• Texas Go Math
  • Big Ideas Math
  • Engageny Math
  • McGraw Hill My Math
  • enVision Math
  • 180 Days of Math
  • Math in Focus Answer Key
  • Math Expressions Answer Key
  • Privacy Policy

CCSS Math Answers

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key

Engage ny eureka math 4th grade module 2 lesson 3 answer key, eureka math grade 4 module 2 lesson 3 problem set answer key.

Question 1. Complete the conversion table.

Explanation: Completed the conversion table as shown above, We know 1 liter  = 10 3 or 1000 ml, So 1 L = 1,000 mL .

5 L = 5,000 mL, 5 L = 5 X 1000 mL= 5,000 mL .

38 L = 38,000 mL, 38 L = 38 X 1000 mL= 38,000 mL .

49 L = 49,000 ml, 49 L = 49 X 1000 mL = 49,000 mL .

54,000 mL = 54 L, as 54,000 mL ÷ 1000 mL = 54 L.

92,000 mL = 92 L, as 92,000 mL ÷ 1000 mL = 92 L.

Question 2. Convert the measurements. a. 2 L 500 mL = ______2,500_______ mL b. 70 L 850 mL = _____70,850________ mL c. 33 L 15 mL = ______33,015_______ mL d. 2 L 8 mL = ______2,008_______ mL e. 3,812 mL = __3___ L ___812____ mL f. 86,003 mL = __86___ L __003_____ mL

a. 2 L 500 mL = 2,500 mL,

Explanation: 2 L 500 mL as 1 L = 10 3 mL = 1000 mL, 2 X 1000 mL + 500 mL= 2,500 mL .

b. 70 L 850 mL = 70,850 mL,

Explanation: 70 L 850 mL as 1 L = 10 3 mL = 1000 mL, 70 X 1000 mL + 850 mL = 70,850 mL .

c. 33 L 15 mL = 33,015 mL,

Explanation: 33 L 15 mL as 1 L = 10 3 mL = 1000 mL, 33 X 1000 mL + 15 mL = 33,015 mL .

d. 2 L 8 mL = 2,008 mL,

Explanation: 2 L 8 mL as 1 L = 10 3 mL = 1000 mL, 2 X 1000 mL + 8 mL = 2,008 mL .

e. 3,812 mL = 3 L 812 mL,

Explanation: 3,812 mL as 1 L = 10 3 mL = 1000 mL, 3,812 ÷ 1000 mL = 3 L 812 mL .

f. 86,003 mL = 86 L 003 mL,

Explanation: 86,003 mL as 1 L = 10 3 mL = 1000 mL, 86,003 ÷ 1000 mL = 86 L 003 mL .

Question 3. Solve. a. 1,760 mL + 40 L b. 7 L – 3,400 mL c. Express the answer in the smaller unit: 25 L 478 mL + 3 L 812 mL d. Express the answer in the smaller unit: 21 L – 2 L 8 mL e. Express the answer in mixed units: 7 L 425 mL – 547 mL f. Express the answer in mixed units: 31 L 433 mL – 12 L 876 mL

Use a tape diagram to model each problem. Solve using a simplifying strategy or an algorithm and write your answer as a statement.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-1

Statement : one thousand seven hundred sixty liter plus forty liter is equal to forty one liter seven hundred sixty milliliter or forty one thousand seven hundred and sixty milliliters.

Explanation: Given  1,760 mL + 40 L = as 40 L = 40 X 1000 mL = 40,000 mL, 40,000 mL +1760 mL 41,760 mL Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement one thousand seven hundred sixty liter plus forty liter is equal to forty one liter seven hundred sixty milliliter or forty one thousand seven hundred and sixty milliliters.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-2

Explanation: Given  7 L – 3,400 mL = as 7 L = 7 X 1000 mL = 7,000 mL, 7,000 mL -3,400 mL 3,600 mL Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement seven liter minus three thousand four hundred milliliter is equal to three thousand six hundred milliliters or three liters six hundred milliliter.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-3

Statement : twenty five liter and four hundred seventy eight milliliters plus three liters eight hundred and twelve milliliters  is equal to twenty nine liter and two ninety milliliter or twenty nine thousand and two ninety milliliters,

