Eureka Math

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

Eureka Math Grade 6 Module 4 Lesson 6 Example Answer Key

Example 1.
Expressions with Only Addition, Subtraction, Multiplication, and Division

What operations are evaluated first?
Answer:
Multiplication and division are evaluated first, from left to right.

What operations are always evaluated last?
Answer:
Addition and subtraction are always evaluated last, from left to right.

Example 2.
Expressions with Four Operations and Exponents
4 + 9² ÷ 3 × 2 – 2

What operation is evaluated first?
Answer:
Exponents (9² = 9 × 9 = 81)

What operations are evaluated next?
Answer:
Multiplication and division, from left to right (81 ÷ 3 = 27; 27 × 2 = 54)

What operations are always evaluated last?
Answer:
Addition and subtraction, from left to right (4 + 54 = 58; 58 – 2 = 56)

What is the final answer?
Answer:
56

Example 3. Expressions with Parentheses
Consider a family of 4 that goes to a soccer game. Tickets are $5.00 each. The mom also buys a soft drink for $2.00. How would you write this expression?
Answer:
4 × 5 + 2

How much will this outing cost?
Answer:
$22

Consider a different scenario: The same family goes to the game as before, but each of the family members wants a drink. How would you write this expression?
Answer:
4 × (5 + 2)

Why would you add the 5 and 2 first?
Answer:
We need to determine how much each person spends. Each person spends $7; then, we multiply by 4 people to figure out the total cost.

How much will this outing cost?
Answer:
$28

How many groups are there?
Answer:
4

What does each group comprise?
Answer:
$5 + $2, or $7

Example 4.
Expressions with Parentheses and Exponents
2 × (3 + 4)²
Which value will we evaluate first within the parentheses? Evaluate.
Answer:
First, evaluate 4² which is 16; then, add 3.The value of the parentheses is 19.
2 × (3 + 4²)
2 × (3 + 16)
2 × 19

Evaluate the rest of the expression.
Answer:
2 × 19 = 38

What do you think will happen when the exponent in this expression is outside of the parentheses?
2 × (3 + 4)²

Will the answer be the same?
Answer:
Answers will vary.

Which should we evaluate first? Evaluate.
Answer:
Parentheses
2 × (3 + 4)²
2 × (7)²

What happened differently here than in our last example?
Answer:
The 4 was not raised to the second power because it did not have an exponent. We simply added the values inside the parentheses.

What should our next step be?

We need to evaluate the exponent next.
Answer:
7² = 7 × 7 = 49

Evaluate to find the final answer.
Answer:
2 × 49
98

What do you notice about the two answers?
Answer:
The final answers were not the same.

What was different between the two expressions?
Answer:
Answers may vary. In the first problem, a value inside the parentheses had an exponent, and that value was evaluated first because it was inside of the parentheses. In the second problem, the exponent was outside of the parentheses, which made us evaluate what was in the parentheses first; then, we raised that value to the power of the exponent.

What conclusions can you draw about evaluating expressions with parentheses and exponents?
Answer:
Answers may vary. Regardless of the location of the exponent in the expression, evaluate the parentheses first. Sometimes there will be values with exponents inside the parentheses. If the exponent is outside the parentheses, evaluate the parentheses first, and then evaluate to the power of the exponent.

Eureka Math Grade 6 Module 4 Lesson 6 Exercise Answer Key

Exercise 1.
4 + 2 × 7
Answer:
4 + 14
18

Exercise 2.
36 ÷ 3 × 4
Answer:
12 × 4
48

Exercise 3.
20 − 5 × 2
Answer:
20 − 10
10

Exercise 4.
90 − 5² × 3
Answer:
90 − 25 × 3
90 − 75
15

Exercise 5.
4³ + 2 × 8
Answer:
64 + 2 × 8
64 + 16
80

Exercise 6.
2 + (9²)
Answer:
2 + (81 – 4)
2 + 77
79

Exercise 7.
2 . (13 + 5 – 14 ÷ (3 + 4)
Answer:
2 . (13 + 5 – 14 ÷ 7)
2 . (13 + 5 – 2)
2 . 16
32

Exercise 8.
7 + (12 – 3²)
Answer:
7 + (12 – 9)
7 + 3
10

Exercise 9.
7 + (12 – 3)²
Answer:
7 + 9²
7 + 81
88

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

Evaluate each expression.

Question 1.
3 × 5 + 2 × 8 + 2
Answer:
15 + 16 + 2
33

Question 2.
($1.75 + 2 × $0.25 + 5 × $0.05) × 24
Answer:
($1.75 + $0.50 + $0.25) × 24
$2.50 × 24
$60.00

Question 3.
(2 × 6) + (8 × 4) + 1
Answer:
12 + 32 + 1
45

Question 4.
((8 × 1.95) + (3 × 2.95) + 10.95) × 1.06
Answer:
(15.6 + 8.85 + 10.99) × 1.06
35.4 × 1.06
37.54

Question 5.
((12 ÷ 3)² – (18 ÷ 3²)) × (4 ÷ 2)
Answer:
(4² − (18 ÷ 9)) × (4 ÷ 2)
(16 – 2) × 2
14 × 2
28

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

Question 1.
Evaluate this expression: 39 ÷ (2 + 1) – 2 × (4 + 1).
Answer:
39 ÷ 3 − 2 × 5
13 − 10
3

Question 2.
Evaluate this expression: 12 × (3 + 2² ÷ 2 − 10
Answer:
12 × (3 + 4) ÷2 − 10
12 × 7 ÷ 2 − 10
84 ÷ 2 − 10
42 − 10
32

Question 3.
Evaluate this expression: 12 × (3 + 2)² ÷ 2 – 10.
Answer:
12 × 5² ÷ 2 – 10
12 × 25 ÷ 2 − 10
300 ÷ 2 − 10
150 − 10
140

Engage NY Eureka Math 6th Grade Module 4 Lesson 5 Answer Key

Eureka Math Grade 6 Module 4 Lesson 5 Example Answer Key

Write each expression in exponential form.
Example 1.
5 × 5 × 5 × 5 × 5 =
Answer:
5⁵

Example 2.
2 × 2 × 2 × 2 =
Answer:
2⁴

Write each expression in expanded form.
Example 3.
8³ =
Answer:
8 × 8 × 8

Example 4.
10⁶ =
Answer:
10 × 10 × 10 × 10 × 10 × 10

Example 5.
g³ =
Answer:
g × g × g

Go back to Examples 1 – 4, and use a calculator to evaluate the expressions.

Example 1.
5 × 5 × 5 × 5 × 5 = 5⁵
Answer:
3,125

Example 2.
2 × 2 × 2 × 2 = 2⁴
Answer:
16

Example 3.
8³ = 8 × 8 × 8
Answer:
512

Example 4.
10⁶ = 10 × 10 × 10 × 10 × 10 × 10
Answer:
1,000,000

Example 5.
What is the difference between 3g and g³?
Answer:
3g = g + g + g or 3 times g; g³ = g × g × g

Example 6.
Write the expression in expanded form, and then evaluate.
(3.8)⁴ =
Answer:
3.8 × 3.8 × 3.8 × 3.8 = 208.5136

Example 7.
Write the expression in exponential form, and then evaluate.
Answer:
2.1 × 2.1 = (2.1)² = 4.41

Example 8.
Write the expression in exponential form, and then evaluate.
0.75 × 0.75 × 0.75
Answer:
= (0.75)³ = 0.421875

The base number can also be a fraction. Convert the decimals to fractions in Examples 7 and 8 and evaluate. Leave your answer as a fraction. Remember how to multiply fractions!

Example 7.
Answer:
\(\frac{21}{10} \times \frac{21}{10}=\left(\frac{21}{10}\right)^{2}=\frac{441}{100}=4 \frac{41}{100}\)

Example 8.
Answer:
\(\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4}=\left(\frac{3}{4}\right)^{3}=\frac{27}{64}\)

Example 9.
Write the expression in exponential form, and then evaluate.
Answer:
\(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}\)

Example 10.
Write the expression in expanded form, and then evaluate.
Answer:
\(\left(\frac{2}{3}\right)^{2}=\frac{2}{3} \times \frac{2}{3}=\frac{4}{9}\)

Eureka Math Grade 6 Module 4 Lesson 5 Exercise Answer Key

Exercise 1.
Fill in the missing expressions for each row. For whole number and decimal bases, use a calculator to find the standard form of the number. For fraction bases, leave your answer as a fraction.

Answer:

Exercise 2.
Write five cubed in all three forms: exponential form, expanded form, and standard form.
Answer:
5³; 5 × 5 × 5; 125

Exercise 3.
Write fourteen and seven-tenths squared in all three forms.
Answer:
(14.7)²; 14.7 × 14.7; 216.09

Exercise 4.
One student thought two to the third power was equal to six. What mistake do you think he made, and how would you help him fix his mistake?
Answer:
The student multiplied the base, 2, by the exponent, 3. This is wrong because the exponent never multiplies the base; the exponent tells how many copies of the base are to be used as factors.

Eureka Math Grade 6 Module 4 Lesson 5 Problem Set Answer Key

Question 1.
Complete the table by filling in the blank cells. Use a calculator when needed.

Answer:

Question 2.
Why do whole numbers raised to an exponent get greater, while fractions raised to an exponent get smaller?
Answer:
As whole numbers are multiplied by themselves, products are larger because there are more groups. As fractions of fractions are taken, the product is smaller. A part of a part is less than how much we started with.

Question 3.
The powers of 2 that are in the range 2 through 1,000 are 2, 4, 8, 16, 32, 64, 128, 256, and 512. Find all the powers of that are in the range 3 through 1,000.
Answer:
3, 9, 27, 81, 243, 729

Question 4.
Find all the powers of 4 in the range 4 through 1,000.
Answer:
4, 16, 64, 256

Question 5.
Write an equivalent expression for n × a using only addition.
Answer:

Question 6.
Write an equivalent expression for w^b using only multiplication.
Answer:

a. Explain what w is in this new expression.
Answer:
w is the factor that will be repeatedly multiplied by itself.

b. Explain what b is in this new expression.
Answer:
b is the number of times w will be multiplied.

Question 7.
What is the advantage of using exponential notation?
Answer:
It is a shorthand way of writing a multiplication expression if the factors are all the same.

Question 8.
What is the difference between 4x and x⁴? Evaluate both of these expressions when x = 2
Answer:
4x means four times x; this is the same as x + x + x + x. On the other hand, x⁴ means x to the fourth power, or x × x × x × x.
When x = 2, 4x = 4 × 2 = 8.
When x = 2, x⁴ = 2 × 2 × 2 × 2 = 16.

Eureka Math Grade 6 Module 4 Lesson 5 Exit Ticket Answer Key

Question 1.
What is the difference between 6z and z⁶?
Answer:
6z = z + z + z + z + z + z or 6 times z; z⁶ = z × z × z × z × z × z

Question 2.
Write 10³ as a multiplication expression having repeated factors.
Answer:
10 × 10 × 10

Question 3.
Write 8 × 8 × 8 × 8 using an exponent.
Answer:
8⁴

Eureka Math Grade 6 Module 4 Lesson 5 Opening Exercise Answer Key

As you evaluate these expressions, pay attention to how you arrive at your answers.

Question 1.
4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4
Answer:
4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 4 × 10
= 4¹⁰

Question 2.
9 + 9 + 9 + 9 + 9
Answer:
9 + 9 + 9 + 9 + 9 = 9 × 5
= 9⁵

Question 3.
10 + 10 + 10 + 10 + 10
Answer:
10 + 10 + 10 + 10 + 10 = 10 × 5
= 10⁵

Eureka Math Grade 6 Module 4 Lesson 5 Multiplication of Decimals Answer Key

Progression of Exercises

Question 1.
0.5 × 0.5 =
Answer:
0.25

Question 2.
0.6 × 0.6 =
Answer:
0.36

Question 3.
0.7 × 0.7
Answer:
0.49

Question 4.
0.5 × 0.6 =
Answer:
0.3

Question 5.
1.5 × 1.5 =
Answer:
0.25

Question 6.
2.5 × 2.5 =
Answer:
6.25

Question 7.
0.25 × 0.25 =
Answer:
0.0625

Question 8.
0.1 × 0.1 =
Answer:
0.01

Question 9.
0.1 × 123.4 =
Answer:
12.34

Question 10.
0.01 × 123.4 =
Answer:
1.234

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

Eureka Math Grade 6 Module 4 Lesson 4 Exercise Answer Key

Exercise 1.
Build subtraction equations using the indicated equations. The first example has been completed for you.

Answer:

Answer:

Exercise 2.
Answer each question using what you have learned about the relationship of division and subtraction.
a. If 12 ÷ x = 3 how many times would x have to be subtracted from 12 in order for the answer to be zero? What is the value of x?
Answer:
3; x = 4

b. 36 − f − f − f − f = 0. Write a division sentence for this repeated subtraction sentence. What is the value
of x?
Answer:
36 ÷ 4 = f or 36 ÷ f = 4; f = 9

c. If 24 ÷ b = 12, which number is being subtracted 12 times in order for the answer to be zero?
Answer:
Two

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

Build subtraction equations using the indicated equations.

Answer:

Answer:

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

Question 1.
Represent 56 ÷ 8 = 7 using subtraction. Explain your reasoning.
Answer:
56 – 7 – 7 – 7 – 7 – 7 – 7 – 7 – 7 = 0 because
56 – 7 = 49; 49 – 7 = 42; 42 – 7 = 35; 35 – 7 = 28; 28 – 7 = 21; 21 – 7 = 14; 14 – 7 = 7; 7 – 7 = 0.

OR

56 – 8 – 8 – 8 – 8 – 8 – 8 – 8 = 0 because
56 – 8 = 48; 48 – 8 = 40; 40 – 8 = 32; 32 – 8 = 24; 24 – 8 = 16; 16 – 8 = 8; 8 – 8 = 0.

