Solution Key is aligned as per the Textbooks and includes Questions from Practice Tests, Assessments, Chapter Tests, Cumulative Practice, Review Tests, etc.

FAQs on Common Core BIM Grade 7th Advanced Solutions

1. Is there any portal that provides the Big Ideas Math 7th Grade Advanced Answers for free?

Yes, Bigideasmathanswer.com is a trusted portal that offers the BIM Grade 7 Advanced Answer Key free of cost.

2. How to download BIM Grade 7 Advanced Answer Key for all Chapters?

You can download the BIM Grade 7 Advanced Solutions for all Chapters by simply clicking on the respective links available on our page.

3. Does practicing from the BIM Book Grade 7 Advanced Textbook Solutions help you score well in exams?

Yes, Practicing from the Big Ideas Math Book 7th Grade Advanced Answers help you score well in exams. However, complement your preparation with trusted preparation sources.

Big Ideas Math Answers Grade 4 | Big Ideas 4th Grade Math Book Answer Key

Big Ideas Math Grade 4 Answers

Students must prepare with Big Ideas Math Answers Grade 4 to get a good knowledge of the concept. You can find plenty of ways to practice Big Ideas Grade 4 problems. Also, we provided practice tests, objective questions, step-by-step explanations, etc., for the best practice of the students. Students can easily clear all their queries by referring to Big Ideas Grade 4 Textbook Solutions. You don’t need to pay a single penny to use Big Ideas Math Answers Grade 4. All the Grade 4 Math concepts are explained by the subject experts as per the latest syllabus.

Big Ideas 4th Grade Math Answers Solutions Pdf | Big Ideas Math Book 4th Grade Answer Key

Chapter-wise quick links are given here where you can find all the concepts and topics deeply. Open every link and read out every chapter to improve your knowledge. Stand out by getting a good score by clearing all assessments on your own. When you stuck while taking the assessments, and then quickly have a look at the concept. Mainly concentrate on important points we have mentioned to easily learn the concept. Get an efficient preparation with the help of the Big Ideas Math Answers Grade 4 Solutions.

Advantages of Big Ideas Common Core Math Answer Key Grade 4

Big Ideas Math 4th Grade Answers will help you to get various benefits. Quick learning and the best practice are in your hands by referring to Big Ideas Math Answer Key Grade 4 Common Core. Check out some of the benefits below.

  • Get the updated Big Ideas Grade 4 Math Solutions Common Core Curriculum.
  • A detailed explanation of Concepts, Chapter-wise explanation, problems with answers, practice tests, chapter tests, worksheets, etc. to test your knowledge.
  • The best way to get success in the exam is by referring to BIM Book Grade 4 Answers. Every concept explained with real-time examples.
  • Improve your math skills easily with the 4th Grade Big Ideas Math Answers. Also, enhance your Problem-Solving Skills with the help of math grade 4 concepts.

FAQs on Elementary School 4th Grade Math Answer Key

1. Where can I get Big Ideas Chapter-wise Math Answers for Grade 4?

You can get the Chapter-wise Big Ideas Grade 4 Math Answers on our page those are organized into categories. Practice every concept and quickly enhance improve your math skills.

2. How to access Big Ideas Common Core 4th Grade Math Answer Key?

Get access to the Big Ideas 4th Grade Math Book Answer Key through the quick links provided on our page. Just click on them and prepare all the available concepts in the links.

3. Does preparing from the Elementary School Big Ideas Math Answers help you get better grades?

Yes, preparing from the Elementary School Big Ideas Math Answers will definitely help you get better grades with the proper practice.

Big Ideas Math Answers Grade 2 Chapter 6 Fluently Subtract within 100

Big Ideas Math Answers Grade 2 Chapter 6

Hello Students!!! Are you looking for Grade 2 Chapter 6 Fluently Subtract within hundred Answers? If yes, then our guess is correct. Here we have given a detailed explanation for every concept of BIM 2nd Grade 6th Chapter Fluently Subtract within 100 Book. Students are advised to practice all the topics covered in this chapter to score good marks in the exam. Also, download Big Ideas Math Answers Grade 2 Chapter 6 Fluently Subtract within 100 Answer Key PDF for making a better preparation plan.

Big Ideas Math Book 2nd Grade Answer Key Chapter 6 Fluently Subtract within 100

This free Big Ideas Math Grade 2 Chapter 6 Fluently Subtract within 100 Solutions helps the students to complete their homework and assignments easily. Refer to Big Ideas Math Book Grade 2 Chapter 6 Fluently Subtract within 100 Answer Key to prove yourself good in the practice tests and surprise tests. After solving this chapter, the students can be able to identify subtraction patterns, show regrouping, model subtraction problems, and explain which strategy is used to find the difference of numbers.

The various topics included in this answer key are Fluently Subtract within 100 Vocabulary, Model and Regroup to Subtract, Use Models to Subtract a One-Digit Number from a Two-Digit number, Use Models to Subtract Two-Digit Numbers, Use Models to Subtract Two-Digit Numbers, Use Addition to Check Subtraction, Practice Two-Digit Subtraction, and More Problem Solving: Subtraction. The performance task included at the end will help you to check your skills. As per your convenience, we have given the links according to the topics given in the textbook. Click on the link and start your preparation.

Vocabulary

Lesson: 1 Model and Regroup to Subtract

Lesson: 2 Use Models to Subtract a One-Digit Number from a Two-Digit numbers

Lesson: 3 Use Models to Subtract Two-Digit Numbers

Lesson: 4 Use Models to Subtract Two-Digit Numbers

Lesson: 5 Use Addition to Check Subtraction 

Lesson: 6 Practice Two-Digit Subtraction

Lesson: 7 More Problem Solving: Subtraction

Performance Task

Fluently Subtract within 100 Vocabulary

Big Ideas Math Answers Grade 2 Chapter 6 Fluently Subtract within 100 1
Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 1.1

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Organize-it

Define It

Match.

Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 2

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Define-it

Lesson 6.1 Model and Regroup to Subtract

Explore and Grow

Model each problem. Make quick sketches to show how you found the difference.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 3
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 3.1

Compare your quick sketches. What step did you use to find 21 − 6 that you did not use to find 19 − 6?

Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 4

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.1-Model- Regroup – Subtract-Explore-Grow.
The step in 21 − 6 that you did not use to find 19 − 6 is regrouping as 6 can be easily subtract from 9 but whereas 6 cannot be subtracted from 1 (21) so we need to regroup to subtract.

Show and Grow

Use models to subtract. Draw to show your work.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 5

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.1-Model- Regroup – Subtract-Show-Grow-Question-1

Apply and Grow: Practice

Use models to subtract. Draw to show your work.

Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 6
Question 2.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 6.1

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.1-Model- Regroup – Subtract-Show-Grow-Question-2

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 7

Answer:Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.1-Model- Regroup – Subtract-Show-Grow-Question-3

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 8

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.1-Model- Regroup – Subtract-Show-Grow-Question-4

Question 5.
Number Sense
Which numbers can you subtract from 44 without regrouping?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 9

Answer:
3 and 4.

Think and Grow: Modeling Real Life

You have 42 craft sticks. You use 8 of them. How many craft sticks are not used?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 10
Subtraction equation:
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 10.1

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.1-Model- Regroup – Subtract-Think-Grow

Show and Grow

Question 6.
There are 74 people in a theater. 7 of them leave. How many people are left?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 12

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.1-Model- Regroup-Subtract- Show-Grow-Question-6
People left from theater are 67.

Question 7.
DIG DEEPER!
You have 50 straws. You use some of them for a project. There are 44 left. How many straws did you use?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 13

Answer:
Number of straws = 50.
Number of straws left = 44.
Number of used straws = 50 – 44. = 6 straws.

Model and Regroup to Subtract Homework & Practice 6.1

Use models to subtract. Draw to show your work.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 14

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- Model-Regroup-Subtract-Homework-Practice 6.1 -Question-1

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 15

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- Model-Regroup-Subtract-Homework-Practice 6.1 -Question-2

Question 3.
YOU BE THE TEACHER
Is Newton correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 16

Answer:
No, The answer is 22. after regrouping one from tens to ones . then in tens we will have 2 tens not 3 tens.

Question 4.
Modeling Real Life
You have 26 magnets. You put 9 of them on your refrigerator. How many magnets are not on the refrigerator?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 17

Answer:
Number of magnets = 26.
Number of magnets on refrigerator = 9.
Number of magnets not on refrigerator = 26 – 9 = 15.

Question 5.
DIG DEEPER!
You have 83 cotton balls. You use some of them. There are 76 left. How many cotton balls do you use?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 18

Answer:

Number of cotton balls = 83
Number of cotton balls left = 76.
Number of cotton balls used = 83 – 76 = 7.

Review & Refresh

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 19

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Model-Regroup-Subtract-Homework-Practice-6.1-Question-6

Question 7.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 20

Answer:

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 21
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Model-Regroup-Subtract-Homework-Practice-6.1-Question-8

Lesson 6.2 Use Models to Subtract a One-Digit Number from a Two-Digit numbers

Explore and Grow

Make a quick sketch to find 41 − 3.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 22

How do you regroup 4 tens and 1 one to subtract 3?
___________________________________
___________________________________
___________________________________

Answer:

We regroup one from tens to ones place then one tens will be subtracted from 4 tens so we have 3 tens remaining and in ones place we added 10 ones + 1 ones earlier .total 11 ones will be in ones place.Later we need to subtract 3 ones from 11 ones so we have 8 ones remaining after subtraction.

Show and Grow

Question 1.
31 – 4 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 23

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.2- Use -Models –to- Subtract- One-Digit Number- from- Two-Digit numbers-Show-Grow-Question-1

Question 2.
7 – 5 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 24

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.2- Use -Models –to- Subtract- One-Digit Number- from- Two-Digit numbers-Show-Grow-Question-2

Question 3.
88 – 9 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 25

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.2- Use -Models –to- Subtract- One-Digit Number- from- Two-Digit numbers-Show-Grow-Question-3

Question 4.
63 – 2 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 26

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.2- Use -Models –to- Subtract- One-Digit Number- from- Two-Digit numbers-Show-Grow-Question-4

Question 5.
90 – 8 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 27

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.2- Use -Models –to- Subtract- One-Digit Number- from- Two-Digit numbers-Show-Grow-Question-5

Question 6.
Number Sense
What one-digit numbers can you subtract from 24 without regrouping? Explain.
Answer:
The one digit which is less or equal to 4 can be used to subtract without regrouping are 1,2,3 and 4.

Think and Grow: Modeling Real Life

You catch 33 fish. You keep the state limit of 25. How many fish do you release?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 28

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.2- Use -Models –to- Subtract- One-Digit Number- from- Two-Digit numbers-Think-Grow

Show and Grow

Question 7.
The temperature this morning was 72 degrees. This afternoon, it is 67 degrees. How much did the temperature drop?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 28.1

Answer:
Temperature in the morning= 72 degrees.
Temperature in the Afternoon= 67 degrees.
The Temperature drop= 72 – 67 =5 degrees.

Question 8.
DIG DEEPER!
You have a bowl of strawberries. You eat 7 of them. There are 19 left. How many strawberries were in the bowl to start?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 29

Answer:
Number of strawberries left =19
Number of strawberries eaten = 7.
Total Number of strawberries = 19 + 7 = 26 strawberries.

Use Models to Subtract a One-Digit Number from a Two-Digit numbers Homework & Practice 6.2

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 30\

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-to-Subtract- One-Digit Number- from -Two-Digit numbers- Homework - Practice 6.2-Question-1

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 31

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-to-Subtract- One-Digit Number- from -Two-Digit numbers- Homework - Practice 6.2-Question-2

Question 3.
YOU BE THE TEACHER
Is Descartes correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 32

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-to-Subtract- One-Digit Number- from -Two-Digit numbers- Homework - Practice 6.2-Question-3

Question 4.
Modeling Real Life
Descartes scores 8 more points than Newton. How many points did Newton score?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 33

Answer:
Score of Descartes = 8 +Score of newton.
Score of Descartes = 97.
Score of newton = 97 – 8 =89.

Question 5.
DIG DEEPER!
There are some piñata toys on the ground. 6 of your friends each pick 1 up. There are17 left. How many toys were there to start?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 34

Answer:
Total pinata toys picked by friends = 6 (6 friends picked 1 so 6 x 1 = 6 toys).
Toys left =17.
Total toys = 17 + 6 =23.

Review & Refresh

Break apart an addend to find the sum.

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 35

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use Models-Subtract- One-Digit Number- Two-Digit numbers-Homework-Practice- 6.2-Question-6

Question 7.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 36

Answer:

Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use Models-Subtract- One-Digit Number- Two-Digit numbers-Homework-Practice- 6.2-Question-7

Lesson 6.3 Use Models to Subtract Two-Digit Numbers

Explore and Grow

Model the problem. Make a quick sketch to find 33 − 17.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 37
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 37.1

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- Lesson 6.3-Use-Models-Subtract-Two-Digit-Numbers-Explore-Grow

Show and Grow

Question 1.
55 – 18 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 38

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.3-Use-Models-Subtract-Two-Digit Numbers-Show-Grow-Question-3

Question 2.
21 – 18 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 39

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- Lesson 6.3-Use-Models-Subtract-Two-Digit-Numbers-Show-Grow-Question-2

Question 3.
44 – 29 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 40

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.3-Use-Models-Subtract-Two-Digit Numbers-Show-Grow-Question-3

Question 4.
60 – 32 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 41

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.3-Use-Models-Subtract-Two-Digit Numbers-Show-Grow-Question-4

Question 5.
75 – 21 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 42

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.3-Use-Models-Subtract-Two-Digit Numbers-Show-Grow-Question-5

Question 6.
Number Sense
Use two of the numbers to write and solve a subtraction problem that requires regrouping
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 43

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.3-Use-Models-Subtract-Two-Digit Numbers-Show-Grow-Question-6

Think and Grow: Modeling Real Life

You have 68 flowers. You give 17 to Newton. Then you give 14 to Descartes. How many flowers do you have left?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 44
Subtraction problems:
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 45

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.3-Use-Models-Subtract-Two-Digit Numbers-Think-Grow-Modeling-Real-Life

Show and Grow

Question 7.
There are 32 pretzels. You eat 11 of them. Your friend eats 12 of them. How many pretzels are left?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 46

Answer:
Number of pretzels = 32.
Number of pretzels Eaten by me = 11.
Number of pretzels Eaten by my friend = 12.
Number of pretzels left = 32 – 11 – 12 = 9

Question 8.
Can you solve the problems on this page using addition and subtraction? Think: How do you know?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 47

Answer:
Yes

Use Models to Subtract Two-Digit Numbers Homework & Practice 6.3

Question 1.
84 – 60 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 48

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models –Subtract- Two-Digit Numbers- Homework-Practice 6.3-Question-1

Question 2.
40 – 15 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 49

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models –Subtract- Two-Digit Numbers- Homework-Practice 6.3-Question-2

Question 3.
Number Sense
Which numbers can you subtract from 55 without regrouping?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 50

Answer:
Numbers that can subtract from 55 without regrouping are 15 , 33 and 24.

Question 4.
Modeling Real Life
You and your friend have 38 cups of lemonade in all. You sell 14 cups. Your friend sells 16. How many cups are left?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 51

Answer:
Total Number of Lemonade cups= 38.
Number of lemonade cups sold by me = 14.
Number of lemonade cups sold by my friend =16.
Number of cups left = 38 – 14 – 16 = 8.

Question 5.
Modeling Real Life
Your principal has 75 awards. She gives 25 to your class. Then she gives 34 to your friend’s class. How many awards are left?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 52

Answer:
Number of Total Awards = 75.
Number of Awards given to my class = 25.
Number of Awards given to my friends class = 34.
Number of Awards left = 75 – 25 – 34 = 16.

Review & Refresh

Question 6.
A bookcase has 5 shelves. There are 3 plants on each shelf. How many plants are there in all?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 53
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 54
Answer:
Each shelf has 3 plants
Total plants = 3 + 3 + 3 + 3 + 3= 15 Plants.

Use Models to Subtract Two-Digit Numbers Homework & Practice 6.4

Make quick sketches to find each difference.

31 – 3 =?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 55

Answer:
31 – 23 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 56

How are the problems the same? How are they different?
_______________________________________
_______________________________________
_______________________________________

Answer:
The Minuend is the same for both the problems that is 31.
The Subtrahend are different for both the problems that is 3 and 23 .
Even the difference of both the problems are different that is 23 and 3 .

Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4

Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-2

Show and Grow

Question 1.
52 – 19 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 57

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Show-Grow-Question-1

Question 2.
46 – 9 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 58

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Show-Grow-Question-2

Question 3.
60 – 21 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 59

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Show-Grow-Question-3

Question 4.
66 – 8 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 60

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Show-Grow-Question-4

Question 5.
84 – 2 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 61

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Show-Grow-Question-5

Question 6.
65 – 38 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 62

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Show-Grow-Question-6

Apply and Grow: Practice

Question 7.
58 – 22 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 63

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Apply-Grow-Practice-Question-7

Question 8.
33 – 4 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 64

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Apply-Grow-Practice-Question-8

Question 9.
22 – 15 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 65

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Apply-Grow-Practice-Question-9

Question 10.
70 – 60 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 66

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Apply-Grow-Practice-Question-10

Question 11.
43 – 26 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 67

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Apply-Grow-Practice-Question-11

Question 12.
95 – 4 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 68

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Apply-Grow-Practice-Question-12

Question 13.
Number Sense
Use the given numbers to complete the problem.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 69

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Apply-Grow-Practice-Question-13

Question 14.
DIG DEEPER!
Subtract a one-digit number from a two-digit number to complete the problem.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 70

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4-Apply-Grow-Practice-Question-14

Think and Grow: Modeling Real Life

You pick 19 yellow flowers and 24 purple flowers. You give 8 flowers away. How many flowers do you have left?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 71
Step 1: Find the total number of flowers you picked.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 72

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models-Subtract-Two-Digit Numbers-Homework-Practice 6.4- Think-Grow-Step1

Step 2: Subtract the number of flowers you give away from your result in Step 1.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 73

Answers:
8 Flowers.

Show and Grow

Question 15.
You bake 36 blueberry muffins and 36 banana nut muffins. You sell 47 muffins. How many muffins do you have left?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 74

Answer:
Number of Blueberry muffins baked = 36.
Number of banana nut muffins baked = 36.
Total number of muffins = 36 + 36 = 72.
Number of muffins sold = 47.
Number of muffins left = 72 – 47 = 25.

Question 16.
There are 54 ladybugs. 7 fly away. Then 25 join. How many ladybugs are there now?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 75

Answer:
Number of ladybugs = 54.
Number of bugs flew away = 7.
Number of bugs joined = 25.
Total number of bugs left = 54 – 7 + 25 =22.

Use Models to Subtract Two-Digit Numbers Homework & Practice 6.4

Question 1.
25 – 7 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 76

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models- Subtract-Two-Digit Numbers-Homework -Practice 6.4-Question-1

Question 2.
65 – 35 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 77

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models- Subtract-Two-Digit Numbers-Homework -Practice 6.4-Question-2

Question 3.
33 – 29 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 78

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models- Subtract-Two-Digit Numbers-Homework -Practice 6.4-Question-3

Question 4.
80 – 53 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 79

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models- Subtract-Two-Digit Numbers-Homework -Practice 6.4-Question-4

Question 5.
92 – 47 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 80

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models- Subtract-Two-Digit Numbers-Homework -Practice 6.4-Question-5

Question 6.
56 – 4 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 81

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models- Subtract-Two-Digit Numbers-Homework -Practice 6.4-Question-6

Question 7.
Writing
Write and solve a subtraction problem using 2 two-digit numbers. Write a story to match.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 82

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use-Models- Subtract-Two-Digit Numbers-Homework -Practice 6.4-Question-7
From the above figure we observe that 86 is the minuend and 24 is the subtrahend.The difference is 62.
The 8 tens from minuend  have to subtract 2 tens from the subtrahend we get 6 tens in the difference . We have 6 ones in the minuend and we have to subtract 4 ones from subtrahend we get 2 ones in difference.
Therefore the difference is 6 tens and 2 ones = 62.

Question 8.
Modeling Real Life
You fill up 17 large water balloons and 24 small water balloons. You break 26 balloons. How many water balloons do you have left?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 83

Answer:
Number of large water balloons = 17
Number of small water balloons = 24
Total number of balloons = 17 + 24 =41
Number of balloons broken = 26.
Number of balloons Remaining = 41 – 26 =15

Question 9.
Modeling Real Life
You have 45 rings. You throw 28 of them. You earn a bonus and get 15 more rings to throw. How many rings do you have now?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 84

Answer:
Total Number of Rings = 45.
Number of rings threw = 28.
Number of Remaining Rings = 45 – 28 = 17.
Number of rings earned as Bonus = 15.
Total Number of Rings = 17 + 15 = 32.

Review & Refresh

Question 10.
How can you make a 10 to find 5 + 8?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 85
Answer:
10 + 3

Lesson 6.5 Use Addition to Check Subtraction

Write the equation shown by each model.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 86

How are the equations related?
__26 – ? =43____________________
__26 + 17 = ?____________________

Explain how you can check whether 24 – 13 = 11 is correct.
Answer:
Here when we add 13 + 11 we should get 24 then the answer is correct.

Show and Grow

Find the difference. Use addition to check your answer.

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 88

Answer:

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 89

Answer:

Apply and Grow: Practice

Find the Difference. Use addition to check your answer.

Question 3.
52 – 27 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 90

Answer:

Question 4.
76 – 58 = ?
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 91

Answer:

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 6 Fluently Subtract within 100 92

Answer:

Question 6.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 93

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- Lesson-6.5-Use-Addition-Check-Subtraction-Show-Grow-Question-6

Question 7.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 94

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- Lesson-6.5-Use-Addition-Check-Subtraction-Show-Grow-Question-7

Question 8.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 95

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.5-Use-Addition-to-Check-Subtraction- Apply-Grow- Practice-Question-8

Question 9.
Reasoning
Newton uses 16 + 37 to check his answer to a subtraction problem. What two subtraction problems could he have solved?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 96

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson-6.5-Use-Addition-Check-Subtraction-Show-Grow-Reasoning.jpg

Think and Grow: Modeling Real Life

A joke book has 96 jokes. 58 are knock-knock jokes. The rest are riddles. How many riddles are there?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 97
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 98

Answer:
The rest riddles are 38.
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- Lesson-6.5-Use-Addition-Check-Subtraction-Think-Grow

Show and Grow

Question 10.
A museum has 71 fossils. 47 are dinosaur fossils. The rest are fish fossils. How many fish fossils are there?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 99

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- Lesson-6.5-Use-Addition-Check-Subtraction-Show-Grow-Question-10

Question 11.
There are 27 more countries in Europe than in North America. There are 50 countries in Europe. How many countries are there in North America?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 100
Answer:
Countries in Europe = 50.
Countries in North America = 50 – 27 ( europe has 27 more than north america).
= 23 .
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- Lesson-6.5-Use-Addition-Check-Subtraction-Show-Grow-Question-11

Use Addition to Check Subtraction Homework & Practice 6.5

Find the difference. Use addition to check your answer.

Question 1.
44 – 21 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 101

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use–Addition-Check-Subtraction-Homework-Practice-6.5-Question-1

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 102

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use–Addition-Check-Subtraction-Homework-Practice-6.5-Question-2

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 103

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use–Addition-Check-Subtraction-Homework-Practice-6.5-Question-3

Question 4.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 104

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use–Addition-Check-Subtraction-Homework-Practice-6.5-Question-4

Question 5.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 105

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use–Addition-Check-Subtraction-Homework-Practice-6.5-Question-5

Question 6.
Number Sense
Descartes subtracts to find 36 − 19. Which addition equation can he use to check his answer?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 106

Answer:
36 -19 = 17
19 + 17 = 36

Question 7.
Modeling Real Life
A theme park has 62 rides. 45 of them are in the amusement park. The rest are in the water park. How many rides are in the water park?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 107

Answer:
Number of rides in Theme park= 62
Number of rides in amusement park =45.
left over rides take place in water park
Number of rides in water park = 62 -45 = 17
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use -Addition –to-Check-Subtraction-Homework-Practice-6.5-Question-7

Question 8.
Modeling Real Life
There are 58 more books in a library than in a classroom. There are 94 books in the library. How many books are in the classroom?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 108

Answer:
Number of books in library = 94.
Number of books in class room = 94 – 58( 58 more books are there in classroom)
Number of books in class room = 36.

Review & Refresh

Use compensation to solve.

Question 9.
18 + 44 = _____
Big Ideas Math Answer Key Grade 2 Chapter 6 Fluently Subtract within 100 109

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use -Addition –to-Check-Subtraction-Homework-Practice-6.5-Question-9

Question 10.
63 + 27 = _____

Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Use -Addition –to-Check-Subtraction-Homework-Practice-6.5-Question-10

Lesson 6.6 Practice Two-Digit Subtraction

Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 110

Work with a partner. Choose different strategies to find 76 − 29.

Compare your strategy to your partner’s. Are your answers the same? Which strategy do you prefer? Why?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 111

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6-Practice-Two-Digit-Subtraction

I prefer Strategy – 2 as we can check the answer, which helps us to know whether the answer is correct or wrong.
Whatever the strategy may be the answer will be the same.

Show and Grow

Use any strategy to find the difference.

Question 1.
52 – 25 = _____

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction-Show-Grow-Question-1

Question 2.
78 – 56 = _____

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction-Show-Grow-Question-2

Question 3.
61 – 33 = ___

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction-Show-Grow-Question-3

Question 4.
83 – 37 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction-Show-Grow-Question-4

Question 5.
72 – 49 = __-
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction-Show-Grow-Question-5

Show and Grow

Question 1.
52 − 25 = __

Question 4.
83 − 37 = __

Question 5.
72 – 49 = __

Question 6.
45 – 39 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction-Show-Grow-Question-6

Question 7.
69 – 58 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction-Show-Grow-Question-7

Question 8.
94 – 45 = __
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction-Show-Grow-Question-8

Question 9.
86 – 48 = __
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction-Show-Grow-Question-9

Apply and Grow: Practice

Use any strategy to find the difference.

Question 10.
95 – 38 = __
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Apply-Grow-Practice -Question-10

Question 11.
66 – 47 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Apply-Grow-Practice -Question-11

Question 12.
58 – 6 = __
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Apply-Grow-Practice -Question-12

Question 13.
81 – 35 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Apply-Grow-Practice -Question-13

Question 14.
43 – 17 = ___
Answer:

Question 15.
37 – 8 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Apply-Grow-Practice -Question-15

Question 16.
28 – 19 = ___
Answer:

Question 17.
72 – 30 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Apply-Grow-Practice -Question-17

Question 18.
82 – 4 = __
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Apply-Grow-Practice -Question-18

Question 19.
DIG DEEPER!
Fill in the missing digits.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 112

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Apply-Grow-Practice -Question-19

Question 20.
YOU BE THE TEACHER
Is Descartes correct? Explain
Big Ideas Math Answer Key Grade 2 Chapter 6 Fluently Subtract within 100 200

Answer:
Yes, You can verify your answer by adding 18 + 23 to get 41. if you wont 41 as your answer the sum is wrong.
18 + 23 = 41.

Think and Grow: Modeling Real Life

Is there a greater difference between the number of guests on Saturday and Sunday or on Sunday and Monday?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 114

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Think-Grow

Show and Grow

Question 21.
Is there a greater difference between the number of bagels sold on Sunday and Monday or on Monday and Tuesday?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 115

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.6- Practice- Two-Digit –Subtraction- Show-Grow-Question-21

Practice Two-Digit Subtraction Homework & Practice 6.6

Use any strategy to find the difference.

Question 1.
34 – 26 = __

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-1

Question 2.
75 – 47 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-2

Question 3.
96 – 48 = __
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-3

Question 4.
51 – 21 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-4

Question 5.
23 – 16 = ___

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-5

Question 6.
47 – 8 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-6

Question 7.
25 – 3 = ___
Answer:

Question 8.
87 – 69 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-8

Question 9.
48 – 29 = ___
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-9

Question 10.
Number Sense
Find each missing number.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 116

Answer:

Question 11.
Structure
Solve 31 − 14 two different ways.
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-11

Question 12.
Modeling Real Life
Is there a greater difference between the number of animals adopted in December and January or in January and February?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 117

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Practice-Two-Digit- Subtraction-Homework- Practice- 6.6 -Question-12

Review & Refresh

Question 13.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 118

Answer:
2 + 8 =10
10 – 8 = 2

Question 14.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 119

Answer:
5 + 6 = 11
11 – 6 = 5

Question 15.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 120
Answer:
9 + 8 = 17
17 – 8 = 9.

Lesson 6.7 More Problem Solving: Subtraction

Explore and Grow

Model the story

Newton has 34 balloons. 18 fly away. How many balloons does Newton have left?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 121Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 122
Answer:
Number of balloons = 34
number of balloons flew=18
Number of balloons left = 34 – 18 =16

Show and Grow

Question 1.
There are 27 more harmonicas than drums in a music room. There are 55 harmonicas. How many drums are there?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 123
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 124

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.7-More-Problem-Solving- Subtraction-Show-Grow

Apply and Grow: Practice

Question 2.
There are 71 fish crackers. You eat 29 of them. Your friend eats 28. How many fish crackers are left?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 125
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 126

Answer:
Number of Fish crackers = 71
Number of fish crackers eaten by me = 29
Number of fish crackers eaten by my friend = 28.
Number of fish crackers left = 71- 29 – 28 =14

Question 3.
Newton solves 16 more math problems than Descartes. Newton solves 34 math problems. How many math problems does Descartes solve?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 127
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 128

Answer:
Problems solved by Newton =34.
Problem solved by Descartes = 34 – 16( as newton solves 16 problems more)
Problem solved by Descartes = 18

Question 4.
Number Sense
You have 78 tokens. Your friend gives you 14 more. You use 35 tokens. Use the given numbers to find how many tokens you have now.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 129
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 130

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.7-More-Problem-Solving- Subtraction-Show-Grow-Question-4

Think and Grow: Modeling Real Life

You see 20 elephants on a safari. You see 7 fewer giraffes than elephants. Then 5 of the giraffes leave the group. How many giraffes are left?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 131
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 131.1

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-Lesson 6.7-More-Problem-Solving- Subtraction-Think-Grow

Show and Grow

Question 5.
Your friend has 54 blocks. You have 18 fewer blocks than your friend. You use 29 blocks to make a tower. How many blocks do you have left?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 132

Answer:
Number of Blocks with my friend = 54
Number of Blocks with me = 54 – 18 = 36
Number of blocks used by me to make a tower = 29.
Number of blocks left = 36 – 29 = 7 .

Question 6.
DIG DEEPER!
You had 19 almonds. You ate some. There are 11 left. Your friend ate 3 fewer almonds than you. How many almonds did your friend eat?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 133

Answer:
Number of Almonds = 19.
Number of Almonds left = 11.
Number of Almonds eaten = 19 – 11 = 8.
Number of Almonds eaten by me = 8 – 3 = 5.
Number of Almonds eaten by my friend = 3.

More Problem Solving: Subtraction Homework & Practice 6.7

Question 1.
Some kids are at a trampoline park. 24 leave. There are 57 left. How many kids were there to start?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 134

Answer:
Total Number of kids at a Trampoline park = X
Number of kids left the park = 24.
Number of kids remaining in the park = 57.
Total number of kids = 24 + 57 =81.

Question 2.
You have 62 dollars saved. You spend 37 dollars on a video game. Then you give 16 dollars to your friend. How many dollars do you have left?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 135

Answer:
Total Number of Dollars = 62.
Number of dollars spend on video game = 37.
Number of dollars given to my friend = 16 .
Number of dollars left = 62 – 37 – 16 =9 .

Question 3.
Structure
You have 50 toothpicks. Your friend has 30 fewer toothpicks than you. Which picture shows how many toothpicks you and your friend have in all?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 136
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 137

Answer:
Picture 2 is correct = 70 Toothpicks.
Number of tooth picks with me = 50.
Number of tooth picks with my friend = 50 – 30 = 20.
Total number of tooth picks = 50 + 20 = 70.

Question 4.
DIG DEEPER!
A museum has 12 dinosaur exhibits and some art exhibits. There are 34 exhibits in all. Then 3 more art exhibits are added. How many art exhibits are there now?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 138

Answer:
Number of Dinosaur Exhibits = 12.
Total Number of Exhibits = 34
Number of Art Exhibits = 34 – 12 = 22
Number of Art Exhibits added = 3.
Total Number of Art Exhibits =22 + 3 = 25 .

Review & Refresh

Break apart the addends to find the sum.

Question 5.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 139

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-More-Problem-Solving- Subtraction-Homework -Practice 6.7-Question-5

Question 6.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 140

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-More-Problem-Solving- Subtraction-Homework -Practice 6.7-Question-6

Fluently Subtract within 100 Performance Task

Question 1.
Your class makes 53 paper snowflakes to decorate your classroom.
a. Your group makes 16 of them. How many snowflakes does the other group make?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 141

Answer:
Total Number of paper snowflakes = 53.
Number of snow flakes made by my group =16.
Number of snowflakes made by other group = 53 – 16 = 37.

b. Your teacher hangs 49 of the snowflakes from the ceiling in an array with 7 equal rows. How many columns of snowflakes are there?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 143

Answer:
Total Number snowflakes = 49.
Number of rows = 7.
Number of Columns = 49 ÷ 7 = 7.

c. Your teacher wants to hang one more equal row of snowflakes. Are there enough snowflakes? If so, how many extra snowflakes are there? If not, how many more snowflakes are needed?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 144

Answer:
Number of snowflakes = 49.
One more row means 8 th row = 7 more snowflakes = 49 + 7 =56 snowflakes.
56 snowflakes are required we have only 49 snowflakes.
Therefore , more 7 snowflakes are required

Question 2.
Another class makes 75 paper snowflakes. 29 are white and some are blue. They make some more blue snowflakes. Now they have 61 blue snowflakes. How many more blue snowflakes did they make?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 145
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 146

Answer:
Number of snowflakes = 75.
Number of white snowflakes = 29 .
Number of blue snowflakes = 75 – 29 = 46 .
Number of blue snowflakes made again = 61.
Total Number of Blue snowflakes = 61 + 46 = 107 .

Fluently Subtract within 100 Activity

To Play: Use Solve and Cover: Subtraction Cards. Place a Difference Card face up on each box. Place the stack of Problem Cards face down. Players take turns. On your turn, flip over a card from the stack. Solve the problem. Place the card on the difference. Play until all of the differences are covered.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 147

Fluently Subtract within 100 Chapter Practice 6

6.1 Model and Regroup to Subtract

Question 1.
Use Models to Subtract. Draw to show your work.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 148
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 149

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-6.1-Model-Regroup- Subtract

6.2 Use Models to Subtract a One-Digit Number from a Two-Digit Number

Question 2.
60 – 8 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 150

Answer:Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-6.2 Use-Models-Subtract-One-Digit Number-Two-Digit Number-Question-2

Question 3.
74 – 3 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 151

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-6.2 Use-Models-Subtract-One-Digit Number-Two-Digit Number-Question-3

6.3 Use Models to Subtract Two-Digit Numbers

Question 4.
45 – 38 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 152

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.3-Use-Model-Subtract-Two-Digit Numbers-Question-4

Question 5.
74 – 27 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 153

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.3-Use-Model-Subtract-Two-Digit Numbers-Question-5

Question 6.
Number Sense
Use two of the numbers to write and solve a subtraction problem that requires regrouping.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 154

Answer:
The two numbers that requires regrouping for subtraction are 63 – 45 = 18

Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.3-Use-Model-Subtract-Two-Digit Numbers-Question-6

6.4 Subtract from a Two-Digit Number

Question 7.
93 – 50 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 155

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-6.4-Subtract-from-a-Two-Digit-Number-Question-7

Question 8.
57 – 28 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 156

Answer:

Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-6.4-Subtract-from-a-Two-Digit-Number-Question-8

Question 9.
67 – 9 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 157

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-6.4-Subtract-from-a-Two-Digit-Number-Question-9

6.5 Use Addition to Check Subtraction

Find the difference. Use addition to check your answer.