Explanation: Given  25 L 478 mL + 3 L 812 mL = As 25 L  478 mL = 25 X 1000 mL + 478 mL = 25000 mL + 478 mL = 25478 mL, 3 L 812 mL = 3 X 1000 mL + 812 mL = 3000 mL + 812 mL = 3812 mL, 25,478 mL +3812 mL 29,290 mL The smaller unit as 1 liter is equal to 1,000 milliliter, So 29,290 mL = 29,290 ÷ 1000 = 29 L 290 mL,  Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement twenty five liter and four hundred seventy eight milliliters plus three liters eight hundred and twelve milliliters  is equal to twenty nine liter and two ninety milliliter or twenty nine thousand and two ninety milliliters.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-4

Explanation: Given  21 L – 2 L 8 mL = As 21 L = 21 X 1000 mL = 21,000 mL, 2 L 8 mL = 2 X 1000 mL + 8 mL = 2000 mL + 8 mL = 2,008 mL, 21,000 mL -2,008 mL 18,992 mL The smaller unit as 1 liter is equal to 1,000 milliliter, So 18,992 mL = 18,992 ÷ 1000 = 18 L 992 mL,  Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement twenty one liter minus two liters and and eight milliliters is equal to eighteen liter and nine hundred ninety two milliliter or eighteen thousand and nine hundred ninety two milliliters.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-5

Explanation: Given  7 L 425 mL – 547 mL = As 7 L 425 mL = 7 X 1000 mL + 425 mL = 7000 mL + 425 mL = 7,425 mL 7,425 mL -547 mL 6,878 mL The answer in mixed units is So 6,878 mL = 6878 ÷ 1000 = 6 L 878 mL,  Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement seven liter four hundred twenty five milliliters minus five hundred and forty seven milliliters is equal to six thousand and eight hundred seventy eight milliliters or six liters and eight hundred seventy eight milliliters.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-6

Explanation: Given 31 L 433 mL – 12 L 876 mL As 31 L 433 mL = 31 X 1000 mL + 433 mL = 31000 mL + 433 mL = 31,433 mL and 12 L 876 mL = 12 X 1000 mL + 876 mL = 12000 mL + 876 mL = 12,876 mL, 31,433 mL -12,876 mL 18,557 mL The answer in mixed units is So 18,557 mL = 18,557 ÷ 1000 = 18 L 557 mL,  Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement  thirty one liter four hundred thirty three milliliters minus twelve liters and eight hundred and seventy six milliliters is equal to  eighteen thousand and five hundred fifty seven milliliters or eighteen liters and five hundred fifty seven milliliters.

Question 4. To make fruit punch, John’s mother combined 3,500 milliliters of tropical drink, 3 liters 95 milliliters of ginger ale, and 1 liter 600 milliliters of pineapple juice. a. Order the quantity of each drink from least to greatest. b. How much punch did John’s mother make?

a. The quantity of order of each drink from least to greatest is 1,600 mL pineapple <  3,095 mL ginger ale < 3,500 mL tropical drink,

Explanation: Given John’s mother combined 3,500 milliliters of tropical drink, 3 liters 95 milliliters of ginger ale and 1 liter 600 milliliters of pineapple juice. As 3 L 95 mL = 3 X 1000 mL + 95 mL = 3,095 mL ginger ale, 1 L 600 mL = 1 X 1000 mL + 600 mL = 1,600 mL pineapple, Now all are in milliliters comparing we get 1,600 mL pineapple < 3,095 mL ginger ale < 3,500 mL tropical drink Therefore, The quantity of order of each drink from least to greatest is 1,600 mL pineapple < 3,095 mL ginger ale < 3,500 mL tropical drink.

b. John’s mother has made 8,195 mL of fruit punch or 8 L 195 mL,

Explanation: Given John’s mother combined 3,500 milliliters of tropical drink, 3 liters 95 milliliters of ginger ale and 1 liter 600 milliliters of pineapple juice. As 3 L 95 mL = 3 X 1000 mL + 95 mL = 3,095 mL ginger ale, 1 L 600 mL = 1 X 1000 mL + 600 mL = 1,600 mL pineapple, now in total we have 3,500 mL + 3,095 mL + 1,600 mL = 3,500 mL 3,095 mL +1,600 mL 8,195 mL or 8,195 ÷ 1000 = 8 L 195 mL . Therefore, John’s mother has made 8,195 mL of fruit punch or 8 L 195 mL .