Question 2.
Explain why 30 ÷ x = 6 is the same as 30 – x – x – x – x – x – x = 0. What is the value of x in this example?
Answer:
30 ÷ 5 = 6, so x = 5. When I subtract 5 from 30 six times, the result is zero. Division is a repeat operation of
subtraction.

Engage NY Eureka Math 6th Grade Module 4 Lesson 3 Answer Key

Eureka Math Grade 6 Module 4 Lesson 3 Exercise Answer Key

Exercise 1.
Write the addition sentence that describes the model and the multiplication sentence that describes the model.

Answer:
5 + 5 + 5 and 3 × 5

Exercise 2.
Write an equivalent expression to demonstrate the relationship of multiplication and addition.
a. 6 + 6
Answer:
2 × 6

b. 3 + 3 + 3 + 3 + 3 + 3
Answer:
6 × 3

c. 4 + 4 + 4 + 4 + 4
Answer:
5 × 4

d. 6 × 2
Answer:
2 + 2 + 2 + 2 + 2+ 2

e. 4 × 6
Answer:
6 + 6 + 6 + 6

f. 3 × 9
Answer:
9 + 9 + 9

g. h + h + h + h + h
Answer:
5h

h. 6y
Answer:
y + y + y + y + y + y

Exercise 3.
Roberto is not familiar with tape diagrams and believes that he can show the relationship of multiplication and addition on a number line. Help Roberto demonstrate that the expression 3 × 2 is equivalent to 2 + 2 + 2 on a number line.
Answer:
Possible answer: The first number line shows that there are 3 groups of 2, resulting in 6. The second number line shows the sum of 2 + 2 + 2, resulting in 6.

Since both number lines start at 0 and end at 6, the expressions are equivalent.

Exercise 4.
Tell whether the following equations are true or false. Then, explain your reasoning.
a. x + 6g − 6g = x
The equation is true because it demonstrates the addition identity.

b. 2f − 4e + 4e = 2f
Answer:
The equation is true because it demonstrates the subtraction identity.

Exercise 5.
Write an equivalent expression to demonstrate the relationship between addition and multiplication.
a. 6 + 6 + 6 + 6 + 6 + 6 + 6
Answer:
4 × 6 + 3 × 4

b. d + d + d + w + w + w + w + w
Answer:
3d + 5w

c. a + a + b + b + b + c + c + c + c
Answer:
2a + 3b + 4c

Eureka Math Grade 6 Module 4 Lesson 3 Problem Set Answer Key

Write an equivalent expression to show the relationship of multiplication and addition.

Question 1.
10 + 10 + 10
Answer:
3 × 10

Question 2.
4 + 4 + 4 + 4 + 4 + 4 + 4
Answer:
7 × 4

Question 3.
8 × 2
Answer:
2 + 2 + 2 + 2 + 2 + 2 + 2 + 2

Question 4.
3 × 9
Answer:
9 + 9 + 9

Question 5.
6m
Answer:
m + m + m + m + m + m

Question 6.
d + d + d + d + d
Answer:
5d

Eureka Math Grade 6 Module 4 Lesson 3 Exit Ticket Answer Key

Write an equivalent expression to show the relationship of multiplication and addition.

Question 1.
8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8
Answer:
9 × 8

Question 2.
4 × 9
Answer:
9 + 9 + 9 + 9

Question 3.
6 + 6 + 6
Answer:
3 × 6

Question 4.
7h
Answer:
h + h + h + h + h + h + h

Question 5.
j + j + j + j + j
Answer:
5j

Question 6.
u + u + u + u + u + u + u + u + u + u
Answer:
10u

Eureka Math Grade 6 Module 4 Lesson 3 Opening Exercise Answer Key

Write two different expressions that can be depicted by the tape diagram shown. One expression should include addition, while the other should include multiplication.

a.

Answer:
Possible answers: 3 + 3 + 3 or 3 × 3

b.

Answer:
Possible answers: 8 + 8 or 2 × 8

c.

Answer:
Possible answers: 5 + 5 + 5 or 3 × 5

Engage NY Eureka Math 6th Grade Module 4 Lesson 2 Answer Key

Eureka Math Grade 6 Module 4 Lesson 2 Problem Set Answer Key

Question 1.
Fill in each blank to make each equation true.
a. 132 ÷ 3 × 3 = __________
Answer:
132

b. _______ ÷ 25 × 25 = 225
Answer:
225

c. 56 × ______ ÷ 8 = 56
Answer:
8

d. 452 × 12 ÷ ___ = 452
Answer:
12

Question 2
How is the relationship of addition and subtraction similar to the relationship of multiplication and division?
Answer:
Possible answer: Both relationships create identities.

Eureka Math Grade 6 Module 4 Lesson 2 Exit Ticket Answer Key

Question 1.
Fill In the blanks to make each equation true.
a. 12 ÷ 3 × __________ = 12
Answer:
3

b. f × h ÷ h = __________
Answer:
f

c. 45 × _________ ÷ 15 = 45
Answer:
15

d. __________ ÷ r × r = p
Answer:
p

Question 2.
Draw a series of tape diagrams to represent the following number sentences.
a. 12 ÷ 3 × 3 = 12
Answer:

b. 4 × 5 ÷ 5 = 4
Answer:

Eureka Math Grade 6 Module 4 Lesson 2 Opening Exercise Answer Key

Draw a pictorial representation of the division and multiplication problems using a tape diagram.

a. 8 ÷ 2
Answer:

b. 3 × 2
Answer:

Eureka Math Grade 6 Module 4 Lesson 2 Exploratory Challenge Answer Key

Work in pairs or small groups to determine equations to show the relationship between multiplication and division. Use tape diagrams to provide support for your findings.

Question 1.
Create two equations to show the relationship between multiplication and division. These equations should be identities and Include variables. Use the squares to develop these equations.
Answer:

Question 2.
Write your equations on large paper. Show a series of tape diagrams to defend each of your equations.
Answer:
Only one number sentence is shown there; the second number sentence and series of tape diagrams are included in the optional Discussion.
Possible answer: a × b ÷ b = a

Possible answer: a ÷ b × b = a

Use the following rubric to critique other posters.
1. Name of the group you are critiquing
2. Equation you are critiquing
3. Whether or not you believe the equations are true and reasons why
Answer:

Eureka Math Grade 6 Module 4 Lesson 2 Division of Fractions Answer Key

Divisions of Fractions – Round 1

Question 1.
9 ones ÷ 3 ones
Answer:
\(\frac{9}{3}\) = 3

Question 2.
9 ÷ 3
Answer:
\(\frac{9}{3}\) = 3

Question 3.
9 tens ÷ 3 tens
Answer:
\(\frac{9}{3}\) = 3

Question 4.
90 ÷ 30
Answer:
\(\frac{9}{3}\) = 3

Question 5.
9 hundreds ÷ 3 hundreds
Answer:
\(\frac{9}{3}\) = 3

Question 6.
900 ÷ 300
Answer:
\(\frac{9}{3}\) = 3

Question 7.
9 halves ÷ 3 halves
Answer:
\(\frac{9}{3}\) = 3

Question 8.
\(\frac{9}{2} \div \frac{3}{2}\)
Answer:
\(\frac{9}{3}\) = 3

Question 9.
9 fourths ÷ 3 fourths
Answer:
\(\frac{9}{3}\)

Question 10.
\(\frac{9}{4} \div \frac{3}{4}\)
Answer:
\(\frac{9}{3}\) = 3

Question 11.
\(\frac{9}{8} \div \frac{3}{8}\)
Answer:
\(\frac{9}{3}\) = 3

Question 12.
\(\frac{2}{3} \div \frac{1}{3}\)
Answer:
\(\frac{2}{1}\) = 2

Question 13.
\(\frac{1}{3} \div \frac{2}{3}\)
Answer:
\(\frac{1}{2}\)

Question 14.
\(\frac{6}{7} \div \frac{2}{7}\)
Answer:
\(\frac{6}{2}\) = 3

Question 15.
\(\frac{5}{7} \div \frac{2}{7}\)
Answer:
\(\frac{5}{2}=2 \frac{1}{2}\)

Question 16.
\(\frac{3}{7} \div \frac{4}{7}\)
Answer:
\(\frac{3}{4}\)

Question 17.
\(\frac{6}{10} \div \frac{2}{10}\)
Answer:
\(\frac{6}{2}\) = 3

Question 18.
\(\frac{6}{10} \div \frac{4}{10}\)
Answer:
\(\frac{6}{4}=1 \frac{1}{2}\)

Question 19.
\(\frac{6}{10} \div \frac{8}{10}\)
Answer:
\(\frac{6}{8}=\frac{3}{4}\)

Question 20.
\(\frac{7}{12} \div \frac{2}{12} \)
Answer:
\(\frac{7}{2}=3 \frac{1}{2}\)

Question 21.
\(\frac{6}{12} \div \frac{9}{12}\)
Answer:
\(\frac{6}{9}=\frac{2}{3}\)

Question 22.
\(\frac{4}{12} \div \frac{11}{12}\)
Answer:
\(\frac{4}{11}\)

Question 23.
\(\frac{6}{10} \div \frac{4}{10}\)
Answer:
\(\frac{6}{4}=1 \frac{1}{2}\)

Question 24.
\(\frac{6}{10} \div \frac{2}{5}=\frac{6}{10} \div \frac{ }{10}\)
Answer:
\(\frac{6}{4}=1 \frac{1}{2}\)

Question 25.
\(\frac{10}{12} \div \frac{5}{12}\)
Answer:
\(\frac{10}{5}=2\)

Question 26.
\(\frac{5}{6} \div \frac{5}{12}=\frac{ }{12} \div \frac{5}{12}\)
Answer:
\(\frac{10}{5}=2\)

Question 27.
\(\frac{10}{12} \div \frac{3}{12}\)
Answer:
\(\frac{10}{3}=3 \frac{1}{3}\)

Question 28.
\(\frac{10}{12} \div \frac{1}{4}=\frac{10}{12} \div \frac{ }{12}\)
Answer:
\(\frac{10}{3}=3 \frac{1}{3}\)

Question 29.
\(\frac{5}{6} \div \frac{3}{12}=\frac{ }{12} \div \frac{3}{12}\)
Answer:
\(\frac{10}{3}=3 \frac{1}{3}\)

Question 30.
\(\frac{5}{10} \div \frac{2}{10}\)
Answer:
\(\frac{5}{2}=2 \frac{1}{2}\)

Question 31.
\(\frac{5}{10} \div \frac{1}{5}=\frac{5}{10} \div \frac{ }{10}\)
Answer:
\(\frac{5}{2}=2 \frac{1}{2}\)

Question 32.
\(\frac{1}{2} \div \frac{2}{10}=\frac{ }{10} \div \frac{2}{10}\)
Answer:
\(\frac{5}{2}=2 \frac{1}{2}\)

Question 33.
\(\frac{1}{2} \div \frac{2}{4}\)
Answer:
\(\frac{2}{2}=1\)

Question 34.
\(\frac{3}{4} \div \frac{2}{8}\)
Answer:
3

Question 35.
\(\frac{1}{2} \div \frac{3}{8}\)
Answer:
\(\frac{4}{3}=1 \frac{1}{3}\)

Question 36.
\(\frac{1}{2} \div \frac{1}{5}=\frac{1}{10} \div \frac{ }{10}\)
Answer:
\(\frac{5}{2}=2 \frac{1}{2}\)

Question 37.
\(\frac{2}{4} \div \frac{1}{3}\)
Answer:
\(\frac{6}{4}=1 \frac{1}{2}\)

Question 38.
\(\frac{1}{4} \div \frac{4}{6}\)
Answer:
\(\frac{3}{8}\)

Question 39.
\(\frac{3}{4} \div \frac{2}{6}\)
Answer:
\(\frac{9}{4}=2 \frac{1}{4}\)

Question 40.
\(\frac{5}{6} \div \frac{1}{4}\)
Answer:
\(\frac{10}{3}=3 \frac{1}{3}\)

Question 41.
\(\frac{2}{9} \div \frac{5}{6}\)
Answer:
\(\frac{4}{15}\)

Question 42.
\(\frac{5}{9} \div \frac{1}{6}\)
Answer:
\(\frac{15}{3}\) = 5

Question 43.
\(\frac{1}{2} \div \frac{1}{7}\)
Answer:
\(\frac{7}{2}=3 \frac{1}{2}\)

Question 44.
\(\frac{5}{7} \div \frac{1}{2}\)
Answer:
\(\frac{10}{7}=1 \frac{3}{7}\)

Divisions of Fractions – Round 2

Question 1.
12 ones ÷ 2 ones
Answer:
\(\frac{12}{2}\) = 6

Question 2.
12 ÷ 2
Answer:
\(\frac{12}{2}\) = 6

Question 3.
12 tens ÷ 2 tens
Answer:
\(\frac{12}{2}\) = 6

Question 4.
120 ÷ 20
Answer:
\(\frac{12}{2}\) = 6

Question 5.
12 hundreds ÷ 2 hundreds
Answer:
\(\frac{12}{2}\) = 6

Question 6.
1,200 ÷ 200
Answer:
\(\frac{12}{2}\) = 6

Question 7.
12 halves ÷ 2 halves
Answer:
\(\frac{12}{2}\) = 6

Question 8.
\(\frac{12}{2} \div \frac{2}{2}\)
Answer:
\(\frac{12}{2}\) =6

Question 9.
12 fourths ÷ 3 fourths
Answer:
\(\frac{12}{3}\) = 4

Question 10.
\(\frac{12}{4} \div \frac{3}{4}\)
Answer:
\(\frac{12}{3}\) = 4

Question 11.
\(\frac{12}{8} \div \frac{3}{8}\)
Answer:
\(\frac{12}{3}\) = 4

Question 12.
\(\frac{2}{4} \div \frac{1}{4}\)
Answer:
\(\frac{2}{1}\) = 2

Question 13.
\(\frac{1}{4} \div \frac{2}{4}\)
Answer:
\(\frac{1}{2}\)

