Eureka Math

Engage NY Eureka Math Grade 6 Module 4 Lesson 33 Answer Key

Eureka Math Grade 6 Module 4 Lesson 33 Example Answer Key

Example 1:
What value(s) does the variable have to represent for the equation or inequality to result in a true number sentence? What value(s) does the variable have to represent for the equation or inequality to result in a false number sentence?

a. y + 6 = 16
Answer:
The number sentence is true when y is 10. The sentence is false when y is any number other than 10.

b. y + 6 > 16
Answer:
The number sentence is true when y is any number greater than 10. The sentence is false when y is 10 or any number less than 10.

c. y + 6 ≥ 16
Answer:
The number sentence is true when y is 10 or any number greater than 10. The sentence is fake when y is a number less than 10.

d. 3g = 15
Answer:
The number sentence is true when g is 5. The number sentence is false when g is any number other than 5.

e. 3g < 15
Answer:
The number sentence is true when g is any number less than 5. The number sentence is false when g is 5 or any number greater than 5.

f. 3g ≤ 15
Answer:
The number sentence is true when g is 5 or any number less than 5. The number sentence is false when g is any number greater than 5.

Example 2:
Which of the following number(s), if any, make the equation or inequality true: (0, 3, 5, 8, 10, 14)?

a. m + 4 = 12
Answer:
m = 8 or (8)

b. m + 4 < 12
Answer:
(0, 3, 5)

→ How does the answer to part (a) compare to the answer to part (b)?
In part (a), 8 is the only number that will result in a true number sentence. But in part (b), any number in the set that is less than 8 will make the number sentence true.

c. f – 4 = 2
Answer:
None of the numbers in the set will result in a true number sentence.

d. f – 4 > 2
Answer:
(8, 10, 14)

→ Is there a number that we could include in the set so that part (c) will have a solution?
Yes. The number 6 will make the equation in part (c) true.

→ Would 6 be part of the solution set in part (d)?
No. The 6 would not make port (d) a true number sentence because 6 – 4 is not greater than 2.

→ How could we change part (d) so that 6 would be part of the solution?
Answers will vary; If the > was changed to a ≥, we could include 6 in the solution set.

e. \(\frac{1}{2}\)h = 8
Answer:
None of the numbers in the set will result in a true number sentence.

f. \(\frac{1}{2}\)h ≥ 8
Answer:
None of the numbers in the set will result in a true number sentence.

→ Which whole numbers, if any, make the inequality in part (f) true?
Answers will vary; 16 and any number greater than 16 will make the number sentence true.

Eureka Math Grade 6 Module 4 Lesson 33 Exercise Answer Key

Exercises:

Choose the number(s), if any, that make the equation or inequality true from the following set of numbers: (0, 1, 5, 8, 11, 17).

Exercise 1.
m + 5 = 6
Answer:
m= 1 or {1)

Exercise 2.
m + 5 ≤ 6
Answer:
(0, 1)

Exercise 3.
5h = 40
Answer:
h = 8 or (8)

Exercise 4.
5h > 40
Answer:
(11, 17)

Exercise 5.
\(\frac{1}{2}\)y = 5
Answer:
There is no solution in the set.

Exercise 6.
\(\frac{1}{2}\)y ≤ 5
Answer:
(0, 1, 5, 8)

Exercise 7.
k – 3 = 20
Answer:
There is no solution in the set.

Exercise 8.
k – 3 > 20
Answer:
There is no solution in the set.

Eureka Math Grade 6 Module 4 Lesson 33 Problem Set Answer Key

Choose the number(s), if any, that make the equation or inequality true from the following set of numbers: (0, 3, 4, 5, 9, 13, 18, 24).

Question 1.
h – 8 = 5
Answer:
h = 13 or (13)

Question 2.
h – 8 < 5
Answer:
(0, 3, 4, 5, 9)

Question 3.
4g = 36
Answer:
g = 9 or (9)

Question 4.
4g ≥ 36
Answer:
(9, 13, 18, 24)

Question 5.
\(\frac{1}{4}\)y = 7
Answer:
There is no number in the set that will make this equation true.

Question 6.
\(\frac{1}{4}\)y > 7
Answer:
There is no number in the set that will make this inequality true.

Question 7.
m – 3 = 10
Answer:
m = 13 or (13)

Question 8.
m – 3 ≤ 10
Answer:
{0, 3, 4, 5, 9, 13}

Eureka Math Grade 6 Module 4 Lesson 33 Exit Ticket Answer Key

Choose the number(s), if any, that make the equation or inequality true from the following set of numbers: {3,4, 7, 9, 12, 18, 32).

Question 1.
\(\frac{1}{3}\)f = 4
Answer:
f = 12 or {12}

Question 2.
\(\frac{1}{3}\)f < 4
Answer:
{3, 4, 7, 9)

Question 3.
m + 7 = 20
Answer:
There is no number in the set that will make this equation true.

Question 4.
m + 7 ≥ 20
Answer:
{18, 32}

Engage NY Eureka Math 6th Grade Module 4 Lesson 19 Answer Key

Eureka Math Grade 6 Module 4 Lesson 19 Example Answer Key

Example 1.

Answer:

a. What if you don’t know how old I am? Let’s use a variable for my age. Let Y = my age In years. Can you
develop an expression to describe how old my sister is?
Answer:
Your sister is Y + 2 years old.

b. Please add that to the last row of the table.
Answer:
My age is Y years. My sister is Y + 2 years old. So, no matter what my age is (or was), my sister’s age in years will always be two years greater than mine.

