Big Ideas Math

Are you looking everywhere to learn about Big Ideas Math 2nd Grade 4th Chapter Fluently Add within 100 Answers? The answer key included topics like related addition questions. Those who are preparing for the BIM Grade 2 Chapter 4 will get the answer key which is extremely useful. You can improve the problem-solving skills by solving all the questions from Big Ideas Math Answers Grade 2 Chapter 4 Fluently Add within 100.

Big Ideas Math Book 2nd Grade Answer Key Chapter 4 Fluently Add within 100

The list of topics in the 2nd Grade 4th Chapter Fluently Add within 100 are given below. Have a look at the answers for topics like Fluently Add within 100 Vocabulary, Use Partial Sums to Add, More Partial Sums, Regroup to Add, Add Two-Digit Numbers, Practice Adding Two-Digit Numbers, Add Up to 3 Two-Digit Numbers, and Add Up to 3 Two-Digit Numbers. after solving this chapter, you will be able to solve addition problems.

Students must have the best preparation strategy to score good marks in the exam. Solve all the problems by checking the Big Ideas Math Answers Grade 2 Chapter 4 Fluently Add within 100. Download Big Ideas Math Book Grade 2 Chapter 4 Fluently Add within 100 PDF for free of cost and practice the problems.

Vocabulary

• Fluently Add within 100 Vocabulary

Lesson 1 Use Partial Sums to Add

• Lesson 4.1 Use Partial Sums to Add
• Use Partial Sums to Add Homework & Practice 4.1

Lesson 2 More Partial Sums

• Lesson 4.2 More Partial Sums
• More Partial Sums Homework & Practice 4.2

Lesson 3 Regroup to Add

• Lesson 4.3 Regroup to Add
• Regroup to Add Homework & Practice 4.3

Lesson 4 Add Two-Digit Numbers

• Lesson 4.4 Add Two-Digit Numbers
• Add Two-Digit Numbers Homework & Practice 4.4

Lesson 5 Practice Adding Two-Digit Numbers

• Lesson 4.5 Practice Adding Two-Digit Numbers
• Practice Adding Two-Digit Numbers Homework & Practice 4.5

Lesson 6 Add Up to 3 Two-Digit Numbers

• Lesson 4.6 Add Up to 3 Two-Digit Numbers
• Add Up to 3 Two-Digit Numbers Homework & Practice 4.6

Lesson 7 Add Up to 3 Two-Digit Numbers

• Lesson 4.7 Add Up to 3 Two-Digit Numbers
• More Problem Solving: Addition Homework & Practice 4.7

Performance Task

• Fluently Add within 100 Performance Task 4

Activity

• Fluently Add within 100 Activity

Chapter Practice

• Fluently Add within 100 Chapter Practice

Cumulative Practice

• Fluently Add within 100 Cumulative Practice 1-4

Fluently Add within 100 Vocabulary

Organize It
Use the review words to complete the graphic organizer.

Answer:
In the above image, the group of rows and columns is known as an array.
The horizontal dots are known as rows.
The vertical dots are known as columns.

Explanation:

Array:
An array means arranging the group rows and columns in a group. Mostly the data will be the same such as integers or strings and these are used to store the collected data. In the above image, the group of rows and columns is known as an array.
Column:
A column is a group of values that are represented vertically and will be in multiple rows and they run from top to bottom.
Row:
A row is a group of values that are represented horizontally which will be lying side by side in a horizontal line. These rows usually arranged in a straight line.

Define it

Use your vocabulary cards to match.

Question 1.
partial sums

Question 2.
regroup

Chapter 4 Vocabulary Cards

Lesson 4.1 Use Partial Sums to Add

Explore and Grow

Model the problem. Make a quick sketch to show how you solved.

Answer:
The sum f 32 + 27 is 59.

Explanation:
To model the given problem 32 + 27 we will model with 50 blocks and 9 blocks, as the sum of 32 + 27 is 59. So we will model with 50 blocks and 9 blocks.

Show and Grow

Question 1.

Answer:
The partial sum of 73 + 12 will be 80 + 5= 85.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 73 + 12 which is 85. So here first we will add tens which are 70 and 10 we will add both the numbers which will be 80 and now we will add ones which are 3 and 2 will be 5. So the total value of 73 + 12 will be 80 + 5= 85.

Question 2.

Answer:
The partial sum of 32 + 24 will be 50 + 6= 56.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with a left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 32 + 24 which is 56. So here first we will add tens which are 30 and 20 we will add both the numbers which will be 50 and now we will add ones which are 2 and 4 will be 6. So the total value of 32 + 24 will be 50 + 6= 56.

Question 3.

Answer:
The partial sum of 37 + 42 will be 70 + 9= 79.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with a left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 37 + 42 which is 79. So here first we will add tens which are 30 and 40 we will add both the numbers which will be 70 and now we will add ones which are 7 and 2 will be 9. So the total value of 37 + 42 will be 70 + 9= 79.

Question 4.

Answer:
The partial sum of 63 + 5 will be 60 + 8= 68.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 63 + 5 which is 68. So here first we will add tens which are 60 and 0 we will add both the numbers which will be 60 and now we will add ones which are 3 and 5 will be 8. So the total value of 63 + 5 will be 60 + 8= 68.

Apply and Grow: Practice

Question 5.

Answer:
The partial sum of 16 + 72 will be 80 + 8= 88.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 16 + 72 which is 88. So here first we will add tens which are 70 and 10 we will add both the numbers which will be 80 and now we will add ones which are 6 and 2 will be 8. So the total value of 16 + 72 will be 80 + 8= 88.

Question 6.

Answer:
The partial sum of 33 + 43 will be 70 + 6= 76.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 33 + 43 which is 76. So here first we will add tens which are 30 and 40 we will add both the numbers which will be 70 and now we will add ones which are 3 and 3 will be 6. So the total value of 33 + 43 will be 70 + 6= 76.

Question 7.

Answer:
The partial sum of 91 + 7 will be 90 + 8= 98.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 91 + 7 which is 98. So here first we will add tens which are 90 and 0 we will add both the numbers which will be 90 and now we will add ones which are 1 and 7 will be 8. So the total value of 91 + 7 will be 90 + 8= 98.

Question 8.

Answer:
The partial sum of 25 + 64 will be 80 + 9= 89.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 25 + 64 which is 89. So here first we will add tens which are 20 and 60 we will add both the numbers which will be 80 and now we will add ones which are 5 and 4 will be 9. So the total value of 25 + 64 will be 80 + 9= 89.

Question 9.
DIG DEEPER!
Find the missing digits.

Answer:
The missing digits are 2, 4, 2.

Explanation:

Here, the missing digits are 2, 4, 2. We had found those values by subtracting the given sum and the addend. So we can get the missing value. In the first image, we can see that the addition of two addends. So here we can see that one number is missing in the second addend. So here we will subtract the addend with the sum which is
46 – 14 and the result will be 32. So we have found out the missing digit which is 2. And we can see in the second addition that one number is missing in the second addend. So here we will subtract the addend with the sum which is 69 – 45 and the result will be 24. So we have found out the missing digit which is 4. And we can see in the third addition that one number is missing in the second addend. So here we will subtract the addend with the sum which is 61 – 87 and the result will be 26. So we have found out the missing digit which is 2.

Think and Grow: Modeling Real Life

You read 34 pages one day and 23 the next day. How many pages do you read in all?

Answer:
The number of pages that were read is 57 pages.

Explanation:

Given that 34 pages were read on one day and the other 23 pages were read on the next day. So to find the number of pages that were read, we will add the number of pages which was read on both days. So the number of pages that were read is 34 + 23 which is 57. And the number of pages that were read is 57 pages.

Show and Grow

Question 10.
You have 57 common trading cards and 11 rare trading cards. How many trading cards do you have in all?

Answer:
The total number of trading cards do we have in all is 57 + 11= 68 cards.

Explanation:
The number of common trading cards is 57 and there are 11 rare trading cards. And the number of trading cards do we have in all is 57 + 11= 68 cards. So the total number of trading cards do we have in all is 57 + 11= 68 cards.

Question 11.
A florist has 7 roses, 6 daisies, and 15 tulips. How many flowers are there in all?

Answer:
The total number of flowers are there in all is 28 flowers.

Explanation:
As the florist has 7 roses, 6 daisies, and 15 tulips. So the total number of flowers will be 7 + 6 + 15= 28 flowers. And the total number of flowers are there in all is 28 flowers.

Use Partial Sums to Add Homework & Practice 4.1

Question 1.

Answer:
The partial sum of 55 + 14 will be 60 + 9= 69.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 55 + 14 which is 69. So here first we will add tens which are 50 and 10 we will add both the numbers which will be 60 and now we will add ones which are 5 and 4 will be 9. So the total value of 55 + 14 will be 60 + 9= 69.

Question 2.

Answer:
The partial sum of 62 + 13 will be 70 + 5= 75.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 62 + 13 which is 75. So here first we will add tens which are 60 and 10 we will add both the numbers which will be 70 and now we will add ones which are 2 and 3 will be 5. So the total value of 62 + 13 will be 70 + 5= 75.

Question 3.

Answer:
The partial sum of 71 + 8 will be 70 + 9= 79.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 71 + 8 which is 79. So here first we will add tens which are 70 and 0 we will add both the numbers which will be 70 and now we will add ones which are 1 and 8 will be 9. So the total value of 71 + 8 will be 70 + 9= 79.

Question 4.

Answer:
The partial sum of 22 + 26 will be 40 + 8= 48.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 22 + 26 which is 48. So here first we will add tens which are 20 and 20 we will add both the numbers which will be 40 and now we will add ones which are 2 and 6 will be 8. So the total value of 22 + 26 will be 40 + 8= 48.

Question 5.
DIG DEEPER!
Find the missing digits. Then find the sum.

Answer:
The missing digits are 2, 5, and the sum of the given values is 39.

Explanation:

In the above image, we can see the missing values and we can see the tens and ones. So the missing values will be 2 in the tens place and 5 in the ones place. As the sum is 20 + 10 and 4 + 5, so the missing digits will be 2 and 5. Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 24 + 15 which is 39. So here first we will add tens which are 20 and 10 we will add both the numbers which will be 30 and now we will add ones which are 4 and 5 will be 9. So the total value of 24 + 15 will be 30 + 9= 39. So the sum of the given values will be 30 + 9= 39

Question 6.
Modeling Real Life
You have 45 balloons. Your friend has 31. How many balloons do you and your friend have in all?

Answer:
The total number of balloons do my friend and mine will be 76 balloons.

Explanation:
As I have 45 balloons and my friend has 31 balloons and the total number of balloons will be 45+31= 76 balloons. So the total number of balloons do my friend and mine will be 76 balloons.

Question 7.
Modeling Real Life
You have 8 toy trains, 4 bouncy balls, and 36 toy soldiers. How many toys do you have in all?

Answer:
The total number of toys do we all have in all will be 48 toys.

Explanation:
As there are 8 toy trains and 4 bouncy balls and 36 toy soldiers. So the number of toys will be 8 + 4 + 36= 48 toys. So the total number of toys do we all have in all will be 48 toys.

Review & Refresh

Question 8.
16 + 10 = ___

Answer:
The addition of 16 and 10 is 26.

Explanation:
On adding 16 and 10 we will get the result as 26.

Question 9.
45 – 10 = ___

Answer:
The difference of 45 – 10 is 35.

Explanation:
On subtracting 10 with 45 we will get the result as 35.

Question 10.
50 – 10 = ___

Answer:
The difference of 50 – 10 is 40.

Explanation:
On subtracting 10 with 50 we will get the result as 40.

Question 11.
63 + 10 = ___

Answer:
The sum of 63 + 10 is 73.

Explanation:
On adding 63 and 10 we will get the result as 73.

Lesson 4.2 More Partial Sums

Make a quick sketch to find 38 + 19.

Answer:
The partial sum of 38 + 19 will be 40 +17= 57.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 38 + 19 which is 57. So here first we will add tens which are 30 and 10 we will add both the numbers which will be 40 and now we will add ones which are 8 and 9 will be 17. So the total value of 38 + 19 will be 40 + 17= 57.

Show and Grow

Question 1.
25 + 19 = ?

Answer:
The partial sum of 25 + 19 will be 30 +14= 44.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 25 + 19 which is 44. So here first we will add tens which are 20 and 10 we will add both the numbers which will be 30 and now we will add ones which are 5 and 9 will be 14. So the total value of 25 + 19 will be 30 + 14= 44.

Question 2.
48 + 33 = ?

Answer:
The partial sum of 48 + 33 will be 70 +11= 81.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 48 + 33 which is 81. So here first we will add tens which are 40 and 30 we will add both the numbers which will be 70 and now we will add ones which are 8 and 3 it will be 11. So the total value of 48 + 33 will be 70 + 11= 81.

Question 3.
57 + 35 = ?

Answer:
The partial sum of 57 + 35 will be 80 +12= 92.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 57 + 35 which is 92. So here first we will add tens which are 50 and 30 we will add both the numbers which will be 80 and now we will add ones which are 7 and 5 it will be 12. So the total value of 57 + 35 will be 80 + 12= 92.

Question 4.
34 + 28 = ?

Answer:
The partial sum of 34 + 28 will be 50 +12= 62.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 34 + 28 which is 62. So here first we will add tens which are 30 and 20 we will add both the numbers which will be 50 and now we will add ones which are 4 and 8 it will be 12. So the total value of 34 + 28 will be 50 + 12= 62.

Question 5.
15 + 76 = ?

Answer:
The partial sum of 15 + 76 will be 80 + 11= 91.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 15 + 76 which is 91. So here first we will add tens which are 10 and 70 we will add both the numbers which will be 80 and now we will add ones which are 5 and 6 it will be 11. So the total value of 15 + 76 will be 80 + 11= 91.

Question 6.
29 + 62 = ?

Answer:
The partial sum of 29 + 62 will be 80 +11= 91.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 29 + 62 which is 91. So here first we will add tens which are 20 and 60 we will add both the numbers which will be 80 and now we will add ones which are 9 and 2 it will be 11. So the total value of 29 + 62 will be 80 + 11= 91.

Question 7.
Number Sense
Which choices are equal to 35 + 27?

Answer:
The choices which are equal to 35 + 27 is 30 + 20 + 5 + 7 and 50 + 12 and 62.

Explanation:
The choices which are equal to 35 + 27 is 30 + 20 + 5 + 7 and 50 + 12 and 62.
As 35 + 27 was divided into tens and ones so the result will be 30 + 20 + 5 + 7 and after adding tens separately and ones separately we will get 50 + 12 by adding both the numbers we will get the result as 62. So the choices which are equal to 35 + 27 is 30 + 20 + 5 + 7 and 50 + 12 and 62.

Question 8.
A giraffe eats 37 pounds of food in the morning and 38 pounds of food in the afternoon. How many pounds of food does the giraffe eat in all?

Answer:
The total number of pounds did the giraffe ate in all is 75 pounds.

Explanation:

Given that the Giraffe eats 37 pounds of food in the morning and 38 pounds of food in the afternoon. So to find the total number of pounds did the giraffe ate in all we will add the food that the giraffe had in the morning and the evening. So here we will perform a partial sum which is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 37 + 38 which is 75. So here first we will add tens which are
30 and 30 we will add both the numbers which will be 60 and now we will add ones which are 7 and 8 it will be 15. So the total value of 37 + 38 will be 60 + 15= 75. So the total number of pounds did the giraffe ate in all is 75 pounds.

Think and Grow: Modeling Real Life

You find 29 items on a scavenger hunt. Your friend finds 17 more than you. How many items does your friend find?

Answer:
The number of items does my friend finds is 46 items.

Explanation:

Given that there are 29 items on a scavenger hunt and my friend finds 17 more than me. So to find how many items my friend finds in the hunt, we will add those items which are 29 + 17= 46 items. So the number of items does my friend finds is 46 items.

Show and Grow

Question 9.
Your friend climbs 48 stairs. You climb 36 more than your friend. How many stairs do you climb?

Answer:
The number of stairs does I climb is 48 + 36= 84 stairs.

Explanation:
Given that my friend climbs 48 stairs and I climb 36 stairs more than my friend. So to find how many stairs did I climb we will add the stairs that my friend climbed and the stairs that I have climbed more stairs. So the number of stairs does I climb is 48 + 36= 84 stairs.

Question 10.
You write 13 fewer words than your friend. You write 39 words. How many words does your friend write?

Answer:
The number of words written by my friend is 26 words.

Explanation:
Given that the number of words written by me is 39 words and my friend wrote 13 fewer words than me so to find that how many words my friend wrote we will subtract the number of words written by me and the fewer words that my friend has written. So the number of words written by my friend is 39 – 13= 26 words.

More Partial Sums Homework & Practice 4.2

Question 1.
27 + 46 = ?

Answer:
The partial sum of 27 + 46 will be 60 +13= 73.

Explanation:

Here we will perform partial sum, the partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 27 + 46 which is 73. So here first we will add tens which are 20 and 40 we will add both the numbers which will be 60 and now we will add ones which are 7 and 6 it will be 13. So the total value of 27 + 46 will be 60 + 13= 73.

Question 2.
54 + 28 = ?

Answer:
The partial sum of 54 + 28 will be 70 +12= 82.

Explanation:

Here we will perform partial sum, the partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 54 + 28 which is 82. So here first we will add tens which are 50 and 20 we will add both the numbers which will be 70 and now we will add ones which are 4 and 8 it will be 12. So the total value of 54 + 28 will be 70 + 12= 82.

Question 3.
18 + 72 = ?

Answer:
The partial sum of 18 + 72 will be 80 +10= 90.

Explanation:

Here we will perform partial sum, the partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 18 + 72 which is 90. So here first we will add tens which are 10 and 70 we will add both the numbers which will be 90 and now we will add ones which are 8 and 2 it will be 10. So the total value of 18 + 72 will be 80 + 10= 90.

Question 4.
DIG DEEPER!
Write the missing digits.

Answer:
The missing digits are 7,3,1 and 3,7,8.

Explanation:

As in the above image, we can see that some of the digits are missing. So to find the missing digits we will parallel check with the result. As we can see that 5 in ones place and 8 in the addend, so by adding 7 we can get the value as 15. So we have got the digit 5 in ones place of the result. Now as we know the value of the addend, we will subtract the addend with the result, so that we can get the other addend. So 75 – 37= 38. So the other missing digit is 3. And the missing value in ones place is 1. Now we will repeat the same process to find the other values. As we can see that 6 in ones place and 9 in the addend, so by adding 7 we can get the value as 16. So we have got the digit 7 in ones place of the result. Now as we know the value of the addend, we will subtract the addend with the result, so that we can get the other addend. So 96 – 57= 39. So the other missing digit is 3. And the missing value in tens place is 8.

Question 5.
You recycle 42 cans and 29 jars. How many items do you recycle in all?

Answer:
The total number of items recycled is 71 items.

Explanation:
Given that 42 cans and 29 jars are recycled and to know that how many items are recycled we will perform the addition, we will add both the cans and jars that are recycled. So 42 cans + 29 jars= 71 items. So the total number of items recycled is 71 items.

Question 6.
Modeling Real Life
Newton plants 63 seeds. Descartes plants 18 more than Newton. How many seeds does Descartes plant?

Answer:
Descartes has planted 81 seeds.

Explanation:
Given that Newton has planted 63 seeds and Descartes has planted 18 more trees than the Newton. That means Descartes has planted 63 + 18 which is 81 seeds. So Descartes has planted 81 seeds.

Question 7.

Answer:
The most favorite instrument is the Drum.

Explanation:
As each emoji represents one student,
so the Drum has six emojis which means 6 × 1= 6 students and
the Triangle has four emojis which means 4 × 1= 4 students.
So the most favorite instrument is the drum as the drum has more number of students which is 6.

Lesson 4.3 Regroup to Add

Explore and Grow

Model the problem. Make a quick sketch to show how you solved.

Show and Grow

Question 1.

Answer:
The addition of regrouping of 46 + 26 is 72.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 46 and 26. So by regrouping, we will carry forward one and the sum of 46 + 26 is 72.

Apply and Grow: Practice

Question 2.

Answer:
The addition of regrouping of 15 + 37 is 52.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 15 and 37. So by regrouping, we will carry forward one and the sum of 15 + 37 is 52.

Question 3.

Answer:
The addition of regrouping of 39 + 32 is 71.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 39 and 32. So by regrouping, we will carry forward one and the sum of 39 + 32 is 71.

Question 4.
DIG DEEPER!
When do you need to regroup to add two numbers?
__________________________________
__________________________________
__________________________________
__________________________________

Answer:
We need to regroup for the addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers.

Think and Grow: Modeling Real Life

You must spell 60 words correctly to win a spelling game. You spell 19 words correctly in Round 1 and36 in Round 2. Do you win?

Answer:
As I spell 55 words correctly, so didn’t win the game.

Explanation:

Given that to spell 60 words correctly to win in a spelling game. I spell out 19 words correctly in Round 1 and 36 words in Round 2. So to find that I had won in the spelling game we will perform regrouping addition. We need to regroup for the addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers.
So the regrouping addition of 19 and 36 is 55, as to win the game we should spell 60 words correctly. So I did not win the spelling game.

Show and Grow

Question 5.
You want to do 80 jumping jacks. You do 45 in the morning and 39 in the evening. Do you reach your goal?

Answer:
As the goal is to reach 80 jumping jacks and I have reached 84 jumping jacks, so I reach the goal.

Explanation:
As the goal is to reach 80 jumping jacks and I did 45 jumping jacks in the morning and 39 jumping jacks in the evening. So the total number of jumping jacks is 45 + 39= 84 jumping jacks. As the goal is to reach 80 jumping jacks and I have reached 84 jumping jacks, so I reach the goal.

Question 6.
You raise $38 selling ham sandwiches and $43 selling turkey sandwiches. Your friend raises $72. Who raises more money?

Answer:
I have raised more money.

Explanation:
As my friend raises $72  money and I raise $38 by selling ham sandwiches and $43 by selling turkey sandwiches, so the total money raised by me by selling the sandwiches is 43 + 38= 81. So I have raised more money.

Regroup to Add Homework & Practice 4.3

Question 1.

Answer:
The addition of regrouping of 36 + 16 is 52.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 36 and 16. So by regrouping, we will carry forward one and the sum of 36 + 16 is 52.

Question 2.

Answer:
The addition of regrouping of 47 + 39 is 86.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 47 and 39. So by regrouping, we will carry forward one and the sum of 47 + 39 is 86.

Question 3.
DIG DEEPER!
Do you have to regroup to add?

Answer:
43 + 29= 72  yes, we have to regroup to add.
54 + 32= 86 no, we can add normally without performing regroup.
33 + 64= 97 no, we can add normally without performing regroup.
17 + 25= 42 yes, we have to regroup to add.

Explanation:
43 + 29= 72  yes, we have to regroup to add. Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 43 and 29. So by regrouping, we will carry forward one and the sum of 43 + 29 is 72.
54 + 32= 86 no, we can add normally without performing regroup. As the place value column is not greater than nine. So here we cannot perform regrouping of addition.
33 + 64= 97 no, we can add normally without performing regroup. As the place value column is not greater than nine. So here we cannot perform regrouping of addition.
17 + 25= 42 yes, we have to regroup to add. Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 43 and 29. So by regrouping, we will carry forward one and the sum of 43 + 29 is 72.

Question 4.
Modeling Real Life
There are 50 words in a word search. You find 25 words in rows and 18 words in columns. Did you find all of the words?

Answer:
No, I have not found all the 50 words.

Explanation:
As there are 50 words in the word search and I have found 25 words in rows and 18 words in columns, so the total number of words did I had found is 25 + 18 which is 43. So I have not found all the 50 words.

Question 5.
Modeling Real Life
You find 15 white shells and 17 spotted shells. Your friend finds 34 shells. Who finds more shells?

Answer:
My friend finds more shells than me.

Explanation:
As I had found 15 white shells and 17 spotted shells, so the total number of shells did I have found is 15 + 17= 32 shells. And my friend finds 34 shells, so my friend finds more shells than me.

Review & Refresh

Compare

Question 6.

Answer:
We will use the “<” sign which is 34 < 80.

Explanation:
In the above image, we can see that 34 is less than 80. So we can represent with < sign as one value is smaller than another we will use < lesser than sign, so 34 < 80.

Question 7.

Answer:
We will use the “>” sign which is 15 > 8.

Explanation:
In the above image, we can see that 15 is less than 8. So we can represent with > sign as one value is greater than another we will use > greater than sign, so 15 > 8.

Question 8.

Answer:
We will use the “=” sign which is 67 = 67.

Explanation:
In the above image, we can see that 67 and the opposite number also 67 which are equal to each other. So we can represent with = sign as one value is equal to each other so we will use = equal to sign, so 67 = 67.

Lesson 4.4 Add Two-Digit Numbers

Explore and Grow

Make a quick sketch to find 38 +24

Answer:
By regrouping addition of 38 and 24 we will get the result as 62.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 38 and 24. So by regrouping, we will carry forward one and the sum of 38 + 24 is 62.

Show and Grow

Question 1.
69 + 22 = ?

Answer:
By regrouping addition of 69 and 22 we will get the result as 91.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 69 and 22. So by regrouping, we will carry forward one and the sum of 69 + 22 is 91.

Question 2.
25 +37 = ?

Answer:
By regrouping addition of 25 and 37 we will get the result as 62.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 25 and 37. So by regrouping, we will carry forward one and the sum of 25 + 37 is 62.

Question 3.
31 + 26 = ?

Answer:
By regrouping addition of 31 and 26 we will get the result as 57.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 31 and 26. So by regrouping, we will carry forward one and the sum of 31 + 26 is 57.

Question 4.
15 + 38 = ?

Answer:
By regrouping addition of 15 and 38 we will get the result as 53.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 15 and 38. So by regrouping, we will carry forward one and the sum of 15 + 38 is 53.

Question 5.
62 + 13 = ?

Answer:
By regrouping addition of 62 and 13 we will get the result as 75.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 62 and 13. So by regrouping, we will carry forward one and the sum of 62 + 13 is 75.

Question 6.
46 + 49 = ?

Answer:
By regrouping addition of 46 and 49 we will get the result as 95.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 46 and 49. So by regrouping, we will carry forward one and the sum of 46 + 49 is 95.

Apply and Grow: Practice

Question 7.
33 + 39 = ?

Answer:
By regrouping addition of 33 and 39 we will get the result as 72.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 33 and 39. So by regrouping, we will carry forward one and the sum of 33 + 39 is 72.

Question 8.
23 + 71 = ?

By regrouping addition of 23 and 71 we will get the result as 94.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 23 and 71. So by regrouping, we will carry forward one and the sum of 23 + 71 is 94.

Question 9.
17 + 64 = ?

Answer:
By regrouping addition of 17 and 64 we will get the result as 81.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 17 and 64. So by regrouping, we will carry forward one and the sum of 17 + 64 is 81.

Question 10.
54 + 25 = ?

Answer:
By regrouping addition of 54 and 25 we will get the result as 79.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 54 and 25. So by regrouping, we will carry forward one and the sum of 54 + 25 is 79.

Question 11.
47 + 39 = ?

Answer:
By regrouping addition of 47 and 39 we will get the result as 86.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 47 and 39. So by regrouping, we will carry forward one and the sum of 47 + 39 is 86.

Question 12.
28 + 26 = ?

Answer:
By regrouping addition of 28 and 26 we will get the result as 54.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 28 and 26. So by regrouping, we will carry forward one and the sum of 28 + 26 is 54.

Question 13.
YOU BE THE TEACHER
Newton finds 26 + 36. Is he correct? Explain.

Answer:
No, Newton is not correct.

Explanation:
No, Newton is not correct. Newton has performed only addition but he did not perform any regroup. So he got the wrong answer. If he perform regrouping, and regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 2 and 36. So by regrouping, we will carry forward one and the sum of 26 + 36 is 62.

Think and Grow: Modeling Real Life

You have 24 gel pens and you buy 36 more. Your friend has 48 and buys 18 more. Who has more gel pens?

Answer:
I have more pens than my friend as 66 is greater than 60.

Explanation:

As I have 24 gel pens and I bought 36 more gel pens so the total number of pens I have bought is 24 + 36= 60 pens and my friend has 48 and buys 18 more, so the total number of pens my friend has bought is 48 + 18= 66 pens. So by comparing we can see that I have more pens than my friend as 66 is greater than 60.

Show and Grow

Question 14.
You have 32 stencils and you buy 16 more. Your friend has 14 and buys 28 more. Who has more stencils?

Answer:
I had more stencils than my friend.

Explanation:
As I have 32 stencils and bought 16 more stencils, so the total number of stencils I had is 32 + 16= 48 stencils. And my friend has 14 stencils and bought 28 stencils, so the total number of stencils my friend had is 14 + 28= 42 stencils. So I had more stencils than my friend.

Question 15.
You have 22 green stars and 17 orange stars. Your friend has 26 blue stars and 12 pink stars. How many stars do you and your friend have in all?

Answer:
The total number of stars I and my friend had is 77 stars.

Explanation:
As I have 22 green stars and 17 orange stars, so the total number of stars I had is 22 + 17= 39 stars. And my friend has 26 blue stars and 12 pink stars, so the total number of stars my friend had is 26 + 12= 38 stars. So the total number of stars I and my friend had is 39 +38= 77 stars.

Add Two-Digit Numbers Homework & Practice 4.4

Question 1.
47 + 36 = ?

Answer:
By regrouping addition of 28 and 26 we will get the result as 54.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 28 and 26. So by regrouping, we will carry forward one and the sum of 28 + 26 is 54.

Question 2.
51 + 28 = ?

Answer:
By regrouping addition of 51 and 28 we will get the result as 79.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 51 and 28. So by regrouping, we will carry forward one and the sum of 51 + 28 is 79.

Question 3.
13 + 79 = ?

Answer:
By regrouping addition of 13 and 79 we will get the result as 92.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 13 and 79. So by regrouping, we will carry forward one and the sum of 13 + 79 is 92.

Question 4.
54 + 42 = ?

Answer:
By regrouping addition of 54 and 42 we will get the result as 96.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 54 and 42. So by regrouping, we will carry forward one and the sum of 54 + 42 is 96.

Question 5.
38 + 23 = ?

Answer:
By regrouping addition of 38 and 23 we will get the result as 61.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 38 and 23. So by regrouping, we will carry forward one and the sum of 38 + 23 is 61.

Question 6.
45 + 44 = ?

