Eureka Math

Engage NY Eureka Math 8th Grade Module 3 End of Module Assessment Answer Key

Eureka Math Grade 8 Module 3 End of Module Assessment Task Answer Key

Question 1.
Use the diagram below to answer the questions that follow.

Answer:

a. Dilate △OPQ from center O and scale factor r=\(\frac{4}{9}\). Label the image △OP’Q’.

b. Find the coordinates of points P’ and Q’.
Answer:
P’ = (6, 2)
Q’ = (6, \(\frac{38}{9}\))
\(\frac{\left(P^{\prime} Q^{\prime}\right)}{|P Q|}\) = \(\frac{4}{9}\)
\(\frac{\left|P^{\prime} Q^{\prime}\right|}{5}\) = \(\frac{4}{9}\)
|P’Q’| = \(\frac{20}{9}\)
\(\frac{20}{9}\) + 2 = \(\frac{20}{9}\) + \(\frac{18}{9}\)
= \(\frac{38}{9}\)

c. Are ∠OQP and ∠OQ’P’ equal in measure? Explain.
Answer:
Yes ∠OQP = ∠OQ’P’ Since D(△OQP) = △OQ’P’ and dilations are degree preserving, then ∠OQP = ∠OQ’P’.
∠OQP & ∠OQ’P’ are corresponding angles of parallel lines PQ & P’Q’, therefore ∠OQR = ∠OQ’P’

d. What is the relationship between segments PQ and P’Q’? Explain in terms of similar triangles.
Answer:
The lines that contain \(\overline{\text { PQ }}\) and \(\overline{p^{\prime} Q^{\prime}}\) are parallel. ∆OPQ ~ ∆OP’Q’ by the AA criterion (∠D = ∠D, ∠OP’Q’ = ∠OPQ), Therefore by the fundamental theorem of similarity \(\overline{P Q}\) || \(\overline{P^{\prime} Q^{\prime}}\)

e. If the length of segment OQ is 9.8 units, what is the length of segment OQ’? Explain in terms of similar triangles.
Answer:
Since ∆OPQ ~ ∆OP’Q’, then the ratios of lengths of corresponding sides will be equal to the scale factor then
\(\frac{\left|O P^{\prime}\right|}{|O P|}\) = \(\frac{\left|O Q^{\prime}\right|}{|O Q|}\) = \(\frac{4}{9}\)
\(\frac{4}{9}\) = \(\frac{\left|O Q^{\prime}\right|}{|9.0|}\)
39.2 = 9(|OQ’|)
4.36 = |OQ’|
|OQ’| is approximately 4.4 units.

Question 2.
Use the diagram below to answer the questions that follow. The length of each segment is as follows: segment OX is 5 units, segment OY is 7 units, segment XY is 3 units, and segment X’Y’ is 12.6 units.

a. Suppose segment XY is parallel to segment X’Y’. Is △OXY similar to △OX’Y’? Explain.
Answer:
Yes, ∆OXY ~ ∆OX’Y’. Since \(\overline{X Y}\) || \(X^{\prime} Y^{\prime}\) and ∠OYX = ∠OY’X’. Because corresponding angles of parallel lines are equal in measure, by AA ∆OXY ~ ∆OX’Y’.

b. What is the length of segment OX’? Show your work.
Answer:
\(\frac{12.6}{3}\) = \(\frac{\left|O X^{\prime}\right|}{|5|}\)
5(12.6) = 3(|OX’|)
63 = 3(|OX’|)
21 = |OX’|
|OX’| is 21 units.

c. What is the length of segment OY’? Show your work.
Answer:
\(\frac{12.6}{3}\) = \(\frac{\left|OY^{\prime}\right|}{|7|}\)
12.6(7) = (3|OY’|)
88.2 = 3(|OY’|)
29.4 = |OY’|
|OY’| is 29.4 units.

Question 3.
Given △ABC ~△A^’ B^’ C’ and △ABC ~△A”B”C” in the diagram below, answer parts (a)–(c).

a. Describe the sequence that shows the similarity for △ABC and △A’ B’ C’.
Answer:
\(\frac{B^{\prime} C^{\prime}}{B C}\) = \(\frac{2}{1}\) = 2 = r
Le D be the dilation from center O and scale factor r=2. Let there be a reflection across the Y-axis. Then the dilation followed by the reflection maps △ABC onto △A’B’C’.

b. Describe the sequence that shows the similarity for △ABC and △A”B”C”.
Answer:
Let D be the dilation from center O and scale factor. 0<r<1. Let there be a rotation of 180° around center O. Then the dilation followed by the rotation maps △ABC onto △A”B”C”.

c. Is △A’B’C’ similar to △A”B”C”? How do you know?
Answer:
Yes △A’B’C’ ~ △A”B”C”. Dilations preserve angle measures and since △ABC ~ △A’B’C and △ABC ~ △A”B”C”, we know ∠A= ∠A’ = ∠A”, ∠B = ∠B’=∠B”, by AA Criterion for similarity △A’B’C’ ~ △ A”B”C”. Also, similarity is Transitive.

Engage NY Eureka Math 4th Grade Module 2 Lesson 5 Answer Key

Eureka Math Grade 4 Module 2 Lesson 5 Sprint Answer Key

A
Convert to Kilograms and Grams

Answer:

Question 1.
2,000 g = 2 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  2,000 g = 2 X 1000 g ÷ 1000 = 2 kg 0 g.

Question 2.
3,000 g = 3 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  3,000 g = 3 X 1000 g ÷ 1000 = 3 kg 0 g.

Question 3.
4,000 g = 4 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  4,000 g = 4 X 1000 g ÷ 1000 = 4 kg 0 g.

Question 4.
9,000 g = 9 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  9,000 g = 9 X 1000 g ÷ 1000 = 9 kg 0 g.

Question 5.
6,000 g = 6 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  6,000 g = 6 X 1000 g ÷ 1000 = 6 kg 0 g.

Question 6.
1,000 g = 1 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  1,000 g = 1 X 1000 g ÷ 1000 = 1 kg 0 g.

Question 7.
8,000 g = 8 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  8,000 g = 8 X 1000 g ÷ 1000 = 8 kg 0 g.

Question 8.
5,000 g = 5 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,000 g = 5 X 1000 g ÷ 1000 = 5 kg 0 g.

Question 9.
7,000 g = 7 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  7,000 g = 7 X 1000 g ÷ 1000 = 7 kg 0 g.

Question 10.
6,100 g = 6 kg 100 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  6,100 g = 6,100 g ÷ 1000 = 6 kg 100 g.

Question 11.
6,110 g = 6 kg 110 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  6,110 g = 6 ,110 g ÷ 1000 = 6 kg 110 g.

Question 12.
6,101 g = 6 kg 101 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  6,101 g = 6,101 g ÷ 1000 = 6 kg 101 g.

Question 13.
6,010 g = 6 kg 10 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  6,010 g = 6,010 g ÷ 1000 = 6 kg 10 g.

Question 14.
6,011 g = 6 kg 11 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  6,011 g = 6,011 g ÷ 1000 = 6 kg 11 g.

Question 15.
6,001 g = 6 kg 1 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  6,001 g = 6,001 g ÷ 1000 = 6 kg 1 g.

Question 16.
8,002 g = 8 kg 2 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  8,002 g = 8,002 g ÷ 1000 = 8 kg 2 g.

Question 17.
8,020 g = 8 kg 20 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  8,002 g = 8,002 g ÷ 1000 = 8 kg 2 g.

Question 18.
8,200 g = 8 kg 200 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  8,200 g = 8,200 g ÷ 1000 = 8 kg 200 g.

Question 19.
8,022 g = 8 kg 22 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  8,022 g = 8,022 g ÷ 1000 = 8 kg 22 g.

Question 20.
8,220 g = 8 kg  220 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  8,220 g = 8,220 g ÷ 1000 = 8 kg 220 g.

Question 21.
8,222 g = 8 kg 222 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  8,222 g = 8,222 g ÷ 1000 = 8 kg 222 g.

Question 22.
7,256 g = 7 kg 256 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  7,256 g = 7,256 g ÷ 1000 = 7 kg 256 g.

Question 23.
3,800 g = 3 kg 800 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  3,800 g = 3,800 g ÷ 1000 = 3 kg 800 g.

Question 24.
4,770 g = 4 kg 770 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  4,770 g = 4,770 g ÷ 1000 = 4 kg 770 g.

Question 25.
4,807 g = 4 kg 807 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  4,807 g = 4,807 g ÷ 1000 = 4 kg 807 g.

Question 26.
5,065 g = 5 kg 065 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,065 g = 5,065 g ÷ 1000 = 5 kg 065 g.

Question 27.
5,040 g = 5 kg 40 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,040 g = 5,040 g ÷ 1000 = 5 kg 40 g.

Question 28.
6,007 g = 6 kg 7 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  6,007 g = 6,007 g ÷ 1000 = 6 kg 7 g.

Question 29.
2,003 g = 2 kg 3 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  2,003 g = 2,003 g ÷ 1000 = 2 kg 3 g.

Question 30.
1,090 g = 1 kg 90 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  1,090 g = 1,090 g ÷ 1000 = 1 kg 90 g.

Question 31.
1,055 g = 1 kg 55 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  1,055 g = 1,055 g ÷ 1000 = 1 kg 55 g.

Question 32.
9,404 g = 9 kg 404 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  9,404 g = 9,404 g ÷ 1000 = 9 kg 404 g.

Question 33.
9,330 g = 9 kg 330 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 9,330 g = 9,330 g ÷ 1000 = 9 kg 330 g.

Question 34.
3,400 g = 3 kg 400 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  3,400 g = 3,400 g ÷ 1000 = 3 kg 400 g.

Question 35.
4,000 g + 2,000 g = 6 kg 0 g,

Explanation:
Given 4,000 g + 2,000 g = 6,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 4,000 g + 2,000 g = 6,000 g = 6,000 g ÷ 1000 = 6 kg 0 g.

Question 36.
5,000 g + 3,000 g = 8 kg 0 g,

Explanation:
Given 5,000 g + 3,000 g = 8,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 5,000 g + 3,000 g = 8,000 g = 8,000 g ÷ 1000 = 8 kg 0 g.

Question 37.
4,000 g + 4,000 g = 8 kg 0 g,

Explanation:
Given 4,000 g + 4,000 g = 8,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  4,000 g + 4,000 g = 8,000 g = 8,000 g ÷ 1000 = 8 kg 0 g.

Question 38.
8 × 7,000 g = 56 kg 0 g,

Explanation:
Given 8 X 7,000 g = 56,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 8 × 7,000 g = 56,000 g = 56,000 g ÷ 1000 = 56 kg 0 g.

Question 39.
49,000 g ÷ 7 = 7 kg 0 g,

Explanation:
Given 49,000 g ÷ 7 = 7,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 49,000 g ÷ 7 = 7,000 g = 7,000 g ÷ 1000 = 7 kg 0 g.

Question 40.
16,000 g × 5 = 80 kg 0 g,

Explanation:
Given 16,000 g X 5 = 80,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 16,000 g X 5 = 80,000 g = 80,000 g ÷ 1000 = 80 kg 0 g.

Question 41.
63,000 g ÷ 7 = 9 kg 0 g,

Explanation:
Given 63,000 g ÷ 7 = 9,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 63,000 g ÷ 7 = 9,000 g = 9,000 g ÷ 1000 = 9 kg 0 g.

Question 42.
17 × 4,000 g = 68 kg 0 g,

Explanation:
Given 17 X 4,000 g  = 68,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 17 X 4,000 g = 68,000 g = 68,000 g ÷ 1000 = 68 kg 0 g.

Question 43.
13,000 g × 5 = 65 kg 0 g,

Explanation:
Given 13,000 g X 5 = 65,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 13,000 g X 5 = 65,000 g = 65,000 g ÷ 1000 = 65 kg 0 g.

Question 44.
84,000 g ÷ 7 = 12 kg 0 g,

Explanation:
Given 84,000 g ÷ 7 = 12,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 84,000 g ÷ 7 = 12,000 g = 12,000 g ÷ 1000 = 12 kg 0 g.

B
Convert to Kilograms and Grams

Answer:

Question 1.
1,000 g = 1 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  1,000 g = 1 X 1000 g ÷ 1000 = 1 kg 0 g.

Question 2.
2,000 g = 2 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  2,000 g = 2 X 1000 g ÷ 1000 = 2 kg 0 g.

Question 3.
3,000 g = 3 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  3,000 g = 3 X 1000 g ÷ 1000 = 3 kg 0 g.

Question 4.
8,000 g = 8 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  8,000 g = 8 X 1000 g ÷ 1000 = 8 kg 0 g.

Question 5.
6,000 g = 6 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  6,000 g = 6 X 1000 g ÷ 1000 = 6 kg 0 g.

Question 6.
9,000 g = 9 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  9,000 g = 9 X 1000 g ÷ 1000 = 9 kg 0 g.

Question 7.
4,000 g = 4 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  4,000 g = 4 X 1000 g ÷ 1000 = 4 kg 0 g.

Question 8.
7,000 g = 7 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  7,000 g = 7 X 1000 g ÷ 1000 = 7 kg 0 g.

Question 9.
5,000 g = 5 kg 0 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,000 g = 5 X 1000 g ÷ 1000 = 5 kg 0 g.

Question 10.
5,100 g = 5 kg 100 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,100 g = 5,100 g ÷ 1000 = 5 kg 100 g.

Question 11.
5,110 g = 5 kg 110 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,110 g = 5 ,110 g ÷ 1000 = 5 kg 110 g.

Question 12.
5,101 g = 5 kg 101 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,101 g = 5,101 g ÷ 1000 = 5 kg 101 g.

Question 13.
5,010 g = 5 kg 10 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,010 g = 5,010 g ÷ 1000 = 5 kg 10 g.

Question 14.
5,011 g = 5 kg 11 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,011 g = 5,011 g ÷ 1000 = 5 kg 11 g.

Question 15.
5,001 g = 5 kg 1 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,001 g = 5,001 g ÷ 1000 = 5 kg 1 g.

Question 16.
7,002 g = 7 kg 2 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  7,002 g = 7,002 g ÷ 1000 = 7 kg 2 g.

Question 17.
7,020 g = 7kg 20 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  7,020 g = 7,020 g ÷ 1000 = 7 kg 20 g.

Question 18.
7,200 g = 7 kg 200 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  7,200 g = 7,200 g ÷ 1000 = 7 kg 200 g.

Question 19.
7,022 g = 7 kg 22 g,
7,022 g = 7 kg 22 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  7,022 g = 7,022 g ÷ 1000 = 7 kg 22 g.