Question 10.
64 – 27 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 158

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.5-Use –Addition-Check- Subtraction-Question-10

Question 11.
86 – 36 = ?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 159

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.5-Use –Addition-Check- Subtraction-Question-11

Question 12.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 160

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.5-Use –Addition-Check- Subtraction-Question-12

Question 13.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 161
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.5-Use –Addition-Check- Subtraction-Question-13

6.6 Practice Two-Digit Subtraction

Use any strategy to find the difference.

Question 14.
55 – 37 = ___
Answer:

Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.6-Practice-Two-Digit-Subtraction-Question-14

Question 15.
60 – 20 = __
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.6—Practice-Two-Digit-Subtraction-Question-15

Question 16.
39 – 3 = __
Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100- 6.6—Practice-Two-Digit-Subtraction-Question-16

6.7 More Problem Solving: Subtraction

Question 17.
Newton has some BINGO chips. He gives 25 of them away. He has 28 left. How many chips did he have to start?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 162
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 162.1

Answer:

Question 18.
Number Sense
You have 35 balls. You use 18 of them. You win a bonus and get 10 more balls. Use the given numbers to find how many balls you have now.
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 163
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 164

Answer:
Big-Ideas-Math-Book-2nd-Grade-Answer-Key-Chapter-6-Fluently-Subtract-100-6.7-More-Problem-Solving-Subtraction- Question-18

Question 19.
Modeling Real Life
A petting zoo has 27 sheep and some goats. There are 41 animals in all. Then 16 more goats are added. How many goats are there now?
Big Ideas Math Answers 2nd Grade Chapter 6 Fluently Subtract within 100 165
Answer:
Total Animals = 41.
The number of sheep = 27.
A number of goat = ?
The number of goats = Total Animals – Number of sheep.
= 41 – 27 = 14
The number of goats added =16.
Total Number of goats= 14 + 16 = 30 goats.

Big Ideas Math 2nd Grade 6th Chapter Fluently Subtract within 100 Solution Key Pdf

The answers given in the Big Ideas Math Book Grade 2 Chapter 6 Fluently Subtract within 100 Answer Key are prepared by the math experts. So, download Big Ideas Math Answers Grade 2 Chapter 6 Fluently Subtract within 100 pdf for free of cost and kick start your preparation. Learn all the chapters of Big Ideas Math Book Grade 2 by keeping in touch with our site.

Parametric Equations of a Parabola Formula, Examples | How to find Parametric Form of Parabola?

Parametric Equations of a Parabola

Parabolas describe many natural phenomena like the motion of objects affected by gravity, increase or decrease in the population, amount of reagents in a chemical equation, etc. At times, you need to evaluate how variables change with respect to time. To track how variables change over time, you can put equations into Parametric Form.

Different Parametric Equations can be used to represent a Parabola. We have listed the simple and easiest way on How to find the Parametric Equations of a Parabola in the below modules. Refer to the Solved Examples on Parametric Equations of Parabola for a better understanding of the concept.

Standard Forms of Parabola and their Parametric Equations

Let us discuss in detail the Parametric Coordinates of a Point on Standard Forms of Parabola and their Parametric Equations

Standard Equation of Parabola y2 = 4ax

  • Parametric Coordinates of the Parabola y2 = 4ax are (at2, 2at)
  • Parametric Equations of Parabola y2 = 4ax are x = at2 and y = 2at

Standard Equation of Parabola y2 = -4ax

  • Parametric Coordinates of the Parabola y2 = -4ax are (-at2, 2at)
  • Parametric Equations of Parabola y2 = -4ax are x = -at2 and y = 2at

Standard Equation of Parabola x2 = 4ay

  • Parametric Coordinates of the Parabola x2 = 4ay are (2at, at2)
  • Parametric Equations of Parabola x2 = 4ay are x = 2at, y = at2

Standard Equation of Parabola x2 = -4ay

  • Parametric Coordinates of the Parabola x2 = 4ay are (2at, -at2)
  • Parametric Equations of Parabola x2 = 4ay are x = 2at, y = -at2

Standard Equation of Parabola (y-k)2 = 4a(x-h)

Parametric Equations of Parabola (y-k)2 = 4a(x-h) are x=h+at2, and y = k+2at

Solved Examples on finding the Parametric Equations of a Parabola

1. Write the Parametric Equations of the Parabola y2 = 16x?

Solution:

Given Equation is in the form of y2 = 4ax

On Comparing the terms we have the 4a = 16

a = 4

The formula for Parametric Equations of the given parabola is x = at2 and y = 2at

Substitute the value of a to get the parametric equations i.e. x = 4t2 and y = 2*4*t = 8t

Therefore, Parametric Equations of Parabola y2 = 16x are x= 4t2 and y = 8t

2. Write the Parametric Equations of Parabola x2 = 12y?

Solution:

Given Equation is in the form of x2 = 4ay

On Comparing the terms we have the 4a = 12

a = 3

The formula for Parametric Equations of the given parabola is x = 2at, and y =  at2

Substitute the value of a to get the parametric equations i.e. x = 2*3*t and y = 3t2

Therefore, Parametric Equations of Parabola x2 = 12y are x = 6t and y = 3t2

3. Write the Parametric Equations of the Parabola (y-3)2 =8(x-2)?

Solution:

Given Equation is in the form of (y-k)2 = 4a(x-2)

Comparing the two equations we have k = 3, h = 2 and 4a = 8 i.e. a =2

The Formula for Parametric Equations of Parabola (y-k)2 = 4a(x-h) are x=h+at2, and y = k+2at

substitute the values of k, a in the formula and obtain the parametric equation

x = 2+2t2 and y = 3+2*2t

x = 2+2t2 and y = 3+4t

Therefore, Parametric Equations of the Parabola (y-3)2 =8(x-2) are x = 2+2t2 and y = 3+4t

Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas

Big Ideas Math Answers Grade 4 Chapter 12

Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas is the best source for students. Students can follow the step-by-step procedure to solve the problems. We have given a detailed explanation for every problem. Also, we included real-time examples to make the student’s preparation perfect. Get the free pdf Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas to learn offline. Important notes provided for every concept where quick preparation is possible. So, immediately, begin your practice now and strengthen your knowledge.

Big Ideas 4th Grade Chapter 12 Use Perimeter and Area Formulas Math Book Answer Key

Easily improve your problem-solving skills with the help of Big Ideas Math Book 4th Grade Answer Key Chapter 12 Use Perimeter and Area Formulas. We have also included an additional practice section where you can test your knowledge and find out the difficult concepts for you. Concentrate on the difficult topics and become the expert on the complete concept. Easy learning and fast solving skills will come at the same place by using Big Ideas Math Grade 4 Chapter 12 Use Perimeter and Area Formulas Solution Key.

Lesson: 1 Perimeter Formula for a Rectangle

Lesson: 2 Area Formula for a Rectangle

Lesson: 3 Find Unknown Measures

Lesson: 4 Problem Solving: Perimeter and Area

Performance Task

Lesson 12.1 Perimeter Formula for a Rectangle

Explore and Grow

Use color tiles to create a rectangle with a perimeter of 12 units. Compare your rectangle to your partner’s. How are they the same? How are they different?
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 1
How do you know that the perimeter of your rectangle is 12 units?

Answer:

The perimeter (P) of a rectangle = P=2l+2w( length=l ; width = w)

GIVEN: Perimeter of rectangle =12
P =2l + 2w
12= 2 (l + w)
12/2=l + w
6= l + w
Hence, the rectangle of my friend and mine are going to be same because perimeter given above is same for both the rectangle of us both.


The rectangle of mine is having a length of 3 units and width of 3 units.
Perimeter of my rectangle= (2 x l) + (2 x w)
P=(2 x 3) + (2 x 3)
P= 6 + 6
P= 12  units.
Here, my rectangle has 12  units perimeter. This is how I know my rectangle is having 12 units as its perimeter.

Structure
How is the perimeter of a rectangle related to its length and width?

Answer:
The perimeter (P) of a rectangle is given by the formula, P=2l+2w , where “l” is the length and “w” is the width of the rectangle.

Think and Grow: Use a Formula for Perimeter

Perimeter is the distance around a figure. A formula is an equation that uses letters and numbers to show how quantities are related. You can use a formula to show how the length, width, and perimeter of a rectangle are related.
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 2
Example
Find the perimeter of the rectangle.
The length is ___ feet and the width is __ feet.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 3
P = (2 × l) + (2 × W) Formula for perimeter of a rectangle
= (2 × ___) + (2 × ___)
= __ + __
= ___
The perimeter is ___ feet.
ANSWER:
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 3
Given: Length of Rectangle = 24 feet
Width of Rectangle = 9 feet
FORMULA: Perimeter = (2 x l) + (2 x w)
P = (2 x 24) + (2 x 9)
P = 48 + 18
P = 66  feet.
Hence, the Perimeter of the rectangle = 66  feet.

Show and Grow

Find the perimeter of the rectangle.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 300
Answer:
GIVEN: Length of rectangle = 16cm
Width of rectangle = 13 cm
Perimeter of the rectangle = (2 x l) + (2 x w)
P = ( 2 x 16) + ( 2 x 13 )
P = 32 + 26
P = 58  cm.
Hence, the perimeter of the rectangle = 58  cm.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 4
Answer:
GIVEN: Length of the rectangle = 8 inches
Width of the rectangle = 4 (1/2) inches =9/2 = 4.5 inches
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2( 8 + 4.5 )
P = 2(12.5 )
P = 25 inches.
Hence, the perimeter of the rectangle = 25 inches.

Apply and Grow: Practice

Find the perimeter of the rectangle.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 5
Answer:
GIVEN: Length of the rectangle = 54 yards
Width of the rectangle = 32 yards
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(54+32)
P= 2 x 86
P= 172 yards.
Hence, the perimeter of the rectangle =172 yards.

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 6
Answer:
GIVEN: Length of the rectangle =87 m
Width of the rectangle = 65 m
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P= 2(87+65)
P= 2 x 153
P= 306  m.
Hence, the perimeter of the rectangle = 306 m.

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 7
Answer:
GIVEN: Length of the rectangle = 49 inches
Width of the rectangle = 18 inches
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(49+18)
P =2 x 67
P =134 inches.
Hence, the perimeter of the rectangle = 134 inches.

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 8
Answer:
GIVEN: Length of the rectangle =11 cm
Width of the rectangle = 7 (3/10) cm = 73/10 cm =7.3 cm
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(11+7.3)
P = 2 x 18.3
P = 36.6  cm.
Hence, the perimeter of the rectangle = 36.6 cm.

Question 7.
You want to string lights around a rectangular room that is 12 feet long and 10 feet wide. How many feet of lights do you need?
Answer:
GIVEN: Length of the rectangle = 12 feet
Width of the rectangle = 10 feet
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(12+10)
P = 2 x 22
P = 44  feet
Hence, 44  feet of light is needed..

Question 8.
YOU BE THE TEACHER
Your friend finds the perimeter of the rectangle. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 9
p = (2 × 13) × (2 × 5)
= 26 × 10
= 260  ft
Answer: GIVEN: Length of the rectangle = 13 feet
Width of the rectangle = 5 feet
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(13+5)
P = 2 x 18
P = 36 feet.
Hence, the perimeter of the rectangle = 36  feet.
Therefore my friend is wrong because his formula for finding perimeter of rectangle is wrong.
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)

Question 9.
DIG DEEPER!
You can use the formula for the perimeter of a rectangle to find the perimeter of the square. What other formula can you use to find the perimeter of the square?
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 10

Answer:
The perimeter of a square is the total length of the four equal sides of the square.
Perimeter of the square = 4 × s = 4 s
The perimeter of a rectangle is the total length of the two lengths and two widths of the rectangle.
If the length and width of a rectangle are l and w, then its perimeter = 2(l + w) units.
We cannot equate the perimeter of a rectangle to find the perimeter of the square because rectangle has length and width whereas square has four sides.
No, other ways are there find the perimeter of the square.

Think and Grow: Modeling Real Life

Example
In a video game, you make a rectangular castle that is 4 times longer than it is wide. What is the perimeter of the castle?
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 11
Multiply 4 and the width of the castle to find the length.
4 × 25 = ___
The length of the castle is ___ yards.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 12
Answer:
GIVEN: Width of the rectangular castle  = 25 yards
Length of the rectangular castle = 4 times longer than it is wide =4 x 25 yards = 100 yards
Perimeter of the rectangle castle = (2 x l) + (2 x w) = 2( l + w)
P = 2 (100+25)
P = 2 x 125
P = 250  yards
Hence, Perimeter of the rectangle  castle= 250 yards.

Show and Grow

Question 10.
A teacher wants to put a border around a rectangular whiteboard. The whiteboard is 2 times longer than it is wide. What is the perimeter of the whiteboard?
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 13
Answer:
GIVEN: Width of the rectangular castle  =  1 m
Length of the rectangular castle = 2 times longer than it is wide = 2 x 1 m = 2 m
Perimeter of the rectangle castle= (2 x l) + (2 x w) = 2( l + w)
P = 2(2+1)
P = 2 x 3
P = 6 m
Hence, Perimeter of the rectangle castle = 6 m.

Question 11.
You want to put a ribbon border around each rectangular card. Which card requires more ribbon? How much more ribbon?
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 14
Answer:
GIVEN: FIRST RECTANGULAR CARD
Width of the rectangular card  =  18 cm
Length of the rectangular card =  18 cm
Perimeter of the rectangle card= (2 x l) + (2 x w) = 2( l + w)
P = 2(18+18)
P = 2 x 36
P = 72 cm.
SECOND RECTANGULAR CARD
Width of the rectangular card  =  14cm
Length of the rectangular card =  21 cm
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(21+14)
P = 2 x 35
P = 70  cm.
The FIRST RECTANGULAR CARD needs the more ribbon than SECOND RECTANGULAR CARD because FIRST RECTANGULAR CARD Perimeter = 72  cm and SECOND RECTANGULAR CARD Perimeter = 70 cm.
That means  difference in perimeter of FIRST RECTANGULAR CARD and the perimeter of SECOND RECTANGULAR CARD.
=72  cm – 70  cm
= 2  cm.

Question 12.
DIG DEEPER!
A rectangular flower bed has a length of 6 feet. The width is 48 inches shorter than the length. What is the perimeter of the flower bed?
Answer:
GIVEN:
Length of the rectangular bed = 6 feet
Width of the rectangular bed = 48 inches shorter than the length.
1 inch = 0.08333 feet
48inches = 48 x 0.0833 =  4 feet
Width of the rectangular bed =  4 feet
The perimeter of the  rectangular flower bed = (2 x l) + (2 x w) = 2( l + w)
P = 2(6+4)
P = 2 x 10
P = 20 feet.
Hence, the perimeter of the  rectangular flower bed = 20 feet.

Perimeter Formula for a Rectangle Homework & Practice 12.1

Find the perimeter of the rectangle.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 15
Answer:
GIVEN: Length of the rectangle = 22 yards
Width of the rectangle = 18 yards
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(22+18)
P = 2 x 40
P = 80 yards.
Hence, the Perimeter of the rectangle = 80 yards.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 16
Answer:
GIVEN: Length of the rectangle = 36 mm
Width of the rectangle = 25 mm
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(36+25)
P = 2 x 61
P =122 mm
Hence, the Perimeter of the rectangle = 122 mm.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 17
Answer:
GIVEN: Length of the rectangle = 30 cm
Width of the rectangle = 14 cm
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(30+14)
P = 2 x 44
P = 88 cm
Hence, the Perimeter of the rectangle = 88 cm.

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 18
Answer:
GIVEN: Length of the rectangle = 12 (1/4) feet = 49/4 feet = 12.25 feet
Width of the rectangle = 8 feet
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(12.25 + 8)
P = 2 x 20.25
P = 40.5 feet
Hence, the Perimeter of the rectangle = 40.5feet.

Question 5.
Number Sense
What is the perimeter of a square tabletop with side lengths of 48 inches?
Answer:
GIVEN: Side of the square tabletop = 48 inches
Perimeter of the square = 4 x sides
P = 4 x 48
P = 192  inches.
Hence, Perimeter of the square tabletop = 192 inches.

Question 6.
Structure
Use the Distributive Property to write P = (2 × l) + (2 × w) another way.
Answer:
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.
Perimeter of rectangle = (2 × l) + (2 × w)
P = 2(l + w)
Hence, According to distributive property,  the another way of the Perimeter of rectangle = 2(l + w).

Question 7.
Open-Ended
Draw a rectangle that has the same perimeter as the one shown, but different dimensions.
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 19
Answer:
GIVEN: Length of the rectangle = 50 m
Width of the rectangle = 25 m
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2( 50+25)
P = 2 x 75
P = 150 m.
MY FIGURE:

Length of the rectangle = 49m
Width of the rectangle = 26 m
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2( 49+26)
P = 2 x 75
P = 150 m.
Hence, the given rectangle Perimeter and my figure  Perimeter are the same of 150 m.

Question 8.
Modeling Real Life
A worker places tape around a rectangular shipping label that is 2 times longer than it is wide. How much tape does the worker need?
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 20
Answer:
GIVEN:
Width of the rectangle shipping label = 74 mm
Length of the rectangle shipping label = 2 times longer than it is wide =  2 x 74mm = 148mm
Perimeter of the rectangle = (2 x l) + (2 x w) = 2( l + w)
P = 2(148+74)
P = 2 x 222
P = 444  mm.
Hence, Perimeter of the rectangle shipping label = 444 mm.

Question 9.
Modeling Real Life
A coach is painting lines around the perimeter of two rectangular fields. Which field requires more paint?
Big Ideas Math Answer Key Grade 4 Chapter 12 Use Perimeter and Area Formulas 21
Answer:
GIVEN: FIRST A COACH
Length of the rectangle field A = 100yards
Width of the rectangle field A =65 yards
Perimeter of the rectangle field A= (2 x l) + (2 x w) = 2( l + w)
P = 2( 100+65)
P =2 x 165
P = 230 yards
Length of the rectangle field B = 120yards
Width of the rectangle field B =53 1/3 yards = 160/3 = 53.3 yards
P = (2 x l) + (2 x w) = 2( l + w)
P = 2( 120+53.3)
P = 2 x 173.3
P = 346.6  yards
Hence, Perimeter of the rectangle field B is more than Perimeter of the rectangle field A
=346.6 – 230 = 116.6  yards.
Therefore, Perimeter of the rectangle field A requires more 116.6 square yards than Perimeter of the rectangle field B.

Review & Refresh

Write the first six numbers in the pattern. Then describe another feature of the pattern.

Question 10.
Rule: Subtract 11.
First number: 99
Answer:
Rule: Subtract 11.
GIVEN: First number = 99  in the pattern
Second number = 99-11= 88
Third number = 88-11= 77
Fourth number = 77-11 = 66
Fifth number = 66-11 = 55
Sixth number = 55-11 =44
Another feature of this pattern is that all the numbers are multiples of table 11.

Question 11.
Rule: Multiply by 5.
First number: 5
Answer: Rule: Multiply by 5.
GIVEN: First number = 5 in the pattern
Second number = 5 x5 =25
Third number = 25 x 5 = 125
Fourth number = 125 x 5= 625
Fifth number = 625 x 5 = 3125
Sixth number = 3125 x 5 = 15625
Another feature of this pattern is that all the numbers are multiples of table 5.

Lesson 12.2 Area Formula for a Rectangle

Explore and Grow

Use color tiles to create a rectangle with an area of 12 square units. Compare your rectangle to your partner’s. How are they the same? How are they different?
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 22
How do you know that the area of your rectangle is 12 square units?
Answer:

Hence, the rectangle of my friend and mine are going to be same because area = 12 square units given above is same for both the rectangle of us both.

Area of the rectangle =l x w , where “l” is the length and “w” is the width.
Length of the rectangle = 4units
Width of the rectangle = 3units
A = l x w
A = 4 x 3
A = 12 square units.
Hence, the Area of the rectangle = 12 square units.

Structure
How is the area of a rectangle related to its length and width?
Answer:
The area “A” of a rectangle is given by the formula, A=l x w , where “l” is the length and “w” is the width.

Think and Grow: Use a Formula for Area

Area is the amount of surface a figure covers. You can use a formula to show how the length, width, and area of a rectangle are related.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 23
Example
Find the area of the rectangle.
The length is ___ inches and the width is ___ inches.
A = l × w Formula for area of a rectangle
= ___ × ___
= ___
The area is ___ square inches.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 24
Answer:
GIVEN:
Length of the rectangle = 19 inches
Width of the rectangle = 14 inches
Area of the rectangle = l x w
A = 19 x 14
A = 266  square inches
Hence, the Area of the rectangle = 266  square inches.

Show and Grow

Find the area of the rectangle

Question 1.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 25
Answer:
GIVEN:
Length of the rectangle = 6 m
Width of the rectangle = 4 m
Area of the rectangle = l x w
A = 6 x 4
A = 24 square m.
Hence, the are of the rectangle = 24 square m.

Question 2.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 26
Answer:
GIVEN:
Length of the rectangle = 20 yards
Width of the rectangle = 12 yards
Area of the rectangle = l x w
A = 20 x 12
A = 240 square yards
Hence, the area of the rectangle = 240 square yards.

Question 3.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 27
Answer:
GIVEN:
Length of the rectangle = 11 cm
Width of the rectangle = 7cm
Area of the rectangle = l x w
A = 11 x 7
A = 77 square cm
Hence, the area of the rectangle = 77 square cm.

Question 4.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 28
Answer:
GIVEN:
Length of the rectangle =  5 feet
Width of the rectangle = 1 (1/2) feet = 3/2 feet = 1.5 feet
Area of the rectangle = l x w
A = 5 x 1.5
A = 7.5 square feet
Hence, the area of the rectangle = 7.5 square feet.

Apply and Grow: Practice

Find the area of the rectangle.

Question 5.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 29
Answer:
GIVEN:
Length of the rectangle =  35 mm
Width of the rectangle = 32 mm
Area of the rectangle = l x w
A = 35 x 32
A = 1120 mm
Hence, the area of the rectangle = 1120 mm.

Question 6.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 30
Answer:
GIVEN:
Length of the rectangle = 9 inches
Width of the rectangle = 4 (2/8) inches = 34/8 = 4.25 inches
Area of the rectangle = l x w
A = 9 x 4.25
A = 38.25 square inches
Hence, the area of the rectangle = 38.25 square inches.

Question 7.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 31
Answer:
GIVEN:
Length of the rectangle =  8 feet
Width of the rectangle = 5 (1/2) feet =11/2 = 5.5 feet
Area of the rectangle = l x w
A = 8 x 5.5
A = 44 square feet
Hence, the area of the rectangle = 44 square feet.

Question 8.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 32
Answer:
GIVEN:
Length of the rectangle = 79 cm
Width of the rectangle = 37 cm
Area of the rectangle = l x w
A = 79 x 37
A = 2923 square cm
Hence, the area of the rectangle = 2923 square cm.

Question 9.
You are installing wall-to-wall carpet in a rectangular bedroom that is 10 feet long and 9 feet wide. How many square feet of carpet do you need?
Answer:
GIVEN:
Length of the wall-to-wall carpet in a rectangular bedroom = 10 feet
Width of the wall-to-wall carpet in a rectangular bedroom = 9 feet
Area of the rectangle = l x w
A = 10 x 9
A = 90 square feet.
Hence, the area of the wall-to-wall carpet in a rectangular bedroom = 90 square feet
Therefore, 90 square feet of carpet is need.

Question 10.
YOU BE THE TEACHER
Newton says the area of the rectangle is 33 square meters. Descartes says the area is 33 meters. Who is correct? Explain.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 33
Answer:
Length of the rectangle = 11 m
Width of the rectangle = 3 m
Area of the rectangle = l x w
A = 11 x 3
A = 33 square m
The area of the rectangle is 33 square meters.
Hence, Newton says correct not Descartes.

Question 11.
DIG DEEPER!
Write a formula for the area of a square that has a side length of s. Then use your formula to find the area of the square shown.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 34
Answer:
Area of the square = 4 x sides = 4 x s
Given : side of the square = 12 yards
Area of the square = 4 x sides = 4 x s
A= 4 x 12
A = 48 square yards
Hence, the Area of the square = 48 square yards.

Think and Grow: Modeling Real Life

Example
The length of the rectangular dance floor is 6 feet longer than the width. What is the area of the dance floor?
Add 6 feet to the width to find the length.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 35
Answer:
GIVEN: Width of rectangular dance floor = 24 feet
Length of the rectangular dance floor = 6 feet longer than the width = 24 +6 = 30 feet
Area of the  rectangular dance floor = l x w
A = 30 X 24
A = 720 square feet
Hence, the Area of the  rectangular dance floor = 720 square feet.

Show and Grow

Question 12.
A designer creates a rectangular advertisement for a website. The length of the advertisement is 1\(\frac{1}{2}\) centimeters longer than the width. What is the area of the advertisement?
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 36
Answer:
Length of the rectangular advertisement = 1\(\ frac {1}{2}\) centimeters longer than the width= 1(1 /2) cm + 7 cm = =3 / 2 cm + 7 cm = (1.5 + 7 )cm =8.5 cm
Width of the rectangular advertisement = 7 cm
Area of the  rectangular advertisement = l x w
A = 8.5 x 7
A = 59.5 cm
Hence, the Area of the  rectangular advertisement = 59.5 cm.

Question 13.
You create a mural using 4 rectangular posters that are each 4\(\frac{1}{4}\) feet long and 2 feet wide. You put the posters next to each other with no gaps or overlaps. What is the area of the mural?
Answer:
GIVEN:
Width of the rectangular posters=  2 feet
Length of the rectangular posters= 4\(\frac{1}{4}\) feet = 4 (1/4) feet = 17/4 feet = 4.25 feet
Area of the mural = l x w
A = 4.25 x 2
A =  8.5  square feet
Hence, the Area of the mural = 8.5 square feet.

Question 14.
DIG DEEPER!
Two rolls of wrapping paper have the same piece. The red roll is 3 feet wide and is 10 yards long when unrolled. The striped roll is 3\(\frac{1}{2}\) feet wide and 8 yards long when unrolled. Which roll is the better buy? Explain.
Answer:
GIVEN:
Width of Red roll =3 feet
CONVERTION : 1 YARD = 3 FEET
Width of Red roll =3 feet = 1 yard
Length of the red roll = 10 yards
Area of the red roll = l x w
A = 10 x 1
A = 10 square yards.
Area of the red roll = 10 square yards.
Width of striped roll = 3\(\frac{1}{2}\) feet = 3 1/2 feet = 3.5 feet.
CONVERTION : 1 YARD = 3 FEET
Width of striped roll = 3.5 feet
I yard = 3.5 feet
Width of striped roll = yard = 3.5 / 1 feet
Width of striped roll = 3.5 yard
Length of striped roll = 8 yards
Area of the striped roll = l x w
A = 3.5 x 8
A = 28 square yards.
Area of the striped roll = 28 square yards.
Its better to buy striped roll than red roll because u are going to get more.

Area Formula for a Rectangle Homework & Practice 12.2

Find the area of the rectangle

Question 1.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 37
Answer:
GIVEN: Length of the rectangle = 8 m
Width of the rectangle = 7 m
Area of the rectangle = l x w
A = 8 x 7
A = 42 square m
Hence, Area of the rectangle = 42 square m.

Question 2.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 38
Answer:
GIVEN: Length of the rectangle = 21 cm
Width of the rectangle = 9 cm
Area of the rectangle = l x w
A = 21 x 9
A = 189 square cm
Hence, Area of the rectangle = 189 square cm.

Question 3.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 39
Answer:
GIVEN: Length of the rectangle = 56 yards
Width of the rectangle = 12 yards
Area of the rectangle = l x w
A = 56 x 12
A = 672 square yards
Hence, Area of the rectangle = 672 square yards.

Question 4.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 40
Answer:
GIVEN:
Length of the rectangle = 10 (3/4) inches = 43/4 = 10.75 inches
Width of the rectangle = 6 inches
Area of the rectangle = l x w
A = 10.75 x 6
A = 64.5 square inches
Hence, Area of the rectangle = 64.5  square inches.

Question 5.
What is the area of the window?
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 41
Answer:
GIVEN:
Length of the window = 48 inches
Width of the window = 18 inches
Area of the window = l x w
A = 48 x 18
A = 864 square inches
Hence, the Area of the window = 864 square inches.

Question 6.
Structure
A rectangle has an area of 40 square feet. The dimensions are whole numbers. What are all of the possible dimensions of the rectangle?
Answer:
GIVEN:
Area of the rectangle = 40 square feet.
The whole numbers are the part of the number system in which it includes all the positive integers from 0 to infinity. These numbers exist in the number line.
Area of the rectangle = l x w
All of the possible dimensions of the rectangle =
40 =1 × 40
40 = 2 × 20
40 = 4 × 10
40 = 5 × 8.

Question 7.
Open-Ended
Draw a rectangle that has the same area as the one shown, but different dimensions.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 42
Answer:
GIVEN:
Length of the rectangle = 40 mm
Width of the rectangle = 30 mm
Area of the rectangle = l x w
A = 40 x30
A = 1200 mm
Area of the rectangle = 1200 mm.

Length of my rectangle = 48 mm
Width of my rectangle = 25 mm
Area of my rectangle = l x w
A = 48 x 25
A = 1200 mm
Hence, the Area of my rectangle = 1200 mm.

Question 8.
Modeling Real Life
An interior designer says that a rug under a dining room table should be 4 feet longer and 4 feet wider than the table. What is the area of a rug a customer should buy for under the table?
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 43
Answer:
GIVEN:
Length of the dinning table = 6 feet
Width of the dinning table = 3feet
AS PER DESIGNER THE DINNING TABLE REQUIRED DIMENSIONS ARE AS BELOW:
Length should be 4 feet longer than table.
Length of the dinning table= 6 feet
Length of the rug = 6 feet + 4 feet = 10 feet
Width should be 4 feet wider than the table.
Width of the dinning table = 3feet
Width of the rug = 3 feet + 4 feet = 7 feet
Area of the rug = l x w
A = 10 x 7
A = 70 square feet.
Hence, the Area of the rug = 70  square feet.

Question 9.
DIG DEEPER!
A wolf hunts within a rectangular area that is 10 miles long and 5 miles wide. A cougar hunts within a rectangular area that is 8 miles long and 6 miles wide. Which animal hunts within a greater area? How much more area does the animal hunt in?
Answer:
GIVEN:
Length of the wolf hunts rectangular area=10 miles
Width of the wolf hunts rectangular area=5 miles
Area the wolf hunts rectangular area= l x w
A = 10 x 5
A = 50 square miles.
Area the wolf hunts rectangular area= 50 square miles.

Length of the cougar hunts rectangular area= 8 miles
Width of the cougar hunts rectangular area= 6 miles
Area of the cougar hunts rectangular area= l x w
A  = 8 x 6
A = 48 square miles
Area of the cougar hunts rectangular area= 48 square miles.

Hence, wolf hunts within a greater area than the cougar hunts.

Area the wolf hunts rectangular area= 50 square miles.
Area of the cougar hunts rectangular area= 48 square miles.
DIFFERENCE:
Area the wolf hunts rectangular area – Area of the cougar hunts rectangular area
= 50 square miles – 48 square miles
= 2 miles.
Therefore, Wolf hunts more of 2 square miles area than the cougar hunts.

Review & Refresh

Find the quotient.

Question 10.
70 ÷ 7 = __
Answer:
70 ÷ 7 = 10.
The quotient of 70 ÷ 7 = 10.

Question 11.
420 ÷ 6 = ___
Answer:
420 ÷ 6 = 70.
The quotient of 420 ÷ 6 = 70.

Question 12.
2,400 ÷ 8 = ___
Answer:
2,400 ÷ 8 = 300.
The quotient of 2,400 ÷ 8 = 300.

Lesson 12.3 Find Unknown Measures

Explore and Grow

For each row of the table, use color tiles to create the rectangle described. Then complete the table.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 44
Answer:
GIVEN:
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 44
Area of the rectangle = 20 square units
length of the rectangle = 5 units
Width of the rectangle = ??
Let the Width of the rectangle be W
Area of the rectangle = l x w
=> 20 square units = 5units x W
=> 20 / 5 = W
=> 4 units =W
Width of the rectangle = 4 units

GIVEN:
Area of the rectangle = 24 square units
Length of the rectangle = ??
Let the Length of the rectangle = l
Width of the rectangle = 4 units
Area of the rectangle = l x w
=> 24 square units = l x 4units
=> 24 / 4= l
=> 6 units =l
Length of the rectangle = 6 units

GIVEN:
Perimeter of the rectangle = 20 units
Length of the rectangle = ??
Let the length of the rectangle be l
Width of the rectangle = 3 units
Perimeter of the rectangle = 2l x 2w = 2 (l + w)
20 = 2(l+ 3)
20 / 2= L + 3
10 = l + 3
10 – 3 =l
l =7 units.
The Length of the rectangle = 7 units.

GIVEN:
Perimeter of the rectangle = 24 units
Length of the rectangle = 8 units
Width of the rectangle = ??
Let the Width of the rectangle = w
Perimeter of the rectangle = 2l x 2w = 2 (l + w)
24 = 2 ( 8 + w)
24/2 =8 + w
12 = 8 + w
12 – 8 = w
w = 4units
Width of the rectangle = 4units.

The required table is as below:

Reasoning
Compare your strategy to your partner’s. How are they the same or different?
Answer:
The strategy of mine and my partner does not vary as both the required answers value is same.

Think and Grow : Find Unknown Measures

Example
The area of the rectangle is 36 square feet. Find the length.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 45
Answer:
GIVEN:
Area of the rectangle = 36 square feet.
Length of the rectangle = l feet
Width of the rectangle = 3 feet
Area of the rectangle = l x w
36 = l x 3
36/3 = l
12 feet = l
Hence, the Length of the rectangle =12 feet

Example
The perimeter of the rectangle is 30 centimeters. Find the width.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 46
Answer:
GIVEN:
Perimeter of the rectangle = 30 centimeters.
Length of the rectangle = 8 cm
Width of the rectangle = w cm
Perimeter of the rectangle = (2 x l) + ( 2 x w )
30 = ( 2 x 8) + (2 x w)
30 = 16 + 2w
30 – 16 = 2w
14 = 2w
14/2 = w
7 = w
Hence, the Width of the rectangle = 7 cm.

 

Show and Grow

Question 1.
Area = 75 square meters
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 410
Answer:
GIVEN:
Area of the rectangle = 75 square meters
Length of the rectangle = l m
Width of the rectangle = 5 m
Area of the rectangle =  l x w
75 = l x 5
75/5 = l
15 = l
Hence, the length of the rectangle = 15 m.

Question 2.
Perimeter = 42 inches
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 411
Answer:
GIVEN:
Perimeter of the rectangle = 42 inches
Length of the rectangle = 12 inches
Width of the rectangle = w inches
Perimeter of the rectangle = (2 x l) + ( 2 x w ) = 2 ( l + w )

Hence, the Width of the rectangle =  3.5 inches.