Example 2.

Answer:

a. How old was I when my sister was 6 years old?
Answer:
4 years old

b. How old was I when my sister was 15 years old?
Answer:
13 years old

c. How do you know?
Answer:
My age is always 2 years less than my sister’s age.

d. Look at the table in Example 2. If you know my sister’s age, can you determine my age?
Answer:
We can subtract two from your sister’s age, and that will equal your age.

e. If we use the variable G for my sister’s age in years, what expression would describe my age in years?
Answer:
G – 2

f. Fill in the last row of the table with the expressions.
Answer:
My age is G – 2 years. My sister is G years old.

g. With a partner, calculate how old I was when my sister was 22, 23, and 24 years old.
Answer:
You were 20, 21, and 22 years old, respectively.

Eureka Math Grade 6 Module 4 Lesson 19 Exercise Answer Key

Exercise 1.
Noah and Carter are collecting box tops for their school. They each bring in 1 box top per day starting on the first day of school. However, Carter had a head start because his aunt sent him 15 box tops before school began. Noah’s grandma saved 10 box tops, and Noah added those on his first day.
a. Fill in the missing values that indicate the total number of box tops each boy brought to school.

Answer:

b. If we let D be the number of days since the new school year began, on day D of school, how many box tops will Noah have brought to school?
Answer:
D + 10 box tops

c. On day D of school, how many box tops will Carter have brought to school?
Answer:
D + 15 box tops

d. On day 10 of school, how many box tops will Noah have brought to school?
Answer:
10 + 10 = 20; On day 10, Noah would have brought in 20 box tops.

e. On day 10 of school, how many box tops will Carter have brought to school?
Answer:
10 + 15 = 25; on day 10, Carter would have brought in 25 box tops.

Exercise 2.
Each week the Primary School recycles 200 pounds of paper. The Intermediate School also recycles the same amount but had another 300 pounds left over from summer school. The Intermediate School custodian added this extra 300 pounds to the first recycle week.
a. Number the weeks, and record the amount of paper recycled by both schools.

Answer:

b. If this trend continues, what will be the total amount collected for each school on Week 10?
Answer:
The Primary School will have collected 2,000 pounds. The Intermediate School will have collected 2,300 pounds.

Exercise 3.
Shelly and Kristen share a birthday, but Shelly is 5 years older.
a. Make a table showing their ages every year, beginning when Kristen was born.

Answer:

b. If Kristen is 16 years old, how old is Shelly?
Answer:
If Kristen is 16 years old, Shelly is 21 years old.

c. If Kristen is K years old, how old is Shelly?
Answer:
If Kristen is K years old, Shelly is K + 5 years old.

d. If Shelly is S years old, how old is Kristen?
Answer:
If Shelly is S years old, Kristen is S – 5 years old.

Eureka Math Grade 6 Module 4 Lesson 19 Problem Set Answer Key

Question 1.
Suellen and Tara are in sixth grade, and both take dance lessons at Twinkle Toes Dance Studio. This Is Suellen’s first year, while this is Tara’s fifth year of dance lessons. Both girls plan to continue taking lessons throughout high school.
a. Complete the table showing the number of years the girls will have danced at the studio.

Answer:

b. If Suellen has been taking dance lessons for Y years, how many years has Tara been taking lessons?
Answer:
Tara has been taking dance lessons for Y + 4 years.

Question 2.
Daejoy and Damian collect fossils. Before they went on a fossil-hunting trip, Daejoy had 25 fossils in her collection, and Damian had 16 fossils In his collection. On a 10-day fossil-hunting trIp, they each collected 2 new fossils each day.
a. Make a table showing how many fossils each person had In their collection at the end of each day.

Answer:

b. If this pattern of fossil finding continues, how many fossils does Damian have when Daejoy has F fossils?
Answer:
When Daejoy has F fossils, Damian has F – 9 fossils.

c. If this pattern of fossil finding continues, how many fossils does Damian have when Daejoy has 55 fossils?
Answer:
55 – 9 = 46
When Daejoy has 55 fossils, Damian has 46 fossils.

Question 3.
A train consists of three types of cars: box cars, an engine, and a caboose. The relationship among the types of cars is demonstrated in the table below.

a. Tom wrote an expression for the relationship depicted in the table as B + 2. Theresa wrote an expression for the same relationship as C – 2. Is It possible to have two different expressions to represent one relationship? Explain.
Answer:
Both expressions can represent the same relationship, depending on the point of view. The expression B + 2 represents the number of box cars plus an engine and a caboose. The expression C – 2 represents the whole car length of the train, less the engine and caboose.

b. What do you think the variable in each student’s expression represents? How would you define them?
Answer:
The variable C would represent the total cars in the train. The variable B would represent the number of box cars.

Question 4.
David was 3 when Marieka was born. Complete the table.

Answer:

Question 5.
Caitlln and Michael are playing a card game. In the first round, Caitlin scored 200 points, and Michael scored 175 points. In each of the next few rounds, they each scored 50 points. Their score sheet is below.

a. If this trend continues, how many points will Michael have when Caitlin has 600 points?
Answer:
600 – 25 = 575
Michael will have 575 points.

b. If this trend continues, how many points will Michael have when Caitlin has C points?
Answer:
Michael will have C – 25 points.

c. If this trend continues, how many points will Caitlin have when Michael has 975 points?
Answer:
975 + 25 = 1000
Caitlin will have 1,000 points.

d. If this trend continues, how many points will Caitlin have when Michael has M points?
Answer:
Caitlin will have M + 25 points.

