lesson 17 homework 3rd grade

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Module 3 Lesson 17 Grade 3 Homework

Module 3 Lesson 17 Grade 3 Homework - Displaying top 8 worksheets found for this concept.

Some of the worksheets for this concept are Grade 3 module 3, Grade 3 module 3, Lesson plan module 3 session 17, Grade 3 math, Teksstaar based lessons, Ks3 science revision work special edition, Grade k module 5 student file a, Grade 3 module 6.

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1. GRADE 3 MODULE 3

2. grade 3 module 3, 3. lesson plan module 3: session 17, 4. grade 3 math -, 5. teks/staar-based lessons, 6. ks3 science revision worksheets special edition, 7. grade k, module 5 student file a, 8. grade 3 module 6.

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Placing Various Fractions On The Number Line

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• Grade 3 Mathematics Module 5, Topic D, Lesson 17

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Does It Make Sense?

Lesson Purpose

The purpose of this lesson is for students to use mental computation and estimation strategies such as rounding to decide if answers to two-step word problems make sense.

Lesson Narrative

Previously, students extended their understanding of addition and subtraction within 1,000 and learned how to round to the nearest ten and hundred. In this lesson, students work with two-step word problems and decide if a given answer for a two-step problem is reasonable. Students estimate answers to two-step problems and determine if each other's solutions make sense after they solve two-step word problems in a way that makes sense to them.

Activity 2: Solve and Reason

Learning Goals

Teacher Facing

• Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
• Solve two-step word problems using addition and subtraction in a way that makes sense to them.

Student Facing

• Let’s decide if our answers make sense.

Required Preparation

Ccss standards.

Building Towards

Lesson Timeline

Teacher reflection questions.

Reflect on students’ assessment. Are they able to flexibly use different strategies, such as considering the answer in context, rounding, or estimating? How can you leverage their previous experiences with sense making and rounding to help them keep reasonableness in mind?

Suggested Centers

• Tic Tac Round (3–5), Stage 1: Nearest Ten or Hundred (Addressing)
• Number Puzzles: Addition and Subtraction (1–4), Stage 5: Within 1,000 (Addressing)

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Eureka TEKS Edition: 3rd Grade Module 3, Lesson 17 (Google Slides)

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This is a Google Slides presentation that follows Module 3, Lesson 17 in the 3rd Grade Eureka: TEKS Edition curriculum. I created these slides for my own 3rd grade class, and have found them to be very helpful! All components of the Eureka Math lesson structure are included in these slides: Fluency, Application Problem, Concept Development, Problem Set, Student Debrief, and Exit Ticket. Corresponding pages from the student edition are also included. These Google Slides have been created after my personal lesson internalization and generally follow the teacher prompts/student responses as written. In order to get the most out of this resource, internalizing the lesson and reviewing slides before instruction is highly recommended!

Questions & Answers

Lakewood math.

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Module 3 Lesson 17 Grade 3 Homework

Displaying top 8 worksheets found for - Module 3 Lesson 17 Grade 3 Homework .

Some of the worksheets for this concept are Grade 3 module 3, Grade 3 module 3, Lesson plan module 3 session 17, Grade 3 math, Teksstaar based lessons, Ks3 science revision work special edition, Grade k module 5 student file a, Grade 3 module 6.

Found worksheet you are looking for? To download/print, click on pop-out icon or print icon to worksheet to print or download. Worksheet will open in a new window. You can & download or print using the browser document reader options.

1. GRADE 3 MODULE 3

2. grade 3 module 3, 3. lesson plan module 3: session 17, 4. grade 3 math -, 5. teks/staar-based lessons, 6. ks3 science revision worksheets special edition, 7. grade k, module 5 student file a, 8. grade 3 module 6.

Math Expressions Grade 3 Unit 1 Lesson 17 Answer Key Play Multiplication and Division Games

Solve the questions in Math Expressions Grade 3 Homework and Remembering Answer Key Unit 1 Lesson 17 Answer Key Play Multiplication and Division Games to attempt the exam with higher confidence.

