lesson 7 homework answer key 5.1

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Eureka Math Grade 5 Module 5 Lesson 7 Answer Key

Engage ny eureka math 5th grade module 5 lesson 7 answer key, eureka math grade 5 module 5 lesson 7 sprint answer key.

Question 1. \(\frac{1}{2}\) × \(\frac{1}{2}\) = Answer: 1/4

Question 2. \(\frac{1}{2}\) × \(\frac{1}{3}\) = Answer: 1/6

Question 3. \(\frac{1}{2}\) × \(\frac{1}{4}\) = Answer: 1/8

Question 4. \(\frac{1}{2}\) × \(\frac{1}{7}\) = Answer: 1/14

Question 5. \(\frac{1}{7}\) × \(\frac{1}{2}\) = Answer: 1/14

Question 6. \(\frac{1}{3}\) × \(\frac{1}{2}\) = Answer: 1/6

Question 7. \(\frac{1}{3}\) × \(\frac{1}{3}\) = Answer: 1/9

Question 8. \(\frac{1}{3}\) × \(\frac{1}{6}\) = Answer: 1/18

Question 9. \(\frac{1}{3}\) × \(\frac{1}{5}\) = Answer: 1/15

Question 10. \(\frac{1}{5}\) × \(\frac{1}{3}\) = Answer: 1/15

Question 11. \(\frac{1}{5}\) × \(\frac{2}{3}\) = Answer: 2/15

Question 12. \(\frac{2}{3}\) × \(\frac{2}{3}\) = Answer: 4/9

Question 13. \(\frac{1}{4}\) × \(\frac{1}{3}\) = Answer: 1/12

Question 14. \(\frac{1}{4}\) × \(\frac{2}{3}\) = Answer: 2/12

Question 15. \(\frac{3}{4}\) × \(\frac{2}{3}\) = Answer: 6/12

Question 16. \(\frac{1}{6}\) × \(\frac{1}{3}\) = Answer: 1/18

Question 17. \(\frac{5}{6}\) × \(\frac{1}{3}\) = Answer: 5/18

Question 18. \(\frac{5}{6}\) × \(\frac{2}{3}\) = Answer: 10/18

Question 19. \(\frac{5}{4}\) × \(\frac{2}{3}\) = Answer: 10/12

Question 20. \(\frac{1}{5}\) × \(\frac{1}{5}\) = Answer: 1/25

Question 21. \(\frac{2}{5}\) × \(\frac{2}{5}\) = Answer: 4/25

Question 22. \(\frac{2}{5}\) × \(\frac{3}{5}\) = Answer: 6/25

Question 23. \(\frac{2}{5}\) × \(\frac{5}{3}\) = Answer: 10/15

Question 24. \(\frac{3}{5}\) × \(\frac{5}{2}\) = Answer: 15/10

Question 25. \(\frac{1}{3}\) × \(\frac{1}{3}\) = Answer: 1/9

Question 26. \(\frac{1}{3}\) × \(\frac{2}{3}\) = Answer: 2/9

Question 27. \(\frac{2}{3}\) × \(\frac{2}{3}\) = Answer: 4/9

Question 28. \(\frac{2}{3}\) × \(\frac{3}{2}\) = Answer: 6/6

Question 29. \(\frac{2}{3}\) × \(\frac{4}{3}\) = Answer: 8/9

Question 30. \(\frac{2}{3}\) × \(\frac{5}{3}\) = Answer: 10/9

Question 31. \(\frac{3}{2}\) × \(\frac{3}{5}\) = Answer: 9/10

Question 32. \(\frac{3}{4}\) × \(\frac{1}{5}\) = Answer: 3/20

Question 33. \(\frac{3}{4}\) × \(\frac{4}{5}\) = Answer: 12/20

Question 34. \(\frac{3}{4}\) × \(\frac{5}{5}\) = Answer: 15/20

Question 35. \(\frac{3}{4}\) × \(\frac{6}{5}\) = Answer: 18/20

Question 36. \(\frac{1}{4}\) × \(\frac{6}{5}\) = Answer: 6/20

Question 37. \(\frac{1}{7}\) × \(\frac{1}{7}\) = Answer: 1/49

Question 38. \(\frac{1}{8}\) × \(\frac{3}{5}\) = Answer: 3/40

Question 39. \(\frac{5}{6}\) × \(\frac{1}{4}\) = Answer: 5/24

Question 40. \(\frac{3}{4}\) × \(\frac{3}{4}\) = Answer: 9/16

Question 41. \(\frac{2}{3}\) × \(\frac{6}{6}\) = Answer: 12/18

Question 42. \(\frac{3}{4}\) × \(\frac{6}{2}\) = Answer: 18/8

Question 43. \(\frac{7}{8}\) × \(\frac{7}{9}\) = Answer: 49/72