Question 5. A family drank 1 liter 210 milliliters of milk at breakfast. If there were 3 liters of milk before breakfast, how much milk is left?

Milk left is 1,790 mL or 1 L 790 mL,

Explanation: Given A family drank 1 liter 210 milliliters of milk at breakfast. If there were 3 liters of milk before breakfast, So milk left is 3 L – 1 L 210 mL = as  3 L = 3 X 1000 mL = 3000 mL and 1 L 210 mL = 1 X 1000 mL + 210 mL = 1,210 mL, now 3000 mL – 1,210 mL = 3000 mL -1,210 mL 1,790 mL or 1,790 ÷ 1000 = 1 L 790 mL, Therefore, Milk left is 1,790 mL or 1 L 790 mL .

Eureka Math Grade 4 Module 2 Lesson 3 Problem Set Answer Key 1

Explanation: Given Petra’s fish tank contains 9 liters 578 milliliters of water. If the capacity of the tank is 12 liters 455 milliliters of water, So more milliliters of water does she needs to fill the tank is 12 L 455 mL – 9 L 578 mL = as 12 L 455 mL = 12 X 1000 mL + 455 mL = 12,455 mL, 9 L 578 mL = 9 X 1000 mL + 578 mL = 9,578 mL, 12,455 mL – 9,578 mL 2,877 mL Therefore, Petra’s need more 2,877 milliliters of water to fill the tank.

Eureka Math Grade 4 Module 2 Lesson 3 Exit Ticket Answer Key

Question 1. Convert the measurements. a. 6 L 127 mL = ____6,127___ mL b. 706 L 220 mL = ___706,220__ mL c. 12 L 9 mL = ____12,009______mL d. ___906___ L __010_____ mL = 906,010 mL

a. 6 L 127 mL 6 L 127 mL = 6,127 mL

Explanation: 6 L 127 mL = 6 X 1000 mL + 127 mL = 6,127 mL .

b. 706 L 220 mL = 706 L 220 mL = 706,220 mL,

Explanation: 706 L 220 mL = 706 X 1000 mL + 220 mL = 706,220 mL .

c. 12 L 9 mL = 12 L 9 mL = 12,009 mL, Explanation:

12 L 9 mL = 12 X 1000 mL + 9 mL = 12000 mL  + 9 mL = 12,009 mL .

d. ____________= 906,010 mL, 906 L 010 mL = 906,010 mL,

Explanation: ____________= 906,010 mL, As 906,010 mL ÷ 1000 = 906 L 010 mL .

Question 2. Solve. 81 L 603 mL – 22 L 489 mL

Use a tape diagram to model the following problem. Solve using a simplifying strategy or an algorithm, and write your answer as a statement.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-7

Explanation: 81 L 603 mL – 22 L 489 mL as 81 L 603 mL = 81 X 1000 mL + 603 mL = 81,603 mL and 22 L 489 mL = 22 X 1000 mL + 489 mL = 22,489 mL 81,603 mL -22,489 mL 59,114 mL  or 59,114 ÷ 1000 = 59 L 114 mL, Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement eighty one liter and six hundred three milliliters minus twenty two liters four hundred eighty nine milliliter is equal to fifty nine thousand one hundred fourteen milliliters or fifty nine liters and one hundred fourteen milliliters.

Question 3. The Smith’s hot tub has a capacity of 1,458 liters. Mrs. Smith put 487 liters 750 milliliters of water in the tub. How much water needs to be added to fill the hot tub completely?

Smith needs to add 970,250 milliliters or 970 L 250 mL of water to completely fill the hot tub,

Explanation: Given the Smith’s hot tub has a capacity of 1,458 liters. Mrs. Smith put 487 liters 750 milliliters of water in the tub. Water needed to be added to fill the hot tub completely is 1,458 liters – 487 liters 750 milliliters = as 1,458 liters = 1,458 X 1000 mL = 1,458,000 mL and 487 liters 750 milliliters = 487 X 1000 mL + 750 mL = 487000 mL + 750 mL = 487,750 mL, So 1,458,000 mL -487,750 mL 970,250 mL or 970,250 ÷ 1000 = 970 L 250 mL, Therefore, Smith needs to add 970,250 milliliters or 970 L 250 mL of water to completely fill the hot tub.