Answer:
By regrouping addition of 45 and 44 we will get the result as 89.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 45 and 44. So by regrouping, we will carry forward one and the sum of 45 + 44 is 89.

Question 7.
Writing
Find the sum. Write an addition story to match.

Answer:
By regrouping addition of 26 and 35 we will get the result as 89.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 26 and 35. So by regrouping, we will carry forward one and the sum of 26 + 35 is 89.

Question 8.
Modeling Real Life
You have 35 dimes and you find 17 more. Your friend has 42 and finds 11 more. Who has more dimes?

Answer:
My friend has more dimes than me

Explanation:
Given that I have 35 dimes and then I found 17 more, so the total number of I had is 52 dimes, and my friend has 42 dimes and then finds 11 more so the total number of dimes my friend had is 42 + 11= 53 dimes. So on comparing, my friend has more dimes than me.

Question 9.
Modeling Real Life
Newton plants 14 red flowers and 14 purple flowers. Descartes plants 26 pink flowers and 26 yellow flowers. How many flowers do Newton and Descartes plant in all?

Answer:
The total number of flowers planted by both Newton and Descartes is 80 flowers.

Explanation:
Given that Newton has planted 14 red flowers and 14 purple flowers, so the total number of plants that was planted by Newton is 14 + 14= 28 flowers. And Descartes has planted 26 pink flowers and 26 yellow flowers so the total number of plants that were planted by Descartes is 26 + 26= 52 flowers. So the total number of flowers planted by both Newton and Descartes is 52 + 28= 80 flowers.

Question 10.
Model 56 two ways.

Answer:
The two models of 56 is 5 tens and 6 six and the other model is 4 tens and 16 ones

Explanation:

To model 56 in two ways we will divide the 56 as 5 tens and 6 ones, and in another way, we will divide the 56 as 4 tens and 16 ones.

Lesson 4.5 Practice Adding Two-Digit Numbers

Use any strategy to find 17 + 23.

Compare your strategy to your partner’s strategy. Are they the same or different? Explain.
_________________________________
_________________________________

Answer:
I have used the partial sum procedure and my friend has used the regrouping of addition procedure. So here the answer will be the same but the procedure is different.

Explanation:
To find 17 + 23 I have used partial sum, here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 17 + 23 which is 40. So here first we will add tens which are 10 and 20 we will add both the numbers which will be 30 and now we will add ones which are 7 and 3 it will be 10. So the total value of 17 + 23 will be 30 + 10= 40. And my friend used regrouping of addition, here, regrouping is defined as the process of making and then carrying out the operation like addition with the two-digit numbers or larger than the two-digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 17 and 23. So by regrouping, we will carry forward one and the sum of 17 + 23 is 40. So here the answer will be the same but the procedure is different.

Show and Grow

Question 1.

Answer:
The partial sum of 43 and 17 is 60.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 43 + 17 which is 60. So here first we will add tens which are 40 and 10 we will add both the numbers which will be 50 and now we will add ones which are 7 and 3 it will be 10. So the total value of 43 + 17 will be 50 + 10= 60.

Question 2.

Answer:
The partial sum of 56 and 25 is 81.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 56 + 25 which is 81. So here first we will add tens which are 50 and 20 we will add both the numbers which will be 70 and now we will add ones which are 6 and 5 it will be 11. So the total value of 56 + 25 will be 70 + 11= 81.

Question 3.

Answer:
The partial sum of 19 and 55 is 74.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 19 + 55 which is 74. So here first we will add tens which are 10 and 50 we will add both the numbers which will be 60 and now we will add ones which are 9 and 5 it will be 14. So the total value of 19 + 55 will be 60 + 14= 74.

Question 4.

Answer:
The partial sum of 41 and 52 is 93.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 41 + 52 which is 93. So here first we will add tens which are 40 and 50 we will add both the numbers which will be 90 and now we will add ones which are 1 and 2 it will be 3. So the total value of 41 + 52 will be 90 + 3= 93.

Question 5.

Answer:
The partial sum of 47 and 26 is 73.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 47 + 26 which is 73. So here first we will add tens which are 40 and 20 we will add both the numbers which will be 60 and now we will add ones which are 7 and 6 it will be 13. So the total value of 47 + 26 will be 60 + 13= 73.

Question 6.

Answer:
The partial sum of 33 and 49 is 82.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 33 + 49 which is 82. So here first we will add tens which are 30 and 40 we will add both the numbers which will be 70 and now we will add ones which are 3 and 9 it will be 12. So the total value of 33 + 49 will be 70 + 12= 82.

Question 7.

Answer:
The partial sum of 29 and 22 is 51.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 29 + 22 which is 51. So here first we will add tens which are 20 and 20 we will add both the numbers which will be 40 and now we will add ones which are 9 and 2 it will be 11. So the total value of 29 + 22 will be 40 + 11= 51.

Question 8.

Answer:
The partial sum of 54 and 44 is 98.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 54 + 44 which is 98. So here first we will add tens which are 50 and 40 we will add both the numbers which will be 90 and now we will add ones which are 4 and 4 it will be 8. So the total value of 54 + 44 will be 90 + 8= 98.

Question 9.

Answer:
The partial sum of 36 and 45 is 81.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 36 + 45 which is 81. So here first we will add tens which are 30 and 40 we will add both the numbers which will be 70 and now we will add ones which are 6 and 5 it will be 11. So the total value of 36 + 45 will be 70 + 11= 81.

Apply and Grow: Practice

Question 10.

Answer:
The partial sum of 24 and 16 is 40.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 24 + 16 which is 40. So here first we will add tens which are 20 and 10 we will add both the numbers which will be 30 and now we will add ones which are 4 and 6 it will be 10. So the total value of 30 + 10 will be 30 + 10= 40.

Question 11.

Answer:
The partial sum of 37 and 46 is 83.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 37 + 46 which is 83. So here first we will add tens which are 30 and 40 we will add both the numbers which will be 70 and now we will add ones which are 7 and 6 it will be 13. So the total value of 70 + 13 will be 70 + 13= 83.

Question 12.

Answer:
The partial sum of 18 and 59 is 77.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 18 + 59 which is 77. So here first we will add tens which are 10 and 50 we will add both the numbers which will be 60 and now we will add ones which are 8 and 9 it will be 17. So the total value of 18 + 59 will be 60 + 17= 77.

Question 13.

Answer:
The partial sum of 61 and 34 is 95.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 61 + 34 which is 95. So here first we will add tens which are 60 and 30 we will add both the numbers which will be 90 and now we will add ones which are 1 and 4 it will be 5. So the total value of 61 + 34 will be 90 + 5= 95.

Question 14.

Answer:
The partial sum of 23 and 28 is 51.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 23 + 28 which is 51. So here first we will add tens which are 20 and 20 we will add both the numbers which will be 40 and now we will add ones which are 3 and 8 it will be 11. So the total value of 23 + 28 will be 40 + 11= 51.

Question 15.

Answer:
The partial sum of 42 and 57 is 99.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 42 + 57 which is 99. So here first we will add tens which are 40 and 50 we will add both the numbers which will be 90 and now we will add ones which are 2 and 7 it will be 9. So the total value of 42 + 57 will be 90 + 9= 99.

Question 16.

Answer:
The partial sum of 73 and 17 is 90.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 73 + 17 which is 60. So here first we will add tens which are 70 and 10 we will add both the numbers which will be 80 and now we will add ones which are 7 and 3 it will be 10. So the total value of 73 + 17 will be 80 + 10= 90.

Question 17.

Answer:
The partial sum of 82 and 12 is 94.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 82 + 12 which is 94. So here first we will add tens which are 80 and 10 we will add both the numbers which will be 90 and now we will add ones which are 2 and 2 it will be 10. So the total value of 82 + 12 will be 90 + 4= 94.

Question 18.

Answer:
The partial sum of 15 and 77 is 92.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 15 + 77 which is 92. So here first we will add tens which are 10 and 70 we will add both the numbers which will be 80 and now we will add ones which are 5 and 7 it will be 12. So the total value of 15 + 77 will be 80 + 12= 92.

Question 19.
YOU BE THE TEACHER
Descartes finds 38 + 53. Is he correct? Explain.

Answer:
No, Descartes is not correct.

Explanation:
No, Descartes is not correct. As Descartes just performed the addition and he placed the carry forward in the result. Descartes should not place the carry forward into the result, it should be added to the next addend. So Descartes was not correct.

Think and Grow: Modeling Real Life

Are there more state parks in North Carolina and Kentucky or in Kentucky and South Carolina?

There are more state parks in ___ and ___.

Answer:
There are more state parks in South Carolina and Kentucky.

Explanation:
Given that the state of North Carolina has 29 state parks and 38 state parks in the state of Kentucky, so the total number of state parks in both North Carlina and Kentucky is 29 + 38= 67 state parks. And the number of state parks in the state of South Carolina is 43, so the total number of state parks in both South Carlina and Kentucky is 43 + 38= 81 state parks. So there are more state parks in South Carolina and Kentucky. On comparing 81 is greater than 67.

Show and Grow

Question 20.
Are there more students in first and second grade or in first and third grade?

There are more students in ___ and __ grade.

Explanation:
Given that the state first grade has 37 students and 53 students in the second grade, so the total number of students in both first grade and second grade is 37 + 53= 90 students. And the number of students in the third grade 49, so the total number of students in both first grade and third grade is 37 + 49= 86. So there are more students in first grade and second grade than first grade and third grade.

Practice Adding Two-Digit Numbers Homework & Practice 4.5

Question 1.

Answer:
The partial sum of 63 and 12 is 75.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 63 + 12 which is 75. So here first we will add tens which are 60 and 10 we will add both the numbers which will be 70 and now we will add ones which are 3 and 2 it will be 5. So the total value of 63 + 12 will be 60 + 5= 65.

Question 2.

Answer:
The partial sum of 32 and 58 is 90.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 32 + 58 which is 90. So here first we will add tens which are 30 and 50 we will add both the numbers which will be 80 and now we will add ones which are 2 and 8 it will be 10. So the total value of 32 + 58 will be 32 + 58= 90.

Question 3.

Answer:
The partial sum of 53 and 38 is 91.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 53 + 38 which is 91. So here first we will add tens which are 50 and 30 we will add both the numbers which will be 80 and now we will add ones which are 3 and 8 it will be 11. So the total value of 53 + 38 will be 80 + 11= 91.

Question 4.

Answer:
The partial sum of 13 and 39 is 52.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 13 + 39 which is 52. So here first we will add tens which are 10 and 30 we will add both the numbers which will be 40 and now we will add ones which are 3 and 9 it will be 12. So the total value of 13 + 39 will be 40 + 12= 52.

Question 5.

Answer:
The partial sum of 62 and 18 is 80.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 62 + 18 which is 80. So here first we will add tens which are 60 and 10 we will add both the numbers which will be 70 and now we will add ones which are 2 and 8 it will be 10. So the total value of 62 + 18 will be 70 + 10= 80.

Question 6.

Answer:
The partial sum of 48 and 16 is 64.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 48 + 16 which is 64. So here first we will add tens which are 40 and 10 we will add both the numbers which will be 50 and now we will add ones which are 8 and 6 it will be 14. So the total value of 48 + 16 will be 50 + 14= 64.

Question 7.

Answer:
The partial sum of 11 and 66 is 77.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 11 + 66 which is 77. So here first we will add tens which are 10 and 60 we will add both the numbers which will be 70 and now we will add ones which are 1 and 6 it will be 77. So the total value of 11 + 66 will be 70 + 7= 77.

Question 8.

Answer:
The partial sum of 64 and 21 is 85.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 64 + 21 which is 85. So here first we will add tens which are 60 and 20 we will add both the numbers which will be 80 and now we will add ones which are 4 and 1 it will be 5. So the total value of 64 + 21 will be 80 + 5= 85.

Question 9.

Answer:
The partial sum of 79 and 14 is 93.

Explanation:
Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 79 + 14 which is 93. So here first we will add tens which are 70 and 10 we will add both the numbers which will be 80 and now we will add ones which are 9 and 4 it will be 13. So the total value of 79 + 14 will be 80 + 13= 93.

Question 10.
DIG DEEPER!
Find the missing digits

Answer:
The missing digits are 6, 8, 6, 2, 2, 2.

Explanation:

To find the missing digits, in the first image we can see that the sum is 1 and one of the addends is 5 so to get the sum as 1 we will place 6, so the sum will be 81 and the addend will be 36. And in the second image, we can see that the sum is 97, and the other addend is 5 so let’s take the other addend to be 2 so the missing digit is 2 and to find the other addend we will subtract the 32 with the sum 97 which is 97 – 32= 65. And in the third image, we can see that the addend is 48 and the other addend with digit 4 so by adding the sum will be 72 so to find the other addend we will subtract the addend 48 with the result 72 so the other addend will be 72 – 48= 24.

Question 11.
Modeling Real Life
Do more people attend the show on Monday and Tuesday or on Tuesday and Wednesday?

Answer:
More people attended the show on Tuesday and Wednesday.

Explanation:
Given that the number of people on Monday is 48 people and the number of people on Tuesday is 26 people, so the number of people on Monday and Tuesday is 48 + 26= 74 people. And the number of people on Wednesday is 56 people, so the number of people on Tuesday and Wednesday is 26 + 56= 82 people. So more people attended the show on Tuesday and Wednesday than Monday and Tuesday.

Review & Refresh

Question 12.
Order from shortest to longest.

Answer:
Purple is Shortest,
Red is Longer,
Purple is Longest.

Explanation:
In the above, we can see that the purple crayon is the shortest crayon and red crayon is a longer crayon and the green crayon is the longest crayon.

Lesson 4.6 Add Up to 3 Two-Digit Numbers

Explore and Grow

Add the circled numbers first. Then find the sum of all three numbers.

Answer:
In the first image, the sum of the circled numbers is 60,
In the second image, the sum of the circled numbers is 50,
In the third image, the sum of the circled numbers is 58,

Explanation:
In the first image, we can see that 34 and 26 are circled, so the sum of circled numbers is 34 + 26= 60. And the number which is not circled is 24, so the sum of all three numbers is 60 + 24= 84.
In the second image, we can see that 26 and 24 are circled, so the sum of circled numbers is 24 + 26= 50. And the number which is not circled is 34, so the sum of all three numbers is 50 + 34= 84.
In the third image, we can see that 34 and 24 are circled, so the sum of circled numbers is 34 + 24= 58. And the number which is not circled is 26, so the sum of all three numbers is 58 + 26= 84.

What is the same? What is different?
_________________________
_________________________
_________________________

Answer:
In the above problem, the result is the same and the circled numbers are different and the sum of circles numbers is different.

Show and Grow

Question 1.

Answer:
On adding 16 + 34 + 21 we will get the result as  71.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 6 + 4 which makes 10 and then we will add the remaining digit which is 1, so 10 + 1= 11. Now we will perform the regrouping of addition and the result will be 16 + 34 + 21= 71.

Question 2.

Answer:
On adding 33 + 15 + 17 we will get the result as  65.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 7 + 3 which makes 10 and then we will add the remaining digit which is 5, so 10 + 5= 15. Now we will perform the regrouping of addition and the result will be 33 + 15 + 17= 65.

Question 3.

Answer:
On adding 31 + 12 + 24 we will get the result as  67.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, so first, we will add 31 + 12 and the result is 43. Now we will perform the addition for all three numbers and the result will be 31 + 12 + 24= 67.

Question 4.

Answer:
On adding 25 + 15 + 13 we will get the result as  53.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 5 + 5 which makes 10 and then we will add the remaining digit which is 3, so 10 + 3= 13. Now we will perform the regrouping of addition and the result will be 25 + 15 + 13= 53.

Question 5.

Answer:
On adding 29 + 22 + 23 we will get the result as 74.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 11 by adding which helps us to perform the addition easily. So we will add the right side digits 9 + 2 which makes 11 and then we will add the remaining digit which is 3, so 11 + 3= 14. Now we will perform the regrouping of addition and the result will be 29 + 22 + 23= 74.

Question 6.

Answer:
On adding 19 + 32 + 11 we will get the result as 62.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 9 + 1 which makes 10 and then we will add the remaining digit which is 2, so 10 + 2= 12. Now we will perform the regrouping of addition and the result will be 19 + 32 + 11= 62.

Question 7.

Answer:
On adding 18 + 28 + 42 we will get the result as  88.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 8 + 2 which makes 10 and then we will add the remaining digit which is 8, so 10 + 8= 18. Now we will perform the regrouping of addition and the result will be 18 + 28 + 42= 88.

Question 8.

Answer:
On adding 53 + 13 + 19 we will get the result as  85.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 12 by adding which helps us to perform the addition easily. So we will add the right side digits 9 + 3 which makes 12 and then we will add the remaining digit which is 3, so 12 + 3= 15. Now we will perform the regrouping of addition and the result will be 53 + 13 + 19= 85.

Question 9.

Answer:
On adding 27 + 27 + 25 we will get the result as 79.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 12 by adding which helps us to perform the addition easily. So we will add the right side digits 7 + 5 which makes 12 and then we will add the remaining digit which is 7, so 12 + 7= 19. Now we will perform the regrouping of addition and the result will be 27 + 27 + 25= 79.

Question 10.

Answer:
On adding 23 + 42 + 17 we will get the result as 82.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 7 + 3 which makes 10 and then we will add the remaining digit which is 2, so 10 + 2= 12. Now we will perform the regrouping of addition and the result will be 23 + 42 + 17= 82.

Question 11.

Answer:
On adding 18 + 34 + 26 we will get the result as 78.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 9 + 3 which makes 12 and then we will add the remaining digit which is 3, so 12 + 3= 15. Now we will perform the regrouping of addition and the result will be 53 + 13 + 19= 85.

Question 12.

Answer:
On adding 51 + 22 + 26 we will get the result as 99.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum by adding two addends which help us to perform the addition easily. So we will add the right side digits 2 + 6 which makes 8 and then we will add the remaining digit which is 1, so 8 + 1= 9. Now we will perform the regrouping of addition and the result will be 51 + 22 + 26= 99.

Question 13.

Answer:
On adding 30 + 45 + 19 we will get the result as 94.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 14 by adding which helps us to perform the addition easily. So we will add the right side digits 9 + 5 which makes 14 and then we will add the remaining digit which is 0, so 14 + 0= 14. Now we will perform the regrouping of addition and the result will be 30 + 45 + 19= 94.

Question 14.

Answer:
On adding 24 + 21 + 28 we will get the result as 73.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 12 by adding which helps us to perform the addition easily. So we will add the right side digits 8 + 4 which makes 12 and then we will add the remaining digit which is 1, so 12 + 1= 13. Now we will perform the regrouping of addition and the result will be 24 + 21 + 28= 73.

Question 15.

Answer:
On adding 39 + 12 + 31 we will get the result as 82.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 9 + 1 which makes 10 and then we will add the remaining digit which is 2, so 10 + 2= 13. Now we will perform the regrouping of addition and the result will be 39 + 12 + 31= 82.

Question 16.

Answer:
On adding 14 + 20 + 35 we will get the result as 69.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 9 by adding which helps us to perform the addition easily. So we will add the right side digits 4 + 5 which makes 9 and then we will add the remaining digit which is 0, so 9 + 0= 9. Now we will perform the regrouping of addition and the result will be 14 + 20 + 35= 69.

Question 17.

Answer:
On adding 46 + 11 +32 we will get the result as 89.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 8 by adding which helps us to perform the addition easily. So we will add the right side digits 6 + 2 which makes 8 and then we will add the remaining digit which is 1, so 8 + 1= 9. Now we will perform the addition and the result will be 46 + 11 +32= 89.

Question 18.

Answer:
On adding 35 + 33 + 29 we will get the result as 97.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 11 by adding which helps us to perform the addition easily. So we will add the right side digits 9 +3 which makes 11 and then we will add the remaining digit which is 11, so 11 + 5= 16. Now we will perform the addition and the result will be 35 + 33 + 29= 97.

Question 19.
Reasoning
You make a 10 to add 16, 38, and 24. Which digits do you add first? Explain.
________________________

________________________

________________________

Answer:

Explanation:

Think and Grow: Modeling Real Life

Newton buys the items shown. How much money does he spend?

Answer:
The total amount spent by the Newton is $51.

Explanation:
Given the cost the items that Newton has bought is, the cost of the first item is $12, the cost of second item is $21 and the cost of third item is $18. So the total money spent by the Newton is 12 + 21 + 18= $51. The total amount spent by the Newton is $51.

Show and Grow

Question 20.
Descartes buys the items shown. How much money does he spend?

Answer:
The total amount spent by the Descartes is $47.

Explanation:
Given of the cost the items that Descartes has bought is, the cost of the first item is $19, the cost of second item is $15 and the cost of third item is $13. So the total money spent by the Descartes is 19 + 15 + 13= $47. The total amount spent by the Descartes is $47.

Question 21.
Newton sells 2 large candles and 1 small candle. How much money does he earn?

Answer:
The total amount earn by the Newton is $66.

Explanation:
Given the cost the large candle that Newton has bought is $26, and the cost of the small candle is $14, so Newton sells two large candles which is 2 × 26= $52 and one small candle which is 1 × 14= $14  . So the total money earn by the Newton is 52 + 14= $66. The total amount earn by the Newton is $66.

Add Up to 3 Two-Digit Numbers Homework & Practice 4.6

Question 1.

Answer:
On adding 11 + 23 + 47 we will get the result as 81.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 7 +3 which makes 10 and then we will add the remaining digit which is 10, so 10 + 1= 11. Now we will perform the addition and the result will be 11 + 23 + 47= 81.

Question 2.

Answer:
On adding 32 + 14 + 28 we will get the result as 74.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 8 +2 which makes 10 and then we will add the remaining digit which is 10, so 10 + 4= 14. Now we will perform the addition and the result will be 32 + 14 + 28= 74.

Question 3.

Answer:
On adding 16 + 37 + 33 we will get the result as 86.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 7 + 3 which makes 10 and then we will add the remaining digit which is 10, so 10 + 6= 16. Now we will perform the addition and the result will be 16 + 37 + 33= 86.

Question 4.

Answer:
On adding 43 + 17 + 37 we will get the result as 97.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 7 + 3 which makes 10 and then we will add the remaining digit which is 10, so 10 + 7= 16. Now we will perform the addition and the result will be 43 + 17 + 37= 97.

Question 5.

Answer:
On adding 15 + 44 + 11 we will get the result as 70.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 9 by adding which helps us to perform the addition easily. So we will add the right side digits 5 + 4 which makes 9 and then we will add the remaining digit which is 9, so 9 + 1= 10. Now we will perform the addition and the result will be 15 + 44 + 11= 70.

Question 6.

Answer:
On adding 16 + 29 + 38 we will get the result as 83.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 15 by adding which helps us to perform the addition easily. So we will add the right side digits 9 + 6 which makes 15 and then we will add the remaining digit which is 8, so 15 + 8= 23. Now we will perform the addition and the result will be 16 + 29 + 38= 83.

Question 7.

Answer:
On adding 31 + 28 + 12 we will get the result as 71.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 8 + 2 which makes 10 and then we will add the remaining digit which is 1, so 10 + 1= 11. Now we will perform the addition and the result will be 31 + 28 + 12= 71.

Question 8.

Answer:
On adding 56 + 26 + 13 we will get the result as 95.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 12 by adding which helps us to perform the addition easily. So we will add the right side digits 6 + 6 which makes 12 and then we will add the remaining digit which is 3, so 12 + 3= 15. Now we will perform the addition and the result will be 56 + 26 + 13= 95.

Question 9.

Answer:
On adding 35 + 23 + 29 we will get the result as 87.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 12 by adding which helps us to perform the addition easily. So we will add the right side digits 9 + 3 which makes 12 and then we will add the remaining digit which is 5, so 12 + 3= 15. Now we will perform the addition and the result will be 35 + 23 + 29= 87.

Question 10.
DIG DEEPER!
Solve two different ways.

Answer:
The sum of the three numbers is 38 + 36 + 22= 96.

Explanation:
The two different ways to add the above problem is, the first method is we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 8 + 2 which makes 10 and then we will add the remaining digit which is 6, so 10 + 6= 16. Now we will perform the regrouping of addition and the result will be 38 + 36 + 22= 96. The other way to find the addition of the numbers is we will add the numbers in any order, first we will add 8 + 6 which is 14 and then we will add the remaining digit which is 14 + 2= 16 and the regrouping of addition and the result will be 38 + 36 + 22= 96.

Question 11.
Modeling Real Life
Descartes buys the items shown. How much money does he spend?

Answer:
The total amount spent by the Descartes is $81.

Explanation:
Given the cost the items that Descartes has bought is, the cost of the first item is $41, the cost of second item is $27 and the cost of third item is $13. So the total money spent by the Descartes is 41 + 27 + 13= $81. The total amount spent by the Descartes is $81.

Question 12.
Modeling Real Life
Your cousin buys the items shown. How much money does she spend?

Answer:
The total amount spent by the my cousin is $67.

Explanation:
Given of the cost the items that my cousin has bought is, the cost of the first item is $24, the cost of second item is $7 and the cost of third item is $36. So the total money spent by my cousin is 24 + 7 + 36= $67. The total amount spent by mu cousin is $67.

Review & Refresh

Is the number even or odd?

Question 13.

Answer:
The number 18 is even number.

Explanation:
The number 18 is even number as the number 18 has two pairs of 9 and the number is divided by 2 and if the number is not divisible by 2 then it will be odd number. As 18 is divisible by 2 so the number 18 is a even number.

Question 14.

Answer:
The number 17 is odd number.

Explanation:
The number 17 is even number as the number 17 is not divisible by 2 and if the number is not divisible by 2 then it will be odd number. As the number 17 is not divisible by 2 so the number 17 is odd number.

Lesson 4.7 More Problem Solving: Addition

Explore and Grow

Model the story.

There are 11 red ants and 14 black ants. 15 more black ants join them. How many ants are there now?

Answer:
The total number of ants is 40 ants.

Explannation:
Given that there are 11 red ants and 14 black ants and 15 more black ants join with them, so to find the number of ants in total we will add the given number of ants, so the total number of ants is 11 + 14 + 15= 40 ants.

Show and Grow

Question 1.
You have 66 marbles. You have 26 fewer marbles than your friend. How many marbles does your friend have?

Answer:
My friend has 40 marbles.

Explanation:
As I have 66 marbles and my friend has 26 fewer marbles than me, so to find how many marbles does my friend had we will subtract 66 – 26= 40 marbles. So my friend has 40 marbles.

Apply and Grow: Practice

Question 2.
You collect 16 red leaves, 21 orange leaves, and 14 yellow leaves. How many leaves do you collect in all?

Answer:
The number of leaves collected by me is 16 + 21 + 14= 51 leaves.

Explanation:
As I have collected 16 red leaves, 21 orange leaves and 14 yellow leaves, so the number of leaves collected by me is 16 + 21 + 14= 51 leaves.

Question 3.
A dentist has 41 toothbrushes. She buys some more. Now she has 85. How many toothbrushes did the dentist buy?

Answer:
The number of toothbrushes did the dentist bought is 85 – 41= 44 toothbrushes.

Explanation:
As dentist has 41 toothbrushes and she bought some more, so now she had 85. So the number of toothbrushes did the dentist bought is 85 – 41= 44 toothbrushes.

Question 4.
You make 17 origami dogs and 13 origami fish. Your friend makes 12 more origami animals than you. How many origami animals does your friend make?

Answer:
The number of origami animals does my friend make is 42 animals.

Explanation:

As I make 17 origami dogs and 13 origami fish so the total origami animals do I make is 17 + 13= 30 animals. And my friend makes 12 more origami animals than me, so the number of origami animals does my friend make is
30 + 12= 42 animals.

Think and Grow: Modeling Real Life

You make a paper chain. Your friend adds 24 links to your chain. Now there are 57. How many links were there to start?

Answer:
The number of links is 33 links.

Explanation:
As I make a paper chain and my friend adds 24 links to my chain. Now there are 57 paper chains, so the number of links there to start is 57 – 24= 33 links.

Show and Grow

Question 5.
You have some stickers. Your friend gives you 32 more stickers. Now you have 58. How many stickers did you have to start?

Answer:
The number of stickers I have to start is 58 – 32 = 26 stickers.

Explanation:
As I have some stickers and my friend gives me 32 more stickers. Now I have 58, so the number of stickers I have to start is 58 – 32 = 26 stickers.

Question 6.
There are 3 buses. There are 29 students on each of the first 2 buses. There are 88 students in all. How many students are on the third bus?

Answer:
The number of students are on the third bus is 30 students.

Explanation:
As there are 3 buses and there are 29 students on each of the first 2 buses so the students in the two buses is
2 × 29= 58 students and there are 88 students in all so the number of students are on the third bus is
88 – 58= 30 students.

More Problem Solving: Addition Homework & Practice 4.7

Question 1.
You study for 31 minutes. You study 11 fewer minutes than your friend. How long does your friend study?

Answer:
My friend studied 20 minutes longer.

Explanation:
As I have studied for 31 minutes and  studied 11 fewer minutes than my friend so how long does your friend studied is 31 – 11= 20 minutes.

Question 2.

You see 23 red cars, 25 black cars, and 15 blue cars. How many cars do you see in all?

Answer:
The number of cars I have seen is 63 cars.

Explanation:
As I saw 23 red cars, 25 black cars, and 15 blue cars, so the number of cars I have seen is 23 + 25 + 15= 63 cars. The number of cars I have seen is 63 cars.

Question 3.
Number Sense
Your friend scores 29 points. You score 16 more points than your friend. Use the given numbers to find how many points you and your friend score in all.

Answer:
The total points I have scored is 45 points and the total points scored by me and my friend is 74 points.

Explanation:

As my friend scores 29 points and I score 16 more points than your friend. So the total points I have scored is
29 + 16= 45 points. And the total points scored by me and my friend is 74 points.

Question 4.
Modeling Real Life
You have some plastic bugs. Your friend gives you 37 more plastic bugs. Now you have 59. How many plastic bugs did you have to start?

Answer:
The number of plastic bugs I do have to start is 22 plastic bugs.

Explanation:
As I have some plastic bugs and my friend gives me 37 more plastic bugs. So now I have 59 plastic bugs, so the number of plastic bugs I do have to start is 59 – 37= 22 plastic bugs.

Question 5.
Modeling Real Life
There are 3 subway cars. There are 36 people on each of the first 2 subway cars. There are 92 people in all. How many people are on the third subway car?