Question 20.
7,220 g = 7 kg 220 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  7,220 g = 7,220 g ÷ 1000 = 7 kg 220 g.

Question 21.
7,222 g = 7 kg 222 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  7,222 g = 7,222 g ÷ 1000 = 7 kg 222 g.

Question 22.
4,378 g = 4 kg 378 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  4,378 g = 4,378 g ÷ 1000 = 4 kg 378 g.

Question 23.
2,700 g = 2 kg 700 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  2,700 g = 2,700 g ÷ 1000 = 2 kg 700 g.

Question 24.
3,660 g = 3 kg 660 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  3,660 g = 3,660 g ÷ 1000 = 3 kg 660 g.

Question 25.
3,706 g = 3 kg 706 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  3,706 g = 3,706 g ÷ 1000 = 3 kg 706 g.

Question 26.
4,095 g = 4 kg 95 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  4,095 g = 4,095 g ÷ 1000 = 4 kg 95 g.

Question 27.
4,030 g = 4 kg 30 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  4,030 g = 4,030 g ÷ 1000 = 4 kg 30 g.

Question 28.
5,006 g = 5 kg 6 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  5,006 g = 5,006 g ÷ 1000 = 5 kg 6 g.

Question 29.
3,004 g = 3 kg 4 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  3,004 g = 3,004 g ÷ 1000 = 3 kg 4 g.

Question 30.
2,010 g = 2 kg 10 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  2,010 g = 2,010 g ÷ 1000 = 2 kg 10 g.

Question 31.
2,075 g = 2 kg 75 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  2,075 g = 2,075 g ÷ 1000 = 2 kg 75 g.

Question 32.
1,504 g = 1 kg 504 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  1,504 g = 1,504 g ÷ 1000 = 1 kg 504 g.

Question 33.
1,440 g = 1 kg 440 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  1,440 g = 1,440 g ÷ 1000 = 1 kg 440 g.

Question 34.
4,500 g = 4 kg 500 g,

Explanation:
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  4,500 g = 4,500 g ÷ 1000 = 4 kg 500 g.

Question 35.
3,000 g + 2,000 g = 5 kg 0 g,

Explanation:
Given 3,000 g + 2,000 g = 5,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So  3,000 g + 2,000 g = 5,000 g
5,000 g ÷ 1000 = 5 kg 0 g.

Question 36.
4,000 g + 3,000 g = 7 kg 0 g,

Explanation:
Given 4,000 g + 3,000 g = 7,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 4,000 g + 3,000 g = 7,000 g = 7,000 g ÷ 1000 = 7 kg 0 g.

Question 37.
5,000 g + 4,000 g = 9 kg 0 g,

Explanation:
Given 5,000 g + 4,000 g = 9,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 5,000 g + 4,000 g = 9,000 g = 9,000 g ÷ 1000 = 9 kg 0 g.

Question 38.
9 × 8,000 g = 72 kg 0 g,

Explanation:
Given 9 X 8,000 g = 72,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 9 × 8,000 g = 72,000 g = 72,000 g ÷ 1000 = 72 kg 0 g.

Question 39.
64,000 g ÷ 8 = 8 kg 0 g,

Explanation:
Given 64,000 g ÷ 8 = 8,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 64,000 g ÷ 8 = 8,000 g = 8,000 g ÷ 1000 = 8 kg 0 g.

Question 40.
17,000 g × 5 = 85 kg 0 g,

Explanation:
Given  17,000 g X 5  = 85,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 17,000 g X 5 = 85,000 g = 85,000 g ÷ 1000 = 85 kg 0 g.

Question 41.
54,000 g ÷ 6 = 9 kg 0 g,

Explanation:
Given 54,000 g ÷ 6 = 9,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 54,000 g ÷ 6 = 9,000 g = 9,000 g ÷ 1000 = 9 kg 0 g.

Question 42.
18,000 g × 4 = 72 kg 0 g,

Explanation:
Given  18,000 g X 4  = 72,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 18,000 g X 4 = 72,000 g = 72,000 g ÷ 1000 = 72 kg 0 g.

Question 43.
14 × 5,000 g = 70 kg 0 g,

Explanation:
Given  14 X 5,000 g  = 70,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 14 X 5,000 g = 70,000 g = 70,000 g ÷ 1000 = 70 kg 0 g.

Question 44.
96,000 g ÷ 8 = 12 kg 0 g,

Explanation:
Given 96,000 g ÷ 8 = 12,000 g
Converted the conversion table as shown above as
We know 1000 g or 10³  = 1 kg,
So 96,000 g ÷ 8 = 12,000 g = 12,000 g ÷ 1000 = 12 kg 0 g.

Eureka Math Grade 4 Module 2 Lesson 5 Problem Set Answer Key

Model each problem with a tape diagram.
Solve and answer with a statement.

Question 1.
The potatoes Beth bought weighed 3 kilograms 420 grams. Her onions weighed 1,050 grams less than the potatoes. How much did the potatoes and onions weigh together?

The potatoes and onions wiegh together 5,790 g or 5 kg 790 g,

Statement: Both the potatoes and onions wiegh together are five thousand seven hundred ninety grams or five kilograms and seven hundred ninety grams,

Explanation:
Given the potatoes Beth bought weighed 3 kilograms 420 grams.
Her onions weighed 1,050 grams less than the potatoes.
So onions wiegh is 3 kg 420 g – 1,050 g =
3 X 1000 g + 420 g = 3,420 g,
now 3,420 g – 1,050 g = 2,370 g,
So the potatoes and onions weigh together are
3,420 g + 2,370 g = 5,790 g or 5,790 ÷ 1,000 = 5 kg 790 g,
Therefore, the potatoes and onions weigh together
5,790 g or 5 kg 790 g,
Modeled  problem with a tape diagram as shown above and
statement is both the potatoes and onions wiegh together are five thousand seven hundred ninety grams or five kilograms and seven hundred ninety grams.

Question 2.
Adele let out 18 meters 46 centimeters of string to fly her kite. She then let out 13 meters 78 centimeters more before reeling back in 590 centimeters. How long was her string after reeling it in?

Adele string is 26,34 centimeters or 26 meters 34 centimeters long after reeling it in.

Statement: Adele string is twenty six thousand thirty centimeters or twenty six meters and thirty centimeters long after reeling it in,

Explanation:
Given Adele let out 18 meters 46 centimeters of string to fly her kite.
She then let out 13 meters 78 centimeters more before reeling back in 590 centimeters. So Adele string length after
reeling it in is 18 m 46 cm + 13 m 78 cm – 590 cm =
as 18 m 46 cm = 18 x 100 cm + 46 cm = 18,00 cm + 46 cm =
18,46 cm and 13 m 78 cm = 13  X 100 cm + 78 cm =
13,00 cm + 78 cm = 13,78 cm.
Now (18,46 cm + 13,78 cm) – 590 cm = 32,24 cm – 590 cm =
26,34 cm or  26,34 cm ÷ 100 = 26 m 34 cm,
Therefore, Adele string is 26,34 centimeters or
26 meters 34 centimeters long after reeling it in.
Modeled  problem with a tape diagram as shown above and statement is Adele string is twenty six thousand thirty centimeters or twenty six meters and thirty centimeters long after reeling it in.

Question 3.
Shyan’s barrel contained 6 liters 775 milliliters of paint. She poured in 1 liter 118 milliliters more. The first day, Shyan used 2 liters 125 milliliters of the paint.
At the end of the second day, there were 1,769 milliliters of paint remaining in the barrel. How much paint did Shyan use on the second day?

Answer:
On the second day Shyan uses 3,999 milliliters or 3 liters 999 milliliters of the paint,

Statement : Shyan on the second day uses three thousand nine hundred ninety nine milliliters or 3 liters and nine hundred ninety nine milliliters of the paint,

Explanation:
Given Shyan’s barrel contained 6 liters 775 milliliters of paint.
She poured in 1 liter 118 milliliters more.
The first day, Shyan used 2 liters 125 milliliters of the paint.
At the end of the second day, there were 1,769 milliliters of paint remaining in the barrel. Total amount of the paint is
6 L 775 mL + 1 L 118 mL = 7 L 893 mL,
Paint left after first day is 7 L 893 mL – 2 L 125 mL =
7 L 893 mL = 7 X 1000 mL + 893 mL = 7,893 mL,
2 L 125 mL = 2 X 1000 mL + 125 mL = 2,125 mL,
So 7,893 mL – 2,125 mL = 5,768 mL paint is left after first day,
Now Second day paint used is 5,768 mL – 1,769 mL = 3,999 mL
or 3,999 mL ÷ 1000 = 3 L 999 mL paint has been
used on the second day by Shyan.
Modeled  problem with a tape diagram as shown above and
statement Shyan on the second day uses three thousand nine hundred ninety nine milliliters or 3 liters and nine hundred ninety nine milliliters of the paint.

Question 4.
On Thursday, the pizzeria used 2 kilograms 180 grams less flour than they used on Friday. On Friday, they used 12 kilograms 240 grams. On Saturday, they used 1,888 grams more than on Friday. What was the total amount of flour used over the three days?

The total amount of flour used over three days is 36,428 grams or 36 kilograms 428 grams,

Statement : The total amount of flour used over three days is thirty six thousand four hundred twenty eight grams or thirty six kilograms and four hundred twenty eight grams,

Explanation:
Given On Thursday, the pizzeria used less 2 kilograms 180 grams less flour than they used on Friday. On Friday, they used 12 kilograms 240 grams.
On Saturday, they used 1,888 grams more than on Friday. So on Thursday they used
12 kg 240 g – 2 kg 180 g =
12 X 1000 g + 240 g = 12,240 g,
2 X 1000 + 180 g = 2,180g,
12,240 g – 2,180 g = 10,060 g on Thursday,
On Saturday they used 1,888 g + 12,240 g = 14,128 g,
Now the total amount of flour used over the three days is
10,060 g + 12,240 g + 14,128 g = 36,428 g or 36,428 g ÷ 1000 =
36 kg 428 g. Therefore the total amount of flour used over three days is
36,428 grams or 36 kilograms 428 grams,
Modeled problem with a tape diagram as shown above and statement is the total amount of flour used over three days is thirty six thousand four hundred twenty eight grams or thirty six kilograms and four hundred twenty eight grams.

Question 5.
The gas tank in Zachary’s car has a capacity of 60 liters. He adds 23 liters 825 milliliters of gas to the tank, which already has 2,050 milliliters of gas. How much more gas can Zachary add to the gas tank?

Answer:
More 34,125 milliliters or 34 L 125 mL gas can Zachary add to the gas tank,

Statement : More thirty four thousand one hundred twenty five milliliters or thirty four liters and one hundred twenty five milliliters gas can Zachary add to the gas tank,

Explanation:
Given the gas tank in Zachary’s car has a capacity of 60 liters.
He adds 23 liters 825 milliliters of gas to the tank,
which already has 2,050 milliliters of gas.
So  the gas tank has  23 L 825 mL + 2,050 mL =
as 23 L 825 mL = 23 X 1000 mL + 825 mL =
23,000 mL + 825 mL = 23,825 mL gas,
Now 23,825 mL + 2,050 mL = 25,875 mL
More gas can Zachary add to the gas tank is 60 L – 25,875 mL =
as 60 L = 60 X 1,000 mL = 60,000 mL,
So 60,000 – 25,875 mL = 34,125 mL or 34,125 mL ÷ 1000 =
34 L 125 mL, therefore, more 34,125 milliliters or
34 L 125 mL gas can Zachary add to the gas tank.
Modeled problem with a tape diagram as shown above and statement is more thirty four thousand one hundred twenty five milliliters or thirty four liters and one hundred twenty five milliliters gas can Zachary add to the gas tank.

Question 6.
A giraffe is 5 meters 20 centimeters tall. An elephant is 1 meter 77 centimeters shorter than a giraffe. A rhinoceros is 1 meter 58 centimeters shorter than an elephant. How tall are the rhinoceros?

Answer:
The rhinoceros is 185 centimeters tall or 1 meter 85 centimeters tall,

Statement: The rhinoceros is one hundred eighty five centimeters tall or one meter eighty five centimeters tall,

Explanation:
Given A giraffe is 5 meters 20 centimeters tall. An elephant is 1 meter 77 centimeters shorter than the giraffe and
A rhinoceros is 1 meter 58 centimeters shorter than the elephant. So height of rhinoceros is first we see height of elephant as 5 m 20 cm – 1 m 77 cm =
as 5 X 100 cm + 20 cm = 5,20 cm and
1 m 77 cm = 1 X 100 cm +77 cm = 1,77 cm,
so 520 cm – 177 cm = 343 cm, Now rhinoceros is
343 cm – 1 m 58 cm =
as 1 m 58 cm = 1 X 100 cm + 58 cm = 158 cm
so 343 cm – 158 cm = 185 cm or 185 cm ÷ 100 = 1 m 85 cm,
therefore the rhinoceros is 185 centimeters tall or 1 meter 85 centimeters tall, modeled problem with a tape diagram as shown above and statement is the
rhinoceros is one hundred eighty five centimeters tall or one meter eighty five centimeters tall.

Eureka Math Grade 4 Module 2 Lesson 5 Exit Ticket Answer Key

Model each problem with a tape diagram. Solve and answer with a statement.

Question 1.
Jeff places a pineapple with a mass of 890 grams on a balance scale. He balances the scale by placing two oranges, an apple, and a lemon on the other side. Each orange weighs 280 grams. The lemon weighs 195 grams less than each orange. What is the mass of the apple?

Answer:
The mass of the apple is 245 grams,

Statement : The mass of the apple is two hundred forty five grams,

Explanation:
Given Jeff places a pineapple with a mass of 890 grams on a balance scale. He balances the scale by placing two oranges, an apple, and a lemon on the other side. Each orange weighs 280 grams. The lemon weighs 195 grams less than each orange.
So the lemon weighs 280 g – 195 g =  85 grams,
Now weigh of two oranges is 280 g + 280 g = 560 grams,
given  2 oranges + lemon + apple = 890 grams, therefore weigh of apple = 890 g – (560g + 85 g) = 890 g – 645 g = 245 g,
therefore, the mass of the apple is 245 g, modeled problem with a tape diagram as shown above and statement is mass of apple is two hundred fortyy five grams.

Question 2.
Brian is 1 meter 87 centimeters tall. Bonnie is 58 centimeters shorter than Brian. Betina is 26 centimeters taller than Bonnie. How tall is Betina?