Question 44. \(\frac{7}{12}\) × \(\frac{9}{8}\) = Answer: 63/96

Question 1. \(\frac{1}{2}\) × \(\frac{1}{3}\) = Answer: 1/6

Question 2. \(\frac{1}{2}\) × \(\frac{1}{4}\) = Answer: 1/8

Question 3. \(\frac{1}{2}\) × \(\frac{1}{5}\) = Answer: 1/10

Question 4. \(\frac{1}{2}\) × \(\frac{1}{9}\) = Answer: 1/18

Question 5. \(\frac{1}{9}\) × \(\frac{1}{2}\) = Answer: 1/18

Question 6. \(\frac{1}{5}\) × \(\frac{1}{2}\) = Answer: 1/10

Question 7. \(\frac{1}{5}\) × \(\frac{1}{3}\) = Answer: 1/15

Question 8. \(\frac{1}{5}\) × \(\frac{1}{7}\) = Answer: 1/35

Question 9. \(\frac{1}{5}\) × \(\frac{1}{3}\) = Answer: 1/15

Question 10. \(\frac{1}{3}\) × \(\frac{1}{5}\) = Answer: 1/15

Question 11. \(\frac{1}{3}\) × \(\frac{2}{5}\) = Answer: 2/15

Question 12. \(\frac{2}{3}\) × \(\frac{2}{5}\) = Answer: 4/15

Question 13. \(\frac{1}{3}\) × \(\frac{1}{4}\) = Answer:1/12

Question 14. \(\frac{1}{3}\) × \(\frac{3}{4}\) = Answer: 3/12

Question 15. \(\frac{2}{3}\) × \(\frac{3}{4}\) = Answer: 6/12

Question 16. \(\frac{1}{3}\) × \(\frac{1}{6}\) = Answer: 1/18

Question 17. \(\frac{2}{3}\) × \(\frac{1}{6}\) = Answer: 2/18

Question 18. \(\frac{2}{3}\) × \(\frac{5}{6}\) = Answer: 10/18

Question 19. \(\frac{3}{2}\) × \(\frac{3}{4}\) = Answer: 9/8

Question 21. \(\frac{3}{5}\) × \(\frac{3}{5}\) = Answer: 9/25

Question 22. \(\frac{3}{5}\) × \(\frac{4}{5}\) = Answer: 12/25

Question 23. \(\frac{3}{5}\) × \(\frac{5}{4}\) = Answer: 15/20

Question 24. \(\frac{4}{5}\) × \(\frac{5}{3}\) = Answer: 20/15

Question 25. \(\frac{1}{4}\) × \(\frac{1}{4}\) = Answer: 1/16

Question 26. \(\frac{1}{4}\) × \(\frac{3}{4}\) = Answer: 3/16

Question 27. \(\frac{3}{4}\) × \(\frac{3}{4}\) = Answer: 9/16

Question 28. \(\frac{3}{4}\) × \(\frac{4}{3}\) = Answer: 12/12

Question 29. \(\frac{3}{4}\) × \(\frac{5}{4}\) = Answer: 15/16

Question 30. \(\frac{3}{4}\) × \(\frac{6}{4}\) = Answer: 18/16

Question 31. \(\frac{4}{3}\) × \(\frac{4}{6}\) = Answer: 16/18

Question 32. \(\frac{2}{3}\) × \(\frac{1}{5}\) = Answer: 2/15

Question 33. \(\frac{2}{3}\) × \(\frac{4}{5}\) = Answer: 8/15

Question 34. \(\frac{2}{3}\) × \(\frac{5}{5}\) = Answer: 10/15

Question 35. \(\frac{2}{3}\) × \(\frac{6}{5}\) = Answer: 12/15

Question 36. \(\frac{1}{3}\) × \(\frac{6}{5}\) = Answer: 6/15

Question 37. \(\frac{1}{9}\) × \(\frac{1}{9}\) = Answer: 1/81

Question 38. \(\frac{1}{5}\) × \(\frac{3}{8}\) = Answer: 3/40

Question 39. \(\frac{3}{4}\) × \(\frac{1}{6}\) = Answer: 3/24

Question 40. \(\frac{2}{3}\) × \(\frac{2}{3}\) = Answer: 4/9

Question 41. \(\frac{3}{4}\) × \(\frac{8}{8}\) = Answer: 24/32

Question 42. \(\frac{2}{3}\) × \(\frac{6}{3}\) = Answer: 12/9

Question 43. \(\frac{6}{7}\) × \(\frac{8}{9}\) = Answer: 48/63

Question 44. \(\frac{7}{12}\) × \(\frac{8}{7}\) = Answer: 72/84

Eureka Math Grade 5 Module 5 Lesson 7 Problem Set Answer Key

Geoffrey builds rectangular planters. Question 1. Geoffrey’s first planter is 8 feet long and 2 feet wide. The container is filled with soil to a height of 3 feet in the planter. What is the volume of soil in the planter? Explain your work using a diagram. Answer:

Volume = length x width x height ‘

V = 8 feet x 2 feet x 3 feet

V = 48 cubic feet

Therefore, the volume of the soil in the planter = 48 cubic feet

Question 2. Geoffrey wants to grow some tomatoes in four large planters. He wants each planter to have a volume of 320 cubic feet, but he wants them all to be different. Show four different ways Geoffrey can make these planters, and draw diagrams with the planters’ measurements on them. Planter A Planter B Planter C Planter D Answer:

Question 3. Geoffrey wants to make one planter that extends from the ground to just below his back window. The window starts 3 feet off the ground. If he wants the planter to hold 36 cubic feet of soil, name one way he could build the planter so it is not taller than 3 feet. Explain how you know. Answer:

Given that, the window starts 3 feet off the ground

36 cubic feet can be shown as 36/2 = 12

So, the measurements of the planter to hold 36 cubic feet of soil

= 4 feet by 3 feet by 3 feet.

Volume : 36 cubic units

Question 4. After all of this gardening work, Geoffrey decides he needs a new shed to replace the old one. His current shed is a rectangular prism that measures 6 feet long by 5 feet wide by 8 feet high. He realizes he needs a shed with 480 cubic feet of storage. a. Will he achieve his goal if he doubles each dimension? Why or why not? b. If he wants to keep the height the same, what could the other dimensions be for him to get the volume he wants? c. If he uses the dimensions in part (b), what could be the area of the new shed’s floor? Answer:

The volume of the current shed =  length x width x height

V = 6 x 5 x 8

V = 240 cubic feet.

According to given condition, if he doubles each dimension.

Then the volume = 12 x 10 x 16

= 1920 cubic feet

Therefore, he did not achieve his goal

To make 480 cubic feet he only needs to double one dimension but not all the three dimensions.

The other dimensions of the shed to calculate the same volume by keeping the height same =

Height = 8 feet and volume = 480 cubic feet.

Other dimension =

length = 12 feet and width = 5 feet

length = 6 feet and width = 8 feet

Area = length x width

According to the dimensions part b

Area = 12 x 5

Therefore, the area of  the new shed’s floor = 60 square feet.

Eureka Math Grade 5 Module 5 Lesson 7 Exit Ticket Answer Key

A storage shed is a rectangular prism and has dimensions of 6 meters by 5 meters by 12 meters. If Jean were to double these dimensions, she believes she would only double the volume. Is she correct? Explain why or why not. Include a drawing in your explanation. Answer:

Volume = length x width x height

V = 6 x 5 x 12

V = 360 cubic metres

According to given condition of doubling the volume, then

Volume = 2880 cubic metres.

288 cubic metres is much larger than 360 cubic metres. So, volume will not be double.