Answer:
The number of people in the third subway car is 20 people.

Explanation:
As there are 3 subway cars and there are 36 people on each of the first 2 subway cars, so the number of people in the 2 subway cars is 2 × 36= 72 people. There are 92 people in all and the number of people in the third subway car is 92 – 72= 20 people.

Review & Refresh

Question 6.
50 – 10 = ___

Answer:
The difference between 50 and 10 is 40.

Explanation:
On subtracting 50 – 10 we will get the result as 40.

Question 7.
90 – 40 = ___

Answer:
The difference between 90 and 40 is 50.

Explanation:
On subtracting 90 – 40 we will get the result as 50.

Fluently Add within 100 Performance Task 4

Question 1.
a. You swim for 37 minutes on Monday. You swim 12 more minutes on Tuesday than on Monday. How many minutes do you swim in all?

Answer:
The total number of minutes do I swim in all is 86 minutes.

Explanation:
As I swim for 37 minutes on Monday and again swim for 12 more minutes on Tuesday than on Monday, so the number of minutes I have a swim on Tuesday is 37 + 12= 49 minutes. And the total number of minutes do I swim in all is 37 + 49= 86 minutes.

b. Do you swim an even or odd number of minutes in all?

Answer:
The number of minutes 86 is an even number.

Explanation:
As the total number of minutes do I swim in all is 86 minutes and 86 is divisible by 2 so the number of minutes 86 is an even number.

Question 2.
a. There are 35 girls and some boys signed up for swim lessons this year. There are 83 kids signed up in all. Then some more boys sign up. Now there are 56 boys signed up. How many more boys signed up for swim lessons?

Answer:
The number of more boys signed up is 8 more boys.

Explanation:
As there are 35 girls and some boys signed up for swim lessons this year. And there are 83 kids signed up in all, so the number of boys is 83 – 35= 48 boys and then some more boys sign up. Now there are 56 boys signed up, so the number of more boys signed up is 56 – 48= 8 more boys.

b. Last year there were 95 kids signed up for swim lessons. 45 were girls. Are there more boys signed up for swim lessons this year or last year?

Answer:
The more boys signed up for swimming is last year than this year.

Explanation:
As in the last year, there were 95 kids signed up for swim lessons and in that 45 were girls and the number of boys is 95 – 45= 50  boys. And the more boys signed up for swimming is last year as the total number of boys in the last year is 56 boys.

Fluently Add within 100 Activity

Solve and Cover: Addition

To Play: Place a Solve and Cover: Addition Sum Card on each box. Players take turns. On your turn, flip over a Solve and Cover: Addition Problem Card. Solve the problem. Place the problem card on the sum. Play until all sums are covered.

Fluently Add within 100 Chapter Practice

4.1 Use Partial Sums to Add

Question 1.

Answer:
The partial sum of 35 + 22 will be 50 + 7= 57.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 35 + 22 which is 57. So here first we will add tens which are 30 and 20 we will add both the numbers which will be 50 and now we will add ones which are 5 and 2 will be 7. So the total value of 35 + 22 will be 50 + 7= 57.

Question 2.

Answer:
The partial sum of 81 + 8 will be 80 + 9= 89.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 81 + 8 which is 89. So here first we will add tens which are 80 and 0 we will add both the numbers which will be 80 and now we will add ones which are 1 and 8 will be 9. So the total value of 81 + 8 will be 81 + 8= 89.

4.2 More Partial Sums

Question 3.
26 + 43 = ?

Answer:
The partial sum of 26 + 43 will be 60 + 9= 69.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 26 + 43 which is 69. So here first we will add tens which are 20 and 40 we will add both the numbers which will be 60 and now we will add ones which are 6 and 3 will be 69. So the total value of 26 + 43 will be 26 + 43= 69.

Question 4.
64 + 19 = ?

Answer:
The partial sum of 64 + 19 will be 70 + 13= 83.

Explanation:

Here partial sum is the sum of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 64 + 19 which is 83. So here first we will add tens which are 60 and 10 we will add both the numbers which will be 70 and now we will add ones which are 4 and 9 will be 13. So the total value of 64 + 19 will be 70 + 13= 83.

4.3 Regroup to Add

Question 5.
Modeling Real Life
You want to complete 40 hours of volunteer work this year. You complete 28 hours during the school year and 13 during the summer. Do you reach your goal?

Answer:
Yes, I have reached my goal.

Explanation:

As I want to complete 40 hours of volunteer work this year and I have complete 28 hours during the school year and 13 during the summer so the total number of hours I have completed is 28 + 13= 41 hours. And I have reached my goal.

4.4 Add Two-Digit Numbers

Question 6.

Answer:
By regrouping the addition of 14 and 77 we will get the result as 91.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 14 and 77. So by regrouping, we will carry forward one and the sum of 14 + 77 is 91.

Question 7.

Answer:
By regrouping the addition of 35 and 35 we will get the result as 70.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 35 and 35. So by regrouping, we will carry forward one and the sum of 35 + 35 is 70.

Question 8.

Answer:
By regrouping the addition of 43 and 49 we will get the result as 92.

Explanation:

Here, regrouping is defined as the process of making and then carrying out the operation like addition with the two digit numbers or larger than the two digit numbers. And we use regrouping in addition when the sum of two digits in the place value column is greater than nine. And Regrouping in subtraction is a method of interchanging one ten into ten ones. This regrouping of subtraction is used to work out the different subtraction problems. Here in the above image, we can see that the addition of 43 and 49. So by regrouping, we will carry forward one and the sum of 43 + 49 is 92.

4.5 Practice Adding Two-Digit numbers

Question 9.

Answer:
The sum of 36 + 38 is 74.

Explanation:
The addition of the two numbers is 36 + 38 is 74.

Question 10.

Answer:
The sum of 62 + 29 is 91.

Explanation:
The addition of the two numbers is 62 + 29 is 91.

Question 11.

Answer:
The sum of 25 + 45 is 70.

Explanation:
The addition of the two numbers is 25 + 45 is 70.

Question 12.
Number Sense
Find the missing digits.

Answer:
The missing digits are 1, 4, 8, 7, 5, 0.

Explanation:

To find the missing digits, in the first image we can see that the sum is 9 and one of the addends is 5 so to get the sum as 9 we will place 4 and place the 1 to get the sum as 4, so the sum will be 49 and the addend will be 34 and 15. And in the second image, we can see that the sum is 6, and the other addend is 8 so let’s take the other addend to be 8 so the missing digit is 8 and by the regrouping of addition we will get the result as 76, so the missing digit is 7. And in the third image, we can see that the addend is 14 and the other addend with digit 6 so by placing the digit as 5 the sum will be 70 and the missing digits will be 5 and 0.

4.6 Add Up to 3 Two-Digit Numbers

Question 13.

Answer:
On adding 12 + 32+ 18 we will get the result as 62.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 8 + 2 which makes 10 and then we will add the remaining digit which is 2, so 10 + 2= 12. Now we will perform the regrouping of addition and the result will be 12 + 32+ 18= 62.

Question 14.

Answer:
On adding 50 + 28 + 18 we will get the result as 96.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 16 by adding which helps us to perform the addition easily. So we will add the right side digits 8 + 8 which makes 16 and then we will add the remaining digit which is 0, so 16 + 0= 16. Now we will perform the regrouping of addition and the result will be 50 + 28 + 18= 96.

Question 15.

Answer:
On adding 50 + 28 + 18 we will get the result as 96.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 16 by adding which helps us to perform the addition easily. So we will add the right side digits 8 + 8 which makes 16 and then we will add the remaining digit which is 0, so 16 + 0= 16. Now we will perform the regrouping of addition and the result will be 50 + 28 + 18= 96.

Question 16.

Answer:
On adding 17 + 26 + 12 we will get the result as 55.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 9 by adding which helps us to perform the addition easily. So we will add the right side digits 7 + 2 which makes 9 and then we will add the remaining digit which is 6, so 9+ 6= 15. Now we will perform the regrouping of addition and the result will be 17 + 26 + 12= 55.

Question 17.

Answer:
On adding 27 + 33 + 18 we will get the result as 78.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 10 by adding which helps us to perform the addition easily. So we will add the right side digits 7+ 3 which makes 10 and then we will add the remaining digit which is 8, so 10 + 8= 18. Now we will perform the regrouping of addition and the result will be 27 + 33 + 18= 78.

Question 18.

Answer:
On adding 15 + 18 + 16 we will get the result as 49.

Explanation:
In the above image, given that three numbers for addition. Here we can add the three numbers by any order, and we can make the sum 14 by adding which helps us to perform the addition easily. So we will add the right side digits 8 + 6 which makes 14 and then we will add the remaining digit which is 5, so 14 + 0= 14. Now we will perform the regrouping of addition and the result will be 15 + 18 + 16= 49.

Question 19.
Modeling Real Life
You, Newton, and Descartes play paddle ball. You record how many times each of you hits the ball in a row. How many times is the ball hit in all?

4.7 More Problem Solving: Addition

Question 20.
Modeling Real Life
You pick 11 berries and 23 apples. Your friend picks 18 more pieces of fruit than you. How many pieces does your friend pick?

Step 1: How many pieces of fruit do you pick?

Answer:
The number of pieces of fruits does I have is 34 fruits.

Explanation:
As I pick 11 berries and 23 apples, so the number of fruits I have picked is 11 + 23= 34 fruits.

Step 2: How many pieces of fruit does your friend pick?

Answer:
The number of fruits that my friend picked is 52 fruits.

Explanation:
As I pick 11 berries and 23 apples and my friend picks 18 more pieces of fruit than me, so the number of fruits that my friend picked is 34 + 18= 52 fruits.

Fluently Add within 100 Cumulative Practice 1-4

Question 1.
Which equation represents the array?

Answer:
The equation that represents the array is 3 + 3 + 3 + 3= 12.

Explanation:
The equation that represents the array is 3 + 3 + 3 + 3= 12 as the array has three rows and four columns, so the equation is 3 + 3+ 3 +3 = 12.

Question 2.
Which expressions are equal to 62 + 24?

Answer:
The expressions which are equal to 62 + 24 is 60 + 20 + 2 + 4, 60 + 26, 80 + 6.

Explanation:
The expressions which are equal to 62 + 24 and the result is 86 so the expressions which are equal to the sum of 86 is 60 + 20 + 2 + 4 and the result is 86, 60 + 26 the result is 86, 80 + 6 the result is 6.

Question 3.
Is each sum equal to 7 + 4?

Answer:
The equation which is equal to 7 + 4 is 10 + 1.

Explanation:
The sum which is equal to 7 + 4 is 10 + 1 as the sum of both equations is 11 so they both are equal.

Question 4.
Write an equation that matches the number line.

Answer:
The number line equation is 10 + 40= 50.

Explanation:
The equation that matches the number line is 10 + 40 and the result is 50. As we start from 10 and then jump from 10 and the size of the jump is 10 and we will take up to 50 jumps. So the equation of the number line is
10 + 40= 50.

Question 5.
Which expressions do you need to regroup to solve?

Answer:
The expressions which we need to regroup is 54 + 38 and the result is 92 and the other expression is 62 + 28 and the result is 90.

Explanation:
The expressions do we need to regroup and is 54 + 38 as by adding 8 + 4 we will get the result as 12, so we will regroup and add 54 + 38 and the result will be 92. And the other expression is 62 + 88 as by adding 2 + 8 we will get the result as 10, so we will regroup and add 62 + 28 and the result will be 90.

Question 6.
Which equation has an even sum greater than 14?

Answer:
The equation that has an even sum whihc is greater than 14 is 8 + 8 which is 16.

Explanation:
In the above, we can see that even sum greater than 14 is 8 + 8 as the sum of 8 + 8 is 16 and 16 is divisible by 2 which is an even number and the number 16 is greater than 14.

Question 7.
There are 4 rows of trees. Each row has 5 trees. How many trees are there in all?

Answer:
The number of trees there in all is 20 trees.

Explanation:
As there are 4 rows of trees and each row has 5 trees, so the number of trees is 4 + 4 + 4 + 4 + 4= 20 trees. So the number of trees is 20 trees.

Question 8.
You have 16 oranges. You give 7 away. How many oranges do you have left?

Answer:
The number of oranges does I have left is 11 oranges.

Explanation:
As I have 16 oranges and I gave 7 away, so the total number of oranges does I have left is 16 – 7= 11 oranges.

Question 9.
Find the sum.

Answer:
The result of the first image is 71 and the result of the second image is 87.

Explanation:
The sum of the first image is 24 + 12 + 35 and the result is 71 and the sum of the second image is 31 + 14 + 42 and the result is 87.

Question 10.
Find the sum. Write the double you used.

Answer:
The sum of the double of the addends is 30.

Explanation:
The sum of 7 + 8 is 15, so the double of the addends is 14 + 16, and the sum of the addends is 14 + 16= 30.

Question 11.
Break apart the addends to find 42 + 37.

Answer:
The Break apart of the addend is 42 + 37= 79.

Explanation:
Here to break apart the addends of part of the sequence and a sequence that sets the numbers in order. This partial sum calculates partial sum which will be working one place value column at a time and then add all the partial sums to find the total. And this partial sum can be added in any order but working with the left to right is the usual procedure. So here we will add tens and add the ones and then we will add the partial sum to find the whole sum of the 42 + 37 which is 79. So here first we will add tens which are 40 and 30 we will add both the numbers which will be 70 and now we will add ones which are 2 and 7 will be 9. So the total value of 70 + 9 will be 70 + 9= 79.

Question 12.
You have 64 craft sticks for a project. 22 are red. The rest are yellow. You buy 39 more yellow craft sticks. How many yellow craft sticks do you have now?

Answer:
The total number of yellow craft sticks do I have is 81 sticks.

Explanation:
As I have 64 craft sticks for a project and there are 22 red sticks so the number of yellow sticks is 64 – 22= 42 yellow sticks and I bought 39 more yellow craft sticks, so the number of yellow sticks is 42 + 39= 81 yellow craft sticks.

Conclusion:

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Big Ideas Math Book 3rd Grade Answer Key Chapter 1 Understand Multiplication and Division: Get the solutions for all the questions in BIM Grade 3 Chapter 1 Understand Multiplication and Division Textbook. This answer key is useful for the students while preparing for the exam or practice test. In the Big Ideas Math 3rd Grade 1st Chapter Solutions PDF, students can see the detailed explanation for each problem that helps them to understand the concept easily.

Big Ideas Math Book Grade 3 Answer Key Chapter 1 Understand Multiplication and Division

Download Big Ideas Math 3rd Grade 1st Chapter Understand Multiplication and Division Answer Key PDF for free of cost. The different lessons in Grade 3 Chapter 1 are Use Equal Groups to Multiply, Use Number Lines to Multiply, Use Arrays to Multiply, Multiply in Any Order, Divide: Size of Equal Groups, Divide: Number of Equal Groups, and Use Number Lines to Divide.

Here, you can get the different methods of solving multiplication and division problems. Solve all the questions from Big Ideas Math Answers Grade 3 Chapter 1 Understand Multiplication and Division. Tap on the links mentioned below to get help while solving multiplication and division problems.

Lesson 1 Use Equal Groups to Multiply

• Lesson 1.1 Use Equal Groups to Multiply
• Use Equal Groups to Multiply Homework & Practice 1.1

Lesson 2 Use Number Lines to Multiply

• Lesson 1.2 Use Number Lines to Multiply
• Use Number Lines to Multiply Homework & Practice 1.2

Lesson 3 Use Arrays to Multiply

• Lesson 1.3 Use Arrays to Multiply
• Use Arrays to Multiply Homework & Practice 1.3

Lesson 4 Multiply in Any Order

• Lesson 1.4 Multiply in Any Order
• Multiply in Any Order Homework & Practice 1.4

Lesson 5 Divide: Size of Equal Groups

• Lesson 1.5 Divide: Size of Equal Groups
• Divide: Size of Equal Groups Homework & Practice 1.5

Lesson 6 Divide: Number of Equal Groups

• Lesson 1.6 Divide: Number of Equal Groups
• Divide: Number of Equal Groups Homework & Practice 1.6

Lesson 7 Use Number Lines to Divide

• Lesson 1.7 Use Number Lines to Divide
• Use Number Lines to Divide Homework & Practice 1.7

Performance Task

• Understand Multiplication and Division Performance Task
• Understand Multiplication and Division Activity
• Understand Multiplication and Division Chapter practice

Lesson 1.1 Use Equal Groups to Multiply

Explore and Grow

Question 1.
Put 24 counters into equal groups. Draw to show your groups.
Answer:
24 counters in four equal groups

Explanation:
To keep 24 counters into an equal group, we will find the factors of 24 and then we can put 24 counters into equal groups. The factors of 24 are 2,3,4,6,8,12. Now we can pick any of the numbers from the factors of 24 and then put these 24 counters in equal groups. Let’s take four equal groups and arrange 24 counters in four equal groups as shown in the below image.

Question 2.
Put 24 counters into a different number of equal groups. Draw to show your groups.
Answer:
7 counters in the first group, 8 counters in the second group, 6 counters in the third group, and 4 counters in the fourth group.

Explanation:
In this, we will take some number of groups and in that, we will place some different number of counters in that group. In the given image, we will take four groups, and we will place 7 counters in the first group, 8 counters in the second group, 6 counters in the third group, and 4 counters in the fourth group.

Question 3.
Structure
Compare your models. How are the models the same? How are they different?
Answer:
In the first model, we can see that there is an equal number of counters in each group, and in the second model, we can see the different number of counters in each group. They are the same with an equal number of groups.

Show and Grow

Use the model to complete the statements.
Question 1.

____ groups of ____
____ + ____ +____ + ____ = ____
____ × ____ = ____
Answer:
4 groups of 5,
5+5+5+5= 20,
4×5= 20.

Explanation:
In the above image, we can see 4 groups and in each group, 5 counters are there, and the product is 4×5= 20. And we can also write by adding 5+5+5+5= 20. There are 4 groups of 5.

Question 2.

____ groups of ____
____ + ____ = ____
____ × ____ = ____
Answer:
2 groups of 6,
6+6= 12,
2×6= 12.

Explanation:
As we can see there are 2 groups of 6 which are two groups with the number 6 in the box. So the product of the two groups is 2×6= 12 and the sum of the two groups is 6+6= 12.

Apply and Grow: Practice

Question 3.
Use the model to complete the statements.

____ groups of ____
____ + ____ = ____
____ × ____ = ____
Answer:
2 groups of 9
9+9= 18,
2×9= 18.

Explanation:
In the above image, we can see 2 groups with 9 counters in each group, and the product is 2×9= 18. By adding these two groups we will get 18, which is 9+9= 18.

Draw equal groups. Then complete the equations.
Question 4.
4 groups of 2
____ + ____ + ____ + ____ =____
____ × ____ = ____
Answer:
4 groups of 2
2+2+2+2= 8,
4×2= 8.

Explanation:
In the above image, we can see 4 groups with 2 counters in each group, and the product is 4×2= 8. By adding these four groups we will get 8, which is 2+2+2+2= 18.

Question 5.
3 groups of 5
____ + ____ + ____ = ____
____ × ____ = ____
Answer:
3 groups of 5,
5+5+5= 15,
3×5= 15.

Explanation:
In the above image, we can see 3 groups with 5 counters in each group, and the product is 3×5= 15. By adding these three groups we will get 15, which is 5+5+5= 15.

Write the addition equation as a multiplication equation.
Question 6.
8 + 8 + 8 = 24
Answer:
3×8= 24.

Explanation:
To write the addition equation as a multiplication equation in the above we can see 3 times 8 which can be written as 3×8= 24.

Question 7.
7 + 7 + 7 + 7 + 7 = 35

Answer:
5×7= 35.

Explanation:
To write the addition equation as a multiplication equation in the above we can see 5 times 7 which can be written as 5×7= 35.

Question 8.
You Be The Teacher
Newton says he circled 3 groups of 4 counters. Is he correct? Explain.

Answer:
No, Newton is not correct.

Explanation:
In the above image, we can see 4 groups of 3 counters. So Newton is not correct.

Question 9.
DIG DEEPER!
You wash 5 cars. How many times do you wash?

Answer: 20 tires.

Explanation:
As the car has 4 tires and there are 5 cars to wash, so there are 5 groups of 4. The number of tires to wash is
5×4= 20. So there are 20 tires to wash.

Think and Grow: Modeling Real Life

Question 1.
You buy 6 packs of 10 trading cards. How many cards do you buy in all?
Complete the statement: ______ groups of ________
Repeated addition equation:
Multiplication equation:
You buy _____ cards in all.

Answer:
6 groups of 10
6×10= 60.

Explanation:
As there are 6 packs of 10 trading cards, which means 6 groups of 10. So 6×10= 60 cards can we buy in all.

Show and Grow

Question 10.
You buy 8 packs of 4 highlighters. How many highlighters do you buy in all?

Answer:
8 groups of 4
8×4= 32

Explanation:
As there 8 packs of 4 highlighters, which means 8 groups of 4. The number of highlighters bought is 8×4= 32.

Question 11.
DIG DEEPER!
You make 5 bracelets. Each of your bracelets has 3 beads. Your friend makes 6 bracelets. Each of your friend’s bracelets has 2 beads. How many beads do you and your friend use in all?

Answer:
27 beads.

Explanation:
In this query, we can see 5 bracelets have 3 beads which are 5×3= 15, and a friend makes 6 bracelets that have 2 beads which are 6×2= 12. So the total number of beads used in all is 15+12= 27 beads.

Use Equal Groups to Multiply Homework & Practice 1.1

Question 1.
Use the model to complete the statements.

___ groups of ____
___ + ___ + ___ = ___
___ × ___ = ___
Answer:
3 groups of 4,
4+4+4= 12,
4×3= 12.

Explanation:
We can see in the above image there are 3 groups with 4 counter in each group, which means 3 groups of 4.
The sum is 4+4+4= 12 and the product is 4×3= 12.

Draw equal groups. Then complete the equations.
Question 2.
2 groups of 8
___ + ___ = ___
___ × ___ = ___
Answer:
8+8= 16
2×8= 16.

Explanation:
2 groups of 8 means, there are 2 groups with 8 counters in each group. So the sum is 8+8= 16 and the product is 2×8= 16.

Question 3.
5 groups of 3
___ + ___ + ___ + ___ + ___ = ___
___ × ___ = ___

Answer:
3+3+3+3+3= 15,
5×3= 15.

Explanation:
5 groups of 3 means, there are 5 groups with 3 counters in each group. So the sum is 3+3+3+3+3= 15, and the product is 5×3= 15.

Write the addition equation as a multiplication equation.
Question 4.
3 + 3 + 3 + 3 + 3 + 3 = 18
Answer:
6×3= 18.

Explanation:
To write the addition equation as a multiplication equation in the above we can see 6 times 3, which can be written as 6×3= 18.

Question 5.
2 + 2 + 2 + 2 + 2 + 2 + 2 = 14
Answer:
7×2= 14.

Explanation:
To write the addition equation as a multiplication equation in the above we can see 7 times 2, which can be written as 7×2= 14.

Question 6.
DIG DEEPER!
You have 16 action figures. Can you put an equal number of figures on 3 shelves? Explain
Answer:
No, we cannot put an equal number of figures on 3 shelves.

Explanation:
As given that we have 16 action figures and asked to determine whether it is possible to an equal number of figures on 3 shelves. As factors of 16 are 1,2,4,8,16 as 16 is not divisible by 3, so we cannot put an equal number of figures on 3 shelves.

Question 7.
Which One Doesn’tBelong?
Which one does not belong with the other two?

Answer:
2 groups of 3 don’t belong with the other two.

Explanation:
As 2 groups of 3 means in 2 groups, there will be 3 counters, which does not belong to the other two. Because in the above image we can see 3 groups of 2, which means 3 groups with 2 counters in each group which is equal to 2+2+2= 6.

Question 8.
Modeling Real Life
You make 7 gift bags for your friends. Each gift bag has 3 pom-pom pets. How many pom-pom pets are there in all?
Answer:
21 pom-pom pets.

Explanation:
As there are 7 gift bags and each gift bag has 3 pom-pom pets, so the number of pom-pom pets are 7×3= 21.

Question 9.
DIG DEEPER
Newton has 2 stacks of 5 books. Descartes has 3 stacks of 4 books. How many books do they have in all?

Answer:
22 books.

Explanation:
As Newton has 2 stacks of 5 books, which is 2×5= 10 books and Descartes has 3 stacks of 4 books which is 3×4= 12 books. So the number of books is 10+12= 22books.

Review & Refresh

Question 10.
50 + 30 = ____

Answer:
80

Explanation:
By adding 50+30 we will get 80.

Question 11.
27 + 40 = ______

Answer:
67

Explanation:
By adding 27+40 we will get 67.

Question 12.
19 + 20 = _____

Answer:
39

Explanation:
By adding 19+20 we will get 39.

Lesson 1.2 Use Number Lines to Multiply

Explore and Grow

Question 1.
Find the sums. Use each sum as the missing addend in the next equation. Model the problems on the number line.

Answer:
0+3= 3,
3+3= 6,
6+3= 9,
9+3= 12.

Explanation:
First, we must begin at 0 and then skip the count by 3’s then make jumps until you get 12.
So 3 x 4 = 12.

Reasoning
How can you use a number line to help you find 4 × 3?

Answer:
4×3= 12.

Explanation:
To find 4×3 using the number line, we will make a jump from 0 to 4 for 3 times.

Think and Grow: Multiplication and Number Lines

Example Find 3 × 4

3 × 4 means 3 groups of 4.
Number of jumps: _____
Size of each jump: ______
Start at 0. Skip count by 4s three times.

3 × 4 ____

Answer:
3 × 4 = 12
Number of jumps: 3
Size of each jump: 4.

Explanation:
To represent the given value in the number line, we will start from 0 and skip the count by 4 three times. So the number of jumps is 3 and the size of each jump is 4.

Show and Grow

Question 1.
Find 2 × 4
Number of jumps: _____
Size of each jump: ______

2 × 4 = _____
Answer:
2 × 4 = 8,
Number of jumps: 2
Size of each jump: 4

Explanation:

To find the product of 2×4 using the number line, we will start from zero and then skip the count by 4s two times. So the number of jumps is 2 and the size of each jump is 4.

Question 2.
Fing 6 × 3
Number of jumps: _____
Size of each jump: ______

6 × 3 = _____
Answer:
6 × 3= 18
Number of jumps: 6
Size of each jump: 3

Explanation:

To find the product of 6 × 3 using the number line, we will start from zero and then skip the count by 3s six times. So the number of jumps is 6 and the size of each jump is 3.

Apply and Grow: Practice
Question 3.
Find 3 × 5
Number of jumps: _____
Size of each jump: ______

3 × 5 = _____
Answer:
3 × 5= 15
Number of jumps: 3
Size of each jump: 5

Explanation:

To find the product of 3 × 5 using the number line, we will start from zero and then skip the count by 5s three times. So the number of jumps is 3 and the size of each jump is 5.

Question 4.
Fing 5 × 4
Number of jumps: _____
Size of each jump: ______

5 × 4 = _____
Answer:
5 × 4= 20
Number of jumps: 5
Size of each jump: 4

Explanation:

To find the product of 5 × 4 using the number line, we will start from zero and then skip the count by 4s five times. So the number of jumps is 5 and the size of each jump is 4.

Question 5.
Find 3 × 8
Number of jumps: _____
Size of each jump: ______

Answer:
3 × 8= 24
Number of jumps: 8
Size of each jump: 3

Explanation:

To find the product of 3 × 8 using the number line, we will start from zero and then skip the count by 3s eight times. So the number of jumps is 8 and the size of each jump is 3.

Question 6.
Structure
Draw jumps to show 4 groups of 6 and 6 groups of 4. Think: How are they the same? How are they different?

Answer:
4 groups of 6
4×6= 24
Number of jumps: 6
Size of each jump: 4
6 groups of 4
6×4= 24
Number of jumps: 4
Size of each jump: 6
The product of 4 groups of 6 and 6 groups of 4 is 24 this is the same in 4 groups of 6 and 6 groups of 4. The number of jumps and the size of the jumps is different.

Explanation:
To find the product of 4 × 6 using the number line, we will start from zero and then skip the count by 4s six times. So the number of jumps is 6 and the size of each jump is 4.

To find the product of 6 × 4 using the number line, we will start from zero and then skip the count by 6s four times. So the number of jumps is 4 and the size of each jump is 6.

Think and Grow: Modeling Real Life
A group of lions is called a pride. There are 2 prides in a savanna. Each pride has 9 lions. How many lions are there in all?

Model:

There are ____ lions in all.

Answer:
2×9= 18 lions.

Explanation:
As there are 2 prides in the savanna and each pride has 9 lions, the total number of lions is 2×9= 18 lions.
To represent this in the number line, we will start from 0 then skip the count by 9 twice. So the number of jumps is 2 and the size of the jump is 9.

Show and Grow

Question 7.
There are 3 bike racks at a park. Each bike rack has 4 bikes. How many bikes are there in all?

Answer:
3×4= 12 bikes.

Explanation:
As there are 3 bike racks and each rack has 4 bikes, so the number of bikes are there is 3×4= 12.
To represent this in the number line, we will start from 0 then skip the count by 4 three times. So the number of jumps is 3 and the size of the jump is 4.

Question 8.
DIG DEEPER!
You dig 8 holes. You plant 2 flower bulbs in each hole. You have5 bulbs left. How many flower bulbs did you have to start?

Answer:
3×5= 15 flower bulbs.

Explanation:
The number of holes dug is 8 and in each hole, 2 flower bulbs are planted, so the number of flower bulbs planted is 2×5= 10 and still, there are 5 bulbs left. So the total number of flower bulbs is 10+5= 15. To represent this in the number line, as there are 5 bulbs added then it will be 3 groups of 5 which is 3×5= 15. So we will start from 0 then skip the count by 5 three times and the number of jumps is 3 and the size of the jump is 5.

Use Number Lines to Multiply Homework & Practice 1.2

Question 1.
Find 3 × 6
Number of jumps: _____
Size of each jump: ______

3 × 6 = _____
Answer:
3 × 6= 18
Number of jumps: 3
Size of each jump: 6.

Explanation:
To represent this in the number line, we will start from 0 then skip the count by 6 three times. So the number of jumps is 3 and the size of the jump is 6.

Question 2.
Find 4 × 5
Number of jumps: _____
Size of each jump: ______

4 × 5 = _____
Answer:
4 × 5= 20
Number of jumps: 4
Size of each jump: 5.