Math Expressions Common Core Grade 3 Unit 1 Lesson 17 Answer Key Play Multiplication and Division Games

Math Expressions Grade 3 Unit 1 Lesson 17 Homework

Write an equation and solve the problem. Show your work

Lesson 17 Answer Key Math Expressions Grade 3 Unit 1 Question 1. Wendy gave 54 apples to her neighbors. She gave away a total of 6 bags of apples. She put the same number of apples in each bag. How many apples were in each bag? Answer: 9 bags

Explanation: Given that, Wendy gave 54 apples to her neighbors. She gave away a total of 6 bags of apples. She put the same number of apples in each bag 54/6 = 9 bags

Lesson 17 Play Multiplication and Division Games Unit 1 Math Expressions Question 2. Dillon had a box of 45 toy trucks. He gave the trucks to his brother but kept 9 trucks for himself. How many trucks did Dillon give to his brother? Answer: 36

Explanation: Dillon gave 36 toys to his brother 45-9=36

Answer Grade 3 Unit 1 Multiplication and Division Games Question 3. Melissa put 18 stickers in her new sticker album. She put them in 6. rows. She put the same number of stickers in each row. How many stickers did she put in each row? Answer: 3 Total 18 stickers She placed in 6 rows 6×3=18

Answer Key Grade 3 Unit 1 Lesson 17 Math Expressions Question 4. Yan took photographs at the zoo. He took 5 photos in each of the 6 animal houses. How many photos did he take? Answer: 30 Yan took 30 photos in the zoo 6x 5=30

Play Multiplication and Division Games Grade 3 Unit 1 Lesson 17 Question 5. Janie stacked some books at the library. She stacked 6 books each in 7 different piles. How many books were in the piles? Answer: 42 books Janie stacked 42 books in piles 6×7=42

Math Expressions Grade 3 Unit 1 Lesson 17 Remembering

Multiply or divide to find the unknown numbers.

Question 1. \(\frac{72}{9}\) = ___ Answer: 72/9 = 6

Question 2. 2 * ___ = 14 Answer: 2 × x = 14 x = 14/2 x = 7

Question 3. 40 ÷ 10 = ___ Answer: 40/10 = 4

Write an equation and solve the problem.

Question 4. Brian has 0 crackers on his plate and divides them among his 5 friends. How many crackers does each friend get? Answer:2 10 divided by 5 is 2

Read each problem and decide what type of problem it is. Write the letter from the list below. Then write an equation and solve the problem.

a. Array Multiplication b. Array Division c. Equal Groups of Multiplication d. Equal Groups Division with Unknown Group Size e. Equal Groups Division with an Unknown Multiplier (number of groups)

Question 5. Tiffany is in the bike shop. She counts 27 wheels on all of the tricycles. How many tricycles are in the bike shop? Answer: Given that, Tiffany is in the bike shop. She counts 27 wheels on all of the tricycles. 27 ÷ 3 = 9 Array Division

Question 6. John buys 4 bags of apples. There are 6 apples in each bag. How many apples does John buy? Answer: Equal Groups of Multiplication

Explanation: Given, John buys 4 bags of apples. There are 6 apples in each bag. 6 × 4 = 24

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Eureka Math Grade 7 Module 3 Lesson 17 Answer Key

Engage ny eureka math 7th grade module 3 lesson 17 answer key, eureka math grade 7 module 3 lesson 17 exploratory challenge answer key.

What is the area of the “rectangle” using the side lengths above? Answer: The area of the “rectangle” is base times height, and, in this case, A = πr ∙ r.

Are the areas of the “rectangle” and the circle the same? Answer: Yes, since we just rearranged pieces of the circle to make the “rectangle,” the area of the “rectangle” and the area of the circle are approximately equal. Note that the more sections we cut the circle into, the closer the approximation.

If the area of the rectangular shape and the circle are the same, what is the area of the circle? Answer: The area of a circle is written as A = πr ∙ r, or A = πr 2 .

Eureka Math Grade 7 Module 3 Lesson 17 Example Answer Key

Question 2. What would be a quicker method for determining the area of the circle other than counting all of the squares in the entire circle? Answer: Count \(\frac{1}{4}\) of the squares needed; then, multiply that by four in order to determine the area of the entire circle.

Question 3. Using the diagram, how many squares were used to cover one-fourth of the circle? Answer: The area of one-fourth of the circle is approximately 79 cm 2 .