Apply and Grow: Practice

Find the unknown measure of the rectangle.

Question 3.
Area = 50 square millimeters
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 47
Answer:
GIVEN:
Area of the rectangle = 50 square millimeters
Length of the rectangle = l mm
Width of the rectangle = 2 mm
Area of the rectangle =  l x w
50 = l x 2
50/2 = l
25 = l
Hence, the length of the rectangle = 25 mm.

Question 4.
Perimeter = 30 centimeters
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 48
Answer:
GIVEN:
Perimeter of the rectangle = 30 centimeters
Length of the rectangle = 9 cm
Width of the rectangle = w cm
Perimeter of the rectangle = (2 x l) + ( 2 x w ) = 2 ( l + w )
30 = 2( 9+w )
30/2 = 9 + w
15 = 9 + w
15 – 9 = w
6 = w
Hence, the width of the rectangle = 6 cm.

Question 5.
Area = 240 square meters
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 49
Answer:
GIVEN:
Area of the rectangle =  240 square meters
Length of the rectangle = 80 m
Width of the rectangle =  w m
Area of the rectangle =  l x w
240 = 80 x w
240/80 = w
3 = w
Hence, the Width of the rectangle =  3 m.

Question 6.
Perimeter = 86 yards
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 50
Answer:
GIVEN:
Perimeter of the rectangle = 86 yards
Length of the rectangle = 23 yards
Width of the rectangle = w yards
Perimeter of the rectangle = (2 x l) + ( 2 x w ) = 2 ( l + w )
86 = 2( 23 + w)
86/2 = 23 +w
43 = 23 + w
43 – 23 = w
20 = w
Hence, the Width of the rectangle = 20 yards.

Question 7.
Perimeter 44 2/4 inches
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 51
Answer:
GIVEN:
Perimeter of the rectangle = 44 (2/4) inches = 90/4 = 22.5 inches
Length of the rectangle = l inches
Width of the rectangle = 8 (1/4) inches = 33/4 =  8.25 inches
Perimeter of the rectangle = (2 x l) + ( 2 x w ) = 2 ( l + w )
22.5 = 2( l+ 8.25)
22.5/2 = l + 8.25
11.25 = l + 8.25
11.25 – 8.25 = l
3.00 = l
Hence, the Length of the rectangle = 3 inches.

Question 8.
Area = 108 square feet
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 52
Answer:
GIVEN:
Area of the rectangle =  108 square feet
Length of the rectangle = l feet
Width of the rectangle =  9 feet
Area of the rectangle =  l x w
108 = l x 9
108/9 = l
12 = l
Hence, the Length of the rectangle = 12 feet.

Question 9.
DIG DEEPER!
What are the dimensions of Newton’s rectangle?
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 53
Answer:
Area of the Newton’s rectangle =  24 square meters
Let the Width of the Newton’s rectangle  be w m.
Length of the Newton’s rectangle = 2 meters longer than the width = 2 + w
Area of the Newton’s rectangle =  l x w
Factors of 24:
1 x 24 = 24 => 1 x ( 1 + 2 ) = 1 x 3= 3
2 x 12 = 24 =>2 x (2+2) =2 x 4 =8
3 x 8 = 24 =>3 x (3+2) = 3 x 5 = 15
4 x 6 = 24 => 4 x (4+2) = 4 x 6 = 24
The dimensions of Newton’s rectangle:
Width of the Newton’s rectangle = 4 m
Length of Newton’s rectangle = 4 + 2= 6 m
CHECK:
Area of of Newton’s rectangle =l x w
A = 4 x 6
A = 24 square meters.
Hence , the The dimensions of Newton’s rectangle = 4 m and 6 m.

 

Question 10.
DIG DEEPER!
The area of a square is 81 square centimeters. What is the perimeter of the square?
Answer:
GIVEN:
Area of a square = 81 square centimeters.
Area of a square  =s2  
81 = s2
√ 81 = s
9 = s
Side of the square = 9 cm
Perimeter of the square = 4x s
P= 4 x 9
P = 36 cm.
Hence, Perimeter of the square = 36 cm.

Think and Grow: Modeling Real Life

Example
The rectangular park has an area of 200 square yards. You kick a soccer ball straight across the width of the park. How far did you kick the soccer ball?
Use a formula to find the width.
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 54
Answer:
GIVEN:
Area of the rectangular park =200 square yards
Length of the rectangular park = 25 yards
Width of the rectangular park =w yards
Area of the rectangular park = l x w
200 = 25 x w
200/25 = w
8 = w
Hence, the Width of the rectangular park =9 yards.
Therefore, the soccer kicks the ball 8 yards far in the rectangular park.
Show and Grow

Question 11.
The rectangular parking spot has an area of 220 square feet. What is the length of the longest car that can fit in the parking spot?
Big Ideas Math Answers 4th Grade Chapter 12 Use Perimeter and Area Formulas 55
Answer:
GIVEN:
Area of the rectangular parking spot = 220 square feet
Width of the rectangular parking spot = 10 feet
Length of the rectangular parking spot = l feet
Area of the rectangular parking spot = l x w
220 = l x 10
220/10 = l
22 = l
Hence, the Length of the rectangular parking spot = 22 feet.
Therefore , the length of the longest car that can fit in the parking spot is 22 feet.

Question 12.
You want to put a frame around the rectangular painting. The painting has a perimeter of 50 inches. How wide should the frame be?
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 56
Answer:
GIVEN:
Perimeter of the rectangular painting frame= 50 inches
Length of the rectangular painting  frame= 14 inches
Width of the rectangular painting frame= w inches
Perimeter of the rectangular painting = 2l + 2w = 2( l + w)
50 = 2(14+w)
50/2 = 14 + w
25 = 14 + w
25 – 14 =w
11 = w
Hence, the Width of the rectangular painting frame= 11 inches.

Question 13.
A rectangular zoo enclosure for a red panda has a perimeter of 116 meters. The length is 50 meters. What is the area of the enclosure?
Answer:
GIVEN:
Perimeter of the rectangular zoo = 116 meters
Length of the rectangular zoo = 50 meters
Width of the rectangular zoo = ??
Let the Width of the rectangular zoo = w meters
Perimeter of the rectangular painting = 2l + 2w = 2( l + w)
116 = 2( 50+ w)
116/2 = 50 + w
58 = 50 + w
58 – 50 = w
8 = w
Hence, the Width of the rectangular zoo = 8 meters
Area of the rectangular zoo =??
Area of the rectangular zoo = l x w
A = 50 x 8
A = 400 square meters
Therefore, the Area of the rectangular zoo = 400 square meters.

Question 14.
DIG DEEPER!
A rectangular patio at a restaurant has an area of 98 square feet. The dimensions of the patio are whole numbers. The length of the patio is 2 times the width. What are the dimensions of the patio?
Answer:
GIVEN:
Area of the rectangular patio = 98 square feet.
The dimensions of the patio are whole numbers.
The length of the patio is 2 times the width.
Let the Width of the rectangular patio be w feet.
=> Length of the the rectangular patio = 2 x w = 2w
Area of the rectangular patio = l x w
98 = (2w) x w
98 = 2w2
98/2 = w2
49 = w2
√ 49 = w
7 = w
Hence, the Width of the rectangular patio = 7 feet.
Length of the the rectangular patio = 2 x w = 2w= 2 x 7 = 14 feet
Therefore, the dimensions of the patio are 14 feet and 7 feet.

Find Unknown Measures Homework & Practice 12.3

Find the unknown measures of the rectangle.

Question 1.
Area = 63 square feet
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 57
Answer:
GIVEN:
Area of the rectangle = 63 square feet
Length of the rectangle = l feet
Width of the rectangle = 7 feet
Area of the rectangle = l x w
63 = l x 7
63/7 = l
9 = l
Hence, the Length of the rectangle = 9 feet.

Question 2.
Perimeter = 26 yards
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 58
Answer:
GIVEN:
Perimeter of the rectangle =  26 yards
Length of the rectangle = l yard
Width of the rectangle = 1 yard
Perimeter of the rectangle = 2l + 2w = 2 ( l + w)
26 = 2( l + 1)
26/2 = l +1
13 = l + 1
13 – 1 = l
12 = l
Hence, the Length of the rectangle = 12 yards.

Find the unknown measure of the rectangle.

Question 3.
Perimeter = 40 centimeters
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 59
Answer:
GIVEN:
Perimeter of the rectangle = 40 centimeters
Length of the rectangle =14 cm
Width of the rectangle = w cm
Perimeter of the rectangle = 2l + 2w = 2 ( l + w)
40 = 2( 14 + w)
40/2 =14 + w
20 = 14 + w
20 – 14= w
6 = w
Hence, the Width of the rectangle = 6 cm.

Question 4.
Area = 88 square millimeters
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 412
Answer:
GIVEN:
Area of the rectangle = 88 square millimeters
Length of the rectangle = l mm
Width of the rectangle = 8 mm
Area of the rectangle = l x w
88 = l x 8
88/8 = l
11 = l
Hence , the Length of the rectangle = 11 mm.

Question 5.
Area = 2,800 square meters
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 60
Answer:
GIVEN:
Area of the rectangle = 2,800 square meters
Length of the rectangle = l m
Width of the rectangle = 40 m
Area of the rectangle = l x w
2800 = l x 40
2800/40 = l
70 = l
Hence, the Length of the rectangle = 70 m.

Question 6.
Perimeter = 41 inches
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 61
Answer:
GIVEN:
Perimeter of the rectangle = 41 inches
Length of the rectangle =12 (4/8) inches = 100/8 = 12.5 inches
Width of the rectangle = w inches
Perimeter of the rectangle = 2l + 2w = 2 ( l + w)
41 = 2( 12.5 +w)
41/2 =12.5 + w
20.5 = 12.5 + w
20.5 – 12.5 = w
8 = w
Hence, the Width of the rectangle = 8 inches.

Question 7.
Structure
A rectangle has an area of 18 square inches and a perimeter of 18 inches. What are the dimensions of the rectangle?
Answer:
GIVEN:
Area of the rectangle = 18 square inches
Perimeter of the rectangle =18 inches.
LET, the Length of the rectangle be l inches
The Width of the rectangle be w inches.
FORMULA:
Area of the rectangle = l x w
18 = l x w
The dimensions of the rectangle :
18 = 1 x 18
18 = 2 x 9
18 = 3 x 6
Hence, the dimensions of the rectangle are (1,18);(2,9);(3,6) inches.

Question 8.
Modeling Real Life
The rectangular fire pit has a perimeter of 176 inches. What is the width of the fire pit?
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 62
Answer:
GIVEN :
Perimeter of the rectangular fire pit = 176 inches
Length of the rectangular fire pit = 50 inches
Width of the rectangular fire pit = ??
Let the Width of the rectangular fire pit be w inches.
Perimeter of the rectangular fire pit = 2l + 2w = 2(l + w)
176 = 2( 50 + w)
176/2 = 50 + w
88 = 50 + w
88 – 50 = w
35 = w
Hence, the Width of the rectangular fire pit =  35 inches.

Question 9.
A painting canvas has an area of 384 square inches. The length and width of the canvas are whole numbers. The length of the canvas is 8 inches greater than the width. What are the dimensions of the canvas?
Answer:
GIVEN:
Area of  the painting canvas = 384 square inches.
The length of the canvas is 8 inches greater than the width.
Let the Width of the painting canvas be w inches.
Area of  the painting canvas = l x w => 384 = l x w
384 = 1x 384 =>1 x (1 + 8) = 1 x 9 = 9
384 = 2 x 192 =>2 x (2+8) = 2 x 10 = 20
384 = 3 x 128 => 3 x (3 + 8) = 3 x 11 = 33
384 = 4 x 96 => 4 x (4 + 8) = 4 x 12 = 48
384= 6 x 64 => 6 x (6 + 8) = 6 x 14 = 84
384 = 8 x 48 => 8 x (8 + 8) =8 x 16 = 128
384 =12 x 32 =>12 x (12 + 8) = 12 x 20 = 240
384 =16 x 24 => 16 x (16 +8) = 16 x 24 = 384
Hence, The dimensions of the canvas=  (24,16) inches
Length of the canvas = 24 inches
Width of the canvas = 16 inches

Review & Refresh

Find the equivalent length.

Question 10.
35 ft = __ in.
Answer:
CONVERTION: 1 feet = 12 inches
35 feet = ?? inches
Let the unknown value be Y inches.
=>35 x 12 = 1 x Y
=>420 = 1 x Y
=> 420/1 = Y
=> 420 = Y
Hence, the unknown value = 420 inches.
35 feet = _420_ inches.

Question 11.
6 mi = __ yd
Answer:
CONVERTION: 1 mile =1760 yard
6 miles = ?? yard
Let the unknown value be X yards.
=> 1 x X = 1760 x 6
=> 1X = 10560
=> X = 10560/1
=> X = 10560
Hence, the unknown value be 10560 yards.
6 mi = _10560_ yd

Question 12.
17 yd = __ ft
Answer:
CONVERTION:
1 yard = 3 feet
17 yd = ?? feet
Let the unknown value be X feet.
=>1 x X = 3 x 17
=>1X = 51
=>X = 51/1
=> X = 51 feet
Hence, the unknown value be 51 feet.
17 yd = _51_ ft

Question 13.
4 yd = __ in.
Answer:
CONVERTION:
1 yard = 36 inches
4 yards = ?? inches
Let the unknown value be Y inches.
=> 1 x Y = 36 x 4
=> 1Y = 144
=> Y = 144/1
=> Y = 144
Hence, the unknown value be 144 inches.
4 yd = _144_ in.

Lesson 12.4 Problem Solving: Perimeter and Area

Explore and Grow

An office has a large rectangular window overlooking a city. Describe two methods for finding the area of the rectangular wall around the window.
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 63
Answer:
GIVEN:
Area of the inside window = l x w
Area of the outside wall = l x w
Area of the rectangular wall around the window = area of the outside wall – Area of the inside window
Hence, there is only one way to find the Area of the rectangular wall around the window.

Make Sense of Problems
Use one of your methods to estimate the area of a wall with a window.
Answer:
The area of the wall can be estimated to be twice the area of the window.

Think and Grow: Problem Solving: Perimeter and Area

Example
A rectangular board has an area of 1,700 square inches. You cut out a rectangular piece that is 10 inches long and 9 inches wide to make a carnival prop similar to the one shown. What is

Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 64
Answer:
GIVEN:
Area of rectangular board =1,700 square inches
Length of the rectangular piece  cut = 10 inches
Width of the rectangular piece cut = 9 inches
Area of the rectangular piece cut = l x w
=>A = 10 X 9
=> A = 90 square inches
Area of the rectangular piece cut = 90 square inches.
Subtract Area of the cut out piece you cut out from the original area.
= 1700 square inches – 90 square inches
= 1610 square inches.
Hence, the Area of the area of the prop = 1610 square inches.

Understand the Problem
What do you know?

  • The original board has an area of 1,700 square inches.
  • The piece you cut out is 10 inches long and 9 inches wide.

What do you need to find?

  • You need to find the area of the carnival prop.

Make a Plan

How will you solve?

  • Find the area of the piece you cut out.
  • Subtract the area of the piece you cut out from the original area.

Solve
Step 1: Find the area of the piece you cut out.
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 65
Step 2: Subtract the area of the you cut out piece you cut out from the original area.
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 66
The area of the prop is __ square inches.
Answer:
Area of rectangular board =1,700 square inches
Length of the rectangular piece  cut = 10 inches
Width of the rectangular piece cut = 9 inches
Area of the rectangular piece cut = l x w
=>A = 10 X 9
=> A = 90 square inches
Area of the rectangular piece cut = 90 square inches.
Subtract Area of the cut out piece you cut out from the original area.
= 1700 square inches – 90 square inches
= 1610 square inches.
Hence, the Area of the area of the prop = 1610 square inches

Show and Grow

Question 1.
Explain how you can check whether your answer above is reasonable.
Answer:
My above answer is reasonable because it satisfies.
Area of the area of the prop + Area of the cut out piece  =The Area of the outside rectangular
=>1610 square inches + 90 square inches
=>1700 square inches.
Hence, proved my answer is reasonable and correct.

Apply and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.
Answer:
GIVEN:
Area of rectangular board =1,700 square inches
Length of the rectangular piece  cut = 10 inches
Width of the rectangular piece cut = 9 inches
This is known in the problem.
You need to find the area of the carnival prop.

Question 2.
A construction worker has 40 feet of caution tape. Is this enough tape to surround a rectangular region that is 120 inches long and 90 inches wide?
Answer:
GIVEN:
A construction worker has 40 feet of caution tape.
Length of the rectangular region = 120 inches
Width of the rectangular region = 90 inches
Perimeter of the rectangular region = 2l + 2 w = 2( l + w)
P = 2( 120 + 90)
P = 2 x 210
P = 420 inches.
CONVERTION: 1 inch = 0.8333 feet  or (1/12) feet
420 inches = ??
Let the unknown value be X feet
=> 1 x X = 420 x 1/12
=> 1X = 420/12
=> 1X = 35 feet.
The required tape to surround the rectangular region = 35 feet.
A construction worker has 40 feet of caution tape.
Therefore, the tape of construction worker bought will be enough to surround a rectangular region that is 120 inches long and 90 inches wide.

Question 3.
One ton of salt de-ices a rectangular section of a road that is 10,500 meters long and 3 meters wide. How many square meters does 6 tons of salt de-ice?
Answer:
GIVEN:
One ton of salt de-ices a rectangular section of a road
Length of the One ton of salt de-ices rectangular section of a road = 10500 meters
Width of the One ton of salt de-ices rectangular section of a road = 3 meters
Area of the One ton of salt de-ices rectangular section of a road = l x w
A = l x w
A = 10500 x 3
A = 31500 square meters
Area of the  One ton of salt de-ices rectangular section of a road = 31500 square meters.
Area of the 6 ton of salt de-ices rectangular section of a road  = ??
Area of the 6 ton of salt de-ices rectangular section of a road  = (Area of the one ton of salt de-ices rectangular section of a road) x 6
=> 31500 x 6
=> 189000 square feet.
Therefore, Area of the 6 ton of salt de-ices rectangular section of a road  = 189000 square feet.

Understand the problem. Then make a plan. How will you solve? Explain.

Question 4.
A worker installs fencing around two rectangular properties. One is 99 feet long and 80 feet wide. The other is95 feet long and 83 feet wide. Which property requires more fencing? How much more?
Answer:
GIVEN:
A worker installs fencing around two rectangular properties.
ONE RECTANGULAR PROPERTY
Length of the  one rectangular property = 99 feet
Width of the one rectangular property = 80 feet
Perimeter of the one rectangular property = 2l + 2w = 2( l + w)
P = 2 ( 99+ 80 )
P = 2 x 179
P = 358 feet
Perimeter of the one rectangular property = 358 feet.
OTHER RECTANGULAR PROPERTY
Length of the  other rectangular property = 95 feet
Width of the other rectangular property = 83 feet
Perimeter of the other rectangular property = 2l + 2w = 2( l + w)
P = 2 ( 95 + 83 )
P = 2 x 178
P = 356 feet
Perimeter of the other rectangular property = 356 feet
DIFFERENCE:
Perimeter of the one rectangular property – Perimeter of the other rectangular property
=> 358 – 356
=> 2 feet.
One rectangular property requires more fencing than the other rectangular property.
Therefore, 2 feet more it requires than the one rectangular property than the other rectangular property.

Question 5.
A roofer covers the rectangular roof with shingles. A chimney occupies a rectangular area that is 4 feet long and 2 feet wide. How many square feet of the roof are with shingles?
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 67
Answer:
GIVEN:
A roofer covers the rectangular roof with shingles.
Length of the rectangular area of chimney = 4 feet
Width of the rectangular area of chimney = 2 feet
Area of the rectangular area of chimney = l x w
A = 4 x 2
A = 8 square feet
Area of the rectangular area of chimney = 8 square feet.
Length of the rectangular roof = 60 feet
Width of the rectangular roof= 30 feet
Area of the rectangular roof = l x w
A = 60 x 30
A = 1800 square feet
Area of the rectangular roof = 1800 square feet.
Area of the roof are with shingles = Area of the rectangular roof – Area of the rectangular area of chimney
=> 1800 square feet – 8 square feet
=> 1792 square feet.
Area of the roof are with shingles = 1792 square feet.

Question 6.
You want to buy a cover for the lid of your laptop. Your laptop is 1\(\frac{1}{3}\) feet long and 1 foot wide. Which cover will fit best on your laptop?
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 68
Answer:
GIVEN:
Length of the Laptop = 1\(\frac{1}{3}\) feet long =1 (1/3) feet = 4/3 feet = 1. 33 feet.
Width of the laptop = 1 feet
Perimeter of the laptop =2l + 2w = 2( l + w)
P = 2( 1.33 + 1)
P =2 x 2.33
P = 4.66 feet
Perimeter of the laptop =4.66 feet.
CONVERTION:
1 feet = 12 inches
4.66 feet =??
=> 4.66 x 12
=> 55.92 inches
Hence, the perimeter of the laptop = 55.92 inches.
FIRST LAPTOP COVER :
Length of the first Laptop cover = 16 inches
Width of the first Laptop cover = 12 inches
Area of the first Laptop cover =  192 square inches
Perimeter of the first Laptop cover = 2l + 2w = 2( l + w)
P = 2( 16 + 12)
P = 2 x 28
P = 56 inches.
Perimeter of the first Laptop cover = 56 inches.
SECOND LAPTOP COVER :
Length of the second Laptop cover = 13 inches
Width of the second Laptop cover = 9 (3/4)inches = 39/4  inches = 9.75 inches
Area of the second Laptop cover =  126 (3/4) square inches
Perimeter of the second Laptop cover = 2l + 2w = 2( l + w)
P = 2( 13 +9.75 )
P = 2 x 22.75
P =  45.5 inches.
Perimeter of the second Laptop cover = 45.5 inches.
THIRD LAPTOP COVER
Length of the third Laptop cover = 14 (1/2) inches = 29/2  inches= 14.5 inches
Width of the third Laptop cover = 11 inches
Area of the third Laptop cover =  159 (1/2) square inches
Perimeter of the third Laptop cover = 2l + 2w = 2( l + w)
P = 2( 14.5 + 11 )
P = 2 x 25.5
P = 51 inches.
Perimeter of the third Laptop cover = 51 inches.
Therefore, first laptop cover will fit correctly to the laptop because cover size is more accurate size of 56 inches than other two laptop covers are lesser in size of the laptop size 55.92 inches.

Think and Grow: Modeling Real Life

Example
A worker wants to cover the miniature golf putting surface with artificial turf. The putting surface is in the shape of two rectangles. How much turf does the worker need?
Think: What do you know? What do you need to find? How will you solve?
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 69
Step 1: Divide the surface into two rectangles. Then find the area of each rectangle.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 70
Answer:
GIVEN:
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 69
Divide the surface into two rectangles. Then find the area of each rectangle.

Length of the Rectangle A = 16 feet
Width of the Rectangle A = `4 feet
Area of the Rectangle A = l x w
A =16 x 4
A = 64 square feet
Area of the Rectangle A = 64 square feet.
Length of the Rectangle B = `10 feet
Width of the Rectangle B = `5 feet
Area of the Rectangle B = l x w
A = 10 x 5
A = 50 square feet
Area of the Rectangle B = 50 square feet.
ADDITION:
Area of the Rectangle A + Area of the Rectangle B
=> 64 square feet + 50 square feet
=> 114 square feet.
Hence, the workers need 114 square feet for artificial turf.

Question 7.
You want to install new carpet in the rectangular bedroom and the rectangular closet. How much carpet do you need to cover the floor?
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 413

Answer:
GIVEN:
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 413
Length of the rectangular bedroom = 20 feet
Width  of the rectangular bedroom = 10 feet
Area of the rectangular bedroom = l x w
A = 20 x 10
A = 200 square feet
Area of the rectangular bedroom = 200 square feet.
Length of the rectangular closet = 10 feet
Width  of the rectangular closet= 4 feet
Area of the rectangular closet = l x w
A = 10 x 4
A = 40 square feet
Area of the rectangular closet = 40 square feet.
ADDITION:
Carpet  needed to cover the floor = Area of the rectangular bedroom + Area of the rectangular closet.
=> 200 square feet + 40 square feet
=> 240 square feet.
Hence, the Carpet  needed to cover the floor = 240 square feet.

Question 8.
A gardener wants to enclose the garden with fencing. The garden is in the shape of two rectangles. How much fencing does the gardener need?
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 414
Answer:
Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas 414
Divide the surface into two rectangles.


Length of the Rectangle A = 8 yard
Width of the Rectangle A = 6 yard
Perimeter of the Rectangle A = 2l x 2w = 2( l + w)
P = 2( 8 + 6 )
P = 2 x 14
P = 28 yards
Perimeter of the Rectangle A = 28 yards.
Length of the Rectangle B = 10 yard
Width of the Rectangle B = 3 yard
Perimeter of the Rectangle B = 2l x 2w = 2( l + w)
P = 2( 10 + 3 )
P = 2 x 13
P = 26 yards
Perimeter of the Rectangle B = 26 yards.
Fencing the gardener needed = Perimeter of the Rectangle A + Perimeter of the Rectangle B
=> 28 yards + 26 yards
=> 54 yards.
Hence, the Fencing the gardener needed = 54 yards.

Problem Solving: Perimeter and Area Homework & Practice 12.4

Understand the problem. Then make a plan. How will you solve? Explain.

Question 1.
An indoor dog park has an area of 50,000 square feet. The owner creates a square welcome center inside the park that is 100 feet long. What is the area of the section that dogs can play in?
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 71
Answer:
GIVEN:
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 71

Area of the indoor dog park = 50,000 square feet.
The owner creates a square welcome center inside the park that is 100 feet long.
=> side of the square welcome center inside the park = 100 feet
Area of the square welcome center inside the park =  s x s
A = 100 x 100
A  = 10,000 square feet
Area of the square welcome center inside the park = 10,000 square feet.
Area of the section that dogs can play in = Area of the indoor dog park – Area of the square welcome center inside the park
=> 50,000 square feet – 10,000 square feet
=> 40,000 square feet.
Hence, the Area of the section that dogs can play in = 40,000 square feet.

Question 2.
You tile a hallway with square tiles that are 12 inches wide. You completely cover the hallway with 3 rows of 5 tiles. What is the area of the hallway?
Answer:
GIVEN:
Side of the square tiles = 12 inches
The hallway completely covered with 3 rows of 5 tiles.
Area of the square tiles hallway= s x s
A = 12 x 12
A = 144 square inches.
Hence, the Area of the square tiles hallway= 144 square inches.

Question 3.
Your friend makes a rectangular poster for a school play. The poster is 4 feet long and 3 feet wide. Ribbon costs $1 per foot. How much does it cost to add a ribbon border to the poster?
Answer:
GIVEN:
Length of the rectangular poster = 4 feet
Width of the rectangular poster = 3 feet
Perimeter of the rectangular poster = 2l +2w = 2(l + w)
P = 2 ( 4 + 3 )
P = 2 x 7
P = 14 feet
Perimeter of the rectangular poster = 14 feet.
Ribbon costs $1 per foot
Cost of the ribbon border to the poster = 14 feet x $1= $14
Therefore, the Cost of the ribbon border to the poster = $14.

Question 4.
Writing
Explain how you know when you need to find the perimeter or the area of a rectangle when solving a word problem.
Answer:
GIVEN:
Perimeter can be thought  as the length of the outline of a shape.
Area can be defined as the space occupied by a flat shape or the surface of an object.
On the above referred points,  I try to calculate the perimeter or the area of the rectangle when solving a word problem.

Question 5.
Modeling Real Life
Your teacher joins two rectangular tables for students to complete a craft. How much newspaper does your teacher need to cover the tops of the tables with no overlap and no paper hanging over the sides?
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 72
Answer:
GIVEN:
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 72
Divide the surface into two rectangles.

Length of the rectangle A = 72 inches
Width of the of the rectangle A = 30 inches
Perimeter of the rectangle A = 2l + 2w = 2( l + w)
P = 2( 72 + 30 )
P =  2 x 102
P = 204 inches
Perimeter of the rectangle A = 204 inches.
Length of the rectangle B = 60 inches
Width of the of the rectangle B = 30 inches
Perimeter of the rectangle B = 2l + 2w = 2( l + w)
P = 2( 60 + 30 )
P = 2 x 90
P = 180 inches
Perimeter of the rectangle B = 180 inches.
Newspaper needed to cover the tops of the tables with no overlap and no paper hanging over the sides = Perimeter of the rectangle A + Perimeter of the rectangle B
= 204 inches + 180 inches
= 384 inches.
Hence, the newspaper needed to cover the tables = 384 inches.

Question 6.
Modeling Real Life
A landscaper buys 2 bags of grass seed. Each bag covers 5,000 square feet. A rectangular lawn is 200 feet long and 40 feet wide. Does the landscaper have enough seed to cover the lawn once? twice? Explain.
Answer:
GIVEN:
A landscaper buys 2 bags of grass seed.
Area of the bags seed covers = 5000 square feet.
Length of the rectangular lawn = 200 feet
Width of the rectangular lawn = 40 feet
Area of the rectangular lawn = l x w
A = 200 x 40
A = 8000 square feet.
Area of the rectangular lawn = 8000 square feet.
The landscaper have enough seed to cover the lawn once only not twice because he has  bags which can cover only 5000 square feet not 8000 square feet rectangular lawn completely.

Review & Refresh

Multiply

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 73
Answer:
GIVEN:
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 73= 2 x 1/4
= 1 x 1/2
= 1/2
= 0.5
Hence, Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 73= 0.5.

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 74
Answer:
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 74= 12/5
= 2.4.
Hence, Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 74 = 2.4.

Question 9.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 75
Answer:
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 75= 6 x 5/4
= 3 x 5/2
=15/2
= 7.5.
Hence, Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 75 = 7.5.

Use Perimeter and Area Formulas Performance Task 12

Stop-motion animation videos are made by taking multiple photographs of an object. Each photograph shows the object in a slightly different position. When all of the photographs are combined into a video, the object appears to be moving.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 76
1. You decide to make a stop-motion video. You make a background with an area of 12 square feet and a perimeter of 14 feet.
a. What are the dimensions of your background?
b. The part of the background in each photograph is 45 inches long and 30 inches wide. What is the perimeter of the background in each photograph?
c. What is the area of the background that is not in each photograph?
2. You take photographs for your video. Your video shows 15 photographs each second. How many photographs are in a 24-second video?
3. You start working on your animation at 3:05 P.M. and finish at 5:20 P.M. You spend equal amounts of time creating your background, taking photographs, and editing your video. How much time do you spend on each activity?
Answer:
GIVEN:
Area of the rectangular background= 12 square feet.
Perimeter  of the rectangular background = 14 feet.
1. a) All dimensions of the rectangular background =
Area of the rectangle = l x w
1 × 12 = 12.
2 × 6 = 12.
3 × 4 = 12.
The dimensions of the rectangular background = (1,12); (2,6);(3,4).
b) The Length of the background in each photograph = 45 inches The Width  of the background in each photograph =30 inches  The perimeter of the background in each photograph = 2l + 2w = 2(l + w)
P = 2 (45 + 30)
P = 2 x 75
P = 150 inches.
Hence, the perimeter of the background in each photograph = 150 inches.
2) 
GIVEN:
15 photograph taken in 1 second
=> ?? in 24 seconds
=> 24 x 15 = photographs taken in 24 seconds
=> 360 = photographs taken in 24 seconds
Hence, photographs taken in 24 seconds = 360.
3)
GIVEN:
Starting time on animation at 3:05 P.M.
Finishing time on animation at 5:20 P.M.
Time taken on each activity = Finishing time on animation – Starting time on animation
=> 5:20 P.M – 3:05 P.M
=> 2: 15 Hours
Hence, Time taken on each activity =2:15 Hours.

Use Perimeter and Area Formulas Activity

Area Roll and Conquer

Directions:

  1. Players take turns rolling two dice.
  2. On your turn, create a rectangle with the numbers on the dice as the length and width. Your rectangle cannot cover another rectangle.
  3. Shade the rectangle in your color. Record the multiplication equation for the rectangle.
  4. If you cannot create a rectangle on the board, then you lose your turn. Play 10 rounds, if possible.
  5. The player with the greatest area covered wins!

Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 77
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 78
Answer:
GIVEN:

  1. Players take turns rolling two dice.

1)
2)

Length of the rectangle = 4 units
Width of the rectangle = 3 units.
3)
Equation of the rectangle = 4 x 3 = 12 square units.
4)   
5) 
The player with the greatest area covered wins!
Given Equation Area = 3 x 2 = 6 square units
My Equation Area= 4 x 3 =12 square units

Use Perimeter and Area Formulas Chapter Practice

12.1 Perimeter Formula for a Rectangle

Find the perimeter of the rectangle.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 79
Answer:
GIVEN:
Length of the rectangle = 15 yard
Width of the rectangle = 13 yard
Perimeter of the rectangle = 2l + 2w = 2( l + w)
P = 2 (15 + 13)
P = 2 x 28
P = 56 yards
Hence, the Perimeter of the rectangle = 56 yards.

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 80
Answer:
GIVEN:
Length of the rectangle = 29 mm
Width of the rectangle = 23 mm
Perimeter of the rectangle = 2l + 2w = 2( l + w)
P = 2 ( 29 + 23)
P = 2 x 52
P = 104 mm
Hence, the Perimeter of the rectangle = 104 mm.

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 81
Answer:
GIVEN:
Length of the rectangle = 55 cm
Width of the rectangle = 46 cm
Perimeter of the rectangle = 2l + 2w = 2( l + w)
P = 2( 55 + 46)
P = 2 x 101
P = 202 cm
Hence, the Perimeter of the rectangle = 202 cm.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 82
Answer:
GIVEN:
Length of the rectangle = 10 (2/12) feet = 20/12 = 1.67 feet
Width of the rectangle = 7 feet
Perimeter of the rectangle = 2l + 2w = 2( l + w)
P = 2(1.67 + 7)
P = 2 x 7.67
P = 15.34 feet
Hence, the Perimeter of the rectangle = 15.34 feet.

12.2 Area Formula for a Rectangle

Find the area of the rectangle.

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 83
Answer:
GIVEN:
Length of the rectangle =  58 m
Width of the rectangle = 37 m
Area of the rectangle = l x w
A = 58 x 37
A = 2146 square m.
Hence, the Area of the rectangle = 2146 square m.

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 84
Answer:
GIVEN:
Length of the rectangle = 13 (1/2) inches = 27/2 = 13.5 inches
Width of the rectangle = 7 inches
Area of the rectangle = l x w
A = 13.5 x 7
A = 94.5 square inches
Hence, the Area of the rectangle = 94.5 square inches.

Question 7.
A rectangle has an area of 60 square feet. The dimensions are whole numbers. What are all of the possible dimensions of the rectangle?
Answer:
GIVEN:
Length of the rectangle = l feet
Width of the rectangle = w feet
Area of the rectangle = 60 square feet
Area of the rectangle = l x w
All possible dimensions of rectangle =
60 = 1 x 60,
60 = 2 x 30,
60 = 3 x 20,
60 = 4 x 15
60 = 5 x 12,
60 = 6 x 10.
All possible dimensions of rectangle =(1,60);(2,30);(3,20);(4,15);(5,12);(6,10).

12.3 Find Unknown Measures

Find the unknown measure of the rectangle.