Question 6.
The high school marching band has 15 drummers this year. The band director Insists that there are to be 5 more trumpet players than drummers at all times.
a. How many trumpet players are in the marching band this year?
Answer:
15 + 5 = 20. There are 20 trumpet players this year.

b. Write an expression that describes the relationship of the number of trumpet players (T) and the number of drummers (D).
Answer:
T = D + 5 or D = T – 5

c. If there are only 14 trumpet players interested in joining the marching band next year, how many drummers will the band director want In the band?
Answer:
14 – 5 = 9.
The band director will want 9 drummers.

Eureka Math Grade 6 Module 4 Lesson 19 Exit Ticket Answer Key

Jenna and Allie work together at a piano factory. They both were hired on January 3, but Jenna was hired in 2005, and Allie was hired in 2009.

a. Fill in the table below to summarize the two workers’ experience totals.

Answer:

b. If both workers continue working at the piano factory, when Allie has A years of experience on the job, how many years of experience will Jenna have on the job?
Answer:
Jenna will have been on the job for A + 4 years.

c. If both workers continue working at the piano factory, when Allie has 20 years of experience on the job, how many years of experience will Jenna have on the job?
Answer:
20 + 4 = 24
Jenna will have been on the job for 24 years.

Eureka Math Grade 6 Module 4 Lesson 19 Opening Exercise Answer Key

My older sister is exactly two years older than I am. Sharing a birthday is both fun and annoying. Every year on our birthday, we have a party, which is fun, but she always brags that she is two years older than I am, which is annoying. Shown below is a table of our ages, starting when I was born:

a. Looking at the table, what patterns do you see? Tell a partner.
Answer:
My sister’s age is always two years more than my age.

b. On the day I turned 8 years old, how old was my sister?
Answer:
10 years old

c. How do you know?
Answer:
Since my sister’s age is always two years more than my age, we just add 2 to my age. 8 + 2 = 10

d. On the day I turned 16 years old, how old was my sister?
Answer:
18 years old

e. How do you know?
Answer:
Since my sister’s age is always two years more than my age, we just add 2 to my age. 16 + 2 = 18

f. Do we need to extend the table to calculate these answers?
Answer:
No; the pattern is to add 2 to your age to calculate your sister’s age.

Eureka Math Grade 6 Module 4 Lesson 19 Subtraction of Decimals Answer Key

Subtraction of Decimals – Round 1
Directions: Evaluate each expression.

Question 1.
55 – 50
Answer:
5

Question 2.
55 – 5
Answer:
50

Question 3.
5.5 – 5
Answer:
0.5

Question 4.
5.5 – 0.5
Answer:
5

Question 5.
88 – 80
Answer:
8

Question 6.
88 – 8
Answer:
80

Question 7.
8.8 – 8
Answer:
0.8

Question 8.
8.8 – 0.8
Answer:
8

Question 9.
33 – 30
Answer:
3

Question 10.
33 – 3
Answer:
30

Question 11.
3.3 – 3
Answer:
0.3

Question 12.
1 – 0.3
Answer:
0.7

Question 13.
1 – 0.03
Answer:
0.97

Question 14.
1 – 0.003
Answer:
0.997

Question 15.
0.1 – 0.03
Answer:
0.07

Question 16.
4 – 0.8
Answer:
3.2

Question 17.
4 – 0.08
Answer:
3.92

Question 18.
4 – 0.008
Answer:
3.992

Question 19.
0.4 – 0.08
Answer:
0.32

Question 20.
9 – 0.4
Answer:
8.6

Question 21.
9 – 0.04
Answer:
8.96

Question 22.
9 – 0.004
Answer:
8.996

Question 23.
9.9 – 5
Answer:
4.9

Question 24.
9.9 – 0.5
Answer:
9.4

Question 25.
0.99 – 0.5
Answer:
0.49

Question 26.
0.99 – 0.05
Answer:
0.94

Question 27.
4.7 – 2
Answer:
2.7

Question 28.
4.7 – 0.2
Answer:
4.5

Question 29.
0.47 – 0.2
Answer:
0.27

Question 30.
0.47 – 0.02
Answer:
0.45

Question 31.
8.4 – 1
Answer:
7.4

Question 32.
8.4 – 0.1
Answer:
8.3

Question 33.
0.84 – 0.1
Answer:
0.74

Question 34.
7.2 – 5
Answer:
2.2

Question 35.
7.2 – 0.5
Answer:
6.7

Question 36.
0.72 – 0.5
Answer:
0.22

Question 37.
0.72 – 0.05
Answer:
0.67

Question 38.
8.6 – 7
Answer:
1.6

Question 39.
8.6 – 0.7
Answer:
7.9

Question 40.
0.86 – 0.7
Answer:
0. 16

Question 41.
0.86 – 0.07
Answer:
0.79

Question 42.
5.1 – 4
Answer:
1.1

Question 43.
5.1 – 0.4
Answer:
4.7

Question 44.
0.51 – 0.4
Answer:
0.11

Subtraction of Decimals – Round 2
Directions: Evaluate each expression.