Answer:
Betina is 155 cm or 1 m 55 cm tall,

Statement: Betina is one hundred fifty five centimeters or one meter fifty five centimeters tall,

Explanation:
Given Brian is 1 meter 87 centimeters tall.
Bonnie is 58 centimeters shorter than Brian means
Bonnie is 1 m 87 cm – 58 = as 1 m 87 cm =
1 X 100 cm + 87 cm = 187 cm, 187 cm – 58 cm = 129 cm,
Bonnie is 129 cm and Betina is 26 centimeters taller than
Bonnie means Betina is 129 cm + 26 cm = 155 cm or
155  cm ÷ 100 = 1 m 55 cm,
So Betina is 155 cm or 1 m 55 cm tall, modeled problem with a tape diagram as shown above and statement is Betina is one hundred fifty five centimeters or one meter fifty five centimeters tall.

Eureka Math Grade 4 Module 2 Lesson 5 Homework Answer Key

Model each problem with a tape diagram. Solve and answer with a statement.

Question 1.
The capacity of Jose’s vase is 2,419 milliliters of water. He poured 1 liter 299 milliliters of water into the empty vase. Then, he added 398 milliliters. How much more water will the vase hold?

Answer:
Jose’s vase can hold more 722 milliliters of water,

Statement : Jose’s vase can hold more seven hundred twenty two milliliters of water,

Explanation:
Given the capacity of Jose’s vase is 2,419 milliliters of water.
He poured 1 liter 299 milliliters of water into the empty vase.
Then, he added 398 milliliters. So now water in the vase is
1 L 299 mL + 398 mL as 1 L 299 mL = 1 X 1000 mL + 299 mL =
1,299 mL +  398 mL = 1,697 mL of water,
Now we can pour more is 2,419 mL – 1,697 mL = 722 mL more, therefore, Jose’s vase can hold more 722 milliliters of water, modeled problem with a tape diagram as shown above and statement is Jose’s vase can hold more seven hundred twenty two milliliters of water.

Question 2.
Eric biked 1 kilometer 125 meters on Monday. On Tuesday, he biked 375 meters less than on Monday. How far did he bike both days?

Answer:
Eric biked both days as 1,875 meters or 1 kilometer 875 meters,

Statement : Eric biked both days as one thousand eight hundred seventy five meters or one kilometer eight hundred seventy five meters,

Explanation:
Given Eric biked 1 kilometer 125 meters on Monday.
On Tuesday, he biked 375 meters less than on Monday.
So on Tuesday he biked 1 k 125 m – 375 m =
as 1 X 1000 m + 125 m = 1,125 m ,
1125 m – 375 m = 750 m,
therefore Eric biked on Monday and Tuesday is
1,125 m + 750 m = 1,875 m or 1,875 m ÷ 1,000 = 1 k 875 m,
therefore, Eric biked both days as 1,875 meters or
1 kilometer 875 meters, modeled problem with a tape diagram as shown above and statement is Eric biked both days as one thousand eight hundred seventy five meters or one kilometer eight hundred seventy five meters.

Question 3.
Zachary weighs 37 kilograms 95 grams. Gabe weighs 4,650 grams less than Zachary. Harry weighs 2,905 grams less than Gabe. How much does Harry weigh?

Answer:
Harry weigh 29,540 grams or 29 kilogram 540 grams,

Statement : Harry weigh’s twenty nine thousand five hundred forty grams or twenty nine kilograms five hundred forty grams,

Explanation:
Given Zachary weighs 37 kilograms 95 grams.
Gabe weighs 4,650 grams less than Zachary and Harry weighs
2,905 grams less than Gabe. So Gabe weigh’s
37 kg 95 g – 4,650 g as 37 kg 95 g =
37 X 1000 g + 95 g = 37,095 g so Gabe is
37,095 g – 4,650 g = 32,445 g,
Now Harry weigh’s 32,445 g – 2,905 g = 29,540 grams or
29,540 g ÷ 1,000 = 29 kilograms 540 grams,
therefore, Harry weigh 29,540 grams or 29 kilogram 540 grams, modeled problem with a tape diagram as shown above and statement is Harry weigh’s twenty-nine thousand five hundred forty grams or twenty nine kilograms five hundredforty grams.

Question 4.
A Springer Spaniel weighs 20 kilograms 490 grams. A Cocker Spaniel weighs 7,590 grams less than a Springer Spaniel. A Newfoundland weighs 52 kilograms 656 grams more than a Cocker Spaniel. What is the difference, in grams, between the weights of the Newfoundland and the Springer Spaniel?

Answer:
The difference in grams, between the weights of the Newfoundland and the Springer Spaniel is 45,066 grams,

Statement: The difference in grams, between the weights of the Newfoundland and the Springer Spaniel is forty five thousand sixty six grams,

Explanation:
Given a Springer Spaniel weighs 20 kilograms 490 grams.
A Cocker Spaniel weighs 7,590 grams less than a Springer Spaniel. A Newfoundland weighs 52 kilograms 656 grams more than a Cocker Spaniel.
So Cocker spaniel weigh’s 20 kg 490 g – 7,590 g
as 20 kg 490 g = 20 X 1000 g + 490 g = 20,490 g,
So 20,490 g – 7,590 g = 12,900 g,
Now Newfoundland weighs 52 kg 656 g +12,900 g as
52 kg 656 g = 52 X 1000 g + 656 g = 52,656 g,
52,656 g + 12,900 g = 65,556 g,
Now the difference in grams, between the weights of the
Newfoundland and the Springer Spaniel is
65,556 g – 20,490 g = 45,066 grams,
modeled problem with a tape diagram as shown above and statement is the difference in grams, between the weights of the Newfoundland and the Springer Spaniel is forty five thousand sixty six grams.

Question 5.
Marsha has three rugs. The first rug is 2 meters 87 centimeters long. The second rug has a length 98 centimeters less than the first. The third rug is 111 centimeters longer than the second rug. What is the difference in centimeters between the length of the first rug and the third rug?

Answer:
The difference in centimeters between the length of the first rug and the third rug is 13 cm,

Statement : The difference in centimeters between the length of the first rug and the third rug is thirteen centimeters,

Explanation:
Given Marsha has three rugs. The first rug is 2 meters 87 centimeters long.
The second rug has a length 98 centimeters less than the first. The third rug is 111 centimeters longer than the second rug so the second rug is
2 m 87 cm – 98 cm as 2 m 87 cm = 2 X 100 cm + 87 cm = 287 cm,
Now 287 cm – 98 cm = 189 cm and the third rug is
189 cm + 111 cm = 300 cm,
first rug is 287 cm, second rug is 189 cm and third rug is 300 cm,
Now the difference in centimeters between the length of the first rug and the third rug is 300 cm – 287 cm = 13 cm,
modeled problem with a tape diagram as shown above and statement is the difference in centimeters between the length of the first rug and the third rug is thirteen centimeters.

Question 6.
One barrel held 60 liters 868 milliliters of sap. A second barrel held 20,089 milliliters more sap than the first. A third barrel held 40 liters 82 milliliters less sap than the second. If the sap from the three barrels was poured into a larger container, how much sap would there be in all ?

Answer:
182,700 milliliters or 182 liters 700 milliliters of sap
would be there in all,

Statement: one lakh eighty two thousand seven hundred milliliters or one hundred eighty two liters and seven hundred milliliters of sap would be there in all,

Explanation:
Given one barrel held 60 liters 868 milliliters of sap.
A second barrel held 20,089 milliliters more sap than the first.
A third barrel held 40 liters 82 milliliters less sap than the second.
Second barrel holds 60 L 868 mL + 20,089 mL as
60 L 868 mL = 60 X 1000 mL + 868 mL = 60,868 mL,
60,868 mL + 20,089 mL = 80,957 mL,
Third barrel holds 80,957 mL – 40 L 82 mL as
40 L 82 mL = 40 X 1000 mL + 82 mL = 40,082 mL,
so 80,957 mL – 40,082 mL = 40,875 mL,
First barrel holds 60,868 mL,
Second barrel holds 80,957 mL,
Third barrel holds 40,875 mL,
Now if the sap from the three barrels was poured into a larger container it will contain
60,868 mL + 80,957 mL + 40,875 mL = 182,700 mL,
modeled problem with a tape diagram as shown above and statement is one lakh eighty two thousand seven hundred milliliters or one hundred eighty two liters and seven hundred milliliters of sap would be there in all.

Eureka Math Grade 4 Module 2 Answer Key

Engage NY Eureka Math Geometry Module 4 Lesson 9 Answer Key

Eureka Math Geometry Module 4 Lesson 9 Example Answer Key

Consider a triangular region in the plane with vertices 0(0, 0), A(5, 2), and B(3, 4). What is the perimeter of the triangular region?
Answer:
Approximately 13.21 units

What is the area of the triangular region?
Answer:
7 square units

Find the general formula for the area of the triangle with vertices 0(0, 0), A(x₁, y₁), and B (x₂, y₂), as shown.

Answer:
Let students struggle with this for a while following the steps modeled above. Scaffold as necessary.
→ What is the area in square units of the rectangle enclosing the triangle?
x₁ y₂

→ What is the area in square units of the surrounding triangles?
Left: \(\frac{1}{2}\)
Right: \(\frac{1}{2}\)(x₁ – x₂) (y₂ – y₁)
Bottom: \(\frac{1}{2}\)(x₁y₁)

→ Write the general formula for the area of ∆OAB.
x₁y₂ – \(\frac{1}{2}\)(x₂y₂) – \(\frac{1}{2}\)(x₁ – x₂) (y₂ – y₁) – \(\frac{1}{2}\)(x₁y₁)

→ Expand the formula and simplify it.
\(\frac{1}{2}\)(x₁y₂ – x₂y₁)

Does the formula work for this triangle?

Answer:
→ Does the formula work for the triangle below?
It does, but students may get either of these formulas: \(\frac{1}{2}\)(x₁y₂ – x₂y₁) or \(\frac{1}{2}\)(x₂y₁ – x₁y₂) depending on which point they labeled (x₁, y₁).

These formulas differ by a minus sign. Let students compare their answer with their classmates. Make the point that the formulas depend on the choice of labeling the points but differ only by a minus sign. Students always get a positive area if they order the points counterclockwise. The method that students should follow is (1) pick a starting point; (2) walk around the figure in a counterclockwise direction doing calculations with adjacent points.

→ Do you think this formula works for any triangle with one vertex at (0, 0)? Does the quadrant matter?
Answers will vary. Yes, it always works no matter the quadrant, and the area is positive as long as (x₁, y₁) is the next point in a counterclockwise direction.
→ Let’s try some problems and see if our formula always works.

Eureka Math Geometry Module 4 Lesson 9 Opening Exercise Answer Key

Find the area of the shaded region.
a.

Answer:
A = 16 units² – 2 units² = 14 units²

b.

Answer:
A = 9π units² – 8 units² ≈ 20.26 units²

Eureka Math Geometry Module 4 Lesson 9 Exercise Answer Key

Find the area of the triangles with vertices listed, first by finding the area of the rectangle enclosing the triangle and subtracting the area of the surrounding triangles, then by using the formula \(\frac{1}{2}\)(x₁y₂ – x₂y₁).
a. 0(0, 0), A(5, 6), B(4, 1)
Answer:
9.5 square units

b. 0(0, 0), A(3, 2), B (- 2, 6)
Answer:
11 square units

c. 0(0, 0), A(5, – 3), B(- 2, 6)
Answer:
12 square units

Eureka Math Geometry Module 4 Lesson 9 Problem Set Answer Key

Question 1.
Use coordinates to compute the perimeter and area of each polygon.
a.

Answer:
Perimeter = 16 units
Area = 15 square units

b.

Answer:
Perimeter ≈ 17.62 units
Area = 10.5 square units

Question 2.
Given the figures below, find the area by decomposing into rectangles and triangles.
a.

Answer:
Area = 8 square units

b.

Answer:
Area = 30 square units

Question 3.
Challenge: Find the area by decomposing the given figure into triangles.

Answer:
11 square units

Question 4.
When using the shoelace formula to work out the area of ABC, we have some choices to make. For example, we can start at any one of the three vertices A, B, or C, and we can move either in a clockwise or counterclockwise direction. This gives six options for evaluating the formula.

Show that the shoelace formula obtained is Identical for the three options that move in a clockwise direction (A to C to B or C to B to A or B to A to C) and identical for the three options in the reverse direction. Verify that the two distinct formulas obtained differ only by a minus sign.

Answer:
Clockwise:
\(\frac{1}{2}\)(x₁y₂ + x₂y₃ + x₃y₁ – y₁x₂ – y₂x₃ – y₃x₁)

Counterclockwise:
\(\frac{1}{2}\)(x₁y₃ + x₃y₂ + x₂y₁ – y₁x₃ – y₃x₂ – y₂x₁)

Question 5.
Suppose two triangles share a common edge. By translating and rotating the triangles, we can assume that the common edge lies along the x-axis with one endpoint at the origin.

a. Show that If we evaluate the shoelace formula for each trIangle, both calculated in the same clockwise direction, then the answers are both negative.
Answer:
Top Triangle: \(\frac{1}{2}\)(0 – bx) = –\(\frac{1}{2}\)bx, where b > 0.
Bottom Triangle: \(\frac{1}{2}\)(xd – 0) = –\(\frac{1}{2}\)dx, where d < 0. Both answers are negative when calculated in the clockwise direction.

b. Show that If we evaluate them both In a counter clock wise direction, then both are positive.
Answer:
Top Triangle: \(\frac{1}{2}\)(xb – 0) = bx, where b > 0.
Bottom Triangle: \(\frac{1}{2}\)(0 – xd) = –\(\frac{1}{2}\)dx, where d < 0.
Both answers aie positive when calculated in the counterclockwise direction.

c. Explain why evaluating one in one direction and the second in the opposite direction, the two values obtained are opposite in sign.
Answer:
If two triangles share a common edge, then evaluating the shoelace formula for each triangle in a consistent direction gives answers that are the same sign; they are either each the positive areas of their respective triangles or each the negative versions of those areas.

Question 6.
A textbook has a picture of a triangle with vertices (3, 6) and (5, 2). Something happened in printing the book, and the coordinates of the third vertex are listed as . The answers in the back of the book give the area of the triangle as 6 square units.
a. What is the y-coordinate of the third vertex?
Answer:
8

b. What if both coordinates were missing, but the area was known? Could you use algebra to find the third coordinate? Explain.
Answer:
No, you would have an equation with two variables, so you could not solve for both algebraically unless you had a second equation with the same unknowns.

Eureka Math Geometry Module 4 Lesson 9 Exit Ticket Answer Key

Given the triangle below with vertices A(4, 3), B(-2, 3), and C(-1, -2).