Eureka Math Grade 4 Module 2 Lesson 3 Homework Answer Key

8 L = 8,000 mL, 8 L = 8 X 1000 mL= 8,000 mL .

27 L = 27,000 mL, 27 L = 27 X 1000 mL= 27,000 mL .

39,000 mL = 39 L, as 39,000 mL ÷ 1000 mL = 39 L.

68 L = 68,000 ml, 68 L = 68 X 1000 mL = 68,000 mL .

102,000 mL = 102 L, as 102,000 mL ÷ 1000 mL = 102 L.

Question 2. Convert the measurements. a. 5 L 850 mL = ______5,850_______ mL b. 29 L 303 mL = _____29,303________ mL c. 37 L 37 mL = _____37,037________ mL d. 17 L 2 mL = ______17,002_______ mL e. 13,674 mL = __13___ L __674____ mL f. 275,005 mL = __275___ L __005____ mL

a. 5 L 850 mL = 5,850 mL,

Explanation: 5 L 850 mL as 1 L = 10 3 mL = 1000 mL, 5 X 1000 mL + 850 mL= 5,850 mL .

b. 29 L 303 mL = 29,303 mL,

Explanation: 29 L 303 mL as 1 L = 10 3 mL = 1000 mL, 29 X 1000 mL + 303 mL = 29 mL .

c. 37 L 37 mL = 37,037 mL,

Explanation: 37 L 37 mL as 1 L = 10 3 mL = 1000 mL, 37 X 1000 mL + 37 mL = 37,037 mL .

d. 17 L 2 mL= 17,002 mL,

Explanation: 17 L 3 mL as 1 L = 10 3 mL = 1000 mL, 17 X 1000 mL + 3 mL = 17,003 mL .

e. 13,674 mL  = 13 L 674 mL,

Explanation: 13,674 mL as 1 L = 10 3 mL = 1000 mL, 13,674 ÷ 1000 mL = 13 L 674 mL .

f. 275,005 mL = 275 L 005 mL,

Explanation: 275,005 mL as 1 L = 10 3 mL = 1000 mL, 275,005 ÷ 1000 mL = 275 L 005 mL .

Question 3. Solve. a. 545 mL + 48 mL b. 8 L – 5,740 mL c. Express the answer in the smaller unit: 27 L 576 mL + 784 mL d. Express the answer in the smaller unit: 27 L + 3,100 mL e. Express the answer in mixed units: 9 L 213 mL – 638 mL f. Express the answer in mixed units: 41 L 724 mL – 28 L 945 mL

Use a tape diagram to model each problem. Solve using a simplifying strategy or an algorithm, and write your answer as a statement.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-8

Explanation: Given  545 mL + 48 mL = 545 mL +48 mL 593 mL Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement five hundred forty five milliliter plus forty eight milliliter is equal to five hundred ninety three milliliters.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-9

Explanation: Given  8 L – 5,740 mL = as 8 L = 8 X 1000 mL = 8,000 mL, 8,000 mL -5,740 mL 2,260 mL or 2,260 ÷ 1000 = 2 L 260 mL, Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement eight liters minus five thousand seven hundred forty milliliter is equal to two thousand two hundred sixty milliliters or two liters two hundred sixty milliliter.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-10

Explanation: Given  27 L 576 mL + 784 mL = As 27 L  576 mL = 27 X 1000 mL + 576 mL = 27000 mL + 576 mL = 27,576 mL, 27,576 mL + 784 mL 28,360 mL The smaller unit as 1 liter is equal to 1,000 milliliter, So 28,360 mL = 28,360 ÷ 1000 = 28 L 360 mL,  Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement twenty seven liter and five hundred seventy six milliliters plus seven hundred eighty four milliliters is equal to twenty eight liter and and three hundred sixty milliliter or twenty eight thousand and three hundred sixty milliliters.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-11

Explanation: Given  27 L + 3,100 mL = As 27 L = 27 X 1000 mL = 27,000 mL, 27,000 mL +3,100 mL 30,100 mL The smaller unit as 1 liter is equal to 1,000 milliliter, So 30,100 mL = 30,100 ÷ 1000 = 30 L 100 mL,  Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement twenty seven liter plus three liters and one hundred milliliters is equal to thirty liter and one hundred milliliters or thirty thousand and one hundred milliliters.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-12

Explanation: Given  9 L 213 mL – 638 mL = As 9 L 213 mL = 9 X 1000 mL + 213 mL = 9000 mL + 213 mL = 9,213 mL 9213 mL – 638 mL 8,575 mL The answer in mixed units is 8,575 mL = 8,575 ÷ 1000 = 8 L 575 mL,  Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement nine liter two hundred thirteen milliliters minus six hundred and thirty eight milliliters is equal to eight thousand and five hundred seventy five milliliters or eight liters and five hundred seventy five milliliters.