Question 1.
66 – 60
Answer:
6

Question 2.
66 – 6
Answer:
60

Question 3.
6.6 – 6
Answer:
0.6

Question 4.
6.6 – 0.6
Answer:
6

Question 5.
99 – 90
Answer:
9

Question 6.
99 – 9
Answer:
90

Question 7.
9.9 – 9
Answer:
0.9

Question 8.
9.9 – 0.9
Answer:
9

Question 9.
22 – 20
Answer:
2

Question 10.
22 – 2
Answer:
20

Question 11.
2.2 – 2
Answer:
0.2

Question 12.
3 – 0.4
Answer:
2.6

Question 13.
3 – 0.04
Answer:
2.96

Question 14.
3 – 0.004
Answer:
2.996

Question 15.
0.3 – 0.04
Answer:
0.26

Question 16.
8 – 0.2
Answer:
7.8

Question 17.
8 – 0.02
Answer:
7.98

Question 18.
8 – 0.0 02
Answer:
7.998

Question 19.
0.8 – 0.02
Answer:
0.78

Question 20.
5 – 0.1
Answer:
4.9

Question 21.
5 – 0.01
Answer:
4.99

Question 22.
5 – 0.001
Answer:
4.999

Question 23.
6.8 – 4
Answer:
2.8

Question 24.
6.8 – 0.4
Answer:
6.4

Question 25.
0.68 – 0.4
Answer:
0.28

Question 26.
0.68 – 0.04
Answer:
0.64

Question 27.
7.3 – 1
Answer:
6.3

Question 28.
7.3 – 0.1
Answer:
7.2

Question 29.
0.73 – 0.1
Answer:
0.63

Question 30.
0.73 – 0.01
Answer:
0.72

Question 31.
9.5 – 2
Answer:
7.5

Question 32.
9.5 – 0.2
Answer:
9.3

Question 33.
0.95 – 0.2
Answer:
0.75

Question 34.
8.3 – 5
Answer:
3.3

Question 35.
8.3 – 0.5
Answer:
7.8

Question 36.
0.83 – 0.5
Answer:
0.33

Question 37.
0.83 – 0.05
Answer:
0.78

Question 38.
7.2 – 4
Answer:
3.2

Question 39.
7.2 – 0.4
Answer:
6.8

Question 40.
0.72 – 0.4
Answer:
0.32

Question 41.
0.72 – 0.04
Answer:
0.68

Question 42.
9.3 – 7
Answer:
2.3

Question 43.
9.3 – 0.7
Answer:
8.6

Question 44.
0.93 – 0.7
Answer:
0.23

Engage NY Eureka Math 6th Grade Module 4 Lesson 18 Answer Key

Eureka Math Grade 6 Module 4 Lesson 18 Example Answer Key

Example 1. The Importance of Being Specific in Naming Variables
When naming variables in expressions, it is important to be very clear about what they represent. The units of measure must be included if something is measured.

Example 2: Writing and Evaluating Addition and Subtraction Expressions
Read each story problem. Identify the unknown quantity, and write the addition or subtraction expression that is described. Finally, evaluate your expression using the information given in column four.


Answer:

Eureka Math Grade 6 Module 4 Lesson 18 Exercise Answer Key

Exercise 1.
Read the variable in the table, and improve the description given, making it more specific.

Answer:
Answers may vary because students may choose a different unit.

Exercise 2.
Read each variable in the table, and improve the description given, making it more specific.

Answer:

Eureka Math Grade 6 Module 4 Lesson 18 Problem Set Answer Key

Question 1.
Read each story problem. Identify the unknown quantity, and write the addition or subtraction expression that is described. Finally, evaluate your expression using the information given in column four.

Answer:
Sample answers are shown. An additional expression can be written for each.

Question 2.
If George went camping 15 times, how could you figure out how many times Dave went camping?
Answer:
Adding 3 to George’s camping trip total (15) would yield an answer of 18 trips for Dave.

Eureka Math Grade 6 Module 4 Lesson 18 Exit Ticket Answer Key

Kathleen lost a tooth today. Now she has lost 4 more than her sister Cara lost.

Question 1.
Write an expression to represent the number of teeth Cara has lost. Let K represent the number of teeth Kathleen lost.
Answer:
Expression: K – 4

Question 2.
Write an expression to represent the number of teeth Kathleen lost. Let C represent the number of teeth Cara lost.
Answer:
Expression: C + 4

Question 3.
If Cara lost 3 teeth, how many teeth has Kathleen lost
Answer:
C + 4; 3 + 4; Kathleen has lost 7 teeth.

Eureka Math Grade 6 Module 4 Lesson 18 Opening Exercise Answer Key

How can we show a number increased by 2?
Answer:
a + 2 or 2 + a

Can you prove this using a model?
Answer:
Yes. I can use a tape diagram.

Engage NY Eureka Math 6th Grade Module 4 Lesson 17 Answer Key

Eureka Math Grade 6 Module 4 Lesson 17 Exercise Answer Key

Exercises

Station One:

1. The sum of a and b
Answer:
a + b

2. Five more than twice a number c
Answer:
5 + 2c or 2c + S

3. Martha bought d number of apples and then ate 6 of them.
Answer:
d – 6

Station Two:

1. 14 decreased by p
Answer:
14 – p

2. The total of d and f, divided by 8
Answer:
\(\frac{d+f}{8}\) or (d + f) ÷ 8

3. Rashod scored 6 less than 3 times as many baskets as Mike. Mike scored b baskets.
Answer:
3b – 6

Station Three:
1. The quotient of c and 6
Answer:
\(\frac{c}{6}\)