Eureka Math Grade 5 Module 5 Lesson 7 Homework Answer Key

Wren makes some rectangular display boxes. Question 1. Wren’s first display box is 6 inches long, 9 inches wide, and 4 inches high. What is the volume of the display box? Explain your work using a diagram. Answer:

Volume = length x width x volume

V = 6 x 9 x 4

V = 216 cubic inches

Therefore, volume of the display box = 216 cubic inches

Question 2. Wren wants to put some artwork into three shadow boxes. She knows they all need a volume of 60 cubic inches, but she wants them all to be different. Show three different ways Wren can make these boxes by drawing diagrams and labeling the measurements. Shadow Box A Shadow Box B Shadow Box C Answer:

Question 3. Wren wants to build a box to organize her scrapbook supplies. She has a stencil set that is 12 inches wide that needs to lay flat in the bottom of the box. The supply box must also be no taller than 2 inches. Name one way she could build a supply box with a volume of 72 cubic inches. Answer:

Give, the height of the supply box should not be taller than 2 inches

72 cubic inches can be shown as

Now, 36/2 = 18

So, the measurements of the box are 18 inches by 2 inches by 2 inches

Therefore, Wren can use a box of 18 by 2 by 2 by inches box

Question 4. After all of this organizing, Wren decides she also needs more storage for her soccer equipment. Her current storage box measures 1 foot long by 2 feet wide by 2 feet high. She realizes she needs to replace it with a box with 12 cubic feet of storage, so she doubles the width. a. Will she achieve her goal if she does this? Why or why not? b. If she wants to keep the height the same, what could the other dimensions be for a 12-cubic-foot storage box? c. If she uses the dimensions in part (b), what is the area of the new storage box’s floor? d. How has the area of the bottom in her new storage box changed? Explain how you know. Answer:

Given, the measurements of the storage box = 1 foot long by 2 feet wide by 2 feet high.

V = 1 x 2 x 2

V = 4 cubic feet

By doubling the width, volume = 1 x 4 x 2

V = 8 cubic feet

Therefore, she did not achieve her goal by doubling the width.

The other dimensions to be for a 12 cubic feet by keeping the height same =

Volume = 3 x 2 x 2 = 12

The area of the storage box =

3 inches x 2 inches

A = 6 square feet

The area of first storage box = 4 square feet and

The area of first storage box = 6 square feet.

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• Grade 5 Eureka - Answer Keys Module 1

Explanation:

b. How many centimeters is it from the pitcher’s mound to home plate

Jules reads that 1 pint is equivalent to 0.473 liters. He asks his teacher how many liters there are in a pint. His teacher responds that there are about 0.47 liters in a pint. He asks his parents, and they say there are about 0.5 liters in a pint. Jules says they are both correct. How can that be true? Explain your answer.

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Chapter 5: Ratios and Proportions

Chapter 5 homework solutions.

5.1-5.3 Extra Practice Worksheet

5.1 Ratios and Rates

5.1 Lesson Preso

5.1 Textbook Exercises: page 167

Ba: 1-6, 11-27 odd;

Avg: 1-6, 11-21 odd, 24-28 even, 29, 31;

Adv: 1-6, 18-38 even

5.2 Proportions

5.2 Textbook Exercises page 174

Ba: 1-4, 5-13 odd, 21-27 odd;

Avg 1-4, 5-13 odd, 22-30 even;

Adv: 1-4, 6-14 even, 22-32 even

5.3 Writing Proportions

5.3 Lesson Preso

5.3 Textbook Exercises page 182

Ba: 1-3, 9, 11, 12, 13-23 odd

Avg: 1-3, 8-14 even, 19-23

Adv: 1-3, 8-24 even, 25

5.4 Solving Proportions

5.4 Lesson Preso

5.4 Textbook Exercises page 190

Ba: 1-3, 5-9 odd, 15-21 odd, 22, 23-27 odd

Avg: 1-3, 5-9 odd, 15-21 odd, 22, 29, 30, 32-35

Adv: 1-3, 4-8 even, 14-38 even

5.5 Lesson Preso

5.5 Textbook Exercises page 196

Ba: 1-3, 5-11 odd, 12, 13-17 odd

Avg: 1-3, 5-11 odd, 12, 14-17

Adv: 1-3, 4-18 even

5.6 Direct Variation

5.6 Lesson Preso

5.6 Textbook Exercises page 202

Ba: 1-3, 7-17 odd, 18, 19-25 odd

Avg: 1-3, 7-17 odd, 18-28 even

Adv: 1-3, 8-28 even

Chapter 5 Review

5.1-5.3 Review Quizizz:

5.4-5.6 Review Quizizz:

Chapter 5 Review Quizizz:

Grade 5 Mathematics Module 1, Topic C, Lesson 7

NYS COMMON CORE MATHEMATICS CURRICULUM

Lesson 7 5 1

Objective: Round a given decimal to any place using place value understanding and the vertical number line.

Suggested lesson structure.

(12 minutes)

(8 minutes)

(30 minutes)

(10 minutes)

Total Time (60 minutes)

Fluency Practice (12 minutes)

 Sprint: Find the Midpoint 5.NBT.4

 Compare Decimal Fractions 5.NBT.3b

 Rename the Units 5.NBT.2

Sprint: Find the Midpoint (7 minutes)

Materials: (S) Find the Midpoint Sprint

(7 minutes)

(2 minutes)

(3 minutes)

MULTIPLE MEANS

OF REPRESENTATION:

Vertical number lines may be a novel representation for students. Their use offers an important scaffold for students’ understanding of rounding in that numbers are quite literally rounded up and down to the nearest multiple rather than left or right as in a horizontal number line. Consider showing both a horizontal and vertical line and comparing their features so that students can see the parallels and gain comfort in the use of the vertical line.

Note: Practicing this skill in isolation helps students conceptually understand the rounding of decimals.

Compare Decimal Fractions (2 minutes)

Materials: (S) Personal white board

Note: This review fluency activity helps students work toward mastery of comparing decimal numbers, a topic introduced in

T: (Write 12.57 ___ 12.75.) On your personal boards, compare the numbers using the greater than, less than, or equal sign.

S: (Write 12.57 < 12.75 on boards.)

Repeat the process and procedure:

4.07 __ forty-seven tenths

OF ENGAGEMENT:

Fluency activities like Compare Decimal

Fractions may be made more active by allowing students to stand and use their arms to make the >, <, and = signs in response to questions on the board.

__ 0.084 328.2 __ 328.099 twenty-four and 9 thousandths___ 3 tens

Lesson 7: Round a given decimal to any place using place value understanding and the vertical number line.

Thi s work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 -Great Minds. eureka math.org

This file derived from G5-M1-TE-1.3.0-06.2015

This work is licensed under a

Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

Rename the Units (3 minutes)

Note: Renaming decimals using various units strengthens student understanding of place value and provides an anticipatory set for rounding decimals in Lessons 7 and 8.

T: (Write 1.5 = ____ tenths.) Fill in the blank.

S: 15 tenths.

T: (Write 1.5 = 15 tenths. Below it, write 2.5 = ____ tenths.) Fill in the blank.

S: 25 tenths.

T: (Write 2.5 = 25 tenths. Below it, write 12.5 = ____ tenths.) Fill in the blank.

S: 125 tenths.

Repeat the process for 17.5, 27.5, 24.5, 24.3, and 42.3.

Application Problem (8 minutes)

Craig, Randy, Charlie, and Sam ran in a 5K race on Saturday. They were the top 4 finishers. Here are their race times:

Craig: 25.9 minutes Randy: 32.2 minutes

Charlie: 32.28 minutes Sam: 25.85 minutes

Who won first place? Who won second place? Third? Fourth?

Note: This Application Problem offers students a quick review of yesterday’s concept before moving toward the rounding of decimals. Students may need reminding that in a race, the lowest number indicates the fastest time.

Concept Development (30 minutes)

Materials: (S) Personal white board, hundreds to thousandths place value chart (Template)

Strategically decompose 155 using multiple units to round to the nearest ten and nearest hundred.

T: Work with your partner to name 155 in unit form. Next, rename 155 using the greatest number of tens possible. Finally, rename 155 using only ones. Record your ideas on your place value chart.