Eureka Math Grade 4 Module 2 Lesson 3 Answer Key-13

Explanation: Given  41 L 724 mL – 28 L 945 mL = As 41 L 724 mL = 41 X 1000 mL + 724 mL = 41000 mL + 724 mL = 41,724 mL and 28 L 945 mL = 28 X 1000 mL + 945 mL = 28000 mL + 945 mL = 28,945 mL, 41,724 mL -28,945 mL 12,779 mL The answer in mixed units is 12,779 mL = 12,779 ÷ 1000 = 12 L 779 mL,  Used a tape diagram to model the problem. Solved using a simplifying strategy and wrote my answer as a statement forty one liter seven hundred twenty four milliliters minus twenty eight liter nine hundred and forty five milliliters is equal to  twelve thousand and seven hundred seventy nine milliliters or twelve liters and seven hundred seventy nine milliliters.

Question 4. Sammy’s bucket holds 2,530 milliliters of water. Marie’s bucket holds 2 liters 30 milliliters of water. Katie’s bucket holds 2 liters 350 milliliters of water. Whose bucket holds the least amount of water?

Answer: Marie’s bucket holds the least amount of water of 2 liters 30 milliliters or 2,030 mL of water,

Explanation: Given Sammy’s bucket holds 2,530 milliliters of water. Marie’s bucket holds 2 liters 30 milliliters of water. Katie’s bucket holds 2 liters 350 milliliters of water. The least amount of water bucket is Sammy’s bucket 2,530 mL, Marie’s bucket 2 L 30 mL means 2 X 1000 mL + 30 mL = 2,030 mL and Katie’s bucket holds 2 L 350 mL = 2 X 1000 mL + 350 mL = 2,350 mL, Now upon comparing 2,030 mL < 2,350 mL < 2,530 mL, Marie < Katie < Sammy’s , So Marie’s bucket holds the least amount of water of 2 liters 30 milliliters or 2,030 mL of water.

Question 5. At football practice, the water jug was filled with 18 liters 530 milliliters of water. At the end of practice, there were 795 milliliters left. How much water did the team drink?

Answer: The team drank 17,735 milliliters or 17 liters 735 milliliters,

Explanation: At football practice, the water jug was filled with 18 liters 530 milliliters of water. At the end of practice, there were 795 milliliters left. So amount of water did the team drank is 18 L 530 mL – 795 mL, 18 L 530 mL = 18 X 1000 mL + 530 mL = 18,530 mL, 18,530 mL –   795 mL 17,735 mL or 17,735 ÷ 1000 = 17 L 735 mL Therefore, the team drank 17,735 milliliters or 17 liters 735 milliliters.

Question 6. 27,545 milliliters of gas were added to a car’s empty gas tank. If the gas tank’s capacity is 56 liters 202 milliliters, how much gas is needed to fill the tank?

Answer: Gas needed to fill the tank is 28,657 milliliters or 28 liters 657 milliliters,

Explanation: Given 27,545 milliliters of gas were added to a car’s empty gas tank. If the gas tank’s capacity is 56 liters 202 milliliters, The gas needed to fill the tank is 56 L 202 mL – 27,545 mL = as 56 L 202 mL = 56 X 1000 mL + 202 mL = 56,202 mL, 56,202 mL -27,545 mL 28,657 mL or 28,657 ÷ 1000 = 28 L 657 mL Therefore, Gas needed to fill the tank is 28,657 milliliters or 28 liters 657 milliliters.

Eureka Math Grade 4 Module 2 Answer Key

Leave a comment cancel reply.

You must be logged in to post a comment.

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

4th grade (Eureka Math/EngageNY)

Unit 1: module 1: place value, rounding, and algorithms for addition and subtraction, unit 2: module 2: unit conversions and problem solving with metric measurement, unit 3: module 3: multi-digit multiplication and division, unit 4: module 4: angle measure and plane figures, unit 5: module 5: fraction equivalence, ordering, and operations, unit 6: module 6: decimal fractions, unit 7: module 7: exploring measurement with multiplication.