Question 14.
\(\frac{4}{5} \div \frac{2}{5}\)
Answer:
\(\frac{4}{2}\) = 2

Question 15.
\(\frac{2}{5} \div \frac{4}{5}\)
Answer:
\(\frac{2}{4}=\frac{1}{2}\)

Question 16.
\(\frac{3}{5} \div \frac{4}{5}\)
Answer:
\(\frac{3}{4}\)

Question 17.
\(\frac{6}{8} \div \frac{2}{8}\)
Answer:
\(\frac{6}{2}\) = 3

Question 18.
\(\frac{6}{8} \div \frac{4}{8}\)
Answer:
\(\frac{6}{4}=1 \frac{1}{2}\)

Question 19.
\(\frac{6}{8} \div \frac{5}{8}\)
Answer:
\(\frac{6}{5}=1 \frac{1}{5}\)

Question 20.
\(\frac{6}{10} \div \frac{2}{10}\)
Answer:
\(\frac{6}{2}\) = 3

Question 21.
\(\frac{7}{10} \div \frac{8}{10}\)
Answer:
\(\frac{7}{8}\)

Question 22.
\(\frac{4}{10} \div \frac{7}{10}\)
Answer:
\(\frac{4}{7}\)

Question 23.
\(\frac{6}{12} \div \frac{4}{12}\)
Answer:
\(\frac{6}{4}=1 \frac{1}{2}\)

Question 24.
\(\frac{6}{12} \div \frac{2}{6}=\frac{6}{12} \div \frac{ }{12}\)
Answer:
\(\frac{6}{4}=1 \frac{1}{2}\)

Question 25.
\(\frac{8}{14} \div \frac{7}{14}\)
Answer:
\(\frac{8}{7}=1 \frac{1}{7}\)

Question 26.
\(\frac{8}{14} \div \frac{1}{2}=\frac{8}{14} \div \frac{ }{14}\)
Answer:
\(\frac{8}{7}=1 \frac{1}{7}\)

Question 27.
\(\frac{11}{14} \div \frac{2}{14}\)
Answer:
\(\frac{11}{2}=5 \frac{1}{2}\)

Question 28.
\(\frac{11}{14} \div \frac{1}{7}=\frac{11}{14} \div \frac{ }{14}\)
Answer:
\(\frac{11}{2}=5 \frac{1}{2}\)

Question 29.
\(\frac{1}{7} \div \frac{6}{14}=\frac{ }{14} \div \frac{6}{14}\)
Answer:
\(\frac{2}{6}=\frac{1}{3}\)

Question 30.
\(\frac{7}{18} \div \frac{3}{18}\)
Answer:
\(\frac{7}{3}=2 \frac{1}{3}\)

Question 31.
\(\frac{7}{18} \div \frac{1}{6}=\frac{7}{18} \div \frac{ }{18}\)
Answer:
\(\frac{7}{3}=2 \frac{1}{3}\)

Question 32.
\(\frac{1}{3} \div \frac{12}{18}=\frac{ }{18} \div \frac{12}{18}\)
Answer:
\(\frac{6}{12}=\frac{1}{2}\)

Question 33.
\(\frac{1}{6} \div \frac{4}{18}\)
Answer:
\(\frac{3}{4}\)

Question 34.
\(\frac{4}{12} \div \frac{8}{6}\)
Answer:
\(\frac{4}{16}=\frac{1}{4}\)

Question 35.
\(\frac{1}{3} \div \frac{3}{15}\)
Answer:
\(\frac{5}{3}=1 \frac{2}{3}\)

Question 36.
\(\frac{2}{6} \div \frac{1}{9}=\frac{1}{18} \div \frac{ }{18}\)
Answer:
\(\frac{6}{2}\)

Question 37.
\(\frac{1}{6} \div \frac{4}{9}\)
Answer:
\(\frac{3}{8}\)

Question 38.
\(\frac{2}{3} \div \frac{3}{4}\)
Answer:
\(\frac{8}{9}\)

Question 39.
\(\frac{1}{3} \div \frac{3}{5}\)
Answer:
\(\frac{5}{9}\)

Question 40.
\(\frac{1}{7} \div \frac{1}{2}\)
Answer:
\(\frac{2}{7}\)

Question 41.
\(\frac{5}{6} \div \frac{2}{9}\)
Answer:
\(\frac{15}{4}=3 \frac{3}{4}\)

Question 42.
\(\frac{5}{9} \div \frac{2}{6}\)
Answer:
\(\frac{10}{6}=1 \frac{2}{3}\)

Question 43.
\(\frac{5}{6} \div \frac{4}{9}\)
Answer:
\(\frac{15}{8}=1 \frac{7}{8}\)

Question 44.
\(\frac{1}{2} \div \frac{4}{5}\)
Answer:
\(\frac{5}{8}\)

Engage NY Eureka Math Grade 6 Module 4 Mid Module Assessment Answer Key

Eureka Math Grade 6 Module 4 Mid Module Assessment Answer Key

Question 1.
Yolanda is planning out her vegetable garden. She decides that her garden will be square. Below are possible sizes of the garden she will create.

a. Complete the table by continuing the pattern.

Answer:

b. Yolanda decides the length of her square vegetable garden will be 17 ft. She calculates that the area of the garden is 34 ft². Determine if Yolanda’s calculation is correct. Explain.
Answer:
A = l.w
A = 17 ft. . 17 ft.
A = 17² ft²
A = 289 ft²

Yolanda is incorrect. Instead of finding 17² (or 17 . 17), she multiplied 17 . 2, the base times the exponent.

Question 2.
Yolanda creates garden cubes to plant flowers. She will fill the cubes with soil and needs to know the amount of soil that will fill each garden cube. The volume of a cube is determined by the following formula: V = s³, where s equals the side length.

a. Represent the volume, in cubic inches, of the garden cube above using a numerical expression.
Answer:
32³ or 32 . 32 . 32

b. Evaluate the expression to determine the volume of the garden cube and the amount of soil, in cubic inches, she will need for each cube.
Answer:
32³ = 32 . 32 . 32 = 32,768

Question 3.
Explain why \(\left(\frac{1}{2}\right)^{4}=\frac{1}{16}\)
Answer:

Question 4.
Yolanda is building a patio in her backyard. She is interested in using both brick and wood for the flooring of the patio. Below is the plan she has created for the patio. All measurements are in feet.

a. Create an expression to represent the area of the patio.

Answer:
12.5(3 + x) or 37.5 + 12.5x

b. Yolanda’s husband develops another plan for the patio because he prefers the patio to be much wider than Yolanda’s plan. Determine the length of the brick section and the length of the wood section. Then, use the dimensions to write an expression that represents the area of the entire patio.

Answer:

24(2x + 4)

Question 5.
The landscaper hired for Yolanda’s lawn suggests a patio that has the same measure of wood as it has brick.

a. Express the perimeter of the patio in terms of x, first using addition and then using multiplication.
Answer:
x + x + x + 2x + 2x + x + x + x = 10x

b. Use substitution to determine if your expressions are equivalent. Explain.
Answer:
The expressions are equivalent.
Let x = 2
x + x + x + 2x + 2x + x + x + x
2 + 2 + 2 + 4 + 4 + 2 + 2 + 2 = 20
OR
10x = 10(2) = 20

Question 6.
Elena and Jorge have similar problems and find the same answer. Each determines that the solution to the problem is 24.
Elena: (14 + 42) ÷ 7 + 4²
Jorge: 14 + (42 ÷ 7) + 4²

a. Evaluate each expression to determine if both Elena and Jorge are correct.
Answer:
Elena: (14 + 42) ÷ 7 + 4²
= 56 ÷ 7 + 4²
= 56 ÷ 7 + 16
= 8 + 16 = 24

Jorge: 14 + (42 ÷ 7) + 4²
= 14 + 6 + 4²
= 14 + 6 + 16
= 20 + 16 = 36

b. Why would each find the solution of 24? What mistakes were made, if any?
Answer:
Elena followed the order of operations correctly. Jorge made a mistake. He added 14 + 42 first and then divided the sum by 7 to get 8. He did not follow the correct order of operations. He should have evaluated the parentheses first.

Question 7.
Jackson gave Lena this expression to evaluate: 14(8 + 12). Lena said that to evaluate the expression was simple; just multiply the factors 14 and 20. Jackson told Lena she was wrong. He solved it by finding the product of 14 and 8 and then adding that to the product of 14 and 12.

a. Evaluate the expression using each student’s method.

Answer:

b. Who was right in this discussion? why?
Answer:
They both were correct. Lena used the order of operations correctly to determine 280. Jackson used the distributive property to determine 280.

Engage NY Eureka Math Grade 6 Module 4 End of Module Assessment Answer Key

Eureka Math Grade 6 Module 4 End of Module Assessment Answer Key

Question 1.
Gertrude is deciding which cell phone plan is the best deal for her to buy. SuperCell charges a monthly fee of $10 and also charges $0.15 per call. She makes a note that the equation is M = 0.15C + 10, where M is the monthly charge, in dollars, and C is the number of calls placed. Global Cellular has a plan with no monthly fee but charges $0.25 per call. She makes a note that the equation is M = 0.25C, where M is the monthly charge, in dollars, and C is the number of calls placed. Both companies offer unlimited text messages.

a. Make a table for both companies showing the cost of service, M, for making from 0 to 200 calls per month. Use multiples of 20.

Answer:

b. Construct a graph for the two equations on the same graph. Use the number of calls, C, as the independent variable and the monthly charge, in dollars, M, as the dependent variable.

Answer:

c. Which cell phone plan is the best deal for Gertrude? Defend your answer with specific examples.
Answer:
The best deal depends on the number of calls placed in a month. The breakpoint is 100 calls per month. Supercell is a better deal if the number of monthly calls is less than 100. Global cellular is a better deal if the number of monthly calls is greater than 100.

Question 2.
Sadie is saving her money to buy a new pony, which costs $600. She has already saved $75. She earns $50 per week working at the stables and wonders how many weeks it will take to earn enough for a pony of her own.

a. Make a table showing the week number, W, and total savings, in dollars, S, in Sadie’s savings account.

Answer:

50w + 75

b. Show the relationship between the number of weeks and Sadie’s savings using an expression.
Answer:
If Sadie works 11 weeks, she will earn $625, which is $25 more than the cost of the pony.

Question 3.
The elevator at the local mall has a weight limit of 1,800 pounds and requires that the maximum person allowance be no more than nine people.

a. Let x represent the number of people. Write an inequality to describe the maximum allowance of people allowed in the elevator at one time.
Answer:
0 ≤ x ≤ 9

b. Draw a number line diagram to represent all possible solutions to part (a).
Answer:

c. Let w represent the amount of weight, in pounds. Write an inequality to describe the maximum weight allowance in the elevator at one time.
Answer:
0 ≤ W ≤ 1,800

d. Draw a number line to represent all possible solutions to part (c).
Answer:

Question 4.
Devin’s football team carpools for practice every week. This week is his parents’ turn to pick up team members and take them to the football field. While still staying on the roads, Devin’s parents always take the shortest route in order to save gasoline. Below is a map of their travels. Each grid line represents a street and the same distance.

Devin’s father checks his mileage and notices that he drove 18 miles between his house and Stop 3.

a. Create an equation and determine the amount of miles each gridline represents.
Answer:
Let G represent the number of grid lines passed on the map.
9G = 18
\(\frac{9G}{9}=\frac{18}{9}\)
C = 2 miles.

b. Using this information, determine how many total miles Devin’s father will travel from home to the football field, assuming he made every stop. Explain how you determined the answer.
Answer:
15 G = miles
15 (2 miles) = 30 miles

c. At the end of practice, Devin’s father dropped off team members at each stop and went back home. How many miles did Devin’s father travel altogether?
Answer:
30 G = miles
30(2 miles) = 60 miles

Question 5.
For a science experiment, Kenneth reflects a beam of a mirror. He is measuring the missing angle created when the light reflects off the mirror. (Note: The figure is not drawn to scale.)

Use an equation to determine the missing angle, labeled x in the diagram.
Answer:
A straight angle measures 180°
51° + x° + 51° = 180°
x° + 102° = 180°
x° + 102° – 102° = 180° – 102°
x° = 78°

Engage NY Eureka Math Grade 6 Module 5 End of Module Assessment Answer Key

Eureka Math Grade 6 Module 5 End of Module Assessment Answer Key

Question 1.
The juice box pictured below is 4 inches high, 3 inches long, and 2 inches wide.

a. In the grid above, the distance between grid lines represents one inch. Use the grid paper to sketch the net of the juice box.
Answer:

b. Find the surface area of the juice box. Show your work.
Answer:
SA = 2(l.w + l.h + w.h)
SA = 2(3 in. . 2 in. + 3 in. . 4 in. + 2 in. . 4 in.)
SA = 2(6 in² + 12 in² + 8 in²)
SA = 2(26 in²)
SA = 52 in²

c. Find the volume of the juice box. Show your work.
Answer:
V = l.w.h
V = 3 in. . 2 in. . 4 in.
V = 24 in²

Question 2.
The Cubic Crystal Company has a new Crystal Cube they want to sell. The packaging manager insists that the cubes be arranged to form a rectangular prism and that the package be designed to hold the Crystal Cubes exactly, with no leftover packaging. Each Crystal Cube measures 1 in. × 1 in. × 1 in. There are 24 Crystal Cubes to be sold in a box.

a. What are the dimensions of the possible box designs in inches?

Answer:

b. Which Crystal Cube box design will use the least amount of cardboard for packaging? Justify your answer as completely as you can.

Answer:

The minimum surface area is found to be on the 2 in × 3 in × 4 in box. That box needs the least amount of cardboard.

c. Another type of cube is the Mini Crystal Cube, which has an edge length of inch. What is the volume in cubic inches of one Mini Crystal Cube? Show your work.
Answer:
V = l.w.h
V = \(\frac{3}{4}\) in . \(\frac{3}{4}\) in . \(\frac{3}{4}\) in
V = \(\frac{27}{64}\) in³

Question 3.
Which of these nets can be folded to form a cube?