Explanation:
To represent this in the number line, we will start from 0 then skip the count by 5 four times. So the number of jumps is 4 and the size of the jump is 5.

Question 3.
Structure
Complete the multiplication equations in two different ways. Model each equation on the number line.
____ × ____ = 12

Answer:
2×6= 12

Explanation:
To represent this in the number line, we will start from 0 then skip the count by 6 two times. So the number of jumps is 2 and the size of the jump is 6.

____ × ____ = 12

Answer:
4×3= 12

Explanation:
To represent this in the number line, we will start from 0 then skip the count by 3 four times. So the number of jumps is 4 and the size of the jump is 3.

Question 4.
Writing
Explain how you can use a number line to find 5 × 3.
Answer:
5×3= 15.

Explanation:
To represent this in the number line, we will start from 0 then skip the count by 3 five times. So the number of jumps is 5 and the size of the jump is 3.

Question 5.
Modeling Real Life
You have6 boxes of blueberry muffins. Each box has 4 muffins. How many muffins do you have in all?

Answer:
6×4= 24.

Explanation:
As we have 6 boxes of blueberry muffins with 4  muffins in each box, which means 6×4 = 24 muffins do we have it all.

Question 6.
DIG DEEPER!
You fill 8 pages of a photo album. Each page has 3 photos. You have one photo left. How many photos did you have to start?
Answer:
25 photos.

Explanation:
The total number of pages in a photo album is 8 and each page contains 3 photos, which means 8×3= 24, so there will be a total of 24 photos. Now 1 photo left, which means 24+1= 25. So there will be a total of 25 photos.

Review & Refresh
Question 7.
9 + 8 + 2 = _____

Answer:
19.

Explanation:
On adding 9+8+2 we will get the sum as 19.

Question 8.
6 + 5 + 3 = _____

Answer:
14.

Explanation:
On adding 6+5+3 we will get the sum as 14.

Question 9.
7 + 4 + 7 = ______

Lesson 1.3 Use Arrays to Multiply

Explore and Grow
Question 1.
Put 24 counters into equal rows. Draw your model.
Answer:
6×4= 24.
rows and 4 columns.

Explanation:
To keep 24 counters into equal rows, we will find the factors of 24 and then we can put 24 counters into equal rows. The factors of 24 are 2,3,4,6,8,12. Now we can pick any of the numbers from the factors of 24 and then put these 24 counters in equal groups. Let’s take six equal rows and arrange 24 counters in four equal rows as shown in the below image.

Question 2.
Put 24 counters into a different number of equal rows. Draw your model.
Answer:

Question 3.
Structure Compare your models. How are the models the same? How are they different?
Answer:

Think and Grow: Multiplication and Arrays

An array is a group of objects organized into rows and columns. Each row has the same number of objects.
Example
How many counters are there in all?

Show and Grow

Question 1.
How many counters are there in all?

Answer:
5 rows and 4 columns,
5+5+5+5= 20,
5×4= 20.
The total number of counters is 20.

Explanation:
From the above image, we can see there are 5 rows and 4 columns. The product is 5×4= 20 and the sum is 5+5+5+5= 20.

Question 2.
Draw an array to multiply 6 × 3.
6 × 3 = ______
Answer:
6 × 3 = 18

Explanation:
To draw an array of  6 × 3, we will place 6 rows and 3 columns. By that, we can form an array of 6×3 which is 18.

Apply and Grow: Practice

Draw an array to multiply.
Question 3.
4 × 8 = _____
Answer:
4 × 8 = 32

Explanation:
To draw an array of  4 × 8, we will place 4 rows and 8 columns. By that, we can form an array of 4×8 which is 32.

Question 4.
3 × 9 = ____
Answer:
3×9= 27

Explanation:
To draw an array of  3×9, we will place 3 rows and 9 columns. By that, we can form an array of 3×9 which is 27.

Question 5.
7 × 3 = _____
Answer:
7 × 3 = 21

Explanation:
To draw an array of  7×3, we will place 7 rows and 3 columns. By that, we can form an array of 7×3 which is 21.

Question 6.
6 × 5 = _____
Answer:
6×5= 30

Explanation:
To draw an array of  6×5, we will place 6 rows and 5 columns. By that, we can form an array of 6×5 which is 30.

Question 7.
Number Sense
Newton has a 2 × 10 array of baseballs. He adds another row. How many baseballs does he add? Write a multiplication equation for his new array.

He adds ______ baseballs
____ × _____ = _____
Answer:
Newton adds 30 baseballs.
3×10= 30.

Explanation:
As Newton has an array of 2 × 10 of baseballs which are 20 baseballs as he adds another row, which means 3×10 of baseballs which are 30 baseballs. So he adds 30 and 20 which is 10 baseballs Newton was added and the multiplication equation for his new array is 3×10 which is 30 baseballs.

Question 8.
DIG DEEPER!
Use 6 counters to make as many different arrays as possible using all of the counters. Draw the arrays. Then write a multiplication equation for each array.

Answer:
2×3= 6,
3×2= 6.

Explanation:
The different possible arrays for 6 counters are 2×3= 6 which has 2 rows and 3 columns and 3×2= 6 which has 3 rows and 2 columns.
This array consists of 2 rows and 3 columns

  This array consists of 3 rows and 2 columns.

Think and Grow: Modeling Real Life

A phone has 6 rows of apps with 4 apps in each row. How many apps are on the phone?
Draw:
Equation:
There are _____ apps on the phone.

Answer:
6×4= 24

Explanation:
As a phone has 6 rows of apps with 4 apps in each row, The number of apps on the phone are 6×4= 24.

Show and Grow

Question 9.
Your classroom has 3 rows of desks with 10 desks in each row. How many desks are in your classroom?
Answer:
3×10= 30 desks.

Explanation:
The number of rows in the classroom is 3 rows with 10 desks in each row, which means 3×10= 30 desks in the classroom.

Question 10.
DIG DEEPER!
A square array has an equal number of rows and columns. A farmer has 9 corn seeds to plant in a square array. Draw the square array the farmer can use to plant all of the seeds. How many rows and columns are there?

Answer:
3×3= 9.

Explanation:
As the farmer has 9 corn seed to plant in a square array, so the possible square array is with 3 rows and 3 columns, which is 3×3= 9.

Use Arrays to Multiply Homework & Practice 1.3

Question 1.
How many counters are there in all?

Answer:
4×9= 36 counters.
4 rows, 9 columns.
9+9+9+9= 36.

Explanation:
In the above image, we can see 4 rows and 9 columns which makes 36 counters. The product is 4×9= 36 and the sum is 9+9+9+9= 36.

Draw an array to multiply.
Question 2.
9 × 2 = _____
Answer:
9×2= 18.

Explanation:
This array contains 9 rows and 2 columns which makes 18 counters. The product is 9×2= 18 and the sum is
9+9= 18.

Question 3.
4 × 5 = _____
Answer:
4×5= 20.

Explanation:
This array contains 4 rows and 5 columns which makes 20 counters. The product is 4×5= 20 and the sum is
5+5+5+5= 20.

Question 4.
YOU BE THE TEACHER
Descartes has 24 counters. He says he can use all the counters to make an array with 3 rows. Is he correct? Explain.

Answer:
With 3 rows and 8 columns, 24 counters can be made.

Explanation:
Yes, he is correct. With 3 rows and 8 columns, he can make 24 counters. The product of the array is 3×8= 24 and the sum of the array is 8+8+8= 24.

Question 5.
Number Sense
Newton has a 4 × 8 array of dominoes. He adds 2 more rows. How many dominoes does he add? Write a multiplication equation for his new array.
He adds _____ dominoes
_____ × ____ = _____
Answer:
Newton adds 16 dominoes,
6×8= 48.

Explanation:
As Newton has a 4×8 array of dominoes which is 32 dominoes and he adds 2 more rows, which is 4+2= 6. Then the dominoes will be a 6×8 array. And the multiplication equation for Newton’s new array is 6×8= 48. So the number of dominoes added is 48-32 which is 16 dominoes he was added.

Question 6.
Modeling Real Life
An art teacher hangs 2 rows of paintings with 10 paintings in each row. How many paintings does she hang?
Answer:
20 paintings.

Explanation:
As an art teacher hangs 2 rows of paintings in each row, so the number of paintings is 2×10= 20 paintings.

Question 7.
DIG DEEPER!
A museum has 16 shark teeth to display in a square array. Draw the square array the museum can use to display all of the teeth. How many rows and columns are there?

Answer:
There are 4 rows and 4 columns.

Explanation:
As the museum has 16 shark teeth to display in a square array, so the array can be written as 4×4= 16 which has 4 rows and 4 columns.

Review & Refresh

Complete the equation
Question 8.
3 + 8 = 8 + ____
Answer:
3+8= 8+3

Explanation:
By the commutative property of addition, we can change the order of the addends which does not change the sum. So 3+8= 8+3.

Question 9.
10 + 0 = ____ + 10
Answer:
10+0= 0+10.

Explanation:
By the commutative property of addition, we can change the order of the addends which does not change the sum. So 10+0= 0+10.

Question 10.
6 + ____ = 7 + 6
Answer:
6+7= 7+6.

Explanation:
By the commutative property of addition, we can change the order of the addends which does not change the sum. So 6+7= 7+6.

Question 11.
____ + 8 = 8 + 9
Answer:
9+8= 8+9.

Explanation:
By the commutative property of addition, we can change the order of the addends which does not change the sum. So 9+8= 8+9.

Lesson 1.4 Multiply in Any Order

Explore and Grow
Question 1.
Write the multiplication equation for the array. Turn your paper and write the equation for the array.

Answer:
5×3= 15.

Explanation:
In the above image, we can see 5 rows and 3 columns. So the product of the array is 5×3= 15.

Structure
Compare the equations. How are they the same? How are they different?

Think and Grow: Commutative Property of Multiplication

Ina multiplication equation, the numbers that are multiplied are called factors. The answer is called the product.

Commutative Property of Multiplication: Changing the order of factors does not change the product.
Example
Complete the statements

Show and Grow

Question 1.
Complete the statements

Answer:
5×2= 2×5.

Explanation:
In the first image, we can see 5 rows and 2 columns, this means 5 rows of 2. And in the second image, we can 2 rows and 5 columns, this means 2 rows of 5. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 5×2= 2×5.

Question 2.
Draw an array to show the Commutative Property of Multiplication. Complete the statements.

____ × ____ = _____
_____ × ____ = _____
So, ____ × _____ = ____ × ____
Answer:
2×3= 6,
3×2= 6.
So, 2×3= 3×2.

Explanation:
In the above image, we can see 2 rows and 3 columns, this means 2 rows of 3. And we can also write as 3 rows and 2 columns as we can see in the given below image. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 2×3= 3×2.

Apply and Grow: Practice

Draw an array to show the Commutative Property of Multiplication. Complete the statements.
Question 3.

Answer:
1×4= 4,
4×1= 4.
So, 1×4= 4×1.

Explanation:
In the above image, we can see 1 row and 4 columns. Which makes 1 row of 4. And we can also write as 4 rows and 1 column as we can see in the given below image. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 1×4= 4×1.

Question 4.

Answer:
2×6= 12,
6×2= 12,
2×6= 6×2.

Explanation:
In the above image, we can see 2 rows and 6 columns. Which makes 2 rows of 6. And we can also write as 6 rows and 2 columns as we can see in the given below image. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 2×6= 6×2.

Question 5.

Answer:
4×5= 20,
5×4= 20.
So, 4×5= 5×4.

Explanation:
In the above image, we can see 4 rows and 5 columns. Which makes 4 rows of 5. And we can also write as 5 rows and 4 columns as we can see in the given below image. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 4×5= 5×4.

Complete the equation.
Question 6.
8 × 3 = 3 × ____
Answer:
8×3= 24,
3×8= 24.
So, 8×3= 3×8.

Explanation:
By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can also write as 3 rows and 8 columns as we can see in the given below image which is 8×3= 3×8.

Question 7.
10 × 2 = ____ × 10
Answer:
10×2 = 20,
2×10= 20.
So, 10×2= 2×10.

Explanation:
By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can also write as 2 rows and 10 columns as we can see in the given below image which is 10×2= 2×10.

Question 8.
1 × ____ = 9 × 1
Answer:
1×9= 9,
9×1= 9.
So, 1×9= 9×1.

Explanation:
By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can also write as 9 rows and 1 column which is 1×9= 9×1.

Question 9.
Structure
Which shape completes the equation?

Answer:
Moon shape.

Explanation:
By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, the missing shape is the moon.

Think and Grow: Modeling Real Life

Your friend makes 7 rows of 6 stickers. You want to put the same number of stickers into 6 rows. How many stickers do you put in each row? Explain.
You put ____ stickers in each row.
Explain.

Answer:
7×6= 42,
6×7= 42.
So, 7×6= 6×7.
You can put 7 stickers in each row.

Explanation:
The number of stickers made is 7 rows of 6 stickers and want to put the same number of stickers into 6 rows. So, by the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can make 6 rows of 7 stickers, which has 6 rows and 7 columns.

Show and Grow

Question 10.
Your friend makes 9 rows of 4 award ribbons. You want to put the same number of award ribbons into 4 rows. How many award ribbons do you put in each row? Explain

Answer:
9×4= 36,
4×9= 36.
So, 9×4= 4×9.
So 9 award ribbons can be put in each row.

Explanation:
Given, 9 rows of 4 award ribbons and we need to put the same number of award ribbons into 4 rows. So by the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So, we can make 4 rows of 9 award ribbons, which has 4 rows and 9 columns and 9 award ribbons can be put in each row.

Question 11.
DIG DEEPER!
You have 2 rows of 8 toy cars. Your friend has 5 rows of 2 toy cars. How can you use the Commutative Property of Multiplication to find how many rows your friend needs to add so that you both have the same number of toy cars?

Answer:

Explanation:

Multiply in Any Order Homework & Practice 1.4

Question 1.
Complete the statements.

Answer:
3 rows of 6,
3×6= 18.
6 rows of 3
6×3= 18,
So, 3×6= 6×3.

Explanation:
In the first image, we can see 3 rows and 6 columns, this means 3 rows of 6. And in the second image, we can 6 rows and 3 columns, this means 6 rows of 3. By the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. So 3×6= 6×3.

Question 2.
Draw an array to show the Commutative Property of Multiplication. Complete the statements.

Answer:
7×4= 28,
4×7= 28.
So, 7×4= 4×7

Explanation:
In the image, we can see 7 rows and 4 columns, this means 7 rows of 4. And by the commutative property of multiplication, we can state that the order in which we multiply the numbers does not change the product. To draw an array by the Commutative Property of Multiplication, we will take 4 rows of 7, which means 4 rows and 7 columns as in the below image. So, 7×4= 4×7.

Which two arrays can you use to show the Commutative Property of Multiplication?

Answer:
The first image and third image shows the commutative property of multiplication.

Explanation:
From the above image, the first and third arrays show the Commutative property of multiplication. Which states that the order in which we multiply the numbers does not change the product. As in the first image, we can see 4 rows of 3 which have 4 rows and 3 columns and in the third image, we can see 3 rows of 4 which has 3 rows and 4 columns.

Question 4.
Precision
Write two equations that show the Commutative Property of Multiplication.
____ × ____ = ____ × _____
____ × _____ = ____ × ____
Answer:
5×6= 6×5,
2×3= 3×2.

Explanation:
The Commutative property of multiplication states that the order in which we multiply the numbers does not change the product. For example, if we take 5×6= 30 then the commutative property is 6×5= 30.
So 5×6= 6×5.

Question 5.
Modeling Real Life
A computer lab has 6 rows of 5 computers. A technology teacher wants to rearrange the computers into5 rows. How many computers does the teacher put in each row? Explain.
Answer:
6 computers do the teacher put in each row.

Explanation:
As a computer lab has 6 rows of 5 computers, and the technology teacher rearranged the computers into 5 rows, so the teacher will put 6 computers by commutative property as it states that the order in which we multiply the numbers does not change the product. So 6×5= 5×6.

Question 6.
DIG DEEPER!
You have 6 rows of 4 pennies. Your friend has 2 rows of 6 pennies. How many rows does your friend need to add so that you both have the same number of pennies?

Answer:
So, a friend needs 2 rows to add then both will have the same number of pennies.

Explanation:
There are 6 rows of 4 pennies, which is 6 rows and 4 columns and the friend has 2 rows of 6 pennies which is 2 rows and 6 columns. So my friend needs to add 2 rows and then it will be 2 rows+2 rows= 4 rows and there will be 6 columns. So, by the commutative property, as it states that the order in which we multiply the numbers does not change the product. So, 6×4= 4×6.

Review & Refresh

Question 7.
Newton hits a ball 5 fewer times than Descartes does. Newton hits the ball 9 times. How many times does Descartes hit the ball?

Answer:

Explanation:

Lesson 1.5 Divide: Size of Equal Groups

Explore and Grow

Question 1.
Put 18 counters in 6 equal groups. Draw to show your groups
Number of counters in each group: _____
Answer:
The number of counters in each group is 3 counters.

Explanation:
To put 18 counters in 6 equal groups, we will divide 18 by 6 then the result is 3. So, we can put counters in 3 counters in 6 equal groups as shown in the given below image.

Question 2.
Put 18 counters in 3 equal groups. Draw to show your groups
Number of counters in each group: _____
Answer:
The number of counters in each group is 6.

Explanation:
To put 18 counters in 3 equal groups, we will divide 18 by 3 then the result is 6. So, we can put counters in 6 counters in 3 equal groups as shown in the given below image.

Structure
How does changing the number of equal groups change the number of counters in each group?
Answer:

Explanation:

Think and Grow: Using Equal Groups to Divide

Division is an operation that gives the size of equal groups or the number of equal groups. When you know the total number of objects and the number of equal groups, you can divide to find the size of each group.3 equal groups
Example
Divide 12 counters into. How many counters are in each group?

Show and Grow

Question 1.
Divide 15 counters into5 equal groups. How many counters are in each group?

15 ÷ 5 = ____

Answer:
The number of counters placed in 5 equal groups is 3 counters.

Explanation:
There are 15 counters and we need to place those 15 counters in 5 equal groups. So we will divide 15 by 5, 15÷5= 3. So we will pace 3 counters in 5 equal groups.

Use the tape diagram to model the equation.

Answer:
The number of counters placed in 5 equal groups is 3 counters.

Explanation:
There are 15 counters and we need to place those 15 counters in 5 equal groups. So we will divide 15 by 5, 15÷5= 3. So we will pace 3 counters in 5 equal groups.

Apply and Grow: Practice

Question 2.
Divide 30 counters into 6 equal groups. How many counters are in each group?

30 ÷ 6 = ____

Answer:
The number of counters placed in 6 equal groups is 5 counters. AS 30÷6= 5.

Explanation:
There are 30 counters and we need to place those 30 counters in 6 equal groups. So we will divide 30 by 6,
30÷6= 5. So we will pace 5 counters in 6 equal groups.

Use the tape diagram to model the equation.

Answer:
The number of counters placed in 6 equal groups is 5 counters. AS 30÷6= 5.

Explanation:
There are 30 counters and we need to place those 30 counters in 6 equal groups. So we will divide 30 by 6,
30÷6= 5. So we will pace 5 counters in 6 equal groups.

Question 3.
Divide 16 counters into 2 equal groups. How many counters are in each group?
16 ÷ 2 = ____
Answer:
16÷2= 8.

Explanation:
There are 16 counters and we need to place those 16 counters in 2 equal groups. So we will divide 16 by 2,
16÷2= 8. So we will pace 8 counters in 2 equal groups.

Question 4.
Divide 9 counters into 3 equal groups. How many counters are in each group?
9 ÷ 3 = ____
Answer:
9 ÷ 3= 3.

Explanation:
There are 9 counters and we need to place those 9 counters in 3 equal groups. So we will divide 9 by 3,
9÷3= 3. So we will pace 3 counters in 3 equal groups.

Question 5.
Structure
Write the division equation that matches the tape diagram.

____ ÷ ____ = ____
Answer:
18÷2= 9.

Explanation:
There are 18 counters and we need to place those 18 counters in 2 equal groups. So we will divide 18 by 2,
18÷2= 9. So we will pace 9 counters in 2 equal groups.

Question 6.
DIG DEEPER!
Newton has a tennis ball collection. He can divide the balls into3 equal groups with none left over. He can also divide the balls into4 equal groups with none left over. How many tennis balls does he have?
Answer:

Think and Grow: Modeling Real Life

You have 30 seashells. You put an equal number of seashells in 5 bags. How many seashells are in each bag?

Model:
Division equation:
There are _____ seashells in each bag.

Answer:
Division equation: 30÷5= 6.
There are 6 seashells in each bag.

Explanation:
As there are 30 seashells and we need to put an equal number of seashells in 5 bags, we will divide 30÷5= 6. So we will 6 seashells in each bag.

Show and Grow

Question 7.
You have 28 rocks. You put an equal number of rocks in 4 piles. How many rocks are in each pile?
Answer:
28÷4= 7.

Explanation:
There are 28 rocks and we should put an equal number of rocks in 4 piles, we will divide 28 by 4. So 28÷4= 7. So there will be 7 rocks in each pile.

Question 8.
DIG DEEPER!
Newton and Descartes each have40 quarters. Newton puts his quarters into5 equal groups. Descartes puts his quarters into4 equal groups. Who has more quarters in each group?

Answer:
Descartes has more quarters than Newton.

Explanation:
As both Newton and Descartes have 40 quarters and Newton puts his quarters into 5 equal groups, which means 40÷5= 8. Newton put his 8 quarters into 5 equal groups and Descartes puts his quarters into 4 equal groups, which means 40÷4= 10. Descartes put his 10 quarters into 4 equal groups. So, Descartes has more quarters than Newton.

Divide: Size of Equal Groups Homework & Practice 1.5

Question 1.
Divide 16 counters into 4 equal groups. How many counters are in each group?

16 ÷ 4 = ____

Answer:
16÷4= 4.

Explanation:
There are 16 counters and we need to place those 16 counters in 4 equal groups. So we will divide 16 by 4,
16÷4= 4. So we will pace 4 counters in 4 equal groups.

Use the tape diagram to model the equation.

Answer:
16÷4= 4.

Explanation:
There are 16 counters and we need to place those 16 counters in 4 equal groups. So we will divide 16 by 4,
16÷4= 4. So we will pace 4 counters in 4 equal groups.

Question 2.
Divide 28 counters into 7 equal groups. How many counters are in each group?
28 ÷ 7 = ____
Answer:
28 ÷ 7= 4.

Explanation:
There are 28 counters and we need to place those 28 counters in 7 equal groups. So we will divide 28 by 7,
28 ÷ 7= 4. So we will pace 4 counters in 7 equal groups.

Question 3.
Divide 27 counters into 3 equal groups. How many counters are in each group?
27 ÷ 3 = ____
Answer:
27 ÷ 3 = 9.

Explanation:
There are 27 counters and we need to place those 27 counters in 3 equal groups. So we will divide 27 by 3,
27 ÷ 3= 9. So we will pace 9 counters in 3 equal groups.

Question 4.
YOU BE THE TEACHER
Newton says you divide 18 counters into 3 equal groups. Is he correct?

Answer:
Yes, Newton is correct.

Explanation:
Yes, Newton is correct. As we can see in the above image he divided 18 counters into 3 equal groups with different numbers of counters in each group.

Question 5.
Precision
A class has 14 boys and 18 girls. Can the teacher divide the class equally into 4 groups with no students remaining? Explain.
Answer:
The teacher places 8 students in 4 equal groups with no students remaining.

Explanation:
As a class has 14 boys and 18 girls, so the total number of students is 14+18= 32 students. As the teacher divided the class equally into 4 groups, 32÷4= 8. So the teacher places 8 students in each group with no students remaining.

Question 6.
Modeling Real Life
You have 14 erasers. You and your friend share them equally. How many erasers do you and your friend each get?

Answer:
7 erasers will get each.

Explanation:
There are 14 erasers and those erasers are shared equally among two of them, so we will divide 14 by 2 which is 14÷2= 7. So 7 erasers will get each.

Question 7.
DIG DEEPER!
Newton and Descartes each have 42 glow-in-the-dark stickers. Newton divides into 6 equal groups. Descartes divides his into7 equal groups. Who has more stickers in each group?
Answer:
Newton has more stickers than Descartes in each group.

Explanation:
As Newton and Descartes, each have 42 glow-in-the-dark stickers and Newton divides them into 6 equal groups, which is 42÷6= 7. So Newton has 7 glow-in-the-dark stickers in 6 equal groups. And Descartes divides his into7 equal groups which is 42÷7= 6. So Descartes has 6 glow-in-the-dark stickers in 7 equal groups. Newton has more stickers than Descartes in each group.

Review & Refresh

Question 8.

Answer:
41

Explanation:
On adding 26+15 we will get 41.

Question 9.

Answer:
87.

Explanation:
On adding 32+55 we will get 87.

Question 10.

Answer:
61.

Explanation:
On adding 49+12 we will get 61.

Question 11.

Answer:
92.

Explanation:
On adding 24+68 we will get 92.

Lesson 1.6 Divide: Number of Equal Groups

Explore and Grow
Question 1.
Put 24 counters in equal groups of 4. Draw to show your groups.
Number of groups: ______

Answer:
The number of groups is 6.

Explanation:
To put 24 counters in equal groups of 4, we will divide 24 by 4, 24÷4= 6. So we will put 4 counters in 6 equal number of group.

Question 2.
Put 24 counters in equal groups of 6. Draw to show your groups.
Number of groups: ______
Answer:
The number of groups is 4.

Explanation:
To put 24 counters in equal groups of 6, we will divide 24 by 6, 24÷6= 4. So we will put 6 counters in 4 equal number of group.

Structure
How does changing the size of the groups change the number of equal groups?

Think and Grow: Using Equal Groups to Divide

When you know the total number of objects and the size of each group, you can divide to find the number of equal groups.
Example
Divide 12 counters into. How many groups are there?

Show and Grow

Question 1.
Divide 10 counters into groups of 2. How many groups are there?

10 ÷ 2 = ____

Answer:
10 ÷ 2 = 5.

Explanation:
To put 10 counters in equal groups of 2, we will divide 10 by 2, 10÷2= 5. So we will put 2 counters in 5 equal number of group.

Use the tape diagram to model the equation.

Answer:
10÷2= 5.

Explanation:
To put 10 counters in equal groups of 2, we will divide 10 by 2, 10÷2= 5. So we will put 2 counters in 5 equal number of group.

Question 2.
Divide 24 counters into groups of 4. How many groups are there?

24 ÷ 4 = ____

Answer:
24÷4= 6.

Explanation:
To put 24 counters in equal groups of 4, we will divide 24 by 4, 24÷4= 6. So we will put 4 counters in 6 equal number of group.

Use the tape diagram to model the equation.

Answer:
24÷4= 6.

Explanation:
To put 24 counters in equal groups of 4, we will divide 24 by 4, 24÷4= 6. So we will put 4 counters in 6 equal number of group.

Apply and Grow: Practice

Question 3.
Divide 30 counters into groups of 6. How many groups are there?

30÷6=

Use the tape diagram to model the equation.

Answer:
30÷6= 5.

Explanation:
To put 30 counters in equal groups of 6, we will divide 30 by 6, 30÷6= 5. So we will put 5 counters in 6 equal number of group.

Question 4.
Divide 15 counters into groups of 5. How many groups are there?
15 ÷ 5 = _____
Answer:
15÷5= 3.

Explanation:
To put 15 counters in equal groups of 5, we will divide 15 by 5, 15÷5= 3. So we will put 3 counters in 5 equal number of group.

Question 5.
Divide 16 counters into groups of 4. How many groups are there?
16 ÷ 4 = _____
Answer:
16÷4= 4.

Explanation:
To put 16 counters in equal groups of 4, we will divide 16 by 4, 16÷4= 4. So we will put 4 counters in 4 equal number of group.

Question 6.
Structure
You want to bake as many loaves of banana bread as possible with 12 eggs. Each loaf of bread requires 2 eggs. Which models can you use to find how many loaves of bread you can make?

Answer:

Explanation:

Think and Grow: Modeling Real life
A florist uses 35 roses to make bouquets. Each bouquet has 7 roses. How many bouquets does the florist make?
Equation:
Model:

The florist makes ______ bouquets

Answer:
Equation: 35÷7= 5

Explanation:
As the florist has 35 roses to make bouquets and each bouquet has35 7 roses, so the florist can make 35÷7= 5 bouquets.

Show and Grow

Question 7.
A farmer puts 48 eggs into cartons. He puts 6 eggs in each carton. How many cartons does he use?
Answer:
8 cartons.

Explanation:
As the farmer puts 48 eggs into cartons and in each carton, there are 6 eggs. So the farmer uses 48÷6= 8 cartons.

Question 8.
DIG DEEPER!
Newton uses his subway pass 3 times each day. Descartes uses his pass 2 times each day. Who will use all of his rides first? Explain.

Would your answer change if Newton and Descartes both use their passes 3 times each day? Explain.
Answer:
Newton will finish his rides first then Descartes.
Descartes will use all of his rides first if Descartes use passes 3 times each day.

Explanation:
As Newton has 24 subway rides left and he uses subway pass 3 times each day, so the number of rides left are
24-3= 21. Descartes has 18 subway rides and he uses his pass 2 times each day, so the number of rides left are
18-2= 16. By using repeated subtraction which starts from 24 as given and then we will subtract 3 until we get 0.
24-3= 21
21-3= 18
18-3= 15
15-3= 12
12-3= 9
9-3= 6
6-3= 3
3-3=0.
So, Newton will complete the subway pass by 8 times.
18-2= 16
16-2= 14
14-2= 12
12-2= 10
10-2= 8
8-2= 6
6-2= 4
4-2= 2
2-2= 0.
And Descartes will complete the subway pass by 9 times. So Newton will use all his rides first.
And if Descartes uses 3 times of his pass then
18-3= 15
15-3= 12
12-3= 9
9-3= 6
6-3= 3
3-3= 0.
Descartes will complete the subway pass by 6 times, So Descartes will complete his pass first.

Divide: Number of Equal Groups Homework & Practice 1.6

Question 1.
Divide 28 counters into groups of 4. How many groups are there?

28 ÷ 4 = ____

Answer:
28 ÷ 4 =7

Explanation:
To put 28 counters in equal groups of 4, we will divide 28 by 4, 28÷4= 7. So we will put 4 counters in 7 equal number of group.

