Eureka Math

Engage NY Eureka Math 8th Grade Module 2 Mid Module Assessment Answer Key

Eureka Math Grade 8 Module 2 Mid Module Assessment Task Answer Key

Question 1.
Translate △XYZ along \(\overrightarrow{A B}\). Label the image of the triangle with X’, Y’, and Z’.
Answer:

b. Reflect △XYZ across the line of reflection, l. Label the image of the triangle with X’, Y’, and Z’.
Answer:

c. Rotate △XYZ around the point (1,0) clockwise 90°. Label the image of the triangle with X’, Y’, and Z’.
Answer:

Question 2.
Use the picture below to answer the questions.
Answer:

a. Can Figure A be mapped onto Figure B using only translation? Explain. Use drawings as needed in your explanation.
Answer:
No, if i translate along vector \(\overrightarrow{A B}\) i can get the longer point of figure A to map onto the lower left point of figure B(one pair of corresponding points). But no other points of the figures coincide.)

b. Can Figure A be mapped onto Figure B using only reflection? Explain. Use drawings as needed in your explanation. Use the graphs below to answer parts (a) and (b).
Answer:
No, when i connect a point of figure A to its image on figure B, the line of reflection should bisect the segment. When i connect midpoints of \(\overline{x x^{\prime}}\) & \(\overline{y y^{\prime}}\) i get a possible line of reflection, but when i check, figure A does not map onto figure B.

Question 3.
Reflect △XYZ over the horizontal line (parallel to the x-axis) through point (0,1). Label the reflected image with X’Y’Z’.

a. Reflect △XYZ over the horizontal line (parallel to the x-axis) through point (0,1). Label the reflected image with X’Y’Z’.
Answer:

b. One triangle in the diagram below can be mapped onto the other using two reflections. Identify the lines of reflection that would map one onto the other. Can you map one triangle onto the other using just one basic rigid motion? If so, explain.
Answer:

A reflection across the x-axis maps △ABC to △A’B’C’ and a reflection across the y-axis maps △A’B’C’ to △A”B”C”.
Since AB || A”B”, BC = B”C”, AC = A”C”, then A 180° rotation about the origin will map △ABC to △A”B”C”.

Engage NY Eureka Math 8th Grade Module 3 Lesson 10 Answer Key

Eureka Math Grade 8 Module 3 Lesson 10 Exercise Answer Key

Exercises 1–2.
Use a protractor to draw a pair of triangles with two pairs of equal angles. Then, measure the lengths of the sides, and verify that the lengths of their corresponding sides are equal in ratio.
Answer:
Sample student work is shown below.

Exercise 2.
Draw a new pair of triangles with two pairs of equal angles. Then, measure the lengths of the sides, and verify that the lengths of their corresponding sides are equal in ratio.
Answer:
Sample student work is shown below.

Exercises 3–5.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.
Answer:
Yes, △ABC~△A’ B’ C’. They are similar because they have two pairs of corresponding angles that are equal, namely, |∠B|=|∠B’|=103°, and |∠A|=|∠A’|=31°.

Exercise 4.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
No, △ABC is not similar to △A’ B’ C’. They are not similar because they do not have two pairs of corresponding angles that are equal, just one, namely, |∠A|=|∠A’|, but |∠B|≠|∠B’ |.

Exercise 5.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
Yes, △ABC~△A’ B’ C’. They are similar because they have two pairs of corresponding angles that are equal. You have to use the triangle sum theorem to find out that |∠B|=60° or |∠C’|=48°. Then, you can see that |∠A|=|∠A’|=72°, |∠B|=|∠B’|=60°, and |∠C|=|∠C’|=48°.

Eureka Math Grade 8 Module 3 Lesson 10 Problem Set Answer Key

Students practice presenting informal arguments to prove whether or not two triangles are similar.

Question 1.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
Yes, △ABC~△A’ B’ C’. They are similar because they have two pairs of corresponding angles that are equal, namely, |∠B|=|∠B’|=103°, and |∠A|=|∠A’|=31°.

Question 2.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
Yes, △ABC~△A’ B’ C’. They are similar because they have two pairs of corresponding angles that are equal. You have to use the triangle sum theorem to find out that |∠B’|=84° or |∠C|=32°. Then, you can see that |∠A|=|∠A’|=64°, |∠B|=|∠B’|=84°, and |∠C|=|∠C’|=32°.

Question 3.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
We do not know if △ABC is similar to △A’ B’ C’. We can use the triangle sum theorem to find out that |∠B|=44°, but we do not have any information about |∠A’| or |∠C’|. To be considered similar, the two triangles must have two pairs of corresponding angles that are equal. In this problem, we only know of one pair of corresponding angles, and that pair does not have equal measure.

Question 4.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
Yes, △ABC~△A’ B’ C’. They are similar because they have two pairs of corresponding angles that are equal, namely, |∠C|=|∠C’|=46°, and |∠A|=|∠A’|=31°.

Question 5.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
Yes, △ABC~△A’ B’ C’. They are similar because they have two pairs of corresponding angles that are equal. You have to use the triangle sum theorem to find out that |∠B|=81° or |∠C’|=29°. Then, you can see that |∠A|=|∠A’|=70°, |∠B|=|∠B’|=81°,
and |∠C|=|∠C’|=29°.

Question 6.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
No, △ABC is not similar to △A’ B’ C’. By the given information, |∠B|≠|∠B’|, and |∠A|≠|∠A’|.

Question 7.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
Yes, △ABC~△A’ B’ C’. They are similar because they have two pairs of corresponding angles that are equal. You have to use the triangle sum theorem to find out that |∠B|=102° or |∠C’|=53°. Then, you can see that |∠A|=|∠A’|=25°, |∠B|=|∠B’|=102°, and |∠C|=|∠C’|=53°.

Eureka Math Grade 8 Module 3 Lesson 10 Exit Ticket Answer Key

Question 1.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
Yes, △ABC~△A’ B’ C’. They are similar because they have two pairs of corresponding angles that are equal. You have to use the triangle sum theorem to find out that |∠B’|=45˚or |∠A|=45˚. Then, you can see that |∠A|=|∠A’|=45˚, |∠B|=|∠B’|=45˚, and |∠C|=|∠C’|=90˚.

Question 2.
Are the triangles shown below similar? Present an informal argument as to why they are or are not similar.

Answer:
No, △ABC is not similar to △A’ B’ C’. They are not similar because they do not have two pairs of corresponding angles that are equal, namely, |∠A|≠|∠A’ |, and |∠B|≠|∠B’|.

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.

Liquid Capacity
L	mL
1	1,000
5	5,000
38	38,000
49	49,000
54	54,000
92	92,000

Explanation:
Completed the conversion table as shown above,
We know 1 liter  = 10³ 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 = 10³ 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 = 10³ 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 = 10³ 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 = 10³ 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 = 10³ 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 = 10³ 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.

a. 1,760 mL + 40 L =
1,760 mL + 40 L = 41 L 760 mL or 41,760 mL,

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.

b. 7 L – 3,400 mL
7 L – 3,400 mL = 3,600 mL or 3 L 600 mL,

Statement : seven liters minus three thousand four
hundred milliliter is equal to three thousand
six hundred milliliters or three liters six hundred milliliter,

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.

c. Express the answer in the smaller unit: 25 L 478 mL + 3 L 812 mL,
25 L 478 mL + 3 L 812 mL = 29 L 290 mL or 29,290 mL,

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.

d. Express the answer in the smaller unit: 21 L – 2 L 8 mL
21 L – 2 L 8 mL = 18 L 992 mL or 18,992 mL,

Statement : twenty one liter minus two liters and eight milliliters is equal to eighteen liter and nine hundred ninety two milliliter or eighteen thousand and
nine hundred ninety two milliliters,

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.

e. Express the answer in mixed units: 7 L 425 mL – 547 mL
7 L 425 mL – 547 mL = 6,878 mL or 6 L 878 mL,

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,

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.

f. Express the answer in mixed units: 31 L 433 mL – 12 L 876 mL
31 L 433 mL – 12 L 876 mL = 18,557 mL or 18 L 557 mL,

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,

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 .

Question 6.
Petra’s fish tank contains 9 liters 578 milliliters of water. If the capacity of the tank is 12 liters 455 milliliters of water, how many more milliliters of water does she need to fill the tank?

Petra’s need more 2,877 milliliters of water to fill the tank,

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.

81 L 603 mL – 22 L 489 mL = 59,114 mL or 59 L 114 mL,

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,

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

Question 1.
Complete the conversion table.

Liquid Capacity
L	mL
1	1,000
8	8,000
27	27,000
39	39,000
68	68,000
102	102,000

Explanation:
Completed the conversion table as shown above,
We know 1 liter  = 10³ or 1000 ml,
So 1 L = 1,000 mL .

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 = 10³ 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 = 10³ 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 = 10³ 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 = 10³ 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 = 10³ 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 = 10³ 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.

a. 545 mL + 48 mL
545 mL + 48 mL = 593 mL,

Statement : five hundred forty five milliliter plus
forty eight milliliter is equal to five hundred ninety three milliliters,

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.

b. 8 L – 5,740 mL
8 L – 5,740 mL = 2,260 mL or 2 L 260 mL,

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,

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.

c. Express the answer in the smaller unit: 27 L 576 mL + 784 mL
27 L 576 mL + 784 mL =  28 L 360 mL or 28,360 mL,

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,

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.

d. Express the answer in the smaller unit: 27 L + 3,100 mL
27 L + 3,100 mL = 30 L 100 mL or 30,100 mL,

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,

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.

e. Express the answer in mixed units: 9 L 213 mL – 638 mL
9 L 213 mL – 638 mL = 8,575 mL or 8 L 575 mL,

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,

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.

f. Express the answer in mixed units: 41 L 724 mL – 28 L 945 mL
41 L 724 mL – 28 L 945 mL = 12,779 mL or 12 L 779 mL,

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,

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

Engage NY Eureka Math 5th Grade Module 3 Lesson 2 Answer Key

Eureka Math Grade 5 Module 3 Lesson 2 Sprint Answer Key

A
Find the Missing Numerator or Denominator

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

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

Question 3.
\(\frac{2}{5}\) = \(\frac{}{10}\)
Answer:
\(\frac{2}{5}\) = \(\frac{4}{10}\)
Explanation :
\(\frac{2}{5}\) = \(\frac{x}{10}\)
x = \(\frac{2}{5}\) × 10
x = \(\frac{20}{5}\)
x = 4 .

Question 4.
\(\frac{3}{5}\) = \(\frac{}{10}\)
Answer:
\(\frac{3}{5}\) = \(\frac{6}{10}\)
Explanation :
\(\frac{3}{5}\) = \(\frac{x}{10}\)
x = \(\frac{3}{5}\) × 10
x = \(\frac{30}{5}\)
x = 6 .

Question 5.
\(\frac{4}{5}\) = \(\frac{}{10}\)
Answer:
\(\frac{4}{5}\) = \(\frac{8}{10}\)
Explanation :
\(\frac{4}{5}\) = \(\frac{x}{10}\)
x = \(\frac{4}{5}\) × 10
x = \(\frac{40}{5}\)
x = 8 .

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

Question 7.
\(\frac{2}{3}\) = \(\frac{}{9}\)
Answer:
\(\frac{2}{3}\) = \(\frac{6}{9}\)
Explanation :
\(\frac{2}{3}\) = \(\frac{x}{9}\)
x = \(\frac{2}{3}\) × 9
x = \(\frac{18}{3}\)
x = 6 .

Question 8.
\(\frac{1}{3}\) = \(\frac{3}{}\)
Answer:
\(\frac{1}{3}\) = \(\frac{3}{9}\)
Explanation :
\(\frac{1}{3}\) = \(\frac{3}{x}\)
x = \(\frac{3}{1}\) × 3
x = 3 × 3
x = 9 .

Question 9.
\(\frac{2}{3}\) = \(\frac{}{9}\)
Answer:
\(\frac{2}{3}\) = \(\frac{6}{9}\)
Explanation :
\(\frac{2}{3}\) = \(\frac{x}{9}\)
x = \(\frac{2}{3}\) × 9
x = \(\frac{18}{3}\)
x = 6 .

Question 10.
\(\frac{1}{4}\) = \(\frac{}{8}\)
Answer:
\(\frac{1}{4}\) = \(\frac{2}{8}\)
Explanation :
\(\frac{1}{4}\) = \(\frac{x}{8}\)
x = \(\frac{1}{4}\) × 8
x = \(\frac{8}{4}\)
x = 2 .