Azha calculated the area using 5 ∙ 6 – \(\frac{1}{2}\)(5 ∙ 1) – \(\frac{1}{2}\)(5 ∙ 5),
while Carson calculated the area using \(\frac{1}{2}\)(4 ∙ 3 + (-2) ∙ (-2) + (-1) ∙ 3 – 3 (-2) – 3 ∙ (-1) – (-2) ∙ 4).
Explain the method each one used.
Answer:
Azha used the decomposition method. She first determined that the area of the rectangle around the triangle is 5 ∙ 6, and then she subtracted the area of the 3 right triangles surrounding the region. Carson used the “shoelace” method that we learned in this lesson.

Engage NY Eureka Math 1st Grade Module 3 Lesson 2 Answer Key

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

Question 1.
Use the paper strip provided by your teacher to measure each picture. Circle the words you need to make the sentence true. Then, fill in the blank.

The basketball bat is the paper strip.

Answer:
The basketball bat is longer than the paper strip.

Explanation:
In the above image, we can see that the basketball bat is the paper strip.
The book is the paper strip.
Answer:
The book is shorter than the paper strip.

Explanation:
In the above image, we can see that the book is a paper strip.

The basketball bat is ____ the book.

Answer:
The basketball bat is longer than the book.

Explanation:
In the above image, we can see that the basketball bat is longer than the book.

Question 2.
Complete the sentences with longer than, shorter than, or the same length as to make the sentences true.
a.
The tube is ____________________________ the cup.

Answer:
The tube is longer than the cup.

Explanation:
In the above image, we can see that the tube is longer than the cup.

b.
The iron is ___ the ironing board.

Answer:
The iron is shorter than the ironing board.

Explanation:
In the above image, we can see that the iron is shorter than the ironing board.

Use the measurements from Problems 1 and 2. Circle the word that makes the sentences true.

Question 3.
The baseball bat is (longer/shorter) than the cup.

Answer:
Longer

Explanation:

Question 4.
The cup is (longer/shorter) than the ironing board.

Answer:
Shorter

Explanation:

Question 5.
The ironing board is (longer/shorter) than the book.

Answer:
Longer.

Explanation:

Question 6.
Order these objects from shortest to longest:
cup, tube, and paper strip

Answer:
Cup, paper strip, tube.

Draw a picture to help you complete the measurement statements. Circle the words that make each statement true.

Question 7.
Sammy is taller than Dion.
Janell is taller than Sammy.
Dion is (taller than/shorter than) Janell.

Answer:
Dion is shorter than Janell.

Explanation:

As given that Sammy is taller than Dion and Janell is taller than Sammy, so Dion is shorter than Janell.

Question 8.
Laura’s necklace is longer than Mihal’s necklace.
Laura’s necklace is shorter than Sarai’s necklace.
Sarai’s necklace is (longer than/shorter than) Mihal’s necklace.

Answer:
Sarai’s necklace is longer than Mihal’s necklace.

Explanation:

As given that Laura’s necklace is longer than Mihal’s necklace and Laura’s necklace is shorter than Sarai’s necklace.

Eureka Math Grade 1 Module 3 Lesson 2 Exit Ticket Answer Key

Draw a picture to help you complete the measurement statements. Circle the words that make each statement true.

Tanya’s doll is shorter than Aline’s doll.
Mira’s doll is taller than Aline’s doll.
Tanya’s doll is (taller than/shorter than) Mira’s doll.

Answer:
Tanya’s doll is shorter than Mira’s doll.

Explanation:
Given that Tanya’s doll is shorter than Aline’s doll and Mira’s doll is taller than Aline’s doll. So Tanya’s doll is shorter than Mira’s doll.

Eureka Math Grade 1 Module 3 Lesson 2 Homework Answer Key

Use the paper strip provided by your teacher to measure each picture. Circle the words you need to make the sentence true. Then, fill in the blank.

Question 1.

The sundae is the paper strip.

Answer:
The sundae is longer than the paper strip.

Explanation:
In the above image, we can see that the sundae is the paper strip.

The spoon is the paper strip.
The spoon is ___ the sundae.

Answer:
The spoon is shorter than the sundae.

Explanation:
The spoon is the sundae.

Question 2.

The
The balloon is ____________________________ the cake.

Answer:
The balloon is shorter than the cake.

Explanation:
In the above image, we can see that the balloon is shorter than the cake.

Question 3.

The ball is shorter than the paper strip.
So, the shoe is ___ the ball.

Answer:
The shoe is longer than the ball.

Explanation:
In the above image, we can see that the shoe is longer than the ball.

Use the measurements from Problems 1–3. Circle the word that makes the sentences true.

Question 4.
The spoon is (longer/shorter) than the cake.

Answer:
The spoon is shorter than the cake.

Question 5.
The balloon is (longer/shorter) than the sundae.

Answer:
The balloon is longer than the sundae.

Question 6.
The shoe is (longer/shorter) than the balloon.

Answer:
The shoe is longer than the balloon.

Question 7.
Order these objects from shortest to longest:
cake, spoon, and paper strip
_______ _________ __________

Answer:
Spoon, cake, and paper strip.

Explanation:
The order of these objects from shortest to longest is Spoon, cake, and paper strip.

Draw a picture to help you complete the measurement statements. Circle the word that makes each statement true.

Question 8.
Marni’s hair is shorter than Wesley’s hair.
Marni’s hair is longer than Bita’s hair.
Bita’s hair is (longer/shorter) than Wesley’s hair

Answer:
Bita’s hair is shorter than Wesley’s hair.

Explanation:
Given that Marni’s hair is shorter than Wesley’s hair and Marni’s hair is longer than Bita’s hair, so Bita’s hair is shorter than Wesley’s hair.

Question 9.
Elliott is shorter than Brady.
Sinclair is shorter than Elliott.
Brady is (taller/shorter) than Sinclair.

Answer:
Brady is taller than Sinclair.

Explanation:
Given that Elliott is shorter than Brady and Sinclair is shorter than Elliott, so Brady is taller than Sinclair.

Engage NY Eureka Math 7th Grade Module 3 Lesson 1 Answer Key

Eureka Math Grade 7 Module 3 Lesson 1 Example Answer Key

Example 1.
Any Order, Any Grouping Property with Addition
a. Rewrite 5x+3x and 5x-3x by combining like terms.
Write the original expressions and expand each term using addition. What are the new expressions equivalent to?
Answer:

Because both terms have the common factor of x, we can use the distributive property to create an equivalent expression.
5x+3x=(5+3)x=8x
5x-3x=(5-3)x=2x

Ask students to try to find an example (a value for x) where 5x+3x≠8x or where
5x-3x≠2x. Encourage them to use a variety of positive and negative rational numbers. Their failure to find a counterexample helps students realize what equivalence means.

In Example 1, part (b), students see that the commutative and associative properties of addition are regularly used in consecutive steps to reorder and regroup like terms so that they can be combined. Because the use of these properties does not change the value of an expression or any of the terms within the expression, the commutative and associative properties of addition can be used simultaneously. The simultaneous use of these properties is referred to as the any order, any grouping property.

b. Find the sum of 2x+1 and 5x.
Answer:

7x+1 Equivalent expression to the given problem
→ Why did we use the associative and commutative properties of addition?
→ We reordered the terms in the expression to group together like terms so that they could be combined.
→ Did the use of these properties change the value of the expression? How do you know?
→ The properties did not change the value of the expression because each equivalent expression includes the same terms as the original expression, just in a different order and grouping.
→ If a sequence of terms is being added, the any order, any grouping property allows us to add those terms in any order by grouping them together in any way.
→ How can we confirm that the expressions (2x+1)+5x and 7x+1 are equivalent expressions?
→ When a number is substituted for the x in both expressions, they both should yield equal results.
The teacher and student should choose a number, such as 3, to substitute for the value of x and together check to see if both expressions evaluate to the same result.

Given Expression
((2x+1)+5x
(2∙3+1)+5∙3
(6+1)+15
(7)+15
22

Equivalent Expression?
7x+1
7∙3+1
21+1
22

→ The expressions both evaluate to 22; however, this is only one possible value of x. Challenge students to find a value for x for which the expressions do not yield the same number. Students find that the expressions evaluate to equal results no matter what value is chosen for x.
→ What prevents us from using any order, any grouping in part (c), and what can we do about it?
→ The second expression, (5a-3), involves subtraction, which is not commutative or associative; however, subtracting a number x can be written as adding the opposite of that number. So, by changing subtraction to addition, we can use any order and any grouping.

c. Find the sum of -3a+2 and 5a-3.
Answer:
(-3a+2)+(5a-3) Original expression
-3a+2+5a+(-3) Add the opposite (additive inverse)
-3a+5a+2+(-3) Any order, any grouping
2a+(-1) Combined like terms (Stress to students that the expression is not yet simplified.)
2a-1 Adding the inverse is subtracting.

→ What was the only difference between this problem and those involving all addition?
→ We first had to rewrite subtraction as addition; then, this problem was just like the others.

Example 2.
Any Order, Any Grouping with Multiplication
Find the product of 2x and 3.
Answer:

→ Why did we use the associative and commutative properties of multiplication?

→ We reordered the factors to group together the numbers so that they could be multiplied.

→ Did the use of these properties change the value of the expression? How do you know?

→ The properties did not change the value of the expression because each equivalent expression includes the same factors as the original expression, just in a different order or grouping.

→ If a product of factors is being multiplied, the any order, any grouping property allows us to multiply those factors in any order by grouping them together in any way.

Example 3.
Any Order, Any Grouping in Expressions with Addition and Multiplication
Use any order, any grouping to write equivalent expressions.
a. 3(2x)
Answer:
(3∙2)x
6x

b. 4y(5)
Answer:
(4∙5)y
20y

c. 4∙2∙z
Answer:
(4∙2)z
8z

d. 3(2x)+4y(5)
Answer:

(3∙2)x+(4∙5)y
6x+20y

e. 3(2x)+4y(5)+4∙2∙z
Answer:

(3∙2)x+(4∙5)y+(4∙2)z
6x+20y+8z

f. Alexander says that 3x+4y is equivalent to (3)(4)+xy because of any order, any grouping. Is he correct? Why or why not?

Encourage students to substitute a variety of positive and negative rational numbers for x and y because in order for the expressions to be equivalent, the expressions must evaluate to equal numbers for every substitution of numbers into all the letters in both expressions.

Alexander is incorrect; the expressions are not equivalent because if we, for example, let x=-2 and let
y=-3, then we get the following:

-18≠18, so the expressions cannot be equivalent.

→ What can be concluded as a result of part (f)?
→ Any order, any grouping cannot be used to mix multiplication with addition. Numbers and letters that are factors within a given term must remain factors within that term.

Eureka Math Grade 7 Module 3 Lesson 1 Problem Set Answer Key

For Problems 1–9, write equivalent expressions by combining like terms. Verify the equivalence of your expression and the given expression by evaluating each for the given values: a=2, b=5, and c=-3.

Question 1.
3a+5a
Answer:
8a
8(2)
16

3(2)+5(2)
6+10
16

Question 2.
8b – 4b
Answer:
4b
4(5)
20

8(5)-4(5)
40-20
20

Question 3.
5c+4c+c
10c
10(-3)
-30

5(-3)+4(-3)+(-3)
-15+(-12)+(-3)
-27+(-3)
-30

Question 4.
3a+6+5a
8a+6
8(2)+6
16+6
22

3(2)+6+5(2)
6+6+10
12+10
22

Question 5.
8b + 8 – 4b
4b+8
4(5)+8
20+8
28

8(5)+8-4(5)
40+8-20
48-20
28

Question 6.
2c
2(-3)
-6

5(-3)-4(-3)+(-3)
-15+(-4(-3))+(-3)
-15+(12)+(-3)
-3+(-3)
-6

Eureka Math Grade 7 Module 3 Lesson 1 Exit Ticket Answer Key

Question 1.
Write an equivalent expression to 2x+3+5x+6 by combining like terms.
Answer:
2x+3+5x+6
2x+5x+3+6
7x+9

Question 2.
Find the sum of (8a+2b-4) and (3b-5).
Answer:
(8a+2b-4)+(3b-5)
8a+2b+(-4)+3b+(-5)
8a+2b+3b+(-4)+(-5)
8a+(5b)+(-9)
8a+5b-9

Question 3.
Write the expression in standard form: 4(2a)+7(-4b)+(3∙c∙5).
Answer:
(4∙2)a+(7∙(-4))b+(3∙5)c
8a+(-28)b+15c
8a-28b+15c

Eureka Math Grade 7 Module 3 Lesson 1 Opening Exercise Answer Key

Each envelope contains a number of triangles and a number of quadrilaterals. For this exercise, let t represent the number of triangles, and let q represent the number of quadrilaterals.
a. Write an expression using t and q that represents the total number of sides in your envelope. Explain what the terms in your expression represent.
Answer:
3t+4q. Triangles have 3 sides, so there will be 3 sides for each triangle in the envelope. This is represented by 3t. Quadrilaterals have 4 sides, so there will be 4 sides for each quadrilateral in the envelope. This is represented by 4q. The total number of sides will be the number of triangle sides and the number of quadrilateral sides together.

b. You and your partner have the same number of triangles and quadrilaterals in your envelopes. Write an expression that represents the total number of sides that you and your partner have. If possible, write more than one expression to represent this total.
Answer:
3t+4q+3t+4q; 2(3t+4q); 6t+8q

c. Each envelope in the class contains the same number of triangles and quadrilaterals. Write an expression that represents the total number of sides in the room.
Answer:
Answer depends on the number of students in the classroom. For example, if there are 12 students in the classroom, the expression would be 12(3t+4q), or an equivalent expression

d. Use the given values of t and q and your expression from part (a) to determine the number of sides that should be found in your envelope.
Answer:
3t+4q
3(4)+4(2)
12+8
20
There should be 20 sides contained in my envelope.

e. Use the same values for t and q and your expression from part (b) to determine the number of sides that should be contained in your envelope and your partner’s envelope combined.

My partner and I have a combined total of 40 sides.

f. Use the same values for t and q and your expression from part (c) to determine the number of sides that should be contained in all of the envelopes combined.
Answer:
Answer will depend on the seat size of your classroom. Sample responses for a class size of 12:

For a class size of 12 students, there should be 240 sides in all of the envelopes combined.

g. What do you notice about the various expressions in parts (e) and (f)?
Answer:
The expressions in part (e) are all equivalent because they evaluate to the same number: 40. The expressions in part (f) are all equivalent because they evaluate to the same number: 240. The expressions themselves all involve the expression 3t+4q in different ways. In part (e), 3t+3t is equivalent to 6t, and 4q+4q is equivalent to 8q. There appear to be several relationships among the representations involving the commutative, associative, and distributive properties.