Answer:

Question 4.
Which box below has the larger surface area?


Answer:
SA = 2(l.w + l.h + w.h)
SA = 2(5 in. . 2 in. + 5 in. . 3in. + 2 in. . 3 in.)
SA = 2(10 in². + 15 in². + 6 in².)
SA = 2(31 in².)
SA = 62 in².

SA = 2(l.w + l.h + w.h)
SA = 2(10 in. . 1 in. + 10 in. 3 in. + 1 in. . 3 in.)
SA = 2(10 in² + 30 in². + 3 in².)
SA = 2(43 in².)
SA = 86 in²
This box has the larger surface area.

Question 5.

a. Draw a polygon in the coordinate plane using the given coordinates.

(4, -4)
(6, -2)
(8, -6)
Answer:

b. Calculate the area of the polygon.
Answer:
Area of Square:
A = l.w or S²
A = 4 u. 4 u or (4 u)²
A = 16 u²

Area of Triangle 1:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) 2 u . 2u
A = 2 u²

Area of Triangle 2:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) 2 u . 4 u
A = 4 u²

Area of Triangle 3:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) . 2 u . 4 u
A = 4 u²

Area of Triangle 4 = 16 u² – 2 u² – 4 u² – 4 u² = 6 u²

Question 6.
Eaglecrest Elementary School is creating a vegetable garden at the school.

a. What is the area of the garden?

Answer:
A = b.h
A = 25 ft . 6 ft
A = 150 ft²

b. After more discussion, Eaglecrest decided to change the location of the garden so that the vegetables can get more sunlight. Below is the new garden.

In which garden can eagle crest students plant more vegetables? Explain your reasoning.
Answer:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\)(7 ft. . 28 ft.)
A = \(\frac{1}{2}\) (194 ft²)
A = 98 ft²

The students of Eaglecrest can plant more vegetables in original garden in the shape of a parallelogram. It has a larger area than the triangular prism.

Engage NY Eureka Math Grade 6 Module 5 Mid Module Assessment Answer Key

Eureka Math Grade 6 Module 5 Mid Module Assessment Answer Key

Question 1.
David is the groundskeeper at Triangle Park, scale shown below.

Answer:

a. David needs to cut the grass four times a month. How many square yards of grass will he cut altogether each month?
Answer:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) 300yd. 50yd
A = \(\frac{1}{2}\) . 15,000 yd²
A = 7,500 yd²
4 . 7,500 yd² = 30,000 yd²

b. During the winter the triangular park and adjacent square parking lot are flooded with water and allowed to freeze so that people can go ice skating. What is the area of the ice?

Answer:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) . 300yd. 50 yd
A = \(\frac{1}{2}\) . 15,000 yd²
A = 7,500 yd²

A = S²
A = (50 yd)²
A = 2,500 yd²

7,500 yd² + 2,500 yd² = 10,000 yd²

Question 2.
Mariska is looking for a new computer table. Below is a sketch of two computer tables she likes when looking at them from above. All measurements are in feet.

a. If Mariska needs to choose the one with the greater area, which one should she choose? Justify your answer with evidence, using coordinates to determine side lengths.
Answer:
Table A:
(7, 15) → (7, 18) = 3 ft.
(7, 18) → (16, 18) = 9 ft.
(16, 18) → (16, 11) = 7 ft.
(16, 11) → (13, 11) = 3 ft.
(13, 11) → (13, 15) = 4 ft.
(13, 15) → (7, 15) = 6 ft.

A = bh
A = 9 ft. . 3 ft.
A = 27 ft²

A = bh
A = 3 ft. . 4 ft.
A = 12 ft²

27 ft² + 12 ft² = 39 ft²

Table B:
(5, 8) → (5, 10) = 2 ft.
(5, 10) → (2, 10) = 7 ft.
(12, 10) → (12, 3) = 7 ft.
(12, 3) → (5, 3) = 7 ft.
(5, 3) → (5, 5) = 2 ft.
(5, 5) → (9, 5) = 4 ft.
(9, 5) → (9, 8) = 3 ft.
(9, 8) → (5, 8) = 4 ft.

A = bh
A = 7 ft. . 2 ft.
A = 14 ft²

A = bh
A = 3 ft. 3 ft
A = 9 ft²

A = bh
A = 7 ft. . 2 ft
A = 14 ft²

14 ft² + 9 ft² + 14 ft² = 37 ft²

Mariska will need to choose Table A because it is the table with the greatest area

b. If Mariska needs to choose the one with the greater perimeter, which one should she choose? Justify your answer with evidence, using coordinates to determine side lengths.

Answer:
Table A:
P = 3 ft. + 9 ft. + 7 ft. + 3 ft. + 4 ft. + 6 ft.
P = 32 ft.

Table B:
P = 2 ft. + 7 ft. + 7 ft. + 7 ft. + 2 ft. + 3 ft. + 4 ft. + 4 ft.
P = 36 ft.
Table B has a large perimeter.

Question 3.
Find the area of the triangular region.

Answer:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) . 5 in. 7 in.
A = \(\frac{1}{2}\) . 35 in²
A = 17.5 in²

Question 4.
The grid below shows a bird’s-eye view of a middle school.

a. Write the coordinates of each point in the table.
Answer:

b. Each space on the grid stands for 10 meters. Find the length of each wall of the school.
Answer:

c. Find the area of the entire building. Show your work.
Answer:
A = bh
A = 100 m . 60 m
A = 6,000 m²

A = bh
A = 30 m . 50 m
A = 1,500 m²

A = bh
A = 20 m . 40 m
A = 800 m²

6,000 m² + 1,500 m² + 800 m² = 8,300 m²

Engage NY Eureka Math Grade 6 Module 5 Lesson 19a Answer Key

Eureka Math Grade 6 Module 5 Lesson 19a Opening Exercise Answer Key

Opening Exercise:

Question 1.
Determine the volume of this aquarium.

Answer:
V = l × w × h
V = 20 in. × 10 in. × 12 in.
V = 2,400 in³

Eureka Math Grade 6 Module 5 Lesson 19a Mathematical Modeling Exercise Answer Key

Mathematical Modeling Exercise: Using Ratios and Unit Rate to Determine Volume

For his environmental science project, Jamie is creating habitats for various wildlife including fish, aquatic turtles, and aquatic frogs. For each of these habitats, he uses a standard aquarium with length, width, and height dimensions measured in inches, identical to the aquarium mentioned in the Opening Exercise. To begin his project, Jamie needs to determine the volume, or cubic inches, of water that can fill the aquarium.

Use the table below to determine the unit rate of gallons/cubic inches.

Answer:

There are 231 cubic inches for every 1 gallon of water. So, the unit rate is 231.

→ Since we determined that for every gallon of water, there are 231 cubic inches, determine how many cubic inches are in the 10 gallons of water that Jamie needs for the fish.
→ How can we determine how many cubic inches are in 10 gallons of water?
We could use a tape diagram or a double number line, or we could find equivalent ratios.
→ Using either of these representations, determine the volume of the aquarium.

Determine the volume of this aquarium.
Answer:
Answers will vary depending on student choice. An example of a tape diagram is below.

→ We determined the volume of this tank is 2,310 in³. This is not the same volume we calculated earlier in the Opening Exercise. Why do you think the volumes are different?
Answers will vary but should include a discussion that there needs to be room for a lid; also, the water level cannot go all the way to the top so that there is room for heaters, filters, and fish, etc., without the water spilling over.

→ Generally, it is suggested that the highest level of water in this tank should be approximately 11.55 inches.
Calculate the volume of the aquarium using this new dimension.
V = l × w × h; V = 20 in. × 10 in. × 11.55 in.; V = 2,310 in³

→ What do you notice about this volume?
This volume is the same as the volume we determined when we found the volume using ratio and unit rates.

→ Let’s use the dimensions 20 in. × 10 in. × 11.55 in. for our exploration.

Eureka Math Grade 6 Module 5 Lesson 19a Exercise Answer Key

Optional Exercise 1:

→ We have determined that the volume for the 10-gallon aquarium with dimensions 20 in. × 10 in. × 11.55 in.
is 2,310 in³.
→ Suppose Jamie needs to fill the aquarium to the top in order to prepare the tank for fish. According to our calculations, if Jamie pours 10 gallons of water into the tank, the height of the water is approximately 11.55 in.
→ Let’s test it. Begin pouring water into the aquarium 1 gallon at a time. Be sure to keep track of the number of gallons. Use a tally system.

→ Measure the height of the water with your ruler.
→ What did you find about our height calculation?
Our calculation was correct. The height is approximately 11.55 in.

Exercise 1:

a. Determine the volume of the tank when filled with 7 gallons of water.
Answer:
231 \(\frac{\text { cubic inches }}{\text { gallon }}\) = 1,617 in³
The volume for 7 gallons of water is 1,617 in³.

b. Work with your group to determine the height of the water when Jamie places 7 gallons of water in the aquarium.
Answer:
1,617 in³ = (20 in. )(10 in. ) h
\(\frac{1,617 \mathrm{in}^{3}}{200 \mathrm{in}^{2}}=\frac{200 \mathrm{in}^{2}}{200 \mathrm{in}^{2}} h\)
8.085 in. = h
The tank should have a water height of 8.085 inches.

Optional Exercise 2:

→ Let’s test it. Begin by pouring water into the aquarium 1 gallon at a time.
→ Be sure to keep track of the number of gallons poured. Use a tally system.
Or, have students mark the height of the water using a wax marker or a dry erase marker on the outside of the tank after each gallon is poured in. Then, students measure the intervals (distance between the marks). Students should notice that the intervals are equal.
→ Test the height at 7 gallons, and record the height measurement.

→ What did you find about our calculation?
Our calculation was correct. The height is about 8 inches.

Exercise 2:

a. Use the table from Example 1 to determine the volume of the aquarium when Jamie pours 3 gallons of water into the tank.
Answer:
The volume of the tank is 231 in³ × 3 = 693 in³.

b. Use the volume formula to determine the missing height dimension.
693 in³ = 20 in.(10 in.) h
\(\frac{693 \mathrm{in}^{3}}{200 \mathrm{in}^{2}}=\frac{200 \mathrm{in}^{2}}{200 \mathrm{in}^{2}} h\)
3.465 in. = h
The tank should have a water height of 3.465 in.

Optional Exercise 3:

→ Let’s test it. Begin by pouring water into the aquarium 1 gallon at a time.
→ Be sure to keep track of the number of gallons poured. Use a tally system.
→ Test the height at 3 gallons, and record the height measurement.

→ What did you find about our calculation?
Our calculation was correct. The height is about 3\(\frac{1}{2}\) inches.

Exercise 3:

a. Using the table of values below, determine the unit rate of liters to gallon.

Answer:

The unit rate is 3.785.

b. Using this conversion, determine the number of liters needed to fill the 10-gallon tank. liters.
Answer:
3.785 \(\frac{\text { liters }}{\text { gallon }}\) × 10 gallons = 37.85 liters

c. The ratio of the number of centimeters to the number of inches is 2.54: 1. What is the unit rate?
Answer:
2.54

d. Using this information, complete the table to convert the heights of the water in inches to the heights of the water in centimeters Jamie will need for his project at home.

Answer:

Exercise 4:

a. Determine the amount of plastic film the manufacturer uses to cover the aquarium faces. Draw a sketch of the aquarium to assist in your calculations. Remember that the actual height of the aquarium is inches.
Answer:
SA = (2lW) + (2lh) + (2wh)
SA = (2.20 in. . 10 in.) + (2.20 in. . 12 in.) + (2.10in. . 12 in.)
SA = 400 in² + 480 in² + 240 in²
SA = 1,120 in²

b. We do not include the measurement of the top of the aquarium since It is open without glass and does not need to be covered with film. Determine the area of the top of the aquarium, and find the amount of film the manufacturer uses to cover only the sides, front, back, and bottom.
Answer:
Area of the top of the aquarium = l . w
Area of the top of the aquarium = 20 in. . 10 in.
Area of the top of the aquarium = 200 in²
SA of aquarium without the top = 1,120 in² – 200 in² = 920 in²

c. Since Jamie needs three aquariums, determine the total surface area of the three aquariums.
Answer:
920 in² + 920 in² + 920 in² = 2.760 in² or 3920 in² = 2,760 in²

Eureka Math Grade 6 Module 5 Lesson 19a Problem Set Answer Key

This Problem Set is a culmination of skills learned in this module. Note that the figures are not drawn to scale.

Question 1.
Calculate the area of the figure below.

Answer:
A = bh
A = (40 ft.) (20 ft.)
A = 800 ft²

Question 2.
Calculate the area of the figure below.

Answer:
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (1.3 m) (1.2 m)
A = 0.78 m²

Question 3.
Calculate the area of the figure below.

Answer:
Area of top rectangle:
A = lw
A = (25 in.)(12 in.)
A = 300 in²

Area of bottom rectangle:
A = lw
A= (7 in.)(20 in.)
A = 140 in²

Total Area = 300 in² + 140 in² = 440 in²

Question 4.
Complete the table using the diagram on the coordinate plane.

Answer:

Question 5.
Plot the points below, and draw the shape. Then, determine the area of the polygon. A(-3, 5), B(4, 3), C(0, -5)

Answer:

Area of Rectangle:
Area = lw
Area = (7 units)(10 units)
Area = 70 units²

Area of Triangle on Left:
Area = \(\frac{1}{2}\)bh
Area = \(\frac{1}{2}\)(3 units)(10 units)
Area = 15 units²

Area of Triangle on Top:
Area on top = \(\frac{1}{2}\)bh
Area = \(\frac{1}{2}\)(7 units) (2 units)
Area = 7 units²

Area of Triangle on Right:
Area = \(\frac{1}{2}\)bh
Area = \(\frac{1}{2}\)(4 units)(8 units)
Area = 16 units²

Total Area = 70 units² – 15 units² – 7 units² – 16 units²
Total Area = 32 units²

Question 6.
Determine the volume of the figure.