Search

  • Sherman Elementary School

Eureka math helps students connect math to the real world and solve problems. Homework Helpers is designed to help parents/guardians with Eureka homework. These Helpers explain, step by step, how to work problems similar to those found in Eureka Math assignments.  There is a homework helper to go with every homework assignment. There are other valuable resources for families at  eureka-math.org . 

4th Grade Links

Grade 4 module 1, grade 4 module 2, grade 4 module 3, grade 4 module 4, grade 4 module 5, grade 4 module 6, grade 4 module 7, 5th grade links, 5th grade module 1, 5th grade module 2, 5th grade module 3, 5th grade module 4, 5th grade module 5, 5th grade module 6, 6th grade links, 6th grade module 1, 6th grade module 2, 6th grade module 3, 6th grade module 4, 6th grade module 5, 6th grade module 6.

  • Questions or Feedback? |
  • Web Community Manager Privacy Policy (Updated) |

Homework Helpers Samples

Homework Helpers provide step-by-step explanations of how (and why!) to work problems similar to those found in Eureka Math homework assignments. Check out a grade-level Homework Helper sample. Click here to purchase a full year of Homework Helpers in print. 

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Samples Grades K-12

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Grade K

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Grade 1

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Grade 2

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Grade 3

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Grade 4

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Grade 5

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Grade 6

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Grade 7

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Grade 8

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Algebra I (Grade 9)

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Geometry (Grade 10)

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Algebra II (Grade 11)

eureka math grade 4 module 2 lesson 3 homework answers

Homework Helpers Sample Pre-Calculus (Grade 12)

You are leaving greatminds.org.

Great Minds partners with a number of third party vendors to provide you with the best manipulatives and workbooks out there. You are about to visit one of our partner sites. Click below to open the site in a new window.

IMAGES

  1. Eureka Math Homework Helpers Grade 4 : Com homework help cc2

    eureka math grade 4 module 2 lesson 3 homework answers

  2. Eureka Math Grade 3 Module 4 Lesson 2 Answer Key

    eureka math grade 4 module 2 lesson 3 homework answers

  3. EngageNY (Eureka Math) Grade 4 Module 2 Answer Key by MathVillage

    eureka math grade 4 module 2 lesson 3 homework answers

  4. Eureka Math Grade 4 Module 2: Homework Answer Keys (Lessons 1-3)

    eureka math grade 4 module 2 lesson 3 homework answers

  5. Eureka Math™, Grade 4, Modules 1–4

    eureka math grade 4 module 2 lesson 3 homework answers

  6. EngageNY (Eureka Math) Grade 4 Module 2 Answer Key by MathVillage

    eureka math grade 4 module 2 lesson 3 homework answers

VIDEO

  1. module 4 Lesson 3 Eureka Math 5th grade Homework

  2. Eureka Math Homework Time Grade 4 Module 7 Lesson 2

  3. Eureka Math Homework Time Grade 4 Module 2 Lesson 5

  4. Eureka Math Homework Time Grade 4 Module 3 Lesson 14

  5. Eureka Math Homework Time Grade 4 Module 6 Lesson 2

  6. Eureka Math Homework Time Grade 4 Module 3 Lesson 21

COMMENTS

  1. Eureka Math Grade 4 Module 2 Lesson 3 Answer Key

    Eureka Math Grade 4 Module 2 Lesson 3 Problem Set Answer Key. Question 1. Complete the conversion table. So 1 L = 1,000 mL . 5 L = 5 X 1000 mL= 5,000 mL . 38 L = 38 X 1000 mL= 38,000 mL . 49 L = 49 X 1000 mL = 49,000 mL . Question 2. Convert the measurements.

  2. Eureka Math Homework Time Grade 4 Module 2 Lesson 3

    It's Homework Time! Help for fourth graders with Eureka Math Module 2 Lesson 3.

  3. Course: G4M3: Multi-Digit Multiplication and Division

    Multi-Digit Multiplication and Division. Eureka Essentials: Grade 4. An outline of learning goals, key ideas, pacing suggestions, and more! Fluency Games. Teach Eureka Lesson Breakdown. Downloadable Resources. Teacher editions, student materials, application problems, sprints, etc. Application Problems.