Question 11.
\(\frac{3}{4}\) = \(\frac{}{8}\)
Answer:
\(\frac{3}{4}\) = \(\frac{6}{8}\)
Explanation :
\(\frac{3}{4}\) = \(\frac{x}{8}\)
x = \(\frac{3}{4}\) × 8
x = \(\frac{24}{4}\)
x = 6 .

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

Question 13.
\(\frac{3}{4}\) = \(\frac{9}{}\)
Answer:
\(\frac{3}{4}\) = \(\frac{9}{12}\)
Explanation :
\(\frac{3}{4}\) = \(\frac{9}{x}\)
x = \(\frac{4}{3}\) × 9
x = \(\frac{36}{3}\)
x = 12 .

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

Question 15.
\(\frac{2}{6}\) = \(\frac{1}{}\)
Answer:
\(\frac{2}{6}\) = \(\frac{1}{3}\)
Explanation :
\(\frac{2}{6}\) = \(\frac{1}{x}\)
x = \(\frac{6}{2}\) × 1
x = \(\frac{6}{2}\)
x = 3 .

Question 16.
\(\frac{2}{10}\) = \(\frac{1}{}\)
Answer:
\(\frac{2}{10}\) = \(\frac{1}{5}\)
Explanation :
\(\frac{2}{10}\) = \(\frac{1}{x}\)
x = \(\frac{10}{2}\) × 1
x = 5 .

Question 17.
\(\frac{4}{10}\) = \(\frac{}{5}\)
Answer:
\(\frac{4}{10}\) = \(\frac{2}{5}\)
Explanation :
\(\frac{4}{10}\) = \(\frac{x}{5}\)
x = \(\frac{4}{10}\) × 5
x = \(\frac{20}{10}\)
x = 2 .

Question 18.
\(\frac{8}{10}\) = \(\frac{}{5}\)
Answer:
\(\frac{8}{10}\) = \(\frac{4}{5}\)
Explanation :
\(\frac{8}{10}\) = \(\frac{x}{5}\)
x = \(\frac{8}{10}\) × 5
x = \(\frac{40}{10}\)
x = 4 .

Question 19.
\(\frac{3}{9}\) = \(\frac{}{3}\)
Answer:
\(\frac{3}{9}\) = \(\frac{1}{3}\)
Explanation :
\(\frac{3}{9}\) = \(\frac{x}{3}\)
x = \(\frac{3}{9}\) × 3
x = \(\frac{9}{9}\)
x = 1 .

Question 20.
\(\frac{6}{9}\) = \(\frac{}{3}\)
Answer:
\(\frac{6}{9}\) = \(\frac{2}{3}\)
Explanation :
\(\frac{6}{9}\) = \(\frac{x}{3}\)
x = \(\frac{6}{9}\) × 3
x = \(\frac{18}{9}\)
x = 2 .

Question 21.
\(\frac{3}{12}\) = \(\frac{1}{}\)
Answer:
\(\frac{3}{12}\) = \(\frac{1}{4}\)
Explanation :
\(\frac{3}{12}\) = \(\frac{1}{x}\)
x = \(\frac{3}{12}\) × 1
x = 4 .

Question 22.
\(\frac{9}{12}\) = \(\frac{}{4}\)
Answer:
\(\frac{9}{12}\) = \(\frac{3}{4}\)
Explanation :
\(\frac{9}{12}\) = \(\frac{x}{4}\)
x = \(\frac{9}{12}\) × 4
x = \(\frac{36}{12}\)
x = 3 .

Question 23.
\(\frac{1}{3}\) = \(\frac{}{12}\)
Answer:
\(\frac{1}{3}\) = \(\frac{4}{12}\)
Explanation :
\(\frac{1}{3}\) = \(\frac{x}{12}\)
x = \(\frac{1}{3}\) × 12
x = \(\frac{12}{3}\)
x = 4 .

Question 24.
\(\frac{2}{3}\) = \(\frac{}{12}\)
Answer:
\(\frac{2}{3}\) = \(\frac{8}{12}\)
Explanation :
\(\frac{2}{3}\) = \(\frac{x}{12}\)
x = \(\frac{2}{3}\) × 12
x = \(\frac{24}{3}\)
x = 8 .

Question 25.
\(\frac{8}{12}\) = \(\frac{}{3}\)
Answer:
\(\frac{8}{12}\) = \(\frac{2}{3}\)
Explanation :
\(\frac{8}{12}\) = \(\frac{x}{3}\)
x = \(\frac{8}{12}\) × 3
x = \(\frac{24}{12}\)
x = 2 .

Question 26.
\(\frac{12}{16}\) = \(\frac{3}{}\)
Answer:
\(\frac{12}{16}\) = \(\frac{3}{4}\)
Explanation :
\(\frac{12}{16}\) = \(\frac{3}{x}\)
x = \(\frac{16}{12}\) × 3
x = 4 .

Question 27.
\(\frac{3}{5}\) = \(\frac{}{25}\)
Answer:
\(\frac{3}{5}\) = \(\frac{15}{25}\)
Explanation :
\(\frac{3}{5}\) = \(\frac{x}{25}\)
x = \(\frac{3}{5}\) × 25
x = 3 × 5
x = 15 .

Question 28.
\(\frac{4}{5}\) = \(\frac{28}{}\)
Answer:
\(\frac{4}{5}\) = \(\frac{28}{35}\)
Explanation :
\(\frac{4}{5}\) = \(\frac{28}{x}\)
x = \(\frac{5}{4}\) × 28
x = 5 × 7
x = 35 .

Question 29.
\(\frac{18}{24}\) = \(\frac{3}{}\)
Answer:
\(\frac{18}{24}\) = \(\frac{3}{4}\)
Explanation :
\(\frac{18}{24}\) = \(\frac{3}{x}\)
x = \(\frac{24}{18}\) × 3
x = \(\frac{24}{6}\)
x = 4 .

Question 30.
\(\frac{24}{30}\) = \(\frac{}{5}\)
Answer:
\(\frac{24}{30}\) = \(\frac{4}{5}\)
Explanation :
\(\frac{24}{30}\) = \(\frac{x}{5}\)
x = \(\frac{24}{30}\) × 5
x = \(\frac{24}{6}\)
x = 4 .

Question 31.
\(\frac{5}{6}\) = \(\frac{35}{}\)
Answer:
\(\frac{5}{6}\) = \(\frac{35}{42}\)
Explanation :
\(\frac{5}{6}\) = \(\frac{35}{x}\)
x = \(\frac{6}{5}\) × 35
x = 6 × 7
x = 42 .

Question 32.
\(\frac{56}{63}\) = \(\frac{}{9}\)
Answer:
\(\frac{56}{63}\) = \(\frac{8}{9}\)
Explanation :
\(\frac{56}{63}\) = \(\frac{x}{9}\)
x = \(\frac{56}{63}\) × 9
x = \(\frac{56}{7}\)
x = 8 .

Question 33.
\(\frac{64}{72}\) = \(\frac{8}{}\)
Answer:
\(\frac{64}{72}\) = \(\frac{8}{9}\)
Explanation :
\(\frac{64}{72}\) = \(\frac{8}{x}\)
x = \(\frac{72}{64}\) × 8
x = \(\frac{72}{8}\)
x = 9 .

Question 34.
\(\frac{5}{8}\) = \(\frac{}{64}\)
Answer:
\(\frac{5}{8}\) = \(\frac{40}{64}\)
Explanation :
\(\frac{5}{8}\) = \(\frac{x}{64}\)
x = \(\frac{5}{8}\) × 64
x = 5 × 8
x = 40 .

Question 35.
\(\frac{5}{6}\) = \(\frac{45}{}\)
Answer:
\(\frac{5}{6}\) = \(\frac{45}{54}\)
Explanation :
\(\frac{5}{6}\) = \(\frac{45}{x}\)
x = \(\frac{6}{5}\) × 45
x = 6 × 9
x = 54 .

Question 36.
\(\frac{45}{81}\) = \(\frac{}{9}\)
Answer:
\(\frac{45}{81}\) = \(\frac{5}{9}\)
Explanation :
\(\frac{45}{81}\) = \(\frac{x}{9}\)
x = \(\frac{45}{81}\) × 9
x = \(\frac{45}{9}\)
x = 5 .

Question 37.
\(\frac{6}{7}\) = \(\frac{48}{}\)
Answer:
\(\frac{6}{7}\) = \(\frac{48}{56}\)
Explanation :
\(\frac{6}{7}\) = \(\frac{48}{x}\)
x = \(\frac{7}{6}\) × 48
x = 7 × 8
x = 56 .

Question 38.
\(\frac{36}{81}\) = \(\frac{}{9}\)
Answer:
\(\frac{36}{81}\) = \(\frac{4}{9}\)
Explanation :
\(\frac{36}{81}\) = \(\frac{x}{9}\)
x = \(\frac{36}{81}\) × 9
x = \(\frac{36}{9}\)
x = 4 .

Question 39.
\(\frac{8}{56}\) = \(\frac{1}{}\)
Answer:
\(\frac{8}{56}\) = \(\frac{1}{7}\)
Explanation :
\(\frac{8}{56}\) = \(\frac{1}{x}\)
x = \(\frac{8}{56}\) × 1
x = 7 .

Question 40.
\(\frac{35}{63}\) = \(\frac{5}{}\)
Answer:
\(\frac{35}{63}\) = \(\frac{5}{9}\)
Explanation :
\(\frac{35}{63}\) = \(\frac{5}{x}\)
x = \(\frac{63}{35}\) × 5
x = \(\frac{63}{7}\)
x = 9 .

Question 41.
\(\frac{1}{6}\) = \(\frac{12}{}\)
Answer:
\(\frac{1}{6}\) = \(\frac{12}{72}\)
Explanation :
\(\frac{1}{6}\) = \(\frac{12}{x}\)
x = \(\frac{6}{1}\) × 12
x = 72 .

Question 42.
\(\frac{3}{7}\) = \(\frac{36}{}\)
Answer:
\(\frac{3}{7}\) = \(\frac{36}{84}\)
Explanation :
\(\frac{3}{7}\) = \(\frac{36}{x}\)
x = \(\frac{7}{3}\) × 36
x = 7 × 12
x = 84 .

Question 43.
\(\frac{48}{60}\) = \(\frac{4}{}\)
Answer:
\(\frac{48}{60}\) = \(\frac{4}{5}\)
Explanation :
\(\frac{48}{60}\) = \(\frac{4}{x}\)
x = \(\frac{60}{48}\) × 4
x = \(\frac{60}{12}\)
x = 5 .

Question 44.
\(\frac{72}{84}\) = \(\frac{}{7}\)
Answer:
\(\frac{72}{84}\) = \(\frac{6}{7}\)
Explanation :
\(\frac{72}{84}\) = \(\frac{x}{7}\)
x = \(\frac{72}{84}\) × 7
x = \(\frac{72}{12}\)
x = 6 .

B
Find the Missing Numerator or Denominator

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

Question 2.
\(\frac{2}{5}\) = \(\frac{}{10}\)
Answer:
\(\frac{2}{5}\) = \(\frac{4}{10}\)
Explanation :
\(\frac{2}{5}\) = \(\frac{x}{10}\)
x = \(\frac{2}{5}\) × 10
x = 4 .

Question 3.
\(\frac{3}{5}\) = \(\frac{}{10}\)
Answer:
\(\frac{3}{5}\) = \(\frac{6}{10}\)
Explanation :
\(\frac{3}{5}\) = \(\frac{x}{10}\)
x = \(\frac{3}{5}\) × 10
x = 6 .

Question 4.
\(\frac{4}{5}\) = \(\frac{}{10}\)
Answer:
\(\frac{4}{5}\) = \(\frac{8}{10}\)
Explanation :
\(\frac{4}{5}\) = \(\frac{x}{10}\)
x = \(\frac{4}{5}\) × 10
x = 8 .