2. Triple the sum of x and 17
Answer:
3(x + 17)

3. Gabrielle had b buttons but then lost 6. Gabrielle took the remaining buttons and split them equally among her 5 friends.
Answer:
\(\frac{b-6}{5}\) or (b – 6) ÷ 5

Station Four:

1. d doubled
Answer:
2d

2. Three more than 4 times a number x
Answer:
4x + 3 or 3 + 4x

3. Mall has c pleas of candy. She doubles the amount of candy she has and then gives away 15 pieces.
Answer:
2c – 15

Station Five:

1. f cubed
Answer:
f³

2. The quantity of 4 increased by a and then the sum is divided by 9.
Answer:
\(\frac{4+a}{9}\) or (4 + a) ÷ 9

3. Tai earned 4 points fewer than double Oden’s points. Oden earned p points.
Answer:
2p – 4

Station Six:

1. The difference between d and 8
Answer:
d – 8

2. 6 less than the sum of d and 9
Answer:
(d + 9) – 6

3. Adalyn has x pants and s shirts. She combined them and sold half of them. How many items did Adalyn sell?
Answer:
\(\frac{x+s}{2}\) or \(\frac{1}{2}\)(x + s)

Eureka Math Grade 6 Module 4 Lesson 17 Problem Set Answer Key

Write an expression using letters and/or numbers for each problem below.

Question 1.
4 less than the quantity of 8 times n
Answer:
8n – 4

Question 2.
6 times the sum of y and 11
Answer:
6(y + 11)

Question 3.
The square of m reduced by 49
Answer:
m² – 49

Question 4.
The quotient when the quantity of 17 plus p is divided by 8
Answer:
\(\frac{17+p}{8}\) or (17 + p) ÷ 8

Question 5.
Jim earned j in tips, and Steve earned s in tips. They combine their tips and then split them equally.
Answer:
\(\frac{\boldsymbol{j}+s}{2}\) or (j + s) ÷ 2

Question 6.
Owen has c collector cards. He quadruples the number of cards he has and then combines them with lan, who has j collector cards.
Answer:
4c + i

Question 7.
Rae runs 4 times as many miles as Madison and Aaliyah combined. Madison runs m miles, and Aaliyah runs a miles.
Answer:
4(m + a)

Question 8.
By using coupons, Mary Jo is able to decrease the retail price of her groceries, g, by $125.
Answer:
g – 125

Question 9.
To calculate the area of a triangle, you find the product of the base and height and then divide by 2
Answer:
\(\frac{b h}{2}\) or bh ÷ 2

Question 10.
The temperature today was 10 degrees colder than twice yesterday’s temperature, t.
Answer:
2t – 10

Eureka Math Grade 6 Module 4 Lesson 17 Exit Ticket Answer Key

Write an expression using letters and/or numbers for each problem below.

Question 1.
d squared
Answer:
d²

Question 2.
A number x increased by 6, and then the sum is doubled.
Answer:
2(x + 6)

Question 3.
The total of h and b is split into 5 equal groups.
Answer:
\(\frac{h+b}{5}\) or (h + b) ÷ 5

Question 4.
Jazmin has increased her $45 by m dollars and then spends a third of the entire amount.
Answer:
\(\frac{45+m}{3}\) or \(\frac{1}{3}\)(45 + m)

Question 5.
Bill has d more than 3 times the number of baseball cards as Frank. Frank has f baseball cards.
Answer:
3f + d or d + 3f

Eureka Math Grade 6 Module 4 Lesson 17 Addition of Decimals Answer Key

Addition of Decimals I – Round 1

Directions: Evaluate each expression.