Answer:
V = l w h
V = (3\(\frac{1}{2}\) m) (\(\frac{7}{8}\) m) (1\(\frac{1}{4}\) m)
V = \(\frac{245}{64}\) m³
V = 3\(\frac{53}{64}\) m³

Question 7.
Give at least three more expressions that could be used to determine the volume of the figure in Problem 6.
Answer:
Answers will vary. Some examples include the following:
(\(\frac{35}{32}\) m²) (3\(\frac{53}{64}\) m)
(1\(\frac{1}{4}\) m) (\(\frac{7}{8}\) m) (3\(\frac{1}{2}\) m)
(\(\frac{49}{16}\) m²) (1\(\frac{1}{4}\) m)

Question 8.
Determine the volume of the irregular figure.

Answer:
Volume of the back Rectangular Prism:
V = l w h
V = (3\(\frac{5}{8}\) ft.) (1\(\frac{1}{3}\) ft) (1\(\frac{1}{4}\) ft.)
V = \(\frac{580}{96}\) ft³

Volume of the front Rectan guiar Prism:
V = l w h
V = (1\(\frac{1}{4}\) ft.) (1\(\frac{1}{6}\) ft.)(1\(\frac{1}{4}\) ft.)
V = \(\frac{175}{96}\) ft³

Total Volume = \(\frac{580}{96}\) ft³ + \(\frac{175}{96}\) ft³
= \(\frac{755}{96}\) ft³ = 7\(\frac{83}{96}\) ft³

Question 9.
Draw and label a net for the following figure. Then, use the net to determine the surface area of the figure.

Answer:

SA = 120 cm² + 84 cm² + 70 cm² + 84 cm² + 120 cm² + 70 cm²
= 548 cm²

Question 10.
Determine the surface area of the figure in Problem 9 using the formula SA = 21w + 21h + 2wh. Then, compare your answer to the solution in Problem 9.
Answer:
SA = 2lw + 2lh + 2wh
SA = 2(10 cm)(7 cm) + 2(10 cm)(12 cm) + 2 (7 cm)(12 cm)
SA = 140 cm² + 240 cm² + 168 cm²
SA= 548 cm²

The answer in Problem 10 is the same as in Problem 9. The formula finds the areas of each pair of equal faces and adds them together, like we did with the net.

Question 11.
A parallelogram has a base of 4.5 cm and an area of 9.495 cm². Tania wrote the equation 4.5x = 9.495 to represent this situation.

a. Explain what x represents in the equation.
Answer:
x represents the height of the parallelogram in centimeters.

b. Solve the equation for x and determine the height of the parallelogram.
Answer:
\(\frac{4.5 x}{4.5}=\frac{9.495}{4.5}\)
x = 2.11
The height of the parallelogram is 2.11 cm.

Question 12.
Triangle A has an area equal to one-third the area of Triangle B. Triangle A has an area of 3\(\frac{1}{2}\) square meters.

a. Gerard wrote the equation \(\frac{B}{3}\) = 3\(\frac{1}{2}\). Explain what B represents in the equation.
Answer:
B represents the area of Triangle B in square meters.

b. Determine the area of Triangle B.
Answer:
\(\frac{B}{3}\) . 3 = 3\(\frac{1}{2}\) . 3
The area of Triangle is 10\(\frac{1}{2}\) square meters.

Eureka Math Grade 6 Module 5 Lesson 19a Exit Ticket Answer Key

Question 1.
What did you learn today? Describe at least one situation in real life that would draw on the skills you used today.
Answer:
Answers will vary.

Engage NY Eureka Math Grade 6 Module 5 Lesson 19 Answer Key

Eureka Math Grade 6 Module 5 Lesson 19 Opening Exercise Answer Key

Opening Exercise:

Question 1.
A box needs to be painted. How many square inches need to be painted to cover the entire surface of the box?

Answer:
SA = 2(15 in.)(12 in.) + 2(15 in.)(6 in.) + 2(12 in.)(6 in.)
SA = 360 in² + 180 in² + 144 in²
SA = 684 in²

Question 2.
A juice box is 4 in. tall, 1 in. wide, and 2 in. long. How much juice fits inside the juice box?
Ans;
V = 1 in. × 2 in. × 4 in. = 8 in³

Question 3.
How did you decide how to solve each problem?
Answer:
I chose to use surface area to solve the first problem because you would need to know how much area the paint would need to cover. I chose to use volume to solve the second problem because you would need to know how much space is inside the juice box to determine how much juice it can hold.

Eureka Math Grade 6 Module 5 Lesson 19 Discussion Answer Key

Discussion:

Answer:

Eureka Math Grade 6 Module 5 Lesson 19 Example Answer Key

Example 1:

Vincent put logs in the shape of a rectangular prism outside his house. However, it is supposed to snow, and Vincent wants to buy a cover so the logs stay dry. If the pile of logs creates a rectangular prism with these measurements: 33 cm long, 12 cm wide, and 48 cm high, what is the minimum amount of material needed to cover the pile of logs?
Answer:

→ Where do we start?
We need to find the size of the cover for the logs, so we need to calculate the surface area. in order to find the surface area, we need to know the dimensions of the pile of logs.

→ Why do we need to find the surface area and not the volume?
We want to know the size of the cover Vincent wants to buy. If we calculated volume, we would not have the information Vincent needs when he goes shopping for a cover.

→ What are the dimensions of the pile of logs?
The length is 33 cm, the width is 12 cm, and the height is 48 cm.

→ How do we calculate the surface area to determine the size of the cover?
We can use the surface area formula for a rectangular prism.
SA = 2(33 cm)(12 cm) + 2(33 cm)(48 cm) + 2(12 cm)(48 cm)
SA = 792 cm2 + 3,168 cm² + 1,152 cm2²
SA = 5,112 cm²

→ What is different about this problem from other surface area problems of rectangular prisms you have encountered? How does this change the answer?
If Vincent just wants to cover the wood to keep it dry, he does not need to cover the bottom of the pile of logs. Therefore, the cover can be smaller.

→ How can we change our answer to find the exact size of the cover Vincent needs?
We know the area of the bottom of the pile of logs has the dimensions 33 cm and 12 cm. We can calculate the area and subtract this area from the total surface area.
The area of the bottom of the pile of logs is 396 cm²; therefore, the total surface area of the cover would need to be 5,112 cm² – 396 cm² = 4,716 cm².

Eureka Math Grade 6 Module 5 Lesson 19 Exercise Answer Key

Exercises:

Use your knowledge of volume and surface area to answer each problem.

Exercise 1.
Quincy Place wants to add a pool to the neighborhood. When determining the budget, Quincy Place determined that it would also be able to install a baby pool that requires less than 15 cubic feet of water. Quincy Place has three different models of a baby pool to choose from.
Choke One: 5 ft. × 5 ft. × 1 ft.
ChokeTwo: 4 ft. × 3 ft. × 1 fi
Choke Three: 4 ft. × 2 ft. × 2 ft.
Which of these chokes is best for the baby pool? Why are the others not good choices?
Answer:
Choice One Volume: 5 ft. × 5 ft. × 1 ft. = 25 ft³
Choice Two Volume: 4 ft. × 3 ft. × 1 ft. = 12 ft³
Choice Three Volume: 4 ft. × 2 ft. × 2 ft. = 16 ft³
Choice Two is within the budget because it holds less than 15 cubic feet of water. The other two choices do not work because they require too much water, and Quincy Place will not be able to afford the amount of water it takes to fill the baby pool.

Exercise 2.
A packaging firm has been hired to create a box for baby blocks. The firm was hired because it could save money by creating a box using the least amount of material. The packaging firm knows that the volume of the box must be 18 cm³.

a. What are possible dimensions for the box if the volume must be exactly 18 cm³?
Answer:
Choice 1: 1 cm × 1 cm × 18 cm
Choice 2: 1 cm × 2 cm × 9 cm
Choice 3: 1 cm × 3 cm × 6 cm
Choice 4: 2 cm × 3 cm × 3 cm

b. Which set of dimensions should the packaging firm choose in order to use the least amount of material? Explain.
Answer:
Choice 1: SA = 2(1 cm)(1 cm) + 2(1 cm)(18 cm) + 2(1 cm)(18 cm) = 74 cm²
Choice 2: SA = 2(1 cm)(2 cm) + 2(1 cm)(9 cm) + 2(2 cm)(9 cm) = 58 cm²
Choice 3: SA = 2(1 cm)(3 cm) + 2(1 cm)(6 cm) + 2(3 cm)(6 cm) = 54 cm²
Choice 4: SA = 2(2 cm)(3 cm) + 2(2 cm)(3 cm) + 2(3 cm)(3 cm) = 42 cm²
The packaging firm should choose Choice 4 because it requires the least amount of material. in order to find the amount of material needed to create a box,, the packaging firm would have to calculate the surface area of each box. The box with the smallest surface area requires the least amount of material.

Exercise 3.
A gift has the dimensions of 50 cm × 35 cm × 5 cm. You have wrapping paper with dimensions of 75 cm × 60 cm. Do you have enough wrapping paper to wrap the gift? Why or why not?
Answer:
Surface Area of the Present SA = 2(50 cm)(35 cm) + 2(50 cm)(5 cm) + 2(35 cm)(5 cm) = 3,500 cm² + 500 cm² + 350 cm² = 4,350cm²
Area of Wrapping Paper: A = 75 cm × 60 cm = 4,500 cm²
I do have enough paper to wrap the present because the present requires 4,350 square centimeters of paper, andi
have 4,500 square centimeters of wrapping paper.

Exercise 4.
Tony bought a flat-rate box from the post office to send a gift to his mother for Mother’s Day. The dimensions of the medium-size box are 14 inches × 12 inches × 3.5 inches. What is the volume of the largest gift he can send to his mother?
Answer:
Volume of the Box: V = 14 in. × 12 in. × 3.5 in. = 588 in³
Tony would have 588 cubic inches of space to fill with a gift for his mother.

Exercise 5.
A cereal company wants to change the shape of its cereal box in order to attract the attention of shoppers. The original cereal box has dimensions of 8 inches × 3 inches × 11 inches. The new box the cereal company is thinking of would have dimensions of 10 inches × 10 inches × 3 inches.

a. Which box holds more cereal?
Answer:
Volume of Original Box: V = 8 in. × 3 in. × 11 in. = 264 in³
Volume of New Box: V= 10 in. × 10 in. × 3 in.= 300 in³
The new box holds more cereal because it has a larger volume.

b. Which box requires more material to make?
Answer:
Surface Area of Original Box: SA = 2(8 in.)(3 in.) + 2(8 in.)(11 in.)+ 2(3 in.)(11 in.) = 48 in² + 176 in² + 66 in² = 290 in²

Surface Area of New Box: SA = 2(10 in.)(10 in.) + 2(10 in.)(3 in.) + 2(10 in.)(3 in.) = 200 in² + 60 in² + 60 in² = 320 in²

The new box requires more material than the original box because the new box has a larger surface area.

Exercise 6.
Cinema theaters created a new popcorn box in the shape of a rectangular prism. The new popcorn box has a length of 6 inches, a width of 3. 5 inches, and a height of 3.5 inches but does not include a lid.

a. How much material is needed to create the box?
Answer:
Surface Area of the Box: SA = 2(6 in.)(3. 5 in.) + 2(6 in.)(3.5 in.) + 2(3.5 in.)(3. 5 in.) = 42 in² + 42 in² + 24.5 in² = 108.5 in²
The box does not have a lid, so we have to subtract the area of the lid from the surface area.
Area of Lid: 6 in. × 3.5 in. = 21 in²
Total Surface Area: 108.5 in² – 21 in² = 87.5 in²
87.5 square inches of material is needed to create the new popcorn box.

b. How much popcorn does the box hold?
Answer:
Volume of the Box: V= 6 in. × 3.5 in. × 3.5 in. = 73.5 in³
The box holds 73.5 in³ of popcorn.

Eureka Math Grade 6 Module 5 Lesson 19 Problem Set Answer Key

Solve each problem below.

Question 1.
Dante built a wooden, cubic toy box for his son. Each side of the box measures 2 feet.

a. How many square feet of wood did he use to build the box?
Answer:
Surface Area of the Box: SA = 6(2 ft)² = 6(4 ft²) = 24 ft²
Dante used 24 square feet of wood to build the box.

b. How many cubic feet of toys will the box hold?
Answer:
Volume of the Box: V = 2 ft. × 21 ft. × 2 ft.= 8 ft³
The toy box would hold 8 cubic feet of toys.

Question 2.
A company that manufactures gift boxes wants to know how many different-sized boxes having a volume of 50 cubic centimeters it can make if the dimensions must be whole centimeters.

a. List all the possible whole number dimensions for the box.
Answer:
Choice One: 1 cm × 1 cm × 50 cm
Choice Two: 1 cm × 2 cm × 25 cm
Choice Three: 1 cm × 5 cm × 10 cm
Choice Four: 2 cm × 5 cm × 5 cm

b. Which possibility requires the least amount of material to make?
Answer:
Choice One: SA = 2(1 cm)(1 cm) + 2(1 cm)(50 cm) + 2(1 cm)(50 cm) = 2 cm² + 100 cm² + 100 cm² = 202 cm²

Choice Two: SA = 2(1 cm)(2 cm) + 2(1 cm)(25 cm) + 2(2 cm)(25 cm) = 4 cm² + 50 cm² + 100 cm² = 154 cm²

Choice Three: SA = 2(1 cm)(5 cm) + 2(1 cm)(10 cm) + 2(5 cm)(10 cm) = 10 cm² + 20 cm² + 100 cm² = 130 cm²

Choice Four: SA = 2(2 cm)(5 cm) + 2(2 cm)(5 cm) + 2(5 cm)(5 cm) = 20 cm² + 20 cm² + 50 cm² = 90 cm²

Choice Four requires the least amount of material because it has the smallest surface area.

c. Which box would you recommend the company use? Why?
Answer:
I would recommend the company use the box with dimensions of 2 cm × 5 cm × 5 cm (Choice Four) because it requires the least amount of material to make, so it would cost the company the least amount of money to make.