Engage NY Eureka Math Kindergarten Module 2 Lesson 9 Answer Key

Eureka Math Kindergarten Module 2 Lesson 9 Problem Set Answer Key

Circle the pictures of the flat shapes with red. Circle the pictures of the solid shapes with green.

Answer:

Explanation:
The shapes circled red are fat shapes and green are solid shapes.

Eureka Math Kindergarten Module 2 Lesson 9 Homework Answer Key

In each row, circle the one that doesn’t belong. Explain your choice to a grown-up.

Answer:

Explanation:
In the first row there is 1 circle and 2 triangles.So, i circled circle.
In the second row there is 1 hexagon and 2 cylindes.So, i circled hexagon.
In the third row there are 2 circles and 1 semicircle.So, i circled the semicircle.
In the fourth row there is 1 solid shape and 2 flat shapes.So, circled solid shape.
In the last row there is a1 flat shape and 2 solid shapes.So, i circled flat shape.

EngageNY Math Grade 4 Module 4 Answer Key | Eureka Math 4th Grade Module 4 Answer Key

Eureka Math Grade 4 Module 4 Angle Measure and Plane Figures

Eureka Math Grade 4 Module 4 Topic A Lines and Angles

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

Eureka Math 4th Grade Module 4 Topic B Angle Measurement

• Eureka Math Grade 4 Module 4 Lesson 5 Answer Key
• Eureka Math Grade 4 Module 4 Lesson 6 Answer Key

Eureka Math Grade 4 Module 4 Mid Module Assessment Answer Key

Engage NY Math 4th Grade Module 4 Topic C Problem Solving with the Addition of Angle Measures

• Eureka Math Grade 4 Module 4 Lesson 7 Answer Key
• Eureka Math Grade 4 Module 4 Lesson 8 Answer Key
• Eureka Math Grade 4 Module 4 Lesson 9 Answer Key
• Eureka Math Grade 4 Module 4 Lesson 10 Answer Key
• Eureka Math Grade 4 Module 4 Lesson 11 Answer Key

EngageNY Math Grade 4 Module 4 Topic D Two-Dimensional Figures and Symmetry

• Eureka Math Grade 4 Module 4 Lesson 12 Answer Key
• Eureka Math Grade 4 Module 4 Lesson 13 Answer Key
• Eureka Math Grade 4 Module 4 Lesson 14 Answer Key
• Eureka Math Grade 4 Module 4 Lesson 15 Answer Key
• Eureka Math Grade 4 Module 4 Lesson 16 Answer Key

Eureka Math Grade 4 Module 4 End of Module Assessment Answer Key

EngageNY Kindergarten Math Module 2 Answer Key | Kindergarten Eureka Math Module 2 Answer Key

Eureka Math Kindergarten Module 2 Two-Dimensional and Three-Dimensional Shapes

Eureka Math Kindergarten Module 2 Topic A Two-Dimensional Flat Shapes

• Eureka Math Kindergarten Module 2 Lesson 1 Answer Key
• Eureka Math Kindergarten Module 2 Lesson 2 Answer Key
• Eureka Math Kindergarten Module 2 Lesson 3 Answer Key
• Eureka Math Kindergarten Module 2 Lesson 4 Answer Key
• Eureka Math Kindergarten Module 2 Lesson 5 Answer Key

Engage NY Math Kindergarten Module 2 Topic B Three-Dimensional Solid Shapes

• Eureka Math Kindergarten Module 2 Lesson 6 Answer Key
• Eureka Math Kindergarten Module 2 Lesson 7 Answer Key
• Eureka Math Kindergarten Module 2 Lesson 8 Answer Key

Kindergarten Eureka Math Module 2 Topic C Two-Dimensional and Three-Dimensional Shapes

• Eureka Math Kindergarten Module 2 Lesson 9 Answer Key
• Eureka Math Kindergarten Module 2 Lesson 10 Answer Key

Eureka Math Kindergarten Module 2 End of Module Assessment Answer Key

Engage NY Eureka Math 5th Grade Module 2 Lesson 16 Answer Key

Eureka Math Grade 5 Module 2 Lesson 16 Sprint Answer Key

A
Divide by Multiples of 10 and 100

Answer:
30/10 = 3, 430/10 = 43, 4300/10 = 430, 50/10 = 5, 43000/100 = 430, 850/10 = 85, 8500/10 = 850, 85000/100 = 850, 600/10 = 60, 60/3 = 20, 600/30 = 20, 4000/200 = 20, 40/2 = 20, 4000/100 = 40, 240/10 = 24, 24/2 = 12, 240/20 = 12, 24/2 = 12, 3600/100 = 36, 36/3 = 12, 3600/300 = 12.

Explanation:
In the above-given question,
given that,
divide by multiples of 10 and 100.
30/10 = 3, 430/10 = 43, 4300/10 = 430, 50/10 = 5, 43000/100 = 430, 850/10 = 85, 8500/10 = 850, 85000/100 = 850, 600/10 = 60, 60/3 = 20, 600/30 = 20, 4000/200 = 20, 40/2 = 20, 4000/100 = 40, 240/10 = 24, 24/2 = 12, 240/20 = 12, 24/2 = 12, 3600/100 = 36, 36/3 = 12, 3600/300 = 12.

Answer:
480/4 = 240, 480/40 = 12, 6300/3 = 2100, 6300/30 = 210, 6300/300 = 23, 8400/2 = 4200, 96000/3 = 32000, 96000/300 = 320, 900/30 = 30, 1200/30 = 40, 1290/30 = 43, 1800/300 = 6, 8000/200 = 40, 12000/200 = 60, 12800/200 = 64, 2240/70 = 32, 18400/800 = 23, 21600/90 = 340, 25200/600 = 42.

Explanation:
In the above-given question,
given that,
divide by multiples of 10 and 100.
480/4 = 240, 480/40 = 12, 6300/3 = 2100, 6300/30 = 210, 6300/300 = 23, 8400/2 = 4200, 96000/3 = 32000, 96000/300 = 320, 900/30 = 30, 1200/30 = 40, 1290/30 = 43, 1800/300 = 6, 8000/200 = 40, 12000/200 = 60, 12800/200 = 64, 2240/70 = 32, 18400/800 = 23, 21600/90 = 340, 25200/600 = 42.

Question 1.
30 ÷ 10 = 3.

Answer:
30/10 = 3.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
30/10 = 3.
zero gets canceled on both sides.
30/10 = 3.

Question 2.
430 ÷ 10 = 43.

Answer:
430/10 = 43.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
430/10 = 43.
zero gets canceled on both sides.
430/10 = 43.

Question 3.
4,300 ÷ 10 = 430.

Answer:
4300/10 = 430.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
4300/10 = 430.
zero gets canceled on both sides.
4300/10 = 430.

Question 4.
4,300 ÷ 100 = 43.

Answer:
4300/100 = 43.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
4300/100 = 43.
zero gets canceled on both sides.
4300/100 = 43.

Question 5.
43,000 ÷ 100 = 430.

Answer:
43000/100 = 430.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
4300/100 = 430.
zero gets canceled on both sides.
4300/100 = 430.

Question 6.
50 ÷ 10 = 5.

Answer:
50/10 = 5.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
50/10 = 5.
zero gets canceled on both sides.
50/10 = 5.

Question 7.
850 ÷ 10 = 85.

Answer:
850/10 = 85.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
850/10 = 85.
zero gets canceled on both sides.
850/10 = 85.

Question 8.
8,500 ÷ 10 = 850.

Answer:
8500/10 = 850.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
8500/10 = 850.
zero gets canceled on both sides.
8500/10 = 850.

Question 9.
8,500 ÷ 100 = 85.

Answer:
8500/100 = 85.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
8500/100 = 85.
zero gets canceled on both sides.
8500/100 = 85.

Question 10.
85,000 ÷ 100 =850.

Answer:
85000/100 = 850.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
85000/100 = 850.
zero gets canceled on both sides.
85000/100 = 850.

Question 11.
600 ÷ 10 = 60.

Answer:
600/10 = 60.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
600/10 = 60.
zero gets canceled on both sides.
600/10 = 60.

Question 12.
60 ÷ 3 = 20.

Answer:
60/3 = 20.

Explanation:
In the above-given question,
given that,
divide by multiples of 3.
60/3 = 20.
zero gets canceled on both sides.
60/3 = 20.

Question 13.
600 ÷ 30 = 20.

Answer:
600/30 = 20.

Explanation:
In the above-given question,
given that,
divide by multiples of 30.
600/30 = 20.
zero gets canceled on both sides.
600/30 = 20.

Question 14.
4,000 ÷ 100 = 40.

Answer:
4000/100 = 40.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
4000/100 = 40.
zero gets canceled on both sides.
4000/100 = 40.

Question 15.
40 ÷ 2 = 20.

Answer:
40/2 = 20.

Explanation:
In the above-given question,
given that,
divide by multiples of 2.
430/10 = 43.
zero gets canceled on both sides.
430/10 = 43.

Question 16.
4,000 ÷ 200 = 40.

Answer:
4000/200 = 40.

Explanation:
In the above-given question,
given that,
divide by multiples of 200.
4000/200 = 40.
zero gets canceled on both sides.
4000/200 = 40.

Question 17.
240 ÷ 10 = 24.

Answer:
240/10 = 24.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
240/10 = 24.
zero gets canceled on both sides.
240/10 = 24.

Question 18
24 ÷ 2 = 12.

Answer:
24/2 = 12.

Explanation:
In the above-given question,
given that,
divide by multiples of 2.
24/2 = 12.
zero gets canceled on both sides.
24/2 = 12.

Question 19.
240 ÷ 20 = 12.

Answer:
240/20 = 12.

Explanation:
In the above-given question,
given that,
divide by multiples of 20.
240/20 = 12.
zero gets canceled on both sides.
240/20 = 12.

Question 20.
3,600 ÷ 100 = 36.

Answer:
3600/100 = 36.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
3600/100 = 36.
zero gets canceled on both sides.
3600/100 = 36.

Question 21.
36 ÷ 3 = 12.

Answer:
36/3 = 12.

Explanation:
In the above-given question,
given that,
divide by multiples of 3.
36/3 = 12.
zero gets canceled on both sides.
36/3 = 12.

Question 22.
3,600 ÷ 300 = 36.

Answer:
3600/300 = 36.

Explanation:
In the above-given question,
given that,
divide by multiples of 300.
3600/300 = 36.
zero gets canceled on both sides.
3600/300 = 36.

Question 23.
480 ÷ 4 = 120.

Answer:
480/4 = 120.

Explanation:
In the above-given question,
given that,
divide by multiples of 4.
480/4 = 120.
zero gets canceled on both sides.
480/4 = 120.

Question 24.
480 ÷ 40 = 12.

Answer:
480/40 = 12.

Explanation:
In the above-given question,
given that,
divide by multiples of 40.
480/40 = 12.
zero gets canceled on both sides.
480/40 = 12.

Question 25.
6,300 ÷ 3 = 2100.

Answer:
6300/3 = 2100.

Explanation:
In the above-given question,
given that,
divide by multiples of 3.
6300/3 = 2100.
zero gets canceled on both sides.
6300/3 = 2100.

Question 26.
6,300 ÷ 30 = 210.

Answer:
6300/30 = 210.

Explanation:
In the above-given question,
given that,
divide by multiples of 30.
6300/30 = 210.
zero gets canceled on both sides.
6300/30 = 210.

Question 27.
6,300 ÷ 300 = 63.

Answer:
6300/300 = 63.

Explanation:
In the above-given question,
given that,
divide by multiples of 300.
6300/300 = 63.
zero gets canceled on both sides.
6300/300 = 63.

Question 28.
8,400 ÷ 2 = 4200.

Answer:
8400/2 = 4200.

Explanation:
In the above-given question,
given that,
divide by multiples of 2.
8400/2 = 4200.
zero gets canceled on both sides.
8400/2 = 4200.

Question 29.
8,400 ÷ 20 = 420.

Answer:
8400/20 = 420.

Explanation:
In the above-given question,
given that,
divide by multiples of 20.
8400/20 = 420.
zero gets canceled on both sides.
8400/20 = 420.

Question 30.
8,400 ÷ 200 = 42.

Answer:
8400/200 = 42.

Explanation:
In the above-given question,
given that,
divide by multiples of 200.
8400/200 = 42.
zero gets canceled on both sides.
8400/200 = 42.

Question 31.
96,000 ÷ 3 = 3200.

Answer:
96000/3 = 3200.

Explanation:
In the above-given question,
given that,
divide by multiples of 3.
96000/3 = 3200.
zero gets canceled on both sides.
96000/3 = 3200.

Question 32.
96,000 ÷ 300 = 320.

Answer:
96000/300 = 320.

Explanation:
In the above-given question,
given that,
divide by multiples of 300.
96000/300 = 320.
zero gets canceled on both sides.
96000/300 = 320.

Question 33.
96,000 ÷ 30 = 3200.

Answer:
96000/30 = 3200.

Explanation:
In the above-given question,
given that,
divide by multiples of 30.
96000/30 = 3200.
zero gets canceled on both sides.
96000/30 = 3200.

Question 34.
900 ÷ 30 = 30.

Answer:
900/30 = 30.

Explanation:
In the above-given question,
given that,
divide by multiples of 30.
900/30 = 30.
zero gets canceled on both sides.
900/30 = 30.

Question 35.
1,200 ÷ 30 = 40.

Answer:
1200/30 = 40.

Explanation:
In the above-given question,
given that,
divide by multiples of 30.
1200/30 = 40.
zero gets canceled on both sides.
1200/30 = 40.

Question 36.
1,290 ÷ 30 = 43.

Answer:
1290/30 = 43.

Explanation:
In the above-given question,
given that,
divide by multiples of 30.
1290/30 = 43.
zero gets canceled on both sides.
1290/30 = 43.

Question 37.
1,800 ÷ 300 = 6.

Answer:
1800/300 = 6.

Explanation:
In the above-given question,
given that,
divide by multiples of 300.
1800/300 = 6.
zero gets canceled on both sides.
1800/300 = 6.

Question 38.
8,000 ÷ 200 = 40.

Answer:
8000/200 = 40.

Explanation:
In the above-given question,
given that,
divide by multiples of 200.
8000/200 = 40.
zero gets canceled on both sides.
8000/200 = 40.

Question 39.
12,000 ÷ 200 = 60.

Answer:
12000/20 = 60.

Explanation:
In the above-given question,
given that,
divide by multiples of 20.
12000/20 = 60.
zero gets canceled on both sides.
12000/20 = 60.

Question 40.
12,800 ÷ 200 = 64.

Answer:
12800/200 = 64.