Question 5.
\(\frac{1}{3}\) = \(\frac{2}{}\)
Answer:
\(\frac{1}{3}\) = \(\frac{2}{6}\)
Explanation :
\(\frac{1}{3}\) = \(\frac{2}{x}\)
x = \(\frac{3}{1}\) × 2
x = 6

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

Question 7.
\(\frac{2}{3}\) = \(\frac{4}{}\)
Answer:
\(\frac{2}{3}\) = \(\frac{4}{12}\)
Explanation :
\(\frac{2}{3}\) = \(\frac{4}{x}\)
x = \(\frac{3}{2}\) × 4
x = 6

Question 8.
\(\frac{1}{3}\) = \(\frac{}{9}\)
Answer:
\(\frac{1}{3}\) = \(\frac{3}{9}\)
Explanation :
\(\frac{1}{3}\) = \(\frac{x}{9}\)
x = \(\frac{1}{3}\) × 9
x = 3

Question 9.
\(\frac{2}{3}\) = \(\frac{6}{}\)
Answer:
\(\frac{2}{3}\) = \(\frac{6}{9}\)
Explanation :
\(\frac{2}{3}\) = \(\frac{6}{x}\)
x = \(\frac{3}{2}\) × 6
x = 9

Question 10.
\(\frac{1}{4}\) = \(\frac{2}{}\)
Answer:
\(\frac{1}{4}\) = \(\frac{2}{8}\)
Explanation :
\(\frac{1}{4}\) = \(\frac{2}{x}\)
x = \(\frac{4}{1}\) × 2
x = 8

Question 11.
\(\frac{3}{4}\) = \(\frac{6}{}\)
Answer:
\(\frac{3}{4}\) = \(\frac{6}{8}\)
Explanation :
\(\frac{3}{4}\) = \(\frac{6}{x}\)
x = \(\frac{4}{3}\) × 6
x = 8

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

Question 13.
\(\frac{3}{4}\) = \(\frac{}{12}\)
Answer:
\(\frac{3}{4}\) = \(\frac{9}{12}\)
Explanation :
\(\frac{3}{4}\) = \(\frac{x}{12}\)
x = \(\frac{3}{4}\) × 12
x = 9

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

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

Question 16.
\(\frac{2}{10}\) = \(\frac{}{5}\)
Answer:
\(\frac{2}{10}\) = \(\frac{1}{5}\)
Explanation :
\(\frac{2}{10}\) = \(\frac{x}{5}\)
x = \(\frac{2}{10}\) × 5
x = 1

Question 17.
\(\frac{4}{10}\) = \(\frac{2}{}\)
Answer:
\(\frac{4}{10}\) = \(\frac{2}{5}\)
Explanation :
\(\frac{4}{10}\) = \(\frac{2}{x}\)
x = \(\frac{10}{4}\) × 2
x = 5

Question 18.
\(\frac{8}{10}\) = \(\frac{4}{}\)
Answer:
\(\frac{8}{10}\) = \(\frac{4}{5}\)
Explanation :
\(\frac{8}{10}\) = \(\frac{4}{x}\)
x = \(\frac{10}{8}\) × 4
x = 5

Question 19.
\(\frac{3}{9}\) = \(\frac{1}{}\)
Answer:
\(\frac{3}{9}\) = \(\frac{1}{3}\)
Explanation :
\(\frac{3}{9}\) = \(\frac{1}{x}\)
x = \(\frac{9}{3}\) × 1
x = 3

Question 20.
\(\frac{6}{9}\) = \(\frac{2}{}\)
Answer:
\(\frac{6}{9}\) = \(\frac{2}{3}\)
Explanation :
\(\frac{6}{9}\) = \(\frac{2}{x}\)
x = \(\frac{9}{6}\) × 2
x = 3

Question 21.
\(\frac{1}{4}\) = \(\frac{}{12}\)
Answer:
\(\frac{1}{4}\) = \(\frac{3}{12}\)
Explanation :
\(\frac{1}{4}\) = \(\frac{x}{12}\)
x = \(\frac{1}{4}\) × 12
x = 3

Question 22.
\(\frac{9}{12}\) = \(\frac{3}{}\)
Answer:
\(\frac{9}{12}\) = \(\frac{3}{}\)
Explanation :
\(\frac{6}{9}\) = \(\frac{2}{x}\)
x = \(\frac{9}{6}\) × 2
x = 3

Question 23.
\(\frac{1}{3}\) = \(\frac{4}{}\)
Answer:
\(\frac{1}{3}\) = \(\frac{4}{12}\)
Explanation :
\(\frac{1}{3}\) = \(\frac{4}{x}\)
x = \(\frac{3}{1}\) × 4
x = 12

Question 24.
\(\frac{2}{3}\) = \(\frac{8}{}\)
Answer:
\(\frac{2}{3}\) = \(\frac{8}{12}\)
Explanation :
\(\frac{2}{3}\) = \(\frac{8}{x}\)
x = \(\frac{3}{2}\) × 8
x = 12

Question 25.
\(\frac{8}{12}\) = \(\frac{2}{}\)
Answer:
\(\frac{8}{12}\) = \(\frac{2}{3}\)
Explanation :
\(\frac{8}{12}\) = \(\frac{2}{x}\)
x = \(\frac{12}{8}\) × 2
x = 3

Question 26.
\(\frac{12}{16}\) = \(\frac{}{4}\)
Answer:
\(\frac{12}{16}\) = \(\frac{3}{4}\)
Explanation :
\(\frac{12}{16}\) = \(\frac{x}{4}\)
x = \(\frac{12}{16}\) × 4
x = 3

Question 27.
\(\frac{3}{5}\) = \(\frac{15}{}\)
Answer:
\(\frac{3}{5}\) = \(\frac{15}{25}\)
Explanation :
\(\frac{3}{5}\) = \(\frac{15}{x}\)
x = \(\frac{5}{3}\) × 15
x = 25

Question 28.
\(\frac{4}{5}\) = \(\frac{}{35}\)
Answer:
\(\frac{4}{5}\) = \(\frac{28}{35}\)
Explanation :
\(\frac{4}{5}\) = \(\frac{x}{35}\)
x = \(\frac{4}{5}\) × 35
x = 28

Question 29.
\(\frac{18}{24}\) = \(\frac{}{4}\)
Answer:
\(\frac{18}{24}\) = \(\frac{3}{4}\)
Explanation :
\(\frac{18}{24}\) = \(\frac{x}{4}\)
x = \(\frac{18}{24}\) × 4
x = 3

Question 30.
\(\frac{24}{30}\) = \(\frac{4}{}\)
Answer:
\(\frac{24}{30}\) = \(\frac{4}{5}\)
Explanation :
\(\frac{24}{30}\) = \(\frac{4}{x}\)
x = \(\frac{30}{24}\) × 4
x = 5

Question 31.
\(\frac{5}{6}\) = \(\frac{}{42}\)
Answer:
\(\frac{5}{6}\) = \(\frac{35}{42}\)
Explanation :
\(\frac{5}{6}\) = \(\frac{x}{42}\)
x = \(\frac{5}{6}\) × 42
x = 35

Question 32.
\(\frac{56}{63}\) = \(\frac{8}{}\)
Answer:
\(\frac{56}{63}\) = \(\frac{8}{9}\)
Explanation :
\(\frac{56}{63}\) = \(\frac{8}{x}\)
x = \(\frac{63}{56}\) × 8
x = 9

Question 33.
\(\frac{64}{72}\) = \(\frac{}{9}\)
Answer:
\(\frac{64}{72}\) = \(\frac{8}{9}\)
Explanation :
\(\frac{64}{72}\) = \(\frac{x}{9}\)
x = \(\frac{64}{72}\) × 9
x = 8

Question 34.
\(\frac{5}{8}\) = \(\frac{40}{}\)
Answer:
\(\frac{5}{8}\) = \(\frac{40}{64}\)
Explanation :
\(\frac{5}{8}\) = \(\frac40}{x}\)
x = \(\frac{8}{5}\) × 40
x = 64

Question 35.
\(\frac{5}{6}\) = \(\frac{}{54}\)
Answer:
\(\frac{5}{6}\) = \(\frac{45}{54}\)
Explanation :
\(\frac{5}{6}\) = \(\frac{x}{54}\)
x = \(\frac{5}{6}\) × 54
x = 45

Question 36.
\(\frac{45}{81}\) = \(\frac{5}{}\)
Answer:
\(\frac{45}{81}\) = \(\frac{5}{9}\)
Explanation :
\(\frac{45}{81}\) = \(\frac{x}{9}\)
x = \(\frac{45}{81}\) × 9
x = 9

Question 37.
\(\frac{6}{7}\) = \(\frac{}{56}\)
Answer:
\(\frac{6}{7}\) = \(\frac{48}{56}\)
Explanation :
\(\frac{6}{7}\) = \(\frac{x}{56}\)
x = \(\frac{7}{6}\) × 56
x = 48

Question 38.
\(\frac{36}{81}\) = \(\frac{4}{}\)
Answer:
\(\frac{36}{81}\) = \(\frac{4}{9}\)
Explanation :
\(\frac{36}{81}\) = \(\frac{4}{x}\)
x = \(\frac{81}{36}\)  × 4
x = 9

Question 39.
\(\frac{8}{56}\) = \(\frac{}{7}\)
Answer:
\(\frac{8}{56}\) = \(\frac{1}{7}\)
Explanation :
\(\frac{8}{56}\) = \(\frac{x}{7}\)
x = \(\frac{8}{56}\) × 7
x = 1

Question 40.
\(\frac{35}{63}\) = \(\frac{}{9}\)
Answer:
\(\frac{35}{63}\) = \(\frac{5}{9}\)
Explanation :
\(\frac{35}{63}\) = \(\frac{x}{9}\)
x = \(\frac{35}{63}\) × 9
x = 5

Question 41.
\(\frac{1}{6}\) = \(\frac{}{72}\)
Answer:
\(\frac{1}{6}\) = \(\frac{12}{72}\)
Explanation :
\(\frac{1}{6}\) = \(\frac{x}{72}\)
x = \(\frac{1}{6}\) × 72
x = 12

Question 42.
\(\frac{3}{7}\) = \(\frac{}{84}\)
Answer:
\(\frac{3}{7}\) = \(\frac{36}{84}\)
Explanation :
\(\frac{3}{7}\) = \(\frac{x}{84}\)
x = \(\frac{3}{7}\) × 84
x = 36

Question 43.
\(\frac{48}{60}\) = \(\frac{}{5}\)
Answer:
\(\frac{48}{60}\) = \(\frac{4}{5}\)
Explanation :
\(\frac{48}{60}\) = \(\frac{x}{5}\)
x = \(\frac{48}{60}\) × 5
x = 4

Question 44.
\(\frac{72}{84}\) = \(\frac{6}{}\)
Answer:
\(\frac{72}{84}\) = \(\frac{6}{7}\)
Explanation :
\(\frac{72}{84}\) = \(\frac{6}{x}\)
x = \(\frac{84}{72}\)  × 6
x = 7

Eureka Math Grade 5 Module 3 Lesson 2 Problem Set Answer Key

Question 1.
Show each expression on a number line. Solve.
a. \(\frac{2}{5}\)+ \(\frac{1}{5}\)
b. \(\frac{1}{3}\) + \(\frac{1}{3}\)+ \(\frac{1}{3}\)
c. \(\frac{3}{10}\) + \(\frac{3}{10}\) + \(\frac{3}{10}\)
d. 2 × \(\frac{3}{4}\) + \(\frac{1}{4}\)
Answer:
a. \(\frac{2}{5}\)+ \(\frac{1}{5}\) = \(\frac{3}{5}\)

Explanation :
Number line is represented from 0 to 1 and is divided into 5 parts .\(\frac{2}{5}\) is marked from 0 and then \(\frac{1}{5}\) is marked from it to get sum of \(\frac{3}{5}\).
b. \(\frac{1}{3}\) + \(\frac{1}{3}\)+ \(\frac{1}{3}\) = \(\frac{3}{3}\) = 1

Explanation :
Number line is represented from 0 to 1 and is divided into 3 parts .\(\frac{1}{3}\) is marked from 0 and then \(\frac{1}{3}\) is marked from it and again \(\frac{1}{3}\) is marked to get sum of \(\frac{3}{3}\) = 1
c. \(\frac{3}{10}\) + \(\frac{3}{10}\) + \(\frac{3}{10}\)= \(\frac{9}{10}\)

Explanation :
Number line is represented from 0 to 1 and is divided into 10 parts .\(\frac{3}{10}\) is marked from 0 and then \(\frac{3}{10}\) is marked from it and again \(\frac{3}{10}\) is marked to get sum of \(\frac{9}{10}\).
d. 2 × \(\frac{3}{4}\) + \(\frac{1}{4}\)
\(\frac{3}{4}\) + \(\frac{3}{4}\) + \(\frac{1}{4}\) = \(\frac{7}{4}\)

Explanation :
Number line is represented from 0 to 2 and is divided into 8 parts .\(\frac{3}{4}\) is marked from 0 and then \(\frac{3}{4}\) is marked from it and again \(\frac{1}{4}\) is marked to get sum of \(\frac{7}{4}\).