T: Which decomposition of 155 helps you round this number to the nearest ten? Turn and talk.

S: 15 tens and 5 ones.  The one that shows 15 tens.

This helps me see that 155 is between 15 tens and 16 tens on the number line. It is exactly halfway, so 155 would round to the next greater ten, which is 16 tens, or 160.

T: Let’s record that on the number line. (Record both of the nearest multiples of ten, the halfway point, and the number being rounded. Circle the correct rounded figure.)

T: Using your chart, which of these representations helps you round 155 to the nearest 100? Turn and talk to your partner about how you will round.

S: The one that shows 1 hundred.  I can see that 155 is between 1 hundred and 2 hundred.  The midpoint between 1 hundred and 2 hundred is 150. 155 is past the midpoint, so 155 is closer to 2 hundreds. It rounds up to 200.

T: Label your number line with the nearest multiples of one hundred, the halfway point, and the number we’re rounding. Then, circle the one to which 155 would round.

Strategically decompose 1.57 to round to the nearest whole and nearest tenth.

T: Work with your partner to name 1.57 in unit form. Next, rename 1.57 using the greatest number of tenths possible.

Finally, rename 1.57 using only hundredths. Record your ideas on your place value chart.

S: (Work and share.)

Ones Tenths Hundredths

T: Which decomposition of 1.57 best helps you to round this number to the nearest tenth? Turn and talk. Label your number line, and circle your rounded number.

S: (Share.)

Bring to students’ attention that this problem parallels conversions between meters and centimeters since different units are being used to name the same quantity: 1.57 meters = 157 centimeters.

Strategically decompose to round 4.381 to the nearest ten, one, tenth, and hundredth.

T: Work with your partner to decompose 4.381 using as many tens, ones, tenths, and hundredths as possible. Record your work on your place value chart.

Thousandths

T: We want to round this number to the nearest ten first. How many tens did you need to name this number?

S: Zero tens.

T: Between what two multiples of 10 will we place this number on the number line? Turn and talk. Draw your number line, and circle your rounded number.

T: Work with your partner to round 4.381 to the nearest one, tenth, and hundredth. Explain your thinking with a number line.

Follow the sequence from above to guide students in realizing that the number 4.381 rounds down to 4 ones, up to 44 tenths (4.4), and down to 438 hundredths (4.38).

Strategically decompose to round 9.975 to the nearest one, ten, tenth, and hundredth.

Follow a sequence similar to the previous problem to lead students in rounding to the given places. This problem can prove to be a problematic rounding case. Naming the number with different units, however, allows students to choose easily between nearest multiples of the given place value.

Repeat this sequence with 99.799. Round to the nearest ten, one, tenth, and hundredth.

Problem Set (10 minutes)

Students should do their personal best to complete the Problem Set within the allotted 10 minutes. For some classes, it may be appropriate to modify the assignment by specifying which problems they work on first.

Some problems do not specify a method for solving. Students solve these problems using the RDW approach used for Application Problems.

On this Problem Set, it is suggested that all students begin with Problems 1, 2, 3, and 5 and possibly leave

Problem 4 until the end if they still have time.

Before circulating while students work, review the Debrief questions relevant to the Problem Set to better guide students to a deeper understanding of, and skill with, the lesson’s objective.

Student Debrief (10 minutes)

Lesson Objective: Round a given decimal to any place using place value understanding and the vertical number line.

The Student Debrief is intended to invite reflection and active processing of the total lesson experience.

Invite students to review their solutions for the Problem

Set. They should check work by comparing answers with a partner before going over answers as a class. Look for misconceptions or misunderstandings that can be addressed in the Debrief. Guide students in a conversation to debrief the Problem Set and process the lesson.

Any combination of the questions below may be used to lead the discussion.

 In Problem 2, which decomposition helps you most if you want to round to the hundredths place? The tens place? Ones place? Why?

 How was Problem 1 different from both Problem

2 and 3? (While students may offer many differences, the salient point here is that Problem

1 is already rounded to the nearest hundredth and tenth.)