Question 3.
A rectangular box of rice is shown below. What is the greatest amount of rice, in cubic inches, that the box can hold?

Answer:
Volume of the Rice Box: V = 15\(\frac{1}{3}\) in. × 7\(\frac{2}{3}\) in. × 6\(\frac{1}{3}\) in. = \(\frac{20,102}{27}\) in³ = 744\(\frac{14}{27}\) in³

Question 4.
The Mars Cereal Company has two different cereal boxes for Mars Cereal. The large box is 8 inches wide, 11 inches high, and 3 inches deep. The small box is 6 inches wide, 10 inches high, and 2.5 inches deep.

a. How much more cardboard is needed to make the large box than the small box?
Answer:
SurfaceArea of the Large Box: SA = 2(8 in.)(11 in.) + 2(8 in.)(3 in.) + 2(11 in.)(3 in.) = 176 in² + 48 in² +66 in² = 290 in²

Surface Area of the Small Box: SA = 2(6 in. )(10 in.) + 2(6 in.)(2. 5 in.) + 2(10 in. )(2. 5 in.) = 120 in² + 30 in² + 50 in² = 200 in²

Difference: 290 in² – 200 in² = 90 in²
The large box requires 90 square inches more cardboard than the small box.

b. How much more cereal does the large box hold than the small box?
Answer:
Volume of the Large Box: V = 8 in. × 11 in. × 3 in. = 264 in³
Volume of the Small Box: V = 6 in. × 10 in. × 2.5 in. = 150 in³
Difference: 264 in³ – 150 in³ = 114 in³
The large box holds 114 cubic inches more cereal than the small box.

Question 5.
A swimming pool is 8 meters long, 6 meters wide, and 2 meters deep. The water-resistant paint needed for the pool costs $6 per square meter. How much will it cost to paint the pool?

a. How many faces of the poo1 do you have to paint?
Answer:
You have to point 5 faces.

b. How much paint (in square meters) do you need to paint the pool?
Answer:
SA = 2(8 m × 6 m) + 2(8 m × 2 m) + 2(6 m × 2 m) = 96 m² + 32 m² + 24 m² = 152 m²
Area of Top of Pool: 8 m × 6 m= 48 m²
Total Point Needed: 152 m² – 48 m² = 104 m²

c. How much will it cost to paint the pool?
Answer:
104 m² × $6/m² = $624
It will cost $624 to point the pool.

Question 6.
Sam is in charge of filling a rectangular hole with cement. The hole is 9 feet long, 3 feet wide, and 2 feet deep. How much cement will Sam need?
Answer:
V = 9 ft. × 3 ft. × 2 ft. = 54 ft³
Sam will need 54 cubic feet of cement to fill the hole.

Question 7.
The volume of Box D subtracted from the volume of Box C is 23.14 cubic centimeters. Box D has a volume of 10.115 cubic centimeters.

a. Let C be the volume of Box C in cubic centimeters. Write an equation that could be used to determine the volume of Box C.
Answer:
C – 10.115 cm³ = 23.14 cm³

b. Solve the equation to determine the volume of Box C.
Answer:
C – 10.115 cm³ + 10.115 cm³ = 23.14 cm³ + 10.115 cm³
C = 33. 255 cm³

c. The volume of Box C is one-tenth of the volume of another box, Box E. Let E represent the volume of Box E in cubic centimeters. Write an equation that could be used to determine the volume of Box E, using the result from part (b).
Answer:
33.255 cm³ = \(\frac{1}{10}\)E

d. Solve the equation to determine the volume of Box E.
Answer:
33.255cm³ ÷ \(\frac{1}{10}\) = \(\frac{1}{10}\)E ÷ \(\frac{1}{10}\)
332.55 cm³ = E

Eureka Math Grade 6 Module 5 Lesson 19 Exit Ticket Answer Key

Solve the word problem below.

Question 1.
Kelly has a rectangular fish aquarium that measures 18 inches long. 8 inches wide, and 12 inches tall.

a. What is the maximum amount of water the aquarium can hold?
Answer:
Volume of the Aquarium: V = 18 in. × 8 in. × 12 in. = 1,728 in³
The maximum amount of water the aquarium can hold is 1, 728 cubic inches.

b. If Kelly wanted to put a protective covering on the four glass walls of the aquarium, how big does the cover have to be?
Answer:
Surface Area of theAquarium: SA = 2(18 in.)(8 in.)+ 2(18 in.)(12 in.)+ 2(8 in.)(12 in.) = 288 in² + 432 in² + 192 in² = 912 in²

We only need to cover the four glass walls, so we can subtract the area of both the top and bottom of the aquarium.
Area of Top: A = 18 in. × 8 in. = 144 in²
Area of Bottom: A = 18 in. × 8 in. = 144 in²
Surface Area of the Four Walls: SA = 912 in² – 144 in² – 144 in² = 624 in²
Kelly would need 624 in² to cover the four walls of the aquarium.

Eureka Math Grade 6 Module 5 Lesson 19 Area of shapes Answer Key

Area of shapes:

Question 1.

Answer:
A = 80 ft²

Question 2.

Answer:
A = 30 m²

Question 3.

Answer:
A = 484 in²

Question 4.

Answer:
A = 1,029 cm²

Question 5.

Answer:
A = 72 ft²

Question 6.

Answer:
A = 156 km²

Question 7.

Answer:
A = 110 in²

Question 8.

Answer:
A = 192 cm²

Question 9.

Answer:
A = 576 m²

Question 10.

Answer:
A = 1,476 ft²

Engage NY Eureka Math Grade 6 Module 5 Lesson 18 Answer Key

Eureka Math Grade 6 Module 5 Lesson 18 Opening Exercise Answer Key

Opening Exercise:

a. What three-dimensional figure does the net create?
Answer:
Rectangular Prism

b. Measure (in inches) and label each side of the figure.
Answer:

c. Calculate the area of each face, and record this value inside the corresponding rectangle.
Answer:

d. How did we compute the surface area of solid figures in previous lessons?
Answer:
To determine surface area, we found the area of each of the faces and then added those areas.

e. Write an expression to show how we can calculate the surface area of the figure above.
Answer:
(4 in. × 1 in.) + (4 in. × 2 in.) + (4 in. × 1 in.) + (4 in. × 2 in.) + (2 in. × 1 in.) + (2 in. × 1 in.) OR
2(4 in. × 1 in.) + 2(4 in. × 2 in.) + 2(2 in. × 1 in.)

f. What does each part of the expression represent?
Answer:
Each part of the expression represents an area of one face of the given figure. We were able to write a more compacted form because there are three pairs of two faces that are identical.

g. What is the surface area of the figure?
Answer:
(4in.× 1 in) + (4 in. × 2 in.) + (4 in. × 1 in.) + (4 in. × 2 in.) + (2 in. × 1 in.) + (2 in. × 1 in.)
2(4 in. × 1 in.) + 2(4 in. × 2 in.) + 2(2 in. × 1 in.)
28 in²

Eureka Math Grade 6 Module 5 Lesson 18 Example Answer Key

Example 1:

Fold the net used in the Opening Exercise to make a rectangular prism. Have the two faces with the largest area be the bases of the prism. Fill in the first row of the table below.

Answer:

→ What do you notice about the areas of the faces?
Pairs of faces have equal areas.

→ What is the relationship between the faces having equal area?
The faces that have the same area are across from each other. The bottom and top have the same area, the front and the back have the same area, and the two sides have the same area.

→ How do we calculate the area of the two bases of the prism?
length × width

→ How do we calculate the area of the front and back faces of the prism?
length × height

→ How do we calculate the area of the right and left faces of the prism?
width × height

→ Using the name of the dimensions, fill in the third row of the table.

Examine the rectangular prism below. Complete the table.


Answer:

→ When comparing the methods to finding surface area of the two rectangular prisms, can you develop a general formula?
SA = l × w + l × w + l × h + l × h + w × h + w × h

→ Since we use the same expression to calculate the area of pairs of faces, we can use the distributive property to write an equivalent expression for the surface area of the figure that uses half as many terms.

→ We have determined that there are two l × w dimensions. Let’s record that as 2 times l times w, or simply 2(l × w). How can we use this knowledge to alter other parts of the formula?
We also have two l × h, so we can write that as 2(l × h), and we can write the two w × h as 2(w × h).

→ Writing each pair in a simpler way, what is the formula to calculate the surface area of a rectangular prism?
SA = 2(l × w) + 2(l × h) + 2(w × h)
→ Knowing the formula to calculate surface area makes it possible to calculate the surface area without a net.

Example 2:

Answer:
→ What are the dimensions of the rectangular prism?
The length is 20 cm, the width is 5 cm, and the height is 9 cm.

→ We use substitution in order to calculate the area. Substitute the given dimensions into the surface area formula.
SA = 2(20 cm)(5 cm) + 2(20 cm)(9 cm) + 2(5 cm)(9 cm)

→ Solve the equation. Remember to use order of operations.
SA = 200cm² + 360 cm² + 90 cm²
SA = 650 cm²

Eureka Math Grade 6 Module 5 Lesson 18 Exercise Answer Key

Exercises 1 – 3:

Exercise 1.
Calculate the surface area of each of the rectangular prisms below.

a.
Answer:
SA = 2(12 in.)(2 in.) + 2(12 in.)(3 in.) + 2(2 in.)(3 in.)
SA = 48 in² + 72 in² + 12 in²
SA = 132 in²

b.
Answer:
SA = 2(8 m)(6 m) + 2(8 m)(22 m) + 2(6 m)(22 m)
SA = 96 m² + 352 m² + 264 m²
SA = 712 m²

c.
Answer:
SA = 2(29 ft. ) (16 ft.) + 2(29 ft.) (23 ft.) + 2(16 ft.) (23 ft.)
SA = 928 ft² + 1,334 ft² + 736 ft²
SA = 2,998 ft²

d.
Answer:
SA = 2(4 cm) (1.2 cm) + 2(4 cm) (2.8 cm) + 2(1.2 cm) (2.8 cm)
SA = 9.6 cm² + 22.4 cm² + 6.72 cm²
SA = 38.72 cm²

Exercise 2.
Calculate the surface area of the cube.

Answer:
SA = 2(5 in.) (5 in.) + 2(5 in.) (5 in.) + 2(5 in.) (5 in.)
SA = 50 in² + 50 in² + 50 in²
SA = 150 in²

Exercise 3.
All the edges of a cube have the same length. Tony claims that the formula SA = 6s2, where s is the length of each side of the cube, can be used to calculate the surface area of a cube.

a. Use the dimensions from the cube in Problem 2 to determine if Tony’s formula is correct.
Answer:
Tony’s formula is correct because SA = 6(5 km)² = 150 km², which is the same surface area when we use the surface area formula for rectangular prisms.

b. Why does this formula work for cubes?
Answer:
Each face is a square, and to find the area of a square, you multiply the side lengths together. However, since the side lengths are the same, you can just square the side length. Also, a cube has 6 identical faces, so after calculating the area of one face, we can just multiply this area by 6 to determine the total surface area of the cube.

c. Becca does not want to try to remember two formulas for surface area, so she is only going to remember the formula for a cube. Is this a good idea? Why or why not?
Answer:
Becca’s idea is not a good idea. The surface area formula for cubes only works for cubes because rectangular prisms do not have 6 identical faces. Therefore, Becca also needs to know the surface area formula for rectangular prisms.

Eureka Math Grade 6 Module 5 Lesson 18 Problem Set Answer Key

Calculate the surface area of each figure below. Figures are not drawn to scale.

Question 1.

Answer:
SA = 2(15 in.)(15 in.) + 2(15 in.)(7 in.) + 2(15 in.)(7 in.)
SA = 450 in² + 210 in² + 210 in²
SA = 870 in²

Question 2.

Answer:
SA = 2(18.7 cm) (2. 3 cm) + 2(18.7 cm) (8. 4cm) + 2(2.3 cm) (8. 4 cm)
SA = 86.02 cm² + 314. 16 cm² + 38.64 cm²
SA = 438.82 cm²

Question 3.

Answer:
SA = 6(2\(\frac{1}{3}\) ft.)²

SA = 6(\(\frac{7}{3}\) ft.)²

SA = 6(\(\frac{49}{9}\) ft.)²

SA = \(\frac{294}{9}\) ft² = 32\(\frac{2}{3}\) ft²

Question 4.

Answer:
SA = 2(32.3 m) (24.7 m) + 2(32.3 m) (7.9 m) + 2(24.7 m) (7.9 m)
SA= 1,595.62 m² + 510.34 m² + 390.26 m²
SA = 2,496.22 m²

Question 5.
Write a numerical expression to show how to calculate the surface area of the rectangular prism. Explain each part of the expression.

Answer:
2(12 ft. × 3 ft.) + 2(12 ft. × 7 ft.) + 2(7 ft. × 3 ft.)
The first part of the expression shows the area of the top and bottom of the rectangular prism. The second part of the expression shows the area of the front and back of the 7 ft. rectangular prism. The third part of the expressin shows the area of the two sides of the rectangular prism. 3 ft. The surface area of the figure is 282 ft².

Question 6.
When Louie was calculating the surface area for Problem 4, he identified the following:
length = 24. 7 m, width = 32.3 m, and height = 7.9 m.
However, when Rocko was calculating the surface area for the same problem, he identified the following:
length = 32.3 m, width = 24.7 m, and height = 7.9 m.
Would Louie and Rocko get the same answer? Why or why not?
Answer:
Louie and Rocko would get the same answer because they are still finding the correct area of all six faces of the rectangular prism.

Eureka Math Grade 6 Module 5 Lesson 18 Exit Ticket Answer Key

Calculate the surface area of each figure below. Figures are not drawn to scale.