Question 2.
Express each fraction as the sum of two or three equal fractional parts. Rewrite each as a multiplication equation. Show Part (a) on a number line.
a. \(\frac{6}{7}\)
b. \(\frac{9}{2}\)
c. \(\frac{12}{10}\)
d. \(\frac{27}{5}\)
Answer:
a. \(\frac{6}{7}\) = \(\frac{2}{7}\) + \(\frac{2}{7}\) + \(\frac{2}{7}\)  = 3× \(\frac{2}{7}\)

Explanation :
Number line is represented from 0 to 1 and is divided into 7 parts .\(\frac{2}{7}\) is marked from 0 and then \(\frac{2}{7}\) is marked from it and again \(\frac{2}{7}\) is marked to get sum of \(\frac{6}{7}\).
b. \(\frac{9}{2}\) = \(\frac{3}{2}\) + \(\frac{3}{2}\) + \(\frac{3}{2}\) .= 3 × \(\frac{3}{2}\)
c. \(\frac{12}{10}\) = \(\frac{6}{10}\) +\(\frac{6}{10}\)= 2×\(\frac{6}{10}\)
d. \(\frac{27}{5}\) = \(\frac{9}{5}\) + \(\frac{9}{5}\) + \(\frac{9}{5}\)  = 3×\(\frac{9}{5}\)

Question 3.
Express each of the following as the sum of a whole number and a fraction. Show Parts (c) and (d) on number lines.
a. \(\frac{9}{7}\)
c. \(\frac{32}{7}\)
d. \(\frac{24}{9}\)
Answer:
a. \(\frac{9}{7}\) =\(\frac{7}{7}\) + \(\frac{2}{7}\) .
c. \(\frac{32}{7}\) = \(\frac{28}{7}\) + \(\frac{4}{7}\) = 4 + \(\frac{4}{7}\).

Explanation :
Number line is represented from 0 to 5 and each whole is divided into 7 parts .4 is marked from 0 and then \(\frac{4}{7}\) is marked from it to get sum of \(\frac{32}{7}\).
d. \(\frac{24}{9}\) = \(\frac{18}{9}\) + \(\frac{6}{9}\) = 2 + \(\frac{6}{9}\) .

Explanation :
Number line is represented from 0 to 3 and each 1 whole divided into 9 parts .2 is marked from 0 and then \(\frac{6}{9}\) is marked from it to get sum of \(\frac{24}{9}\).

Question 4.
Marisela cut four equivalent lengths of ribbon. Each was 5 eighths of a yard long. How many yards of ribbon did she cut? Express your answer as the sum of a whole number and the remaining fractional units. Draw a number line to represent the problem.
Answer:

Explanation :
Length of each ribbon = \(\frac{5}{8}\).
Number of ribbons = 4
Total Length of ribbon = \(\frac{5}{8}\) + \(\frac{5}{8}\) + \(\frac{5}{8}\) + \(\frac{5}{8}\) = \(\frac{20}{8}\) .
\(\frac{20}{8}\) = 2 + \(\frac{4}{8}\) .

Eureka Math Grade 5 Module 3 Lesson 2 Exit Ticket Answer Key

Question 1.
Show each expression on a number line. Solve.
a. \(\frac{5}{5}\) + \(\frac{2}{5}\)
b. \(\frac{6}{3}\)+ \(\frac{2}{3}\)
Answer:
a. \(\frac{5}{5}\) + \(\frac{2}{5}\) = \(\frac{7}{5}\)

Explanation :
Number line is represented from 0 to 2 and each 1 whole divided into 5 parts .1 is marked from 0 and then \(\frac{2}{5}\) is marked from it to get sum of \(\frac{7}{5}\).
b. \(\frac{6}{3}\)+ \(\frac{2}{3}\) = 2 + \(\frac{2}{3}\) = \(\frac{8}{3}\)

Explanation :
Number line is represented from 0 to 3 and each 1 whole divided into 3 parts .2 is marked from 0 and then \(\frac{2}{3}\) is marked from it to get sum of \(\frac{8}{3}\).

Question 2.
Express each fraction as the sum of two or three equal fractional parts. Rewrite each as a multiplication equation. Show Part (b) on a number line.
a. \(\frac{6}{9}\)
b. \(\frac{15}{4}\)
Answer:
a. \(\frac{6}{9}\) = \(\frac{3}{9}\) + \(\frac{3}{9}\) = 2 × \(\frac{3}{9}\)
b. \(\frac{15}{4}\) = \(\frac{4}{4}\) + \(\frac{4}{4}\) + \(\frac{4}{4}\) + \(\frac{3}{4}\) = 3 + \(\frac{3}{4}\) .

Explanation :
Number line is represented from 0 to 4 and each 1 whole divided into 4 parts .3 is marked from 0 and then \(\frac{3}{4}\) is marked from it to get sum of \(\frac{15}{4}\).

Eureka Math Grade 5 Module 3 Lesson 2 Homework Answer Key

Question 1.
Show each expression on a number line. Solve.
a. \(\frac{4}{9}\) + \(\frac{1}{9}\)
b. \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\)
c. \(\frac{2}{7}\) + \(\frac{2}{7}\) + \(\frac{2}{7}\)
d. 2 × \(\frac{3}{5}\) + \(\frac{1}{5}\)
Answer:
a. \(\frac{4}{9}\) + \(\frac{1}{9}\) = \(\frac{5}{9}\) .

Explanation :
Number line is represented from 0 to 1 and each 1 whole divided into 9 parts .\(\frac{4}{9}\) is marked from 0 and then \(\frac{1}{9}\) is marked from it to get sum of \(\frac{5}{9}\).
b. \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) = \(\frac{4}{4}\) = 1

Explanation :
Number line is represented from 0 to 1 and each 1 whole divided into 4 parts .\(\frac{1}{4}\) is marked from 0 and then \(\frac{1}{4}\) is marked again  \(\frac{1}{4}\) and \(\frac{1}{4}\) from it to get sum of \(\frac{4}{4}\).
c. \(\frac{2}{7}\) + \(\frac{2}{7}\) + \(\frac{2}{7}\) = \(\frac{6}{7}\)

Explanation :
Number line is represented from 0 to 1 and each 1 whole divided into 7 parts .\(\frac{2}{7}\) is marked from 0 and then \(\frac{2}{7}\) is marked and again \(\frac{2}{7}\) is marked from it to get sum of \(\frac{6}{7}\).
d. 2 × \(\frac{3}{5}\) + \(\frac{1}{5}\) = \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{1}{5}\) = \(\frac{7}{5}\) .

Explanation :
Number line is represented from 0 to 2 and each 1 whole divided into 5 parts .\(\frac{3}{5}\) is marked from 0 and then \(\frac{3}{5}\) is marked and \(\frac{1}{5}\) from it to get sum of \(\frac{7}{5}\).

Question 2.
Express each fraction as the sum of two or three equal fractional parts. Rewrite each as a multiplication equation. Show Part (a) on a number line.
a. \(\frac{6}{11}\)
b. \(\frac{9}{4}\)
c. \(\frac{12}{8}\)
d. \(\frac{27}{10}\)
Answer:
a. \(\frac{6}{11}\) = \(\frac{3}{11}\) + \(\frac{3}{11}\) = 2 × \(\frac{3}{11}\) .

b. \(\frac{9}{4}\) = \(\frac{3}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) = 3 × \(\frac{3}{4}\)
c. \(\frac{12}{8}\) =\(\frac{4}{8}\) + \(\frac{4}{8}\) + \(\frac{4}{8}\)= 3 × \(\frac{4}{8}\)
d. \(\frac{27}{10}\) = \(\frac{9}{10}\) + \(\frac{9}{10}\) + \(\frac{9}{10}\) = 3 × \(\frac{9}{10}\) .

Question 3.
Express each of the following as the sum of a whole number and a fraction. Show Parts (c) and (d) on number lines.
a. \(\frac{9}{5}\)
b. \(\frac{7}{2}\)
c. \(\frac{25}{7}\)
d. \(\frac{21}{9}\)
Answer:
a. \(\frac{9}{5}\) = \(\frac{5}{5}\) + \(\frac{4}{5}\) = 1 + \(\frac{4}{5}\) .
b. \(\frac{7}{2}\) = \(\frac{2}{2}\) + \(\frac{2}{2}\) + \(\frac{2}{2}\) + \(\frac{1}{2}\) =
1+ 1 + 1 + \(\frac{1}{2}\) =  3 + \(\frac{1}{2}\) .
c. \(\frac{25}{7}\) = \(\frac{7}{7}\) + \(\frac{7}{7}\) + \(\frac{7}{7}\) + \(\frac{4}{7}\) =      1+ 1+ 1+ \(\frac{4}{7}\) =3 + \(\frac{4}{7}\) .

Explanation :
Number line is represented from 0 to 4 and each 1 whole divided into 7 parts .3 1′ are marked from 0 and then \(\frac{4}{7}\) is marked from it to get sum of \(\frac{25}{7}\).
d. \(\frac{21}{9}\) = \(\frac{9}{9}\) + \(\frac{9}{9}\) + \(\frac{3}{9}\) = 1+ 1+ \(\frac{3}{9}          \) = 2 + \(\frac{3}{9}\) .

Explanation :
Number line is represented from 0 to 3 and each 1 whole divided into 9 parts .2 1’s are marked from 0 and then \(\frac{3}{9}\) is marked from it to get sum of \(\frac{21}{9}\).

Question 4.
Natalie sawed five boards of equal length to make a stool. Each was 9 tenths of a meter long. What is the total length of the boards she sawed? Express your answer as the sum of a whole number and the remaining fractional units. Draw a number line to represent the problem.
Answer:

Explanation :
Number of boards = 5
Length of each board = 9 tenths  = \(\frac{9}{10}\) .
Total Length of board = 5 × \(\frac{9}{10}\) = \(\frac{45}{10}\).
\(\frac{45}{10}\) = \(\frac{10}{10}\) + \(\frac{10}{10}\) + \(\frac{10}{10}\) + \(\frac{10}{10}\) + \(\frac{5}{10}\) = 1+ 1+ 1+ 1+ \(\frac{5}{10}\) = 4 + \(\frac{5}{10}\) .

Engage NY Eureka Math 1st Grade Module 2 Lesson 13 Answer Key

Eureka Math Grade 1 Module 2 Lesson 13 Problem Set Answer Key

Solve. Use 5-group rows, and cross out to show your work.

Question 1.
Mike has 10 cookies on a plate and 3 cookies in a box. He eats 9 cookies from the plate. How many cookies are left?

Mike has __ cookies left.
Answer:

Mike has 4 cookies left.
Explanation:
Mike has 10 cookies on a plate and 3 cookies in a box. He eats 9 cookies from the plate. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of 13 can be written as ten and three. Mike ate nine cookies from the plate. Cross out nine cookies. Subtract nine cookies from thirteen cookies then we got four cookies. Mike has four cookies left.

Question 2.
Fran has 10 crayons in a box and 5 crayons on the desk. Fran lends Bob 9 crayons from the box. How many crayons does Fran have to use?

Fran has __ crayons to use.
Answer:

Fran has 6 crayons to use.
Explanation:
Fran has 10 crayons in a box and 5 crayons on the desk. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of 15 can be written as ten and five. Fran lends Bob 9 crayons from the box. So cross out nine crayons. Subtract nine crayons from fifteen crayons then we got six crayons. Fran has six crayons to use.

Question 3.
10 ducks are in the pond, and 7 ducks are on the land. 9 of the ducks in the pond are babies, and all the rest of the ducks are adults. How many adult ducks are there?

There are __ adult ducks.
Answer:

There are 8 adult ducks.
Explanation:
There are 10 ducks in the pond, and 7 ducks are on the land. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of seventeen can be written as ten and seven. Nine of the ducks in the pond are babies, and all the rest of the ducks are adults. So cross out nine baby ducks. Subtract nine ducks from seventeen ducks then we got eight ducks. There are eight adult ducks.

With a partner, create your own stories to match, and solve the number sentences.
Make a number bond to show the whole as 10 and some ones. Draw 5-group rows to match your story. Write the complete number sentence on the line.

Question 4.

______
Answer:

16 – 9 = 7
Explanation:
There were  10 pigs lying in the mud and 6 pigs eating by the trough outside. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of sixteen can be written as ten and six. Nine of the muddy pigs went inside the barn. So cross out nine pigs lying in the mud. Subtract nine pigs from sixteen pigs then we got seven pigs. There are seven pigs stayed outside.

Question 5.

______
Answer:

12 – 9 = 3
Explanation:
Sunitha has 12 carrots. Ten are on her plate, and two are in the bag. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of twelve can be written as ten and two. She ate 9 of the carrots on her plate. So cross out nine carrots on her plate. Subtract nine carrots from twelve carrots then we got three carrots. Sunitha has three carrots now.