Use the tape diagram to model the equation.

Answer:
28÷4= 7.

Explanation:
To put 28 counters in equal groups of 4, we will divide 28 by 4, 28÷4= 7. So we will put 4 counters in 7 equal number of group.

Question 2.
Divide 25 counters into groups of 5. How many groups are there?
25 ÷ 5 = _____
Answer:
25 ÷ 5 = 5.

Explanation:
To put 25 counters in equal groups of 5, we will divide 25 by 5, 25÷5= 5. So we will put 5 counters in 5 equal number of group.

Question 3.
Divide 12 counters into groups of 6. How many groups are there?
12 ÷ 6 = _____
Answer:
12 ÷ 6 = 2.

Explanation:
To put 12 counters in equal groups of 6, we will divide 12 by 6, 12÷6= 2. So we will put 2 counters in 6 equal number of group.

Question 4.
Writing
Write and solve a problem in which you need to find the number of equal groups.
Answer:
Divide 22 counters into groups of 11. How many groups are there?

Explanation:
To put 22 counters in equal groups of 11, we will divide 22 by 11, 22÷11= 2. So we will put 2 counters in 11 equal number of group.

Question 5.
Reasoning
Your classroom has 30 chairs that need to be stacked with 5 chairs in each stack. Your teacher already made 2 stacks. How many stacks of chairs still need to be made?
Answer:
The number of stacks of chairs that still need to be made is 4.

Explanation:
As a classroom has 30 chairs and that needed to be stacked with 5 chairs in each stack, so 30÷5= 6 stacks. And the teacher already made 2 stacks, which means 6-2= 4. So the number of stacks of chairs that still need to be made are 4 stacks.

Question 6.
DIG DEEPER!
You have more than 30 and fewer than 40 piñata toys. You divide them into groups with 8 in each group. How many groups do you make?

Answer:
There will be 4 groups with 8 pinata toys in each group.

Explanation: As there are more than 30 and fewer than 40 pinata toys, and we need to divide them into groups with 8 in each group. So we need to choose the number between 30-40. As we can see fewer than 40 and more than 30, so we will choose 32 as it will divide them into groups with 8 in each group, 32÷8= 4. So there will be 4 groups with 8 pinata toys in each group.

Question 7.
A street vendor puts 42 apples into baskets. She puts 6 apples in each basket. How many baskets does she use?
Answer:
7 baskets.

Explanation:
As a street vendor puts 42 apples into baskets and she puts 6 apples in each basket, so she needs 42÷6= 7 baskets.

Question 8.
DIG DEEPER!
Newton and Descartes are at an amusement park. Newton uses 2 tickets to ride each roller coaster. Descartes uses 3 tickets to ride at the front of each roller coaster. Who runs out of tickets first? Explain.

Newton Descartes, They run out of tickets at the same time

Answer:
Newton will run out of tickets first, as he has fewer tickets.

Explanation:
As Newton has 14 coaster tickets and uses 2 tickets to ride each roller coaster, so there will 14-2= 12 tickets remaining. And Descartes has 21 coaster tickets and uses 3 tickets to ride, so the remaining tickets are 21-3= 18 tickets. Newton will run out of tickets first, as he has fewer tickets.

Review & Refresh

Find the difference. Use addition to check your answer.
Question 9.

Answer:
96-58= 38,
38+58= 96.

Explanation:
On subtracting 96-58 we will get 38 and by adding 58 and 38 we will get 96. So by adding we can check your answer.

Question 10.

Answer:
48-32= 16,
16+32= 48.

Explanation:
On subtracting 48-32 we will get 16 and by adding 32 and 16 we will get 48. So by adding we can check your answer.

Lesson 1.7 Use Number Lines to Divide

Explore and Grow

Question 1.
Find the difference. Use each difference as the starting number in the next equation. Model the problems on the number line.

Answer:
12-3= 9,
9-3= 6,
6-3= 3,
3-3= 0.

Explanation:
To represent a number line, we will use repeated subtraction which starts from 12 as given and then we will subtract 3 until we get 0. So the number of jumps is 4 and the size of the jump is 3.

Question 2.
Structure
How can you use a number line to help you find 12 ÷ 3?
Answer:
12÷3= 4.

Explanation:
To use the number line for the given value 12÷3 which is 4, we will start from 0 then skip the count by 4 three times. So the number of jumps is 4 and the size of the jump is 3. And we can also use the subtraction method, which starts with 12 and subtract 3 until we get 0.

Think and Grow: Number Lines and Repeated Subtraction

Example Find 20 ÷ 4.

Another Way:
Use repeated subtraction. Start with 20. Subtract 4 until you reach 0.

Show and Grow

Complete the equations.
Question 1.
24 ÷ 6 = ______

Answer:
24 ÷ 6 = 4.

Explanation:
By using the number line, we will countback by 6s from 24 until we reach 0,
24÷6= 4, so there are 4 groups of 6.

Question 2.
16 ÷ 8 = _____
16 – 8 = _____
____ – 8 = 0
Answer:
16 ÷ 8 = 3,
16-8= 8,
8-8= 0.

Explanation:
We will use repeated subtraction which starts from 16 as given and then we will subtract 8 until we get 0. So the number of jumps is 2 and the size of the jump is 8.
16 ÷ 8 = 3,
16-8= 8,
8-8= 0.

Question 3.
27 ÷ 9 = _____
27 – 9 = _____
____ – 9 = ____
_____ – 9 = 0
Answer:
27 ÷ 9 = 3,
27-9= 18,
18-9= 9,
9-9= 0.

Explanation:
We will use repeated subtraction which starts from 27 as given and then we will subtract 9 until we get 0. So the number of jumps is 3 and the size of the jump is 9.
27 ÷ 9 = 3,
27-9= 18,
18-9= 9,
9-9= 0.

Apply and Grow: Practice

Complete the equations.
Question 4.
25 ÷ 5 = _____

Answer:
25 ÷ 5 = 5.

Explanation:
By using the number line, we will countback by 5s from 25 until we reach 0,
25÷5= 5, so there are 5 groups of 5.

Question 5.
21 – 7 = _____
____ – 7 = _____
_____ – 7 = 0
____ ÷ _____ = ____
Answer:
21-7= 14,
14-7= 7,
7-7= 0,
21÷7= 3.

Explanation:
we will use repeated subtraction which starts from 21 as given and then we will subtract 7 until we get 0.
21-7= 14,
14-7= 7,
7-7= 0,
21÷7= 3.

Question 6.
36 – 9 = _____
____ – 9 = ____
____ – 9 = ____
_____ – 9 = 0
____ ÷ ____ = ____
Answer:
36-9= 27,
27-9= 18,
18-9= 9,
9-9= 0.
36÷9= 4.

Explanation:
we will use repeated subtraction which starts from 36 as given and then we will subtract 9 until we get 0.
36-9= 27,
27-9= 18,
18-9= 9,
9-9= 0.
36÷9= 4.

Question 7.
YOU BE THE TEACHER
Descartes uses a number line to find 18 ÷ 2. Is he correct? Explain.

Explanation:
No, Descartes is not correct. As 18÷2= 9 and here the given result 8. So Descartes is incorrect.

Think and Grow: Modeling Real life

Each age group is divided into teams with 6 players on each team. Each team receives a trophy at the end of the season. How many trophies are needed?

Division equations:
Addition equations:

_____ trophies are needed

Answer:
7 trophies are needed.

Explanation:
As each age group is divided into teams with 6 players on each team and each team receives a trophy at the end of the season, so the number of trophies needed for age group 6-7 years old is 18÷6= 3 trophies. And for the age group 8-9 years old is 24÷6= 4 trophies.

Show and Grow

Question 8.
Each age group is put into cabins with 8 campers in each cabin. How many cabins are needed?

The two age groups are combined into one group. Does the total number of cabins that are needed change? Explain

Answer:
The number of cabins needed for the age group of 5-7 years is 4 cabins and the number of cabins needed for the age group of 8-10 years old is 3 cabins. There will be no change in the total number of cabins.

Explanation:
For the age group of 5-7 years old number of cabins needed are 32÷8= 4 cabins. And for the age group 8- 10 years old number of cabins needed are 24÷8= 3 cabins. If two age groups are combined into one group then the total number of campers are 24+32= 56. So the total number of cabins needed are 56÷8= 7 cabins. So there will be no change.

Use Number Lines to Divide Homework & Practice 1.7

Complete the equations

Question 1.
9 ÷ 3 = _____

Answer:
9÷3= 3.

Explanation:
By using the number line, we will countback by 3s from 9 until we reach 0,
9÷3= 3, so there are 3 groups of 3.

Question 2.
20 – 5 = ____
____ – 5 = ____
____ – 5 = ____
____ – 5 = 0
____ ÷ ____ = ____
Answer:
20-5= 15,
15-5= 10,
10-5= 5,
5-5= 0,
20÷5= 4.

Explanation:
we will use repeated subtraction which starts from 20 as given and then we will subtract 5 until we get 0.
20-5= 15,
15-5= 10,
10-5= 5,
5-5= 0,
20÷5= 4.

Question 3.
10 – 2 = ____
____ – 2 = ____
____ – 2 = ____
____ – 2 = ____
____ – 2 = 0
____ ÷ ____ = ____
Answer:
10-2 = 8,
8-2= 6,
6-2= 4,
4-2= 2,
2-2= 0,
10÷5= 5.

Explanation:
we will use repeated subtraction which starts from 10 as given and then we will subtract 2 until we get 0.
10-2 = 8,
8-2= 6,
6-2= 4,
4-2= 2,
2-2= 0,
10÷5= 5.

Question 4.
YOU BE THE TEACHER
Descartes uses repeated subtraction to find 15 ÷ 5. Is he correct? Explain.

Answer: Yes, he is correct.

Explanation:
As Descartes uses repeated subtraction which starts from 15 as given and then we will subtract 5 until we get 0. So Descartes is correct.

Question 5.
DIG DEEPER!
Find Newton’s missing number. Explain how you solved.

Answer:
18-6= 12,
12-6= 6,
6-6 = 0.

Explanation:
Here, given subtract 6 from a number 3 times, and reach 0, so we will multiply 6 by 3 and we will get a number,
which is 6×3= 18. So the number is 18 if we subtract 6, 3 times we will reach 0.
18-6= 12,
12-6= 6,
6-6 = 0.

Question 6.
Modeling Real Life
Each age group is divided into groups of 7 swimmers. How many groups are there in each age group?

The two age groups are combined into one group. Does the total number of groups change? Explain.
Answer:
By combining two age groups into one group the total number of groups does not change, as the total number of groups and the combined two age group is equal.

Explanation:
As each age was divided into groups of 7 swimmers, so the number of groups in the 6-8 years old age group is 28÷7= 4 groups. And the number of groups in the 9-11 years old age group is 14÷7= 2 groups. And the total number of groups is 4+2= 6. As the two age groups are combined into one group, so 28+14= 42. So the total number of groups will be 42÷7= 6 groups.

Review & Refresh

Question 7.
Circle the values of the underlined digit.

Answer:
8 tens.

Explanation:
The value of 8 in the digit 581 is 8 tens.

Understand Multiplication and Division Performance Task

Question 1.
a. Your science teacher gives you 71 picture cards to sort into categories, living and nonliving. You sort 11 cards into the nonliving category. How many cards do you sort into the living category?

b. You sort the living cards into 2 categories, plants, and animals. The numbers of cards in each category are equal. How many cards are in the animal category?
c. You divide the animal cards into 6 equal groups. How many animal cards are in each group?
d. The animals in 5 of the groups have backbones, and the animals in the other group do not have backbones. How many more cards have animals with backbones? Explain.

Answer:
a) 60.
b) 30 cards in the animal category.
c) 5 animal cards.
d)

Explanation:
a) As there are 71 picture cards and 11 of them are sorted into the non-living category, so the remaining cards in the living category are 71-11= 60.
b) As the living cards are 60 and those are divided into two equal categories 60÷2= 30, which is 30 plants and 30 animal cards. So there are 30 cards in the animal category.
c) To divide the animal cards into 6 equal groups, we will divide 30 by 6 and there will be 30÷6= 5  animal cards are in each group.
d)

Understand Multiplication and Division Activity

Hooray Array!

Getting Started: Fill in your board with each number from the Number List. You may write each number in any square. Each square can only have one number.
Directions:
1. Choose a player to be the caller. The caller selects a Hooray Array Equation Card and reads the equation.
2. All players solve the equation and place a counter on the answer. Cover only 1 number per turn.
3. Repeat the process with players taking turns as the caller.
4. The winner is the first player who creates a 3 × 3 array on the board and yells, “HOORAY ARRAY!”

Answer:

Understand Multiplication and Division Chapter practice

1.1 Use Equal Groups to Multiply

Question 1.
Use the model to complete the statements.

Answer:
2 groups of 3,
3+3= 6,
2×3= 6.

Explanation:
Here, we can see 2 groups with 3 counters in each group.
So, 2 groups of 3 which is 6, which means 2×3= 6.
And if we add 2 groups of the counter, we will get the  same result,
which is 3+3= 6.

Draw equal groups. Then complete the equations.

Question 2.
3 groups of 6

Answer:
3 groups of 6,
2+2+2= 6,
3×6= 18.

Explanation:
Here, we can see 3 groups with 6 counters in each group.
So, 3 groups of 6 which are 18, which means 3×6= 18.
And if we add 3 groups of the counter, we will get the  same result,
which is 2+2+2= 6.

Question 3.
4 groups of 5

Answer:
4 groups of 5,
5+5+5+5= 20,
4×5= 20.

Explanation:
Here, we can see 4 groups with 5 counters in each group.
So, 4 groups of 5 which is 20, which means 4×5= 20.
And if we add 4 groups of the counter, we will get the  same result,
which is 5+5+5+5= 20.

1.2 Use Number Lines to Multiply

Question 4.
Find 8 × 3
Number of jumps: ___
Size of each jump: ___

8 × 3 = ____
Answer:
8×3= 24.

Explanation:
Here, we will start at 0 and then we will skip count by 3s eight times. So, the number of jumps is 8, and the size of each jump is 3. which is 8×3= 24.

1.3 Use Arrays to Multiply

Draw an array to multiply.

Question 5.
2 × 8 = ___

Answer:
2 × 8 = 16.

Explanation:
To draw an array of 2×8, we will take 2 rows and 8 columns to build an array.

Question 6.
7 × 4 = ___
Answer:
7 × 4 = 28

Explanation:
To draw an array of 7×4, we will take 7 rows and 4 columns to build an array.

Question 7.
YOU BE THE TEACHER
Newton has 32 counters. He says that he can use all the counters to make an array with 6 rows. Is he correct? Explain.

Answer: No, Newton is not correct.

Explanation:
No, as Newton has 32 counters which will not make an array with 6 rows. As 32 is not divisible 6, so he is not correct.

1.4 Multiply in Any Order

Question 8.
Draw an array to show the Commutative Property of Multiplication. Complete the statements.

____ × ____ = ____
____ × ____ = ____
So, ___ × ___ = ____ × ___

Answer:
9×5 = 45,
5×9 = 45,
So, 9×5 = 5×9.

Explanation:
In the above image, we can see 9 rows of 5,
which means 9×5 = 45,
so by the commutative property of multiplication,
9×5 = 5×9,
which is 45.

Complete the equation.

Question 9.
4 × 10 = 10 × ___
Answer: 4.

Explanation:
By the commutative property of multiplication, which means changing the order of factors which does not change the product. So,
4 × 10 = 10 × 4.

Question 10.
3 × 9 = ____ × 3
Answer: 9

Explanation:
By the commutative property of multiplication, which means changing the order of factors which does not change the product. So,
3 × 9 = 9 × 3.

Question 11.
8 × ___ = 4 × 8.
Answer: 4

Explanation:
By the commutative property of multiplication, which means changing the order of factors which does not change the product. So,
8 × 4= 4 × 8.

1.5 Divide: Size of Equal Groups

Question 12.
Divide 27 counters into 3 equal groups. How many counters are in each group?

Use the tape diagram to model the equation.

27 ÷ 3 = ___
Answer: 9 counters in each group.

Explanation:
Given Counters is 27,
now, we will divide 27 counters into groups of 3,
which means
27÷3= 9.
On dividing 27 counters by 3,
we will get the result as 9 groups.

Question 13
Divide 32 counters into4 equal groups. How many counters are in each group?
32 ÷ 4 = ___
Answer: 8 counters in each group.

Explanation:
Given Counters is 32,
now, we will divide 32 counters into groups of 4,
which means
32÷4= 8.
On dividing 32 counters by 4,
we will get the result as 8 counters in each group.

Question 14.
Divide 12 counters into4 equal groups. How many counters are in each group?
12 ÷ 4 = ___
Answer: 3 counters in each group.

Explanation:
Given Counters is 12,
now, we will divide 12 counters into groups of 4,
which means
12÷4= 3.
On dividing 12 counters by 4,
we will get the result as 3 counters in each group.

1.6 Divide: Number of Equal Groups

Question 15.
Divide 15 counters into groups of 3. How many groups are there?
15 ÷ 3 = ___
Answer: 5 groups.

Explanation:
Given Counters is 15,
now, we will divide 15 counters into groups of 3
which means
15÷3= 5.
On dividing 15 counters by 3,
we will get the result in 5 groups.

Question 16.
Divide 20 counters into groups of 2. How many groups are there?
20 ÷ 2 = __
Answer: 10 groups.

Explanation:
Given Counters is 20,
now, we will divide 20 counters into groups of 2,
which means
20÷2= 10.
On dividing 20 counters by 2,
we will get the result as 10 groups.

Question 17.
Modeling Real Life
Newton and Descartes are trying a new music app. Newton uses 4 credits a day to hear songs without commercials. Descartes uses 2 credits a day to hear songs with commercials. Who runs out of credits first? Explain.


Answer:
Newton will run out of credits first than Descartes. No, they will not run out of credits at the same time.

Explanation:
The total credits do Newton has in a new music app are 24 and he uses 4 credits a day to hear songs without commercials, so the remaining credits are 24-4= 20. And the total credits Descartes had are 14 and he uses 2 credits a day to hear songs with commercials, which is 14-2= 12. we will use repeated subtraction which starts from 24 as given and then we will subtract 4 until we get 0 which is
24-4= 20
20-4= 16
16-4= 12
12-4= 8
8-4= 4
4-4= 0
so, Newton will run out of credits by his 6 uses. And we will use repeated subtraction for Descartes’s credits, which is
14-2= 12
12-2= 10
10-2= 8
8-2= 6
6-2= 4
4-2= 2
2-2= 0
so, Descartes will run out of credits by his 7 uses. So Newton will run out of credits earlier than Descartes.

1.7 Use Number Lines to Divide

Complete the equations.

Question 18.
28 ÷ 7 = ___

Answer:
28 ÷ 7 = 4.

Explanation:
By using the number line, we will countback by 7s from 28 until we reach 0,
28÷7= 4, so there are 4 groups of 7.

Question 19.
18 – 9 = ___
___ – 9 = 0
___ ÷ ___ = ___
Answer:
18-9= 0,
9-9= 0
18÷9= 2

Explanation:
We will use repeated subtraction which starts with 18 as given and then we will subtract 9 until we get 0. So the number of jumps is 2 and the size of the jump is 9.

Question 20.
12 – 4 = ___
___ – 4 = ___
___ – 4 = 0
___ ÷ ___ = ___
Answer:
12-4= 8
8-4= 4
4-4= 0
12÷4= 3.

Explanation:
We will use repeated subtraction which starts with 12 as given and then we will subtract 4 until we get 0. So the number of jumps is 3 and the size of the jump is 4.

Final Words:

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Big Ideas Math Book Algebra 2 Answer Key Chapter 5 Rational Exponents and Radical Functions

Students can get detailed solutions for all questions included in BigIdeas Math Book Algebra 2 Chapter 5 Answer Key. The BIM Textbook Answers Algebra 2 Ch 5 Rational Exponents and Radical Functions covers various preparation resources like Topic-wise exercise questions, chapter tests, reviews, practices, quiz, cumulative assessment, etc. to understand the concepts and learn efficiently. Therefore, click on the links available below and quickly access the respective exercise questions of BIM Algebra 2 Ch 5 Rational Exponents and Radical Functions for better preparation.

• Rational Exponents and Radical Functions Maintaining Mathematical Proficiency – Page 235
• Rational Exponents and Radical Functions Mathematical Practices – Page 236
• Lesson 5.1 nth Roots and Rational Exponents – Page(238-242)
• nth Roots and Rational Exponents 5.1 Exercises – Page(241-242)
• Lesson 5.2 Properties of Rational Exponents and Radicals – Page(244-250)
• Properties of Rational Exponents and Radicals 5.2 Exercises – Page(248-250)
• Lesson 5.3 Graphing Radical Functions – Page(252-258)
• Graphing Radical Functions 5.3 Exercises – Page(256-258)
• Rational Exponents and Radical Functions Study Skills – Page 259
• Rational Exponents and Radical Functions 5.1 – 5.3 Quiz – Page 260
• Lesson 5.4 Solving Radical Equations and Inequalities – Page(262-268)
• Solving Radical Equations and Inequalities 5.4 Exercises – Page(266-268)
• Lesson 5.5 Performing Function Operations – Page(270-274)
• Performing Function Operations 5.5 Exercises – Page(273-274)
• Lesson 5.6 Inverse of a Function – Page(276-284)
• Inverse of a Function 5.6 Exercises – Page(281-284)
• Rational Exponents and Radical Functions Performance Task – Page 285
• Rational Exponents and Radical Functions Chapter Review – Page(286-288)
• Rational Exponents and Radical Functions Chapter Test – Page 289
• Rational Exponents and Radical Functions Cumulative Assessment – Page(290-291)

Rational Exponents and Radical Functions Maintaining Mathematical Proficiency

Simplify the expression

Question 1.
y⁶ • y

Question 2.
\(\frac{n^{4}}{n^{3}}\)

Question 3.
\(\frac{x^{5}}{x^{6} \cdot x^{2}}\)

Question 4.
\(\frac{x^{-6}}{x^{5}}\) • 3x^2/sup>

Question 5.
\(\left(\frac{4 w^{3}}{2 z^{2}}\right)^{3}\)

Question 6.
\(\left(\frac{m^{7} \cdot m}{z^{2} \cdot m^{3}}\right)^{2}\)

Solve the literal equation for y.

Question 7.
4x + y = 2

Question 8.
x − \(\frac{1}{3}\)y = −1

Question 9.
2y − 9 = 13x

Question 10.
2xy + 6y = 10

Question 11.
8x − 4xy = 3

Question 12.
6x + 7xy = 15

Question 13.
ABSTRACT REASONING Is the order in which you apply properties of exponents important? Explain your reasoning.

Rational Exponents and Radical Functions Mathematical Practices

Monitoring Progress

Question 1.
Use the Pythagorean Theorem to find the exact lengths of a, b, c, and d in the figure.

Question 2.
Use a calculator to approximate each length to the nearest tenth of an inch.

Question 3.
Use a ruler to check the reasonableness of your answers.

Lesson 5.1 nth Roots and Rational Exponents

Essential Question

How can you use a rational exponent to represent a power involving a radical?
Previously, you learned that the nth root of a can be represented as
\(\sqrt[n]{a}\) = \(a^{1 / n}\) Definition of rational exponent
for any real number a and integer n greater than 1.

EXPLORATION 1
Exploring the Definition of a Rational Exponent
Work with a partner. Use a calculator to show that each statement is true.
a. \(\sqrt{9}\) = \(9^{1 / 2}\)
b. \(\sqrt{2}\) = \(2^{1 / 2}\)
c. \(\sqrt[3]{8}\) = \(8^{1 / 3}\)
d. \(\sqrt[3]{3}\) = \(3^{1 / 3}\)
e. \(\sqrt[4]{16}\) = \(16^{1 / 4}\)
f. \(\sqrt[4]{12}\) = \(12^{1 / 4}\)

EXPLORATION 2
Writing Expressions in Rational Exponent Form

Work with a partner. Use the definition of a rational exponent and the properties of exponents to write each expression as a base with a single rational exponent. Then use a calculator to evaluate each expression. Round your answer to two decimal places.

a. \((\sqrt{5})^{3}\)
b. \((\sqrt[4]{4})^{2}\)
c. \((\sqrt[3]{9})^{2}\)
d. \((\sqrt[5]{10})^{4}\)
e. \((\sqrt{15})^{3}\)
f. \((\sqrt[3]{27})^{4}\)

EXPLORATION 3
Writing Expressions in Radical Form
Work with a partner. Use the properties of exponents and the definition of a rational exponent to write each expression as a radical raised to an exponent. Then use a calculator to evaluate each expression. Round your answer to two decimal places.
Sample \(5^{2 / 3}\) = (\left(5^{1 / 3}\right)\(\))² = (\((\sqrt[3]{5})\))² ≈ 2.92
a. \(8^{2 / 3}\)
b. \(6^{5 / 2}\)
c. \(12^{3 / 4}\)
d. \(10^{3 / 2}\)
e. \(16^{3 / 2}\)
f. \(20^{6 / 5}\)

Communicate Your Answer

Question 4.
How can you use a rational exponent to represent a power involving a radical?

Question 5.
Evaluate each expression without using a calculator. Explain your reasoning.
a. \(4^{3 / 2}\)
b. \(32^{4 / 5}\)
c. \(625^{3 / 4}\)
d. \(49^{3 / 2}\)
e. \(125^{4 / 3}\)
f. \(100^{6 / 3}\)

5.1 Lesson

Monitoring Progress

Question 1.
n = 4, a = 16

Question 2.
n = 2, a = −49

Question 3.
n = 3, a = −125

Question 4.
n = 5, a = 243

Evaluate the expression without using a calculator.

Question 5.
\(4^{5 / 2}\)

Question 6.
\(9^{-1 / 2}\)

Question 7.
\(81^{3 / 4}\)

Question 8.
\(1^{7 / 8}\)

Evaluate the expression using a calculator. Round your answer to two decimal places when appropriate.

Question 9.
\(6^{2 / 5}\)

Question 10.
\(64^{-2 / 3}\)

Question 11.
\((\sqrt[4]{16})^{5}\)

Question 12.
\((\sqrt[3]{-30})^{2}\)

Find the real solution(s) of the equation. Round your answer to two decimal places when appropriate.

Question 13.
8x³ = 64

Question 14.
\([\frac{1}{2}/latex]x⁵ = 512

Question 15.
(x + 5)⁴ = 16

Question 16.
(x − 2)³ = −14

Question 17.
WHAT IF? In Example 5, what is the annual depreciation rate when the salvage value is $6000?

nth Roots and Rational Exponents 5.1 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY Rewrite the expression [latex]a^{-s / t}\) in radical form. Then state the index of the radical.
Answer:

Question 2.
COMPLETE THE SENTENCE For an integer n greater than 1, if bⁿ = a, then bis a(n) ___________ of a.
Answer:

Question 3.
WRITING Explain how to use the sign of a to determine the number of real fourth roots of a and the number of real fifth roots of a.
Answer:

Question 4.
WHICH ONE DOESN’T BELONG? Which expression does not belong with the other three? Explain your reasoning.

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, find the indicated real nth root(s) of a.

Question 5.
n = 3, a = 8
Answer:

Question 6.
n = 5, a = −1
Answer:

Question 7.
n = 2, a = 0
Answer:

Question 8.
n = 4, a = 256
Answer:

Question 9.
n = 5, a = −32
Answer:

Question 10.
n = 6, a = −729
Answer:

In Exercises 11–18, evaluate the expression without using a calculator.

Question 11.
\(64^{1 / 6}\)
Answer:

Question 12.
\(8^{1 / 3}\)
Answer:

Question 13.
\(25^{3 / 2}\)
Answer:

Question 14.
\(81^{3 / 4}\)
Answer:

Question 15.
\((-243)^{1 / 5}\)
Answer:

Question 16.
\((-64)^{4 / 3}\)
Answer:

Question 17.
\(8^{-2 / 3}\)
Answer:

Question 18.
\(16^{-7 / 4}\)
Answer:

ERROR ANALYSIS In Exercises 19 and 20, describe and correct the error in evaluating the expression.

Question 19.

Answer:

Question 20.

Answer:

USING STRUCTURE In Exercises 21–24, match the equivalent expressions. Explain your reasoning.

Question 21.
\((\sqrt[3]{5})^{4}\) A. \(5^{-1 / 4}\)
Answer:

Question 22.
\((\sqrt[4]{5})^{3}\) B. \(5^{4 / 3}\)
Answer:

Question 23.
\(\frac{1}{\sqrt[4]{5}}\) C. \(-5^{1 / 4}\)
Answer:

Question 24.
\(-\sqrt[4]{5}\) D. \(5^{3 / 4}\)
Answer:

In Exercises 25–32, evaluate the expression using a calculator. Round your answer to two decimal places when appropriate.

Question 25.
\(\sqrt[5]{32,768}\)
Answer:

Question 26.
\(\sqrt[7]{1695}\)
Answer:

Question 27.
\(25^{-1 / 3}\)
Answer:

Question 28.
\(85^{1 / 6}\)
Answer:

Question 29.
\(20,736^{4 / 5}\)
Answer:

Question 30.
\(86^{-5 / 6}\)
Answer:

Question 31.
\((\sqrt[4]{187})^{3}\)
Answer:

Question 32.
\((\sqrt[5]{-8})^{8}\)
Answer:

MATHEMATICAL CONNECTIONS In Exercises 33 and 34, find the radius of the figure with the given volume.

Question 33.
V = 216 ft³

Answer:

Question 34.
V = 1332 cm³

Answer:

In Exercises 35–44, find the real solution(s) of the equation. Round your answer to two decimal places when appropriate.

Question 35.
x³ = 125
Answer:

Question 36.
5x³ = 1080
Answer:

Question 37.
(x + 10)⁵ = 70
Answer:

Question 38.
(x − 5)⁴ = 256
Answer:

Question 39.
x⁵ = −48
Answer:

Question 40.
7x⁴ = 56
Answer:

Question 41.
x⁶ + 36 = 100
Answer:

Question 42.
x³ + 40 = 25
Answer:

Question 43.
\(\frac{1}{3}\)x⁴ = 27
Answer:

Question 44.
\(\frac{1}{6}\)x³ = −36
Answer:

Question 45.
MODELING WITH MATHEMATICS When the average price of an item increases from p₁ to p₂ over a period of n years, the annual rate of inflation r (in decimal form) is given by r = \(\left(\frac{p_{2}}{p_{1}}\right)^{1 / n}\) − 1. Find the rate of inflation for each item in the table.