Question 1.

Answer:
SA = 2lw + 2lh + 2wh
SA = 2(12 ft.) (2 ft.) + 2(12 ft ) (10 ft.) + 2(2 ft) (10 ft.)
SA = 48 ft² + 240 ft² + 40 ft²
SA = 328 ft²

Question 2.

Answer:
SA = 6S²
SA = 6(8 cm)²
SA = 6(64 cm²)
SA = 384 cm²

Engage NY Eureka Math 8th Grade Module 4 Lesson 17 Answer Key

Eureka Math Grade 8 Module 4 Lesson 17 Exercise Answer Key

Exercise 1.
Find at least three solutions to the equation y=2x, and graph the solutions as points on the coordinate plane. Connect the points to make a line. Find the slope of the line.

Answer:
The slope of the line is 2; m=2.

Exercise 2.
Find at least three solutions to the equation y=3x-1, and graph the solutions as points on the coordinate plane. Connect the points to make a line. Find the slope of the line.

Answer:
The slope of the line is 3; m=3.

Exercise 3.
Find at least three solutions to the equation y=3x+1, and graph the solutions as points on the coordinate plane. Connect the points to make a line. Find the slope of the line.

Answer:
The slope of the line is 3; m=3.

Exercises 4–11
Students work independently or in pairs to identify the slope from an equation and to transform the standard form of an equation into slope-intercept form.

Exercise 4.
The graph of the equation y=7x-3 has what slope?
Answer:
The slope is 7.

Exercise 5.
The graph of the equation y=-\(\frac{3}{4}\) x-3 has what slope?
Answer:
The slope is –\(\frac{3}{4}\).

Exercise 6.
You have $20 in savings at the bank. Each week, you add $2 to your savings. Let y represent the total amount of money you have saved at the end of x weeks. Write an equation to represent this situation, and identify the slope of the equation. What does that number represent?
Answer:
y=2x+20
The slope is 2. It represents how much money is saved each week.

Exercise 7.
A friend is training for a marathon. She can run 4 miles in 28 minutes. Assume she runs at a constant rate. Write an equation to represent the total distance, y, your friend can run in x minutes. Identify the slope of the equation. What does that number represent?
Answer:
\(\frac{y}{x}\)=\(\frac{4}{28}\)
y=\(\frac{4}{28}\) x
y=\(\frac{1}{7}\) x
The slope is \(\frac{1}{7}\). It represents the rate at which my friend can run, one mile in seven minutes.

Exercise 8.
Four boxes of pencils cost $5. Write an equation that represents the total cost, y, for x boxes of pencils. What is the slope of the equation? What does that number represent?
Answer:
y=\(\frac{5}{4}\) x
The slope is \(\frac{5}{4}\). It represents the cost of one box of pencils, $1.25.

Exercise 9.
Solve the following equation for y, and then identify the slope of the line: 9x-3y=15.
Answer:
9x-3y=15
9x-9x-3y=15-9x
-3y=15-9x
\(\frac{-3}{-3}\) y=\(\frac{15-9 x}{-3}\)
y=\(\frac{15}{-3}\)–\(\frac{9 x}{-3}\)
y=-5+3x
y=3x-5
The slope of the line is 3.

Exercise 10.
Solve the following equation for y, and then identify the slope of the line: 5x+9y=8.
Answer:
5x+9y=8
5x-5x+9y=8-5x
9y=8-5x
\(\frac{9}{\mathbf{9}}\) y=\(\frac{8-5 x}{9}\)
y=\(\frac{8}{9}\)–\(\frac{5}{9}\)x
y=-\(\frac{5}{9}\)x+\(\frac{8}{9}\)
The slope of the line is –\(\frac{5}{9}\).

Question 11.
Solve the following equation for y, and then identify the slope of the line: ax+by=c.
Answer:
ax+by=c
ax-ax+by=c-ax
by=c-ax
\(\frac{b}{b}\) y=\(\frac{c-a x}{b}\)
y=\(\frac{c}{b}\)–\(\frac{ax}{b}\)
y=\(\frac{c}{b}\)–\(\frac{a}{b}\) x
y=-\(\frac{a}{b}\) x+\(\frac{c}{b}\)
The slope of the line is –\(\frac{a}{b}\).

Eureka Math Grade 8 Module 4 Lesson 17 Problem Set Answer Key

Students practice transforming equations from standard form into slope-intercept form and showing that the line joining two distinct points of the graph y=mx+b has slope m. Students graph the equation and informally note the y-intercept.

Question 1.
Solve the following equation for y: -4x+8y=24. Then, answer the questions that follow.
Answer:
-4x+8y=24
-4x+4x+8y=24+4x
8y=24+4x
\(\frac{8}{8}\) y=\(\frac{24}{8}\)+\(\frac{4}{8}\) x
y=3+\(\frac{1}{2}\) x
y=\(\frac{1}{2}\) x+3

a. Based on your transformed equation, what is the slope of the linear equation -4x+8y=24?
Answer:
The slope is \(\frac{1}{2}\).

b. Complete the table to find solutions to the linear equation.

Answer:

c. Graph the points on the coordinate plane.
Answer:

d. Find the slope between any two points.
Answer:
Using points (0,3) and (2,4):
m=\(\frac{4-3}{2-0}\)
=\(\frac{1}{2}\)

e. The slope you found in part (d) should be equal to the slope you noted in part (a). If so, connect the points to make the line that is the graph of an equation of the form y=mx+b that has slope m.

f. Note the location (ordered pair) that describes where the line intersects the y-axis.
Answer:
(0,3) is the location where the line intersects the y-axis.

Question 2.
Solve the following equation for y: 9x+3y=21. Then, answer the questions that follow.
Answer:
9x+3y=21
9x-9x+3y=21-9x
3y=21-9x
\(\frac{3}{3}\) y=\(\frac{21}{3}\)–\(\frac{9}{3}\) x
y=7-3x
y=-3x+7

a. Based on your transformed equation, what is the slope of the linear equation 9x+3y=21?
Answer:
The slope is -3.

b. Complete the table to find solutions to the linear equation.

Answer:

c. Graph the points on the coordinate plane.
Answer:

d. Find the slope between any two points.
Answer:
Using points (1,4) and (2,1):
m=\(\frac{4-1}{1-2}\)
=\(\frac{3}{-1}\)
=-3

e. The slope you found in part (d) should be equal to the slope you noted in part (a). If so, connect the points to make the line that is the graph of an equation of the form y=mx+b that has slope m.

f. Note the location (ordered pair) that describes where the line intersects the y-axis.
Answer:
(0,7) is the location where the line intersects the y-axis.

Question 3.
Solve the following equation for y: 2x+3y=-6. Then, answer the questions that follow.
Answer:
2x+3y=-6
2x-2x+3y=-6-2x
3y=-6-2x
\(\frac{3}{3}\) y=\(\frac{-6}{3}\)–\(\frac{2}{3}\) x
y=-2-\(\frac{2}{3}\) x
y=-\(\frac{2}{3}\) x-2

a. Based on your transformed equation, what is the slope of the linear equation 2x+3y=-6?
Answer:
The slope is –\(\frac{2}{3}\).

b. Complete the table to find solutions to the linear equation.

Answer:

c. Graph the points on the coordinate plane.
Answer:

d. Find the slope between any two points.
Answer:
Using points (-6,2) and (3,-4):
m=\(\frac{2-(-4)}{-6-3}\)
=\(\frac{6}{-9}\)
=-\(\frac{2}{3}\)

e. The slope you found in part (d) should be equal to the slope you noted in part (a). If so, connect the points to make the line that is the graph of an equation of the form y=mx+b that has slope m.

f. Note the location (ordered pair) that describes where the line intersects the y-axis.
Answer:
(0,-2) is the location where the line intersects the y-axis.

Question 4.
Solve the following equation for y: 5x-y=4. Then, answer the questions that follow.
Answer:
5x-y=4
5x-5x-y=4-5x
-y=4-5x
y=-4+5x
y=5x-4

a. Based on your transformed equation, what is the slope of the linear equation 5x-y=4?
Answer:
The slope is 5.

b. Complete the table to find solutions to the linear equation.

Answer:

c. Graph the points on the coordinate plane.

d. Find the slope between any two points.
Answer:
Using points (0,-4) and (1,1):
m=\(\frac{-4-1}{0-1}\)
=\(\frac{-5}{-1}\)
=5

e. The slope you found in part (d) should be equal to the slope you noted in part (a). If so, connect the points to make the line that is the graph of an equation of the form y=mx+b that has slope m.

f. Note the location (ordered pair) that describes where the line intersects the y-axis.
Answer:
(0,-4) is the location where the line intersects the y-axis.

Eureka Math Grade 8 Module 4 Lesson 17 Exit Ticket Answer Key

Question 1.
Solve the following equation for y: 35x-7y=49.
Answer:
35x-7y=49
35x-35x-7y=49-35x
-7y=49-35x
\(\frac{-7}{-7}\) y=\(\frac{49}{-7}\)–\(\frac{35}{-7}\) x
y=-7-(-5x)
y=5x-7

Question 2.
What is the slope of the equation in Problem 1?
Answer:
The slope of y=5x-7 is 5.

Question 3.
Show, using similar triangles, why the graph of an equation of the form y=mx is a line with slope m.

Answer:

Solutions will vary. A sample solution is shown below.
The line shown has slope 2. When we compare the corresponding side lengths of the similar triangles, we have the ratios \(\frac{2}{1}\)=\(\frac{4}{2}\)=2. In general, the ratios would be \(\frac{x}{1}\)=\(\frac{y}{m}\), equivalently y=mx, which is a line with slope m.

Engage NY Eureka Math 8th Grade Module 4 Lesson 18 Answer Key

Eureka Math Grade 8 Module 4 Lesson 19 Opening Exercise Answer Key

Examine each of the graphs and their equations. Identify the coordinates of the point where the line intersects the y-axis. Describe the relationship between the point and the equation y=mx+b.
a. y=\(\frac{1}{2}\) x+3

b. y=-3x+7

c. y=-\(\frac{2}{3}\) x-2

d. y=5x-4

Answer:
A point is noted in each graph above where the line intersects the y-axis:
y=\(\frac{1}{2}\) x+3, (0,3)
y=-3x+7, (0,7)
y=-\(\frac{2}{3}\) x-2, (0,-2)
y=5x-4, (0,-4)
In each equation, the number b was the y-coordinate of the point where the line intersected the y-axis.

Eureka Math Grade 8 Module 4 Lesson 19 Example Answer Key

Graph an equation in the form of y=mx+b.

Example 1.
Graph the equation y=\(\frac{2}{3}\) x+1. Name the slope and y-intercept point.
Answer:
The slope is m=\(\frac{2}{3}\), and the y-intercept point is (0,1).

→ To graph the equation, we must begin with the known point. In this case, that is the y-intercept point. We cannot begin with the slope because the slope describes the rate of change between two points. That means we need a point to begin with. On a graph, we plot the point (0,1).

→ Next, we use the slope to find the second point. We know that m=\(\frac{|Q R|}{|P Q|}\) =\(\frac{2}{3}\). The slope tells us exactly how many units to go to the right of P to find point Q and then how many vertical units we need to go from Q to find point R. How many units will we go to the right in order to find point Q? How do you know?
→ We need to go 3 units to the right of point P to find Q. We go 3 units because |PQ|=3.

→ How many vertical units from point Q must we go to find point R? How do you know?
→ We need to go 2 units from point Q to find R. We go 2 units because |QR|=2.
→ Will we go up from point Q or down from point Q to find R? How do you know?
→ We need to go up because the slope is positive. That means that the line will be left-to-right inclining.

→ Since we know that the line joining two distinct points of the form y=mx+b has slope m, and we specifically constructed points P and R with the slope in mind, we can join the points with a line.

Example 2.
Graph the equation y=-\(\frac{3}{4}\) x-2. Name the slope and y-intercept point.
Answer:
The slope is m=-\(\frac{3}{4}\), and the y-intercept point is (0,-2).

→ How do we begin?
→ We must begin by putting a known point on the graph, (0,-2).

→ We know that m=\(\frac{|Q R|}{|P Q|}\) =-\(\frac{3}{4}\). How many units will we go to the right in order to find point Q? How do you know?
→ We need to go 4 units to the right of point P to find Q. We go 4 units because |PQ|=4.

→ How many units from point Q must we go to find point R? How do you know?
→ We need to go 3 units from point Q to find R. We go 3 units because |QR|=3.
→ Will we go up from point Q or down from point Q to find R? How do you know?
→ We need to go down from point Q to point R because the slope is negative. That means that the line will be left-to-right declining.

→ Now we draw the line through the points P and R.

Example 3
Graph the equation y=4x-7. Name the slope and y-intercept point.
Answer:
The slope is m=4, and the y-intercept point is (0,-7).

→ Graph the equation y=4x-7. Name the slope and y-intercept point.
→ The slope is m=4, and the y-intercept point is (0,-7).
→ How do we begin?
→ We must begin by putting a known point on the graph, (0,-7).

→ Notice this time that the slope is the integer 4. In the last two examples, our slopes have been in the form of a fraction so that we can use the information in the numerator and denominator to determine the lengths of |PQ| and |QR|. Since m=\(\frac{|Q R|}{|P Q|}\) =4, what fraction can we use to represent slope to help us graph?
→ The number 4 is equivalent to the fraction \(\frac{4}{1}\).
→ Using m=\(\frac{|Q R|}{|P Q|}\) =\(\frac{4}{1}\), how many units will we go to the right in order to find point Q? How do you know?
→ We need to go 1 unit to the right of point P to find Q. We go 1 unit because |PQ|=1.

→ How many vertical units from point Q must we go to find point R? How do you know?
→ We need to go 4 units from point Q to find R. We go 4 units because |QR|=4.
→ Will we go up from point Q or down from point Q to find R? How do you know?
→ We need to go up from point Q to point R because the slope is positive. That means that the line will be left-to-right inclining.