Question 6.

______
Answer:

19 – 9 = 10
Explanation:
Satya has 19 apples. Ten are on her plate, and nine are in the bag. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of nineteen apples can be written as ten apples and nine apples. She ate 9 of the apples on her plate. So cross out nine apples. Subtract nine apples from nineteen apples then we got ten apples. Satya has ten apples now.

Eureka Math Grade 1 Module 2 Lesson 13 Exit Ticket Answer Key

Solve. Fill in the number bond. Use 5-group rows, and cross out to show your work.
Gabriela has 4 hair clips in her hair and 10 hair clips in her bedroom. She gives 9 of the hair clips in her room to her sister. How many hair clips does Gabriela have now?

Gabriela has __ hair clips.
Answer:

Gabriela has 5 hair clips.
Explanation:
Gabriela has 4 hair clips in her hair and 10 hair clips in her bedroom. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of fourteen clips can be written as ten clips and four clips. She gave 9 of the hair clips in her room to her sister. So cross out nine clips in her room. Subtract nine clips from fourteen clips then we got five clips. Gabriela has five hair clips.

Eureka Math Grade 1 Module 2 Lesson 13 Homework Answer Key

Solve. Use 5-group rows, and cross out to show your work. Write number sentences.

Question 1.
In a park, 10 dogs are running on the grass, and 1 dog is sleeping under the tree. 9 of the running dogs leave the park. How many dogs are left in the park?

There are __ dogs left in the park.
Answer:

11 – 9 = 2
There are 2 dogs left in the park.
Explanation:
In a park, 10 dogs are running on the grass, and 1 dog is sleeping under the tree. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of eleven dogs can be written as ten dogs and one dog. Nine of the running dogs leave the park. Cross out nine circles from the ten because those stand for the running dogs leave the park. Subtract nine dogs from eleven dogs then we got two dogs. There are two dogs left in the park.

Question 2.
Alejandro had 9 rocks in his yard and 10 rocks in his room. 9 of the rocks in his room are gray rocks, and the rest of the rocks are white. How many white rocks does Alejandro have?

Alejandro has __ white rocks.
Answer:

19  –  9  =  10
Alejandro has 10 white rocks.
Explanation:
Alejandro had 9 rocks in his yard and 10 rocks in his room. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of nineteen rocks can be written as ten rocks and nine rocks. Nine of the rocks in his room are gray rocks. Subtract nine gray rocks from nineteen rocks then we got ten rocks. Alejandro has ten white rocks.

Question 3.
Sophia has 8 toy cars in the kitchen and 10 toy cars in her bedroom. 9 of the toy cars in the bedroom are blue. The rest of her cars are red. How many red cars does Sophia have?

Sophia has __ red cars.
Answer:

18  –  9  =  9
Sophia has 9 red cars.
Explanation
Sophia has 8 toy cars in the kitchen and 10 toy cars in her bedroom. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of eighteen toy cars can be written as ten toy cars and eight toy cars. 9 of the toy cars in the bedroom are blue. Subtract nine toy cars from eighteen toy cars in order to get red cars. Sophia has nine red cars.

Question 4.
Complete the number bond, and fill in the math story. Use 5-group rows, and cross out to show your work. Write number sentences.

There were ____ birds splashing in a puddle and ____ birds walking on the dry grass. 9 of the splashing birds flew away. How many birds are left?
There are __ birds left.
Answer:

14  –  9  =  5
There were 10 birds splashing in a puddle and 4 birds walking on the dry grass. 9 of the splashing birds flew away.
There are 5 birds left.
Explanation:
There were 10 birds splashing in a puddle and 4 birds walking on the dry grass. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of fourteen birds can be written as ten birds splashing in a puddle and four birds walking on the dry glass. Nine of the splashing birds flew away. Cross out nine circles from the ten because those stand for the splashing birds flew away. Subtract nine birds from fourteen birds then we got two five birds. There are five birds left.

Engage NY Eureka Math Geometry Module 5 Lesson 9 Answer Key

Eureka Math Geometry Module 5 Lesson 9 Example Answer Key

Example 1.
a. What is the length of the arc that measures 60° in a circle of radius 10 cm?

Answer:
Arc length = \(\frac{1}{6}\) (2π×10)
Arc length = \(\frac{10 \pi}{3}\)
The marked arc length is \(\frac{10 \pi}{3}\) cm.

b. Given the concentric circles with center A and with m∠A = 60°, calculate the arc length intercepted by ∠A on each circle. The inner circle has a radius of 10, and each circle has a radius 10 units greater than the previous circle.

Answer:
Arc length of circle with radius \(\overline{A B}\) = (\(\frac{60}{360}\))(2π)(10) = \(\frac{10 \pi}{3}\)
Arc length of circle with radius \(\overline{A C}\) = (\(\frac{60}{360}\))(2π)(20) = \(\frac{20 \pi}{3}\)
Arc length of circle with radius \(\overline{A D}\) = (\(\frac{60}{360}\))(2π)(30) = \(\frac{30 \pi}{3}\) = 10π

c. An arc, again of degree measure 60°, has an arc length of 5π cm. What is the radius of the circle on which the arc sits?
Answer:
\(\frac{1}{6}\) (2π×r) = 5π
2πr = 30π
r = 15
The radius of the circle on which the arc sits is 15 cm.

d. Give a general formula for the length of an arc of degree measure x° on a circle of radius r.
Answer:
Arc length = (\(\frac{x}{360}\))2πr

e. Is the length of an arc intercepted by an angle proportional to the radius? Explain.
Answer:
Yes, the arc length is a constant \(\frac{2 \pi x}{360}\) times the radius when x is a constant angle measure, so it is proportional to the radius of an arc intercepted by an angle.

SECTOR: Let \(\widehat{A B}\) be an arc of a circle with center O and radius r. The union of all segments \(\overline{O P}\), where P is any point of \(\widehat{A B}\), is called a sector.

Answer:
→ We can use the constant of proportionality \(\frac{\pi}{180}\) to define a new angle measure, a radian. A radian is the measure of the central angle of a sector of a circle with arc length of one radius length. Say that with me.
→ A radian is the measure of the central angle of a sector of a circle with arc length of one radius length.
→ So, 1° = \(\frac{\pi}{180}\) radians. What does 180° equal in radian measure?
π radians
→ What does 360° or a rotation through a full circle equal in radian measure?
2π radians
→ Notice, this is consistent with what we found above. You will learn more about radian measure and why it was developed in Algebra II and Calculus.

Example 2.
a. Circle O has a radius of 10 cm. What is the area of the circle? Write the formula.
Answer:
Area = π(10 cm)² = 100π cm²

b. What is the area of half of the circle? Write and explain the formula.
Answer:
Area = \(\frac{1}{2}\) (π(10 cm)²) = 50π cm². 10 cm is the radius of the circle, and \(\frac{1}{2}\) = \(\frac{180}{360}\), which is the fraction of the circle.

c. What is the area of a quarter of the circle? Write and explain the formula.
Answer:
Area = \(\frac{1}{4}\) (π(10 cm)²) = 25π cm². 10 cm is the radius of the circle, and \(\frac{1}{4}\) = \(\frac{90}{360}\), which is the fraction of the circle.

d. Make a conjecture about how to determine the area of a sector defined by an arc measuring 60°.
Answer:
Area(sector AOB) = \(\frac{60}{360}\) (π(10 cm)²) = \(\frac{1}{6}\) (π(10 cm)²); the area of the circle times the arc measure divided by 360
Area(sector AOB) = \(\frac{50\pi}{3}\) cm²
The area of the sector AOB is \(\frac{50\pi}{3}\) cm².

e. Circle O has a minor arc \(\widehat{A C}\) with an angle measure of 60°. Sector AOB has an area of 24π. What is the radius of circle O?
Answer:
24π = \(\frac{1}{6}\) (πr²)
144π = (πr²)
r = 12
The radius has a length of 12 units.

f. Give a general formula for the area of a sector defined by an arc of angle measure x° on a circle of radius r.
Answer:
Area of sector = (\(\frac{x}{360}\))πr²

Eureka Math Geometry Module 5 Lesson 9 Exercise Answer Key

Exercise 1.
The radius of the following circle is 36 cm, and the m∠ABC = 60°.

a. What is the arc length of \(\widehat{A C}\)?
The degree measure of \(\widehat{A C}\) is 120°. Then the arc length of \(\widehat{A C}\) is calculated by
Arc length = \(\frac{1}{3}\) (2π ∙ 36)
Arc length = 24π.
The arc length of \(\widehat{A C}\) is 24π cm.

b. What is the radian measure of the central angle?
Answer:
Arc length = (angle measure of central angle in radians)(radius)
Arc length = (angle measure of central angle in radians)(36)
24π = 36(angle measure of central angle in radians)
(angle measure of central angle in radians) = \(\frac{24\pi}{36}\) = \(\frac{2\pi}{3}\)
The measure of the central angle is \(\frac{2\pi}{3}\) radians.

Exercises 2–3
Exercise 2.
The area of sector AOB in the following image is 28π cm². Find the measurement of the central angle labeled x°.

Answer:
28π = \(\frac{x}{360}\) (π(12)² )
x = 70
The central angle has a measurement of 70°.

Exercise 3.
In the following figure of circle O, m∠AOC = 108° and \(\widehat{A B}\) = \(\widehat{A C}\) = 10 cm.

a. Find m∠OAB.
Answer:
36°

b. Find \(m\widehat{B C}\).
Answer:
144°

c. Find the area of sector BOC.
Answer:
Area(sector BOC) = \(\frac{144}{360}\) (π(5.305)²)
Area(sector BOC) ≈ 35.37
The area of sector BOC is 35.37 cm².

Eureka Math Geometry Module 5 Lesson 9 Problem Set Answer Key

Question 1.
P and Q are points on the circle of radius 5 cm, and the measure of \(\widehat{P Q}\) is 72°. Find, to one decimal place, each of the following.

a. The length of \(\widehat{P Q}\)
Answer:
Arc length(\(\widehat{P Q}\)) = \(\frac{72}{360}\) (2π×5)
Arc length(\(\widehat{P Q}\)) = 2π
The arc length of \(\widehat{P Q}\) is 2π cm or approximately 6.3 cm.

b. The ratio of the arc length to the radius of the circle
Answer:
\(\frac{\pi}{180}\) ∙ 72 = \(\frac{2\pi}{5}\) radians

c. The length of chord \(\overline{P Q}\)
Answer:

The length of \(\overline{P Q}\) is twice the value of x in △OQR.
x = 5 sin⁡36°
PQ = 2x = 10 sin⁡36°
Chord \(\overline{P Q}\) has a length of 10 sin⁡36 cm or approximately 5.9 cm.

d. The distance of the chord \(\overline{P Q}\) from the center of the circle
Answer:

The distance of chord \(\overline{P Q}\) from the center of the circle is labeled as y in △OQR.
y = 5 cos⁡36°
The distance of chord \(\overline{P Q}\) from the center of the circle is 5 cos⁡36 cm, or approximately 4 cm.

e. The perimeter of sector POQ
Answer:
Perimeter(sector POQ) = 5 + 5 + 2π
Perimeter(sector POQ) = 10 + 2π
The perimeter of sector POQ is (10 + 2π) cm, or approximately 16.3 cm.

f. The area of the wedge between the chord \(\overline{P Q}\)and \(\widehat{P Q}\)
Answer:
Area(wedge) = Area(sector POQ) – Area(△POQ)
Area(△POQ) = \(\frac{1}{2}\) (10 sin⁡36 )(5 cos⁡36 )
Area(sector POR) = \(\frac{72}{360}\) (π(5)²)
Area(wedge) = \(\frac{72}{360}\) (π(5)² ) – \(\frac{1}{2}\) (10 sin⁡36 )(5 cos⁡36 )
The area of wedge between chord \(\overline{P Q}\)and the arc PQ is approximately 3.8 cm².

g. The perimeter of this wedge
Answer:
Perimeter(wedge) = 2π + 10 sin⁡36
The perimeter of the wedge is approximately 12.2 cm.

Question 2.
What is the radius of a circle if the length of a 45° arc is 9π?
Answer:
9π = \(\frac{45}{360}\) (2πr)
r = 36
The radius of the circle is 36.