Question 8.
Area = 48 square yards
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 85
Answer:
GIVEN:
Area of the rectangle = 48 square yards
Length of the rectangle = l yards
Width of the rectangle = 3 yards
Area of the rectangle = l x w
48 = l x 3
48/3 = l
16 = l
Hence, the Length of the rectangle = 16 yards.

Question 9.
Perimeter = 90 centimeters
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 86
Answer:
GIVEN:
Perimeter of the rectangle = 90 centimeters
Length of the rectangle = l cm
Width of the rectangle = 12 cm
Perimeter of the rectangle = 2l + 2w = 2( l + w)
90 = 2( l + 12)
90/2 = l + 12
45 = l + 12
45 – 12 = l
33 = l
Hence, the Length of the rectangle = 33 cm.

Question 10.
Logic
What are the dimensions of Descartes’s rectangle?
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 87
Answer:
GIVEN:
Perimeter of the rectangle = 10 meters
Length and Width have a product = 6 square meters
=> l x w = 6 square meters
=> Area of the rectangle = l x w = 6 square meters.
All possible dimensions of the rectangle =
6 = 1 x 6
6 = 2 x 3.
Hence, All possible dimensions of the rectangle =(1,6);(2,3).

12.4 Problem Solving: Perimeter and Area

Question 11.
Modeling Real Life
You want to paint the wall. What is the area of the wall you will paint?
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 88
Answer:
GIVEN:
Length of the outer rectangle = 14 feet
Width of the outer rectangle = 10 feet
Area of the outer rectangle = l x w
A = 14 x 10
A = 140 square feet.
Area of the outer rectangle = 140 square feet.
Length of the inside rectangle = 4 feet
Width of the inside rectangle = 1(1/2) feet = 3/2 = 1.5 feet
Area of the inside rectangle = l x w
A = 4 x 1.5
A = 6 square feet.
Area of the inside rectangle = 6 square feet.
SUBTRACTION:
Area of the wall for paint = Area of the outer rectangle  – Area of the inside rectangle
= 140 square feet – 6 square feet.
= 134 square feet.
Hence, the Area of the wall for paint = 134 square feet.

Question 12.
A park director orders 360 feet of fencing. Does he have enough to surround the tennis court?
Big Ideas Math Solutions Grade 4 Chapter 12 Use Perimeter and Area Formulas 89
Answer:
GIVEN:
Fencing length park director ordered = 360 feet.
Length of the rectangle = 40 yards
Width of the rectangle = 20 yards
Perimeter of the rectangle = 2l + 2w = 2 ( l + w)
P= 2( 40 + 20)
P = 2 x 60
P = 120 yards
CONVERTION:
1 yard = 3 feet
120 yards = ??
=> 120 x 3
=> 360 feet.
Hence, the director ordered fencing  360 feet is enough for fencing the tennis court because the required court perimeter is same 360 feet.

Conclusion:

I wish the information provided in the Big Ideas Math Answers Grade 4 Chapter 12 Use Perimeter and Area Formulas is helpful for you all. Share the pdf links with your friends and help them to overcome the difficulties in maths. If you any doubts feel free to post the comments in the below-mentioned comment box. Bookmark our page to get the updates of Big Ideas Math Grade 4 Answer Key for all Chapters.

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 is designed to build knowledge when they are not in class. Our aim is to provide the best materials to students to enhance their math skills and to get interested in the subject. There are many ways to solve the problems know how to solve the problems in a simple manner on this page. With the help of the Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 you can score good marks in the exams.

Big Ideas Math Book Grade K Answer Key Chapter 3 Count and Write Numbers 6 to 10

To excel in the exam you have to choose the best material to practice for the exams. Elementary School students can find detailed explanations for all the questions in the Big Ideas Math Book Grade K Solution Key Ch 3 Count and Write Numbers 6 to 10. The Chapter Count and Write Numbers 6 to 10 chapter includes Model and Count 6, Understand and Write 6, Model and Count 7, Understand and Write 7, and so on. Thus Download Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 free pdf and start your preparation.

Vocabulary

Lesson: 1 Model and Count 6

Lesson: 2 Understand and Write 6

Lesson: 3 Model and Count 7

Lesson: 4 Understand and Write 7

Lesson: 5 Model and Count 8

Lesson: 6 Understand and Write 8

Lesson: 7 Model and Count 9

Lesson: 8 Understand and Write 9

Lesson: 9 Model and Count 10

Lesson: 10 Understand and Write 10

Lesson: 11 Count and Order Numbers to 10

Performance Tasks

Count and Write Numbers 6 to 10 Vocabulary

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 1
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 2

Directions:

  • Count the objects. Say the number. Write the number.
  • Compare the number of lions to the number of cars. Circle the number that is greater than the other number.
    Answer:
    Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10

Chapter 3 Vocabulary Cards

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 3
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 4

Lesson 3.1 Model and Count 6

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 5
Directions:
Place 6 counters on the parking lot. Slide the counters to the frame.
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 Model and Count 6

Think and Grow
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 6
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Think and Grow

Directions:
Count the objects. Color the boxes to show how many.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 7
Answer:
There are 6 objects
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Apply and Grow-Practice1.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 8
Answer:
There are 5 objects
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Apply and Grow-Practice2.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 9
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Apply and Grow-Practice3.

Directions:
1 – 3 Count the objects. Color the boxes to show how many.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 10
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 10.1
Directions:
Count the objects in the picture. Color the boxes to show how many.
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Think and Grow-Modeling Real Life-1

Model and Count 6 Homework & Practice 3.1

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 12
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 6 Homework & Practice 3.1.1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 13
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 6 Homework & Practice 3.1.2

Directions:
1 and 2 Count the objects. Color the boxes to show how many.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 14
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 6 Homework & Practice 3.1.3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 15
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 6 Homework & Practice 3.1.4

Question 5.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 16
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 6 Homework & Practice 3.1.5

Directions:
3 and 4 Count the objects. Color the boxes to show how many. 5 Count the objects in the picture. Color the boxes to show how many.

Lesson 3.2 Understand and Write 6

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 17
Directions:
Use counters to show how many drums are in the story Music Class. Write how many drums are in the story.

Answer:
Number of Drums in the story = 2

Explanation:
blocks

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 18
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 19

Directions:

  • Count the objects. Say the number. Trace and write the number.
  • Count the instruments. Say the number. Write the number.
    Answer:
    Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Think and GrowBig-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Think and Grow...

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 20
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Apply and Grow-Practice.1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 21

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Apply and Grow-Practice.2

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 22

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Apply and Grow-Practice.3

Directions:
1 – 4 Count the objects. Say the number. Write the number.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 23
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Think and Grow-Modeling Real Life

Directions:
Count the instruments in the picture. Say the number. Write the number.

Understand and Write 6 Homework & Practice 3.2

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 24
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Understand and Write 6 Homework & Practice 3.2.1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 25
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Understand and Write 6 Homework & Practice 3.2.2

Directions:
1 and 2 Count the dots. Say the number. Write the number.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 26
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Understand and Write 6 Homework & Practice 3.2.3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 27
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Understand and Write 6 Homework & Practice 3.2.4

Question 5.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 28
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.1 -Understand and Write 6 Homework & Practice 3.2.5

Directions:
3 and 4 Count the objects. Say the number. Write the number. 5 Count the instruments in the picture. Say the number. Write the number.

Lesson 3.3 Model and Count 7

Explore and Grow
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 29
Directions:
Place 7 counters in the forest. Slide the counters to the frame.
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.3 Model and Count 7

Think and Grow
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 70
Directions:
Count the objects. Color the boxes to show how many.
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.3 Model and Count 7-Think and Grow

Apply and Grow: Practice

Question 1.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 31
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.3 Model and Count 7-Apply and Grow-Practice1

Question 2.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 32
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.3 Model and Count 7-Apply and Grow-Practice2

Question 3.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 33
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.3 Model and Count 7-Apply and Grow-Practice3

Directions:
1 – 3 Count the animals. Color the boxes to show how many.

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 34
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.3 Model and Count 7-Think and Grow-Modeling Real Life

Directions:
Count the animals in the picture. Color the boxes to show how many.

Model and Count 7 Homework & Practice 3.3

Question 1.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 36
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 7 Homework & Practice 3.3.1

Question 2.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 37
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 7 Homework & Practice 3.3.2

Directions:
1 and 2 Count the animals. Color the boxes to show how many.

Question 3.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 38
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 7 Homework & Practice 3.3.3

Question 4.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 39
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 7 Homework & Practice 3.3.4

Question 5.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 40
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Model and Count 7 Homework & Practice 3.3.5

Directions:
3 and 4 Count the animals. Color the boxes to show how many. 5 Count the animals in the picture. Color the boxes to show how many.

Lesson 3.4 Understand and Write 7

Explore and Grow

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 41
Directions:
Use counters to show how many raindrops are in the story Rainy Day. Write how many raindrops are in the story.
Answer:
Number of Rain drops in the story = one or 1

Explanation:
Rain drops in the story =
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4 Understand and Write 7

Think and Grow

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 42
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4 Understand and Write 7-Think and Grow
Directions:

  • Count the objects. Say the number. Trace and write the number.
  • Count the objects. Say the number. Write the number.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 43
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Apply and Grow-Practice1

Question 2.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 44
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Apply and Grow-Practice2

Question 3.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 45
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Apply and Grow-Practice3

Question 4.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 46
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Apply and Grow-Practice4

Directions:
1 – 4 Count the objects. Say the number. Write the number.

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 47
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Think and Grow-Modeling Real Life

Directions:
Count the objects in the picture. Say the number. Write the number.

Understand and Write 7 Homework & Practice 3.4

Question 1.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 48
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Understand and Write 7 Homework & Practice 3.4.1

Question 2.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 49
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Understand and Write 7 Homework & Practice 3.4.2

Directions:
1 and 2 Count the dots. Say the number. Write the number.

Question 3.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 50
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Understand and Write 7 Homework & Practice 3.4.3

Question 4.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 51
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Understand and Write 7 Homework & Practice 3.4.4

Question 5.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 52
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Understand and Write 7 Homework & Practice 3.4.5

Directions:
3 and 4 Count the objects. Say the number. Write the number. 5 Count the objects in the picture. Say the number. Write the number.

Lesson 3.5 Model and Count 8

Explore and Grow

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 53
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.4-Lesson 3.5 Model and Count 8

Directions:
Place 8 counters in the desert. Slide the counters to the frame.

Think and Grow

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 54

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 55
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5 Model and Count 8-Think and Grow

Directions:
Count the objects. Color the boxes to show how many.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 57.1
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5 Model and Count 8-Apply and Grow-Practice1

Question 2.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 57.2
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5 Model and Count 8-Apply and Grow-Practice2

Question 3.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 57.3
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5 Model and Count 8-Apply and Grow-Practice3

Directions:
1 – 3 Count the objects. Color the boxes to show how many.

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 58
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5-Think and Grow-Modeling Real Life
Directions:
Count the objects in the picture. Color the boxes to show how many.

Model and Count 8 Homework & Practice 3.5

Question 1.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 59
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5-Model and Count 8 Homework & Practice 3.5.1

Question 2.
Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 60
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5-Model and Count 8 Homework & Practice 3.5.2

Directions:
1 and 2 Count the objects. Color the boxes to show how many.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 61
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5-Model and Count 8 Homework & Practice 3.5.3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 62
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5-Model and Count 8 Homework & Practice 3.5.4

Question 5.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 63
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.5-Model and Count 8 Homework & Practice 3.5.5

Directions:
3 and 4 Count the objects. Color the boxes to show how many. 5 Count the objects in the picture. Color the boxes to show how many.

Lesson 3.6 Understand and Write 8

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 65
Answer:
There are zero bugs on the Ladybug in the story….

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 Understand and Write 8
Directions:
Use counters to show how many spots are on the ladybug in the story Bugs, Bugs, Bugs. Write how many spots are on the ladybug.

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 66

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 67
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Think and Grow
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Think and Grow...

Directions:

  • Count the objects. Say the number. Trace and write the number.
  • Count the bugs. Say the number. Write the number.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 69
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Apply and Grow-Practice1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 700
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Apply and Grow-Practice2

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 71
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Apply and Grow-Practice3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 72
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Apply and Grow-Practice4

Directions:
1 – 4 Count the objects. Say the number. Write the number.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 73
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Think and Grow-Modeling Real Life

Directions:
Count the objects in the picture. Say the number. Write the number.

Understand and Write 8 Homework & Practice 3.6

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 74
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Understand and Write 8 Homework & Practice 3.6.1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 75
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Understand and Write 8 Homework & Practice 3.6.2

Directions:
1 and 2 Count the dots. Say the number. Write the number.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 76
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Understand and Write 8 Homework & Practice 3.6.3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 77
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Understand and Write 8 Homework & Practice 3.6.4

Question 5.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 78
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.6 -Understand and Write 8 Homework & Practice 3.6.5

Directions:
3 and 4 Count the objects. Say the number. Write the number. 5 Count the objects in the picture. Say the number. Write the number.

Lesson 3.7 Model and Count 9

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 79
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9

Directions:
Place 9 counters on the beach. Slide the counters to the frame.

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 80
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Think and Grow

Directions:
Count the objects. Color the boxes to show how many.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 83
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Apply and Grow-Practice1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 84
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Apply and Grow-Practice2

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 85
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Apply and Grow-Practice3

Directions:
1 – 3 Count the objects. Color the boxes to show how many.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 86
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 87
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Think and Grow-Modeling Real Life

Directions:
Count the objects in the picture. Color the boxes to show how many.

Model and Count 9 Homework & Practice 3.7

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 89
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Model and Count 9 Homework & Practice 3.7.1

Question 2.

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 90
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Model and Count 9 Homework & Practice 3.7.2

Directions:
1 and 2 Count the objects. Color the boxes to show how many.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 91
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Model and Count 9 Homework & Practice 3.7.3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 92
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Model and Count 9 Homework & Practice 3.7.4

Question 5.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 93
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Model and Count 9 Homework & Practice 3.7.5

Directions:
3 and 4 Count the objects. Color the boxes to show how many. 5 Count the animals in the picture. Color the boxes to show how many.

Lesson 3.8 Understand and Write 9

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 94

Directions:
Use counters to show how many baseballs are in the story My Baseball Game. Write how many baseballs are in the story.
Answer:
Number of balls in the story My Baseball Game = One or 1.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Lesson 3.8 Understand and Write 9

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 95

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 96
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Lesson 3.8 Understand and Write 9-Think and Grow

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson 3.7 Model and Count 9-Lesson 3.8 Understand and Write 9-Think and Grow....

Directions:

  • Count the objects. Say the number. Trace and write the number.
  • Count the objects. Say the number. Write the number.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 97
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Apply and Grow-Practice1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 97.1
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Apply and Grow-Practice2

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 97.2
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Apply and Grow-Practice3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 97.3
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Apply and Grow-Practice4

Directions:
1 – 4 Count the objects. Say the number. Write the number.

Think and Grow: Modeling Real Life
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 98
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Think and Grow-Modeling Real Life
Directions:
Count the balls in the picture. Say the number. Write the number.

Understand and Write 9 Homework & Practice 3.8

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 99
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Understand and Write 9 Homework & Practice 3.8.1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 100
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Understand and Write 9 Homework & Practice 3.8.2

Directions:
1 and 2 Count the dots. Say the number. Write the number.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 101
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Understand and Write 9 Homework & Practice 3.8.3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 102
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Understand and Write 9 Homework & Practice 3.8.4

Question 5.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 103
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.8 Understand and Write 9-Understand and Write 9 Homework & Practice 3.8.5

Directions:
3 and 4 Count the objects. Say the number. Write the number. 5 Count the bowling balls in the picture. Say the number. Write the number.

Lesson 3.9 Model and Count 10

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 104
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10

Directions:
Place 10 counters in the soil. Slide the counters to the frame.

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 105

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 106
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Think and Grow

Directions:
Count the objects. Color the boxes to show how many.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 107
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Apply and Grow-Practice1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 108
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Apply and Grow-Practice2

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 109
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Apply and Grow-Practice3

Directions:
1 – 3 Count the objects. Color the boxes to show how many.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 110
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Think and Grow-Modeling Real Life

Directions:
Count the objects in the picture. Color the boxes to show how many.

Model and Count 10 Homework & Practice 3.9

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 111
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Model and Count 10 Homework & Practice 3.9.1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 112
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Model and Count 10-Homework & Practice 3.9.2

Directions:
1 and 2 Count the objects. Color the boxes to show how many.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 113
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Model and Count 10-Homework & Practice 3.9.3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 114
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Model and Count 10-Homework & Practice 3.9.4

Question 5.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 115
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.9 Model and Count 10-Model and Count 10-Homework & Practice 3.9.5

Directions:
3 and 4 Count the objects. Color the boxes to show how many. 5 Count the objects in the picture. Color the boxes to show how many.

Lesson 3.10 Understand and Write 10

Explore and Grow
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 116
Answer:
Number of Star fishes in the story In the water = One or 1.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10
Directions:
Use counters to show how many starfish are in the story In the Water. Write how many starfish are in the story.

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 117
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Think and Grow

Directions:

  • Count the objects. Say the number. Trace and write the number.
  • Count the sea creatures. Say the number. Write the number.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 119
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Apply and Grow-Practice1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 120
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Apply and Grow-Practice2

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 121
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Apply and Grow-Practice3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 122
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Apply and Grow-Practice4

Directions:
1 – 4 Count the objects. Say the number. Write the number.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 123
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Think and Grow-Modeling Real Life

Directions:
Count the sea creatures in the picture. Say the number. Write the number.

Understand and Write 10 Homework & Practice 3.10

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 124
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Understand and Write 10 Homework & Practice 3.10.1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 125
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Understand and Write 10 Homework & Practice 3.10.2

Directions:
1 and 2 Count the dots. Say the number. Write the number.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 126
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Understand and Write 10 Homework & Practice 3.10.3

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 127
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Understand and Write 10 Homework & Practice 3.10.4

Question 5.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 128
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Understand and Write 10 Homework & Practice 3.10.5

Directions:
3 and 4 Count the objects. Say the number. Write the number. 5 Count the sea creatures in the picture. Say the number. Write the number.

Lesson 3.11 Count and Order Numbers to 10

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 129
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Lesson 3.11 Count and Order Numbers to 10

Directions:
Place counters in the ten frame as you count forward to 10. Trace and write the missing numbers.

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 130
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Think and Grow

Directions:
Count the dots in each ten frame. Say the number. Write the number. Write the numbers in order. Start with the given number.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 131
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Apply and Grow-Practice1

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 132
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Apply and Grow-Practice2

Directions:
1 and 2 Count the dots in each ten frame. Say the number. Write the number. Then write the numbers in order.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 133
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Think and Grow-Modeling Real Life

Directions:
Count backward from the number on the timer. Write the numbers.

Count and Order Numbers to 10 Homework & Practice 3.11

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 134
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Order Numbers to 10 Homework & Practice 3.11.1

Directions:
1 Count the dots. Say the number. Write the number. Then write the numbers in order.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 135
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Order Numbers to 10 Homework & Practice 3.11.2

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 136
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Order Numbers to 10 Homework & Practice 3.11.3

Directions:
2 Count the dots on each domino. Say the number. Write the number. Write the numbers in order. Start with the number 7. 3 You are playing hide-and-seek. Count forward from 1. Write the numbers.

Count and Write Numbers 6 to 10 Performance Task

Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 137

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 138
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Performance Task

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 139
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Performance Task2

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 140
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Performance Task3

Directions:
1 Count the animals in the picture. Say the number. Write the number. Then write the numbers in order. 2 Draw dots to show 10 birds in the tree. 3 Show how many birds in another way.

Count and Write Numbers 6 to 10 Activity

Number Land
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 141
Directions:
Put the Subitizing Cards 5–10 into a pile. Start at Descartes. Take turns drawing a card and moving your piece to that matching number. Repeat this process until you have gone around the board and back to Descartes.

Count and Write Numbers 6 to 10 Chapter Practice 3

3.1 Model and Count 6

Question 1.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 142
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.1

3.2 Understand and Write 6

Question 2.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 143
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.2

3.3 Model and Count 7

Question 3.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 144
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.3

Directions:
1 Count the train engines. Color the boxes to show how many. 2 Count the music notes. Say the number. Write the number. 3 Count the foxes. Color the boxes to show how many.

3.4 Understand and Write 7

Question 4.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 145
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.4

3.5 Model and Count 8

Question 5.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 146
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.5

3.6 Understand and Write 8

Question 6.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 147
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.6

Directions:
4 Count the lightning bolts. Say the number. Write the number.
5 Count the grapes. Color the boxes to show how many.
6 Count the grasshoppers. Say the number. Write the number.

3.7 Model and Count 9

Question 7.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 148
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.7

3.8 Understand and Write 9

Question 8.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 149
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.8

3.9 Model and Count 10

Question 9.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 150
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.9

Directions:
7 Count the shovels. Color the boxes to show how many.
8 Count the softballs. Say the number. Write the number.
9 Count the potatoes. Color the boxes to show how many.

3.10 Understand and Write 10

Question 10.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 151
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.10

3.11 Count and Order Numbers to 10

Question 11.
Big Ideas Math Answer Key Grade K Chapter 3 Count and Write Numbers 6 to 10 152
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter 3-Count-and-Write-Numbers-6 to 10-Lesson-Lesson 3.10 Understand and Write 10-Count and Write Numbers 6 to 10 Chapter Practice 3.11

Directions:
10 Count the shells. Say the number. Write the number. 11 Count the dots on each domino. Say the number. Write the number. Then write the numbers in order.

Conclusion:

We wish the details prevailed in the above article regarding Big Ideas Math Answers Grade K Chapter 3 Count and Write Numbers 6 to 10 is helpful for you. So, the students of Kth Grade can make use of the above links and score good marks in the exams. Stay tuned to our page to get the latest updates and solutions for all Big Ideas Math Grade K Chapters from 1 to 13.

Big Ideas Math Answers Grade 5 Chapter 8 Add and Subtract Fractions

Big Ideas Math Answers Grade 5 Chapter 8 Add and Subtract Fractions

Big Ideas Math Answers Grade 5 Chapter 8 Add and Subtract Fractions is the best source to learn Add and Subtract Fractions concepts. Get free access to download Big Ideas Math Answers Grade 5 Chapter 8 Add and Subtract Fractions pdf. Learn how to solve a problem in different ways. We have provided a step-by-step process to solve the problems with real-time examples. Various tips and tricks are given to learn the concepts deeply. Follow the important notes for the quick learning of maths. Improve your math skills with the help of the Big Ideas Math Grade 5 Chapter 8 Add and Subtract Fractions Answers.

Big Ideas Math Book 5th Grade Chapter 8 Add and Subtract Fractions Answer Key

It is mandatory for the students to go through all the topics included in this chapter to kickstart their preparation. The list of topics is mentioned below along with the quick links. Understand each topic by referring to every link and solving every problem available below. Find the various methods to solve the questions and choose the best one to practice well for the exam. All the topics such as Simplest form, Subtract Fractions with Unlike Denominators, Find Common Denominators, Estimate Sums, and Differences of Fractions, Add Fractions with Unlike Denominators are prepared with a clear explanation in BIM Grade 5 Ch 8 Add and Subtract Fractions Solution key.

Lesson: 1 Simplest Form

Lesson: 2 Estimate Sums and Differences of Fractions

Lesson: 3 Find Common Denominators

Lesson: 4 Add Fractions with Unlike Denominators

Lesson: 5 Subtract Fractions with Unlike Denominators

Lesson: 6 Add Mixed Numbers

Lesson: 7 Subtract Mixed Numbers

Lesson: 8 Problem Solving: Fractions

Chapter 8 – Add and Subtract Fractions 

Lesson 8.1 Simplest Form

Explore and Grow

Use the model to write as many fractions as possible that are equivalent to \(\frac{36}{72}\) but have numerators less than 36 and denominators less than 72.
Big Ideas Math Answers Grade 5 Chapter 8 Add and Subtract Fractions 1
Which of your fractions has the fewest equal parts? Explain.

Construct Arguments
When might it be helpful to write \(\frac{48}{72}\) as \(\frac{2}{3}\) in a math problem?

Think and Grow: Simplest Form

Key Idea
When the numerator and denominator of a fraction have no common factors other than 1, the fraction is in simplest form. To write a fraction in simplest form, divide the numerator and the denominator by the greatest of their common factors.
Example
simplify \(\frac{6}{8}\) in the simplest form.
Big Ideas Math Answers Grade 5 Chapter 8 Add and Subtract Fractions 2
Step 1: Find the common factors of 6 and 8.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 3
The common factors of 6 and 8 are 1 and 2.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 4

Answer:
Big-Ideas-Math-Solutions-Grade-5-Chapter-8-Add-and-Subtract-Fractions-4

Show and Grow

Question 1.
Use the model to write \(\frac{2}{4}\) in simplest form.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 5
Answer:
Step 1: Find the common factors of 2 and 4.
Factors of 2:   1, 2
Factors of 4:   1, 2, 4
The common factors of 2 and 4 are 1 and 2.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
                              Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 5
Because 1 and 2 have no common factors other than 1, \(\frac{2}{4}\) is in simplest form.

Question 2.
Write \(\frac{8}{12}\) in simplest form
Answer:
Step 1: Find the common factors of 8 and 12.
Factors of 8:    1, 2, 4, 8
Factors of 12:  1, 2, 3, 4, 6, 12
The common factors of 8 and 12 are 1, 2 and 4.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.

Because 2 and 3 have no common factors other than 1, \(\frac{8}{12}\) is in simplest form.

Apply and Grow: Practice

Use the model to write the fraction in simplest form.

Question 3.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 6
Answer:
Step 1: Find the common factors of 8 and 10.
Factors of 8:    1, 2, 4, 8
Factors of 10:  1, 2, 5, 10
The common factors of 8 and 10 are 1 and 2.

Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.

\small \frac{8}{10}= \frac{8 \div 2}{10 \div 2} = \frac{4}{5}

Because 4 and 5 have no common factors other than 1, \(\frac{8}{10}\) is in simplest form.

Question 4.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 7
Answer:
Step 1: Find the common factors of 5 and 15.
Factors of 5: 1, 5
Factors of 15: 1, 3, 5,15
The common factors of 5 and 15 are 1 and 5.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{5}{15}= \frac{5 \div 5}{15 \div 5} = \frac{1}{3}
Because 1 and 3 have no common factors other than 1, \(\frac{5}{15}\) is in simplest form.

Question 5.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 8

Write the fraction in simplest form.
Answer:
Step 1: Find the common factors of 10 and 12.
Factors of 10:  1, 2, 5, 10
Factors of 12:  1, 2, 3, 4, 6, 12
The common factors of 10 and 12 are 1 and 2.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{10}{12}= \frac{10 \div 2}{12 \div 2} = \frac{5}{6}
Because 5 and 6 have no common factors other than 1, \(\frac{10}{12}\) is in simplest form.

Question 6.
\(\frac{3}{6}\)
Answer:
Step 1: Find the common factors of 3 and 6.
Factors of 3:  1, 3
Factors of 6:  1, 2, 3, 6
The common factors of 3 and 6 are 1 and 3.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{3}{6}= \frac{3 \div 3}{6 \div 3} = \frac{1}{2}
Because 1 and 2 have no common factors other than 1, \(\frac{3}{6}\) is in simplest form.

Question 7.
\(\frac{2}{10}\)
Answer:
Step 1: Find the common factors of 2 and 10.
Factors of 2:  1, 2
Factors of 10:  1, 2, 5, 10
The common factors of 2 and 10 are 1 and 2.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{2}{10}= \frac{2 \div 2}{10 \div 2} = \frac{1}{5}
Because 1 and 5 have no common factors other than 1, \(\frac{2}{10}\) is in simplest form.

Question 8.
\(\frac{6}{8}\)
Answer:
Step 1: Find the common factors of 6 and 8.
Factors of 6:  1, 2, 3, 6
Factors of 8:  1, 2, 4, 8
The common factors of 6 and 8 are 1 and 2.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{6}{8}= \frac{6 \div 2}{8 \div 2} = \frac{3}{4}
Because 3 and 4 have no common factors other than 1, \(\frac{6}{8}\) is in simplest form.

Question 9.
\(\frac{7}{14}\)
Answer:
Step 1: Find the common factors of 7 and 14.
Factors of 7:  1, 7
Factors of 14:  1, 2, 7, 14
The common factors of 7 and 14 are 1 and 7.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{7}{14}= \frac{7 \div 7}{14 \div 7} = \frac{1}{2}
Because 1 and 2 have no common factors other than 1, \(\frac{7}{14}\) is in simplest form.

Question 10.
\(\frac{10}{100}\)
Answer:
Step 1: Find the common factors of 10 and 100.
Factors of 10: 1, 2, 5, 10
Factors of 100:  1, 2, 4, 5, 10, 20, 25, 50, 100
The common factors of 10 and 100 are 1, 2, 5 and 10.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{10}{100}= \frac{10 \div 10}{100 \div 10} = \frac{1}{10}
Because 1 and 10 have no common factors other than 1, \(\frac{10}{100}\) is in simplest form.

Question 11.
\(\frac{12}{4}\)
Answer:
Step 1: Find the common factors of 12 and 4.
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 4:  1, 2, 4
The common factors of 12 and 4 are 1, 2 and 4.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{12}{4}= \frac{12 \div 4}{4 \div 4} = \frac{3}{1}
Because 3 and 1 have no common factors other than 1, \(\frac{12}{4}\) is in simplest form.

Question 12.
Three out of nine baseball players are in the outfield. In simplest form, what fraction of the players are in the outfield?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 9
Answer:
Step 1: Find the common factors of 3 and 9.
Factors of 3: 1, 3
Factors of 9:  1, 3, 9
The common factors of 3 and 9 are 1 and 3
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{3}{9}= \frac{3 \div 3}{9 \div 3} = \frac{1}{3}
Because 1 and 3 have no common factors other than 1.
Therefore, players are in the outfield.

Question 13.
YOU BE THE TEACHER
Your friend writes \(\frac{2}{6}\) in simplest form. Is your friend correct? Explain
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 10

Answer: No, the answer is wrong.
The numerator and the denominator has to divide by the greatest of the common factors. You have divided only the denominator.

Explanation for \(\frac{2}{6}\) in simplest form.
Step 1: Find the common factors of 2 and 6.
Factors of 2: 1, 2
Factors of 6:  1, 2, 3, 6
The common factors of 2 and 6 are 1 and 2.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{2}{6}= \frac{2 \div 2}{6 \div 2} = \frac{1}{3}
Because 1 and 3 have no common factors other than 1, \(\frac{2}{6}\) is in simplest form.

Question 14.
Reasoning
The numerator and denominator of a fraction have 1, 2, and 4 as common factors. After you divide the numerator and denominator by 2, the fraction is still not in simplest form. Why?
Answer:
Given that, common factors are 1, 2 and 4
For the fraction to be in the simplest form, the numerator and denominator has to divide by the greatest of the common factors.
Here 4 is the greatest common factor. So, divide both the numerator and denominator by 4 to get the simplest form.
For example 4 and 8
Factors for 4: 1, 2, 4
Factors for 8: 1, 2, 4, 8
common factors: 1, 2 and 4
Simplest form:    \small \frac{4}{8}= \frac{4 \div 4}{8 \div 4} = \frac{1}{2}

Think and Grow: Modeling Real Life

Example
A quarterback passes the ball 45 times during a game. The quarterback completes 35 passes. What fraction of the passes, in simplest form, does the quarterback complete?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 11
Find the number of passes that are not completed by subtracting the pass completions from the total number of passes.
45 – 35 = 10
Write a fraction for the passes the quarterback does not complete.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 12
Find common factors of 10 and 45. Then write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 13
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 14
The quarterback does not complete __ of the passes.

Answer:
Big-Ideas-Math-Solutions-Grade-5-Chapter-8-Add-and-Subtract-Fractions-14
The quarterback does not complete 2/9 of the passes.

Show and Grow

Question 15.
There are 24 students in your class. Four of the students have blue eyes. What fraction of the class, in simplest form, do not have blue eyes?
Answer:
Given that,
Total no. of students in the class = 24
Students have blue eyes = 4
Students do not have blue eyes = 24 – 4 = 20
Step 1: Find the common factors of 24 and 20.
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Factors of 20:  1, 2, 4, 5, 10, 20
The common factors of 24 and 20 are 1, 2 and 4.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.

\small \frac{20}{24}= \frac{20 \div 4}{24 \div 4} = \frac{5}{6}

Because 5 and 6 have no common factors other than 1.
Therefore, \small \frac{5}{6}th of the class do not have blue eyes.

Question 16.
DIG DEEPER!
A student answers 4 out of 12 questions on a test incorrectly. What fraction of the questions, in simplest form, does the student answer incorrectly? Interpret the fraction.
Answer:
Given that,
Total no. of questions = 12
No. of incorrect answers = 4
Step 1: Find the common factors of 12 and 4.
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 4:  1, 2, 4
The common factors of 12 and 4 are 1, 2 and 4.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{4}{12} = \frac{4 \div 4}{12 \div 4} = \frac{1}{3}
Because 1 and 3 have no common factors other than 1.
Therefore, \small \frac{1}{3} of the questions student answered incorrectly.

Lesson 8.1 Simplest Form Homework & Practice 8.1

Use the model to write the fraction in simplest form.

Question 1.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 16
Answer:
Step 1: Find the common factors of 6 and 9.
Factors of 6: 1, 2, 3, 6
Factors of 9:  1, 3, 9
The common factors of 6 and 9 are 1 and 3.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.

\small \frac{6}{9} = \frac{6 \div 3}{9 \div 3} = \frac{2}{3}
Because 2 and 3 have no common factors other than 1, \small \frac{6}{9} is in simplest form.

Question 2.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 17
Answer:
Step 1: Find the common factors of 3 and 12.
Factors of 3: 1, 3
Factors of 12:  1, 3, 4, 6, 12
The common factors of 3 and 12 are 1 and 3.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{3}{12} = \frac{3 \div 3}{12 \div 3} = \frac{1}{4}
Because 1 and 4 have no common factors other than 1, \small \frac{3}{12} is in simplest form.

Question 3.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 18

Write the fraction in simplest form
Answer:
Step 1: Find the common factors of 5 and 10.
Factors of 5: 1, 5
Factors of 10:  1, 2, 5, 10
The common factors of 5 and 10 are 1 and 5.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{5}{10} = \frac{5 \div 5}{10 \div 5} = \frac{1}{2}
Because 1 and 2 have no common factors other than 1, \small \frac{5}{10} is in the simplest form.

Question 4.
\(\frac{4}{8}\)
Answer:
Step 1: Find the common factors of 4 and 8.
Factors of 4: 1, 2, 4
Factors of 8:  1, 2, 4, 8
The common factors of 4 and 8 are 1, 2 and 4.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{4}{8} = \frac{4 \div 4}{8 \div 4} = \frac{1}{2}
Because 1 and 2 have no common factors other than 1, \(\frac{4}{8}\) is in simplest form.

Question 5.
\(\frac{5}{100}\)
Answer:
Step 1: Find the common factors of 5 and 100.
Factors of 5: 1, 5
Factors of 100:  1, 2, 4, 5, 10, 20, 25, 50, 100
The common factors of 5 and 100 are 1 and 5.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{5}{100} = \frac{5 \div 5}{100 \div 5} = \frac{1}{20}
Because 1 and 20 have no common factors other than 1, \(\frac{5}{100}\) is in simplest form.

Question 6.
\(\frac{20}{15}\)
Answer:
Step 1: Find the common factors of 20 and 15.
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 15:  1, 3, 5, 15
The common factors of 20 and 15 are 1 and 5.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{20}{15} = \frac{20 \div 5}{15 \div 5} = \frac{4}{3}
Because 4 and 3 have no common factors other than 1, \(\frac{20}{15}\) is in simplest form.