Explanation:
In the above-given question,
given that,
divide by multiples of 200.
12800/200 = 64.
zero gets canceled on both sides.
12800/200 = 64.

Question 41.
2,240 ÷ 70 = 32.

Answer:
2240/70 = 32.

Explanation:
In the above-given question,
given that,
divide by multiples of 70.
2240/70 = 32.
zero gets canceled on both sides.
2240/70 = 32.

Question 42.
18,400 ÷ 800 = 23.

Answer:
18400/800 = 23.

Explanation:
In the above-given question,
given that,
divide by multiples of 800.
18400/800 = 23.
zero gets canceled on both sides.
18400/800 = 23.

Question 43.
21,600 ÷ 90 = 240.

Answer:
21600/90 = 240.

Explanation:
In the above-given question,
given that,
divide by multiples of 90.
21600/90 = 240.
zero gets canceled on both sides.
21600/90 = 240.

Question 44.
25,200 ÷ 600 = 42.

Answer:
25200/600 = 42.

Explanation:
In the above-given question,
given that,
divide by multiples of 600.
25200/600 = 42.
zero gets canceled on both sides.
25200/600 = 42.

B
Divide by Multiples of 10 and 100

Answer:
20/10 = 2, 420/10 = 42, 4200/10 = 420, 4200/100 = 42, 40/10 = 4, 840/10 = 840, 8400/10 = 840, 8400/100 = 84, 84000/100 = 840, 900/10 = 90, 90/3 = 30, 900/30 = 30, 6000/100 = 60, 60/2 = 30, 6000/200 = 30, 240/10 = 24, 24/2 = 12, 240/20 = 12, 6300/100 = 63, 63/3 = 21, 6300/300 = 21.

Explanation:
In the above-given question,
given that,
divide by multiples of 10 and 100.
20/10 = 2, 420/10 = 42, 4200/10 = 420, 4200/100 = 42, 40/10 = 4, 840/10 = 840, 8400/10 = 840, 8400/100 = 84, 84000/100 = 840, 900/10 = 90, 90/3 = 30, 900/30 = 30, 6000/100 = 60, 60/2 = 30, 6000/200 = 30, 240/10 = 24, 24/2 = 12, 240/20 = 12, 6300/100 = 63, 63/3 = 21, 6300/300 = 21.

Answer:
840/4 = 210, 840/40 = 21, 3600/3 = 1200, 3600/30 = 120, 3600/300 = 12, 4800/2 = 2400, 4800/20 = 240, 4800/200 = 24, 69000/3 = 23000, 69000/300 = 230, 69000/30 = 2300, 800/40 = 20, 1200/40 = 30, 1280/40 = 32, 1600/400 = 4, 8000/200 = 40, 14000/200 = 70, 14600/200 = 73, 2560/80 = 32, 16,100/700 = 23, 14400/60 = 240, 37800/900 = 42.

Explanation:
In the above-given question,
given that,
divide by multiples of 10 and 100.
840/4 = 210, 840/40 = 21, 3600/3 = 1200, 3600/30 = 120, 3600/300 = 12, 4800/2 = 2400, 4800/20 = 240, 4800/200 = 24, 69000/3 = 23000, 69000/300 = 230, 69000/30 = 2300, 800/40 = 20, 1200/40 = 30, 1280/40 = 32, 1600/400 = 4, 8000/200 = 40, 14000/200 = 70, 14600/200 = 73, 2560/80 = 32, 16,100/700 = 23, 14400/60 = 240, 37800/900 = 42.

Question 1.
20 ÷ 10 = 2.

Answer:
20/10 = 2.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
20/10 = 2.
zero gets canceled on both sides.
20/10 = 2.

Question 2.
420 ÷ 10 = 42.

Answer:
420/10 = 42.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
420/10 = 42.
zero gets canceled on both sides.
420/10 = 42.

Question 3.
4,200 ÷ 10 = 420.

Answer:
4200/10 = 420.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
4200/10 = 420.
zero gets canceled on both sides.
4200/10 = 420.

Question 4.
4,200 ÷ 100 = 42.

Answer:
4200/100 = 42.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
4200/100 = 42.
zero gets canceled on both sides.
4200/100 = 42.

Question 5.
42,000 ÷ 100 = 420.

Answer:
42000/100 = 420.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
42000/100 = 420.
zero gets canceled on both sides.
42000/100 = 420.

Question 6.
40 ÷ 10 = 4.

Answer:
40/10 = 4.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
40/10 = 4.
zero gets canceled on both sides.
40/10 = 4.

Question 7.
840 ÷ 10 = 84.

Answer:
840/10 = 84.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
840/10 = 84.
zero gets canceled on both sides.
840/10 = 84.

Question 8.
8,400 ÷ 10 = 840.

Answer:
8400/10 = 840.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
8400/10 = 840.
zero gets canceled on both sides.
8400/10 = 840.

Question 9.
8,400 ÷ 100 = 84.

Answer:
8400/100 = 84.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
8400/100 = 84.
zero gets canceled on both sides.
84000/100 = 84.

Question 10.
84,000 ÷ 100 = 840.

Answer:
8400/100 = 840.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
8400/100 = 840.
zero gets canceled on both sides.
8400/100 = 840.

Question 11.
900 ÷ 10 = 90.

Answer:
900/10 = 90.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
900/10 = 90.
zero gets canceled on both sides.
900/10 = 90.

Question 12.
90 ÷ 3 = 30.

Answer:
90/3 = 30.

Explanation:
In the above-given question,
given that,
divide by multiples of 3.
90/3 = 30.
zero gets canceled on both sides.
90/3 = 30.

Question 13.
900 ÷ 30 =30.

Answer:
900/30 = 30.

Explanation:
In the above-given question,
given that,
divide by multiples of 30.
900/30 = 30.
zero gets canceled on both sides.
900/30 = 30.

Question 14.
6,000 ÷ 100 = 60.

Answer:
6000/100 = 60.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
6000/100 = 60.
zero gets canceled on both sides.
6000/100 = 60.

Question 15.
60 ÷ 2 = 30.

Answer:
60/2 = 30.

Explanation:
In the above-given question,
given that,
divide by multiples of 2.
60/2 = 30.
zero gets canceled on both sides.
60/2 = 30.

Question 16.
6,000 ÷ 200 = 30.

Answer:
6000/200 = 30.

Explanation:
In the above-given question,
given that,
divide by multiples of 200.
6000/200 = 30.
zero gets canceled on both sides.
6000/200 = 30.

Question 17.
240 ÷ 10 = 24.

Answer:
240/10 = 24.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
240/10 = 24.
zero gets canceled on both sides.
240/10 = 24.

Question 18.
24 ÷ 2 = 12.

Answer:
24/2 = 12.

Explanation:
In the above-given question,
given that,
divide by multiples of 2.
24/2 = 12.
zero gets canceled on both sides.
24/2 = 12.

Question 19.
240 ÷ 20 = 12.

Answer:
240/20 = 12.

Explanation:
In the above-given question,
given that,
divide by multiples of 20.
240/20 = 12.
zero gets canceled on both sides.
240/20 = 12.

Question 20.
6,300 ÷ 100 = 63.

Answer:
6300/100 = 63.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
6300/100 = 63.
zero gets canceled on both sides.
6300/100 = 63.

Question 21.
63 ÷ 3 = 21.

Answer:
63/3 = 21.

Explanation:
In the above-given question,
given that,
divide by multiples of 3.
63/3 = 21.
zero gets canceled on both sides.
63/3 = 21.

Question 22.
6,300 ÷ 300 = 21.

Answer:
6300/300 = 21.

Explanation:
In the above-given question,
given that,
divide by multiples of 300.
6300/300 = 21.
zero gets canceled on both sides.
6300/300 = 21.

Question 23.
840 ÷ 4 = 210.

Answer:
840/4 = 210.

Explanation:
In the above-given question,
given that,
divide by multiples of 4.
840/4 = 210.
zero gets canceled on both sides.
840/4 = 210.

Question 24.
840 ÷ 40 = 21.

Answer:
840/40 = 21.

Explanation:
In the above-given question,
given that,
divide by multiples of 40.
840/40 = 21.
zero gets canceled on both sides.
840/40 = 21.

Question 25.
3,600 ÷ 3 = 1200.

Answer:
3600/3 = 1200.

Explanation:
In the above-given question,
given that,
divide by multiples of 3.
3600/3 = 1200.
zero gets canceled on both sides.
3600/3 = 1200.

Question 26.
3,600 ÷ 30 =120.

Answer:
3600/30 = 120.

Explanation:
In the above-given question,
given that,
divide by multiples of 30.
3600/30 = 120.
zero gets canceled on both sides.
3600/30 = 120.

Question 27.
3,600 ÷ 300 = 12.

Answer:
3600/300 = 12.

Explanation:
In the above-given question,
given that,
divide by multiples of 300.
3600/300 = 12.
zero gets canceled on both sides.
3600/300 = 12.

Question 28.
4,800 ÷ 2 = 2400.

Answer:
4800/2 = 2400.

Explanation:
In the above-given question,
given that,
divide by multiples of 2.
4800/2 = 2400.
zero gets canceled on both sides.
4800/2 = 2400.

Question 29.
4,800 ÷ 20 = 240.

Answer:
4800/20 = 240.

Explanation:
In the above-given question,
given that,
divide by multiples of 20.
4800/20 = 240.
zero gets canceled on both sides.
4800/20 = 240.

Question 30.
4,800 ÷ 200 = 24.

Answer:
4800/200 = 24.

Explanation:
In the above-given question,
given that,
divide by multiples of 200.
4800/200 = 24.
zero gets canceled on both sides.
4800/200 = 24.

Question 31.
69,000 ÷ 3 = 23000.

Answer:
69000/3 = 23000.

Explanation:
In the above-given question,
given that,
divide by multiples of 3.
69000/3 = 23000.
zero gets canceled on both sides.
69000/3 = 23000.

Question 32.
69,000 ÷ 300 = 230.

Answer:
69000/300 = 230.

Explanation:
In the above-given question,
given that,
divide by multiples of 300.
69000/300 = 230.
zero gets canceled on both sides.
69000/300 = 230.

Question 33.
69,000 ÷ 30 = 2300.

Answer:
69000/30 = 2300.

Explanation:
In the above-given question,
given that,
divide by multiples of 30.
69000/30 = 2300.
zero gets canceled on both sides.
69000/30 = 2300.

Question 34.
800 ÷ 40 = 20.

Answer:
800/40 = 20.

Explanation:
In the above-given question,
given that,
divide by multiples of 40.
800/40 = 20.
zero gets canceled on both sides.
800/40 = 20.

Question 35.
1,200 ÷ 40 = 30.

Answer:
1200/40 = 30.

Explanation:
In the above-given question,
given that,
divide by multiples of 40.
1200/40 = 30.
zero gets canceled on both sides.
1200/40 = 30.

Question 36.
1,280 ÷ 40 = 32.

Answer:
1280/40 = 32.

Explanation:
In the above-given question,
given that,
divide by multiples of 40.
1280/40 = 32.
zero gets canceled on both sides.
1280/40 = 32.

Question 37.
1,600 ÷ 400 = 4.

Answer:
1600/400 = 4.

Explanation:
In the above-given question,
given that,
divide by multiples of 400.
1600/400 = 4.
zero gets canceled on both sides.
1600/400 = 4.

Question 38.
8,000 ÷ 200 =40.

Answer:
8000/200 = 40.

Explanation:
In the above-given question,
given that,
divide by multiples of 20.
8000/200 = 400.
zero gets canceled on both sides.
8000/200 = 400.

Question 39.
14,000 ÷ 200 = 70.

Answer:
14000/200 = 70.

Explanation:
In the above-given question,
given that,
divide by multiples of 200.
14000/200 = 70.
zero gets canceled on both sides.
14000/200 = 70.

Question 40.
14,600 ÷ 200 = 73.

Answer:
14600/200 = 73.

Explanation:
In the above-given question,
given that,
divide by multiples of 200.
14600/200 = 73.
zero gets canceled on both sides.
14600/200 = 73.

Question 41.
2,560 ÷ 80 =32.

Answer:
2560/80 = 32.

Explanation:
In the above-given question,
given that,
divide by multiples of 80.
2560/80 = 32.
zero gets canceled on both sides.
2560/80 = 32.

Question 42.
16,100 ÷ 700 = 23.

Answer:
16100/700 = 23.

Explanation:
In the above-given question,
given that,
divide by multiples of 700.
16100/700 = 23.
zero gets canceled on both sides.
16100/700 = 23.

Question 43.
14,400 ÷ 60 = 240.

Answer:
14400/60 = 240.

Explanation:
In the above-given question,
given that,
divide by multiples of 60.
14400/60 = 240.
zero gets canceled on both sides.
14400/60 = 240.

Question 44.
37,800 ÷ 900 = 42.

Answer:
37800/900 = 42.

Explanation:
In the above-given question,
given that,
divide by multiples of 900.
37800/900 = 42.
zero gets canceled on both sides.
37800/900 = 42.

Eureka Math Grade 5 Module 2 Lesson 16 Problem Set Answer Key

Question 1.
Divide. Draw place value disks to show your thinking for (a) and (c). You may draw disks on your personal white board to solve the others if necessary.
a. 500 ÷ 10

Answer:
500/10 = 50.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
500/10 = 50.
zero gets canceled on both sides.
500/10 = 50.

b. 360 ÷ 10

Answer:
360/10 = 36.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
360/10 = 36.
zero gets canceled on both sides.
360/10 = 36.

c. 12,000 ÷ 100

Answer:
12000/100 = 120.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
12000/100 = 120.
zero gets canceled on both sides.
12000/100 = 120.

d. 450,000 ÷ 100

Answer:
450000/100 = 4500.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
450000/100 = 4500.
zero gets canceled on both sides.
450000/100 = 4500.

e. 700,000 ÷ 1,000

Answer:
700000/1000 = 700.

Explanation:
In the above-given question,
given that,
divide by multiples of 1000.
700000/1000 = 7000.
zero gets canceled on both sides.
700000/1000 = 7000.

f. 530,000 ÷ 100

Answer:
530,000/100 = 5300.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
530,000/100 = 5300.
zero gets canceled on both sides.
530,000/100 = 5300.

Question 2.
Divide. The first one is done for you.
a. 12,000 ÷ 30
= 12,000 ÷ 10 ÷ 3
= 1,200 ÷ 3
= 400

Answer:
12000/30 = 400.

Explanation:
In the above-given question,
given that,
12000/30.
12000/10/3.
1200/3.
400.

b. 12,000 ÷ 300

Answer:
12000/300 = 40.

Explanation:
In the above-given question,
given that,
12000/300.
12000/100/3.
120/3.
40.

c. 12,000 ÷ 3,000

Answer:
12000/3000 = 4.