Question 3.
\(\widehat{A B}\) and \(\widehat{C D}\) both have an angle measure of 30°, but their arc lengths are not the same. OB = 4 and BD = 2.

a. What are the arc lengths of \(\widehat{A B}\) and \(\widehat{C D}\)?
Arc length(\(\widehat{A B}\) ) = \(\frac{30}{360}\) (2π)(4)
Arc length(\(\widehat{A B}\) ) = \(\frac{2}{3}\) π
The arc length of \(\widehat{A B}\) is \(\frac{2}{3}\) π.
Arc length(\(\widehat{C D}\) ) = \(\frac{30}{360}\) (2π)(6)
Arc length(\(\widehat{C D}\) ) = π
The arc length of \(\widehat{C D}\) is π.

b. What is the ratio of the arc length to the radius for both of these arcs? Explain.
Answer:
\(\frac{30\pi}{180}\) = \(\frac{\pi}{6}\) radians. The angle is constant, so the ratio of arc length to radius will be the angle measure, 30°, multiplied by \(\frac{\pi}{180}\).

c. What are the areas of the sectors AOB and COD?
Answer:
Area(sector AOB) = \(\frac{30}{360}\) (π(4)²)
Area(sector AOB) = \(\frac{4}{3}\) π
The area of the sector AOB is \(\frac{4}{3}\) π.
Area(sector COD) = \(\frac{30}{360}\) (π(6)²)
The area of the sector COD is 3π.

Question 4.
In the circles shown, find the value of x. Figures are not drawn to scale.
a. The circles have central angles of equal measure.

Answer:
x = (4)(\(\frac{\pi}{6}\)) = \(\frac{2\pi}{3}\)
\(\frac{2\pi}{3}\) radians

b.

Answer:
x = \(\frac{\pi}{6}\)
\(\frac{\pi}{6}\) radians

c.

Answer:
x = \(\frac{18}{5}\)

d.

Answer:
x = \(\frac{2\pi}{45}\)

Question 5.
The concentric circles all have center A. The measure of the central angle is 45°. The arc lengths are given.

a. Find the radius of each circle.
Answer:
Radius of inner circle: \(\frac{\pi}{2}\) = \(\frac{45\pi}{180}\) r, r = 2
Radius of middle circle: \(\frac{5\pi}{4}\) = \(\frac{45\pi}{180}\) r, r = 5
Radius of outer circle: \(\frac{9\pi}{4}\) = \(\frac{45\pi}{180}\) r, r = 9

b. Determine the ratio of the arc length to the radius of each circle, and interpret its meaning.
Answer:
\(\frac{\pi}{4}\) is the ratio of the arc length to the radius of each circle. It is the measure of the central angle in radians.

Question 6.
In the figure, if the length of \(\widehat{P Q}\) is 10 cm, find the length of \(\widehat{Q R}\).

Answer:
Since 6° is \(\frac{1}{15}\) of 90°, then the arc length of \(\widehat{Q R}\) is \(\frac{1}{15}\) of 10 cm; the arc length of \(\widehat{Q R}\) is \(\frac{2}{3}\) cm.

Question 7.
Find, to one decimal place, the areas of the shaded regions.
a.

Answer:
Shaded Area = Area of sector – Area of Triangle
(or \(\frac{1}{4}\) (Area of circle) – Area of triangle)
Shaded Area = \(\frac{90}{360}\) (π(5)² ) – \(\frac{1}{2}\)(5)(5)
Shaded Area = 6.25π – 12.5

The shaded area is approximately 7.13.

b. The following circle has a radius of 2.

Answer:

Shaded Area = \(\frac{3}{4}\) (Area of circle) + Area of triangle
Note: The triangle is a 45° – 45° – 95° triangle with legs of length 2 (the legs are comprised by the radii, like the triangle in the previous question).
Shaded Area = \(\frac{3}{4}\) (π(2)² ) + \(\frac{1}{2}\)(2)(2)
Shaded Area = 3π + 2
The shaded area is approximately 11.4.

c.

Answer:

Shaded Area = (Area of 2 sectors) + (Area of 2 triangles)
Shaded Area = 2(\(\frac{98}{3}\) π) + 4(\(\frac{49 \sqrt{3}}{2}\))
Shaded Area = \(\frac{196}{3}\) π + 98\(\sqrt{3}\)
The shaded area is approximately 374.99.

Eureka Math Geometry Module 5 Lesson 9 Exit Ticket Answer Key

Question 1.
Find the arc length of \(\widehat{P Q R}\).
Answer:
Arc length(\(\widehat{P R}\)) = \(\frac{162}{360}\) (2π)(15)
Arc length(\(\widehat{P R}\)) = 13.5π
Circumference(circle O) = 30π
The arc length of \(\widehat{P Q R}\) is (30π – 13.5π) cm or 16.5π cm.

Question 2.
Find the area of sector POR.
Answer:
Area(sector POR) = \(\frac{162}{360}\) (π(15)² )
Area(sector POR) = 101.25π
The area of sector POR is 101.25π cm².

Engage NY Eureka Math 3rd Grade Module 5 Lesson 22 Answer Key

Eureka Math Grade 3 Module 5 Lesson 22 Problem Set Answer Key

Write the shaded fraction of each figure on the blank. Then, draw a line to match the equivalent fractions.

Answer :

Explanations :
The fraction of shaded parts = Number of shaded parts ÷ Total Number of Parts .
All the fractions are written and the respective fractions are matched .

Question 2.
Write the missing parts of the fractions.

Answer :

Explanations :
The fraction of shaded parts = Number of shaded parts ÷ Total Number of Parts .
The Shaded fractions are written .

Question 3.
Why does it take 2 copies of \(\frac{1}{8}\) to show the same amount as 1 copy of \(\frac{1}{4}\)? Explain your answer in words and pictures.
Answer :

Explanation :
Because, the parts in 1/8 is doubled than 1/4, so, we need to double the copies.
Since, by the Above diagram,

Thus, the parts in 1/8 is doubled than 1/4, so, we need to double the copies.

Question 4.
How many sixths does it take to make the same amount as \(\frac{1}{3}\) ? Explain your answer in words and pictures.
Answer :

Explanation :
The Rectangular strip is divided into 3 parts and 6 parts . The first strip is marked \(\frac{1}{3}\) and the second is marked \(\frac{2}{6}\) which means \(\frac{1}{3}\) is equal to \(\frac{2}{6}\).
It takes two one-sixths to make a third.

Question 5.
Why does it take 10 copies of 1 sixth to make the same amount as 5 copies of 1 third? Explain your answer in words and pictures.
Answer :

Explanation :
The Sixths have as many as units as thirds .
So, \(\frac{10}{6}\) are as equal as \(\frac{5}{3}\) copies as shown in above figure .

Eureka Math Grade 3 Module 5 Lesson 22 Exit Ticket Answer Key

Question 1.
Draw and label two models that show equivalent fractions.
Answer :

Explanation :
From The above figure we notice
The First rectangular strip is divided into 2 equal parts and fraction shaded is \(\frac{1}{2}\) .
The Second rectangular strip is divided into 6 equal parts and the fraction shaded is \(\frac{3}{6}\) .
Both show the equivalent fraction \(\frac{1}{2}\) = \(\frac{3}{6}\) .

Question 2.
Draw a number line that proves your thinking about Problem 1.
Answer :

Explanation :
The number is represented for the above equivalent Fraction \(\frac{1}{2}\) = \(\frac{3}{6}\) .

Eureka Math Grade 3 Module 5 Lesson 22 Homework Answer Key

Question 1.
Write the shaded fraction of each figure on the blank. Then, draw a line to match the equivalent fractions.

Answer :

Explanation :
The fraction of shaded parts = Number of shaded parts ÷ Total Number of Parts .
All the fractions are written and the respective fractions are matched .

Question 2.
Complete the fractions to make true statements.
Answer :
Answer :

Explanation :
The Figures are divided into double and the shaded Fractions are compared .

Question 3.
Why does it take 3 copies of \(\frac{1}{6}\) to show the same amount as 1 copy of \(\frac{1}{2}\)? Explain your answer in words and pictures.
Answer :

Explanation :
The Two tape Diagram Shows 3 copies of \(\frac{3}{6}\) is the same length as 1 copy of \(\frac{1}{2}\) .

Question 4.
How many ninths does it take to make the same amount as \(\frac{1}{3}\)? Explain your answer in words and pictures.
Answer :

Explanation :
The Two tape Diagram Shows 3 ninths ( \(\frac{3}{9}\)) is the same length as 1Thirds (\(\frac{1}{3}\)) .

Question 5.
A pie was cut into 8 equal slices. If Ruben ate \(\frac{3}{4}\) of the pie, how many slices did he eat? Explain your answer using a number line and words.
Answer :

Explanation :
\(\frac{3}{4}\) is the same length as \(\frac{6}{8}\)
Ruben ate 6 slices which is \(\frac{6}{8}\) = \(\frac{3}{4}\).

Engage NY Eureka Math Grade 6 Module 4 Lesson 31 Answer Key

Eureka Math Grade 6 Module 4 Lesson 31 Example Answer Key

Example 1:

Marcus reads for 30 minutes each night. He wants to determine the total number of minutes he will read over the course of a month. He wrote the equation t = 30d to represent the total amount of time that he has spent reading, where t represents the total number of minutes read and d represents the number of days that he read during the month. Determine which variable is independent and which is dependent. Then, create a table to show how many minutes he has read in the first seven days.
Independent Variable: ________
Dependent Variable: _________

Answer:
Independent Variable: Number of days
Dependent Variable: Total minutes read

Example 2:

Kira designs websites. She can create three different websites each week. Kira wants to create an equation that will give her the total number of websites she can design given the number of weeks she works. Determine the independent and dependent variables. Create a table to show the number of websites she can design over the first 5 weeks. Finally, write an equation to represent the number of websites she can design when given any number of weeks.
Independent Variable: ________
Dependent Variable: _________
Equation: _______

Answer:
Independent variable: # of weeks worked Worked (w)
Dependent variable: # of websites designed

Equation d = 3w, where w is the number of weeks worked and d is the number of websites designed.

Example 3:

Priya streams movies through a company that charges her a $5 monthly fee plus $1. 50 per movie. Determine the independent and dependent variables, write an equation to model the situation, and create a table to show the total cost per month given that she might stream between 4 and 10 movies in a month.
Independent Variable: ________
Dependent Variable: _________
Equation: _______

Answer:
Independent variable: # of movies watched per month
Dependent variable: Total cost per month, in dollars
Equation: c = 1.5m + 5 or c = 1.50m + 5

Eureka Math Grade 6 Module 4 Lesson 31 Exercise Answer Key

Exercises:

Exercise 1.
Sarah is purchasing pencils to share. Each package has 12 pencils. The equation n = 12p, where n is the total number of pencils and p is the number of packages, can be used to determine the total number of pencils Sarah purchased. Determine which variable is dependent and which is independent. Then, make a table showing the number of pencils purchased for 3 – 7 packages.

Answer:
The number of packages, p, is the independent variable.
The total number of pencils, n, is the dependent variable.

Exercise 2.
Charlotte reads 4 books each week. Let b be the number of books she reads each week, and let w be the number of weeks that she reads. Determine which variable is dependent and which is independent. Then, write an equation to model the situation, and make a table that shows the number of books read in under 6 weeks.

Answer:
The number of weeks, w, is the independent variable.
The number of books, b, is the dependent variable.
b = 4w

Exercise 3.
A miniature golf course has a special group rate. You can pay $20 plus $3 per person when you have a group of 5 or more friends. Let f be the number of friends and c be the total cost. Determine which variable is independent and which is dependent, and write an equation that models the situation. Then, make a table to show the cost for 5 to 12 friends.

Answer:
The number of friends, f, is the independent variable.
The total cost in dollars, c, is the dependent variable.
c = 3f +20

Exercise 4.
Carlos is shopping for school supplies. He bought a pencil box for $3, and he also needs to buy notebooks. Each notebook is $2. Let t represent the total cost of the supplies and n be the number of notebooks Carlos buys. Determine which variable is independent and which is dependent, and write an equation that models the situation. Then, make a table to show the cost for 1 to 5 notebooks.

Answer:
The total number of notebooks, n, is the independent variable.
The total cost in dollars, t, is the dependent variable.
t = 2n + 3

Eureka Math Grade 6 Module 4 Lesson 31 Problem Set Answer Key

Question 1.
Jazlyah sells 3 houses each month. To determine the number of houses she can sell in any given number of months, she uses the equation t = 3m, where t is the total number of houses sold and m is the number of months. Name the independent and dependent variables. Then, create a table to show how many houses she sells in fewer than 6 months.

Answer:
The independent variable is the number of months. The dependent variable is the total number of houses sold.

Question 2.
Joshua spends 25 minutes of each day reading. Let d be the number of days that he reads, and let m represent the total minutes of reading. Determine which variable is independent and which is dependent. Then, write an equation that models the situation. Make a table showing the number of minutes spent reading over 7 days.