Question 4. What is the area of the entire circle? Answer: A ≈ 4 ∙ 79 cm 2 A ≈ 316 cm 2

What information is needed to solve the problem? Answer: The formula to find the area of a circle is A = πr 2 . If the radius is 12 ft., then A = π ∙ (12 ft.) 2 = 144π ft 2 , or approximately 452 ft 2 .

Question 1. What information is needed to solve the problem? Answer: The area of the circle and the area of the square are needed so that we can subtract the area of the circle from the area of the square to determine the amount of waste.

Question 2. What information do we need to determine the area of the square and the circle? Answer: Circle: just radius because A = πr 2 Square: one side length

Question 3. How will we determine the waste? Answer: The waste is the area left over from the square after cutting out the circular region. The area of the circle is A = π ∙ (3 ft.) 2 = 9π ft 2 ≈ 28.26 ft 2 . The area of the square is found by first finding the diameter of the circle, which is the same as the side of the square. The diameter is d = 2r; so, d = 2 ∙ 3 ft. or 6 ft. The area of a square is found by multiplying the length and width, so A = 6 ft. ∙ 6 ft. = 36 ft 2 . The solution is the difference between the area of the square and the area of the circle, so 36 ft 2 -28.26 ft 2 ≈ 7.74 ft 2 .

Question 4. Does your solution answer the problem as stated? Answer: Yes, the amount of waste is 7.74 ft 2 .

Eureka Math Grade 7 Module 3 Lesson 17 Exercise Answer Key

Exercises 1–3 Solve the problem below individually. Explain your solution.

Exercise 1. Find the radius of a circle if its circumference is 37.68 inches. Use π ≈ 3.14. Answer: If C = 2πr, then 37.68 = 2πr. Solving the equation for r, 37.68 = 2πr (\(\frac{1}{2 \pi}\))37.68 = (\(\frac{1}{2 \pi}\))2πr \(\frac{1}{6.28}\) (37.68) ≈ r 6 ≈ r The radius of the circle is approximately 6 in.

Exercise 3. Find the length of a rectangle if the area is 27 cm 2 and the width is 3 cm. Answer: If the area of the rectangle is Area = length ∙ width, then 27 cm 2 = l ∙ 3 cm \(\frac{1}{3}\) ∙ 27 cm 2 = \(\frac{1}{3}\) ∙ l ∙ 3 cm 9 cm = l

Exercises 4–6

Exercise 4. A circle has a radius of 2 cm. a. Find the exact area of the circular region. Answer: A = π ∙ (2 cm) 2 = 4π cm 2

b. Find the approximate area using 3.14 to approximate π. Answer: A = 4 cm 2 ∙ π ≈ 4 cm 2 ∙ 3.14 ≈ 12.56 cm 2

Exercise 5. A circle has a radius of 7 cm. a. Find the exact area of the circular region. Answer: A = π ∙ (7 cm) 2 = 49π cm 2

b. Find the approximate area using \(\frac{22}{7}\) to approximate π. Answer: A = 49 ∙ π cm 2 ≈ (49 ∙ \(\frac{22}{7}\) ) cm 2 ≈ 154 cm 2

c. What is the circumference of the circle? Answer: C = 2π ∙ 7 cm = 14π cm ≈ 43.96 cm

Eureka Math Grade 7 Module 3 Lesson 17 Problem Set Answer Key

Question 2. A circle has a diameter of 20 inches. a. Find the exact area, and find an approximate area using π ≈ 3.14. Answer: If the diameter is 20 in., then the radius is 10 in. If A = πr 2 , then A = π ∙ (10 in.) 2 or 100π in 2 . A ≈ (100 ∙ 3.14) in 2 ≈ 314 in 2 .

b. What is the circumference of the circle using π ≈ 3.14 ? Answer: If the diameter is 20 in., then the circumference is C = πd or C ≈ 3.14 ∙ 20 in. ≈ 62.8 in.

Question 3. A circle has a diameter of 11 inches. a. Find the exact area and an approximate area using π ≈ 3.14. Answer: If the diameter is 11 in., then the radius is \(\frac{11}{2}\) in. If A = πr 2 , then A = π ∙ (\(\frac{11}{2}\) in.) 2 or \(\frac{121}{4}\) π in 2 . A ≈ (\(\frac{121}{4}\) ∙ 3.14) in 2 ≈ 94.985 in 2

b. What is the circumference of the circle using π ≈ 3.14? Answer: If the diameter is 11 inches, then the circumference is C = πd or C ≈ 3.14 ∙ 11 in. ≈ 34.54 in.