  4. 4th Grade Math (Eureka Math/EngageNY)

    Unit 1: Module 1: Place value, rounding, and algorithms for addition and subtraction. 0/2000 Mastery points. Topic A: Place value of multi-digit whole numbers Topic B: Comparing multi-digit whole numbers Topic C: Rounding multi-digit whole numbers. Topic D: Multi-digit whole number addition Topic E: Multi-digit whole number subtraction.

  5. Course: G4M2: Unit Conversions and Problem Solving with Metric Measurement

    Grade 4. Gr4General. Gr4Mod1. Gr4Mod2. Grade 4 Module 2. ... Lesson 1. Lesson 2. Lesson 3. Topic B: Application of Metric Unit Conversions. Lesson 4. Lesson 5. End-of-Module Reivew. Gr4Mod3. Gr4Mod4. ... This work by EMBARC.Online based upon Eureka Math and is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 ...

  6. Printed Materials

    As the creator of Engage NY Math and Eureka Math, Great Minds is the only place where you can get print editions of the PK-12 curriculum.Our printed materials are available in two configurations: Learn, Practice, Succeed, or student workbooks, teacher editions, assessment and fluency materials. The Learn, Practice, Succeed configuration is available for grades K-8 and offers teachers ...

  7. Eureka Math Grade 4 Module 3 Lesson 2

    EngageNY/Eureka Math Grade 4 Module 3 Lesson 2For more Eureka Math (EngageNY) videos and other resources, please visit http://EMBARC.onlinePLEASE leave a mes...

  8. PDF Grade 4 • Module 4

    Mathematics Curriculum GRADE 4 • MODULE 4 Module 4: Angle Measure and Plane Figures Answer Key GRADE 4 • MODULE 4 Angle Measure and Plane Figures A STORY OF UNITS ... 2 4•Lesson 3 Answer Key Homework 1. Perpendicular lines accurately traced 2. Answers will vary. 3.

  9. Homework Helpers / Home

    Eureka math helps students connect math to the real world and solve problems. Homework Helpers is designed to help parents/guardians with Eureka homework. These Helpers explain, step by step, how to work problems similar to those found in Eureka Math assignments. ... Grade 4 Module 2. Comments (-1) Grade 4 Module 3. Comments (-1) Grade 4 Module ...

  10. PDF Module 3 Lessons 1-38

    Module 3 Lessons 1-38 Eureka Math™ Homework Helper 2015-2016. 2015-16 ... 4•3 A Story of Units G4-M3-Lesson 2 1. A rectangular pool is 2 feet wide. It is 4 times as long as it is wide. a. Label the diagram with the dimensions of the pool. ... Sample Answer: The perimeter of the bedroom rug is 𝟏𝟏𝟏𝟏 ...

  11. PDF Eureka Math

    10 9 8 7 6 5 4 3 2 1 Eureka Math™ Grade 3, Module 4 Student File_A Contains copy-ready classwork and homework ... Explain your answer. _____ squares G3 ... 3Lesson 2 Homework 4 Name Date 1. Each is a square unit. Count to find the area of each rectangle. Then, circle all the rectangles

  12. Great Minds Resources

    Homework Helpers Samples. Homework Helpers provide step-by-step explanations of how (and why!) to work problems similar to those found in Eureka Math homework assignments. Check out a grade-level Homework Helper sample. Click here to purchase a full year of Homework Helpers in print. Read Morekeyboard_arrow_down.

  13. Eureka Math Homework Time Grade 4 Module 4 Lesson 3

    It's Homework Time! Help for fourth graders with Eureka Math Module 4 Lesson 3.

  14. Eureka Math Grade 4 Module 3 Lesson 38 Answer Key

    Question 2. 35 × 53 Answer: Explanation: Solved 35 X 53 = 1,855 using the multiplication algorithm as shown above. Eureka Math Grade 4 Module 3 Lesson 38 Homework Answer Key. Question 1. Express 26 × 43 as two partial products using the distributive property. Solve. 26 × 43 = (__6_ forty-threes) + (_20___ forty-threes) Answer: Explanation:

  15. Eureka Math Grade 2, Module 3, Lesson 4 Instructional Video

    Walks viewers through the problem set and explains each type of question on the homework set from Module 3, Lesson 4 from the Second Grade Eureka Math curric...