→ Now we join the points P and R to make the line.

Eureka Math Grade 8 Module 4 Lesson 19 Exercise Answer Key

Exercises 1–4 (5 minutes)
Students complete Exercises 1–4 individually or in pairs.

Exercise 1.
Graph the equation y=\(\frac{5}{2}\) x -4.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=\(\frac{5}{2}\), and the y-intercept point is (0,-4).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Exercise 2.
Graph the equation y=-3x+6.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=-3, and the y-intercept point is (0,6).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 3.
The equation y=1x+0 can be simplified to y=x. Graph the equation y=x.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=1, and the y-intercept point is (0,0).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 4.
Graph the point (0,2).
Answer:

a. Find another point on the graph using the slope, m=\(\frac{2}{7}\).

b. Connect the points to make the line.

c. Draw a different line that goes through the point (0,2) with slope m=\(\frac{2}{7}\). What do you notice?
Answer:
Only one line can be drawn through the given point with the given slope.

Exercises 5–6
Students complete Exercises 5–6 individually or in pairs.

Exercise 5.
A bank put $10 into a savings account when you opened the account. Eight weeks later, you have a total of $24. Assume you saved the same amount every week.
a. If y is the total amount of money in the savings account and x represents the number of weeks, write an equation in the form y=mx+b that describes the situation.
Answer:
24=m(8)+10
14=8m
\(\frac{14}{8}\)=m
\(\frac{7}{4}\)=m
y=\(\frac{7}{4}\) x+10

b. Identify the slope and the y-intercept point. What do these numbers represent?
Answer:
The slope is \(\frac{7}{4}\), and the y-intercept point is (0,10). The y-intercept point represents the amount of money the bank gave me, in the amount of $10. The slope represents the amount of money I save each week, \(\frac{7}{4}\)=$1.75.

c. Graph the equation on a coordinate plane.
Answer:

d. Could any other line represent this situation? For example, could a line through point (0,10) with slope \(\frac{7}{5}\) represent the amount of money you save each week? Explain.
Answer:
No, a line through point (0,10) with slope \(\frac{7}{5}\) cannot represent this situation. That line would show that at the end of the 8 weeks I would have $21.20, but I was told that I would have $24 by the end of the 8 weeks.

Exercise 6.
A group of friends are on a road trip. After 120 miles, they stop to eat lunch. They continue their trip and drive at a constant rate of 50 miles per hour.
a. Let y represent the total distance traveled, and let x represent the number of hours driven after lunch. Write an equation to represent the total number of miles driven that day.
Answer:
y=50x+120

b. Identify the slope and the y-intercept point. What do these numbers represent?
Answer:
The slope is 50 and represents the rate of driving. The y-intercept point is 120 and represents the number of miles they had already driven before driving at the given constant rate.

c. Graph the equation on a coordinate plane.
Answer:

d. Could any other line represent this situation? For example, could a line through point (0,120) with slope 75 represent the total distance the friends drive? Explain.
Answer:
No, a line through point (0,120) with a slope of 75 could not represent this situation. That line would show that after an hour, the friends traveled a total distance of 195 miles. According to the information given, the friends would only have traveled 170 miles after one hour.

Eureka Math Grade 8 Module 4 Lesson 19 Exit Ticket Answer Key

Mrs. Hodson said that the graphs of the equations below are incorrect. Find the student’s errors, and correctly graph the equations.

Question 1.
Student graph of the equation y=\(\frac{1}{2}\) x+4:

Error:
Answer:
The student should have gone up 1 unit when finding |QR| since the slope is positive.
Correct graph of the equation:

Answer:

Question 2.
Student graph of the equation y=-\(\frac{3}{5}\) x-1:

Error:
Answer:
The student did not find the y-intercept point correctly. It should be the point (0,-1).
Correct graph of the equation:

Answer:

Eureka Math Grade 8 Module 4 Lesson 19 Problem Set Answer Key

Students practice graphing equations using y-intercept point and slope. Students need graph paper to complete the Problem Set. Optional Problem 11 has students show that there is only one line passing through a point with a given negative slope.

Graph each equation on a separate pair of x- and y-axes.

Question 1.
Graph the equation y=\(\frac{4}{5}\) x-5.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=\(\frac{4}{5}\), and the y-intercept point is (0,-5).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 2.
Graph the equation y=x+3.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=1, and the y-intercept point is (0,3).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 3.
Graph the equation y=-\(\frac{4}{3}\) x+4.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=-\(\frac{4}{3}\), and the y-intercept point is (0,4).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 4.
Graph the equation y=\(\frac{5}{2}\) x.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=\(\frac{5}{2}\), and the y-intercept point is (0,0).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 5.
Graph the equation y=2x-6.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=2, and the y-intercept point is (0,-6).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 6.
Graph the equation y=-5x+9.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=-5, and the y-intercept point is (0,9).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 7.
Graph the equation y=\(\frac{1}{3}\) x+1.
a. Name the slope and the y-intercept point.
Answer:
The slope is m=\(\frac{1}{3}\), and the y-intercept point is (0,1).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 8.
Graph the equation 5x+4y=8. (Hint: Transform the equation so that it is of the form y=mx+b.)
a. Name the slope and the y-intercept point.
Answer:
5x+4y=8
5x-5x+4y=8-5x
4y=8-5x
\(\frac{4}{4}\) y=\(\frac{8}{4}\)–\(\frac{5}{4}\) x
y=2-\(\frac{5}{4}\) x
y=-\(\frac{5}{4}\) x+2
The slope is m=-\(\frac{5}{4}\), and the y-intercept point is (0,2).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 9.
Graph the equation -2x+5y=30.
a. Name the slope and the y-intercept point.
Answer:
-2x+5y=30
2x+2x+5y=30+2x
5y=30+2x
\(\frac{5}{5}\) y=\(\frac{30}{5}\)+\(\frac{2}{5}\) x
y=6+\(\frac{2}{5}\) x
y=\(\frac{2}{5}\) x+6
The slope is m=\(\frac{2}{5}\), and the y-intercept point is (0,6).

b. Graph the known point, and then use the slope to find a second point before drawing the line.
Answer:

Question 10.
Let l and l’ be two lines with the same slope m passing through the same point P. Show that there is only one line with a slope m, where m<0, passing through the given point P. Draw a diagram if needed.
Answer:
First, assume that there are two different lines l and l’ with the same negative slope passing through P. From point P, I mark a point Q one unit to the right. Then, I draw a line parallel to the y-axis through point Q. The intersection of this line and line l and l’ are noted with points R and R’, respectively. By definition of slope, the lengths |QR| and |QR’ | represent the slopes of lines l and l’, respectively. We are given that the lines have the same slope, which means that lengths |QR| and |QR’| are equal. Since that is true, then points R and R’coincide and so do lines l and l’. Therefore, our assumption that they are different lines is false; l and l’ must be the same line. Therefore, there is only one line with slope m passing through the given point P.

Engage NY Eureka Math 8th Grade Module 4 Lesson 16 Answer Key

Eureka Math Grade 8 Module 4 Lesson 16 Example Answer Key

Example 1.
Using what you learned in the last lesson, determine the slope of the line with the following graph.

Answer:
The slope of the line is 3.

Example 2.
Using what you learned in the last lesson, determine the slope of the line with the following graph.

Answer:
The slope of this line is 2.

Example 3.
What is different about this line compared to the last two examples?

Answer:
This time, if we choose two points on the line that have a horizontal distance at 1, we cannot precisely determine the slope of the line because the vertical change is not an integer. It is some fractional amount.

→ Make a conjecture about how you could find the slope of this line.

Have students write their conjectures and share their ideas about how to find the slope of the line in this example; then, continue with the Discussion that follows.

Eureka Math Grade 8 Module 4 Lesson 16 Exercise Answer Key

Exercise
Let’s investigate concretely to see if the claim that we can find slope between any two points is true.

a. Select any two points on the line to label as P and R.
Answer:
Sample points are selected on the graph.

b. Identify the coordinates of points P and R.
Answer:
Sample points are labeled on the graph.

c. Find the slope of the line using as many different points as you can. Identify your points, and show your work below.
Answer:
Points selected by students will vary, but the slope should always equal 2. Students could choose to use points (0,5), (-1,3), (-2,1), (-3,-1), (-4,-3), and (-5,-5).

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key

Students practice finding slope between any two points on a line. Students also see that m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\) yields the same result as m=\(\frac{r_{2}-p_{2}}{r_{1}-p_{1}}\).

Question 1.
Calculate the slope of the line using two different pairs of points.

m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{30-(-10)}{-10-30}\)
=\(\frac{40}{-40}\)
=\(\frac{1}{1}\)
=-1
m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{10-(-10)}{10-30}\)
=\(\frac{20}{-20}\)
=-1

Question 2.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{3-2}{-2-2}\)
=\(\frac{1}{-4}\)
=-\(\frac{1}{4}\)

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{1-2}{6-2}\)
=\(\frac{-1}{4}\)
=-\(\frac{1}{4}\)

Question 3.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{1-5}{5-6}\)
=\(\frac{-4}{-1}\)
=\(\frac{4}{1}\)
=4

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{-3-5}{4-6}\)
=\(\frac{-8}{-2}\)
=4

Question 4.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{5-1}{3-5}\)
=\(\frac{4}{-2}\)
=\(-\frac{2}{1}\)
=-2

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{3-1}{4-5}\)
=\(\frac{2}{-1}\)
=-2

Question 5.
Calculate the slope of the line using two different pairs of points.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{2-1}{1-6}\)
=\(\frac{1}{-5}\)
=-\(\frac{1}{5}\)

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\)
=\(\frac{0-1}{11-6}\)
=\(\frac{-1}{5}\)
=-\(\frac{1}{5}\)

Question 6.
Calculate the slope of the line using two different pairs of points.

a. Select any two points on the line to compute the slope.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{-5-(-2)}{-7-(-5)}\)
=\(\frac{-3}{-2}\)
=\(\frac{3}{2}\)

b. Select two different points on the line to calculate the slope.
Answer:
Let the two new points be (-3,1) and (-1,4).
m=\(\frac{q_{2}-s_{2}}{q_{1}-s_{1}}\)
=\(\frac{1-4}{-3-(-1)}\)
=\(\frac{-3}{-2}\)
=\(\frac{3}{2}\)

c. What do you notice about your answers in parts (a) and (b)? Explain.
Answer:
The slopes are equal in parts (a) and (b). This is true because of what we know about similar triangles.
The slope triangle that is drawn between the two points selected in part (a) is similar to the slope triangle that is drawn between the two points in part (b) by the AA criterion. Then, because the corresponding sides of similar triangles are equal in ratio, the slopes are equal.

Question 7.
Calculate the slope of the line in the graph below.

Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{-4-2}{-7-2}\)
=\(\frac{-6}{-9}\)
=\(\frac{2}{3}\)

Question 8.
Your teacher tells you that a line goes through the points (-6, \(\frac{1}{2}\)) and (-4,3).
a. Calculate the slope of this line.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{\frac{1}{2}-3}{-6-(-4)}\)
=\(\frac{-\frac{5}{2}}{-2}\)
= \(\frac{\frac{5}{2}}{2}\)
=\(\frac{5}{2}\)÷2
=\(\frac{5}{2}\)×\(\frac{1}{2}\)
= \(\frac{5}{4}\)

b. Do you think the slope will be the same if the order of the points is reversed? Verify by calculating the slope, and explain your result.
The slope should be the same because we are joining the same two points.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
= \(\frac{3-\frac{1}{2}}{-4-(-6)}\)
= \(\frac{\frac{5}{2}}{2}\)
= \(\frac{5}{4}\)
Since the slope of a line can be computed using any two points on the same line, it makes sense that it does not matter which point we name as P and which point we name as R.

Question 9.
Use the graph to complete parts (a)–(c).

a. Select any two points on the line to calculate the slope.
Answer:
m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\)
=\(\frac{1-(-3)}{-3-0}\)
=\(\frac{4}{-3}\)
=-\(\frac{4}{3}\)

b. Compute the slope again, this time reversing the order of the coordinates.
Answer:
m=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\)
=\(\frac{-3-1}{0-(-3)}\)
=\(\frac{-4}{3}\)
=-\(\frac{4}{3}\)

c. What do you notice about the slopes you computed in parts (a) and (b)?
Answer:
The slopes are equal.

d. Why do you think m=\(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\).?
Answer:
If I multiply the first fraction by \(\frac{-1}{-1}\), then I get the second fraction:
\(\frac{-1}{-1}\)×\(\left(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\right)\)=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\).
I can do the same thing to the second fraction to obtain the first:
\(\frac{-1}{-1}\)×\(\left(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\right)\)=\(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)
Also, since I know that I can find the slope between any two points, it should not matter which point I pick first.

Question 10.
Each of the lines in the lesson was non-vertical. Consider the slope of a vertical line, x=2. Select two points on the line to calculate slope. Based on your answer, why do you think the topic of slope focuses only on non-vertical lines?

Answer:
Students can use any points on the line x=2 to determine that the slope is undefined. The computation of slope using the formula leads to a fraction with zero as its denominator. Since the slope of a vertical line is undefined, there is no need to focus on them.

Challenge:

Question 11.
A certain line has a slope of \(\frac{1}{2}\). Name two points that may be on the line.
Answer:
Answers will vary. Accept any answers that have a difference in y-values equal to 1 and a difference of x-values equal to 2. Points (6,4) and (4,3) may be on the line, for example.

Eureka Math Grade 8 Module 4 Lesson 16 Exit Ticket Answer Key

Find the rate of change of the line by completing parts (a) and (b).

a. Select any two points on the line to label as P and R. Name their coordinates.
Answer:
Answers will vary. Other points on the graph may have been chosen.
P(-1,0) and R(5,3)

b. Compute the rate of change of the line.
Answer:
m = \(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)
m=\(\frac{0-3}{-1-5}\)
=\(\frac{-3}{-6}\)
=\(\frac{1}{2}\)