Question 5. A path bounds a circular lawn at a park. If the inner edge of the path is 132 ft. around, approximate the amount of area of the lawn inside the circular path. Use π ≈ \(\frac{22}{7}\) . Answer: The length of the path is the same as the circumference. Find the radius from the circumference; then, find the area. C = 2πr 132 ft. ≈ 2 ∙ \(\frac{22}{7}\) ∙ r 132 ft. ≈ \(\frac{44}{7}\) r \(\frac{7}{44}\) ∙ 132 ft. ≈ \(\frac{7}{44}\) ∙ \(\frac{44}{7}\) r 21 ft. ≈ r A ≈ \(\frac{22}{7}\) ∙ (21 ft.) 2 A ≈ 1386 ft 2

Question 6. The area of a circle is 36π cm 2 . Find its circumference. Answer: Find the radius from the area of the circle; then, use it to find the circumference. A = πr 2 36π cm 2 = πr 2 \(\frac{1}{\pi}\) ∙ 36π cm 2 = \(\frac{1}{\pi}\) ∙ πr 2 36 cm 2 = r 2 6 cm = r C = 2πr C = 2π ∙ 6 cm C = 12π cm

Question 7. Find the ratio of the area of two circles with radii 3 cm and 4 cm. Answer: The area of the circle with radius 3 cm is 9π cm 2 . The area of the circle with the radius 4 cm is 16π cm 2 . The ratio of the area of the two circles is 9π:16π or 9:16.

Question 8. If one circle has a diameter of 10 cm and a second circle has a diameter of 20 cm, what is the ratio of the area of the larger circle to the area of the smaller circle? Answer: The area of the circle with the diameter of 10 cm has a radius of 5 cm. The area of the circle with the diameter of 10 cm is π ∙ (5 cm) 2 , or 25π cm 2 . The area of the circle with the diameter of 20 cm has a radius of 10 cm. The area of the circle with the diameter of 20 cm is π ∙ (10 cm) 2 or 100π cm 2 . The ratio of the diameters is 20 to 10 or 2:1, while the ratio of the areas is 100π to 25π or 4:1.

Question 9. Describe a rectangle whose perimeter is 132 ft. and whose area is less than 1 ft 2 . Is it possible to find a circle whose circumference is 132 ft. and whose area is less than 1 ft 2 ? If not, provide an example or write a sentence explaining why no such circle exists. Answer: A rectangle that has a perimeter of 132 ft. can have a length of 65.995 ft. and a width of 0 .005 ft. The area of such a rectangle is 0.329975 ft 2 , which is less than 1 ft 2 . No, because a circle that has a circumference of 132 ft. has a radius of approximately 21 ft. A = πr 2 = π(21) 2 = 1387.96 ≠ 1

Question 10. If the diameter of a circle is double the diameter of a second circle, what is the ratio of the area of the first circle to the area of the second? Answer: If I choose a diameter of 24 cm for the first circle, then the diameter of the second circle is 12 cm. The first circle has a radius of 12 cm and an area of 144π cm 2 . The second circle has a radius of 6 cm and an area of 36π cm 2 . The ratio of the area of the first circle to the second is 144π to 36π , which is a 4 to 1 ratio. The ratio of the diameters is 2, while the ratio of the areas is the square of 2, or 4.

Eureka Math Grade 7 Module 3 Lesson 17 Exit Ticket Answer Key

Question 3. The circumference of the circle is _______________________. Answer: The circumference of the circle is 2πr .

Question 4. The _________________ of the ___________________ is 2r. Answer: The diameter of the circle is 2r.

Question 5. The ratio of the circumference to the diameter is ______. Answer: The ratio of the circumference to the diameter is π .

Question 6. Area (circle) = Area of (_____________) = \(\frac{1}{2}\) ∙ circumference ∙ r = \(\frac{1}{2}\) (2πr) ∙ r = π ∙ r ∙ r = _____________. Answer: Area (circle) = Area of ( rectangle ) = \(\frac{1}{2}\) ∙ circumference ∙ r = \(\frac{1}{2}\) (2πr) ∙ r = π ∙ r ∙ r = πr 2 .

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