Explanation:
In the above-given question,
given that,
12000/3000.
12000/1000/3.
12/3.
4.

d. 560,000 ÷ 70

Answer:
560000/70 = 8000.

Explanation:
In the above-given question,
given that,
560000/70.
560000/10/7.
56000/7.
8000.

e. 560,000 ÷ 700

Answer:
560000/700 = 800.

Explanation:
In the above-given question,
given that,
560000/700.
560000/100/7.
5600/7.
800.

f. 560,000 ÷ 7,000

Answer:
560000/7000 = 80.

Explanation:
In the above-given question,
given that,
560000/7000.
566000/1000/7.
560/7.
80.

g. 28,000 ÷ 40

Answer:
28000/40 = 700.

Explanation:
In the above-given question,
given that,
28000/40.
28000/10/4.
2800/4.
700.

h. 450,000 ÷ 500

Answer:
450000/500 = 900.

Explanation:
In the above-given question,
given that,
450000/500.
450000/100/5.
4500/5.
900.

i. 810,000 ÷ 9,000

Answer:
810000/9000 = 90.

Explanation:
In the above-given question,
given that,
810000/9000.
810000/1000/9.
810/9.
90.

Question 3.
The floor of a rectangular banquet hall has an area of 3,600 m². The length is 90 m.
a. What is the width of the banquet hall?

Answer:
The width of the banquet hall = 40 m.

Explanation:
In the above-given question,
given that,
The floor of a rectangular banquet hall has an area of 3600 sq m.
the length is 90 m.
3600/90 = 40.
the width of the banquet hall = 40 m.

b. A square banquet hall has the same area. What is the length of the room ?

Answer:
The length of the banquet hall = 40 m.

Explanation:
In the above-given question,
given that,
The floor of a square banquet hall has an area of 3600 sq m.
the width of the room is  40 m.
3600/90 = 40.
the width of the banquet hall = 40 m.

c. A third rectangular banquet hall has a perimeter of 3,600 m. What is the width if the length is 5 times the width?

Answer:
The width of the banquet hall = 90 m.

Explanation:
In the above-given question,
given that,
The third rectangular banquet hall has an area of 3600 sq m.
the length is 90 m.
3600/40 = 90.
the width of the banquet hall = 90 m.

Question 4.
Two fifth graders solved 400,000 divided by 800. Carter said the answer is 500, while Kim said the answer is 5,000.
a. Who has the correct answer? Explain your thinking.

Answer:
400000/800 = 500.

Explanation:
In the above-given question,
given that,
Two fifth graders solved 400000 divided by 800.
400000/800 = 500.
Carter was correct.

b. What if the problem is 4,000,000 divided by 8,000? What is the quotient?

Answer:
500.

Explanation:
In the above-given question,
given that,
4000000/8000 = 500.

Eureka Math Grade 5 Module 2 Lesson 16 Exit Ticket Answer Key

Divide. Show your thinking.

a. 17,000 ÷ 100

Answer:
17000/100 =170.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
17000/100 = 170.
zero gets canceled on both sides.
17000/100 = 170.

b. 59,000 ÷ 1,000

Answer:
59000/1000 = 59.

Explanation:
In the above-given question,
given that,
divide by multiples of 1000.
59000/1000 = 59.
zero gets canceled on both sides.
59000/1000 = 59.

c. 12,000 ÷ 40

Answer:
12000/40 = 300.

Explanation:
In the above-given question,
given that,
divide by multiples of 40.
12000/40 = 300.
zero gets canceled on both sides.
12000/40 = 300.

d. 480,000 ÷ 600

Answer:
480000/600 = 800.

Explanation:
In the above-given question,
given that,
divide by multiples of 600.
480000/600 = 800.
zero gets canceled on both sides.
480000/600 = 800.

Eureka Math Grade 5 Module 2 Lesson 16 Homework Answer Key

Question 1.
Divide. Draw place value disks to show your thinking for (a) and (c). You may draw disks on your personal white board to solve the others if necessary.
a. 300 ÷ 10

Answer:
300/10 = 30.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
300/10 = 30.
zero gets canceled on both sides.
300/10 = 30.

b. 450 ÷ 10

Answer:
450/10 = 45.

Explanation:
In the above-given question,
given that,
divide by multiples of 10.
450/10 = 45.
zero gets canceled on both sides.
450/10 = 45.

c. 18,000 ÷ 100

Answer:
18000/100 = 180.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
18000/100 = 180.
zero gets canceled on both sides.
18000/100 = 180.

d. 730,000 ÷ 100

Answer:
730000/100 = 7300.

Explanation:
In the above-given question,
given that,
divide by multiples of 100.
730000/100 = 7300.
zero gets canceled on both sides.
730000/100 = 7300.

e. 900,000 ÷ 1,000

Answer:
9600000/1000 = 9600.

Explanation:
In the above-given question,
given that,
divide by multiples of 1000.
9600000/1000 = 9600.
zero gets canceled on both sides.
9600000/1000 = 9600.

f. 680,000 ÷ 1,000

Answer:
680000/1000 = 680.

Explanation:
In the above-given question,
given that,
divide by multiples of 1000.
680000/1000 = 680.
zero gets canceled on both sides.
680000/1000 = 680.

Question 2.
Divide. The first one is done for you.
a. 18,000 ÷ 20
= 18,000 ÷ 10 ÷ 2
= 1,800 ÷ 2
= 900

Answer:
18000/20 = 900.

Explanation:
In the above-given question,
given that,
18000/20.
18000/10/2.
1800/2.
900.

b. 18,000 ÷ 200

Answer:
18000/200 = 90.

Explanation:
In the above-given question,
given that,
18000/200.
18000/100/2.
180/2.
90.

c. 18,000 ÷ 2,000

Answer:
18000/2000 = 9.

Explanation:
In the above-given question,
given that,
18000/2000.
18000/1000/2.
18/2.
9.

d. 420,000 ÷ 60

Answer:
420000/60 = 7000.

Explanation:
In the above-given question,
given that,
420000/60.
420000/10/6.
42000/6.
7000.

e. 420,000 ÷ 600

Answer:
420000/600 = 700.

Explanation:
In the above-given question,
given that,
420000/600.
420000/100/6.
4200/6.
700.

f. 420,000 ÷ 6,000

Answer:
420000/6000 = 70.

Explanation:
In the above-given question,
given that,
420000/6000.
420000/1000/6.
420/6.
70.

g. 24,000 ÷ 30

Answer:
24000/30 = 800.

Explanation:
In the above-given question,
given that,
24000/30.
24000/10/3.
2400/3.
800.

h. 560,000 ÷ 700

Answer:
560000/700 = 800.

Explanation:
In the above-given question,
given that,
560000/700.
560000/100/7.
5600/7.
800.

i. 450,000 ÷ 9,000

Answer:
450000/9000 = 50.

Explanation:
In the above-given question,
given that,
450000/9000.
450000/1000/9.
450/9.
50.

Question 3.
A stadium holds 50,000 people. The stadium is divided into 250 different seating sections. How many seats are in each section?

Answer:
The seats are in each section = 200.

Explanation:
In the above-given question,
given that,
A stadium holds 50,000 people.
The stadium is divided into 250 different seating sections.
50000/250 = 200.

Question 4.
Over the course of a year, a tractor trailer commutes 160,000 miles across America.
a. Assuming a trucker changes his tires every 40,000 miles, and that he starts with a brand new set of tires, how many sets of tires will he use in a year?

Answer:
The sets of tires will he use in a year = 4 miles.

Explanation:
In the above-given question,
given that,
Over the course of a year, a tractor trailer commutes 160,000 miles across America.
Assuming a trucker changes his tires every 40,000 miles.
160,000/40,000 = 4.

b. If the trucker changes the oil every 10,000 miles, and he starts the year with a fresh oil change, how many times will he change the oil in a year?

Answer:
The sets of tires will he use in a year = 16 miles.

Explanation:
In the above-given question,
given that,
Over the course of a year, a tractor trailer commutes 160,000 miles across America.
Assuming a trucker changes his tires every 10,000 miles.
160,000/10,000 = 16.

Engage NY Eureka Math 5th Grade Module 2 Lesson 14 Answer Key

Eureka Math Grade 5 Module 2 Lesson 14 Problem Set Answer Key

Question 1.
Solve. The first one is done for you.

a. Convert days to weeks.
28 days = 28 × (1 day)
= 28 × (\(\frac{1}{7}\) week)
= \(\frac{28}{7}\) week
= 4 weeks

Answer:
28 days = 4 weeks,

Explanation:
In the above-given question,
given that,
28 days = 28 x (1 day)
28 x (\(\frac{1}{7}\) week)
\(\frac{28}{7}\) week.
28/7 = 4.

b. Convert quarts to gallons.
20 quarts = ____20_______ × ( 1 quart)
= ______20_____ × (\(\frac{1}{4}\) gallon)
= _____20/4______ gallons
= _____5______ gallons

Answer:
20 quarts = 5 gallons,

Explanation:
In the above-given question,
given that,
20 quarts = 28 x (1 gallon)
20 x (\(\frac{1}{4}\) gallon)
\(\frac{20}{4}\) gallon.
20/4 = 5.

c. Convert centimeters to meters.
920 cm = ___920________ × ( ____1_______ cm)
= ____920_______ × ( ____0.01_______ m)
= _____9.2______ m

Answer:
920 cm = 92000 m.

Explanation:
In the above-given question,
given that,
920 cm = 920 x (1 m)
920 x (\(\frac{1}{0.01}\) meter)
\(\frac{920}{0.01}\) meter.
920/0.01 = 92000.

d. Convert meters to kilometers.
1,578 m = _____1578______ × ( _____1______ m)
= ______1578_____ × (0.001 km)
= ___1578000________ km

Answer:
1578 m = 1578000 km,

Explanation:
In the above-given question,
given that,
1578 = 1578 x (1 km)
1578 x (\(\frac{1}{0.001}\) kilometers)
\(\frac{1}{0.001}\) kilometers.
1578/0.001 = 1578000.

e. Convert grams to kilograms.
6,080 g = 60,80,000 kg.

Answer:
6080 g = 60,80,000,

Explanation:
In the above-given question,
given that,
6080 g = 6080 x (1 gram)
6080 x (\(\frac{1}{0.001}\) kilograms)
\(\frac{6080}{0.001}\) kilograms.
6080/0.001 = 60,80,000.

f. Convert milliliters to liters.
509 mL = 509000 liters.

Answer:
509 ml = 509000 liters,

Explanation:
In the above-given question,
given that,
509 ml = 509 x (1 ml)
509 x (\(\frac{1}{0.001}\) liters)
\(\frac{509}{0.001}\) liters.
509/0.001 = 509000 liters.

Question 2.
After solving, write a statement to express each conversion. The first one is done for you.
a. The screen measures 24 inches. Convert 24 inches to feet.
24 inches = 24 × (1 inch)
= 24 × (\(\frac{1}{12}\)feet)
= \(\frac{24}{12}\) feet
= 2 feet
The screen measures 24 inches or 2 feet.

Answer:
24 inches =2 feet,

Explanation:
In the above-given question,
given that,
24 inches = 24 x (1 inch)
24 x (\(\frac{1}{0.0833}\) feet)
\(\frac{24}{0.0833}\) feet.
24/0.0833 = 288.11 feet.

b. A jug of syrup holds 12 cups. Convert 12 cups to pints.

Answer:
12 cups = 24 pints,

Explanation:
In the above-given question,
given that,
12 cups = 12 x (cup)
12 x (\(\frac{1}{0.5}\) pints)
\(\frac{12}{0.5}\) pints.
12/0.5 = 24.

c. The length of the diving board is 378 centimeters. What is its length in meters?

Answer:
378 cm = 37800 meters,

Explanation:
In the above-given question,
given that,
378 cm = 378 x (1 cm)
378 x (\(\frac{1}{0.01}\) meters)
\(\frac{378}{0.01}\) meters.
378/0.01 = 37800 meters.

d. The capacity of a container is 1,478 milliliters. Convert this to liters.

Answer:
1478 ml = 1478000 liters,

Explanation:
In the above-given question,
given that,
1478 ml = 1478 x (1 ml)
1478 x (\(\frac{1}{0.001}\) liters)
\(\frac{1478}{0.001}\) liters.
1478/0.001 = 1478000 liters.

e. A truck weighs 3,900,000 grams. Convert the truck’s weight to kilograms.

Answer:
3900000 g = 3.9 kilograms.

Explanation:
In the above-given question,
given that,
3900000 g = 3900000 x (1 gram)
3900000 x (\(\frac{1}{0.001}\) kilograms)
\(\frac{3900000}{0.001}\) kilograms.
3900000/0.001 = 3.9 kilograms.

f. The distance was 264,040 meters. Convert the distance to kilometers.

Answer:
264,040 m =  264040000km,

Explanation:
In the above-given question,
given that,
264,040 = 264,040 x (1 km)
264,040 x (\(\frac{1}{0.001}\) kilometers)
\(\frac{164040}{0.001}\) kilometers.
264040/0.001 = 264040000.

Eureka Math Grade 5 Module 2 Lesson 14 Exit Ticket Answer Key

Question 1.
Convert days to weeks by completing the number sentences.
35 days = _____35______ × ( ____1_______ day)
= _______35____ × ( _____7______ week)
= 35/7.
= 5.

Answer:
35 days = 5 weeks,

Explanation:
In the above-given question,
given that,
35 days = 35 x (1 day)
35 x (\(\frac{1}{7}\) week)
\(\frac{35}{7}\) week.
35/7 = 5.

Question 2.
Convert grams to kilograms by completing the number sentences.
4,567 grams = ____4567_______ × __1__gram_______
= __4567_________ × _____0.001 kg______
=4567/0.001
=4567000 kg

Answer:
4567 g = 4567000 kilograms.

Explanation:
In the above-given question,
given that,
4567 g = 4567 x (1 gram)
4567 x (\(\frac{1}{0.001}\) kilograms)
\(\frac{4567}{0.001}\) kilograms.
4567/0.001 = 4567000 kilograms.