Question 7.
There are 18 students in your class. Six of the students pack their lunch. In simplest form, what fraction of the students in your class pack their lunch?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 19
Answer:
Total students in the class = 18
No. of students pack their lunch = 6
Step 1: Find the common factors of 6 and 18.
Factors of 6: 1, 2, 3, 6
Factors of 18:  1, 2, 3, 6, 9, 18
The common factors of 6 and 18 are 1, 2, 3 and 6
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{6}{18} = \frac{6 \div 6}{18 \div 6} = \frac{1}{3}
Because 1 and 3 have no common factors other than 1.
Therefore, \small \frac{1}{3} of the students pack their lunch.

Question 8.
Reasoning
Why do you have to divide a numerator and a denominator by the greatest of their common factors to write a fraction in simplest form?
Answer:
To simplify a fraction to lowest terms, divide both the numerator and the denominator by their common factors. Repeat as needed until the only common factor is 1.

Question 9.
Writing
Explain how you know when a fraction is in simplest form.
Answer:
If the fraction has no common factors other than 1, then it is said to be the simplest form of the fraction.

Question 10.
Open-Ended
Write a fraction in which the numerator and the denominator have 1, 2, 4, and 8 as common factors. Then write the fraction in the simplest form.
Answer:
The fraction in which the numerator and the denominator is \small \frac{8}{16}.
Step 1: Find the common factors of 8 and 16.
Factors of 8: 1, 2, 4, 8
Factors of 16:  1, 2, 4, 8, 16
The common factors of 8 and 16 are 1, 2, 4 and 8.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{8}{16} = \frac{8 \div 8}{16 \div 8} = \frac{1}{2}
Because 1 and 2 have no common factors other than 1.

Question 11.
Modeling Real Life
A flight attendant has visited 30 of the 50 states. What fraction of the states, in simplest form, has he not visited?
Answer:
Given that,
No. of states = 50
A flight attendant has visited 30 states.
The no. of states he has not visited = 50 – 30 = 20
Step 1: Find the common factors of 20 and 50.
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 50: 1, 2, 5, 10, 25, 50
The common factors of 20 and 50 are 1, 2, 5 and 10.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{20}{50} = \frac{20 \div 10}{50 \div 10} = \frac{2}{5}
Because 2 and 5 have no common factors other than 1.
So the flight attendant has not visited \small \frac{2}{5} of the states.

Question 12.
DIG DEEPER!
A bin has red, orange, yellow, green, blue, and purple crayons. There are 4 of each color in the bin. In simplest form, what fraction of the crayons are red, orange, yellow, or green?
Answer:
Given that, a bin has 6 colors(red, orange, yellow, green, blue, and purple) of crayons.
There are 4 crayons in the each color = 4 x 6 = 24
If the bin having only 3 colors(red, orange, yellow or green) = 4 x 3 = 12
Step 1: Find the common factors of 12 and 24.
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
The common factors of 12 and 24 are 1, 2, 3, 4, 6 and 12.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{12}{24} = \frac{12 \div 12}{24 \div 12} = \frac{1}{2}
Because 1 and 2 have no common factors other than 1.
\small \frac{1}{2} of the crayons are red, orange, yellow or green.

Review & Refresh

Estimate the sum or difference.

Question 13.
598.44 – 45.61 = 
Answer: 550
The number round to 598.44 is 600.
The number round to 45.61 is 50.
600 – 50 = 550
Thus the estimated difference is 550.

Question 14.
93.8 + 4.3 =
Answer: 98
The number round to 93.8 is 94.
The number round to 4.3 is 4.
94 + 4 = 98
Thus the estimated addition is 98.

Lesson 8.2 Estimate Sums and Differences of Fractions

Explore and Grow

Plot \(\frac{7}{12}\), \(\frac{5}{6}\) and \(\frac{1}{10}\) on the number line.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 20
How can you estimate \(\frac{7}{12}\) + \(\frac{5}{6}\) ?
How can you estimate \(\frac{2}{3}\) – \(\frac{1}{10}\)?

Reasoning
Write two fractions that have a sum of about \(\frac{1}{2}\). Then write two fractions that have a difference of about \(\frac{1}{2}\). Explain your reasoning.

Think and Grow: Estimate Sums and Differences

You have used the benchmarks \(\frac{1}{2}\) and 1 to compare fractions. You can use the benchmarks 0, \(\frac{1}{2}\), and 1 to estimate sums and differences of fractions.
Example
Estimate \(\frac{1}{6}\) + \(\frac{5}{8}\)
Step 1: Use a number line to estimate each fraction.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 21
Step 2: Estimate the sum.
An estimate of \(\frac{1}{6}\) + \(\frac{5}{8}\) is __ + __ = ___
Example
Estimate \(\frac{9}{10}\) – \(\frac{2}{5}\).
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 22

Show and Grow

Estimate the sum or difference

Question 1.
\(\frac{1}{3}\) + \(\frac{11}{12}\)
Answer:
Step 1: Estimate each fraction.
\small \frac{1}{3} is between 0 and \small \frac{1}{2} , but is closer to \small \frac{1}{2}
\small \frac{11}{12} is between \small \frac{1}{2} and 1, but is closer to 1
Step 2: Estimate the sum.
An estimate of \(\frac{1}{3}\) + \(\frac{11}{12}\) is   \small \frac{1}{2} +  1 =  \small \frac{3}{2}

Question 2.
\(\frac{3}{5}\) + \(\frac{5}{6}\)
Answer:
Step 1: Estimate each fraction
\small \frac{3}{5} is between  \small \frac{1}{2} and 1 , but is closer to \small \frac{1}{2}
\small \frac{5}{6} is between \small \frac{1}{2} and 1, but is closer to 1
Step 2: Estimate the sum.
An estimate of \(\frac{3}{5}\) + \(\frac{5}{6}\) is \small \frac{1}{2} +  1 =  \small \frac{3}{2}

Question 3.
\(\frac{15}{16}\) – \(\frac{7}{8}\)
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{15}{16} is about
Think : The numerator is about the same as the denominator.
\small \frac{7}{8} is about
Think : The numerator is about the same as the denominator.
Step 2: Estimate the difference.
An estimate of \small \frac{15}{16}\small \frac{7}{8}  is 1 – 1 = 0.

Apply and Grow: Practice

Estimate the sum or difference.

Question 4.
\(\frac{1}{6}\) + \(\frac{3}{5}\)
Answer:
Step 1: Estimate each fraction.
\small \frac{1}{6} is between 0 and \small \frac{1}{2}, but is closer to 0.
\small \frac{3}{5} is between \small \frac{1}{2} and 1, but is closer to \small \frac{1}{2}.
Step 2: Estimate the sum.
An estimate of \(\frac{1}{6}\) + \(\frac{3}{5}\) = 0 + \small \frac{1}{2} = \small \frac{1}{2}

Question 5.
\(\frac{4}{5}\) – \(\frac{5}{12}\)
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{4}{5} is about
Think: The numerator is about the same as the denominator.
\small \frac{5}{12} is about
Think: The numerator is about half of the denominator.
Step 2: Estimate the difference.
An estimate of \small \frac{15}{16}\small \frac{7}{8}  is 1 – \small \frac{1}{2}  = \small \frac{1}{2}

Question 6.
\(\frac{13}{16}\) + \(\frac{5}{6}\)
Answer:
Step 1: Estimate each fraction.
\small \frac{13}{16} is between \small \frac{1}{2} and 1, but is closer to 1.
\small \frac{5}{6} is between \small \frac{1}{2} and 1, but is closer to 1.
Step 2: Estimate the sum.
An estimate of \(\frac{13}{16}\) + \(\frac{5}{6}\) = 1+1 = 2

Question 7.
\(\frac{3}{6}\) – \(\frac{1}{8}\)
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{3}{6} is about
Think: The numerator is about half of the denominator.
\small \frac{1}{8} is about
Think: The numerator is near to zero.
Step 2: Estimate the difference.
An estimate of \small \frac{3}{6}\small \frac{1}{8} is  \small \frac{1}{2}  – 0 = \small \frac{1}{2} .

Question 8.
\(\frac{1}{14}\) + \(\frac{98}{100}\)
Answer:
Step 1: Estimate each fraction.
\small \frac{1}{14} is closer to 0
\small \frac{98}{100} is closer to 1.
Step 2: Estimate the sum.
An estimate of \(\frac{1}{14}\) + \(\frac{98}{100}\) = 0 +1 = 1

Question 9.
\(\frac{11}{12}\) – \(\frac{1}{8}\)
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{11}{12} is about
Think: The numerator is about the same as the denominator.
\small \frac{1}{8} is about
Think: The numerator is near to zero.
Step 2: Estimate the difference.
An estimate of  \small \frac{11}{12} – \small \frac{1}{8} is  1  – 0 = 1.

Question 10.
You walk \(\frac{1}{10}\) mile to your friend’s house and then you both walk \(\frac{2}{5}\) mile. Estimate how much farther you walk with your friend than you walk alone.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 23
Answer:
To find how much farther I walk with my friend than I walk alone, subtract the distance that I walk alone from we both walk.
Step 1: Use mental math to estimate each fraction.
\small \frac{1}{10} is about____
Think: The numerator is near to zero.
\small \frac{2}{5} is about____
Think: The numerator is about half of the denominator.
Step 2: Estimate the difference.
An estimate of  \small \frac{2}{5}\small \frac{1}{10} = \small \frac{1}{2} – 0 = \small \frac{1}{2}
So the distance I walk with my friend than I walk alone is \small \frac{1}{2} mile.

Question 11.
A carpenter has two wooden boards. One board is \(\frac{3}{4}\) foot long and the other board is \(\frac{1}{6}\) foot long. To determine whether the total length of the boards is 1 foot, should the carpenter use an estimate, or is an exact answer required? Explain.
Answer:
Given,
A carpenter has two wooden boards. One board is \(\frac{3}{4}\) foot long and the other board is \(\frac{1}{6}\) foot long.
\(\frac{3}{4}\) + \(\frac{1}{6}\)
The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.
LCM = 12
\(\frac{3}{4}\) × \(\frac{3}{3}\) = \(\frac{9}{12}\)
\(\frac{1}{6}\) × \(\frac{2}{2}\) = \(\frac{2}{12}\)
\(\frac{9}{12}\) + \(\frac{2}{12}\) = \(\frac{11}{12}\)
\(\frac{11}{12}\) is approximately equal to 1 foot.

Question 12.
Number Sense
A fraction has a numerator of 1 and a denominator greater than 4. Is the fraction closer to 0, \(\frac{1}{2}\), or 1? Explain.
Answer:
Given,
A fraction has a numerator of 1 and a denominator greater than 4.
\(\frac{1}{4}\) = 0.25
If the denominator is greater than 4. Let’s consider 8.
\(\frac{1}{8}\) = 0.125
The fraction will be close to 0.

Think and Grow: Modeling Real Life

Example
In the human body, the small intestine is about 20\(\frac{1}{12}\) feet long. The large intestine is about 4\(\frac{5}{6}\) feet long. About how long are the intestines in the human body?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 24
To find the total length of the intestines, estimate 20\(\frac{1}{12}\) + 4\(\frac{5}{6}\).
Step 1: Use mental math to round each mixed number to the nearest whole number.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 25

Answer:
Big-Ideas-Math-Solutions-Grade-5-Chapter-8-Add-and-Subtract-Fractions-25

Show and Grow

Question 13.
A bullfrog jumps 5\(\frac{11}{12}\) feet. A leopard frog jumps 4\(\frac{1}{3}\) feet. About how much farther does the bullfrog jump than the leopard frog?
Answer:
Step 1: Use mental math to round each mixed number to the nearest whole number.
5\small \frac{11}{12} is about,  \small \frac{11}{12} is closer to 1 than 0.
4\small \frac{1}{3} is about, \small \frac{1}{3} is closer to 0 than 1.
Step 2: Estimate the difference
An estimate of 5\small \frac{11}{12}  – 4\small \frac{1}{3} = 1 – 0 = 1
So, bullfrog jumps 1 feet farther than the leopard frog.
Question 14.
DIG DEEPER!
A cell phone has 32 gigabytes of storage. The amounts of storage used by photos, songs, and apps are shown. About how many gigabytes of storage are left?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 26
Answer:
Given that,
No. of gigabytes of storage in cellphone = 32
Step 1: Use mental math to round each mixed number to the nearest whole number.
Photos —>  8\small \frac{4}{5} is about,  \small \frac{4}{5} is closer to 1 than 0
Songs —>   2\small \frac{3}{100} is about,  \small \frac{3}{100} is closer to 0 than 1
Apps —>     6\small \frac{7}{10}  is about,  \small \frac{7}{10} is closer to 1 than 0
Step 2: Storage left in the phone = Total storage – storage(photos + songs + apps)
= 32 – (1 + 0 + 1)
Therefore, storage left = 30 gigabytes.

Question 15.
DIG DEEPER!
Use two different methods to estimate how many cups of nut medley the recipe makes. Which estimate do you think is closer to the actual answer? Explain.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 27
Answer:
1 \(\frac{3}{8}\) + \(\frac{5}{8}\) + 2 \(\frac{1}{3}\)
First add the whole numbers
1 + 2 = 3
\(\frac{3}{8}\) + \(\frac{5}{8}\) + \(\frac{1}{3}\)
1 \(\frac{1}{3}\)
3 + 1 \(\frac{1}{3}\) = 4 \(\frac{1}{3}\)
The fraction is 4 \(\frac{1}{3}\)
4 is equal to the actual answer.

Estimate Sums and Differences of Fractions Homework & Practice 8.2

Estimate the sum or difference

Question 1.
\(\frac{11}{12}\) – \(\frac{5}{6}\)
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{11}{12} is about
Think : The numerator is about the same as the denominator.
\small \frac{5}{6} is about
Think : The numerator is about same as the denominator.
Step 2: Estimate the difference.
An estimate of  \small \frac{11}{12}\small \frac{5}{6}  is  1  – 1 = 0.

Question 2.
\(\frac{17}{20}\) + \(\frac{13}{20}\)
Answer:
Step 1: Estimate each fraction.
\small \frac{17}{20} is between \small \frac{1}{2} and 1, but is closer to 1.
\small \frac{13}{20} is between \small \frac{1}{2} and 1, but is closer to \small \frac{1}{2}.
Step 2: Estimate the sum.
An estimate of \(\frac{17}{20}\) + \(\frac{13}{20}\) = 1 + \small \frac{1}{2} = \small \frac{3}{2}.

Question 3.
\(\frac{3}{8}\) – \(\frac{1}{6}\)
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{3}{8} is about
Think : The numerator is about half of the denominator.
\small \frac{1}{6} is about
Think : The numerator is near to zero.
Step 2: Estimate the difference.
An estimate of  \small \frac{3}{8}  – \small \frac{1}{6}   is  \small \frac{1}{2}  – 0 = \small \frac{1}{2}.

Question 4.
\(\frac{7}{12}\) + \(\frac{2}{5}\)
Answer:
Step 1: Estimate each fraction.
\small \frac{7}{12} is between \small \frac{1}{2} and 1, but is closer to \small \frac{1}{2}.
\small \frac{2}{5} is between 0 and \small \frac{1}{2}, but is closer to \small \frac{1}{2}.
Step 2: Estimate the sum.
An estimate of \(\frac{7}{12}\) + \(\frac{2}{5}\) = \small \frac{1}{2} + \small \frac{1}{2} = 1.

Question 5.
\(\frac{4}{5}\) – \(\frac{7}{12}\)
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{4}{5} is about
Think: The numerator is about the same as the denominator.
\small \frac{7}{12} is about
Think: The numerator is about half of the denominator.
Step 2: Estimate the difference.
An estimate of  \small \frac{4}{5}  – \small \frac{7}{12}   is  1 –  \small \frac{1}{2}  = \small \frac{1}{2}

Question 6.
\(\frac{1}{5}\) + 1\(\frac{10}{21}\)
Answer:
Step 1: Use mental math to round each mixed number to the nearest whole number.
\(\frac{1}{5}\) is closer to 0
1\(\frac{10}{21}\) is close to 1.
Step 2: Estimate the sum.
An estimate of \(\frac{1}{5}\) + 1\(\frac{10}{21}\)
= 0 + 1 = 1

Question 7.
3\(\frac{5}{8}\) – \(\frac{1}{10}\)
Answer:
Step 1: Use mental math to round each mixed number to the nearest whole number.
3\small \frac{5}{8}  is about, \small \frac{5}{8} is closer to 1 than 0
\small \frac{1}{10} is near to 0
Step 2: Estimate the difference
An estimate of 3\small \frac{5}{8}  – \small \frac{1}{10} = 1 – 0 = 1.

Question 8.
6\(\frac{1}{3}\) + 2\(\frac{4}{6}\)
Answer:
Step 1: Use mental math to round each mixed number to the nearest whole number.
\(\frac{1}{3}\) is close to 0 than 1.
\(\frac{4}{6}\) is to 1 than 0.
Step 2: Estimate the sum
An estimate of 6\(\frac{1}{3}\) is 6.
An estimate of 2\(\frac{4}{6}\) is 3.
6 + 3 = 9

Question 9.
5\(\frac{7}{8}\) – 4\(\frac{49}{100}\)
Answer:
Step 1: Use mental math to round each mixed number to the nearest whole number.
5\small \frac{7}{8}  is about, \small \frac{7}{8} is closer to 1 than 0
4\small \frac{49}{100}  is about, \small \frac{49}{100} is closer to 0 than 1
Step 2: Estimate the difference
An estimate of 5\small \frac{7}{8}  – 4\small \frac{49}{100}  = 1 – 0 =  1.

Question 10.
You make a bag of trail mix with \(\frac{2}{3}\) cup of raisins and \(\frac{9}{8}\) cups of peanuts. About how much trail mix do you make?
Answer:
Raisins = \small \frac{2}{3} cups
Peanuts = \small \frac{9}{8} cups
Step 1: Estimate each fraction.
\small \frac{2}{3} is between \small \frac{1}{2} and 1, but is closer to \small \frac{1}{2}.
\small \frac{9}{8} is closer to 1.
Step 2: Estimate the sum.
An estimate of \(\frac{2}{3}\) + \(\frac{9}{8}\) = \small \frac{1}{2} + 1 = \small \frac{3}{2}.
So trail mix = \small \frac{3}{2}.

Question 11.
You have \(\frac{2}{3}\) cup of flour in a bin and \(\frac{7}{8}\) cup of flour in a bag. To determine whether you have enough flour for a recipe that needs 1\(\frac{3}{4}\) cups of flour, should you use an estimate, or is an exact answer required? Explain.
Answer:
Given,
You have \(\frac{2}{3}\) cup of flour in a bin and \(\frac{7}{8}\) cup of flour in a bag.
\(\frac{2}{3}\) + \(\frac{7}{8}\) = 1 \(\frac{13}{24}\)
Th estimated fraction of 1 \(\frac{13}{24}\) is 1\(\frac{3}{4}\)

Question 12.
Writing
Explain how you know \(\frac{9}{10}\) – \(\frac{3}{5}\) is about \(\frac{1}{2}\).
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{9}{10} is about
Think: The numerator is about the same as the denominator.
\small \frac{3}{5} is about
Think : The numerator is about half of the denominator.
Step 2: Estimate the difference.
An estimate of \small \frac{9}{10}\small \frac{3}{5}  is 1 – \small \frac{1}{2}  = \small \frac{1}{2}.

Question 13.
Precision
Your friend says \(\frac{5}{8}\) + \(\frac{7}{12}\) is about 2. Find a closer estimate. Explain why your estimate is closer.
Answer:
\(\frac{5}{8}\) + \(\frac{7}{12}\) = 1 \(\frac{5}{24}\)
\(\frac{5}{24}\) is closer to 1 than 0.
1 \(\frac{5}{24}\) is about 2.

Question 14.
Modeling Real Life
About how much taller is Robot A than Robot B?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 28
Answer:
Step 1: Use mental math to round each mixed number to the nearest whole number.
1 \small \frac{6}{10} is about, \small \frac{6}{10} is closer to 1 than 0.
1 \small \frac{1}{5}  is about, \small \frac{1}{5} is closer to 0 than 1.
Step 2: Estimate the difference
An estimate of 1 \small \frac{6}{10}  – 1 \small \frac{1}{5}   = 1 – 0 =  1.
Robot A is 1 meter taller than Robot B.

Question 15.
Modeling Real Life
A class makes a paper chain that is 5\(\frac{7}{12}\) feet long. The class adds another 3\(\frac{5}{6}\) feet to the chain. About how long is the chain now?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 29
Answer:
Step 1: Use mental math to round each mixed number to the nearest whole number.
5\small \frac{7}{12} is about,  \small \frac{7}{12} is closer to 1 than 0.
3\small \frac{5}{6} is about, \small \frac{5}{6} is closer to 1 than 0.
Step 2: Estimate the sum
An estimate of 5\small \frac{7}{12}  + 3\small \frac{5}{6} = 1 + 1 = 2
Therefore, the chain is now 2 feet long.

Review & Refresh

Find the product. Check whether your answer is reasonable.

Question 16.
509 × 5 = ___
Answer:
The number 509 round to hundred is 500.
500 × 5 = 2500
509 × 5 = 2545
So the answer is reasonable.

Question 17.
7,692 × 6 = ___
Answer:
The number 7692 round to hundred is 7700.
7700 × 6 = 46200
7,692 × 6 = 46152
So the answer is reasonable.

Question 18.
31,435 × 7 = ___
Answer:
The number 31435 round to hundred is 31,400.
31,400 × 7 = 219800
31,435 × 7 = 220045
So the answer is reasonable.

Lesson 8.3 Find Common Denominators

Explore and Grow

You cut a rectangular pan of vegetable lasagna into equal-sized pieces. You serve \(\frac{1}{2}\) of the lasagna to a large table and \(\frac{1}{3}\) of the lasagna to a small table. Draw a diagram that shows how you cut the lasagna.
What fraction of the lasagna does each piece represent? How does the denominator of the fraction compare to the denominators of \(\frac{1}{2}\) and \(\frac{1}{3}\) ?

Reasoning
Is there another way you can cut the lasagna? Explain your reasoning.

Think and Grow: Find Common Denominators

Key Idea
Fractions that have the same denominator are said to have a common denominator. You can find a common denominator either by finding a common multiple of the denominators or by finding the product of the denominators.
Example
Use a common denominator to write equivalent fractions for \(\frac{1}{2}\) and \(\frac{5}{8}\).
List multiples of the denominators.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 30

Answer:
Big-Ideas-Math-Solutions-Grade-5-Chapter-8-Add-and-Subtract-Fractions-30

Example
Use a common denominator to write equivalent fractions for \(\frac{2}{3}\) and \(\frac{1}{4}\). Use the product of the denominators: 3 × 4 = __.
Write equivalent fractions with denominators of 12.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 31

Answer:
Big-Ideas-Math-Solutions-Grade-5-Chapter-8-Add-and-Subtract-Fractions-31

Show and Grow

Use a common denominator to write an equivalent fraction for each fraction.

Question 1.
\(\frac{2}{3}\) and \(\frac{1}{6}\)
Answer:
Use the product of the denominators : 3 \small \times 6 = 18
Write equivalent fractions with denominators of 18
\small \frac{2}{3} = \frac{2 \times 6}{3 \times 6} = \frac{12}{18}
\small \frac{1}{6} = \frac{1 \times 3}{6 \times 3} = \frac{3}{18}
Therefore, equivalent fractions are \small \frac{12}{18}  and \small \frac{3}{18}.

Question 2.
\(\frac{5}{6}\) and \(\frac{3}{4}\)
Answer:
Step 1: Use the product of the denominators : 6 \small \times 4 = 24
Step 2: Write equivalent fractions with denominators of 24
\small \frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24}
\small \frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}
Therefore, equivalent fractions are \small \frac{20}{24}  and \small \frac{18}{24}.

Apply and Grow: Practice

Use a common denominator to write an equivalent fraction for each fraction.

Question 3.
\(\frac{2}{3}\) and \(\frac{5}{6}\)
Answer:
Step 1: Use the product of the denominators : 3 \small \times 6 = 18
Step 2: Write equivalent fractions with denominators of 18

\small \frac{2}{3} = \frac{2 \times 6}{3 \times 6} = \frac{12}{18}

\small \frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18}
Therefore, equivalent fractions are \small \frac{12}{18}  and \small \frac{15}{18}.

Question 4.
\(\frac{3}{4}\) and \(\frac{1}{2}\)
Answer:
Step 1: Use the product of the denominators : 4 \small \times 2 = 8
Step 2: Write equivalent fractions with denominators of 8
\small \frac{3}{4} = \frac{3 \times 2}{4 \times 2} = \frac{6}{8}
\small \frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}
Therefore, equivalent fractions are \small \frac{6}{8}  and \small \frac{4}{8}.

Question 5.
\(\frac{5}{9}\) and \(\frac{2}{3}\)
Answer:
Step 1: Use the product of the denominators: 9 \small \times 3 = 27
Step 2: Write equivalent fractions with denominators of 27
\small \frac{5}{9} = \frac{5 \times 3}{9 \times 3} = \frac{15}{27}
\small \frac{2}{3} = \frac{2 \times 9}{3 \times 9} = \frac{18}{27}
Therefore, equivalent fractions are \small \frac{15}{27}  and \small \frac{18}{27}.

Question 6.
\(\frac{8}{21}\) and \(\frac{3}{7}\)
Answer:
Step 1: Use the product of the denominators: 21 \small \times 7 = 147
Step 2: Write equivalent fractions with denominators of 147
\small \frac{8}{21} = \frac{8 \times 7}{21 \times 7} = \frac{56}{147}
\small \frac{3}{7} = \frac{3 \times 21}{7 \times 21} = \frac{63}{147}
Therefore, equivalent fractions are \small \frac{56}{147}  and \small \frac{63}{147}.

Question 7.
\(\frac{1}{5}\) and \(\frac{1}{2}\)
Answer:
Step 1: Use the product of the denominators: 5 \small \times 2 = 10
Step 2: Write equivalent fractions with denominators of 10
\small \frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10}
\small \frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}
Therefore, equivalent fractions are \small \frac{2}{10}  and \small \frac{5}{10}.

Question 8.
\(\frac{3}{4}\) and \(\frac{1}{6}\)
Answer:
Step 1: Use the product of the denominators: 4 \small \times 6 = 24
Step 2: Write equivalent fractions with denominators of 24
\small \frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24}
\small \frac{1}{6} = \frac{1 \times 4}{6 \times 4} = \frac{4}{24}
Therefore, equivalent fractions are \small \frac{18}{24}  and \small \frac{4}{24}.

Question 9.
\(\frac{3}{7}\) and \(\frac{2}{9}\)
Answer:
Step 1: Use the product of the denominators: 7 \small \times 9 = 63
Step 2: Write equivalent fractions with denominators of 63
\small \frac{3}{7} = \frac{3 \times 9}{7 \times 9} = \frac{27}{63}
\small \frac{2}{9} = \frac{2 \times 7}{9 \times 7} = \frac{14}{63}
Therefore, equivalent fractions are \small \frac{27}{63}  and \small \frac{14}{63}.

Question 10.
\(\frac{3}{8}\) and \(\frac{5}{11}\)
Answer:
Step 1: Use the product of the denominators: 8 \small \times 11 = 88
Step 2: Write equivalent fractions with denominators of 88
\small \frac{3}{8} = \frac{3 \times 11}{8 \times 11} = \frac{33}{88}
\small \frac{5}{11} = \frac{5 \times 8}{11 \times 8} = \frac{40}{88}
Therefore, equivalent fractions are \small \frac{33}{88} and \small \frac{40}{88}.

Question 11.
You walk your dog \(\frac{3}{4}\) mile on Saturday and \(\frac{5}{8}\) mile on Sunday. Use a common denominator to write an equivalent fraction for each fraction.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 32
Answer:
Step 1: Use the product of the denominators: 4 \small \times 8 = 32
Step 2: Write equivalent fractions with denominators of 32
\small \frac{3}{4} = \frac{3 \times 8}{4 \times 8} = \frac{24}{32}
\small \frac{5}{8} = \frac{5 \times 4}{8 \times 4} = \frac{20}{32}
Dog walk on Saturday, equivalent fraction = \small \frac{24}{32}
Dog walk on Sunday, equivalent fraction = \small \frac{20}{32}

Question 12.
Writing
Explain how to use the models to find a common denominator for \(\frac{1}{2}\) and \(\frac{3}{5}\). Then write an equivalent fraction for each fraction.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 33
Answer:
Initially find the product of the denominators and that product value is the denominator for both fractions.
Step 1: Use the product of the denominators: 2 \small \times 5 = 10
Step 2: Write equivalent fractions with denominators of 10
\small \frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}
\small \frac{3}{5} = \frac{3 \times 2}{5 \times 2} = \frac{6}{10}
Therefore, equivalent fractions are \small \frac{5}{10} and \small \frac{6}{10}.

Question 13.
Number Sense
Which pairs of fractions are equivalent to \(\frac{1}{2}\) and \(\frac{2}{3}\) ?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 34
Answer:
\small \frac{6}{12} and \small \frac{8}{12}

\small \frac{3}{6} and \small \frac{4}{6}
These two fractions are equivalent to \(\frac{1}{2}\) and \(\frac{2}{3}\).

Think and Grow: Modeling Real Life

Example
You and your friend make woven key chains. Your key chain is \(\frac{2}{4}\) foot long. Your friend’s is \(\frac{3}{6}\) foot long. Are the key chains the same length?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 35
Use a common denominator to write equivalent fractions for the lengths of the key chains. Use the product of the denominators.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 36
Write equivalent fractions with denominators of 24.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 37
Compare the lengths of the key chains.
So, the key chains __ the same length.

Answer:
4 × 6 = 24
Write equivalent fractions with denominators of 24.
Big-Ideas-Math-Solutions-Grade-5-Chapter-8-Add-and-Subtract-Fractions-37
So, the key chains has the same length.

Show and Grow

Question 14.
Your hamster weighs \(\frac{13}{16}\) ounce. Your friend’s hamster weighs \(\frac{6}{8}\) ounce. Do the hamsters weigh the same amount?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 38
Answer:
Use a common denominator to write equivalent fractions for the hamsters weight.
Step 1: Use the product of the denominators: 16 \small \times 8 = 128
Step 2: Write equivalent fractions with denominators of 128

\small \frac{13}{16} = \frac{13 \times 8}{16 \times 8} = \frac{104}{128}

\small \frac{6}{8} = \frac{6 \times 16}{8 \times 16} = \frac{96}{128}
Hamsters weigh the different amount. One is \small \frac{104}{128} ounce and the other one is \small \frac{96}{128} ounce.

Question 15.
DIG DEEPER!
You have three vegetable pizzas of the same size. One has 4 equal slices. The second has 8 equal slices. The third has 6 equal slices. You cut the pizzas until all of them have the same number of slices. How many slices does each pizza have?
Answer:
Given,
You have three vegetable pizzas of the same size.
One has 4 equal slices. The second has 8 equal slices.
The third has 6 equal slices. You cut the pizzas until all of them have the same number of slices.
4 + 6 + 8 = 18
one has \(\frac{4}{18}\)
second has \(\frac{6}{18}\)
Third has \(\frac{8}{18}\)
Total there are 18 slices.

Find Common Denominators Homework & Practice 8.3

Use a common denominator to write an equivalent fraction for each fraction.

Question 1.
\(\frac{1}{2}\) and \(\frac{3}{8}\)
Answer:
Step 1: Use the product of the denominators: 2 \small \times 8 = 16
Step 2: Write equivalent fractions with denominators of 16

\small \frac{1}{2} = \frac{1 \times 8}{2 \times 8} = \frac{8}{16}

\small \frac{3}{8} = \frac{3 \times 2}{8 \times 2} = \frac{6}{16}
Therefore, equivalent fractions are \small \frac{8}{16}  and \small \frac{6}{16}.

Question 2.
\(\frac{7}{9}\) and \(\frac{2}{3}\)
Answer:
Step 1: Use the product of the denominators: 9 \small \times 3 = 27
Step 2: Write equivalent fractions with denominators of 27.

\small \frac{7}{9} = \frac{7 \times 3}{9 \times 3} = \frac{21}{27}

\small \frac{2}{3} = \frac{2 \times 9}{3 \times 9} = \frac{18}{27}
Therefore, equivalent fractions are \small \frac{21}{27}  and \small \frac{18}{27}.

Question 3.
\(\frac{5}{6}\) and \(\frac{1}{2}\)
Answer:
Step 1: Use the product of the denominators: 6 \small \times 2 = 12
Step 2: Write equivalent fractions with denominators of 12.

\small \frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}

\small \frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}
Therefore, equivalent fractions are \small \frac{10}{12}  and \small \frac{6}{12}.

Question 4.
\(\frac{3}{4}\) and \(\frac{5}{16}\)
Answer:
Step 1: Use the product of the denominators: 4 \small \times 16 = 64
Step 2: Write equivalent fractions with denominators of 64.

\small \frac{3}{4} = \frac{3 \times 16}{4 \times 16} = \frac{48}{64}

\small \frac{5}{16} = \frac{5 \times 4}{16 \times 4} = \frac{20}{64}
Therefore, equivalent fractions are \small \frac{48}{64}  and \small \frac{20}{64}.

Question 5.
\(\frac{18}{24}\) and \(\frac{5}{6}\)
Answer:
Step 1: Use the product of the denominators: 24 \small \times 6 = 144
Step 2: Write equivalent fractions with denominators of 144.

\small \frac{18}{24} = \frac{18 \times 6}{24 \times 6} = \frac{108}{144}

\small \frac{5}{6} = \frac{5 \times 24}{6 \times 24} = \frac{120}{144}
Therefore, equivalent fractions are \small \frac{108}{144}  and \small \frac{120}{144}.

Question 6.
\(\frac{1}{3}\) and \(\frac{1}{5}\)
Answer:
Step 1: Use the product of the denominators: 3 \small \times 5 = 15
Step 2: Write equivalent fractions with denominators of 15.

\small \frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}

\small \frac{1}{5} = \frac{1 \times 3}{5 \times 3} = \frac{3}{15}
Therefore, equivalent fractions are \small \frac{5}{15}  and \small \frac{3}{15}.

Question 7.
\(\frac{3}{5}\) and \(\frac{4}{7}\)
Answer:
Step 1: Use the product of the denominators: 5 \small \times 7= 35
Step 2: Write equivalent fractions with denominators of 35.

\small \frac{3}{5} = \frac{3 \times 7}{5 \times 7} = \frac{21}{35}

\small \frac{4}{7} = \frac{4 \times 5}{7 \times 5} = \frac{20}{35}
Therefore, equivalent fractions are \small \frac{21}{35}  and \small \frac{20}{35}.

Question 8.
\(\frac{5}{8}\) and \(\frac{2}{9}\)
Answer:
Step 1: Use the product of the denominators: 8 \small \times 9 = 72
Step 2: Write equivalent fractions with denominators of 72.

\small \frac{5}{8} = \frac{5 \times 9}{8 \times 9} = \frac{45}{72}

\small \frac{2}{9} = \frac{2 \times 8}{9 \times 8} = \frac{16}{72}
Therefore, equivalent fractions are \small \frac{45}{72}  and \small \frac{16}{72}.

Question 9.
A mint plant grows \(\frac{7}{8}\) inch in 1 week and \(\frac{13}{16}\) inch the next week. Use a common denominator to write an equivalent fraction for each fraction.
Answer:
Step 1: Use the product of the denominators : 8 \small \times 16 = 128
Step 2: Write equivalent fractions with denominators of 128.