Answer:

Question 46.
HOW DO YOU SEE IT? The graph of y = xⁿ is shown in red. What can you conclude about the value of n? Determine the number of real nth roots of a. Explain your reasoning.

Answer:

Question 47.
NUMBER SENSE Between which two consecutive integers does \(\sqrt[4]{125}\) lie? Explain your reasoning.
Answer:

Question 48.
THOUGHT PROVOKING In 1619, Johannes Kepler published his third law, which can be given by d³ = t², where d is the mean distance (in astronomical units) of a planet from the Sun and t is the time (in years) it takes the planet to orbit the Sun. It takes Mars 1.88 years to orbit the Sun. Graph a possible location of Mars. Justify your answer. (The diagram shows the Sun at the origin of the xy-plane and a possible location of Earth.)

Answer:

Question 49.
PROBLEM SOLVING A weir is a dam that is built across a river to regulate the flow of water. The flow rate Q (in cubic feet per second) can be calculated using the formula Q= 3.367ℓ\(h^{3 / 2}\), where ℓ is the length (in feet) of the bottom of the spillway and his the depth (in feet) of the water on the spillway. Determine the flow rate of a weir with a spillway that is 20 feet long and has a water depth of 5 feet.

Answer:

Question 50.
REPEATED REASONING The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain river normally flows at a speed of 1 meter per second. What must its speed be in order to transport particles that are twice as massive as usual? 10 times as massive? 100 times as massive?
Answer:

Maintaining Mathematical Proficiency

Simplify the expression. Write your answer using only positive exponents. (Skills Review Handbook)

Question 51.
5 • 5⁴

Answer:

Question 52.
\(\frac{4^{2}}{4^{7}}\)
Answer:

Question 53.
\(\left(z^{2}\right)^{-3}\)
Answer:

Question 54.
\(\left(\frac{3 x}{2}\right)^{4}\)
Answer:

Write the number in standard form. (Skills Review Handbook)

Question 55.
5 × 10³

Answer:

Question 56.
4 × 10⁻²

Answer:

Question 57.
8.2 × 10⁻¹

Answer:

Question 58.
6.93 × 10⁶

Answer:

Lesson 5.2 Properties of Rational Exponents and Radicals

Essential Question
How can you use properties of exponents to simplify products and quotients of radicals?

EXPLORATION 1
Reviewing Properties of Exponents
Work with a partner. Let a and b be real numbers. Use the properties of exponents to complete each statement. Then match each completed statement with the property it illustrates.

EXPLORATION 2
Simplifying Expressions with Rational Exponents
Work with a partner. Show that you can apply the properties of integer exponents to rational exponents by simplifying each expression. Use a calculator to check your answers.
a. \(5^{2 / 3}\) • \(5^{4 / 3}\)
b. \(3^{1 / 5}\) • \(3^{4 / 5}\)
c. \(\left(4^{2 / 3}\right)^{3}\)
d. \(\frac{\sqrt{98}}{\sqrt{2}}\)
e. \(\frac{\sqrt[4]{4}}{\sqrt[4]{1024}}\)
f. \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\)

EXPLORATION 3
Simplifying Products and Quotients of Radicals
Work with a partner. Use the properties of exponents to write each expression as a single radical. Then evaluate each expression. Use a calculator to check your answers.

a. \(\sqrt{3}\) • \(\sqrt{12}\)
b. \(\sqrt[3]{5}\) • \(\sqrt[3]{25}\)
c. \(\sqrt[4]{27}\) • \(\sqrt[4]{3}\)
d. \(\frac{\sqrt{98}}{\sqrt{2}}\)
e. \(\frac{\sqrt[4]{4}}{\sqrt[4]{1024}}\)
f. \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\)

Communicate Your Answer

Question 4.
How can you use properties of exponents to simplify products and quotients of radicals?

Question 5.
Simplify each expression.
a. \(\sqrt{27}\) • \(\sqrt{6}\)
b. \(\frac{\sqrt[3]{240}}{\sqrt[3]{15}}\)
c. (\(5^{1 / 2}\) \(16^{1 / 4}\))²

5.2 Lesson

Monitoring Progress

Simplify the expression.

Question 1.
\(2^{3 / 4}\) • \(2^{1 / 2}\)

Question 2.
\(\frac{3}{3^{1 / 4}}\)

Question 3.
\(\left(\frac{20^{1 / 2}}{5^{1 / 2}}\right)^{3}\)

Question 4.
(\(5^{1 / 3}\) • \(7^{1 / 4}\))³

Simplify the expression

Question 5.
\(\sqrt[4]{27}\) • \(\sqrt[4]{3}\)

Question 6.
\(\frac{\sqrt[3]{250}}{\sqrt[3]{2}}\)

Question 7.
\(\sqrt[3]{104}\)

Question 8.
\(\sqrt[5]{\frac{3}{4}}\)

Question 9.
\(\frac{3}{6-\sqrt{2}}\)

Question 10.
\(7 \sqrt[5]{12}\) – \(\sqrt[5]{12}\)

Question 11.
4(\(9^{2 / 3}\)) + (\(9^{2 / 3}\))

Question 12.
\(\sqrt[3]{5}\) + \(\sqrt[3]{40}\)

Simplify the expression. Assume all variables are positive.

Question 13.
\(\sqrt[3]{27 q^{9}}\)

Question 14.
\(\sqrt[5]{\frac{x^{10}}{y^{5}}}\)

Question 15.
\(\frac{6 x y^{3 / 4}}{3 x^{1 / 2} y^{1 / 2}}\)

Question 16.
\(\sqrt{9} w^{5}\) – \(w \sqrt{w^{3}}\)

Properties of Rational Exponents and Radicals 5.2 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING How do you know when a radical expression is in simplest form?
Answer:

Question 2.
WHICH ONE DOESN’T BELONG? Which radical expression does not belong with the other three? Explain your reasoning.

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, use the properties of rational exponents to simplify the expression.

Question 3.
\(\left(9^{2}\right)^{1 / 3}\)
Answer:

Question 4.
\(\left(12^{2}\right)^{1 / 4}\)
Answer:

Question 5.
\(\frac{6}{6^{1 / 4}}\)
Answer:

Question 6.
\(\frac{7}{7^{1 / 3}}\)
Answer:

Question 7.
\(\left(\frac{8^{4}}{10^{4}}\right)^{-1 / 4}\)
Answer:

Question 8.
\(\left(\frac{9^{3}}{6^{3}}\right)^{-1 / 3}\)
Answer:

Question 9.
(\(3^{-2 / 3}\) • \(3^{1 / 3}\))^-1

Answer:

Question 10.
(\(5^{1 / 2}\) • \(5^{-3 / 2}\))^-1/4

Answer:

Question 11.
\(\frac{2^{2 / 3} \cdot 16^{2 / 3}}{4^{2 / 3}}\)
Answer:

Question 12.
\(\frac{49^{3 / 8} \cdot 49^{7 / 8}}{7^{5 / 4}}\)
Answer:

In Exercises 13–20, use the properties of radicals to simplify the expression.

Question 13.
\(\sqrt{2}\) • \(\sqrt{72}\)
Answer:

Question 14.
\(\sqrt[3]{16}\) • \(\sqrt[3]{32}\)
Answer:

Question 15.
\(\sqrt[4]{6}\) • \(\sqrt[4]{8}\)
Answer:

Question 16.
\(\sqrt[4]{8}\) • \(\sqrt[4]{8}\)
Answer:

Question 17.
\(\frac{\sqrt[5]{486}}{\sqrt[5]{2}}\)
Answer:

Question 18.
\(\frac{\sqrt{2}}{\sqrt{32}}\)
Answer:

Question 19.
\(\frac{\sqrt[3]{6} \cdot \sqrt[3]{72}}{\sqrt[3]{2}}\)
Answer:

Question 20.
\(\frac{\sqrt[3]{3} \cdot \sqrt[3]{18}}{\sqrt[6]{2} \cdot \sqrt[6]{2}}\)
Answer:

In Exercises 21–28, write the expression in simplest form.

Question 21.
\(\sqrt[4]{567}\)
Answer:

Question 22.
\(\sqrt[5]{288}\)
Answer:

Question 23.
\(\frac{\sqrt[3]{5}}{\sqrt[3]{4}}\)
Answer:

Question 24.
\(\frac{\sqrt[4]{4}}{\sqrt[4]{27}}\)
Answer:

Question 25.
\(\sqrt{\frac{3}{8}}\)
Answer:

Question 26.
\(\sqrt[3]{\frac{7}{4}}\)
Answer:

Question 27.
\(\sqrt[3]{\frac{64}{49}}\)
Answer:

Question 28.
\(\sqrt[4]{\frac{1296}{25}}\)
Answer:

In Exercises 29–36, write the expression in simplest form.

Question 29.
\(\frac{1}{1+\sqrt{3}}\)
Answer:

Question 30.
\(\frac{1}{2+\sqrt{5}}\)
Answer:

Question 31.
\(\frac{5}{3-\sqrt{2}}\)
Answer:

Question 32.
\(\frac{11}{9-\sqrt{6}}\)
Answer:

Question 33.
\(\frac{9}{\sqrt{3}+\sqrt{7}}\)
Answer:

Question 34.
\(\frac{2}{\sqrt{8}+\sqrt{7}}\)
Answer:

Question 35.
\(\frac{\sqrt{6}}{\sqrt{3}-\sqrt{5}}\)
Answer:

Question 36.
\(\frac{\sqrt{7}}{\sqrt{10}-\sqrt{2}}\)
Answer:

In Exercises 37–46, simplify the expression.

Question 37.
\(9 \sqrt[3]{11}\) + \(3 \sqrt[3]{11}\)
Answer:

Question 38.
\(8 \sqrt[6]{5}\) – \(12 \sqrt[6]{5}\)
Answer:

Question 39.
\(3\left(11^{1 / 4}\right)\) + \(9\left(11^{1 / 4}\right)\)
Answer:

Question 40.
\(13\left(8^{3 / 4}\right)\) – \(4\left(8^{3 / 4}\right)\)
Answer:

Question 41.
\(5 \sqrt{12}\) – \(19 \sqrt{3}\)
Answer:

Question 42.
\(27 \sqrt{6}\) + \(7 \sqrt{150}\)
Answer:

Question 43.
\(\sqrt[5]{224}\) + \(3 \sqrt[5]{7}\)
Answer:

Question 44.
\(7 \sqrt[3]{2}\) – \(\sqrt[3]{128}\)
Answer:

Question 45.
\(5\left(24^{1 / 3}\right)\) – 4(\(\left(3^{1 / 3}\right)\))
Answer:

Question 46.
\(5^{1 / 4}\) + 6(\([405^{1 / 4}/latex])
Answer:

Question 47.
ERROR ANALYSIS Describe and correct the error in simplifying the expression.

Answer:

Question 48.
MULTIPLE REPRESENTATIONS Which radical expressions are like radicals?
A. [latex]\left(5^{2 / 9}\right)^{3 / 2}\)
B. \(\frac{5^{3}}{(\sqrt[3]{5})^{8}}\)
C. \(\sqrt[3]{625}\)
D. \(\sqrt[3]{5} 145\) – \(\sqrt[3]{875}\)
E. \(\sqrt[3]{5}\) + \(3 \sqrt[3]{5}\)
F. \(7 \sqrt[4]{80}\) – \(2 \sqrt[4]{405}\)
Answer:

In Exercises 49–54, simplify the expression.

Question 49.
\(\sqrt[4]{81 y^{8}}\)
Answer:

Question 50.
\(\sqrt[3]{64 r^{3} t^{6}}\)
Answer:

Question 51.
\(\sqrt[5]{\frac{m^{10}}{n^{5}}}\)
Answer:

Question 52.
\(\sqrt[4]{\frac{k^{16}}{16 z^{4}}}\)
Answer:

Question 53.
\(\sqrt[6]{\frac{g^{6} h}{h^{7}}}\)
Answer:

Question 54.
\(\sqrt[8]{n^{2} p^{-1}}\)
Answer:

Question 55.
ERROR ANALYSIS Describe and correct the error in simplifying the expression.

Answer:

Question 56.
OPEN-ENDED Write two variable expressions involving radicals, one that needs absolute value in simplifying and one that does not need absolute value. Justify your answers.
Answer:

In Exercises 57–64, write the expression in simplest form. Assume all variables are positive.

Question 57.
\(\sqrt{81 a^{7} b^{12} c^{9}}\)
Answer:

Question 58.
\(\sqrt[3]{125 r^{4} s^{9} t^{7}}\)
Answer:

Question 59.
\(\sqrt[5]{\frac{160 m^{6}}{n^{7}}}\)
Answer:

Question 60.
\(\sqrt[4]{\frac{405 x^{3} y^{3}}{5 x^{-1} y}}\)
Answer:

Question 61.
\(\frac{\sqrt[3]{w} \cdot \sqrt{w^{5}}}{\sqrt{25 w^{16}}}\)
Answer:

Question 62.
\(\frac{\sqrt[4]{v^{6}}}{\sqrt[7]{v^{5}}}\)
Answer:

Question 63.
\(\frac{18 w^{1 / 3} v^{5 / 4}}{27 w^{4 / 3} v^{1 / 2}}\)
Answer:

Question 64.
\(\frac{7 x^{-3 / 4} y^{5 / 2} z^{-2 / 3}}{56 x^{-1 / 2} y^{1 / 4}}\)
Answer:

In Exercises 65–70, perform the indicated operation. Assume all variables are positive.

Question 65.
\(12 \sqrt[3]{y}\) + \(9 \sqrt[3]{y}\)
Answer:

Question 66.
\(11 \sqrt{2 z}\) – \(5 \sqrt{2 z}\)
Answer:

Question 67.
\(3 x^{7 / 2}\) – 5\(x^{7 / 2}\)
Answer:

Question 68.
\(7 \sqrt[3]{m^{7}}\) + \(3 m^{7 / 3}\)
Answer:

Question 69.
\(\sqrt[4]{16 w^{10}}\) + \(2 w \sqrt[4]{w^{6}}\)
Answer:

Question 70.
(p^1/2 • p^1/4) – \(\sqrt[4]{16 p^{3}}\)
Answer:

MATHEMATICAL CONNECTIONS In Exercises 71 and 72, find simplified expressions for the perimeter and area of the given figure.

Question 71.

Answer:

Question 72

Answer:

Question 73.
MODELING WITH MATHEMATICS The optimum diameter d (in millimeters) of the pinhole in a pinhole camera can be modeled by d = 1.9[(5.5 × 10⁻⁴)ℓ]^1/2, where ℓ is the length (in millimeters) of the camera box. Find the optimum pinhole diameter for a camera box with a length of 10 centimeters.

Answer:

Question 74.
MODELING WITH MATHEMATICS The surface area S(in square centimeters) of a mammal can be modeled by S = km^2/3, where m is the mass (in grams) of the mammal and k is a constant. The table shows the values of k for different mammals.

a. Find the surface area of a bat whose mass is 32 grams.
b. Find the surface area of a rabbit whose mass is 3.4 kilograms (3.4 × 10³ grams).
c. Find the surface area of a human whose mass is 59 kilograms.
Answer:

Question 75.
MAKING AN ARGUMENT Your friend claims it is not possible to simplify the expression 7\(\sqrt{11}\) − 9 \(\sqrt{44}\) because it does not contain like radicals. Is your friend correct? Explain your reasoning.
Answer:

Question 76.
PROBLEM SOLVING The apparent magnitude of a star is a number that indicates how faint the star is in relation to other stars. The expression f(x) = \(\frac{2.512^{m_{1}}}{2.512^{m_{2}}}\) tells how many times fainter a star with apparent magnitude m₁ is than a star with apparent magnitude m₂.

a. How many times fainter is Altair than Vega?
b. How many times fainter is Deneb than Altair?
c. How many times fainter is Deneb than Vega?

Answer:

Question 77.
CRITICAL THINKING Find a radical expression for the perimeter of the triangle inscribed in the square shown. Simplify the expression.

Answer:

Question 78.
HOW DO YOU SEE IT? Without finding points, match the functions f(x) = \(\sqrt{64 x^{2}}\) and g(x) = \(\sqrt[3]{64 x^{6}}\) with their graphs. Explain your reasoning.

Answer:

Question 79.
REWRITING A FORMULA You have filled two round balloons with water. One balloon contains twice as much water as the other balloon.
a. Solve the formula for the volume of a sphere, V = \(\frac{4}{3}\)πr³, for r.
b. Substitute the expression for r from part (a) into the formula for the surface area of a sphere, S = 4πr². Simplify to show that S = (4π)^1/3(3V)^2/3.
c. Compare the surface areas of the two water balloons using the formula in part (b).
Answer:

Question 80.
THOUGHT PROVOKING Determine whether the expressions (x²)^1/6 and (x^1/6)² are equivalent for all values of x.
Answer:

Question 81.
DRAWING CONCLUSIONS Substitute different combinations of odd and even positive integers for m and n in the expression \(\sqrt[n]{x^{m}}\). When you cannot assume x is positive, explain when absolute value is needed in simplifying the expression.
Answer:

Maintaining Mathematical Proficiency

Identify the focus, directrix, and axis of symmetry of the parabola. Then graph the equation. (Section 2.3)

Question 82.
y = 2x²
Answer:

Question 83.
y² = −x
Answer:

Question 84.
y² = 4x
Answer:

Write a rule for g. Describe the graph of g as a transformation of the graph of f. (Section 4.7)

Question 85.
f(x) = x⁴ − 3x² − 2x, g(x) = −f(x)
Answer:

Question 86.
f(x) = x³ − x, g(x) = f(x) − 3
Answer:

Question 87.
f(x) = x³ − 4, g(x) = f(x − 2)
Answer:

Question 88.
f(x) = x⁴ + 2x³ − 4x², g(x) = f(2x)
Answer:

Lesson 5.3 Graphing Radical Functions

Essential Question

How can you identify the domain and range of a radical function?

EXPLORATION 1
Identifying Graphs of Radical Functions
Work with a partner. Match each function with its graph. Explain your reasoning. Then identify the domain and range of each function.
a. f(x) = \(\sqrt{x}\)
b. f(x) = \(\sqrt[3]{x}\)
c. f(x) = \(\sqrt[4]{x}\)
d. f(x) = \(\sqrt[5]{x}\)

EXPLORATION 2
Identifying Graphs of Transformations
Work with a partner. Match each transformation of f(x) = \(\sqrt{x}\) with its graph. Explain your reasoning. Then identify the domain and range of each function.
a. g(x) = \(\sqrt{x+2}\)
b. g(x) = \(\sqrt{x-2}\)
c. g(x) = \(\sqrt{x}+2-2\)
d. g(x) = −\(\sqrt{x+2}\)

Communicate Your Answer

Question 3.
How can you identify the domain and range of a radical function?

Question 4.
Use the results of Exploration 1 to describe how the domain and range of a radical function are related to the index of the radical.

5.3 Lesson

Monitoring Progress

Question 1.
Graph g(x) = \(\sqrt{x+1}\). Identify the domain and range of the function.

Question 2.
Describe the transformation of f(x) = \(\sqrt[3]{x}\) represented by g(x) = −\(\sqrt[3]{x}\) − 2. Then graph each function.

Question 3.
WHAT IF? In Example 3, the function N(d ) = 2.4 • E(d) approximates the number of seconds it takes a dropped object to fall d feet on the Moon. Write a rule for N. How long does it take a dropped object to fall 25 feet on the Moon?

Question 4.
In Example 4, is the transformed function the same when you perform the translation followed by the horizontal shrink? Explain your reasoning.

Question 5.
Use a graphing calculator to graph −4y² = x + 1. Identify the vertex and the direction that the parabola opens.

Question 6.
Use a graphing calculator to graph x² + y² = 25. Identify the radius and the intercepts.

Graphing Radical Functions 5.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE Square root functions and cube root functions are examples of __________ functions.
Answer:

Question 2.
COMPLETE THE SENTENCE When graphing y = a\(\sqrt[3]{x-h}\) + k, translate the graph of y = a\(\sqrt[3]{x}\)h units __________ and k units __________.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, match the function with its graph.

Question 3.
f(x) = \(\sqrt{x}+3\)
Answer:

Question 4.
h(x) = \(\sqrt{x}\) + 3
Answer:

Question 5.
f(x) = \(\sqrt{x-3}\)
Answer:

Question 6.
g(x) = \(\sqrt{x}\) − 3
Answer:

Question 7.
h(x) = \(\sqrt{x+3}\) − 3
Answer:

Question 8.
f(x) = \(\sqrt{x-3}\) + 3
Answer:

In Exercises 9–18, graph the function. Identify the domain and range of the function.

Question 9.
h(x) = \(\sqrt{x}\) + 4
Answer:

Question 10.
g(x) = \(\sqrt{x}\) − 5
Answer:

Question 11.
g(x) = − \(\sqrt[3]{2 x}\)
Answer:

Question 12.
f(x) = \(\sqrt[3]{-5 x}\)
Answer:

Question 13.
g(x) = \(\frac{1}{5} \sqrt{x}-3\)
Answer:

Question 14.
f(x) = \(\frac{1}{2} \sqrt[3]{x}+6\)
Answer:

Question 15.
f(x) = \((6 x)^{1 / 2}\) + 3
Answer:

Question 16.
g(x) = −3(x + 1)^1/3

Answer:

Question 17.
h(x) = −\(\sqrt[4]{x}\)
Answer:

Question 18.
h(x) = \(\sqrt[5]{2 x}\)
Answer:

In Exercises 19–26, describe the transformation of f represented by g. Then graph each function.

Question 19.
f(x) = \(\sqrt{x}\), g(x) = \(\sqrt{x+1}\) + 8
Answer:

Question 20.
f(x) = \(\sqrt{x}\), g(x) = 2\(\sqrt{x-1}\)
Answer:

Question 21.
f(x) = \(\sqrt[3]{x}\), g(x) = −\(\sqrt[3]{x}\) − 1
Answer:

Question 22.
f(x) = \(\sqrt[3]{x}\), g(x) = \(\sqrt[3]{x+4}\) − 5
Answer:

Question 23.
f(x) = \(x^{1 / 2}\), g(x) = \(\frac{1}{4}(-x)^{1 / 2}\)
Answer:

Question 24.
f(x) = \(x^{1 / 3}\), g(x) = \(\frac{1}{3} x^{1 / 3}\) + 6
Answer:

Question 25.
f(x) = \(\sqrt[4]{x}\), g(x) = \(2 \sqrt[4]{x+5}\) − 4
Answer:

Question 26.
f(x) = \(\sqrt[5]{x}\), g(x) = \(\sqrt[5]{-32 x}\) + 3
Answer:

Question 27.
ERROR ANALYSIS Describe and correct the error in graphing f(x) = \(\sqrt{x-2}\) − 2.

Answer:

Question 28.
ERROR ANALYSIS Describe and correct the error in describing the transformation of the parent square root function represented by g(x) = \(\sqrt{\frac{1}{2} x}\) + 3.

Answer:

USING TOOLS In Exercises 29–34, use a graphing calculator to graph the function. Then identify the domain and range of the function.

Question 29.
g(x) = \(\sqrt{x^{2}+x}\)
Answer:

Question 30.
h(x) = \(\sqrt{x^{2}-2 x}\)
Answer:

Question 31.
f(x) = \(\sqrt[3]{x^{2}+x}\)
Answer:

Question 32.
f(x) = \(\sqrt[3]{3 x^{2}-x}\)
Answer:

Question 33.
f(x) = \(\sqrt{2 x^{2}+x+1}\)
Answer:

Question 34.
h(x) = \(\sqrt[3]{\frac{1}{2} x^{2}-3 x+4}\)
Answer:

ABSTRACT REASONING In Exercises 35–38, complete the statement with sometimes, always, or never.

Question 35.
The domain of the function y = a\(\sqrt{x}\) is ______ x ≥ 0.
Answer:

Question 36.
The range of the function y = a\(\sqrt{x}\) is ______ y ≥ 0.
Answer:

Question 37.
The domain and range of the function y = \(\sqrt[3]{x-h}\) + k are ________ all real numbers.
Answer:

Question 38.
The domain of the function y = a\(\sqrt{-x}\) + k is ________ x ≥ 0.
Answer:

Question 39.
PROBLEM SOLVING The distance (in miles) a pilot can see to the horizon can be approximated by E(n) = 1.22\(\sqrt{n}\), where n is the plane’s altitude (in feet above sea level) on Earth. The function M(n) = 0.75E(n) approximates the distance a pilot can see to the horizon n feet above the surface of Mars. Write a rule for M. What is the distance a pilot can see to the horizon from an altitude of 10,000 feet above Mars?

Answer:

Question 40.
MODELING WITH MATHEMATICS The speed (in knots) of sound waves in air can be modeled by
v(K) = 643.855\(\sqrt{\frac{K}{273.15}}\)
where K is the air temperature (in kelvin). The speed (in meters per second) of sound waves in air can be modeled by s(K) = \(\frac{v(K)}{1.944}\)
Write a rule for s. What is the speed (in meters per second) of sound waves when the air temperature is 305 kelvin?
Answer:

In Exercises 41–44, write a rule for g described by the transformations of the graph of f.

Question 41.
Let g be a vertical stretch by a factor of 2, followed by a translation 2 units up of the graph of f(x) = \(\sqrt{x}\) + 3.
Answer:

Question 42.
Let g be a reflection in the y-axis, followed by a translation 1 unit right of the graph of f(x) = \(2 \sqrt[3]{x-1}\).
Answer:

Question 43.
Let g be a horizontal shrink by a factor of \(\frac{2}{3}\), followed by a translation 4 units left of the graph of f(x) = \(\sqrt{6 x}\).
Answer:

Question 44.
Let g be a translation 1 unit down and 5 units right, followed by a reflection in the x-axis of the graph of f(x) = −\(-\frac{1}{2} \sqrt[4]{x}+\frac{3}{2}\)
Answer:

In Exercises 45 and 46, write a rule for g.

Question 45.

Answer:

Question 46.

Answer:

In Exercises 47–50, write a rule for g that represents the indicated transformation of the graph of f.

Question 47.
f(x) = 2\(\sqrt{x}\), g(x) = f(x + 3)
Answer:

Question 48.
f(x) = \(\frac{1}{3} \sqrt{x-1}\), g(x) = −f(x) + 9
Answer:

Question 49.
f(x) = −\(\sqrt{x^{2}-2}\), g(x) = −2f(x + 5)
Answer:

Question 50.
f(x) = \(\sqrt[3]{x^{2}+10 x}\), g(x) =\(\frac{1}{4}\)f(−x) + 6
Answer:

In Exercises 51–56, use a graphing calculator to graph the equation of the parabola. Identify the vertex and the direction that the parabola opens.

Question 51.
\(\frac{1}{4}\)y² = x
Answer:

Question 52.
3y² = x
Answer:

Question 53.
−8y² + 2 = x
Answer:

Question 54.
2y² = x − 4
Answer:

Question 55.
x + 8 = \(\frac{1}{5}\)y²

Answer:

Question 56.
\(\frac{1}{2}\)x = y² − 4
Answer:

In Exercises 57–62, use a graphing calculator to graph the equation of the circle. Identify the radius and the intercepts.

Question 57.
x² + y² = 9
Answer:

Question 58.
x² + y² = 4
Answer:

Question 59.
1 − y² = x²

Answer:

Question 60.
64 − x² = y²

Answer:

Question 61.
−y² = x² − 36
Answer:

Question 62.
x² = 100 − y²

Answer:

Question 63.
MODELING WITH MATHEMATICS The period of a pendulum is the time the pendulum takes to complete one back-and-forth swing. The period T (in seconds) can be modeled by the function T = 1.11\(\sqrt{\ell}\), where ℓ is the length (in feet) of the pendulum. Graph the function. Estimate the length of a pendulum with a period of 2 seconds. Explain your reasoning.

Answer:

Question 64.
HOW DO YOU SEE IT? Does the graph represent a square root function or a cube root function? Explain. What are the domain and range of the function?

Answer:

Question 65.
PROBLEM SOLVING For a drag race car with a total weight of 3500 pounds, the speed s(in miles per hour) at the end of a race can be modeled by s = 14.8\(\sqrt[3]{p}\), where p is the power (in horsepower). Graph the function.
a. Determine the power of a 3500-pound car that reaches a speed of 200 miles per hour.
b. What is the average rate of change in speed as the power changes from 1000 horsepower to 1500 horsepower?
Answer:

Question 66.
THOUGHT PROVOKING The graph of a radical function f passes through the points (3, 1) and (4, 0). Write two different functions that could represent f(x + 2) + 1. Explain.
Answer:

Question 67.
MULTIPLE REPRESENTATIONS The terminal velocity v_t(in feet per second) of a skydiver who weighs 140 pounds is given by v_t = 33.7\(\sqrt{\frac{140}{\Lambda}}\)
where A is the cross-sectional surface area (in square feet) of the skydiver. The table shows the terminal velocities (in feet per second) for various surface areas (in square feet) of a skydiver who weighs 165 pounds.

a. Which skydiver has a greater terminal velocity for each value of A given in the table?
b. Describe how the different values of A given in the table relate to the possible positions of the falling skydiver.
Answer:

Question 68.
MATHEMATICAL CONNECTIONS The surface area S of a right circular cone with a slant height of 1 unit is given by S = πr + πr², where r is the radius of the cone.

a. Use completing the square to show that r = \(\frac{1}{\sqrt{\pi}} \sqrt{S+\frac{\pi}{4}}-\frac{1}{2}\).
b. Graph the equation in part (a) using a graphing calculator. Then find the radius of a right circular cone with a slant height of 1 unit and a surface area of \(\frac{3 \pi}{4}\) square units.
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check your solutions. (Skills Review Handbook)

Question 69.
| 3x + 2 | = 5
Answer:

Question 70.
∣4x + 9 ∣ = −7
Answer:

Question 71.
| 2x − 6 ∣ = ∣ x ∣
Answer:

Question 72.
| x + 8 | = | 2x + 2 |
Answer:

Solve the inequality. (Section 3.6)

Question 73.
x² + 7x + 12 < 0
Answer:

Question 74.
x² − 10x + 25 ≥ 4
Answer:

Question 75.
2x² + 6 > 13x
Answer:

Question 76.
\(\frac{1}{8}\)x² + x ≤ −2
Answer:

Rational Exponents and Radical Functions Study Skills : Analyzing Your Errors

5.1–5.3 What Did You Learn?