Question 1.
5.1 + 6
Answer:
11.1

Question 2.
5.1 + 0.6
Answer:
5.7

Question 3.
5.1 + 0.06
Answer:
5. 16

Question 4.
5.1 + 0.006
Answer:
5. 106

Question 5.
5.1 + 0.0006
Answer:
5. 1006

Question 6.
3 + 2.4
Answer:
5.4

Question 7.
0.3 + 2.4
Answer:
2.7

Question 8.
0.03 + 2.4
Answer:
2.43

Question 9.
0.003 + 2.4
Answer:
2.403

Question 10.
0.0003 + 2.4
Answer:
2. 4003

Question 11.
24 + 0.3
Answer:
24.3

Question 12.
2 + 0.3
Answer:
2.3

Question 13.
0.2 + 0.03
Answer:
0.23

Question 14.
0.02 + 0.3
Answer:
0.32

Question 15.
0.2 + 3
Answer:
3.2

Question 16.
2 + 0.03
Answer:
2.03

Question 17.
5 + 0.4
Answer:
5.4

Question 18.
0.5 + 0.04
Answer:
0.54

Question 19.
0.05 + 0.4
Answer:
0.45

Question 20.
0.5 + 4
Answer:
4.5

Question 21.
5 + 0.04
Answer:
5.04

Question 22.
0.5 + 0.4
Answer:
0.9

Question 23.
3.6 + 2.1
Answer:
5.7

Question 24.
3.6 + 0.21
Answer:
3.81

Question 25.
3.6 + 0.021
Answer:
3.621

Question 26.
0.36 + 0.02 1
Answer:
0.381

Question 27.
0.036 + 0.021
Answer:
0.057

Question 28.
1.4 + 42
Answer:
43.4

Question 29.
1.4 + 4.2
Answer:
5.6

Question 30.
1.4 + 0.42
Answer:
1.82

Question 31.
1.4 + 0.042
Answer:
1.442

Question 32.
0.14 + 0.042
Answer:
0. 182

Question 33.
0.0 14 + 0.042
Answer:
0.056

Question 34.
0.8 + 2
Answer:
2.8

Question 35.
0.8 + 0.2
Answer:
1

Question 36.
0.08 + 0.02
Answer:
0. 1

Question 37.
0.008 + 0.002
Answer:
0.01

Question 38.
6 + 0.4
Answer:
6.4

Question 39.
0.6 + 0.4
Answer:
1

Question 40.
0.06 + 0.04
Answer:
0. 1

Question 41.
0.006 + 0.004
Answer:
0.01

Question 42.
0.1 + 9
Answer:
9.1

Question 43.
0.1 + 0.9
Answer:
1

Question 44.
0.01 + 0.09
Answer:
0.1

Addition of Decimals I – Round 2

Directions: Evaluate each expression.

Question 1.
3.2 + 5
Answer:
8.2

Question 2.
3.2 + 0.5
Answer:
3.7

Question 3.
3.2 + 0.05
Answer:
3.25

Question 4.
3.2 + 0.005
Answer:
3.205

Question 5.
3.2 + 0.0005
Answer:
3.2005

Question 6.
4 + 5.3
Answer:
9.3

Question 7.
0.4 + 5.3
Answer:
5.7

Question 8.
0.04 + 5.3
Answer:
5.34

Question 9.
0.004 + 5.3
Answer:
5.304

Question 10.
0.0004 + 5.3
Answer:
5.3004

Question 11.
4 + 0.53
Answer:
4.53

Question 12.
6 + 0.2
Answer:
6.2

Question 13.
0.6 + 0.02
Answer:
0.62

Question 14.
0.06 + 0.2
Answer:
0.26

Question 15.
0.6 + 2
Answer:
2.6

Question 16.
2 + 0.06
Answer:
2.06

Question 17.
1 + 0.7
Answer:
1.7

Question 18.
0.1 + 0.07
Answer:
0. 17

Question 19.
0.01 + 0.7
Answer:
0.71

Question 20.
0.1 + 7
Answer:
7.1

Question 21.
1 + 0.07
Answer:
1.07

Question 22.
0.1 + 0.7
Answer:
0.8

Question 23.
4.2 + 5.5
Answer:
9.7

Question 24.
4.2 + 0.55
Answer:
4.75

Question 25.
4.2 + 0.055
Answer:
4.255

Question 26.
0.42 + 0.055
Answer:
0.475

Question 27.
0.042 + 0.055
Answer:
0.097

Question 28.
2.7 + 12
Answer:
14.7

Question 29.
2.7 + 1.2
Answer:
3.9

Question 30.
2.7 + 0.12
Answer:
2.82

Question 31.
2.7 + 0.012
Answer:
2.712

Question 32.
0.27 + 0.012
Answer:
0.282

Question 33.
0.027 + 0.012
Answer:
0.039

Question 34.
0.7 + 3
Answer:
3.7

Question 35.
0.7 + 0.3
Answer:
1

Question 36.
0.07 + 0.03
Answer:
0. 1

Question 37.
0.007 + 0.003
Answer:
0.01

Question 38.
5 + 0.5
Answer:
5.5

Question 39.
0.5 + 0.5
Answer:
1

Question 40.
0.05 + 0.05
Answer:
0. 1

Question 41.
0.005 + 0.005
Answer:
0.01

Question 42.
0.2 + 8
Answer:
8.2

Question 43.
0.2 + 0.8
Answer:
1

Question 44.
0.02 + 0.08
Answer:
0. 1

Engage NY Eureka Math 6th Grade Module 4 Lesson 16 Answer Key

Eureka Math Grade 6 Module 4 Lesson 16 Exercise Answer Key

Mark the text by underlining key words, and then write an expression using variables and/or numbers for each statement.

Exercise 1.
b decreased by c squared
Answer:
b decreased by c squared
b – c²

Exercise 2.
24 divided by the product of 2 and a
Answer:
24 divided by the product of 2 and a
\(\frac{24}{2 a}\) or 24 ÷ (2a)

Exercise 3.
150 decreased by the quantity of 6 plus b
Answer:
150 decreased by the quantity of 6 plus b
150 – (6 + b)

Exercise 4.
The sum of twice c and 10
Answer:
The sum of twice c and 10
2c + 10

Exercise 5.
Marlo had $35 but then spent Sm
Answer:
Mario had $35 but then spent $m.
35 – m

Exercise 6.
Samantha saved her money and was able to quadruple the original amount, m.
Answer:
Samantha saved her money and was able to quadruple the original amount, m.
4m

Exercise 7.
Veronica increased her grade, g, by 4 points and then doubled it.
Answer:
Veronica increased her grade, g, by 4 points and then doubled it.
2(g + 4)

Exercise 8.
Adbell had m pieces of candy and ate 5 of them. Then, he split the remaining candy equally among 4 friends.
Answer:
Adbell had m pieces of candy ate 5 of them. Then, he split the remaining candy equally among 4 friends.
\(\frac{m-5}{4}\) or (m – 5) ÷ 4

Exercise 9.
To find out how much paint is needed, Mr. Jones must square the side length, s, of the gate and then subtract 15.
Answer:
To find out how much paint is needed, Mr. Jones must square the side length, s, of the gate and then subtract 15.
s² – 15

Exercise 10.
Luis brought x cans of cola to the party, Faith brought d cans of cola, and De’Shawn brought h cans of cola. How many cans of cola did they bring altogether?
Answer:
Luis brought x cans of cola to the party, Faith brought d cans of cola, and De’Shawn brought h cans of cola. How many cans of cola did they bring altogether?
x + d + h

Eureka Math Grade 6 Module 4 Lesson 16 Problem Set Answer Key

Mark the text by underlining key words, and then write an expression using variables and numbers for each of the statements below.