\small \frac{7}{8} = \frac{7 \times 16}{8 \times 16} = \frac{112}{128}

\small \frac{13}{16} = \frac{13 \times 8}{16 \times 8} = \frac{104}{128}
Therefore, equivalent fractions are \small \frac{112}{128}  and \small \frac{104}{128}.

Question 10.
Which One Doesn’t Belong? Which pair of fractions is not equivalent to \(\frac{2}{5}\) and \(\frac{1}{10}\)?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 39
Answer:
So, \small \frac{6}{15} and \small \frac{5}{15} is not equivalent to \(\frac{2}{5}\) and \(\frac{1}{10}\) and remaining all the pairs are equivalent.

Question 11.
YOU BE THE TEACHER
Your friend says she used a common denominator to find fractions equivalent to \(\frac{2}{3}\) and \(\frac{8}{9}\). Is your friend correct? Explain.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 39.1
Answer:
No, she is wrong because 9 and 12 are not common denominators.
Step 1: Use the product of the denominators : 3 \small \times 9 = 27
Step 2: Write equivalent fractions with common denominator of 27.

\small \frac{2}{3} = \frac{2 \times 9}{3 \times 9} = \frac{18}{27}

\small \frac{8}{9} = \frac{8 \times 3}{9 \times 3} = \frac{24}{27}
Therefore, equivalent fractions are \small \frac{18}{27}  and \small \frac{24}{27}.

Question 12.
Modeling Real Life
Some friends spend \(\frac{1}{3}\) hour collecting sticks and \(\frac{5}{6}\) hour building a fort. Do they spend the same amount of time on each? Explain.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 40
Answer:
No, they can not spend the same time on each because \small \frac{1}{3} is not equivalent to \small \frac{5}{6}.
\small \frac{1}{3} is equivalent to \small \frac{2}{6}.

Question 13.
DIG DEEPER!
Use a common denominator to write an equivalent fraction for each fraction. Which two students are the same distance from the school? Are they closer to or farther from the school than the other student?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 41
Answer:
Step 1: Use the LCM of the denominators. LCM of 12, 8 and 6 = 24
Step 2: Write equivalent fractions with common denominator of 24.
Student A –>  \small \frac{10}{12} = \frac{10 \times 2}{12 \times 2} = \frac{20}{24}
Student B –>  \small \frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24}
Student C –>  \small \frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24}
Therefore, student A and student C are the same distance from the school i.e. \small \frac{20}{24} mile.
They are closer to student B.

Review & Refresh

Find the value of the expression.

Question 14.
102
Answer: 100

Question 15.
8 × 104
Answer: 80,000

Question 16.
6 × 103
Answer: 6000

Question 17.
9 × 105
Answer: 9,00,000

Lesson 8.4 Add Fractions with Unlike Denominators

Use a model to find the sum.

Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 42

Explain how you can use a model to add fifths and tenths.

Construct Arguments
How can you add two fractions with unlike denominators without using a model? Explain why your method makes sense.

Think and Grow: Add Fractions with Unlike Denominators

You can use equivalent fractions to add fractions that have unlike denominators.
Example
Find \(\frac{1}{4}\) + \(\frac{3}{8}\)
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 8 is a multiple of 4, so rewrite \(\frac{1}{4}\) with a denominator of 8.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 43

Answer:
Big-Ideas-Math-Solutions-Grade-5-Chapter-8-Add-and-Subtract-Fractions-43
Example
Find \(\frac{7}{8}\) + \(\frac{1}{6}\) Estimate __
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 8 is not a multiple of 6, so rewrite each fraction with a denominator of 8 × 6 = 48.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 44

Answer:
Big-Ideas-Math-Solutions-Grade-5-Chapter-8-Add-and-Subtract-Fractions-44

Show and Grow

Add.

Question 1.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 45
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 6 is a multiple of 3, so rewrite it with a denominator of 6.
Rewrite \small \frac{2}{3} as \small \frac{2 \times 2}{3 \times 2} = \small \frac{4}{6}

\small \frac{5}{6} + \small \frac{2}{3}  =  \small \frac{5}{6} + \small \frac{4}{6}

= \small \frac{9}{6} or \small \frac{3}{2}

Question 2.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 46
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 5 is not a multiple of 4, so rewrite each fraction with a denominator of 5 \small \times 4 = 20

Rewrite \small \frac{1}{5}  as \small \frac{1 \times 4}{5 \times 4} = \small \frac{4}{20}  and  \small \frac{3}{4} as \small \frac{3 \times 5}{4 \times 5} = \small \frac{15}{20}

\small \frac{1}{5} + \small \frac{3}{4} = \small \frac{4}{20} + \small \frac{15}{20}

= \small \frac{19}{20}

Question 3.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 47
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 6 is not a multiple of 4, so rewrite each fraction with a denominator of 6 \small \times 4 = 24
Rewrite \small \frac{1}{6}  as \small \frac{1 \times 4}{6 \times 4} = \small \frac{4}{24}  and  \small \frac{1}{4} as \small \frac{1 \times 6}{4 \times 6} = \small \frac{6}{24}

\small \frac{1}{6} + \small \frac{1}{4} = \small \frac{4}{24} + \small \frac{6}{24}

= \small \frac{10}{24} or \small \frac{5}{12}

Apply and Grow: Practice

Add.

Question 4.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 48
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 8 is a multiple of 4, so rewrite it with a denominator of 8.
Rewrite \small \frac{1}{4} as \small \frac{1 \times 2}{4 \times 2} = \small \frac{2}{8}

\small \frac{5}{8} + \small \frac{1}{4} = \small \frac{5}{8} + \small \frac{2}{8}

= \small \frac{7}{8}

Question 5.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 49
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 12 is a multiple of 3, so rewrite it with a denominator of 12.
Rewrite \small \frac{2}{3} as \small \frac{2 \times 4}{3 \times 4} = \small \frac{8}{12}

\small \frac{2}{3} + \small \frac{7}{12}\small \frac{8}{12} + \small \frac{7}{12}

= \small \frac{15}{12} or \small \frac{5}{4}

Question 6.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 50
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 15 is a multiple of 5 , so rewrite it with a denominator of 15.
Rewrite \small \frac{2}{5} as \small \frac{2 \times 3}{5 \times 3} = \small \frac{6}{15}

\small \frac{2}{5} + \small \frac{10}{15} = \small \frac{6}{15} + \small \frac{10}{15}

= \small \frac{16}{15}

Question 7.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 51
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 8 is not a multiple of 6, so rewrite each fraction with a denominator of 8 × 6 = 48.
Rewrite \small \frac{1}{6} as \small \frac{1 \times 8}{6 \times 8} = \small \frac{8}{48}  and  \small \frac{4}{8} as \small \frac{4 \times 6}{8 \times 6} = \small \frac{24}{48}

\small \frac{1}{6} + \small \frac{4}{8} = \small \frac{8}{48} + \small \frac{24}{48}

= \small \frac{32}{48} or \small \frac{2}{3}

Question 8.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 52
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 12 is not a multiple of 5, so rewrite each fraction with a denominator of 12 × 5 = 60.
Rewrite \small \frac{11}{12} as \small \frac{11 \times 5}{12 \times 5} = \small \frac{55}{60}  and  \small \frac{3}{5} as \small \frac{3 \times 12}{5 \times 12} = \small \frac{36}{60}

\small \frac{11}{12} + \small \frac{3}{5} = \small \frac{55}{60} + \small \frac{36}{60}

= \small \frac{91}{60}

Question 9.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 53
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 9 is a multiple of 3 , so rewrite it with a denominator of 9.
Rewrite \small \frac{4}{3} as \small \frac{4 \times 3}{3 \times 3} = \small \frac{12}{9}

\small \frac{2}{9} + \small \frac{4}{3} + \small \frac{5}{9}  =  \small \frac{2}{9} + \small \frac{12}{9} + \small \frac{5}{9}

= \small \frac{19}{9}

Question 10.
Your friend buys \(\frac{1}{8}\) pound of green lentils and \(\frac{3}{4}\) pound of brown lentils. What fraction of a pound of lentils does she buy?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 54
Answer:
Given that,
pound of green lentils = \small \frac{1}{8}
Pound of brown lentils = \small \frac{3}{4}
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 8 is a multiple of 4, so rewrite \small \frac{3}{4} with a denominator of 8.
Rewrite \small \frac{3}{4} as \small \frac{3 \times 2}{4 \times 2} = \small \frac{6}{8}
Pound of lentils = \small \frac{1}{8} + \small \frac{3}{4}
= \small \frac{1}{8} + \small \frac{6}{8}
Fraction of pound of lentils = \small \frac{7}{8}

Question 11.
Reasoning
Newton and Descartes find \(\frac{1}{2}\) + \(\frac{1}{6}\). Newton says the sum is \(\frac{4}{6}\). Descartes says the sum is \(\frac{2}{3}\). Who is correct? Explain.
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
\small \frac{1}{2} + \small \frac{1}{6}
Think: 6 is a multiple of 2, so rewrite \small \frac{1}{2} with a denominator of 6.
Rewrite \small \frac{1}{2} as \small \frac{1 \times 3}{2 \times 3} = \small \frac{3}{6}
\small \frac{1}{2} + \small \frac{1}{6} = \small \frac{3}{6} + \small \frac{1}{6}
= \small \frac{4}{6} or \small \frac{2}{3}
Therefore, both Newton and Descartes answers are correct.

Question 12.
DIG DEEPER!
Write two fractions that have a sum of 1 and have different denominators.
Answer:
\(\frac{1}{2}\) + \(\frac{3}{6}\)
= \(\frac{1}{2}\) × \(\frac{3}{3}\) + \(\frac{3}{6}\)
= \(\frac{3}{6}\) + \(\frac{3}{6}\)
= \(\frac{6}{6}\)
= 1

Think and Grow: Modeling Real Life

Example
About \(\frac{17}{15}\) of Earth’s surface is covered by ocean water.
About \(\frac{3}{100}\) of Earth’s surface is covered by other water resources.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 55
About how much of Earth’s surface is covered by water?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 56

Answer:
Big-Ideas-Math-Solutions-Grade-5-Chapter-8-Add-and-Subtract-Fractions-56

Show and Grow

Question 13.
The George Washington Bridge links Manhattan, NY, to FortLee, NJ. The part of the bridge in New Jersey is about \(\frac{1}{2}\) mile long. The part in New York is about \(\frac{2}{5}\) mile long. About how long is the George Washington Bridge?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 57
Answer:
Given that,
New Jersey bridge = \(\frac{1}{2}\) mile long
New York bridge = \(\frac{2}{5}\) mile long
Add \small \frac{1}{2} and \small \frac{2}{5} to find how long is the George Washington Bridge
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 5 is not a multiple of 2, so rewrite each fraction with a denominator of 5 \small \times 2 = 10
Rewrite \small \frac{1}{2} as \small \frac{1 \times 5}{2 \times 5} = \small \frac{5}{10}  and \small \frac{2}{5} as \small \frac{2 \times 2}{5 \times 2} = \small \frac{4}{10}
\small \frac{1}{2} + \small \frac{2}{5}  = \small \frac{5}{10} + \small \frac{4}{10}
= \small \frac{9}{10}
So, George Washington Bridge is about \small \frac{9}{10} mile long.

Question 14.
DIG DEEPER!
Your goal is to practice playing the saxophone for at least 2 hours in 1 week. Do you reach your goal? Explain.
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 58
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
\small \frac{3}{4} + \small \frac{1}{2} + \small \frac{2}{3}
Think: Rewrite the denominators as 4 \small \times 2 \small \times 3 = 24
Rewrite \small \frac{3}{4} as \small \frac{3 \times 6}{4 \times 6} = \small \frac{18}{24}
\small \frac{1}{2} as \small \frac{1 \times 12}{2 \times 12} = \small \frac{12}{24}
\small \frac{2}{3} as \small \frac{2 \times 8}{3 \times 8} = \small \frac{16}{24}
\small \frac{3}{4} + \small \frac{1}{2} + \small \frac{2}{3} = \small \frac{18}{24} + \small \frac{12}{24} + \small \frac{16}{24}
= \small \frac{46}{24} or \small \frac{23}{12}
Total practice time in a week = \small \frac{23}{12} = 1.91 hours
So, goal does not reached.

Add Fractions with Unlike Denominators Homework & Practice 8.4

Add

Question 1.
\(\frac{1}{9}\) + \(\frac{2}{3}\) = ___
Answer:
\small \frac{1}{9} + \small \frac{2}{3}
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 9 is a multiple of 3, so rewrite it with a denominator of 9
Rewrite \small \frac{2}{3} as \small \frac{2 \times 3}{3 \times 3} = \small \frac{6}{9}
\small \frac{1}{9} + \small \frac{2}{3} = \small \frac{1}{9} + \small \frac{6}{9}
= \small \frac{7}{9}
\(\frac{1}{9}\) + \(\frac{2}{3}\) = \small \frac{7}{9}

Question 2.
\(\frac{1}{2}\) + \(\frac{3}{4}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 4 is a multiple of 2, so rewrite it with a denominator of 4
Rewrite \small \frac{1}{2} as \small \frac{1 \times 2}{2 \times 2} = \frac{2}{4}
\small \frac{1}{2} + \small \frac{3}{4} = \small \frac{2}{4} + \small \frac{3}{4}
= \small \frac{5}{4}
\(\frac{1}{2}\) + \(\frac{3}{4}\) = \small \frac{5}{4}

Question 3.
\(\frac{4}{6}\) + \(\frac{5}{12}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 12 is a multiple of 6, so rewrite it with a denominator of 12
Rewrite \small \frac{4}{6} as \small \frac{4 \times 2}{6 \times 2} = \small \frac{8}{12}
\small \frac{4}{6} + \small \frac{5}{12} = \small \frac{8}{12} + \small \frac{5}{12}
= \small \frac{13}{12}
\(\frac{4}{6}\) + \(\frac{5}{12}\) = \small \frac{13}{12}

Question 4.
\(\frac{1}{3}\) + \(\frac{1}{4}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 4 is not a multiple of 3, so rewrite each fraction with a denominator of 4 \small \times 3 = 12
Rewrite \small \frac{1}{3} as \small \frac{1 \times 4}{3 \times 4} = \small \frac{4}{12}  and \small \frac{1}{4} as \small \frac{1 \times 3}{4 \times 3} = \small \frac{3}{12}
\small \frac{1}{3} + \small \frac{1}{4}  = \small \frac{4}{12} + \small \frac{3}{12}
= \small \frac{7}{12}
\(\frac{1}{3}\) + \(\frac{1}{4}\) = \small \frac{7}{12}

Question 5.
\(\frac{3}{2}\) + \(\frac{4}{5}\) = __
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 5 is not a multiple of 2, so rewrite each fraction with a denominator of 5 \small \times 2 = 10
Rewrite \small \frac{3}{2} as \small \frac{3 \times 5}{2 \times 5} = \small \frac{15}{10}  and \small \frac{4}{5} as \small \frac{4 \times 2}{5 \times 2} = \small \frac{8}{10}
\small \frac{3}{2} + \small \frac{4}{5}  = \small \frac{15}{10} + \small \frac{8}{10}
= \small \frac{23}{10}
\(\frac{3}{2}\) + \(\frac{4}{5}\) = \small \frac{23}{10}

Question 6.
\(\frac{6}{8}\) + \(\frac{9}{10}\) + \(\frac{1}{8}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 10 is not a multiple of 8, so rewrite each fraction with a denominator of 10 \small \times 8 = 80
Rewrite \small \frac{6}{8} as \small \frac{6 \times 10}{8 \times 10} = \small \frac{60}{80}
\small \frac{9}{10} as \small \frac{9 \times 8}{10 \times 8} = \small \frac{72}{80}
\small \frac{1}{8} as \small \frac{1 \times 10}{8 \times 10} = \small \frac{10}{80}
\small \frac{6}{8} + \small \frac{9}{10} + \small \frac{1}{8} = \small \frac{60}{80} + \small \frac{72}{80} + \small \frac{10}{80} = \small \frac{142}{80}
\(\frac{6}{8}\) + \(\frac{9}{10}\) + \(\frac{1}{8}\) = \small \frac{142}{80}

Question 7.
You use beads to make a design. Of the beads, \(\frac{1}{3}\) are red and \(\frac{1}{6}\) are blue. The rest are white.What fraction of the beads are red or blue?
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 6 is a multiple of 3, so rewrite it with a denominator of 6
Rewrite \small \frac{1}{3} as \small \frac{1 \times 2}{3 \times 2} = \small \frac{2}{6}
\small \frac{1}{3} + \small \frac{1}{6} = \small \frac{2}{6} + \small \frac{1}{6}
= \small \frac{3}{6}
The fraction of beads are red or blue = \small \frac{3}{6} = \small \frac{1}{2}
Rest are white = \small \frac{1}{2}

Question 8.
YOU BE THE TEACHER
Your friend says the sum of \(\frac{1}{5}\) and \(\frac{9}{10}\) is \(\frac{10}{15}\). Is your friend correct? Explain.
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 10 is a multiple of 5, so rewrite it with a denominator of 10
Rewrite \small \frac{1}{5} as \small \frac{1 \times 2}{5 \times 2} = \small \frac{2}{10}
\small \frac{1}{5} + \small \frac{9}{10} = \small \frac{2}{10} + \small \frac{9}{10}
= \small \frac{11}{10}
Therefore, the sum of \(\frac{1}{5}\) and \(\frac{9}{10}\) is \small \frac{11}{10}.
So, my friend answer is wrong.

Question 9.
Reasoning
Which expressions are equal to \(\frac{14}{15}\)?
Big Ideas Math Solutions Grade 5 Chapter 8 Add and Subtract Fractions 59
Answer:
\small \frac{3}{5} + \small \frac{1}{3} and \small \frac{1}{5} + \small \frac{11}{15} are equal to \(\frac{14}{15}\) and these two expressions only having denominator of 15.

Question 10.
Modeling Real Life
There are 100 senators in the 115th Congress. Democrats make up of the senators, and Republicans make up \(\frac{13}{25}\) of the 25 senators. The rest are Independents. What fraction of the senators are Democrat or Republican?
Answer:
Given,
There are 100 senators in the 115th Congress.
Democrats make up of the senators, and Republicans make up \(\frac{13}{25}\) of the 25 senators.
\(\frac{100}{115}\) + \(\frac{13}{25}\)
= \(\frac{100}{115}\) × \(\frac{5}{5}\) + \(\frac{13}{25}\) × \(\frac{23}{23}\)
= \(\frac{799}{575}\)
= 1 \(\frac{224}{575}\)

Question 11.
Modeling Real Life
Your friend needs 1 cup of homemade orange juice. He squeezes \(\frac{1}{2}\) cup of orange juice from one orange and \(\frac{3}{8}\) cup from another orange. Does your friend need to squeeze another orange? Explain.
Answer:
\small \frac{1}{2} + \small \frac{3}{8}
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 8 is a multiple of 2, so rewrite it with a denominator of 8
Rewrite \small \frac{1}{2} as \small \frac{1 \times 4}{2 \times 4} = \small \frac{4}{8}
\small \frac{1}{2} + \small \frac{3}{8} = \small \frac{4}{8} + \small \frac{3}{8} = \small \frac{7}{8}
Juice from 2 oranges = \small \frac{7}{8}
My friend needs 1 cup of orange juice = 1 – \small \frac{7}{8} = \small \frac{1}{8}
So, my friend needs to squeeze \small \frac{1}{8} cup from another orange.

Question 12.
DIG DEEPER!
Of all the atoms in caffeine, \(\frac{1}{12}\) are oxygen atoms, \(\frac{1}{6}\) are nitrogen atoms, and \(\frac{1}{3}\) are carbon atoms. The rest of the atoms are hydrogen. What fraction of the atoms in caffeine are oxygen, nitrogen, or hydrogen?
Answer:
From the given information, hydrogen atoms = 1- (\small \frac{1}{12} + \small \frac{1}{6} + \small \frac{1}{3}) = \small \frac{5}{12}
\small \frac{1}{12} + \small \frac{1}{6} + \small \frac{5}{12} = ?
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 12 is a multiple of 6, so rewrite it with a denominator of 12
Rewrite \small \frac{1}{6} as \small \frac{1 \times 2}{6 \times 2} = \small \frac{2}{12}
\small \frac{1}{12} + \small \frac{1}{6} + \small \frac{5}{12} = \small \frac{1}{12} + \small \frac{2}{12} + \small \frac{5}{12}
= \small \frac{8}{12}
= \small \frac{2}{3}
So \small \frac{2}{3} of the atoms in caffeine are oxygen, nitrogen, or hydrogen.

Review & Refresh

Use properties to find the sum or product.

Question 13.
5 × 84
Answer: 420

Explanation:
We can find the product by using the distributive property.
5 × 84 = 5 × (80 + 4)
= (5 × 80) + (5 × 4)
= 400 + 20
= 420

Question 14.
521 + 0 + 67
Answer: 588

Explanation:
We can find the sum of the given expression using the additive identity.
521 + 0 + 67 = 521 + 67
= 588

Question 15.
25 × 8 × 4
Answer: 800

Explanation:
25 × 8 × 4
= 25 × 32
= 800

Lesson 8.5 Subtract Fractions with Unlike Denominators

Explore and Grow

Use a model to find the difference.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 60
Explain how you can use a model to subtract fourths from twelfths.

Construct Arguments
How can you subtract two fractions with unlike denominators without using a model? Explain why your method makes sense.

Think and Grow: Subtract Fractions with Unlike Denominators

You can use equivalent fractions to subtract fractions that have unlike denominators.
Example
Find \(\frac{9}{10}\) – \(\frac{1}{2}\).
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 10 is a multiple of 2, so rewrite \(\frac{1}{2}\) with a denominator of 10.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 61

Answer:
Big-Ideas-Math-Answers-5th-Grade-Chapter-8-Add-and-Subtract-Fractions-61

Example
Find \(\frac{4}{3}\) – \(\frac{1}{4}\). Estimate _____
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 4 is not a multiple of 3, so rewrite each fraction with a denominator of 3 × 4 = 12.

Show and Grow

Subtract.

Question 1.
\(\frac{1}{2}\) – \(\frac{1}{4}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 4 is a multiple of 2, so rewrite \(\frac{1}{2}\) with a denominator of 4.
Rewrite \small \frac{1}{2} as \small \frac{1 \times 2}{2 \times 2} = \small \frac{2}{4}
\small \frac{1}{2}\small \frac{1}{4} = \small \frac{2}{4}\small \frac{1}{4}
= \(\frac{1}{4}\)
\(\frac{1}{2}\) – \(\frac{1}{4}\) = \(\frac{1}{4}\)

Question 2.
\(\frac{7}{9}\) – \(\frac{2}{3}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 9 is a multiple of 3, so rewrite \(\frac{2}{3}\) with a denominator of 9.
Rewrite \small \frac{2}{3} as \small \frac{2 \times 3}{3 \times 3} = \small \frac{6}{9}
\small \frac{7}{9}\small \frac{2}{3} = \small \frac{7}{9}\small \frac{6}{9}
= \small \frac{1}{9}
\(\frac{7}{9}\) – \(\frac{2}{3}\) = \(\frac{1}{9}\)

Question 3.
\(\frac{6}{5}\) – \(\frac{3}{8}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 8 is not a multiple of 5, so rewrite each fraction with a denominator of 5 \small \times 8 = 40
Rewrite \small \frac{6}{5} as \small \frac{6 \times 8}{5 \times 8} = \small \frac{48}{40}
\small \frac{3}{8} as \small \frac{3 \times 5}{8 \times 5} = \small \frac{15}{40}
\small \frac{6}{5}\small \frac{3}{8} = \small \frac{48}{40} – \small \frac{15}{40}
= \small \frac{33}{40}
\(\frac{6}{5}\) – \(\frac{3}{8}\) = \small \frac{33}{40}

Apply and Grow: Practice

Subtract.

Question 4.
\(\frac{10}{12}\) – \(\frac{3}{4}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 12 is a multiple of 4, so rewrite \(\frac{3}{4}\) with a denominator of 12.
Rewrite \small \frac{3}{4} as \small \frac{3 \times 3}{4 \times 3} = \small \frac{9}{12}
\small \frac{10}{12}\small \frac{3}{4} = \small \frac{10}{12}\small \frac{9}{12}
=\small \frac{1}{12}
\(\frac{10}{12}\) – \(\frac{3}{4}\) = \(\frac{1}{12}\)

Question 5.
\(\frac{1}{3}\) – \(\frac{1}{6}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 6 is a multiple of 3, so rewrite \(\frac{1}{3}\) with a denominator of 6.
Rewrite \small \frac{1}{3} as \small \frac{1 \times 2}{3 \times 2} = \small \frac{2}{6}
\small \frac{1}{3}\small \frac{1}{6} = \small \frac{2}{6}\small \frac{1}{6}
= \small \frac{1}{6}
\(\frac{1}{3}\) – \(\frac{1}{6}\) = \(\frac{1}{6}\)

Question 6.
\(\frac{9}{10}\) – \(\frac{2}{5}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 10 is a multiple of 5, so rewrite \(\frac{2}{5}\) with a denominator of 10.
Rewrite \small \frac{2}{5} as \small \frac{2 \times 2}{5 \times 2} = \small \frac{4}{10}
\small \frac{9}{10}\small \frac{2}{5} = \small \frac{9}{10}\small \frac{4}{10}
= \small \frac{5}{10} or \small \frac{1}{2}
\(\frac{9}{10}\) – \(\frac{2}{5}\) = \(\frac{1}{2}\)

Question 7.
\(\frac{5}{4}\) – \(\frac{2}{5}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 5 is not a multiple of 4, so rewrite each fraction with a denominator of 4 \small \times 5 = 20
Rewrite \small \frac{5}{4} as \small \frac{5 \times 5}{4 \times 5} = \small \frac{25}{20}
\small \frac{2}{5} as \small \frac{2 \times 4}{5 \times 4} = \small \frac{8}{20}
\small \frac{5}{4}\small \frac{2}{5} = \small \frac{25}{20}\small \frac{8}{20}
= \small \frac{17}{20}
\(\frac{5}{4}\) – \(\frac{2}{5}\) = \(\frac{17}{20}\)

Question 8.
\(\frac{13}{16}\) – \(\frac{3}{16}\) – \(\frac{5}{8}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 16 is a multiple of 8, so rewrite \(\frac{5}{8}\) with a denominator of 16.
Rewrite \small \frac{5}{8} as \small \frac{5 \times 2}{8 \times 2} = \small \frac{10}{16}
\small \frac{13}{16}\small \frac{3}{16}\small \frac{10}{16} = 0
\(\frac{13}{16}\) – \(\frac{3}{16}\) – \(\frac{5}{8}\) = 0

Question 9.
\(\frac{8}{9}\) – (\(\frac{2}{3}\) + \(\frac{1}{6}\)) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator.
common denominator for 9, 3 and 6 = 18
Rewrite \small \frac{8}{9} as \small \frac{16}{18}
\small \frac{2}{3} = \small \frac{12}{18}
\small \frac{1}{6} = \small \frac{3}{18}
\small \frac{8}{9}\small \frac{2}{3} + \small \frac{1}{6} = \small \frac{16}{18}\small \frac{12}{18} + \small \frac{3}{18} = \small \frac{1}{18}
\(\frac{8}{9}\) – (\(\frac{2}{3}\) + \(\frac{1}{6}\)) = \(\frac{1}{18}\)

Question 10.
You have \(\frac{1}{3}\) yard of wire. You use \(\frac{1}{3}\) yard to make an electric circuit. How much wire do you have left?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 63
Answer: Left with 0 wire
\small \frac{1}{3}\small \frac{1}{3} = 0

Question 11.
Your friend finds \(\frac{5}{8}\) – \(\frac{2}{5}\). Explain why his answer is unreasonable. What did he do wrong?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 64
Answer: For subtracting two fractions, the denominators must be same. Here, denominators are different.
So, the answer is wrong
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 8 is not a multiple of 5, so rewrite each fraction with a denominator of 8 \small \times 5 = 40
Rewrite \small \frac{5}{8} as \small \frac{5 \times 5}{8 \times 5} = \small \frac{25}{40}
\small \frac{2}{5} as \small \frac{2 \times 8}{5 \times 8} = \small \frac{16}{40}
\small \frac{5}{8}\small \frac{2}{5} = \small \frac{25}{40}\small \frac{16}{40}
= \small \frac{9}{40}
\(\frac{5}{8}\) – \(\frac{2}{5}\) = \(\frac{9}{40}\)

Question 12.
Number Sense
Which two fractions have a difference of \(\frac{1}{8}\) ?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 65
Answer:
\small \frac{1}{2} and \small \frac{3}{8}
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 8 is a multiple of 2, so rewrite \small \frac{1}{2} with a denominator of 8.
Rewrite \small \frac{1}{2} as \small \frac{1 \times 4}{2 \times 4} = \small \frac{4}{8}
\small \frac{1}{2}\small \frac{3}{8} = \small \frac{4}{8}\small \frac{3}{8}
= \small \frac{1}{8}
So, fractions \small \frac{1}{2} and \small \frac{3}{8} have the difference of \small \frac{1}{8}.

Think and Grow: Modeling Real Life

Example
A geologist needs \(\frac{1}{2}\) cup of volcanic sand to perform an experiment. She has \(\frac{3}{2}\) cups of quartz sand. She has \(\frac{2}{3}\) cup more quartz sand than volcanic sand. Can she perform the experiment?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 66
Find how many cups of volcanic sand the geologist has by subtracting \(\frac{2}{3}\) from \(\frac{3}{2}\).
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 3 is not a multiple of 2, so rewrite each fraction with a denominator 2 × 3 = 6.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 67

Answer:
Big-Ideas-Math-Answers-5th-Grade-Chapter-8-Add-and-Subtract-Fractions-67

Show and Grow

Question 13.
The world record for the longest dog tail is \(\frac{77}{100}\) meter. The previous record was \(\frac{1}{20}\) meter. shorter than the current record. Was the previous record longer than \(\frac{3}{4}\) meter?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 68
Answer:
Given that,
World record for the longest dog tail is \small \frac{77}{100} meter.
Previous record was \small \frac{1}{20} meter shorter than the current record
So subtract \small \frac{1}{20} meter from the current record to find previous record.
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Rewrite \small \frac{1}{20} as \small \frac{5}{100}
Previous record = \small \frac{77}{100}\small \frac{5}{100} = \small \frac{72}{100} = \small \frac{18}{25} meter
So the previous record is not longer than \small \frac{3}{4} meter.

Question 14.
DIG DEEPER!
A woodworker has 1 gallon of paint for a tree house. He uses \(\frac{3}{8}\) gallon to paint the walls and \(\frac{1}{5}\) gallon to paint the ladder. He needs \(\frac{1}{4}\) gallon to paint the roof. Does he have enough paint? Explain.
Answer:
Given that,
Woodworker has 1 gallon of paint
\small \frac{3}{8} gallon is used to paint the walls.
\small \frac{1}{5} gallon is used to paint the ladder.
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
\small \frac{3}{8} = \small \frac{15}{40}
\small \frac{1}{5} = \small \frac{8}{40}
1 – \small \frac{15}{40}\small \frac{8}{40} = \small \frac{17}{40}
Therefore, he has more than \small \frac{1}{4} gallon to paint the roof.

Subtract Fractions with Unlike Denominators Homework & Practice 8.5

Subtract

Question 1.
\(\frac{3}{4}\) – \(\frac{1}{8}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 8 is a multiple of 4, so rewrite \(\frac{3}{4}\) with a denominator of 8
Rewrite \small \frac{3}{4} as \small \frac{3 \times 2}{4 \times 2} = \small \frac{6}{8}
\small \frac{3}{4}\small \frac{1}{8} = \small \frac{6}{8}\small \frac{1}{8}
= \small \frac{5}{8}
\(\frac{3}{4}\) – \(\frac{1}{8}\) = \small \frac{5}{8}

Question 2.
\(\frac{4}{5}\) – \(\frac{6}{15}\) = __
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 15 is a multiple of 5, so rewrite \(\frac{4}{5}\) with a denominator of 15
Rewrite \small \frac{4}{5} as \small \frac{4 \times 3}{5 \times 3} = \small \frac{12}{15}
\small \frac{4}{5}\small \frac{6}{15} = \small \frac{12}{15}\small \frac{6}{15}
= \small \frac{6}{15}
\(\frac{4}{5}\) – \(\frac{6}{15}\) = \small \frac{6}{15}

Question 3.
\(\frac{1}{2}\) – \(\frac{1}{8}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 8 is a multiple of 2, so rewrite \(\frac{1}{2}\) with a denominator of 8
Rewrite \small \frac{1}{2} as \small \frac{1 \times 4}{2 \times 4} = \small \frac{4}{8}
\small \frac{1}{2}\small \frac{1}{8} = \small \frac{4}{8}\small \frac{1}{8}
= \small \frac{3}{8}
\(\frac{1}{2}\) – \(\frac{1}{8}\) = \small \frac{3}{8}

Question 4.
\(\frac{5}{3}\) – \(\frac{3}{4}\) = _____
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 4 is not a multiple of 3, so rewrite each fraction with a denominator of 4 x 3 =12
Rewrite \small \frac{5}{3} as \small \frac{5 \times 4}{3 \times 4} = \small \frac{20}{12}
\small \frac{3}{4} as \small \frac{3 \times 3}{4 \times 3} = \small \frac{9}{12}
\small \frac{5}{3}\small \frac{3}{4} = \small \frac{20}{12}\small \frac{9}{12}
= \small \frac{11}{12}
\(\frac{5}{3}\) – \(\frac{3}{4}\) = \small \frac{11}{12}

Question 5.
\(\frac{6}{8}\) – \(\frac{7}{10}\) = _____
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 10 is not a multiple of 8, so rewrite each fraction with a denominator of 10 x 8 =80
Rewrite \small \frac{6}{8} as \small \frac{6 \times 10}{8 \times 10} = \small \frac{60}{80}
\small \frac{7}{10} as \small \frac{7 \times 8}{10 \times 8} = \small \frac{56}{80}
\small \frac{6}{8}\small \frac{7}{10} = \small \frac{60}{80}\small \frac{56}{80}
= \small \frac{4}{80}
\(\frac{6}{8}\) – \(\frac{7}{10}\) = \small \frac{4}{80} = \small \frac{1}{20}

Question 6.
\(\frac{5}{6}\) – \(\frac{1}{4}\) – \(\frac{3}{12}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 12 is a multiple of 4 and 6, so rewrite each fraction with a denominator of 12
Rewrite \small \frac{5}{6} as \small \frac{5 \times 2}{6 \times 2} = \small \frac{10}{12}
\small \frac{1}{4} as \small \frac{1 \times 3}{4 \times 3} = \small \frac{3}{12}
\small \frac{5}{6}\small \frac{1}{4}\small \frac{3}{12} = \small \frac{10}{12}\small \frac{3}{12}\small \frac{3}{12}
= \small \frac{4}{12}
\(\frac{5}{6}\) – \(\frac{1}{4}\) – \(\frac{3}{12}\) = \small \frac{4}{12}

Question 7.
You eat \(\frac{1}{12}\) of a vegetable casserole. Your friend eats \(\frac{1}{6}\) of the same casserole. How much more does your friend eat than you?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 69
Answer:
\small \frac{1}{6}\small \frac{1}{12}
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 12 is a multiple of 6, so rewrite \small \frac{1}{6} with a denominator of 12
Rewrite \small \frac{1}{6} as \small \frac{1 \times 2}{6 \times 2} = \small \frac{2}{12}
\small \frac{1}{6}\small \frac{1}{12} = \small \frac{2}{12}\small \frac{1}{12} = \small \frac{1}{12}
So, my friend eats \small \frac{1}{12} of a vegetable casserole than me.