Core Vocabulary

Core Concepts
Section 5.1
Real nth Roots of a, p. 238
Rational Exponents, p. 239

Section 5.2
Properties of Rational Exponents, p. 244
Properties of Radicals, p. 245

Section 5.3
Parent Functions for Square Root and Cube Root Functions, p. 252
Transformations of Radical Functions, p. 253

Mathematical Practices

Question 1.
How can you use definitions to explain your reasoning in Exercises 21–24 on page 241?

Question 2.
How did you use structure to solve Exercise 76 on page 250?

Question 3.
How can you check that your answer is reasonable in Exercise 39 on page 257?

Question 4.
How can you make sense of the terms of the surface area formula given in Exercise 68 on page 258?

Study Skills
Analyzing Your Errors
Application Errors
What Happens: You can do numerical problems, but you struggle with problems that have context.

How to Avoid This Error: Do not just mimic the steps of solving an application problem. Explain out loud what the question is asking and why you are doing each step. After solving the problem, ask yourself, “Does my solution make sense?”

Rational Exponents and Radical Functions 5.1 – 5.3 Quiz

5.1–5.3 Quiz

Find the indicated real nth root(s) of a. (Section 5.1)

Question 1.
n = 4, a = 81

Question 2.
n = 5, a = −1024

Question 3.
Evaluate (a) 16^3/4 and (b) 125^2/3 without using a calculator. Explain your reasoning. (Section 5.1)

Find the real solution(s) of the equation. Round your answer to two decimal places.(Section 5.1)

Question 4.
2x⁶ = 1458

Question 5.
(x + 6)³ = 28

Simplify the expression.(Section 5.2)

Question 6.
\(\left(\frac{48^{1 / 4}}{6^{1 / 4}}\right)^{6}\)

Question 7.
\(\sqrt[4]{3}\) • \(\sqrt[4]{432}\)

Question 8.
\(\frac{1}{3+\sqrt{2}}\)

Question 9.
\(\sqrt[3]{16}\) – \(5 \sqrt[3]{2}\)

Question 10.
Simplify \(\sqrt[8]{x^{9} y^{8} z^{16}}\). (Section 5.2)

Write the expression in simplest form. Assume all variables are positive.(Section 5.2)

Question 11.
\(\sqrt[3]{216 p^{9}}\)

Question 12.
\(\frac{\sqrt[5]{32}}{\sqrt[5]{m^{3}}}\)

Question 13.
\(\sqrt[4]{n^{4} q}\) + \(7 n \sqrt[4]{q}\)

Question 14.
Graph f(x) = 2\(2 \sqrt[3]{x}\) + 1. Identify the domain and range of the function. (Section 5.3)

Describe the transformation of the parent function represented by the graph of g. Then write a rule for g.(Section 5.3)

Question 15.

Question 16.

Question 17.

Question 18.
Use a graphing calculator to graph x = 3y² − 6. Identify the vertex and direction the parabola opens. (Section 5.3)

Question 19.
A jeweler is setting a stone cut in the shape of a regular octahedron. A regular octahedron is a solid with eight equilateral triangles as faces, as shown. The formula for the volume of the stone is V= 0.47s³, where s is the side length (in millimeters) of an edge of the stone. The volume of the stone is 161 cubic millimeters. Find the length of an edge of the stone. (Section 5.1)

Question 20.
An investigator can determine how fast a car was traveling just prior to an accident using the model s = 4\(\sqrt{d}\), where s is the speed (in miles per hour) of the car and d is the length (in feet) of the skid marks. Graph the model. The length of the skid marks of a car is 90 feet. Was the car traveling at the posted speed limit prior to the accident? Explain your reasoning. (Section 5.3)

Lesson 5.4 Solving Radical Equations and Inequalities

Essential Question
How can you solve a radical equation?

EXPLORATION 1
Solving Radical Equations
Work with a partner. Match each radical equation with the graph of its related radical function. Explain your reasoning. Then use the graph to solve the equation, if possible. Check your solutions.
a. \(\sqrt{x-1}\) – 1 = 0
b. \(\sqrt{2 x+2}\) – \(\sqrt{x+4}\) = 0
c. \(\sqrt{9-x^{2}}\) = 0
d. \(\sqrt{x+2}\) – x = 0
e. \(\sqrt{-x+2}\) – x = 0
f. \(\sqrt{3 x^{2}+1}\) = 0

EXPLORATION 2
Solving Radical Equations
Work with a partner. Look back at the radical equations in Exploration 1. Suppose that you did not know how to solve the equations using a graphical approach.
a. Show how you could use a numerical approach to solve one of the equations. For instance, you might use a spreadsheet to create a table of values.

b. Show how you could use an analytical approach to solve one of the equations. For instance, look at the similarities between the equations in Exploration 1. What first step may be necessary so you could square each side to eliminate the radical(s)? How would you proceed to find the solution?

Communicate Your Answer

Question 3.
How can you solve a radical equation?

Question 4.
Would you prefer to use a graphical, numerical, or analytical approach to solve the given equation? Explain your reasoning. Then solve the equation.
\(\sqrt{x+3}\) – \(\sqrt{x-2}\) = 1

Monitoring Progress

Solve the equation. Check your solution.

Question 1.
\(\sqrt[3]{x}\) – 9 = -6

Question 2.
\(\sqrt{x+25}\) = 2

Question 3.
2\(\sqrt[3]{x-3}\) = 4

Question 4.
WHAT IF? Estimate the air pressure at the center of the hurricane when the mean sustained wind velocity is 48.3 meters per second.

Solve the equation. Check your solution(s).

Question 5.
\(\sqrt{10 x+9}\) = x + 3

Question 6.
\(\sqrt{2 x+5}\) = \(\sqrt{x+7}\)

Question 7.
\(\sqrt{x+6}\) – 2 = \(\sqrt{x-2}\)

Solve the equation. Check your solution(s).

Question 8.
(3x)^1/3 = −3

Question 9.
(x + 6)^1/2 = x

Question 10.
(x + 2)^3/4 = 8

Question 11.
Solve
(a) 2\(\sqrt{x}\) − 3 ≥ 3 and
(b) 4\(\sqrt[3]{x+1}\) < 8

Solving Radical Equations and Inequalities 5.4 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY Is the equation 3x − \(\sqrt{2}\) = \(\sqrt{6}\) a radical equation? Explain your reasoning.
Answer:

Question 2.
WRITING Explain the steps you should use to solve \(\sqrt{x}\) + 10 < 15
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation. Check your solution.

Question 3.
\(\sqrt{5 x+1}\) = 6
Answer:

Question 4.
\(\sqrt{3 x+10}\) = 8
Answer:

Question 5.
\(\sqrt[3]{x-16}\) = 2
Answer:

Question 6.
\(\sqrt[3]{x}\) − 10 = −7
Answer:

Question 7.
−2\(\sqrt{24 x}\) + 13 = −1
Answer:

Question 8.
8\(\sqrt[3]{10 x}\) − 15 = 17
Answer:

Question 9.
\(\frac{1}{5} \sqrt[3]{3 x}\) = 8
Answer:

Question 10.
\(\sqrt{2 x}-\frac{2}{3}\) = 0
Answer:

Question 11.
\(2 \sqrt[5]{x}\) + 7 = 15
Answer:

Question 12.
\(\sqrt[4]{4 x}\) − 13 = −15
Answer:

Question 13.
MODELING WITH MATHEMATICS Biologists have discovered that the shoulder height h (in centimeters) of a male Asian elephant can be modeled by h = 62.5\(\sqrt[3]{t}\) + 75.8, where t is the age (in years) of the elephant. Determine the age of an elephant with a shoulder height of 250 centimeters.

Answer:

Question 14.
MODELING WITH MATHEMATICS In an amusement park ride, a rider suspended by cables swings back and forth from a tower. The maximum speed v (in meters per second) of the rider can be approximated by v = \(\sqrt{2 g h}\), where h is the height (in meters) at the top of each swing and g is the acceleration due to gravity (g ≈ 9.8 m/sec²). Determine the height at the top of the swing of a rider whose maximum speed is 15 meters per second.
Answer:

In Exercises 15–26, solve the equation. Check your solution(s).

Question 15.
x− 6 = \(\sqrt{3 x}\)
Answer:

Question 16.
x − 10 = \(\sqrt{9 x}\)
Answer:

Question 17.
\(\sqrt{44-2 x}\) = x − 10
Answer:

Question 18.
\(\sqrt{2 x+30}\) = x + 3
Answer:

Question 19.
\(\sqrt[3]{8 x^{3}-1}\) = 2x − 1
Answer:

Question 20.
\(\sqrt[4]{3-8 x^{2}}\) = 2x
Answer:

Question 21.
\(\sqrt{4 x+1}\) = \(\sqrt{x+10}\)
Answer:

Question 22.
\(\sqrt{3 x-3}\) – \(\sqrt{x+12}\) = 0
Answer:

Question 23.
\(\sqrt[3]{2 x-5}\) – \(\sqrt[3]{8 x+1}\) = 0
Answer:

Question 24.
\(\sqrt[3]{x+5}\) = 2\(\sqrt[3]{2 x+6}\)
Answer:

Question 25.
\(\sqrt{3 x-8}\) + 1 = \(\sqrt{x+5}\)
Answer:

Question 26.
\(\sqrt{x+2}\) = 2 – \(\sqrt{x}\)
Answer:

In Exercises 27–34, solve the equation. Check your solution(s).

Question 27.
2x^2/3 = 8
Answer:

Question 28.
4x^3/2 = 32
Answer:

Question 29.
x^1/4 + 3 = 0
Answer:

Question 30.
2x^3/4 − 14 = 40
Answer:

Question 31.
(x + 6)^1/2 = x
Answer:

Question 32.
(5 − x)^1/2 − 2x = 0
Answer:

Question 33.
2(x + 11)^1/2 = x + 3
Answer:

Question 34.
(5x² − 4)^1/4 = x
Answer:

ERROR ANALYSIS In Exercises 35 and 36, describe and correct the error in solving the equation.

Question 35.

Answer:

Question 36.

Answer:

In Exercises 37–44, solve the inequality.

Question 37.
\(2 \sqrt[3]{x}\) − 5 ≥ 3
Answer:

Question 38.
\(\sqrt[3]{x-4}\) ≤ 5
Answer:

Question 39.
\(4 \sqrt{x-2}\) > 20
Answer:

Question 40.
7\(\sqrt{x}\) + 1 < 9
Answer:

Question 41.
2\(\sqrt{x}\) + 3 ≤ 8
Answer:

Question 42.
\(\sqrt[3]{x+7}\) ≥ 3
Answer:

Question 43.
\(-2 \sqrt[3]{x+4}\) < 12
Answer:

Question 44.
−0.25\(\sqrt{x}\) − 6 ≤ −3
Answer:

Question 45.
MODELING WITH MATHEMATICS The length ℓ (in inches) of a standard nail can be modeled by ℓ = 54d^3/2, where d is the diameter (in inches) of the nail. What is the diameter of a standard nail that is 3 inches long?
Answer:

Question 46.
DRAWING CONCLUSIONS “Hang time” is the time you are suspended in the air during a jump. Your hang time t (in seconds) is given by the function t = 0.5\(\sqrt{h}\), where h is the height (in feet) of the jump. Suppose a kangaroo and a snowboarder jump with the hang times shown.

a. Find the heights that the snowboarder and the kangaroo jump.
b. Double the hang times of the snowboarder and the kangaroo and calculate the corresponding heights of each jump.
c. When the hang time doubles, does the height of the jump double? Explain.
Answer:

USING TOOLS In Exercises 47–52, solve the nonlinear system. Justify your answer with a graph.

Question 47.
y² = x − 3
y = x − 3
Answer:

Question 48.
y² = 4x + 17
y = x + 5
Answer:

Question 49.
x² + y² = 4
y = x – 2
Answer:

Question 50.
x² + y² = 25
y = \(-\frac{3}{4}\)x + \(\frac{25}{4}\)
Answer:

Question 51.
x² + y² = 1
y = \(\frac{1}{2}\)x² – 1
Answer:

Question 52.
x² + y² = 4
y² = x + 2
Answer:

Question 53.
PROBLEM SOLVING The speed s (in miles per hour) of a car can be given by s = \(\sqrt{30 f d}\), where f is the coefficient of friction and d is the stopping distance (in feet). The table shows the coefficient of friction for different surfaces.

a. Compare the stopping distances of a car traveling 45 miles per hour on the surfaces given in the table.
b. You are driving 35 miles per hour on an icy road when a deer jumps in front of your car. How far away must you begin to brake to avoid hitting the deer? Justify your answer.
Answer:

Question 54.
MODELING WITH MATHEMATICS The Beaufort wind scale was devised to measure wind speed. The Beaufort numbers B, which range from 0 to 12, can be modeled by B = 1.69\(\sqrt{s+4.25}\) − 3.55, where s is the wind speed (in miles per hour).

a. What is the wind speed for B = 0? B = 3?
b. Write an inequality that describes the range of wind speeds represented by the Beaufort model.
Answer:

Question 55.
USING TOOLS Solve the equation x − 4 = \(\sqrt{2 x}\). Then solve the equation x − 4 = − \(\sqrt{2 x}\).
a. How does changing \(\sqrt{2 x}\) to −\(\sqrt{2 x}\) change the solution(s) of the equation?
b. Justify your answer in part (a) using graphs.
Answer:

Question 56.
MAKING AN ARGUMENT Your friend says it is impossible for a radical equation to have two extraneous solutions. Is your friend correct? Explain your reasoning.
Answer:

Question 57.
USING STRUCTURE Explain how you know the radical equation \(\sqrt{x+4}\) = −5 has no real solution without solving it.
Answer:

Question 58.
HOW DO YOU SEE IT? Use the graph to find the solution of the equation 2\(\sqrt{x-4}\) = −\(\sqrt{x-1}\) + 4. Explain your reasoning.

Answer:

Question 59.
WRITING A company determines that the price p of a product can be modeled by p = 70 − \(\sqrt{0.02 x+1}\), where x is the number of units of the product demanded per day. Describe the effect that raising the price has on the number of units demanded.
Answer:

Question 60.
THOUGHT PROVOKING City officials rope off a circular area to prepare for a concert in the park. They estimate that each person occupies 6 square feet. Describe how you can use a radical inequality to determine the possible radius of the region when P people are expected to attend the concert.

Answer:

Question 61.
MATHEMATICAL CONNECTIONS The Moeraki Boulders along the coast of New Zealand are stone spheres with radii of approximately 3 feet. A formula for the radius of a sphere is
r = \(\frac{1}{2} \sqrt{\frac{S}{\pi}}\)
where S is the surface area of the sphere. Find the surface area of a Moeraki Boulder.
Answer:

Question 62.
PROBLEM SOLVING You are trying to determine the height of a truncated pyramid, which cannot be measured directly. The height h and slant heightℓof the truncated pyramid are related by the formula below.

In the given formula, b1 and b2 are the side lengths of the upper and lower bases of the pyramid, respectively. When ℓ = 5, b₁ = 2, and b₂ = 4, what is the height of the pyramid?
Answer:

Question 63.
REWRITING A FORMULA A burning candle has a radius of r inches and was initially h0inches tall. After t minutes, the height of the candle has been reduced to h inches. These quantities are related by the formula
r = \(\sqrt{\frac{k t}{\pi\left(h_{0}-h\right)}}\)
where k is a constant. Suppose the radius of a candle is 0.875 inch, its initial height is 6.5 inches, and k = 0.04.
a. Rewrite the formula, solving for h in terms of t.
b. Use your formula in part (a) to determine the height of the candle after burning 45 minutes.
Answer:

Maintaining Mathematical Proficiency

Perform the indicated operation. (Section 4.2 and Section 4.3)

Question 64.
(x³ − 2x² + 3x + 1) + (x⁴ − 7x)
Answer:

Question 65.
(2x⁵ + x⁴ − 4x²) − (x⁵ − 3)
Answer:

Question 66.
(x³ + 2x² + 1)(x² + 5)
Answer:

Question 67.
(x⁴ + 2x³ + 11x² + 14x − 16) ÷ (x + 2)
Answer:

Let f(x) = x³ – 4x² + 6. Write a rule for g. Describe the graph of g as a transformation of the graph of f.(Section 4.7)

Question 68.
g(x) = f(−x) + 4
Answer:

Question 69.
g(x) = \(\frac{1}{2}\)f(x) − 3
Answer:

Question 70.
g(x) = −f(x − 1) + 6
Answer:

Lesson 5.5 Performing Function Operations

Essential Question
How can you use the graphs of two functions to sketch the graph of an arithmetic combination of the two functions?
Just as two real numbers can be combined by the operations of addition, subtraction, multiplication, and division to form other real numbers, two functions can be combined to form other functions. For example, the functions f(x) = 2x − 3 and g(x) = x² − 1 can be combined to form the sum, difference, product, or quotient of f and g.
f(x) + g(x) = (2x − 3) + (x² − 1) = x² + 2x − 4 sum
f(x) − g(x) = (2x − 3) − (x² − 1) = −x² + 2x − 2 difference
f(x) • g(x) = (2x − 3)(x² − 1) = 2x³ − 3x² − 2x + 3 product

EXPLORATION 1
Graphing the Sum of Two Functions
Work with a partner. Use the graphs of f and g to sketch the graph of f + g. Explain your steps.
Sample Choose a point on the graph of g. Use a compass or a ruler to measure its distance above or below the x-axis. If above, add the distance to the y-coordinate of the point with the same x-coordinate on the graph of f. If below, subtract the distance. Plot the new point. Repeat this process for several points. Finally, draw a smooth curve through the new points to obtain the graph of f + g.

Communicate Your Answer

Question 2.
How can you use the graphs of two functions to sketch the graph of an arithmetic combination of the two functions?

Question 3.
Check your answers in Exploration 1 by writing equations for f and g, adding the functions, and graphing the sum.

5.5 Lesson

Monitoring Progress

Question 1.
Let f(x) = −2x^2/3 and g(x) = 7x^2/3. Find (f + g)(x) and (f − g)(x) and state the domain of each. Then evaluate (f + g)(8) and (f − g)(8).

Question 2.
Let f(x) = 3x and g(x) = x^1/5. Find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate (fg)(32) and (\(\frac{f}{g}\))(32).

Question 3.
Let f(x) = 8x and g(x) = 2x^5/6. Use a graphing calculator to evaluate (f + g)(x), (f − g)(x), (fg)(x), and (\(\frac{f}{g}\)) (x) when x = 5. Round your answers to two decimal places.

Question 4.
In Example 5, explain why you can evaluate (f + g)(3), (f − g)(3), and (fg)(3) but not (\(\frac{f}{g}\))(3).

Question 5.
Use the answer in Example 6(a) to find the total number of heartbeats over the lifetime of a white rhino when its body mass is 1.7 × 10⁵ kilograms.

Performing Function Operations 5.5 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING Let f and g be any two functions. Describe how you can use f, g, and the four basic operations to create new functions.
Answer:

Question 2.
WRITING What x-values are not included in the domain of the quotient of two functions?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, find (f + g)(x) and (f – g)(x) and state the domain of each. Then evaluate f + g and f – g for the given value of x.

Question 3.
f(x) = \(-5 \sqrt[4]{x}\), g(x) = 19\(\sqrt[4]{x}\); x = 16
Answer:

Question 4.
f(x) = \(\sqrt[3]{2 x}\), g(x) = −11\(\sqrt[3]{2 x}\) ; x = −4
Answer:

Question 5.
f(x) = 6x − 4x²− 7x³, g(x) = 9x²− 5x; x = −1
Answer:

Question 6.
f(x) = 11x + 2x², g(x) = −7x − 3x² + 4; x = 2
Answer:

In Exercises 7–12, find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate fg and \(\frac{f}{g}\) for the given value of x.

Question 7.
f(x) = 2x³, g(x) = \(\sqrt[3]{x}\) ; x = −27
Answer:

Question 8.
f(x) = x⁴, g(x) = \(3 \sqrt{x}\) ; x = 4
Answer:

Question 9.
f(x) = 4x, g(x) = 9x^1/2; x = 9
Answer:

Question 10.
f(x) = 11x³, g(x) = 7x^7/3; x = −8
Answer:

Question 11.
f(x) = 7x^3/2, g(x) =−14x^1/3; x = 64
Answer:

Question 12.
f(x) = 4x^5/4, g(x) = 2x^1/2; x = 16
Answer:

USING TOOLS In Exercises 13–16, use a graphing calculator to evaluate (f + g)(x), (f − g)(x), (fg)(x), and (\(\frac{f}{g}\))(x) when x = 5. Round your answers to two decimal places.

Question 13.
f(x) = 4x⁴; g(x) = 24x^1/3

Answer:

Question 14.
f(x) = 7x^5/3; g(x) = 49x^2/3

Answer:

Question 15.
f(x) =−2x^1/3; g(x) = 5x^1/2

Answer:

Question 16.
f(x) = 4x^1/2; g(x) = 6x^3/4

Answer:

ERROR ANALYSIS In Exercises 17 and 18, describe and correct the error in stating the domain.

Question 17.

Answer:

Question 18.

Answer:

Question 19.
MODELING WITH MATHEMATICS From 1990 to 2010, the numbers (in millions) of female F and male M employees from the ages of 16 to 19 in the United States can be modeled by F(t) =−0.007t² + 0.10t + 3.7 and M(t) = 0.0001t³ − 0.009t² + 0.11t + 3.7, where t is the number of years since 1990.
a. Find (F + M)(t).
b. Explain what (F + M)(t) represents.
Answer:

Question 20.
MODELING WITH MATHEMATICS From 2005 to 2009, the numbers of cruise ship departures (in thousands) from around the world W and Florida F can be modeled by the equations
W(t) = −5.8333t³ + 17.43t² + 509.1t + 11496
F(t) = 12.5t³ − 60.29t² + 136.6t + 4881
where t is the number of years since 2005.
a. Find (W − F )(t).
b. Explain what (W − F )(t) represents.
Answer:

Question 21.
MAKING AN ARGUMENT Your friend claims that the addition of functions and the multiplication of functions are commutative. Is your friend correct? Explain your reasoning.
Answer:

Question 22.
HOW DO YOU SEE IT? The graphs of the functions f(x) = 3x² − 2x − 1 and g(x) = 3x + 4 are shown. Which graph represents the function f + g? the function f − g? Explain your reasoning.

Answer:

Question 23.
REASONING The table shows the outputs of the two functions f and g. Use the table to evaluate (f + g)(3), (f − g)(1), (fg)(2), and (\(\frac{f}{g}\))(0).

Answer:

Question 24.
THOUGHT PROVOKING Is it possible to write two functions whose sum contains radicals, but whose product does not? Justify your answers.
Answer:

Question 25.
MATHEMATICAL CONNECTIONS A triangle is inscribed in a square, as shown. Write and simplify a function r in terms of x that represents the area of the shaded region.

Answer:

Question 26.
REWRITING A FORMULA For a mammal that weighs w grams, the volume b (in milliliters) of air breathed in and the volume d (in milliliters) of “dead space” (the portion of the lungs not filled with air) can be modeled by
b(w) = 0.007w and d(w) = 0.002w.
The breathing rate r (in breaths per minute) of a mammal that weighs w grams can be modeled by
r(w) = \(\frac{1.1 w^{0.734}}{b(w)-d(w)}\).
Answer:

Question 27.
PROBLEM SOLVING A mathematician at a lake throws a tennis ball from point A along the water’s edge to point B in the water, as shown. His dog, Elvis, first runs along the beach from point A to point D and then swims to fetch the ball at point B.

a. Elvis runs at a speed of about 6.4 meters per second. Write a function r in terms of x that represents the time he spends running from point A to point D. Elvis swims at a speed of about 0.9 meter per second. Write a function s in terms of x that represents the time he spends swimming from point D to point B.
b. Write a function t in terms of x that represents the total time Elvis spends traveling from point A to point D to point B.
c. Use a graphing calculator to graph t. Find the value of x that minimizes t. Explain the meaning of this value.
Answer:

Maintaining Mathematical Proficiency

Solve the literal equation for n. (Skills Review Handbook)

Question 28.
3xn − 9 = 6y
Answer:

Question 29.
5z = 7n + 8nz
Answer:

Question 30.
3nb = 5n − 6z
Answer:

Question 31.
\(\frac{3+4 n}{n}\) = 7b
Answer:

Determine whether the relation is a function. Explain. (Skills Review Handbook)

Question 32.
(3, 4), (4, 6), (1, 4), (2, −1)
Answer:

Question 33.
(−1, 2), (3, 7), (0, 2), (−1, −1)
Answer:

Question 34.
(1, 6), (7, −3), (4, 0), (3, 0)
Answer:

Question 35.
(3, 8), (2, 5), (9, 5), (2, −3)
Answer:

Lesson 5.6 Inverse of a Function

Essential Question
How can you sketch the graph of the inverse of a function?

EXPLORATION 1
Graphing Functions and Their Inverses
Work with a partner. Each pair of functions are inverses of each other. Use a graphing calculator to graph f and g in the same viewing window. What do you notice about the graphs?

a. f(x) = 4x + 3
g(x) = \(\frac{x-3}{4}\)
b. f(x) = x³ + 1
g(x) = \(\sqrt[3]{x-1}\)
c. f(x) = \(\sqrt{x-3}[latex]
g(x) = x² + 3, x ≥ 0
d. f(x) = [latex]\frac{4 x+4}{x+5}\)
g(x) = \(\frac{4-5 x}{x-4}\)

EXPLORATION 2
Sketching Graphs of Inverse Functions
Work with a partner. Use the graph of f to sketch the graph of g, the inverse function of f, on the same set of coordinate axes. Explain your reasoning.

Communicate Your Answer

Question 3.
How can you sketch the graph of the inverse of a function?

Question 4.
In Exploration 1, what do you notice about the relationship between the equations of f and g? Use your answer to find g, the inverse function of
f(x) = 2x − 3.
Use a graph to check your answer.

5.6 Lesson

Monitoring Progress

Solve y = f(x) for x. Then find the input(s) when the output is 2.

Question 1.
f(x) = x − 2

Question 2.
f(x) = 2x²

Question 3.
f(x) = −x³ + 3

Find the inverse of the function. Then graph the function and its inverse.

Question 4.
f(x) = 2x

Question 5.
f(x) = −x + 1

Question 6.
f(x) = \(\frac{1}{3}\)x − 2

Find the inverse of the function. Then graph the function and its inverse.

Question 7.
f(x) = −x², x ≤ 0

Question 8.
f(x) = −x³ + 4

Question 9.
f(x) = \(\sqrt{x+2}\)

Determine whether the functions are inverse functions.

Question 10.
f(x) = x + 5, g(x) = x − 5

Question 11.
f(x) = 8x³, g(x) = \(\sqrt[3]{2 x}\)

Question 12.
The distance d (in meters) that a dropped object falls in t seconds on Earth is represented by d = 4.9t². Find the inverse of the function. How long does it take an object to fall 50 meters?

Inverse of a Function 5.6 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY In your own words, state the definition of inverse functions.
Answer:

Question 2.
WRITING Explain how to determine whether the inverse of a function is also a function.
Answer:

Question 3.
COMPLETE THE SENTENCE Functions f and g are inverses of each other provided that f(g(x)) = ____ and g(f(x)) = ____.
Answer:

Question 4.
DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, solve y = f(x) for x. Then find the input(s) when the output is -3.

Question 5.
f(x) = 3x + 5
Answer:

Question 6.
f(x) = −7x − 2
Answer:

Question 7.
f(x) = \(\frac{1}{2}\)x − 3
Answer:

Question 8.
f(x) = −\(\frac{2}{3}\)x + 1
Answer:

Question 9.
f(x) = 3x³

Answer:

Question 10.
f(x) = 2x⁴ − 5
Answer:

Question 11.
f(x) = (x − 2)² − 7
Answer:

Question 12.
f(x) = (x − 5)³ − 1
Answer:

In Exercises 13–20, find the inverse of the function. Then graph the function and its inverse.

Question 13.
f(x) = 6x
Answer:

Question 14.
f(x) = −3x
Answer:

Question 15.
f(x) = −2x + 5
Answer:

Question 16.
f(x) = 6x − 3
Answer:

Question 17.
f(x) = −\(\frac{1}{2}\)x + 4
Answer:

Question 18.
f(x) = \(\frac{1}{3}\)x − 1
Answer:

Question 19.
f(x) = \(\frac{2}{3}\)x − \(\frac{1}{3}\)
Answer:

Question 20.
f(x) = −\(\frac{4}{5}\)x + \(\frac{1}{5}\)
Answer:

Question 21.
COMPARING METHODS Find the inverse of the function f(x) = −3x + 4 by switching the roles of x and y and solving for y. Then find the inverse of the function f by using inverse operations in the reverse order. Which method do you prefer? Explain.
Answer:

Question 22.
REASONING Determine whether each pair of functions f and g are inverses. Explain your reasoning.

Answer:

In Exercises 23–28, find the inverse of the function. Then graph the function and its inverse.

Question 23.
f(x) = 4x², x ≤ 0
Answer:

Question 24.
f(x) = 9x², x ≤ 0
Answer:

Question 25.
f(x) = (x − 3)³

Answer:

Question 26.
f(x) = (x + 4)³

Answer:

Question 27.
f(x) = 2x⁴, x ≥ 0
Answer:

Question 28.
f(x) = −x⁶, x ≥ 0
Answer:

ERROR ANALYSIS In Exercises 29 and 30, describe and correct the error in finding the inverse of the function.

Question 29.

Answer:

Question 30.

Answer:

USING TOOLS In Exercises 31–34, use the graph to determine whether the inverse of f is a function. Explain your reasoning.

Question 31.

Answer:

Question 32.

Answer:

Question 33.

Answer:

Question 34.

Answer:

In Exercises 35–46, determine whether the inverse of f is a function. Then find the inverse.