Question 1.
Justin can type w words per minute. Melvin can type 4 times as many words a Justin. Write an expression that represents the rate at which Melvin can type.
Answer:
Justin can type w words per minute. Melvin can type 4 times as many words as justin. Write an expression that represents the rate at which Melvin can type.
4w

Question 2.
Yohanna swam y yards yesterday. Sheylin swam 5 yards less than half the amount of yards as Yohanna. Write an expression that represents the number of yards Sheylin swam yesterday.
Answer:
Yohanna swam y yards yesterday. Sheylin swam 5 yards less than half the amount of yards as Yohanna. Write an expression that represents the number of yards Sheylin swam yesterday.
\(\frac{y}{2}\) – 5 or y ÷ 2 – 5 or \(\frac{1}{2}\)y – 5

Question 3.
A number d is decreased by 5 and then doubled.
Answer:
A number d is decreased by 5 and then doubled.
2(d – 5)

Question 4.
Nahom had n baseball cards, and Semir had s baseball cards. They combined their baseball cards and then sold 10 of them.
Answer:
Nahom had n baseball cards, and Semir had s baseball cards. They combined their baseball cards and then sold 10 of them.
n + s – 10

Question 5.
The sum of 25 and h is divided by f cubed.
Answer:
The sum of 25 and h is divided by f cubed.
\(\frac{25+h}{f^{3}}\) or (25 + h) ÷ f³

Eureka Math Grade 6 Module 4 Lesson 16 Exit Ticket Answer Key

Mark the text by underlining key words, and then write an expression using variables and/or numbers for each of the statements below.

Question 1.
Omaya picked x amount of apples, took a break, and then picked y more. Write the expression that models the total number of apples Omaya picked.
Answer:
Omaya picked x amount of apples, took a break, and then picked v more.
x + v

Question 2.
A number h is tripled and then decreased by 8.
Answer:
A number h is tripled and then decreased by 8.
3h – 8

Question 3.
Sidney brought s carrots to school and combined them with Jenan’s j carrots. She then split them equally among 8 friends.
Answer:
Sidney brought s carrots to school and combined them with Jenan’s j carrots. She then split them equally among 8 friends.
\(\frac{s+j}{8}\) or (s + j) ÷ 8

Question 4.
15 less than the quotient of e and d
Answer:
15 less than the quotient of e and d
\(\frac{e}{d}\) – 15 ore ÷ d – 15

Question 5.
Marissa’s hair was 10 inches long, and then she cut h inches.
Answer:
Marissa’s hair was 10 inches long, and then she cut h inches.
10 – h

Eureka Math Grade 6 Module 4 Lesson 16 Opening Exercise Answer Key

Underline the key words in each statement.

a. The sum of twice b and 5
Answer:
The sum of twice b and 5

b. The quotient of c and d
Answer:
The quotient of C and d

c. a raised to the fifth power and then increased by the product of 5 and c
Answer:
a raised to the fifth power and then increased by the product of 5 and c

d. The quantity of a plus b divided by 4
Answer:
The quantity of a plus b divided by 4

e. 10 less than the product of 15 and c
Answer:
10 less than the product of 15 and c

f. 5 times d and then increased by 8
Answer:
5 times d and then increased by 8

Eureka Math Grade 6 Module 4 Lesson 16 Mathematical Model Exercise Answer Key

Mathematical Modeling Exercise 1

Model how to change the expressions given in the Opening Exercise from words to variables and numbers.

a. The sum of twice band 5
Answer:
2b + 5

b. The quotient of c and d
Answer:
\(\frac{c}{d}\)

c. a raised to the fifth power and then increased by the product of 5 and c
Answer:
a⁵ + 5c

d. The quantity of a plus b divided by 4
Answer:
\(\frac{a+b}{4}\)

e. 10 less than the product of 15 and c
Answer:
15c – 10

f. 5 times d and then increased by 8
Answer:
5d + 8

Mathematical Modeling Exercise 2

Model how to change each real-world scenario to an expression using variables and numbers. Underline the text to show the key words before writing the expression.

Marcus has 4 more dollars than Yaseen. If y is the amount of money Yaseen has, write an expression to show how much money Marcus has.
Answer:
→ Underline key words.
Marcus has 4 more dollars than Yaseen.
→ If Yaseen had $7, how much money would Marcus have?
$11
→ How did you get that?
Added 7 + 4
→ Write an expression using y for the amount of money Yaseen has.
y + 4

Mario is missing half of his assignments. If a represents the number of assignments, write an expression to show how many assignments Mario is missing.
Answer:
→ Underline key words.
Mario is missing half of his assignments.
→ If Mario was assigned 10 assignments, how many is he missing?
5
→ How did you get that?
10 ÷ 2
→ Write an expression using a for the number of assignments Mario was assigned.
\(\frac{a}{2}\) or a ÷ 2