Question 8.
Writing
Why do fractions need a common denominator before you can add or subtract them?
Answer:
In order to add fractions, the fractions must have a common denominator. We need the pieces of each fraction to be the same size to combine them together. These two fractions have the same denominator, so the equal parts that the whole has been split into are the same size.

Question 9.
Logic
Find a.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 70
Answer:
a = \small \frac{7}{10}\small \frac{1}{2}
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 10 is a multiple of 2, so rewrite \small \frac{1}{2} with a denominator of 10
Rewrite \small \frac{1}{2} as \small \frac{1 \times 5}{2 \times 5} = \small \frac{5}{10}
a = \small \frac{7}{10}\small \frac{1}{2} = \small \frac{7}{10}\small \frac{5}{10}
a = \small \frac{2}{10} = \small \frac{1}{5}

Question 10.
DIG DEEPER!
Write and solve an equation to find the difference between Length A and Length B on the ruler.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 71
Answer:
\(\frac{1}{10}\) × \(\frac{2}{2}\) = \(\frac{2}{20}\)
\(\frac{9}{10}\) × \(\frac{2}{2}\) = \(\frac{18}{20}\)
\(\frac{18}{20}\) – \(\frac{2}{20}\) = \(\frac{16}{20}\)

Question 11.
Modeling Real Life
You want to stack cups in \(\frac{1}{4}\) minute. Your first attempt takes \(\frac{1}{2}\) minute. Your second attempt takes \(\frac{3}{10}\) minute less than your first attempt. Do you meet your goal?
Answer:
\small \frac{1}{2}\small \frac{3}{10}
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 10 is a multiple of 2, so rewrite \small \frac{1}{2} with a denominator of 10
Rewrite \small \frac{1}{2} as \small \frac{1 \times 5}{2 \times 5} = \small \frac{5}{10}
\small \frac{1}{2}\small \frac{3}{10} = \small \frac{5}{10}\small \frac{3}{10}
= \small \frac{2}{10}
= \small \frac{1}{5}
So my second attempt takes \small \frac{1}{5} minute and I did not meet my goal.

Question 12.
Modeling Real Life
You and your friend each have a canvas of the same size. You divide your canvas into 5 sections and paint 3 of them. Your friend divides her canvas into 7 sections and paints 4 of them. Who paints more? How much more?
Answer:
My canvas = \small \frac{3}{5}
My friend canvas = \small \frac{4}{7}
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 7 is a multiple of 5, so rewrite both with a denominator of 7 x 5 = 35
Rewrite \small \frac{3}{5} as \small \frac{3 \times 7}{5 \times 7} = \small \frac{21}{35}
\small \frac{4}{7} as \small \frac{4 \times 5}{7 \times 5} = \small \frac{20}{35}
\small \frac{3}{5}\small \frac{4}{7} = \small \frac{21}{35}\small \frac{20}{35} = \small \frac{1}{35}
I paint \small \frac{1}{35} more canvas than my friend.

Review & Refresh

Evaluate. Check whether your answer is reasonable.

Question 13.
1.7 + 5 + 4.3 = ___
Answer: 11

Question 14.
15.24 + 6.13 – 7 = ___
Answer: 14.37

Lesson 8.6 Add Mixed Numbers

Explore and Grow

Use a model to find the sum.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 72

Construct Arguments
How can you add mixed numbers with unlike denominators without using a model? Explain why your method makes sense.

Think and Grow: Add Mixed Numbers

Key Idea
A proper fraction is a fraction less than 1. An improper fraction is a fraction greater than 1. A mixed number represents the sum of a whole number and a proper fraction. You can use equivalent fractions to add mixed numbers.
Example
Find 1\(\frac{1}{2}\) + 2\(\frac{5}{6}\)
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 73

Show and Grow

Add.

Question 1.
2\(\frac{2}{3}\) + 2\(\frac{1}{6}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
2 \small \frac{2}{3}  = 2 \small \frac{4}{6}
2 \small \frac{1}{6}  = 2 \small \frac{1}{6}
2 \small \frac{4}{6}  + 2 \small \frac{1}{6} = 4 \small \frac{5}{6}

Question 2.
1\(\frac{5}{12}\) + 3\(\frac{3}{4}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
1 \small \frac{5}{12} = 1 \small \frac{5}{12}
3 \small \frac{3}{4} = 3 \small \frac{9}{12}
1 \small \frac{5}{12} + 3 \small \frac{9}{12} = 4 \small \frac{14}{12} = 4 \small \frac{7}{6}

Apply and Grow: Practice

Add.

Question 3.
5\(\frac{4}{9}\) + 1\(\frac{2}{3}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
5 \small \frac{4}{9} = 5 \small \frac{4}{9}
1 \small \frac{2}{3} = 1 \small \frac{6}{9}
5 \small \frac{4}{9} + 1 \small \frac{6}{9} = 6 \small \frac{10}{9} = \small \frac{64}{9} = 7 \small \frac{1}{9}

Question 4.
3\(\frac{1}{2}\) + \(\frac{5}{12}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
3 \small \frac{1}{2} = 3 \small \frac{6}{12}
3 \small \frac{1}{2} + \small \frac{5}{12} = 3 \small \frac{6}{12} + \small \frac{5}{12}
= 3 \small \frac{11}{12}

Question 5.
4\(\frac{5}{6}\) + 3\(\frac{5}{12}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
4 \small \frac{5}{6} = 4 \small \frac{10}{12}
3 \small \frac{5}{12} = 3 \small \frac{5}{12}
4 \small \frac{10}{12} + 3 \small \frac{5}{12} = 7 \small \frac{15}{12} = \small \frac{99}{12} = 8 \small \frac{3}{12}

Question 6.
\(\frac{4}{5}\) + 8\(\frac{7}{20}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
Rewrite \small \frac{4}{5} as \small \frac{16}{20}
\small \frac{4}{5} + 8 \small \frac{7}{20} = \small \frac{16}{20} + 8 \small \frac{7}{20}
= 8 \small \frac{23}{20}

Question 7.
2\(\frac{1}{3}\) + \(\frac{1}{6}\) + 3\(\frac{2}{3}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
2 \small \frac{1}{3} = 2 \small \frac{2}{6}
3 \small \frac{2}{3} = 3 \small \frac{4}{6}
2 \small \frac{1}{3} + \small \frac{1}{6} + 3 \small \frac{2}{3} = 2 \small \frac{2}{6} + \small \frac{1}{6} + 3 \small \frac{4}{6}
= 5 \small \frac{7}{6}

Question 8.
5\(\frac{1}{2}\) + 4\(\frac{3}{4}\) + 6\(\frac{5}{8}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
5 \small \frac{1}{2} = 5 \small \frac{4}{8}
4 \small \frac{3}{4} = 4 \small \frac{6}{8}
5 \small \frac{1}{2} + 4 \small \frac{3}{4}  + 6 \small \frac{5}{8} = 5 \small \frac{4}{8} + 4 \small \frac{6}{8} + 6 \small \frac{5}{8}
= 15 \small \frac{15}{8}

Question 9.
Your science class makes magic milk using 1\(\frac{1}{8}\) cups of watercolor paint and 1\(\frac{3}{4}\) cups of milk. How many cups of magic milk does your class make?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 74
Answer:
Given data,
Watercolor paint = 1 \small \frac{1}{8} cups
Milk = 1 \small \frac{3}{4} cups
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
1 \small \frac{3}{4} = 1 \small \frac{6}{8}
1 \small \frac{1}{8} + 1 \small \frac{6}{8} = 2 \small \frac{7}{8} = \small \frac{23}{8}
So the class makes 2 \small \frac{7}{8} cups of magic milk

Question 10.
Structure
Find 2\(\frac{3}{10}\) + 4\(\frac{2}{5}\) two different ways.
Answer:
Method 1:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
4 \small \frac{2}{5} = 4 \small \frac{4}{10}
2 \small \frac{3}{10} + 4 \small \frac{2}{5} = 2 \small \frac{3}{10} + 4 \small \frac{4}{10}
= 6 \small \frac{7}{10}
Method 2:
Write the mixed numbers as improper fractions with a common denominator and then add.
2 \small \frac{3}{10} = 2 + \small \frac{3}{10} = \small \frac{20}{10} + \small \frac{3}{10} = \small \frac{23}{10}
4 \small \frac{2}{5} = 4 + \small \frac{2}{5} = \small \frac{22}{5} = \small \frac{44}{10}
2 \small \frac{3}{10} + 4 \small \frac{2}{5} = \small \frac{23}{10} + \small \frac{44}{10}
= \small \frac{67}{10}
= 6 \small \frac{7}{10}

Question 11.
DIG DEEPER!
Find the missing numbers.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 75
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
Rewrite 2 \small \frac{3}{4}  as 2 \small \frac{6}{8}
we can write 4 \small \frac{3}{8} as \small \frac{35}{8} = 3 \small \frac{11}{8}
2 \small \frac{6}{8} + 1 \small \frac{5}{8} = 3 \small \frac{11}{8} = 4 \small \frac{3}{8}
So the missing numbers are 1 and 5.
2 \small \frac{6}{8} + 1 \small \frac{5}{8} = 4 \small \frac{3}{8}

Think and Grow: Modeling Real Life

Example
You kayak 1\(\frac{8}{10}\) miles and then take a break. You kayak 1\(\frac{1}{4}\) more miles. How many miles do you kayak altogether?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 76
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 77

Answer:
Big-Ideas-Math-Answers-5th-Grade-Chapter-8-Add-and-Subtract-Fractions-77

Show and Grow

Question 12.
You listen to a song that is 2\(\frac{3}{4}\) minutes long. Then you listen to a song that is 3\(\frac{1}{3}\) minutes long. How many minutes do you spend listening to the two songs altogether?
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator 4 x 3 = 12.
2 \small \frac{3}{4} = 2 \small \frac{9}{12}
3 \small \frac{1}{3} = 3 \small \frac{4}{12}
2 \small \frac{3}{4} + 3 \small \frac{1}{3} = 2 \small \frac{9}{12} + 3 \small \frac{4}{12}
= 5 \small \frac{13}{12} min
So I spend 5 \small \frac{13}{12} min listening to the two songs altogether.

Question 13.
DIG DEEPER!
A beekeeper collects 3\(\frac{3}{4}\) more pounds of honey from Hive 3 than Hive 1. Which hive produces the most honey? Explain.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 79
Answer:
From the given information,
Honey from Hive 3 = Hive 1 honey + 3 \small \frac{3}{4}
= 23 \small \frac{5}{8} + 3 \small \frac{3}{4}
= 23 \small \frac{5}{8} + 3 \small \frac{6}{8}
= 26 \small \frac{11}{8}
Use a common denominator for all the hives
Hive 1 honey = 23 \small \frac{5}{8}
Hive 2 honey = 27 \small \frac{1}{2} = 27 \small \frac{4}{8}
Hive 3 honey = 26 \small \frac{11}{8}
Therefore, Hive 2 produces the most honey.

Add Mixed Numbers Homework & Practice 8.6

Add

Question 1.
6\(\frac{2}{5}\) + 1\(\frac{3}{10}\)
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
6 \small \frac{2}{5}= 6 \small \frac{4}{10}
6 \small \frac{2}{5} + 1 \small \frac{3}{10} = 6 \small \frac{4}{10} + 1 \small \frac{3}{10}
= 7 \small \frac{7}{10}

Question 2.
2\(\frac{2}{3}\) + 5\(\frac{3}{6}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator.
2 \small \frac{2}{3}= 2 \small \frac{4}{6}
2 \small \frac{2}{3} + 5 \small \frac{3}{6} = 2 \small \frac{4}{6} + 5 \small \frac{3}{6}
= 7 \small \frac{7}{6}

Question 3.
\(\frac{1}{4}\) + 3\(\frac{2}{5}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator 4 x 5 = 20
\small \frac{1}{4} = \small \frac{5}{20}
3 \small \frac{2}{5} = 3 \small \frac{8}{20}
\small \frac{1}{4} + 3 \small \frac{2}{5}  = \small \frac{5}{20} + 3 \small \frac{8}{20} = 3 \small \frac{13}{20}

Question 4.
9\(\frac{5}{7}\) + \(\frac{2}{3}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator 7 x 3 = 21
9 \small \frac{5}{7} = 9 \small \frac{15}{21}
\small \frac{2}{3} = \small \frac{14}{21}
9 \small \frac{5}{7} + \small \frac{2}{3} = 9 \small \frac{15}{21} + \small \frac{14}{21} = 9 \small \frac{29}{21}

Question 5.
2\(\frac{1}{2}\) + 1\(\frac{3}{4}\) + \(\frac{1}{2}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator
2 \small \frac{1}{2} = 2 \small \frac{2}{4}
1 \small \frac{3}{4} = 1 \small \frac{3}{4}
\small \frac{1}{2} = \small \frac{2}{4}
2 \small \frac{1}{2} + 1 \small \frac{3}{4}  + \small \frac{1}{2} = 2 \small \frac{2}{4} +1 \small \frac{3}{4} + \small \frac{2}{4} = 3 \small \frac{7}{4}

Question 6.
2\(\frac{2}{3}\) + 4\(\frac{1}{2}\) + 3\(\frac{5}{6}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator
2 \small \frac{2}{3} = 2 \small \frac{4}{6}
4 \small \frac{1}{2} = 4 \small \frac{3}{6}
2 \small \frac{2}{3} + 4 \small \frac{1}{2}  + 3 \small \frac{5}{6} = 2 \small \frac{4}{6} + 4 \small \frac{3}{6}  + 3 \small \frac{5}{6} = 9 \small \frac{12}{6}
2 \small \frac{2}{3} + 4 \small \frac{1}{2}  + 3 \small \frac{5}{6} = 11

Question 7.
A veterinarian spends 3\(\frac{3}{4}\) hours helping cats and 5\(\frac{1}{2}\) hours helping dogs. How many hours does she spend helping cats and dogs altogether?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 80
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator
3 \small \frac{3}{4}
5 \small \frac{1}{2} = 5 \small \frac{2}{4}
3 \small \frac{3}{4} + 5 \small \frac{1}{2} = 3 \small \frac{3}{4} + 5 \small \frac{2}{4}
= 8 \small \frac{5}{4}
So veterinarian spends 8 \small \frac{5}{4} hours helping cats and dogs altogether.

Question 8.
Writing
How is adding mixed numbers with unlike denominators similar to adding fractions with unlike denominators? How is it different?
Answer:
For both adding mixed numbers and adding fractions we have to use a common denominator.
How it is different?
For adding fractions, directly add the fractions with common denominator.
But for adding mixed numbers, add the fractional parts and add the whole number parts.

Question 9.
Logic
Can you add two mixed numbers and get a sum of 2? Explain.
Answer:
Yes, the sum of two mixed numbers can be equal to 2 but only of one of the mixed numbers is negative.

Question 10.
Structure
Shade the model to represent the sum. Then write an equation to represent your model.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 81
Answer: 4 \(\frac{1}{4}\)

Question 11.
Modeling Real Life
An emperor tamarin has a body length of 9\(\frac{5}{10}\) inches and a tail length of 14\(\frac{1}{4}\) inches. How long is the emperor tamarin?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 82
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator 10 x 4 = 40
9 \small \frac{5}{10} = 9 \small \frac{20}{40}
14 \small \frac{1}{4} = 14 \small \frac{10}{40}
9 \small \frac{5}{10} + 14 \small \frac{1}{4} = 9 \small \frac{20}{40} + 14 \small \frac{10}{40}
= 23 \small \frac{30}{40}
So, an emperor tamarin is 23 \small \frac{3}{4} inches long.

Question 12.
DIG DEEPER!
A long jumper jumps 1\(\frac{2}{3}\) feet farther on her third attempt than her second attempt. On which attempt does she jump the farthest? Explain.
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 83
Answer:
Third attempt = 1 \small \frac{2}{3} + 13 \small \frac{3}{4}
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator 3 x 4 = 12
1 \small \frac{2}{3} = 1 \small \frac{8}{12}
13 \small \frac{3}{4} = 13 \small \frac{9}{12}
1 \small \frac{2}{3} + 13 \small \frac{3}{4} = 1 \small \frac{8}{12} + 13 \small \frac{9}{12} = 14 \small \frac{17}{12}
Therefore, she jumps the farthest on her first attempt.

Review & Refresh

Find the product. Check whether your answer is reasonable.

Question 13.
354 × 781
Answer:
The number close to 354 is 350
The number close to 781 is 800.
350 × 800 = 280000
354 × 781 = 276474
Yes, the answer is reasonable.

Question 14.
4,029 × 276
Answer:
The number close to 4029 is 4000
The number close to 276 is 300
4000 × 300 = 1200000
4029 × 276 = 1112004
Yes, the answer is reasonable.

Question 15.
950 × 326
Answer:
The number close to 950 is 1000
The number close to 326 is 300
1000 × 300 = 300000
950 × 326 = 309700
Yes, the answer is reasonable.

Lesson 8.7 Subtract Mixed Numbers

Explore and Grow

Use a model to find the difference
3\(\frac{5}{6}\) – 2\(\frac{1}{3}\)

Construct Arguments
How can you subtract mixed numbers with unlike denominators without using a model? Explain why your method makes sense.

Think and Grow: Subtract Mixed Numbers

You can use equivalent fractions to subtract mixed numbers that have fractional parts with unlike denominators.
Example
Find 3\(\frac{1}{4}\) – 1\(\frac{1}{2}\).

Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 84

Show and Grow

Subtract.

Question 1.
1\(\frac{4}{5}\) – 1\(\frac{3}{10}\)
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
1 \small \frac{4}{5} = 1 \small \frac{8}{10}
1 \small \frac{4}{5} – 1 \small \frac{3}{10} = 1 \small \frac{8}{10} – 1 \small \frac{3}{10} = 0 + \small \frac{5}{10}
= \small \frac{1}{2}

Question 2.
5\(\frac{7}{12}\) – 3\(\frac{5}{6}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
3 \small \frac{5}{6} = 3 \small \frac{10}{12}
5 \small \frac{7}{12} – 3 \small \frac{5}{6} = 5 \small \frac{7}{12} – 3 \small \frac{10}{12}
5 \small \frac{7}{12}  = 4 + \small \frac{12}{12} + \small \frac{7}{12} = 4 \small \frac{19}{12}
5 \small \frac{7}{12} – 3 \small \frac{10}{12} = 4 \small \frac{19}{12} – 3 \small \frac{10}{12}
= 1 \small \frac{9}{12}

Apply and Grow: Practice

Subtract.

Question 3.
8\(\frac{11}{12}\) – 5\(\frac{2}{3}\) = _____
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
5 \small \frac{2}{3} = 5 \small \frac{8}{12}
8 \small \frac{11}{12} – 5 \small \frac{2}{3} = 8 \small \frac{11}{12} – 5 \small \frac{8}{12}
8\(\frac{11}{12}\) – 5\(\frac{2}{3}\) = 3 \small \frac{3}{12} = 3 \small \frac{1}{4}

Question 4.
6 – 4\(\frac{3}{4}\) = ___
Answer:
6 – 4 \small \frac{3}{4} = 6 – \small \frac{19}{4} = \small \frac{5}{4}

Question 5.
21\(\frac{2}{9}\) – 10\(\frac{1}{3}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
10 \small \frac{1}{3} = 10 \small \frac{3}{9}
21 \small \frac{2}{9} = 20 + \small \frac{9}{9} + \small \frac{2}{9} = 20 \small \frac{11}{9}
21 \small \frac{2}{9} – 10 \small \frac{1}{3} = 20 \small \frac{11}{9} – 10 \small \frac{3}{9}
= 10 \small \frac{8}{9}

Question 6.
7\(\frac{1}{2}\) – \(\frac{5}{8}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
7 \small \frac{1}{2} = 7 \small \frac{4}{8} = 6 + \small \frac{8}{8} + \small \frac{4}{8} = 6 \small \frac{12}{8}
7 \small \frac{1}{2}\small \frac{5}{8} = 6 \small \frac{12}{8}\small \frac{5}{8}
= 6 \small \frac{7}{8}

Question 7.
9\(\frac{7}{20}\) – 1\(\frac{3}{5}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
1 \small \frac{3}{5} = 1 \small \frac{12}{20}
9 \small \frac{7}{20} = 8 + \small \frac{20}{20} + \small \frac{7}{20} = 8 \small \frac{27}{20}
9 \small \frac{7}{20} – 1 \small \frac{3}{5} = 8 \small \frac{27}{20} – 1 \small \frac{12}{20}
= 7 \small \frac{15}{20}
= 7 \small \frac{3}{4}

Question 8.
7\(\frac{5}{6}\) – 1\(\frac{1}{6}\) – 2\(\frac{2}{3}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
2 \small \frac{2}{3} = 2 \small \frac{4}{6}
7 \small \frac{5}{6} = 6 + \small \frac{6}{6} + \small \frac{5}{6} = 6 \small \frac{11}{6}
7 \small \frac{5}{6} – 1 \small \frac{1}{6} – 2 \small \frac{2}{3} = 6 \small \frac{11}{6} – 1 \small \frac{1}{6} – 2 \small \frac{4}{6}
= 3 \small \frac{6}{6}
= 3 + 1
7\(\frac{5}{6}\) – 1\(\frac{1}{6}\) – 2\(\frac{2}{3}\) = 4

Question 9.
A volunteer at a food bank buys 3\(\frac{3}{4}\) pounds of cheese to make sandwiches. She uses 2\(\frac{7}{8}\) pounds. How much cheese does she have left?
Big Ideas Math Answers 5th Grade Chapter 8 Add and Subtract Fractions 87
Answer:
Given that,
A volunteer buys 3 \small \frac{3}{4} pounds of cheese
She uses 2 \small \frac{7}{8} pounds
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
3 \small \frac{3}{4} = 3 \small \frac{6}{8} = 2 + \small \frac{8}{8} + \small \frac{6}{8} = 2 \small \frac{14}{8}
Cheese left = 2 \small \frac{14}{8} – 2 \small \frac{7}{8} = 0 + \small \frac{7}{8}
So she left with \small \frac{7}{8} pounds of cheese.

Question 10.
Writing
How is adding mixed numbers the same as subtracting mixed numbers? How is it different?
Answer:
Adding and subtracting mixed numbers is same because we need to use a common denominator for both.
Adding and subtracting mixed numbers is different because
For subtracting mixed numbers, the first fraction should be greater than the second number.

Question 11.
Number Sense
Write the words as an expression. Then evaluate.
Subtract the sum of four and three-fourths and two and five-eighths from eleven and seven-eighths.
Answer:
11 \small \frac{7}{8} – (4 \small \frac{3}{4} + 2 \small \frac{5}{8})
To subtract or add the fractional parts, use a common denominator.
4 \small \frac{3}{4} = 4 \small \frac{6}{8}
4 \small \frac{3}{4} + 2 \small \frac{5}{8} = 4 \small \frac{6}{8} + 2 \small \frac{5}{8} = 6 \small \frac{11}{8}
11 \small \frac{7}{8} – (4 \small \frac{3}{4} + 2 \small \frac{5}{8}) = 11 \small \frac{7}{8} – 6 \small \frac{11}{8}
11 \small \frac{7}{8} = 10 + \small \frac{8}{8} + \small \frac{7}{8} = 10 \small \frac{15}{8}
11 \small \frac{7}{8} – (4 \small \frac{3}{4} + 2 \small \frac{5}{8}) = 10 \small \frac{15}{8} – 6 \small \frac{11}{8}
11 \small \frac{7}{8} – (4 \small \frac{3}{4} + 2 \small \frac{5}{8}) = 4 \small \frac{4}{8} = 4 \small \frac{1}{2}

Question 12.
DIG DEEPER!
Find the missing number.
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 88
Answer:
To subtract the fractional parts, use a common denominator.
3 \small \frac{1}{4} = 3 \small \frac{3}{12}

3 \small \frac{1}{4} – 1 \small \frac{1}{12} = 3 \small \frac{3}{12} – 1 \small \frac{1}{12}
= 2 \small \frac{2}{12}
3 \small \frac{1}{4} – 1 \small \frac{1}{12} = 2 \small \frac{1}{6}
So the missing number is 1.

Think and Grow: Modeling Real Life

Example
A dragonfly is 1\(\frac{1}{2}\) inches long. How much longer is the walking leaf than the dragonfly?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 89
To find how much longer the walking leaf is than the dragonfly, subtract the length of the dragonfly from the length of the walking leaf.
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 90
The walking leaf is __ inches longer than the dragonfly.

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-8-Add-and-Subtract-Fractions-90
The walking leaf is 1 \(\frac{1}{6}\) inches longer than the dragonfly.

Show and Grow

Question 13.
You volunteer 5\(\frac{3}{4}\) hours in 1 month. You spend 3\(\frac{1}{3}\) hours volunteering at an animal shelter. You spend the remaining hours picking up litter on the side of the road. How many hours do you spend picking up litter?
Answer:
Given that,
Volunteering hours in 1 month = 5 \small \frac{3}{4}
Time spent at an animal shelter = 3 \small \frac{1}{3}
Remaining hours in a month are for picking up litter = 5 \small \frac{3}{4} – 3 \small \frac{1}{3}
To subtract the fractional parts, use a common denominator 4 x 3 =12
5 \small \frac{3}{4} = 5 \small \frac{9}{12}
3 \small \frac{1}{3} = 3 \small \frac{4}{12}
Time spend for picking up litter = 5 \small \frac{9}{12} – 3 \small \frac{4}{12} = 2 \small \frac{5}{12}.

Question 14.
A professional basketball player is 6\(\frac{3}{4}\) feet tall. Your friend is 4\(\frac{5}{6}\) feet tall. How much taller is the basketball player than your friend?
Answer:
To find how much taller is the basketball player than the friend, subtract the height of the friend from the height of the basketball player.
Given that,
Basketball player is 6 \small \frac{3}{4} feet tall.
My friend is 4 \small \frac{5}{6} feet tall.
To subtract the fractional parts, use a common denominator 4 x 6 = 24
6 \small \frac{3}{4} = 6 \small \frac{18}{24} = 5 + \small \frac{24}{24} + \small \frac{18}{24} = 5 \small \frac{42}{24}
4 \small \frac{5}{6} = 4 \small \frac{20}{24}
6 \small \frac{3}{4} – 4 \small \frac{5}{6} = 5 \small \frac{42}{24} – 4 \small \frac{20}{24}
= 1 \small \frac{22}{24}
= 1 \small \frac{11}{12}
The basketball player is 1 \small \frac{11}{12} feet taller than my friend.

Question 15.
Your rain gauge has 2\(\frac{1}{2}\) inches of water. After a rainstorm, your rain gauge has 1\(\frac{3}{4}\) more inches of water. It is sunny for a week. Now your rain gauge has 2\(\frac{2}{3}\) inches of water. How many inches of water evaporated?
Answer:
After a rainstorm, water in the rain gauge = 2 \small \frac{1}{2} + 1 \small \frac{3}{4}
To subtract or add the fractional parts, use a common denominator.
2 \small \frac{1}{2} = 2 \small \frac{2}{4}
Water in the rain gauge = 2 \small \frac{1}{2} +1 \small \frac{3}{4} = 2 \small \frac{2}{4} + 1 \small \frac{3}{4} = 3 \small \frac{5}{4} inches
Water evaporated = 3 \small \frac{5}{4} – 2 \small \frac{2}{3}
3 \small \frac{5}{4} = 3 \small \frac{15}{12}
2 \small \frac{2}{3} = 2 \small \frac{8}{12}

3 \small \frac{5}{4} – 2 \small \frac{2}{3} = 3 \small \frac{15}{12} – 2 \small \frac{8}{12} = 1 \small \frac{7}{12}
So, 1 \small \frac{7}{12} inches of water evaporated.

Subtract Mixed Numbers Homework & Practice 8.7

Subtract

Question 1.
9\(\frac{5}{6}\) – 4\(\frac{1}{2}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
4 \small \frac{1}{2} = 4 \small \frac{3}{6}
9 \small \frac{5}{6} – 4 \small \frac{1}{2} = 9 \small \frac{5}{6} – 4 \small \frac{3}{6}
= 5 \small \frac{2}{6}
9\(\frac{5}{6}\) – 4\(\frac{1}{2}\) = 5 \small \frac{2}{6} = 5 \small \frac{1}{3}

Question 2.
3\(\frac{2}{3}\) – \(\frac{1}{9}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
3 \small \frac{2}{3} = 3 \small \frac{6}{9}
3 \small \frac{2}{3}\small \frac{1}{9} = 3 \small \frac{6}{9}\small \frac{1}{9} = 3 \small \frac{5}{9}

Question 3.
6\(\frac{1}{3}\) – 1\(\frac{11}{12}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
6 \small \frac{1}{3} = 6 \small \frac{4}{12} = 5 \small \frac{16}{12}
6 \small \frac{1}{3} – 1 \small \frac{11}{12} = 5 \small \frac{16}{12} – 1 \small \frac{11}{12}
= 4 \small \frac{5}{12}

Question 4.
12\(\frac{5}{6}\) – 7\(\frac{3}{10}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator 6 x 10 = 60
12 \small \frac{5}{6} = 12 \small \frac{50}{60}
7 \small \frac{3}{10} = 7 \small \frac{18}{60}
12 \small \frac{5}{6} – 7 \small \frac{3}{10} = 12 \small \frac{50}{60} – 7 \small \frac{18}{60}
= 5 \small \frac{32}{60}
12\(\frac{5}{6}\) – 7\(\frac{3}{10}\) = 5 \small \frac{32}{60} = 5 \small \frac{8}{15}

Question 5.
5 – 2\(\frac{3}{4}\) = ___
Answer:
5 – 2 \small \frac{3}{4} = 5 – \small \frac{11}{4} = \small \frac{9}{4}

Question 6.
4\(\frac{1}{5}\) – 2\(\frac{1}{4}\) = __
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator 5 x 4 = 20
4 \small \frac{1}{5} = 4 \small \frac{4}{20} = 3 \small \frac{24}{20}
2 \small \frac{1}{4} = 2 \small \frac{5}{20}

4 \small \frac{1}{5} – 2 \small \frac{1}{4} = 3 \small \frac{24}{20} – 2 \small \frac{5}{20} = 1 \small \frac{19}{20}

Subtract.

Question 7.
7\(\frac{5}{8}\) – 1\(\frac{5}{6}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator 8 x 6 = 48
7 \small \frac{5}{8} = 7 \small \frac{30}{48} = 6 \small \frac{78}{48}
1 \small \frac{5}{6} = 1 \small \frac{40}{48}
7 \small \frac{5}{8} – 1 \small \frac{5}{6} = 6 \small \frac{78}{48} – 1 \small \frac{40}{48} = 5 \small \frac{38}{48}
= 5 \small \frac{19}{24}

Question 8.
8\(\frac{1}{9}\) – 6\(\frac{7}{8}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator 8 x 9 = 72
8 \small \frac{1}{9} = 8 \small \frac{8}{72} = 7 \small \frac{80}{72}
6 \small \frac{7}{8} = 6 \small \frac{63}{72}
8 \small \frac{1}{9} – 6 \small \frac{7}{8} = 7 \small \frac{80}{72} – 6 \small \frac{63}{72}
= 1 \small \frac{17}{72}

Question 9.
1\(\frac{6}{7}\) + 5\(\frac{13}{14}\) – 2\(\frac{1}{2}\) = ___
Answer:
To subtract or add the fractional parts, use a common denominator
1 \small \frac{6}{7} = 1 \small \frac{12}{14}
2 \small \frac{1}{2} = 2 \small \frac{7}{14}

1 \small \frac{6}{7} + 5 \small \frac{13}{14} – 2 \small \frac{1}{2} = 1 \small \frac{12}{14} + 5 \small \frac{13}{14} – 2 \small \frac{7}{14}
= 4 \small \frac{18}{14}
= 4 \small \frac{9}{7}

Question 10.
Your friend says the difference of 8 and 3\(\frac{7}{10}\) is 5\(\frac{7}{10}\). Is your friend correct? Explain.
Answer:
8 – 3 \small \frac{7}{10} = 8 – \small \frac{37}{10} = \small \frac{43}{10}

Question 11.
DIG DEEPER!
Use a symbol card to complete the equation. Then find b.
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 91
Answer:
4 \small \frac{1}{4} – 1 \small \frac{17}{20} – b = 1 \small \frac{1}{2}
4 \small \frac{1}{4} = 4 \small \frac{5}{20} = 2 \small \frac{45}{20}
1 \small \frac{1}{2} = 1 \small \frac{10}{20}
b = 4 \small \frac{1}{4} – 1 \small \frac{17}{20} – 1 \small \frac{1}{2}
= 2 \small \frac{45}{20} – 1 \small \frac{17}{20} – 1 \small \frac{10}{20}
= \small \frac{18}{20}
b = \small \frac{18}{20} = \small \frac{9}{10}

Question 12.
Modeling Real Life
The world record for the heaviest train pulled with a human beard is 2\(\frac{3}{4}\) metric tons. The world record for the heaviest train pulled by human teeth is 4\(\frac{1}{5}\) metric tons. How much heavier is the train pulled by teeth than the train pulled with a beard?
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator 4 x 5 = 20
2 \small \frac{3}{4} = 2 \small \frac{15}{20}
4 \small \frac{1}{5} = 4 \small \frac{4}{20} = 3 \small \frac{24}{20}
4 \small \frac{1}{5} – 2 \small \frac{3}{4} = 3 \small \frac{24}{20} – 2 \small \frac{15}{20}
= 1 \small \frac{9}{20}
The train pulled by teeth 1 \small \frac{9}{20} metric tons heavier than the train pulled with a beard.

Question 13.
Modeling Real Life
Your friend’s hair is 50\(\frac{4}{5}\) centimeters long. Your hair is 8\(\frac{9}{10}\) centimeters long. How much longer is your friend’s hair than yours?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 92
Answer:
My friend’s hair = 50 \small \frac{4}{5} cm
My hair = 8 \small \frac{9}{10} cm
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator
50 \small \frac{4}{5} = 50 \small \frac{8}{10} = 49 \small \frac{18}{10}
50 \small \frac{4}{5} – 8 \small \frac{9}{10} = 49 \small \frac{18}{10} – 8 \small \frac{9}{10} = 41 \small \frac{9}{10}
My friend’s hair is 41 \small \frac{9}{10} cms longer than my hair.

Review & Refresh

Question 14.
Round 6.294.
Nearest whole number:
Nearest tenth:
Nearest hundredth:
Answer:
Nearest whole number: 6
Nearest tenth: 60
Nearest hundredth: 600

Question 15.
Round 10.571.
Nearest whole number:
Nearest tenth:
Nearest hundredth:
Answer:
Nearest whole number: 11
Nearest tenth: 110
Nearest hundredth: 1100

Lesson 8.8 Problem Solving: Fractions

Explore and Grow

Make a plan to solve the problem.