Question 35.
f(x) = x³ − 1
Answer:

Question 36.
f(x) = −x³ + 3
Answer:

Question 37.
f(x) = \(\sqrt{x+4}\)
Answer:

Question 38.
f(x) = \(\sqrt{x-6}\)
Answer:

Question 39.
f(x) = \(2 \sqrt[3]{x-5}\)
Answer:

Question 40.
f(x) = 2x² − 5
Answer:

Question 41.
f(x) = x⁴ + 2
Answer:

Question 42.
f(x) = 2x³ − 5
Answer:

Question 43.
f(x) = \(3 \sqrt[3]{x+1}\)
Answer:

Question 44.
f(x) = \(-\sqrt[3]{\frac{2 x+4}{3}}\)
Answer:

Question 45.
f(x) = \(\frac{1}{2}\)x⁵

Answer:

Question 46.
f(x) = \(-3 \sqrt{\frac{4 x-7}{3}}\)
Answer:

Question 47.
WRITING EQUATIONS What is the inverse of the function whose graph is shown?
A. g(x) = \(\frac{3}{2}\)x − 6
B. g(x) = \(\frac{3}{2}\)x + 6
C. g(x) = \(\frac{2}{3}\)x − 6
D. g(x) = \(\frac{2}{3}\)x + 12

Answer:

Question 48.
WRITING EQUATIONS What is the inverse of f(x) = −\(-\frac{1}{64}\)x³?
A. g(x) = −4x³
B. g(x) = 4\(\sqrt[3]{x}\)
C. g(x) = −4\(\sqrt[3]{x}\)
D. g(x) = \(\sqrt[3]{-4 x}\)
Answer:

In Exercises 49–52, determine whether the functions are inverses.

Question 49.
f(x) = 2x − 9, g(x) = x — 2 + 9
Answer:

Question 50.
f(x) = \(\frac{x-3}{4}\), g(x) = 4x + 3
Answer:

Question 51.
f(x) = \(\sqrt[5]{\frac{x+9}{5}}\), g(x) = 5x⁵ − 9
Answer:

Question 52.
f(x) = 7x^3/2 − 4, g(x) = (\(\frac{x+4}{7}\))^3/2

Answer:

Question 53.
MODELING WITH MATHEMATICS The maximum hull speed v (in knots) of a boat with a displacement hull can be approximated by v = 1.34\(\sqrt{\ell}\), where ℓ is the waterline length (in feet) of the boat. Find the inverse function. What waterline length is needed to achieve a maximum speed of 7.5 knots?

Answer:

Question 54.
MODELING WITH MATHEMATICS Elastic bands can be used for exercising to provide a range of resistance. The resistance R (in pounds) of a band can be modeled by R = \(\frac{3}{8}\)L − 5, where L is the total length (in inches) of the stretched band. Find the inverse function. What length of the stretched band provides 19 pounds of resistance?

Answer:

ANALYZING RELATIONSHIPS In Exercises 55–58, match the graph of the function with the graph of its inverse.

Question 55.

Answer:

Question 56.

Answer:

Question 57.

Answer:

Question 58.

Answer:

Question 59.
REASONING You and a friend are playing a number-guessing game. You ask your friend to think of a positive number, square the number, multiply the result by 2, and then add 3. Your friend’s final answer is 53. What was the original number chosen? Justify your answer.
Answer:

Question 60.
MAKING AN ARGUMENT Your friend claims that every quadratic function whose domain is restricted to nonnegative values has an inverse function. Is your friend correct? Explain your reasoning.
Answer:

Question 61.
PROBLEM SOLVING When calibrating a spring scale, you need to know how far the spring stretches for various weights. Hooke’s Law states that the length a spring stretches is proportional to the weight attached to it. A model for one scale is ℓ = 0.5w + 3, whereℓ is the total length (in inches) of the stretched spring and w is the weight (in pounds) of the object.

a. Find the inverse function. Describe what it represents.
b. You place a melon on the scale, and the spring stretches to a total length of 5.5 inches. Determine the weight of the melon.
c. Verify that the function ℓ = 0.5w + 3 and the inverse model in part (a) are inverse functions.
Answer:

Question 62.
THOUGHT PROVOKING Do functions of the form y = x^m/n, where m and n are positive integers, have inverse functions? Justify your answer with examples.
Answer:

Question 63.
PROBLEM SOLVING At the start of a dog sled race in Anchorage, Alaska, the temperature was 5°C. By the end of the race, the temperature was −10°C. The formula for converting temperatures from degrees Fahrenheit F to degrees Celsius C is C = \(\frac{5}{9}\)(F − 32).
a. Find the inverse function. Describe what it represents.
b. Find the Fahrenheit temperatures at the start and end of the race.

c. Use a graphing calculator to graph the original function and its inverse. Find the temperature that is the same on both temperature scales.
Answer:

Question 64.
PROBLEM SOLVING The surface area A (in square meters) of a person with a mass of 60 kilograms can be approximated by A = 0.2195h^0.3964, where h is the height (in centimeters) of the person.
a. Find the inverse function. Then estimate the height of a 60-kilogram person who has a body surface area of 1.6 square meters.
b. Verify that function A and the inverse model in part (a) are inverse functions.
Answer:

USING STRUCTURE In Exercises 65–68, match the function with the graph of its inverse.

Question 65.
f(x) = \(\sqrt[3]{x-4}\)
Answer:

Question 66.
f(x) = \(\sqrt[3]{x+4}\)
Answer:

Question 67.
f(x) = \(\sqrt{x+1}\) – 3
Answer:

Question 68.
f(x) = \(\sqrt{x-1}\) + 3
Answer:

Question 69.
DRAWING CONCLUSIONS Determine whether the statement is true or false. Explain your reasoning.
a. If f(x) = xⁿ and n is a positive even integer, then the inverse of f is a function.
b. If f(x) = xⁿ and n is a positive odd integer, then the inverse of f is a function.
Answer:

Question 70.
HOW DO YOU SEE IT? The graph of the function f is shown. Name three points that lie on the graph of the inverse of f. Explain your reasoning.

Answer:

Question 71.
ABSTRACT REASONING Show that the inverse of any linear function f(x) = mx + b, where m ≠ 0, is also a linear function. Identify the slope and y-intercept of the graph of the inverse function in terms of m and b.
Answer:

Question 72.
CRITICAL THINKING Consider the function f(x) = −x.
a. Graph f(x) = −x and explain why it is its own inverse. Also, verify that f(x) = −x is its own inverse algebraically.
b. Graph other linear functions that are their own inverses. Write equations of the lines you graphed.
c. Use your results from part (b) to write a general equation describing the family of linear functions that are their own inverses.
Answer:

Maintaining Mathematical Proficiency

Simplify the expression. Write your answer using only positive exponents.(Skills Review Handbook)

Question 73.
(−3)⁻³

Answer:

Question 74.
2³ • 2²

Answer:

Question 75.
\(\frac{4^{5}}{4^{3}}\)
Answer:

Question 76.
(\(\frac{2}{3}\))⁴

Answer:

Describe the x-values for which the function is increasing, decreasing, positive, and negative.

Question 77.

Answer:

Question 78.

Answer:

Question 79.

Answer:

Rational Exponents and Radical Functions Performance Task: Turning the Tables

5.4–5.6 What Did You Learn?

Core Vocabulary
radical equation, p. 262
extraneous solutions, p. 263
inverse functions, p. 277

Core Concepts
Section 5.4
Solving Radical Equations, p. 262
Solving Radical Inequalities, p. 265

Section 5.5
Operations on Functions, p. 270

Section 5.6
Exploring Inverses of Functions, p. 276
Inverses of Nonlinear Functions, p. 278
Horizontal Line Test, p. 278

Mathematical Practices

Question 1.
How did you find the endpoints of the range in part (b) of Exercise 54 on page 267?

Question 2.
How did you use structure in Exercise 57 on page 268?

Question 3.
How can you evaluate the reasonableness of the results in Exercise 27 on page 274?

Question 4.
How can you use a graphing calculator to check your answers in Exercises 49–52 on page 282?

Performance Task

Turning the Tables

In this chapter, you have used properties of rational exponents and functions to find an answer to the problem. Using those same properties, can you find a problem to the answer? How many problems can you find?

To explore the answers to these questions and more, go to BigIdeasMath.com

Rational Exponents and Radical Functions Chapter Review

Evaluate the expression without using a calculator.

Question 1.
8^7/3

Question 2.
9^5/2

Question 3.
(−27)^−2/3

Find the real solution(s) of the equation. Round your answer to two decimal places when appropriate.

Question 4.
x⁵ + 17 = 35

Question 5.
7x³ = 189

Question 6.
(x + 8)⁴ = 16

Simplify the expression.

Question 7.
(\(\frac{6^{1 / 5}}{6^{2 / 5}}\))³

Question 8.
\(\sqrt[4]{32} \cdot \sqrt[4]{8}\)

Question 9.
\(\frac{1}{2-\sqrt[4]{9}}\)

Question 10.
\(4 \sqrt[5]{8}\) + \(3 \sqrt[5]{8}\)

Question 11.
\(2 \sqrt{48}\) – \(\sqrt{3}\)

Question 12.
(\(5^{2 / 3} \cdot 2^{3 / 2}\))^1/2

Simplify the expression. Assume all variables are positive.

Question 13.
\(\sqrt[3]{125 z^{9}}\)

Question 14.
\(\frac{2^{1 / 4} z^{5 / 4}}{6 z}\)

Question 15.
\(\sqrt{10 z^{5}}-z^{2} \sqrt{40 z}\)

Describe the transformation of f represented by g. Then graph each function.

Question 16.
f(x) = \(\sqrt{x}\), g(x) = −2\(\sqrt{x}\)

Question 17.
f(x) = \(\sqrt[3]{x}\), g(x) = \(\sqrt[3]{-x}\) − 6

Question 18.
Let the graph of g be a reflection in the y-axis, followed by a translation 7 units to the right of the graph of f(x) = \(\sqrt[3]{x}\). Write a rule for g.

Question 19.
Use a graphing calculator to graph 2y² = x − 8. Identify the vertex and the direction that the parabola opens.

Question 20.
Use a graphing calculator to graph x² + y² = 81. Identify the radius and the intercepts.

Solve the equation. Check your solution.

Question 21.
\(4 \sqrt[3]{2 x+1}\) = 20

Question 22.
\(\sqrt{4 x-4}\) = \(\sqrt{5 x-1}\) – 1

Question 23.
(6x)^2/3 = 36

Solve the inequality.

Question 24.
\(5 \sqrt{x}\) + 2 > 17

Question 25.
\(2 \sqrt{x-8}\) < 24

Question 26.
7\(\sqrt[3]{x-3}\) ≥ 21

Question 27.
In a tsunami, the wave speeds (in meters per second) can be modeled by s(d ) = \(\sqrt{9.8 d}\), where d is the depth (in meters) of the water. Estimate the depth of the water when the wave speed is 200 meters per second.

Question 28.
Let f(x) = 2\(\sqrt{3-x}\) and g(x) = 4\(\sqrt[3]{3-x}\). Find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate(fg)(2) and (\(\frac{f}{g}\))(2).

Question 29.
Let f(x) = 3x² + 1 and g(x) = x + 4. Find (f + g)(x) and (f − g)(x) and state the domain of each. Then evaluate (f + g)(−5) and (f − g)(−5).

Question 30.
f(x) = −\(\frac{1}{2}\)x + 10

Question 31.
f(x) = x² + 8, x ≥ 0

Question 32.
f(x) = −x³ − 9

Question 33.
f(x) = 3\(\sqrt{x}\) + 5

Determine whether the functions are inverse functions.

Question 34.
f(x) = 4(x − 11)², g(x) = \(\frac{1}{4}\)(x + 11)²

Question 35.
f(x) = −2x + 6, g(x) = −\(\frac{1}{2}\)x + 3

Question 36.
On a certain day, the function that gives U.S. dollars in terms of British pounds is d = 1.587p, where d represents U.S. dollars and p represents British pounds. Find the inverse function. Then find the number of British pounds equivalent to 100 U.S. dollars.

Rational Exponents and Radical Functions Chapter Test

Question 1.
Solve the inequality \(\) − 2 ≤ 13 and the equation 5\(\) − 2 = 13. Describe the similarities and differences in solving radical equations and radical inequalities.

Describe the transformation of f represented by g. Then write a rule for g.

Question 2.
f(x) = \(\sqrt{x}\)

Question 3.
f(x) = \(\sqrt[3]{x}\)

Question 4.
f(x) = \(\sqrt[5]{x}\)

Simplify the expression. Explain your reasoning.

Question 5.
64^2/3

Question 6.
(−27)^5/3

Question 7.
\(\sqrt[4]{48 x y^{11} z^{3}}\)

Question 8.
\(\frac{\sqrt[3]{256}}{\sqrt[3]{32}}\)

Question 9.
Write two functions whose graphs are translations of the graph of y = \(\sqrt{x}\). The first function should have a domain of x ≥ 4. The second function should have a range of y ≥ −2.

Question 10.
In bowling, a handicap is a change in score to adjust for differences in the abilities of players. You belong to a bowling league in which your handicap h is determined using the formula h = 0.9(200 − a), where a is your average score. Find the inverse of the model. Then find the average for a bowler whose handicap is 36.

Question 11.
The basal metabolic rate of an animal is a measure of the amount of calories burned at rest for basic functioning. Kleiber’s law states that an animal’s basal metabolic rate R (in kilocalories per day) can be modeled by R = 73.3w^3/4, where w is the mass (in kilograms) of the animal. Find the basal metabolic rates of each animal in the table.

Question 12.
Let f(x) = 6x^3/5 and g(x) = −x^3/5. Find (f + g)(x) and (f − g)(x) and state the domain of each. Then evaluate (f + g)(32) and (f − g)(32).

Question 13.
Let f(x) = \(\frac{1}{2}\)x^3/4 and g(x) = 8x. Find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate (fg)(16) and (\(\frac{f}{g}\))(16).

Question 14.
A football player jumps to catch a pass. The maximum height h (in feet) of the player above the ground is given by the function h = \(\frac{1}{64}\)s², where s is the initial speed (in feet per second) of the player. Find the inverse of the function. Use the inverse to find the initial speed of the player shown. Verify that the functions are inverse functions.

Rational Exponents and Radical Functions Cumulative Assessment

Question 1.
Identify three pairs of equivalent expressions. Assume all variables are positive. Justify your answer.

Question 2.
The graph represents the function . Choose the correct values to complete the function.

Question 3.
In rowing, the boat speed s (in meters per second) can be modeled by s= 4.62\(\sqrt[9]{n}\), where n is the number of rowers.
a. Find the boat speeds for crews of 2 people, 4 people, and 8 people.
b. Does the boat speed double when the number of rowers doubles? Explain.
c. Find the time (in minutes) it takes each crew in part (a) to complete a 2000-meter race.

Question 4.
A polynomial function fits the data in the table. Use finite differences to find the degree of the function and complete the table. Explain your reasoning.

Question 5.
The area of the triangle is 42 square inches. Find the value of x.

Question 6.
Which equations are represented by parabolas? Which equations are functions? Place check marks in the appropriate spaces. Explain your reasoning.

Question 7.
What is the solution of the inequality 2\(\sqrt{x+3}\) − 1 < 3?
A. x < 1
B. −3 < x < 1
C. −3 ≤ x < 1
D. x ≥ −3

Question 8.
Which function does the graph represent? Explain your reasoning.

Question 9.
Your friend releases a weather balloon 50 feet from you. The balloon rises vertically. When the balloon is at height h, the distance d between you and the balloon is given by d = \(\sqrt{2500+h^{2}}\), where h and d are measured in feet. Find the inverse of the function. What is the height of the balloon when the distance between you and the balloon is 100 feet?

Question 10.
The graphs of two functions f and g are shown. Are f and g inverse functions? Explain your reasoning.

Looking across the web for a friendly site that caters to all your needs regarding the Big Ideas Math Geometry Concepts? Don’t worry we are with you in this and we will provide you the Complete Big Ideas Math Geometry Answers Chapter 1 Basics of Geometry. All the Solutions covered in the BIM Geometry Ch 1 Basics of Geometry Answer Key are as per the latest Common Core State Standards guidelines. Practice using the BigIdeas Math Chapter 1 Solutions via quick links available and clear the exams with flying colors.

Big Ideas Math Book Geometry Answer Key Chapter 1 Basics of Geometry

Access the Big Ideas Math Geometry Ch 1 Answers for all the topics and prepare accordingly. Find complete assistance on Geometry Chapter 1 including questions from Lessons 1.1-1.6, Performance Tests, Review Tests, Cumulative Practice, Assessment Tests, etc. All you have to do is simply tap on the quick links available and clear your ambiguities in no time. To make it easy for you we have compiled the BIM Geometry Chapter 1 Basics of Geometry Textbook Solutions aligned as per the BIM Textbooks.

• Basics of Geometry Maintaining Mathematical Proficiency – Page 1
• Basics of Geometry Mathematical Practices – Page 2
• 1.1 Points, Lines, and Planes – Page(3-10)
• Lesson 1.1 Points, Lines, and Planes – Page(4-7)
• Exercise 1.1 Points, Lines, and Planes – Page(8-10)
• 1.2 Measuring and Constructing Segments – Page(11-18)
• Lesson 1.2 Measuring and Constructing Segments – Page(12-15)
• Exercise 1.2 Measuring and Constructing Segments – Page(16-18)
• 1.3 Using Midpoint and Distance Formulas – Page(19-26)
• Lesson 1.3 Using Midpoint and Distance Formulas – Page(20-23)
• Exercise 1.3 Using Midpoint and Distance Formulas -Page(24-26)
• Study Skills: Keeping Your Mind Focused – Page 27
• 1.1 – 1.3 Quiz – Page 28
• 1.4 Perimeter and Area in the Coordinate Plane – Page(28-36)
• Lesson 1.4 Perimeter and Area in the Coordinate Plane – Page(30-33)
• Exercise 1.4 Perimeter and Area in the Coordinate Plane – Pae(34-36)
• 1.5 Measuring and Constructing Angles – Page(37-46)
• Lesson 1.5 Measuring and Constructing Angles – Page(38-43)
• Exercise 1.5 Measuring and Constructing Angles – Page(43-46)
• 1.6 Describing Pairs of Angles – Page(47-54)
• Lesson 1.6 Describing Pairs of Angles – Page(48-51)
• Exercise 1.6 Describing Pairs of Angles – Page(52-54)
• 1.4 – 1.6 Performance Task: Comfortable Horse Stalls – Page 55
• Basics of Geometry Chapter Review – Page(56-58)
• Basics of Geometry Chapter Test – Page 59
• Basics of Geometry Cumulative Assessment – Page(60-61)

Basics of Geometry Maintaining Mathematical Proficiency

Simplify the expression.

Question 1.
|8 – 12|
Answer:
The given absolute value expression is:
| 8 – 12|
We know that,
| x | = x for x > 0
| -x | = x for x > 0
So,
| 8 – 12 |= | -4| = 4
Hence, from the above,
We can conclude that the value of the given absolute value expression is: 4

Question 2.
|- 6 – 5|
Answer:
The given absolute value expression is:
| -6 – 5|
We know that,
| x | = x for x > 0
| -x | = x for x > 0
So,
| -6 – 5| = | -11 | = 11
Hence, from the above,
We can conclude that the value of the given absolute value expression is: 11

Question 3.
|4 + (-9)|
Answer:
The given absolute value expression is:
| 4 + (-9) |
We know that,
| x | = x for x > 0
| -x | = x for x > 0
So,
| 4 + (-9) | = | 4 – 9 |
= | -5 | = 5
Hence, from the above,
We can conclude that the value of the given absolute value expression is: 5

Question 4.
|13 + (-4)|
Answer:
The given absolute value expression is:
| 13 + (-4) |
We know that,
| x | = x for x > 0
| -x | = x for x > 0
So,
| 13 + (-4) | = | 13 – 4 |
= | 9 | = 9
Hence, from the above,
We can conclude that the value of the given absolute value expression is: 9

Question 5.
|6 – (- 2)|
Answer:
The given absolute value expression is:
| 6 – (-2) |
We know that,
| x | = x for x > 0
| -x | = x for x > 0
So,
| 6 – (-2) | = | 6 + 2 |
= | 8 | = 8
Hence, from the above,
We can conclude that the value of the given absolute value expression is: 8

Question 6.
|5 – (- 1)|
Answer:
The given absolute value expression is:
| 5 – (-1) |
We know that,
| x | = x for x > 0
| -x | = x for x > 0
So,
| 5 – (-1) | = | 5 + 1 |
= | 6 | = 6
Hence, from the above,
We can conclude that the value of the given absolute value expression is: 6

Question 7.
|- 8 – (- 7)|
Answer:
The given absolute value expression is:
| -8 – (-7) |
We know that,
| x | = x for x > 0
| -x | = x for x > 0
So,
| -8 – (-7) | = | -8 + 7 |
= | -1 | = 1
Hence, from the above,
We can conclude that the value of the given absolute value expression is: 1

Question 8.
|8 – 13|
Answer:
The given absolute value expression is:
| 8 – 13 |
We know that,
| x | = x for x > 0
| -x | = x for x > 0
So,
| 8 – 13 | = | -5 | = 5
Hence, from the above,
We can conclude that the value of the given absolute value expression is: 5

Question 9.
|- 14 – 3|
Answer:
The given absolute value expression is:
| -14 – 3 |
We know that,
| x | = x for x > 0
| -x | = x for x > 0
So,
| -14 – 3 | = | -17 | = 17
Hence, from the above,
We can conclude that the value of the given absolute value expression is: 17

Find the area of the triangle.

Question 10.

Answer:
The given figure is:

We know that,
The area of the triangle (A) is given as:
A = 1/2 ×Base × Height
So,
The area of the given triangle is:
A = 1/2 × 14m × 22m
= 11 × 14
= 154 m²
Hence, from the above,
We can conclude that the area of the given triangle is: 154 m²

Question 11.

Answer:
The given figure is:

We know that,
The area of the triangle (A) is given as:
A = \(\frac{1}{2}\) ×Base × Height
So,
The area of the given triangle is:
A = \(\frac{1}{2}\) × 7yd × 24yd
= \(\frac{1}{2}\) × \(\frac{24}{1}\) × 7
= 12 × 7
= 84 yd²
Hence, from the above,
We can conclude that the area of the given triangle is: 84 yd²

Question 12.

Answer:
The given figure is:

We know that,
The area of the triangle (A) is given as:
A = \(\frac{1}{2}\) ×Base × Height
So,
The area of the given triangle is:
A = \(\frac{1}{2}\) × 16 in × 25 in
= \(\frac{1}{2}\) × \(\frac{16}{1}\) × 25
= 8 × 25
= 200 in²
Hence, from the above,
We can conclude that the area of the given triangle is: 200 in²

Question 13.
ABSTRACT REASONING
Describe the possible values for x and y when |x – y| > 0. What does it mean when |x – y| = 0 ? Can |x – y| < 0? Explain your reasoning.
Answer:
We know that,
The value of the absolute expression must be  greater than or equal to 0 but not less than 0
So,
The values for | x – y | do not exist
Now,
The possible values of | x – y | > 0 should be greater than 0 and maybe x > y and x < y
The possible values of | x – y | = 0 should be only one value i.e., 0 as x and y must be equal to make the difference value 0

Basics of Geometry Mathematical Practices

Monitoring Progress

Question 1.
Find the area of the polygon using the specified units. Round your answer to the nearest hundredth.

Answer:
The given figure is:

We know that,
Area of the triangle = \(\frac{1}{2}\) × Base × Height
So,
From the figure,
The value of the Base is: 2 cm
The value of Height is: 2 cm
We know that,
1 cm = \(\frac{25}{64}\) inch = 0.3937 inch
So,
2cm = \(\frac{25}{32}\) inch = 0.7874 inch
So,
The area of the given triangle = \(\frac{1}{2}\) × \(\frac{25}{32}\) × \(\frac{25}{32}\)
= \(\frac{1}{2}\) × \(\frac{625}{1024}\)
= \(\frac{625}{2,048}\) inch²
= 0.3051 inch²
Hence, from the above,
We can conclude that the area of the given triangle in square inches is: 0.3051 inch²

Question 2.
parallelogram (square centimeters)

Answer:
The given figure is:

WE know hat,
The area of the parallelogram = Base × Height
So,
From the given figure,
The base of the parallelogram is: 2.5 in
The height of the parallelogram = 2 in
We know that,
1 inch = 2.54 cm
So,
2 inch = 2.08 cm
2.5 inch = 6.35 cm
So,
The area of the given parallelogram = 2.08 × 6.35
= 13.208 cm²
Hence, from the above,
We can conclude that the area of the given parallelogram in square cm is: 13.208 cm²

Question 3.
The distance between the two cities is 120 miles. What is the distance in kilometers? Round your answer to the nearest whole number.
Answer:
It is given that the distance between the two cities is 120 miles.
Now,
We know that,
1 mile = 1.609 kilometer
So,
The distance between the two cities in kilometers = 120 × 1.609
= 193 kilometers
Hence, from the above,
We can conclude that the distance between the two cities in kilometers is: 193 kilometers

1.1 Points, Lines, and Planes

Exploration 1

Using Dynamic Geometry Software

Work with a partner: Use dynamic geometry software to draw several points. Also draw some lines, line segments, and rays. What is the difference between a line, a line segment, and a ray?
Sample

Answer:

From the above,
The differences between a line, a ray, and a line segment are:
A Ray has no starting and ending points
A line has a starting point but no ending point
A line segment has both starting point and an ending point

Exploration 2

Intersections of Lines and Planes

Work with a partner:

a. Describe and sketch the ways in which two lines can intersect or not intersect. Give examples of each using the lines formed by the walls. floor. and ceiling in your classroom.
Answer:

Examples of non-intersecting lines are: Floor and ceiling
Examples of intersecting lines are: Walls and floor

b. Describe and sketch the ways in which a line and a plane can intersect or not intersect. Give examples of each using the walls. floor, and ceiling in your classroom.
Answer:

Examples of the intersection of a plane and a line are: Walls and ceiling
Examples of the non-intersection of a plane and a line are:  Floor and a blackboard

c. Describe and sketch the ways in which two planes can intersect or not intersect. Give examples of each using the walls. floor, and ceiling in your classroom.
Answer:

Examples of the intersection of planes are: Floor and benches
Examples of the non-intersection of planes are: Floor and ceiling

Exploration 3

Exploring Dynamic Geometry Software

UNDERSTANDING MATHEMATICAL TERMS
To be proficient in math, you need to understand definitions and previously established results. An appropriate tool, such as a software package, can sometimes help.

Work with a partner. Use dynamic geometry software to explore geometry. Use the software to find a term or concept that is unfamiliar to you. Then use the capabilities of the software to determine the meaning of the term or concept.
Answer:
The dynamic geometry software used is “GeoGebra”
The URL for “GeoGebra – Geometry” is:
https://www.geogebra.org/calculator
So,
Using the above URL,
Check the term that you are unfamiliar  to you and find the meaning of that term using the above URL

Communicate Your Answer

Question 4.
How can you use dynamic geometry software to visualize geometric concepts?
Answer:
Some of the uses dynamic geometry software to visualize geometric concepts are:
a. Request dispatching for cheap energy prices in cloud data centers
b. Springerlink training kit
c. Luminosity measurements at Hadron colliders
d. From word embeddings to document distances

Lesson 1.1 Points, Lines, and Planes

Monitoring Progress

Question 1.
Use the diagram in Example 1. Give two other names for . Name a point that is not coplanar with points Q. S, and T.

Answer:
The given plane is:

The definition of a ray is:
A ray has no starting and ending points
So,
The other names for \(\overline{S T}\) are: Line m and \(\overline{T S}\)
We know that,
“Co-planar points” are the points if all of them lie in the same plane
Hence, among the points Q, S, and T,
S and T are co-planar since they lie in the same plane and Q is not co-planar

Question 2.
Give another name for \(\overline{K L}\).

Answer:
The given figure is:

So,
The other name of \(\overline{K L}\) is: \(\overline{L K}\)

Question 3.
Are \(\vec{K}\)P and \(\vec{P}\)K the same ray? Are \(\vec{N}\)P and \(\vec{N}\)M the same ray? Explain.

Answer:
The given figure is:

So,
From the above figure,
We can observe that
\(\vec{K}\)P and \(\vec{P}\)K are not the same ray
Reason:
\(\overline{K L}\) is the ray and KP is the line segment
Now,
We can also observe that
\(\vec{N}\)P and \(\vec{N}\)M are in the same ray
Reason:
Since the points N, P, and M are collinear, the ray \(\vec{N}\)M and \(\vec{N}\)P are in the same plane

Question 4.
Sketch two different lines that intersect a plane at the same point.
Answer:
The representation of the two different lines that intersect a plane at the same point is:

Question 5.
Name the intersection of  and line k.

Answer:
The given figure is:

From the above figure,
We can observe that the intersection of \(\overline{P Q}\) and line k is: M

Question 6.
Name the intersection of plane A and plane B.

Answer:
The given figure is:

From the above figure,
We can observe that the intersection of plane A and plane B is: Line k

Question 7.
Name the intersection of line k and plane A.

Answer:
The given figure is:

From the given figure,
We can observe that the intersection of line k and plane A is: Line k

Monitoring Progress

Use the diagram that shows a molecule of phosphorus pentachloride.

Question 8.
Name two different planes that contain line s.
Answer:
The given figure is:

We know that,
A ‘Plane” can be formed by using any three non-collinear points on the same plane
Now,
From the given figure,
We can observe that
The different planes that contain line s are: JHG and KLI

Question 9.
Name three different planes that contain point K.
Answer:
The given figure is:

From the above figure,
The three different planes that contain pint K are: HGK, HKL, and KLI

Question 10.
Name two different planes that contain \(\vec{H}\)J.
Answer:
The given figure is:

From the above figure,
We can say that
The 2 different planes that contain \(\vec{H}\)J are: HJI and HJL

Exercise 1.1 Points, Lines, and Planes

Vocabulary and Core Concept Check

Question 1.
WRITING
Compare collinear points and coplanar points.
Answer:

Question 2.
WHICH ONE DOES DOESN’T BELONG?
Which term does not belong with the other three? Explain your reasoning.

Answer:
The given expressions are:
a. \(\overline{A B}\)
b. plane CDE
c. \(\vec{FG}\)
d. \(\vec{H}\)I
So,
From the above four expressions,
We can observe that the three expressions are 2-dimensional geometrical expressions whereas 1 figure is a 3-dimensional geometrical expression
The 2-dimensional geometrical expressions from the given expressions are: a), c), and d)
Hence, from the above,
We can conclude that expression c. does not belong with the other three

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, use the diagram.

Question 3.
Name four points.
Answer:

Question 4.
Name two lines.
Answer:
The given figure is:

Hence,
From the above figure,
We can conclude that the two lines are: \(\vec{B}\)C and \(\vec{D}\)E