 Unit choice is the foundation of the current lesson. Problem 3 on the Problem Set offers an opportunity to discuss how the choice of unit affects the result of rounding. Be sure to allow time for these important understandings to be articulated by asking the following: If a number rounds up when rounded to the nearest tenth, does it follow that it will round up when rounded to the nearest hundredth? Thousandth? Why or why not? How do we decide about rounding up or down ? How does the unit we are rounding to affect the position of the number relative to the midpoint?

 Problem 3 also offers a chance to discuss how 9 numbers often round to the same number regardless of the unit to which they are rounded.

Point out that decomposing to smaller units makes this type of number easier to round. The decompositions make it simpler to identify which numbers to use as endpoints on the number line.

Extension: Problem 5 offers an opportunity to discuss the effect rounding to different places has on the accuracy of a measurement. Which rounded value is closest to the actual measurement? Why?

In this problem, does that difference in accuracy matter? In another situation, might those differences in accuracy be more important? What should be considered when deciding to round, and to which place one might round? (For some students, this may lead to an interest in significant digits and their role in measurement in other disciplines.)

Exit Ticket (3 minutes)

After the Student Debrief, instruct students to complete the Exit Ticket. A review of their work will help with assessing students’ understanding of the concepts that were presented in today’s lesson and planning more effectively for future lessons. The questions may be read aloud to the students.

Lesson 7 Sprint 5 1

Number Correct: _______ A

Find the Midpoint

Number Correct: _______

Improvement: _______

Lesson 7 Problem Set 5 1

Fill in the table, and then round to the given place. Label the number lines to show your work. Circle the rounded number.

3.1 a. Hundredths b. Tenths c. Tens

Tens Ones Tenths Hundredths Thousandths a. Hundredths b. Ones c. Tens

Tens Ones Tenths Hundredths Thousandths

Tens Ones Tenths Hundredths Thousandths a.

Hundredths b. Tenths c. Ones d. Tens

For open international competition, the throwing circle in the men’s shot put must have a diameter of

2.135 meters. Round this number to the nearest hundredth. Use a number line to show your work.

Jen’s pedometer said she walked 2.549 miles. She rounded her distance to 3 miles. Her brother rounded her distance to 2.5 miles. When they argued about it, their mom said they were both right. Explain how that could be true. Use number lines and words to explain your reasoning.

Lesson 7 Exit Ticket 5 1

Use the table to round the number to the given places. Label the number lines, and circle the rounded value.

Hundredths b. Tens

Lesson 7 Homework 5 1

Hundredths b. Tenths c. Ones

Hundredths b. Ones c. Tens

On a Major League Baseball diamond, the distance from the pitcher’s mound to home plate is 18.386 meters. a.

Round this number to the nearest hundredth of a meter. Use a number line to show your work. b.

How many centimeters is it from the pitcher’s mound to home plate?

Jules reads that 1 pint is equivalent to 0.473 liters. He asks his teacher how many liters there are in a pint.

His teacher responds that there are about 0.47 liters in a pint. He asks his parents, and they say there are about 0.5 liters in a pint. Jules says they are both correct. How can that be true? Explain your answer.

Lesson 7 Template 5 1

 hundreds to thousandths place value chart

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Home > CC2 > Chapter 5 > Lesson 5.1.1

Lesson 5.1.1, lesson 5.1.2, lesson 5.2.1, lesson 5.2.2, lesson 5.2.3, lesson 5.2.4, lesson 5.2.5, lesson 5.2.6, lesson 5.3.1, lesson 5.3.2, lesson 5.3.3, lesson 5.3.4, lesson 5.3.5.

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Math Expressions Grade 5 Unit 7 Lesson 2 Answer Key Simplify Expressions

Solve the questions in Math Expressions Grade 5 Homework and Remembering Answer Key Unit 7 Lesson 2 Answer Key Simplify Expressions to attempt the exam with higher confidence. https://mathexpressionsanswerkey.com/math-expressions-grade-5-unit-7-lesson-2-answer-key/

Math Expressions Common Core Grade 5 Unit 7 Lesson 2 Answer Key Simplify Expressions

Math Expressions Grade 5 Unit 7 Lesson 2 Homework

Unit 7 Lesson 2 Simplify Expressions  Question 1. Follow the Order of Operations to simplify 27 ÷ (3 • 3) + 17 Step I Perform operations inside parentheses. ___________ Step 2 Multiply and divide from left to right. ___________ Step 3 Add and subtract from left to right. __________________ Answer:

Step 1: 27 ÷ (9) + 17

Step 2: 3 + 17

Simplify. Follow the Order of Operations.