Answer:
The number of days, d, is the independent variable.
The total number of minutes of reading, m, is the dependent variable.
m = 25d

Question 3.
Each package of hot dog buns contains 8 buns. Let p be the number of packages of hot dog buns and b be the total number of buns. Determine which variable is independent and which is dependent. Then, write an equation that models the situation, and make a table showing the number of hot dog buns in 3 to 8 packages.

Answer:
The number of packages, p, is the independent variable.
The total number of hot dog buns, b, is the dependent variable.
b = 8p

Question 4.
Emma was given 5 seashells. Each week she collected 3 more. Let w be the number of weeks Emma collects seashells and s be the number of seashells she has total. Which variable is independent, and which is dependent? Write an equation to model the relationship, and make a table to show how many seashells she has from week 4 to week 10.

Answer:
The number of weeks, w, is the independent variable.
The total number of seashells, s, is the dependent variable.
s = 3w + 5

Question 5.
Emilia is shopping for fresh produce at a farmers market. She bought a watermelon for $5, and she also wants to buy peppers. Each pepper is $0. 75. Let t represent the total cost of the produce and n be the number of peppers bought. Determine which variable Is independent and which is dependent, and write an equation that models the situation. Then, make a table to show the cost for 1 to 5 peppers.

Answer:
The number of peppers, n, is the independent variable.
The total cost in dollars, t, is the dependent variable.
t = 0.75n + 5

Question 6.
A taxicab service charges a flat fee of $7 plus an additional $1.25 per mile driven. Show the relationship between the total cost and the number of miles driven. Which variable is independent, and which is dependent? Write an equation to model the relationship, and make a table to show the cost of 4 to 10 miles.

Answer:
The number of miles driven, m, is the independent variable.
The total cost in dollars, c, is the dependent variable.
c = 125m + 7

Eureka Math Grade 6 Module 4 Lesson 31 Exit Ticket Answer Key

For each problem, determine the independent and dependent variables, write an equation to represent the situation, and then make a table with at least 5 values that models the situation.

Question 1.
Kyla spends 60 minutes of each day exercising. Let d be the number of days that Kyla exercises, and let m represent the total minutes of exercise in a given time frame. Show the relationship between the number of days that Kyla exercises and the total minutes that she exercises.
Independent variable: _______
Dependent variable: ________
Equation: _______

Answer:
Tables may vary.
Independent variable: Number of Days
Dependent variable: Total Number of Minutes
Equation: m = 60d

Question 2.
A taxicab service charges a flat fee of $8 plus an additional $1.50 per mile. Show the relationship between the total cost and the number of miles driven.
Independent variable: _______
Dependent variable: ________
Equation: _______

Answer:
Tables may vary.
Independent variable: Number of miles
Dependent variable: Total cost, in dollars
Equation: c = 1.50m + 8

Engage NY Eureka Math 3rd Grade Module 2 Lesson 20 Answer Key

Eureka Math Grade 3 Module 2 Lesson 20 Sprint Answer Key

A
Round to the Nearest Hundred

Question 1.
201 ≈

Answer:
201 ≈ 200

Question 2.
301 ≈

Answer:
301 ≈ 300

Question 3.
401 ≈

Answer:
401 ≈ 400

Question 4.
801 ≈

Answer:
801 ≈ 800

Question 5.
1,801 ≈

Answer:
1801 ≈ 1800

Question 6.
2,801 ≈

Answer:
2801 ≈ 2800

Question 7.
3,801 ≈

Answer:
3801 ≈ 3800

Question 8.
7,801 ≈

Answer:
7801 ≈ 7800

Question 9.
290 ≈

Answer:
290 ≈ 300

Question 10.
390 ≈

Answer:
390 ≈ 400

Question 11.
490 ≈

Answer:
490 ≈ 500

Question 12.
890 ≈

Answer:
890 ≈ 900

Question 13.
1,890 ≈

Answer:
1890 ≈ 1900

Question 14.
2,890 ≈

Answer:
2890 ≈ 2900

Question 15.
3,890 ≈

Answer:
3890 ≈ 3900

Question 16.
7,890 ≈

Answer:
7890 ≈ 7900

Question 17.
512 ≈

Answer:
512 ≈ 500

Question 18.
2,512 ≈

Answer:
2512 ≈ 2500

Question 19.
423 ≈

Answer:
423 ≈ 400

Question 20.
3,423 ≈

Answer:
3423 ≈ 3400

Question 21.
677 ≈

Answer:
677 ≈ 700

Question 22.
4,677 ≈

Answer:
4677 ≈ 4700

Question 23.
350 ≈

Answer:
350 ≈ 400

Question 24.
1,350 ≈

Answer:
1350 ≈ 1400

Question 25.
450 ≈

Answer:
450 ≈ 500

Question 26.
5,450 ≈

Answer:
5540 ≈ 5500

Question 27.
850 ≈

Answer:
850 ≈ 900

Question 28.
6,850 ≈

Answer:
6850 ≈ 6900

Question 29.
649 ≈

Answer:
649 ≈ 600

Question 30.
651 ≈

Answer:
651 ≈ 700

Question 31.
691 ≈

Answer:
691 ≈ 700

Question 32.
791 ≈

Answer:
791 ≈ 800

Question 33.
891 ≈

Answer:
891 ≈ 900

Question 34.
991 ≈

Answer:
991 ≈ 1000

Question 35.
995 ≈

Answer:
995 ≈ 1000

Question 36.
998 ≈

Answer:
998 ≈ 1000

Question 37.
9,998 ≈

Answer:
9998 ≈ 10000

Question 38.
7,049 ≈

Answer:
7049 ≈ 7000

Question 39.
4,051 ≈

Answer:
4051 ≈ 4100

Question 40.
8,350 ≈

Answer:
8350 ≈ 8400

Question 41.
3,572 ≈

Answer:
3572 ≈ 3600

Question 42.
9,754 ≈

Answer:
9754 ≈ 9800

Question 43.
2,915 ≈

Answer:
2915 ≈ 2900

Question 44.
9,996 ≈

Answer:
9996 ≈ 10000

B
Round to the Nearest Hundred

Question 1.
101 ≈

Answer:
101 ≈ 100

Question 2.
201 ≈

Answer:
201 ≈ 200

Question 3.
301 ≈

Answer:
301 ≈ 300

Question 4.
701 ≈

Answer:
701 ≈ 700

Question 5.
1,701 ≈

Answer:
1701 ≈ 1700

Question 6.
2,701 ≈

Answer:
2701 ≈ 2700

Question 7.
3,701 ≈

Answer:
3701 ≈ 3700

Question 8.
8,701 ≈

Answer:
8701 ≈ 8700

Question 9.
190 ≈

Answer:
190 ≈ 200

Question 10.
290 ≈

Answer:
290 ≈ 300

Question 11.
390 ≈

Answer:
390 ≈ 400

Question 12.
790 ≈

Answer:
790 ≈ 800

Question 13.
1,790 ≈

Answer:
1790 ≈ 1800

Question 14.
2,790 ≈

Answer:
2790 ≈ 2800

Question 15.
3,790 ≈

Answer:
3790 ≈ 3800

Question 16.
8,790 ≈

Answer:
8790 ≈ 8800

Question 17.
412 ≈

Answer:
412 ≈ 400

Question 18.
2,412 ≈

Answer:
2412 ≈ 2400

Question 19.
523 ≈

Answer:
523 ≈ 500

Question 20.
3,523 ≈

Answer:
3523 ≈ 3500

Question 21.
877 ≈

Answer:
877 ≈ 900

Question 22.
4,877 ≈

Answer:
4877 ≈ 4900

Question 23.
250 ≈

Answer:
250 ≈ 300

Question 24.
1,250 ≈

Answer:
1250 ≈ 1300

Question 25.
350 ≈

Answer:
350 ≈ 400

Question 26.
5,350 ≈

Answer:
5350 ≈ 5400

Question 27.
750 ≈

Answer:
750 ≈ 800

Question 28.
6,750 ≈

Answer:
6750 ≈ 6800

Question 29.
649 ≈

Answer:
649 ≈ 600

Question 30.
652 ≈

Answer:
652 ≈ 700

Question 31.
692 ≈

Answer:
692 ≈ 700

Question 32.
792 ≈

Answer:
792 ≈ 800

Question 33.
892 ≈

Answer:
892 ≈ 900

Question 34.
992 ≈

Answer:
992 ≈ 1000

Question 35.
996 ≈

Answer:
996 ≈ 1000

Question 36.
999 ≈

Answer:
999 ≈ 1000

Question 37.
9,999 ≈

Answer:
9999 ≈ 10000

Question 38.
4,049 ≈

Answer:
4049 ≈ 4000

Question 39.
2,051 ≈

Answer:
2051 ≈ 2100

Question 40.
7,350 ≈

Answer:
7350 ≈ 7400

Question 41.
4,572 ≈

Answer:
4572 ≈ 4600

Question 42.
8,754 ≈

Answer:
8754 ≈ 8800

Question 43.
3,915 ≈

Answer:
3915 ≈ 3900

Question 44.
9,997 ≈

Answer:
9997 ≈ 10000

Eureka Math Grade 3 Module 2 Lesson 20 Problem Set Answer Key

Question 1.
a. Find the actual differences either on paper or using mental math. Round each total and part to the nearest hundred and find the estimated differences.

Answer:

b. Look at the differences that gave the most precise estimates. Explain below what they have in common. You might use a number line to support your explanation.

Answer:

Explanation:
In the differences that gave the most precise estimates both numbers either rounded down or both numbers rounded up.So, its different for subtractions than for additions.
When the numbers round the same way the distance is staying about the same.When we round in opposite directions the distance gets either much longer or shorter!

Question 2.
Camden uses a total of 372 liters of gas in two months. He uses 184 liters of gas in the first month. How many liters of gas does he use in the second month?
a. Estimate the amount of gas Camden uses in the second month by rounding each number as you think best.

Answer:
372 L ≈ 400 L
184 L ≈ 200 L
400L-200L=200L
Camden uses about 200 liters of gas in the second month.

b. How many liters of gas does Camden actually use in the second month? Model the problem with a tape diagram.

Answer:

Explanation:
Camden actually uses 188 L of gas in the second month.

Eureka Math Grade 3 Module 2 Lesson 20 Exit Ticket Answer Key

Question 3.
The weight of a pear, apple, and peach are shown to the right. The pear and apple together weigh 372 grams. How much does the peach weigh?
a. Estimate the weight of the peach by rounding each number as you think best. Explain your choice.

Answer:

Explanation:
The peach weighs about 130g.I decided to round 372g to the nearest 10g and do mental math.500g is already rounded to the nearest 10g.

b. How much does the peach actually weigh? Model the problem with a tape diagram.

Answer:

Explanation:
The peach weighs 128grams.

Question:4
Kathy buys a total of 416 grams of frozen yogurt for herself and a friend. She buys 1 large cup and 1 small cup.

Large Cup	363 grams
Small Cup	? grams

a. Estimate how many grams are in the small cup of yogurt by rounding.

Answer:
416 g ≈ 400 g
363 g ≈ 400 g
400-400=0 g
Therefore the estimated weight of small cup of yogurt by rounding to nearest hundreds is 0

b. Estimate how many grams are in the small cup of yogurt by rounding in a different way.

Answer:
416 g ≈ 420 g
363 g ≈ 360 g
420-360=60 g
Therefore the estimated weight of small cup of yogurt by rounding to nearest tens is 60 g.

c. How many grams are actually in the small cup of yogurt?

Answer:

d. Is your answer reasonable? Which estimate was closer to the exact weight? Explain why.

Answer:
Yes my answer was reasonable.The estimate where we rounded the numbers to the nearest tens is close to the exact weight as the difference between the estimated and the exact weight is just 7 g.

Eureka Math Grade 3 Module 2 Lesson 20 Homework Answer Key

Question 1.
Melissa and her mom go on a road trip. They drive 87 kilometers before lunch. They drive 59 kilometers after lunch.
a. Estimate how many more kilometers they drive before lunch than after lunch by rounding to the nearest 10 kilometers.

Answer:
87 km ≈  90 km
59 km ≈  60 km
90-60=30km
Melissa and her mom drove 30 km before lunch than after lunch.

b. Precisely how much farther do they drive before lunch than after lunch?

Answer:
87-59=28 km

c. Compare your estimate from (a) to your answer from (b). Is your answer reasonable? Write a sentence to explain your thinking.

Answer:
30 km is near to 28 km.
Yes my answer is reasonable as their is only 2 km difference ib between the estimated and the actual answer.