Eureka Math Grade 5 Module 2 Lesson 14 Homework Answer Key

Question 1.
Solve. The first one is done for you.

a. Convert days to weeks.
42 days = 42 × (1 day)
= 42 × (\(\frac{1}{7}\) week)
= \(\frac{42}{7}\) week
= 6 weeks

Answer:
42 days = 6 weeks,

Explanation:
In the above-given question,
given that,
42 days = 42 x (1 day)
42 x (\(\frac{1}{7}\) week)
\(\frac{42}{7}\) week.
42/7 = 6.

b. Convert quarts to gallons.
36 quarts = ______36_____ × ( 1 quart)
= ______36_____ × (1/4 “gallon” )
= ___________ gallons
= ___________ gallons

Answer:
36 quarts = 9 gallons,

Explanation:
In the above-given question,
given that,
36 quarts = 36 x (1 gallon)
36 x (\(\frac{36}{4}\) gallon)
\(\frac{36}{4}\) gallon.
36/4 = 9.

c. Convert centimeters to meters.
760 cm = _____760______ × ( ___1________ cm)
= _____760______ × ( ___0.01________ m)
= ___768000________ m

Answer:
760 cm = 768000 meters,

Explanation:
In the above-given question,
given that,
760 cm = 760 x (1 cm)
760 x (\(\frac{1}{0.01}\) meters)
\(\frac{760}{0.01}\) meters.
760/0.01 = 76000 meters.

d. Convert meters to kilometers.
2,485 m = ____2485_______ × ( ____1_______ m)
= ___2485________ × (0.001 km)
= ____2485000_______ km

Answer:
2,485 m =  2485,000km,

Explanation:
In the above-given question,
given that,
2485 = 2485 x (1 km)
2485 x (\(\frac{1}{0.001}\) kilometers)
\(\frac{2485}{0.001}\) kilometers.
2485/0.001 = 24,85,000.

e. Convert grams to kilograms.
3,090 g = 3090000 kg.

Answer:
3090 g = 3090000 kilograms.

Explanation:
In the above-given question,
given that,
3090 g = 3090 x (1 gram)
3090 x (\(\frac{1}{0.001}\) kilograms)
\(\frac{3090}{0.001}\) kilograms.
3090/0.001 = 3090000 kilograms.

f. Convert milliliters to liters.
205 mL = 205000 liters.

Answer:
205 ml = 205000 liters,

Explanation:
In the above-given question,
given that,
205 ml = 205 x (1 ml)
205 x (\(\frac{1}{0.001}\) liters)
\(\frac{205}{0.001}\) liters.
205/0.001 = 205000 liters.

Question 2.
After solving, write a statement to express each conversion. The first one is done for you.

a. The screen measures 36 inches. Convert 36 inches to feet.
36 inches = 36 × (1 inch)
= 36 × (\(\frac{1}{12}\) feet)
= \(\frac{36}{12}\) feet
= 3 feet
The screen measures 36 inches or 3 feet.

Answer:
36 inches = 432.17 feet,

Explanation:
In the above-given question,
given that,
36 inches = 36 x (1 inch)
36 x (\(\frac{1}{0.0833}\) feet)
\(\frac{36}{0.0833}\) feet.
36/0.0833 = 432.17 feet.

b. A jug of juice holds 8 cups. Convert 8 cups to pints.

Answer:
8 cups = 16 pints,

Explanation:
In the above-given question,
given that,
8 cups = 8 x (cup)
8 x (\(\frac{1}{0.5}\) pints)
\(\frac{8}{0.5}\) pints.
8/0.5 = 16.

c. The length of the flower garden is 529 centimeters. What is its length in meters?

Answer:
529 cm = 52900 meters,

Explanation:
In the above-given question,
given that,
529 cm = 529 x (1 cm)
529 x (\(\frac{1}{0.01}\) meters)
\(\frac{529}{0.01}\) meters.
529/0.01 = 52900 meters.

d. The capacity of a container is 2,060 milliliters. Convert this to liters.

Answer:
2060 ml = 2060000 liters,

Explanation:
In the above-given question,
given that,
2060 ml = 2060 x (1 ml)
2060 x (\(\frac{1}{0.001}\) liters)
\(\frac{2060}{0.001}\) liters.
2060/0.001 = 2060000 liters.

e. A hippopotamus weighs 1,560,000 grams. Convert the hippopotamus’ weight to kilograms.

Answer:
1560000 g = 1.56 kilograms.

Explanation:
In the above-given question,
given that,
1560000 g = 1560000 x (1 gram)
1560000 x (\(\frac{1}{0.001}\) kilograms)
\(\frac{1560000}{0.001}\) kilograms.
1560000/0.001 = 1.56 kilograms.

f. The distance was 372,060 meters. Convert the distance to kilometers.

Answer:
372060 m =  372060,000km,

Explanation:
In the above-given question,
given that,
372060 = 372060 x (1 km)
372060 x (\(\frac{1}{0.001}\) kilometers)
\(\frac{372060}{0.001}\) kilometers.
372060/0.001 = 372060,000.

Engage NY Eureka Math 5th Grade Module 2 Lesson 13 Answer Key

Eureka Math Grade 5 Module 2 Lesson 13 Problem Set Answer Key

Question 1.
Solve. The first one is done for you.
a. Convert weeks to days.
8 weeks = 8 × (1 week)
= 8 × (7 days)
= 56 days

Answer:
8 weeks = 56 days.

Explanation:
In the above-given question,
given that,
8 weeks = 8 x (1 week).
8 x (7 days).
1 week = 7 days.
8 x 7 = 56.
8 weeks = 56 days.

b. Convert years to days.
4 years = _____4______ × ( ____1_______ year)
= ______4_____ × ( ___365________ days)
= ____1460_______ days

Answer:
4 years = 1460 days.

Explanation:
In the above-given question,
given that,
4 years = 4 x (1 year).
4 x (365 days).
1 year = 365 days.
4 x 365 = 1460.
4 years = 1460 days.

c. Convert meters to centimeters.
9.2 m = ____9.2_______ × ( _____1______ m)
= ___9.2________ × ( ____100_______ cm)
= _____920______ cm

Answer:
9.2 meters = 920 cm.

Explanation:
In the above-given question,
given that,
9.2 meters = 9.2 x (1 meter).
9.2 x (100 cms).
1 meter = 100 cms.
9.2 x 100 = 920.
9.2 meters = 920 cm.

d. Convert yards to feet.
5.7 yards

Answer:
5.7 yards = 17.1feet.

Explanation:
In the above-given question,
given that,
5.7 yards = 5.7 x (1 yard).
5.7 x (3 feet).
5.7 yards =  17.1 feet.
5.7 x 7 = 17.1.
5.7 yards = 17.1 feet.

e. Convert kilograms to grams.
6.08 kg

Answer:
6.08 kg =6080 grams.

Explanation:
In the above-given question,
given that,
6.08 kg = 6.08 x (1 kg).
6.08 x (1000 grams).
1 kg = 1000 grams.
6.08 x 1000 = 6080 grams.
6.08 kg = 6080 grams.

f. Convert pounds to ounces.
12.5 pounds

Answer:
12.5 pounds = 200 ounces.

Explanation:
In the above-given question,
given that,
12.5 pounds = 12.5 x (1 pound).
12.5 x (16 ounces).
1 pound = 16 ounces.
12.5 x 16 = 200.
12.5 pounds = 200 ounces.

Question 2.
After solving, write a statement to express each conversion. The first one is done for you.

a. Convert the number of hours in a day to minutes.
24 hours = 24 × (1 hour)
= 24 × (60 minutes)
= 1,440 minutes
One day has 24 hours, which is the same as 1,440 minutes.

Answer:
24 hours = 1440 minutes.

Explanation:
In the above-given question,
given that,
24 hours = 24 x (1 hour).
1 x (60 minutes).
1 hour = 60 minutes.
24 x 60 = 1440.
24 hours = 1440 minutes.

b. A small female gorilla weighs 68 kilograms. How much does she weigh in grams?

Answer:
68 kg = 6800 grams.

Explanation:
In the above-given question,
given that,
68 kg = 68 x (1 kg).
68 x (1000 grams).
1 kg = 1000 grams.
68 x 1000 = 6800 grams.
68 kg = 6800 grams.

c. The height of a man is 1.7 meters. What is his height in centimeters?

Answer:
1.7 meters = 170 cm.

Explanation:
In the above-given question,
given that,
1.7 meters = 1.7 x (1 meter).
1.7 x (100 cms).
1 meter = 100 cms.
1.7 x 100 = 170.
1.7 meters = 170 cm.

d. The capacity of a syringe is 0.08 liters. Convert this to milliliters.

Answer:
0.08 liters = 80 ml.

Explanation:
In the above-given question,
given that,
0.08 liters = 0.08 x (1 liter).
0.08 x (1000 ml).
1 liter = 1000 ml.
0.08 x 1000 = 80 ml.
0.08 liters = 80 ml.

e. A coyote weighs 11.3 pounds. Convert the coyote’s weight to ounces.

Answer:
11.3 pounds = 180.8 ounces.

Explanation:
In the above-given question,
given that,
11.3 pounds = 11.3 x (1 pound).
11.3 x (16 ounces).
1 pound = 16 ounces.
11.3 x 16 = 180.8.
11.3 pounds = 180.8 ounces.

f. An alligator is 2.3 yards long. What is the length of the alligator in inches?

Answer:
2.3 yards = 6.9 feet.

Explanation:
In the above-given question,
given that,
2.3 yards = 2.3 x (1 yard).
2.3 x (3 feet).
2.3 yards =  6.9 feet.
2.3 x 3 = 6.9.
2.3 yards = 6.9 feet.

Eureka Math Grade 5 Module 2 Lesson 13 Exit Ticket Answer Key

Solve.
a. Convert pounds to ounces.
(1 pound = 16 ounces)
14 pounds = _____14____ × (1 pound)
= ___14______ × ( ____16_____ ounces)
= _____224____ ounces

Answer:
14 pounds = 224 ounces.

Explanation:
In the above-given question,
given that,
14 pounds = 14 x (1 pound).
14 x (16 ounces).
1 pound = 16 ounces.
14 x 16 = 224.
14 pounds = 224 ounces.

b. Convert kilograms to grams.
18.2 kilograms = ____18.2_____ × ( ____1 kg_____ )
= ____18.2_____ × ( _1000________ )
= ___18,200______ grams

Answer:
18.2 kg = 18200 grams.

Explanation:
In the above-given question,
given that,
18.2 kg = 18200 x (1 kg).
18.2 x (1000 grams).
1 kg = 1000 grams.
18.2 x 1000 = 18200 grams.
18.2 kg = 18200 grams.

Eureka Math Grade 5 Module 2 Lesson 13 Homework Answer Key

Question 1.
Solve. The first one is done for you.
a. Convert weeks to days.
6 weeks = 6 × (1 week)
= 6 × (7 days)
= 42 days

Answer:
6 weeks = 42 days.

Explanation:
In the above-given question,
given that,
6 weeks = 6 x (1 week).
6 x (7 days).
1 week = 7 days.
6 x 7 = 42.
6 weeks = 42 days.

b. Convert years to days.
7 years = ____7_______ × ( ____1_______ year)
= ______7_____ × ( ____365_______ days)
= _____2555______ days

Answer:
7 years = 2555 days.

Explanation:
In the above-given question,
given that,
7 years = 7 x (1 year).
7 x (365 days).
1 year = 365 days.
7 x 365 = 2555.
7 years = 2555 days.

c. Convert meters to centimeters.
4.5 m = ____4.5_______ × ( ____1_______ m)
= ____4.5_______ × ( ____100_______ cm)
= _____450______ cm

Answer:
4.5 meters = 450 cm.

Explanation:
In the above-given question,
given that,
4.5 meters = 4.5 x (1 meter).
4.5 x (100 cms).
1 meter = 100 cms.
4.5 x 100 = 450.
4.5 meters = 450 cm.

d. Convert pounds to ounces.
12.6 pounds

Answer:
12.6 pounds = 201.6 ounces.

Explanation:
In the above-given question,
given that,
12.6 pounds = 12.6 x (1 pound).
12.6 x (16 ounces).
1 pound = 16 ounces.
12.6 x 16 = 201.6.
12.6 pounds = 201.6 ounces.

e. Convert kilograms to grams.
3.09 kg

Answer:
3.09 kg = 3090 grams.

Explanation:
In the above-given question,
given that,
3.09 kg = 3.09 x (1 kg).
3.09 x (1000 grams).
1 kg = 1000 grams.
3.09 x 1000 = 3090 grams.
3.09 kg = 3090 grams.

f. Convert yards to inches.
245 yd

Answer:
245 yards = 8820 inches.

Explanation:
In the above-given question,
given that,
245 yards = 245 x (1 yard).
245 x (36 inches).
245 yards =  8820.
245 x 36 = 8820.
245 yards = 8820 inches.

Question 2.
After solving, write a statement to express each conversion. The first one is done for you.
a. Convert the number of hours in a day to minutes.
24 hours = 24 × (1 hour)
= 24 × (60 minutes)
= 1,440 minutes
One day has 24 hours, which is the same as 1,440 minutes.

Answer:
24 hours = 1440 minutes.

Explanation:
In the above-given question,
given that,
24 hours = 24 x (1 hour).
1 x (60 minutes).
1 hour = 60 minutes.
24 x 60 = 1440.
24 hours = 1440 minutes.

b. A newborn giraffe weighs about 65 kilograms. How much does it weigh in grams?

Answer:
65 kg = 6500 grams.

Explanation:
In the above-given question,
given that,
65 kg = 3.09 x (1 kg).
65 x (1000 grams).
1 kg = 1000 grams.
65 x 1000 = 6500 grams.
65 kg = 6500 grams.

c. The average height of a female giraffe is 4.6 meters. What is her height in centimeters?

Answer:
4.6 meters = 460 cm.

Explanation:
In the above-given question,
given that,
4.6 meters = 4.6 x (1 meter).
4.6 x (100 cms).
1 meter = 100 cms.
4.6 x 100 = 460.
4.6 meters = 460 cm.

d. The capacity of a beaker is 0.1 liter. Convert this to milliliters.

Answer:
0.1 liters = 100 ml.

Explanation:
In the above-given question,
given that,
0.1 liters = 0.1 x (1 liter).
0.1 x (1000 ml).
1 liter = 1000 ml.
0.1 x 1000 = 100 ml.
0.1 liters = 100 ml.

e. A pig weighs 9.8 pounds. Convert the pig’s weight to ounces.

Answer:
9.8 pounds = 156.8 ounces.

Explanation:
In the above-given question,
given that,
9.8 pounds = 9.8 x (1 pound).
9.8 x (16 ounces).
1 pound = 16 ounces.
9.8 x 16 = 156.8.
9.8 pounds = 156.8 ounces.

f. A marker is 0.13 meters long. What is the length in millimeters?

Answer:
0.13 meters = 130 mm.

Explanation:
In the above-given question,
given that,
0.13 meters = 0.13 x (1 meter).
0.13 x (1000 mm).
1 meter = 1000 mm.
0.13 x 1000 = 130.
0.13 meters = 130 mm.