Math Expressions Grade 5 Simplify  Question 2. 54 – 200 ÷ 4 Answer:

54 – 200 ÷ 4

54 – 50

Unit 7 Answer Key Math Expressions Lesson 2 Simplify  Question 3. 0.8 ÷ (0.07 – 0.06) Answer:

0.8 ÷ ( 0.01 )

Unit 7 Lesson 2 Math Expressions Question 4. 3 • 8 – 6 ÷ 2 Answer:

3 • 8 – 6 ÷ 2

3 . 8 – 3

24 – 3

Unit 7 Lesson 2 Answer Key Math Expressions Question 5. (\(\frac{3}{8}\) + \(\frac{1}{4}\)) • 16 Answer:

3/8 + 1/4 . 16

Math Expressions Common Core Grade 5 Simplify  Question 6. 64 + 46 – 21 + 29 Answer:

64 + 46 – 21 + 29

139 – 21

Simplify Answers Math Expressions Unit 7 Lesson 2 Question 7. 72 ÷ (7 – 1) • 3 Answer:

72 ÷ (7 – 1) • 3

Question 8. 0.8 – 0.5 ÷ 5 + 0.2 Answer:

0.8 – 0.5 ÷ 5 + 0.2

0.8 + 0.2 – 0.1

1 – 0.1

Question 9. \(\frac{5}{6}\) – 4 • \(\frac{1}{12}\)

5/6 – 4 . 1/12

5/6 – 1/3

Question 10. 5 • 15 ÷ 3 Answer:

Question 11. 32 ÷ (2.3 + 1.7) • 3 Answer:

32 ÷ (2.3 + 1.7) • 3

Question 12. (1\(\frac{1}{2}\) – \(\frac{3}{4}\)) × \(\frac{1}{4}\) Answer:

1 1/2 – 3/4 x 1/4

Question 13. (6.3 – 5.1) • (0.7 + 0.3) Answer:

(6.3 – 5.1) • (0.7 + 0.3)

Question 14. 12 ÷ 0.1 + 12 ÷ 0.01 Answer:

Question 15. \(\frac{1}{2}\) • \(\frac{1}{2}\) ÷ \(\frac{1}{2}\) Answer:

1/2 x 1/2 ÷  1/2

Question 16. 10 – 4 + 2 – 1 Answer:

10 – 4 + 2 – 1

10 – 6 – 1

Math Expressions Grade 5 Unit 7 Lesson 2 Remembering

Write an equation to solve the problem. Draw a model if you need to.

Question 4. The students of Turner Middle School are going on a field trip. There are 540 students attending. A bus can hold 45 students. How many buses are needed for the field trip? Answer:

Given, Total number of students = 540

Number of students a bus can hold = 45

Now, 540/45

Therefore, 12 busses are needed for the field trip.

Question 5. The area of a rectangular court is 433.37 square meters, and the length of the court is 28.7 meters. What is the width of the court? Answer:

Area of the rectangular court = 433.37 sq.m

Length = 28.7 sq. m

Area = length x width

433.37 = 28.7 x n

n = 433.37/28.7

n = 15.1 meters.

Write the computation in words.

Question 6. 5 ÷ \(\frac{1}{8}\) _____________________ Answer:

5 divided by 1/8

Question 7. 2.4 ÷ 0.6 + 0.2 __________________________ Answer:

Divide the sum of  0.6 and 0.2 by 2.4

Question 8. Stretch Your Thinking Write step-by-step instructions for simplifying the following expression. 10 • [60 ÷ (11 + 4)] – 3 Answer:

Step 1 : Add 11 + 4

10. [60 ÷ 15]-3

Step 2: Divide 60 by 15

10. [ 4 ] – 3

Step 3: Multiply 10 by 4

40 – 3

Step 4: Subtract 3 from 40

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