Question 2.
Amy measures ribbon. She measures a total of 393 centimeters of ribbon and cuts it into two pieces. The first piece is 184 centimeters long. How long is the second piece of ribbon?
a. Estimate the length of the second piece of ribbon by rounding in two different ways.

Answer:
Round of to nearest hundreds
393 cm ≈ 400 cm

184 cm ≈ 200 cm
400-200cm=200cm
The second piece of ribbon is about 200cm.

Rund of to nearest tens.
393 cm ≈ 390 cm
184 cm ≈ 180 cm
390-180=210 cm
The second piece of ribbon is about 210cm.

b. Precisely how long is the second piece of ribbon? Explain why one estimate was closer.

Answer:

The actual length of the second piece of ribbon is 199cm.
My first estimate when rounded to nearest hundreds is very close to the actual answer just by 1cm difference.

Question 3.
The weight of a chicken leg, steak, and ham are shown to the right. The chicken leg and the steak together weigh 341 grams. How much does the ham weigh?
a. Estimate the weight of the ham by rounding.

Answer:
989 g ≈ 1000 g
341 g ≈ 300 g
1000-300=600 g
The estimated weight of the ham by rounding is 600 g.

b. How much does the ham actually weigh?
989-341=648 g

Question 4.
Kate uses 506 liters of water each week to water plants. She uses 252 liters to water the plants in the greenhouse. How much water does she use for the other plants?
a. Estimate how much water Kate uses for the other plants by rounding.

Answer:
Round of to nearest hundreds
506 L ≈ 500 L
252 L ≈ 300 L
500-300=200 L
Kate uses 200 L of water for the other plants by rounding.

b. Estimate how much water Kate uses for the other plants by rounding a different way.

Answer:
Round of to nearest tens.
506 L ≈ 500 L
252 L ≈  250 L
500-250=250 L
Kate uses 250 L of water for the other plants by rounding.

c. How much water does Kate actually use for the other plants? Which estimate was closer? Explain why.

Answer:
506 L-252 L=254 L

My estimate where i rounded the numbers to the nearest tens is closer as the difference between the estimated and the actual answer is 4 L only.

Engage NY Eureka Math 1st Grade Module 2 Lesson 7 Answer Key

Eureka Math Grade 1 Module 2 Lesson 7 Problem Set Answer Key

to show how you made ten to help you solve.

Question 1.
John has 8 tennis balls. Toni has 5. How many tennis balls do they have in all?

8 and ______ make ______.
10 and ______ make ______.
John and Toni have ______ tennis balls in all.
Answer:

8 and   5   make  13.
10 and   3   make  13 .
John and Toni have  13 tennis balls in all.
Explanation:
John has 8 tennis balls. Toni has 5 tennis balls as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD eight balls of John with two balls of Toni results ten. ADD ten balls with three balls of Toni which makes thirteen. John and Toni have thirteen balls in all.

Question 2.
Bob has 8 raisins, and Jenny has 4. How many raisins do they have altogether?

8 and ______ make ______.
10 and ______ make ______.
Bob and Jenny have ______ raisins altogether.
Answer:

8 and  4  make  12.
10 and  2 make  12 .
Bob and Jenny have 12 raisins altogether.
Explanation:
Bob has 8 raisins, and Jenny has 4 raisins as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD eight raisins of Bob with two raisins of Jenny results ten. ADD ten raisins with two raisins of Jenny which makes twelve. Bob and Jenny have 12 raisins altogether.

Question 3.
There are 3 chairs on the right side of the classroom and 8 on the left side. How many total chairs are in the classroom?

8 and ______ make ______.
10 and ______ make ______.
There are ______ total chairs.
Answer:

8 and  3  make  11.
10 and  1  make  11.
There are 11 total chairs.
Explanation:
There are 3 chairs on the right side of the classroom and 8  chairs on the left side of the classroom as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD eight chairs on the left side of the classroom with two chairs on the right side of the classroom which makes ten.  ADD ten chairs with one chair on the right side of the classroom results eleven chairs. Total number of chairs in the classroom are eleven.

Question 4.
There are 7 children sitting on the rug and 8 children standing. How many children are there in all?

8 and ______ make ______.
10 and ______ make ______.
There are ______ children in all.
Answer:

8 and   7   make  15.
10 and  5  make  15.
There are 15  children in all.
Explanation:
There are 7 children sitting on the rug and 8 children standing as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD two children sitting on the rug and eight children standing on the rug which results ten. ADD ten children with five children sitting on the rug which results fifteen. There are fifteen children in all.

Eureka Math Grade 1 Module 2 Lesson 7 Exit Ticket Answer Key

Draw, label, and to show how you made ten to help you solve.
Write the number sentences you used to solve.
Nick picks some peppers. He picks 5 green peppers and 8 red peppers. How many peppers does he pick in all?

8 and ______ make ______.
10 and ______ make ______.
Nick picks ____ peppers.
Answer:

8 and  5  make 13.
10 and  3 make 13.
Nick picks 13 peppers.
Explanation:
Nick picks some peppers. He picks 5 green peppers and 8 red peppers as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD two green peppers and eight red peppers which results ten.  Draw a circle for ten peppers. ADD ten peppers with three green peppers which results thirteen. Nick picks thirteen peppers.

Eureka Math Grade 1 Module 2 Lesson 7 Homework Answer Key

Draw, label, and to show how you made ten to help you solve.
Write the number sentences you used to solve.

Question 1.
Meg gets 8 toy animals and 4 toy cars at a party. How many toys does Meg get in all?

8 + 4 = ____
10 + ____ = ____
Meg gets _____ toys.
Answer:

8 + 4 = 12
10 + 2 = 12
Meg gets 12 toys.
Explanation:
Meg gets 8 toy animals and 4 toy cars at a party as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD eight toy animals with two toy cars which results ten. Draw a circle for ten toy animals. ADD ten toy animals with two toy cars which results twelve. Meg gets twelve toys.

Question 2.
John makes 6 baskets in his first basketball game and 8 baskets in his second. How many baskets does he make altogether?
____ + ____ = ____
____ + ____ = ____
John makes _____ baskets.
Answer:

6 + 8 = 14
10 + 4 = 14
John makes 14 baskets.
Explanation:
John makes 6 baskets in his first basketball game and 8 baskets in his second basketball game as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD two baskets in first basketball game of John with eight baskets in second basketball game which results ten. Draw a circle for ten baskets in basketball game. ADD ten baskets with four baskets which results fourteen. John makes fourteen baskets altogether.

Question 3.
May has a party. She invites 7 girls and 8 boys. How many friends does she invite in all?
____ + ____ = ____
____ + ____ = ____
May invites _____ friends.
Answer:

7 + 8 = 15
10 + 5 = 15
May invites 15 friends.
Explanation:
May has a party. She invites 7 girls and 8 boys as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD two girls with eight boys which results ten. Draw a circle for ten friends. ADD ten friends with five girls which results fifteen. May invites fifteen friends.

Question 4.
Alec collects baseball hats. He has 9 Mets hats and 8 Yankees hats. How many hats are in his collection?
____ + ____ = ____
____ + ____ = ____
Alec has _____ hats.
Answer:

9 + 8 = 17
10 + 7 = 17
Alec has 17 hats.
Explanation:
Alec collects baseball hats. He has 9 Mets hats and 8 Yankees hats as we can observe in the above image. An addition sentence is a mathematical expression that shows two or more values added together and their sum. ADD nine Mets hats with one Yankees hat  which results ten hats. Draw a circle for ten hats. ADD ten hats with seven Yankees hats which results seventeen. Alec has seventeen hats.

Eureka Math Grade 1 Module 2 Lesson 7 Fluency Template 1 Answer Key

Answer:
9 + 2 = 11             3 + 9 = 12
9 + 4 = 13             5 + 9 = 14
9 + 6 = 15             7 + 9 = 16
9 + 8 = 17             9 + 9 = 18
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe number sentences.
By adding nine with two results eleven.
By adding three with nine results twelve.
By adding nine with four results thirteen.
By adding five with nine results fourteen.
By adding nine with six results fifteen.
By adding seven with nine results sixteen.
By adding nine with eight results seventeen.
By adding nine with nine results eighteen.

Eureka Math Grade 1 Module 2 Lesson 7 Fluency Template 2 Answer Key

Friendly Fact Go Around: Make It Equal

Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together to get their sum.
In the above image, we can observe 9+ 1 number sentence. By adding nine with one results ten. The number sentence 9 + 1 is equal to 10 + 0.
In the above image, we can observe 9+ 3 number sentence. By adding nine with three results twelve. The number sentence 9 + 3 is equal to 10 + 2.
In the above image, we can observe 9+ 5 number sentence. By adding nine with five results fourteen. The number sentence 9 + 5 is equal to 10 + 4.
In the above image, we can observe 9+ 4 number sentence. By adding nine with four results thirteen. The number sentence 9 + 4 is equal to 10 + 3.
In the above image, we can observe 9+ 7 number sentence. By adding nine with seven results sixteen. The number sentence 9 + 7 is equal to 10 + 6.
In the above image, we can observe 9+ 6 number sentence. By adding nine with six results fifteen. The number sentence 9 + 6 is equal to 10 + 5.

Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together to get their sum.
In the above image, we can observe 3 + 9 number sentence. By adding three with nine results twelve. The number sentence 3 + 9 is equal to 10 + 2.
In the above image, we can observe 2 + 9 number sentence. By adding two with nine results eleven. The number sentence 2 + 9 is equal to 10 + 1.
In the above image, we can observe 8 +9 number sentence. By adding eight with nine results seventeen. The number sentence 8 + 9 is equal to 10 + 7.
In the above image, we can observe 5 + 9 number sentence. By adding five with nine results fourteen. The number sentence 5 + 9 is equal to 10 + 4.
In the above image, we can observe 4 + 9 number sentence. By adding four with nine results thirteen. The number sentence 4 + 9 is equal to 10 + 3.
In the above image, we can observe 9+ 9 number sentence. By adding nine with nine results eighteen. The number sentence 9 + 9 is equal to 10 + 8.

Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together to get their sum.
In the above image, we can observe 9+ 4 number sentence. By adding nine with four results thirteen. The number sentence 9 + 4 is equal to 3 + 10.
In the above image, we can observe 9+ 6 number sentence. By adding nine with six results fifteen. The number sentence 9 + 6 is equal to 5 + 10.
In the above image, we can observe 9+ 5 number sentence. By adding nine with five results fourteen. The number sentence 9 + 5 is equal to 4 + 10.
In the above image, we can observe 9+ 2 number sentence. By adding nine with two results eleven. The number sentence 9 + 2 is equal to 1+ 10.
In the above image, we can observe 9+ 7 number sentence. By adding nine with seven results sixteen. The number sentence 9 + 7 is equal to 6 + 10.
In the above image, we can observe 9+ 9 number sentence. By adding nine with nine results eighteen. The number sentence 9 + 9 is equal to 8 + 10.
In the above image, we can observe 9+ 6 number sentence. By adding nine with six results fifteen. The number sentence 9 + 6 is equal to 10 + 5.
In the above image, we can observe 9+ 8 number sentence. By adding nine with eight results seventeen. The number sentence 9 + 8 is equal to 10 + 7.
In the above image, we can observe 9+ 9 number sentence. By adding nine with nine results eighteen. The number sentence 9 + 9 is equal to 10 + 8.
In the above image, we can observe 9+ 4 number sentence. By adding nine with four results thirteen. The number sentence 9 + 4 is equal to 10 + 3.
In the above image, we can observe 9+ 5 number sentence. By adding nine with five results fourteen. The number sentence 9 + 5 is equal to 10 + 4.
In the above image, we can observe 9+ 7 number sentence. By adding nine with seven results sixteen. The number sentence 9 + 7 is equal to 10 + 6.

Engage NY Eureka Math Kindergarten Module 2 Lesson 2 Answer Key

Eureka Math Kindergarten Module 2 Lesson 2 Problem Set Answer Key

Find the triangles, and color them blue. Put an X on shapes that are not triangles.

Answer:

Explanation:
In the above picture i coloured the triangle shapes and i put a cross mark against the shapes that are not triangles.

Draw some triangles.
Answer:

Explanation:
In the above picture i drew some triangles.

Eureka Math Kindergarten Module 2 Lesson 2 Homework Answer Key

Color the triangles red and the other shapes blue.

Answer:

Explanation:
In the above picture i coloured the triangles red and the other shapes blue.

Draw 2 different triangles of your own.
Answer:

Explanation:
In the above picture i drew two different types of triangles.