At a state park, every \(\frac{1}{10}\) mile of walking trail is marked. Every \(\frac{1}{4}\) mile of a separate biking trail is marked. The table shows the number of mileage markers you and your friend pass while walking and biking on the trails. Who travels farther? How much farther?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 93
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 94

Make Sense of Problems
You decide to walk farther and you pass 4 more mileage markers on the walking trail. Does this change your plan to solve the problem? Explain.
Answer:
At a state park, every \(\frac{1}{10}\) mile of the walking trail is marked.
8 × \(\frac{1}{10}\) = \(\frac{8}{10}\)
Every \(\frac{1}{4}\) mile of a separate biking trail is marked.
9 × \(\frac{1}{4}\) = 2 \(\frac{1}{4}\)
\(\frac{8}{10}\) + \(\frac{4}{10}\) = \(\frac{12}{10}\)
= 1 \(\frac{2}{10}\)
Your friend travels more farther.

Think and Gow: Problem Solving: Fractions

Example
To repair a skate ramp, you cut a piece of wood from a 9\(\frac{1}{2}\)-foot-long board. Then you cut the remaining piece in half. Each half is 3\(\frac{5}{12}\) feet long. How long is the first piece you cut?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 95

Understand the Problem

What do you know?

  • The board is 9 feet long.2first piece you cut.
  • You cut a piece from the board.
  • You cut the rest into two pieces that are each 3\(\frac{5}{12}\) feet long.

What do you need tofind?

  • You need to find the length of the first piece you cut.

Make a Plan

How will you solve?
Write and solve an equation: Subtract the sum of the lengths of the last two pieces you cut from the total length of the board.

Solve
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 96
Let g represent the length of the first piece you cut.
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 97
So, the length of the first piece you cut is __ feet.

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-8-Add-and-Subtract-Fractions-97
So, the length of the first piece you cut is 2 \(\frac{2}{3}\) feet.

Show and Grow

Question 1.
Explain how you can check your answer in the example above.
Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-8-Add-and-Subtract-Fractions-97

Apply and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.

Question 2.
A racehorse eats 38\(\frac{1}{2}\) pounds of food each day. He eats 22\(\frac{3}{4}\) pounds of hay and 7\(\frac{1}{2}\) pounds of grains. How many pounds of his daily diet is not hay or grains?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 98
Answer:
Given,
A racehorse eats 38\(\frac{1}{2}\) pounds of food each day.
He eats 22\(\frac{3}{4}\) pounds of hay and 7\(\frac{1}{2}\) pounds of grains.
38\(\frac{1}{2}\) – 22\(\frac{3}{4}\)
= 15 \(\frac{3}{4}\)
Thus 15 \(\frac{3}{4}\) pounds of his daily diet is not grains.

Question 3.
In 2015, American Pharoah won all of the horse races shown in the table. How many kilometers did American Pharoah run in the races altogether?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 99

Understand the problem. Then make a plan. How will you solve? Explain.
Answer:
Add all the lengths to find how many kilometers did American Pharoah run in the races altogether.
2 + 1 \(\frac{9}{10}\) + 2 \(\frac{2}{5}\)
First add all the whole numbers.
2 + 1 + 2 = 5
Now add the fractions
\(\frac{9}{10}\) + \(\frac{4}{10}\) = 1 \(\frac{3}{10}\)
5 + 1 \(\frac{3}{10}\) = 6 \(\frac{3}{10}\)

Question 4.
You have 2\(\frac{1}{2}\) cups of blueberries. You use 1\(\frac{1}{4}\) cups for pancakes and \(\frac{1}{2}\) cup for muffins. What fraction of a cup of blueberries do you have left?
Answer:
Given,
You have 2\(\frac{1}{2}\) cups of blueberries.
You use 1\(\frac{1}{4}\) cups for pancakes and \(\frac{1}{2}\) cup for muffins.
1\(\frac{1}{4}\) + \(\frac{1}{2}\) = 1 \(\frac{3}{4}\)
2\(\frac{1}{2}\) – 1 \(\frac{3}{4}\) = \(\frac{3}{4}\)

Question 5.
A customer orders 2 pounds of cheese at a deli. The deli worker places some cheese in a bowl and weighs it. The scale shows 1\(\frac{1}{4}\) pounds. The bowl weighs \(\frac{1}{8}\) pound. What fraction of a pound of cheese does the worker need to add to the bowl?
Answer:
Given,
A customer orders 2 pounds of cheese at a deli. The deli worker places some cheese in a bowl and weighs it.
The scale shows 1\(\frac{1}{4}\) pounds. The bowl weighs \(\frac{1}{8}\) pound.
1\(\frac{1}{4}\) + \(\frac{1}{8}\) = 1 \(\frac{3}{8}\)
2 – 1 \(\frac{3}{8}\) = \(\frac{5}{8}\)

Question 6.
Reasoning
Student A is 8\(\frac{1}{2}\) inches shorter than Student B. Student B is 3\(\frac{1}{4}\) inches taller than Student C. Student C is 56\(\frac{3}{8}\) inches tall. How tall is Student A? Student B?
Answer:
Student A is 8\(\frac{1}{2}\) inches shorter than Student B.
Student B is 3\(\frac{1}{4}\) inches taller than Student C.
Student C is 56\(\frac{3}{8}\) inches tall.
Student B is 56\(\frac{3}{8}\) + 3\(\frac{1}{4}\) = 59 \(\frac{5}{8}\)
Thus the height of student B is 59 \(\frac{5}{8}\) inches.
Student A is 59 \(\frac{5}{8}\) – 8\(\frac{1}{2}\) = 51 \(\frac{1}{8}\)
Thus the height of Student A is 51 \(\frac{1}{8}\) inches.

Question 7.
DIG DEEPER!
A police dog spends \(\frac{1}{8}\) of his workday in a police car, \(\frac{3}{4}\) of his workday in public, and the rest of his workday at the police station. What fraction of the dog’s day is spent at the police station?
Answer:
Given,
A police dog spends \(\frac{1}{8}\) of his workday in a police car, \(\frac{3}{4}\) of his workday in public, and the rest of his workday at the police station.
\(\frac{1}{8}\) + \(\frac{3}{4}\) = \(\frac{7}{8}\)
1 – \(\frac{7}{8}\) = \(\frac{1}{8}\)
Thus \(\frac{1}{8}\) fraction of the dog’s day is spent at the police station

Think and Grow: Modeling Real Life

Example
The Magellan spacecraft, launched by the United States, spent 5\(\frac{5}{12}\) years in space before it burned in Venus’s atmosphere. Its first 4 cycles around Venus each lasted \(\frac{2}{3}\) year. The remaining cycles around Venus lasted a total of 1\(\frac{1}{2}\) years. How long did it take to travel from Earth to Venus?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 100
Think: What do you know? What do you need to find? How will you solve?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 101

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-8-Add-and-Subtract-Fractions-101

Show and Grow

Question 8.
You have one of each euro coin shown. Your friend has four euro coins that have a total weight of 21\(\frac{3}{10}\) grams. Whose coins weigh more? How much more?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 102
Answer:
2 \(\frac{3}{10}\) + 5 \(\frac{3}{4}\) + 7 \(\frac{4}{5}\) + 7 \(\frac{1}{2}\)
= 23 \(\frac{7}{20}\)
23 \(\frac{7}{20}\) – 21\(\frac{3}{10}\)
= 2 \(\frac{1}{20}\)

Problem Solving: Fractions Homework & Practice 8.8

Understand the problem. What do you know? What do you need to find? Explain.

Question 1.
Your goal is to exercise for 15 hours this month. You exercise for 3\(\frac{1}{2}\) hours the first week and 3\(\frac{3}{4}\) hours the next week. How many more hours do you need to exercise to reach your goal?
Answer:
Given,
Your goal is to exercise for 15 hours this month.
You exercise for 3\(\frac{1}{2}\) hours the first week and 3\(\frac{3}{4}\) hours the next week.
3\(\frac{1}{2}\) + 3\(\frac{3}{4}\) = 7 \(\frac{1}{4}\)
15 – 7 \(\frac{1}{4}\) = 7 \(\frac{3}{4}\) hours

Question 2.
A taxi driver travels 4\(\frac{5}{8}\) miles to his first stop. He travels 1\(\frac{3}{4}\) miles less to his second stop. How many miles does the taxi driver travel for the two stops?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 103

Understand the problem. Then make a plan. How will you solve? Explain.
Answer:
Given,
A taxi driver travels 4\(\frac{5}{8}\) miles to his first stop. He travels 1\(\frac{3}{4}\) miles less to his second stop.
4\(\frac{5}{8}\) + 1\(\frac{3}{4}\) = 7 \(\frac{1}{4}\)

Question 3.
During the U.S. Civil War, \(\frac{5}{9}\) of the states fought for the Union, and \(\frac{11}{36}\) of the states fought for the Confederacy. The rest of the states were border states. What fraction of the states were border states?
Answer:
Given,
During the U.S. Civil War, \(\frac{5}{9}\) of the states fought for the Union, and \(\frac{11}{36}\) of the states fought for the Confederacy. The rest of the states were border states.
\(\frac{5}{9}\) + \(\frac{11}{36}\) = \(\frac{31}{36}\)

Question 4.
You have 6\(\frac{3}{4}\) pounds of clay. You use 4\(\frac{1}{6}\) pounds to make a medium-sized bowl and 1\(\frac{1}{2}\) pounds to make a small bowl. How many pounds of clay do you have left?
Answer:
Given,
You have 6\(\frac{3}{4}\) pounds of clay.
You use 4\(\frac{1}{6}\) pounds to make a medium-sized bowl and 1\(\frac{1}{2}\) pounds to make a small bowl.
4\(\frac{1}{6}\) + 1\(\frac{1}{2}\) = 5 \(\frac{2}{3}\)
6\(\frac{3}{4}\) – 5 \(\frac{2}{3}\) = 1 \(\frac{1}{12}\)

Question 5.
DIG DEEPER!
Newton and Descartes have a 70-day summer vacation. They go to camp for \(\frac{23}{70}\) of their vacation, and they travel for \(\frac{6}{35}\) of their vacation. They stay home the rest of their vacation. How many weeks do Newton and Descartes spend at home?
Answer:
Given,
Newton and Descartes have a 70-day summer vacation. They go to camp for \(\frac{23}{70}\) of their vacation, and they travel for \(\frac{6}{35}\) of their vacation.
\(\frac{23}{70}\) + \(\frac{6}{35}\) = \(\frac{1}{2}\)
That means Newton and Descartes spend 5 weeks at home.

Question 6.
Modeling Real Life
A farmer plants beets in a square garden with side lengths of 12\(\frac{2}{3}\) feet. He plants squash in a garden with a perimeter of 50\(\frac{1}{2}\) feet. Which garden has a greater perimeter? How much greater is it?
Answer:
Given,
A farmer plants beets in a square garden with side lengths of 12\(\frac{2}{3}\) feet.
The perimeter of the square = 4s
P = 4 × 12\(\frac{2}{3}\)
P = 50 \(\frac{2}{3}\)
50 \(\frac{2}{3}\) is greater than 50\(\frac{1}{2}\)

Question 7.
DIG DEEPER!
Which grade uses more leafy greens daily for its classroom rabbits? How much more does it use?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 104
Answer: 4th Grade uses more leafy greens.

Review & Refresh

Find the product.

Question 1.
0.43 × 1,000 = 430
Answer:
43 x 10-2 x 103 = 43 x 101
= 430

Question 2.
25.8 × 0.1 = 2.58
Answer:
258 x 10-1​​​​​​​ x 10-1= 2.58

Add and Subtract Fractions Performance Task 8

Many historic landmarks are located in Washington, D.C.

Question 1.
Initial construction of the Washington Monument began in 1848. When the height of the monument reached 152 feet, construction halted due to lack of funds. How many feet were added to the height of the monument when construction resumed 23 years later?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 105
Answer:
The initial construction of the Washington Monument began in 1848. When the height of the monument reached 152 feet, construction halted due to lack of funds.
554 \(\frac{3}{4}\) – 152 = 402 \(\frac{3}{4}\)

Question 2.
You visit several historic landmarks. You start at the Capitol Building and walk to the Washington Monument, then the Lincoln Memorial, then the White House, and then back to the Capitol Building.
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 106
a. You walk 3 miles each hour. It takes you \(\frac{1}{2}\) hour to walk from the Lincoln Memorial to the White House. What is the distance from the Lincoln Memorial to the White House? Label the map.

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-8-Add-and-Subtract-Fractions-106
b. What is the total distance you walk visiting the landmarks?
Answer: 1 \(\frac{1}{5}\) + \(\frac{4}{5}\) + 1 \(\frac{4}{5}\) + 1 \(\frac{3}{5}\)
= 5 \(\frac{2}{5}\)

Question 3.
A law in Washington, D.C., restricts a new building’s height to no more than 20 feet taller than the width of the street it faces. You design a building with stories that are each 15 feet tall for a street that is 88\(\frac{2}{3}\) feet wide. What is the greatest number of stories your building can have? How much shorter is your building than the height restriction?
Answer:
Given,
A law in Washington, D.C., restricts a new building’s height to no more than 20 feet taller than the width of the street it faces.
20 × 88\(\frac{2}{3}\) = 1773 \(\frac{1}{3}\)
You design a building with stories that are each 15 feet tall for a street that is 88\(\frac{2}{3}\) feet wide.
15 × 88\(\frac{2}{3}\) = 1330
1773 \(\frac{1}{3}\) – 1330 = 443 \(\frac{1}{3}[/latex

Add and Subtract Fractions Activity

Mixed Number Number
Subtract and Add
Directions:

  1. Each player flips four Mixed Number Cards.
  2. Each player arranges the cards to create two differences that will have the greatest possible sum.
  3. Each player records the two differences, and then adds the differences.
  4. Players repeat Steps 1–3.
  5. Each player adds Sum A and Sum B to find the total. The player with the greatest total wins!

Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 107

Answer:
Big-Ideas-Math-Answer-Key-Grade-5-Chapter-8-Add-and-Subtract-Fractions-107

Add and Subtract Fractions Performance Chapter Practice

8.1 Simplest Form

Write the fraction in simplest form.

Question 1.
[latex]\frac{2}{12}\)
Answer:
Step 1: Find the common factors of 2 and 12.
Factors of 2:    1, 2
Factors of 12:  1, 2, 3, 4, 6, 12
The common factors of 2 and 12 are 1 and 2.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{2}{12} = \frac{2 \div 2}{12 \div 2} = \frac{1}{6}
Because 1 and 6 have no common factors other than 1, \(\frac{2}{12}\) is in simplest form.

Question 2.
\(\frac{15}{30}\)
Answer:
Step 1: Find the common factors of 15 and 30.
Factors of 15:    1, 3, 5, 15
Factors of 30:  1, 2, 3, 5, 6, 10, 15, 30
The common factors of 15 and 30 are 1, 3, 5 and 15.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.

\small \frac{15}{30} = \frac{15 \div 15}{30 \div 15} = \frac{1}{2}
Because 1 and 2 have no common factors other than 1, \(\frac{15}{30}\) is in simplest form.

Question 3.
\(\frac{16}{24}\)
Answer:
Step 1: Find the common factors of 16 and 24.
Factors of 16:    1, 2, 4, 8, 16
Factors of 24:  1, 2, 3, 4, 6, 8, 12, 24
The common factors of 16 and 24 are 1, 2, 4 and 8.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{16}{24} = \frac{16 \div 8}{24 \div 8} = \frac{2}{3}
Because 2 and 3 have no common factors other than 1, \(\frac{16}{24}\) is in simplest form.

Question 4.
\(\frac{18}{36}\)
Answer:
Step 1: Find the common factors of 18 and 36.
Factors of 18:  1, 2, 3, 6, 9, 18
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
The common factors of 18 and 36 are 1, 2, 3, 6, 9 and 18.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.

\small \frac{18}{36} = \frac{18 \div 18}{36 \div 18} = \frac{1}{2}
Because 1 and 2 have no common factors other than 1, \(\frac{18}{36}\) is in simplest form.

Question 5.
\(\frac{8}{32}\)
Answer:
Step 1: Find the common factors of 8 and 32.
Factors of 8:  1, 2, 4, 8
Factors of 32: 1, 2, 4, 8, 16, 32
The common factors of 8 and 32 are 1, 2, 4 and 8.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{8}{32} = \frac{8 \div 8}{32 \div 8} = \frac{1}{4}
Because 1 and 4 have no common factors other than 1, \(\frac{8}{32}\) is in simplest form.

Question 6.
\(\frac{25}{10}\)
Answer:
Step 1: Find the common factors of 25 and 10.
Factors of 25:  1, 5, 25
Factors of 10: 1, 2, 5, 10
The common factors of 25 and 10 are 1 and 5.
Step 2: Write an equivalent fraction by dividing the numerator and the denominator by the greatest of the common factors.
\small \frac{25}{10} = \frac{25 \div 5}{10 \div 5} = \frac{5}{2}
Because 5 and 2 have no common factors other than 1, \(\frac{25}{10}\) is in simplest form.

8.2 Estimate Sums and Differences of Fractions

Estimate the sum or difference.

Question 7.
\(\frac{7}{8}\) – \(\frac{1}{5}\)
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{7}{8} is about
Think: The numerator is about the same as the denominator.
\small \frac{1}{5} is about
Think: The numerator is near to zero.
Step 2: Estimate the difference.
An estimate of \small \frac{7}{8}\small \frac{1}{5}  is 1 – 0 = 1.

Question 8.
\(\frac{5}{6}\) + \(\frac{9}{10}\)
Answer:
Step 1: Estimate each fraction.
\small \frac{5}{6} is between \small \frac{1}{2} and 1, but is closer to 1.
\small \frac{9}{10} is between \small \frac{1}{2} and 1, but is closer to 1.
Step 2: Estimate the sum.
An estimate of \(\frac{5}{6}\) + \(\frac{9}{10}\) = 1 + 1 = 2.

Question 9.
\(\frac{11}{12}\) – \(\frac{89}{100}\)
Answer:
Step 1: Use mental math to estimate each fraction.
\small \frac{11}{12} is about
Think : The numerator is about the same as the denominator.
\small \frac{89}{100} is about
Think : The numerator is closer to denominator.
Step 2: Estimate the difference.
An estimate of \small \frac{11}{12}\small \frac{89}{100}  is 1 – 1 = 0.

Question 10.
Precision
Your friend says \(\frac{7}{8}\) – \(\frac{5}{12}\) is about 0. Find a closer estimate. Explain why your estimate is closer.
Answer:
\(\frac{7}{8}\) – \(\frac{5}{12}\) = \(\frac{11}{24}\)
\(\frac{11}{24}\) = 0.45
The number 0.45 is close to 0.

8.3 Find Common Denominators

Use a common denominator to write an equivalent fraction for each fraction.

Question 11.
\(\frac{1}{4}\) and \(\frac{1}{2}\)
Answer:
Use the product of the denominators: 4 \small \times 2 = 8
Write equivalent fractions with denominators of 8
\small \frac{1}{4} = \frac{1\times 2}{4\times 2} = \frac{2}{8}
\small \frac{1}{2} = \frac{1 \times 4}{2 \times 4} = \frac{4}{8}
Therefore, equivalent fractions are \small \frac{2}{8}  and \small \frac{4}{8}.

Question 12.
\(\frac{2}{3}\) and \(\frac{2}{9}\)
Answer:
Use the product of the denominators : 3 \small \times 9 = 27
Write equivalent fractions with denominators of 27
\small \frac{2}{3} = \frac{2 \times 9}{3 \times 9} = \frac{18}{27}
\small \frac{2}{9} = \frac{2 \times 3}{9 \times 3} = \frac{6}{27}
Therefore, equivalent fractions are \small \frac{18}{27}  and \small \frac{6}{27}.

Question 13.
\(\frac{2}{3}\) and \(\frac{5}{6}\)
Answer:
Use the product of the denominators : 3 \small \times 6 = 18
Write equivalent fractions with denominators of 18
\small \frac{2}{3} = \frac{2 \times 6}{3 \times 6} = \frac{12}{18}
\small \frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18}
Therefore, equivalent fractions are \small \frac{12}{18}  and \small \frac{15}{18}.

Question 14.
\(\frac{4}{5}\) and \(\frac{1}{3}\)
Answer:
Use the product of the denominators : 5 \small \times 3 = 15
Write equivalent fractions with denominators of 15
\small \frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}
\small \frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}
Therefore, equivalent fractions are \small \frac{12}{15}  and \small \frac{5}{15}.

Question 15.
\(\frac{5}{6}\) and \(\frac{8}{9}\)
Answer:
Use the product of the denominators : 6 \small \times 9 = 54
Write equivalent fractions with denominators of 54
\small \frac{5}{6} = \frac{5 \times 9}{6 \times 9} = \frac{45}{54}
\small \frac{8}{9} = \frac{8 \times 6}{9 \times 6} = \frac{48}{54}
Therefore, equivalent fractions are \small \frac{45}{54}  and \small \frac{48}{54}.

Question 16.
\(\frac{4}{5}\) and \(\frac{3}{4}\)
Answer:
Use the product of the denominators : 5 \small \times 4 = 20
Write equivalent fractions with denominators of 20
\small \frac{4}{5} = \frac{4 \times 4}{5 \times 4} = \frac{16}{20}
\small \frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}
Therefore, equivalent fractions are \small \frac{16}{20}  and \small \frac{15}{20}.

8.4 Add Fractions with Unlike Denominators

Add

Question 17.
\(\frac{2}{15}\) + \(\frac{2}{3}\) = __
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 15 is a multiple of 3, so rewrite \(\frac{2}{3}\) with a denominator of 15.
Rewrite \small \frac{2}{3}  as \small \frac{2 \times 5}{3 \times 5} = \small \frac{10}{15}
\small \frac{2}{15} + \small \frac{2}{3} = \small \frac{2}{15} + \small \frac{10}{15}
= \small \frac{12}{15}

Question 18.
\(\frac{3}{4}\) + \(\frac{1}{8}\) = __
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 8 is a multiple of 4, so rewrite \(\frac{3}{4}\) with a denominator of 8.
Rewrite \small \frac{3}{4}  as \small \frac{3 \times 2}{4 \times 2} = \small \frac{6}{8}
\small \frac{3}{4} + \small \frac{1}{8} = \small \frac{6}{8} + \small \frac{1}{8}
= \small \frac{7}{8}

Question 19.
\(\frac{7}{2}\) + \(\frac{1}{6}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 6 is a multiple of 2, so rewrite \(\frac{7}{2}\) with a denominator of 6.
Rewrite \small \frac{7}{2}  as \small \frac{7 \times 3}{2 \times 3} = \small \frac{21}{6}

\small \frac{7}{2} + \small \frac{1}{6} = \small \frac{21}{6} + \small \frac{1}{6}
= \small \frac{22}{6}
= \small \frac{11}{3}

Question 20.
\(\frac{5}{9}\) + \(\frac{1}{2}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 9 is not a multiple of 2, so rewrite each fraction with a denominator of 9 x 2 = 18.
Rewrite \small \frac{5}{9}  as \small \frac{5 \times 2}{9 \times 2} = \small \frac{10}{18}
\small \frac{1}{2} as \small \frac{1 \times 9}{2 \times 9} = \small \frac{9}{18}
\small \frac{5}{9} + \small \frac{1}{2} = \small \frac{10}{18} + \small \frac{9}{18}
= \small \frac{19}{18}

Question 21.
\(\frac{7}{10}\) + \(\frac{5}{6}\) = ___
Answer:

Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 10 is not a multiple of 6, so rewrite each fraction with a denominator of 10 x 6 = 60.

Rewrite \small \frac{7}{10}  as \small \frac{7 \times 6}{10 \times 6} = \small \frac{42}{60}

\small \frac{5}{6} as \small \frac{5 \times 10}{6 \times 10} = \small \frac{50}{60}

\small \frac{7}{10} + \small \frac{5}{6} = \small \frac{42}{60} + \small \frac{50}{60}

= \small \frac{92}{60}

Question 22.
\(\frac{1}{6}\) + \(\frac{11}{12}\) + \(\frac{4}{6}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the sum.
Think: 12 is a multiple of 6, so rewrite each fraction with a denominator of 12.

Rewrite \small \frac{1}{6}  as \small \frac{1 \times 2}{6 \times 2} = \small \frac{2}{12}

\small \frac{4}{6} as \small \frac{4 \times 2}{6 \times 2} = \small \frac{8}{12}

\small \frac{1}{6} + \small \frac{11}{12} + \small \frac{4}{6} = \small \frac{2}{12} + \small \frac{11}{12} + \small \frac{8}{12}

= \small \frac{21}{12}

= \small \frac{7}{4}

8.5 Subtract Fractions with Unlike Denominators

Subtract

Question 23.
\(\frac{1}{4}\) – \(\frac{1}{8}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 8 is a multiple of 4, so rewrite \(\frac{1}{4}\) with a denominator of 8.
Rewrite \small \frac{1}{4}  as \small \frac{1 \times 2}{4 \times 2} = \small \frac{2}{8}
\small \frac{1}{4}\small \frac{1}{8} = \small \frac{2}{8}\small \frac{1}{8}
= \small \frac{1}{8}

Question 24.
\(\frac{3}{2}\) – \(\frac{7}{10}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 10 is a multiple of 2, so rewrite \(\frac{3}{2}\) with a denominator of 10.
Rewrite \small \frac{3}{2}  as \small \frac{3 \times 5}{2 \times 5} = \small \frac{15}{10}
\small \frac{3}{2}\small \frac{7}{10} = \small \frac{15}{10}\small \frac{7}{10}
= \small \frac{8}{10}
= \small \frac{4}{5}

Question 25.
\(\frac{15}{16}\) – \(\frac{7}{8}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 16 is a multiple of 8, so rewrite \(\frac{7}{8}\) with a denominator of 16.
Rewrite \small \frac{7}{8}  as \small \frac{7 \times 2}{8 \times 2} = \small \frac{14}{16}
\small \frac{15}{16}\small \frac{7}{8} = \small \frac{15}{16}\small \frac{14}{16}
= \small \frac{1}{16}

Question 26.
\(\frac{4}{3}\) – \(\frac{2}{5}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 5 is not a multiple of 3, so rewrite each fraction with a denominator of 5 x 3 = 15.
Rewrite \small \frac{4}{3}  as \small \frac{4 \times 5}{3 \times 5} = \small \frac{20}{15}
\small \frac{2}{5} as \small \frac{2 \times 3}{5 \times 3} = \small \frac{6}{15}
\small \frac{4}{3}\small \frac{2}{5} = \small \frac{20}{15}\small \frac{6}{15}
= \small \frac{14}{15}

Question 27.
\(\frac{5}{6}\) – \(\frac{3}{4}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 6 is not a multiple of 4, so rewrite each fraction with a denominator of 6 x 4 = 24.
Rewrite \small \frac{5}{6}  as \small \frac{5 \times 4}{6 \times 4} = \small \frac{20}{24}
\small \frac{3}{4} as \small \frac{3 \times 6}{4 \times 6} = \small \frac{18}{24}
\small \frac{5}{6}\small \frac{3}{4} = \small \frac{20}{24}\small \frac{18}{24}
= \small \frac{2}{24}
= \small \frac{1}{12}

Question 28.
\(\frac{7}{10}\) – \(\frac{2}{5}\) + \(\frac{11}{20}\) = ___
Answer:
Use equivalent fractions to write the fractions with a common denominator. Then find the difference.
Think: 20 is a multiple of 10 and 5, so rewrite each fraction with a denominator of 20.
Rewrite \small \frac{7}{10}  as \small \frac{7 \times 2}{10 \times 2} = \small \frac{14}{20}
\small \frac{2}{5} as \small \frac{2 \times 4}{5 \times 4} = \small \frac{8}{20}

\small \frac{7}{10}\small \frac{2}{5} + \small \frac{11}{20} = \small \frac{14}{20}\small \frac{8}{20} + \small \frac{11}{20}

8.6 Add Mixed Numbers

Add

Question 29.
1\(\frac{3}{4}\) + 7\(\frac{5}{8}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator
1 \small \frac{3}{4} = 1 \small \frac{6}{8}
1 \small \frac{3}{4} + 7 \small \frac{5}{8} = 1 \small \frac{6}{8} + 7 \small \frac{5}{8}
= 8 \small \frac{11}{8}

Question 30.
3\(\frac{3}{10}\) + 2\(\frac{7}{20}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator
3 \small \frac{3}{10} = 3 \small \frac{6}{20}
3 \small \frac{3}{10} + 2 \small \frac{7}{20} = 3 \small \frac{6}{20} + 2 \small \frac{7}{20}
= 5 \small \frac{13}{20}

Question 31.
\(\frac{1}{3}\) + 6\(\frac{4}{5}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator 3 x 5 = 15
\small \frac{1}{3} = \small \frac{5}{15}
6 \small \frac{4}{5} = 6 \small \frac{12}{15}
\small \frac{1}{3} + 6 \small \frac{4}{5}  = \small \frac{5}{15} + 6 \small \frac{12}{15}
= 6 \small \frac{17}{15}

Question 32.
5\(\frac{8}{9}\) + \(\frac{5}{6}\) = _____
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator 9 x 6 = 54
5 \small \frac{8}{9} = 5 \small \frac{48}{54}
\small \frac{5}{6} = \small \frac{45}{54}
5 \small \frac{8}{9} + \small \frac{5}{6}  = 5 \small \frac{48}{54} + \small \frac{45}{54}
= 5 \small \frac{93}{54}

Question 33.
2\(\frac{2}{3}\) + \(\frac{4}{9}\) + 4\(\frac{1}{3}\) = ___
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator
2 \small \frac{2}{3} = 2 \small \frac{6}{9}
4 \small \frac{1}{3} = 4 \small \frac{3}{9}
2 \small \frac{2}{3} + \small \frac{4}{9} + 4 \small \frac{1}{3}   = 2 \small \frac{6}{9} + \small \frac{4}{9} + 4 \small \frac{3}{9}
= 6 \small \frac{13}{9}

Question 34.
5\(\frac{1}{2}\) + 2\(\frac{5}{8}\) + 3\(\frac{3}{4}\) = _____
Answer:
Add the fractional parts and add the whole number parts.
To add the fractional parts, use a common denominator
5 \small \frac{1}{2} = 5 \small \frac{4}{8}
3 \small \frac{3}{4} = 3 \small \frac{6}{8}
5 \small \frac{1}{2} + 2 \small \frac{5}{8} + 3 \small \frac{3}{4}   =5 \small \frac{4}{8} + 2 \small \frac{5}{8} + 3 \small \frac{6}{8}
= 10 \small \frac{15}{8}

8.7 Subtract Mixed Numbers

Subtract

Question 35.
8\(\frac{7}{10}\) – 1\(\frac{2}{5}\) = ___
Answer:

Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
1 \small \frac{2}{5} = 1 \small \frac{4}{10}
8 \small \frac{7}{10} – 1 \small \frac{2}{5}  = 8 \small \frac{7}{10} – 1 \small \frac{4}{10} = 7 \small \frac{3}{10}

Question 36.
15\(\frac{97}{100}\) – 10\(\frac{7}{20}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
10 \small \frac{7}{20} = 10 \small \frac{35}{100}
15 \small \frac{97}{100} – 10 \small \frac{7}{20}  = 15 \small \frac{97}{100} – 10 \small \frac{35}{100} = 5 \small \frac{62}{100}

Question 37.
4 – 3\(\frac{5}{6}\) = ___
Answer:
4 – 3 \small \frac{5}{6} = 4 – \small \frac{23}{6} = \small \frac{1}{6}

Question 38.
5\(\frac{1}{3}\) – 2\(\frac{1}{2}\) = _____
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator 3 x 2 = 6.
5 \small \frac{1}{3} = 5 \small \frac{2}{6} = 4 \small \frac{8}{6}
2 \small \frac{1}{2} = 2 \small \frac{3}{6}
5 \small \frac{1}{3} – 2 \small \frac{1}{2} = 4 \small \frac{8}{6} – 2 \small \frac{3}{6}
= 2 \small \frac{5}{6}

Question 39.
9\(\frac{2}{5}\) – 6\(\frac{3}{4}\) = ___
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator 5 x 4 = 20.
9 \small \frac{2}{5} = 9 \small \frac{8}{20} = 8 \small \frac{28}{20}
6 \small \frac{3}{4} = 6 \small \frac{15}{20}
9 \small \frac{2}{5} – 6 \small \frac{3}{4} = 8 \small \frac{28}{20} – 6 \small \frac{15}{20}
= 2 \small \frac{13}{20}

Question 40.
2\(\frac{3}{8}\) + 7\(\frac{1}{2}\) – 1\(\frac{11}{16}\) = ____
Answer:
To add or subtract the fractional parts, use a common denominator.
2 \small \frac{3}{8} = 2 \small \frac{6}{16}
7 \small \frac{1}{2} = 7 \small \frac{8}{16}
2 \small \frac{3}{8} + 7 \small \frac{1}{2}  – 1 \small \frac{11}{16} = 2 \small \frac{6}{16} + 7 \small \frac{8}{16} – 1 \small \frac{11}{16}
= 8 \small \frac{3}{16}

Question 41.
Modeling Real Life
A family adopts a puppy that weighs 7\(\frac{7}{8}\) pounds. They take him to the vet 2 weeks later, and he weighs 12\(\frac{3}{16}\) pounds. How much weight did the puppy gain?
Answer:
Subtract the fractional parts and subtract the whole number parts.
To subtract the fractional parts, use a common denominator.
7 \small \frac{7}{8} = 7 \small \frac{14}{16}
12 \small \frac{3}{16} – 7 \small \frac{7}{8} = 12 \small \frac{3}{16}  – 7 \small \frac{14}{16}
= 11 \small \frac{19}{16} – 7 \small \frac{14}{16}
= 4 \small \frac{5}{16}
So the puppy gains 4 \small \frac{5}{16} pounds weight.

8.8 Problem Solving: Fractions

Question 42.
A radio station plays three commercials between two songs. The commercials play for 2 minutes altogether. The first commercial is \(\frac{1}{2}\) minute, and the second commercial is 1\(\frac{1}{4}\) minutes. How long is the third commercial?
Answer:
Given,
A radio station plays three commercials between two songs.
The commercials play for 2 minutes altogether. The first commercial is \(\frac{1}{2}\) minute, and the second commercial is 1\(\frac{1}{4}\) minutes.
\(\frac{1}{2}\) + 1\(\frac{1}{4}\) = 1 \(\frac{3}{4}\)
2 – 1 \(\frac{3}{4}\) = \(\frac{1}{4}\)

Question 43.
Your friend plants a tree seedling on Earth Day that is 1\(\frac{1}{3}\) feet tall. In 1 year, the tree grows 1\(\frac{5}{6}\) feet. After 2 years, the tree is 4\(\frac{11}{12}\) feet tall. How much did the tree grow in the second year?
Big Ideas Math Answer Key Grade 5 Chapter 8 Add and Subtract Fractions 109
Answer:
Given,
Your friend plants a tree seedling on Earth Day that is 1\(\frac{1}{3}\) feet tall.
In 1 year, the tree grows 1\(\frac{5}{6}\) feet. After 2 years, the tree is 4\(\frac{11}{12}\) feet tall.
4\(\frac{11}{12}\) – 1\(\frac{5}{6}\)  = 3 \(\frac{1}{12}\)
3 \(\frac{1}{12}\) – 1\(\frac{1}{3}\) = 1 \(\frac{3}{4}\)
Thus the growth of the tree in the second year is 1 \(\frac{3}{4}\) feet.

Conclusion:

Enhance your math skills and performance skills by referring to our Big Ideas Math Grade 5 Answer Key for Chapter 8 Add and Subtract Fractions. Make use of the links to complete your homework and assignments. Practice the given questions number of times to score the highest marks in the exams. Also, keep in touch with our site to get the Solution Key for all Big Ideas Math Grade 5 Chapters from 1 to 14.