Kamilah’s weight has tripled since her first birthday. If w represents the amount Kamilah weighed on her first birthday, write an expression to show how much Kamilah weighs now.
Answer:
→ Underline key words.
Kamilah’s weight has tripled since her first birthday.
→ If Kamilah weighed 20 pounds on her first birthday, how much does she weigh flow?
60 pounds
→ How did you get that?
Multiplied 3 by 20
→ Write an expression using w for Kamilah’s weight on her first birthday.
3w

Nathan brings cupcakes to school and gives them to his five best friends, who share them equally. If c represents the number of cupcakes Nathan brings to school, write an expression to show how many cupcakes each of his friends receive.
Answer:
→ Underline key words.
Nathan brings cupcakes to school and gives them to his five best friends, who share them equally.
→ If Nathan brings 15 cupcakes to school, how many will each friend receive?
3
→ How did you determine that?
15 ÷ 5
→ Write an expression using c to represent the number of cupcakes Nathan brings to school.
\(\frac{c}{5}\) or c ÷ 5

Mrs. Marcus combines her atlases and dictionaries and then divides them among 10 different tables. If a represents the number of atlases and d represents the number of dictionaries Mrs. Marcus has, write an expression to show how many books would be on each table.
Answer:
→ Mrs. Marcus combines her atlases and dictionaries and then divides them among 10 different tables.
→ If Mrs. Marcus had 8 atlases and 12 dictionaries, how many books would be at each table?
2
→ How did you determine that?
Added the atlases and dictionaries together and then divided by lo.
→ Write an expression using a for atlases and d for dictionaries to represent how many books each table would receive.
\(\frac{a+d}{10}\) or (a + d) ÷ 10

To improve in basketball, Ivan’s coach told him that he needs to take four times as many free throws and four times as many jump shots every day. If f represents the number of free throws and j represents the number of jump shots Ivan shoots daily, write an expression to show how many shots he will need to take in order to improve in basketball.
Answer:
→ Underline key words.
To improve in basketball, Ivan needs to shoot 4 times more free throws and jump shots daily.
→ If Ivan shoots S free throws and 10 jump shots, how many will he need to shoot in order to improve in basketball?
60
→ How did you determine that?
Added the free throws and jump shots together and then multiplied by 4
→ Write an expression using f for free throws and j for jump shots to represent how many shots Ivan will have to take in order to improve in basketball.
4(f + j) or 4f + 4j

Engage NY Eureka Math 6th Grade Module 4 Lesson 15 Answer Key

Eureka Math Grade 6 Module 4 Lesson 15 Example Answer Key

Example 1
Write an expression using words.
a. a – b
Answer:
Possible answers: a minus b; the difference of a and b; a decreased by b; b subtracted from a

b. xy
Answer:
Possible answers: the product of x and y; x multiplied by y; x times y

c. 4f + p
Answer:
Possible answers: p added to the product of 4 and f; 4 times f plus p; the sum of 4 multiplied by f and p

d. d – b³
Answer:
Possible answers: d minus b cubed; the difference of d and the quantity b to the third power

e. 5(u – 10) + h
Answer:
Possible answers: Add h to the product of 5 and the difference of u and 10; 5 times the quantity of u minus 10 added to h.

f. \(\frac{3}{d+f}\)
Answer:
Possible answers: Find the quotient of 3 and the sum of d and f; 3 divided by the quantity d plus f.

Eureka Math Grade 6 Module 4 Lesson 15 Exercise Answer Key

Circle all the vocabulary words that could be used to describe the given expression.

Exercise 1.
6h – 10
ADDITION            SUBTRACTION           MULTIPLICATION           DIVISION
Answer:

Exercise 2.
\(\frac{5 d}{6}\)
SUM           DIFFERENCE           PRODUCT           QUOTIENT
Answer:

Exercise 3.
5(2 + d) – 8
ADD           SUBTRACT           MULTIPLY           DIVIDE
Answer:

Exercise 4.
abc
MORE THAN           LESS THAN           TIMES           EACH
Answer:

Write an expression using vocabulary to represent each given expression.

Exercise 5.
8 – 2g
Answer:
Possible answers: 8 minus the product of 2 and g; 2 times g subtracted from 8; 8 decreased by g doubled

Exercise 6.
15(a + c)
Answer:
Possible answers: 15 times the quantity of a increased by c; the product of 15 and the sum of a and c; 15 multiplied by the total of a and c

Exercise 7.
\(\frac{m+n}{5}\)
Answer:
Possible answers: the sum of m and n divided by 5; the quotient of the total of m and n, and 5; m plus n split into 5 equal groups

Exercise 8.
b³ – 18
Answer:
Possible answers: b cubed minus 18; b to the third power decreased by 18

Exercise 9.
f – \(\frac{d}{2}\)
Answer:
Possible answers: f minus the quotient of d and 2; d split into 2 groups and then subtracted from f; d divided by 2 less than f

Exercise 10.
\(\frac{u}{x}\)
Answer:
Possible answers: u divided by X; the quotient of u and X; u divided into x parts

Eureka Math Grade 6 Module 4 Lesson 15 Problem Set Answer Key

Question 1.
List five different vocabulary words that could be used to describe each given expression.
a. a – d + c
Answer:
Possible answers: sum, add, total, more than, increase, decrease, difference, subtract, less than

b. 20 – 3c
Answer:
Possible answers: difference, subtract, fewer than, triple, times, product

c. \(\frac{b}{d+2}\)
Answer:
Possible answers: quotient, divide, split, per, sum, add, increase, more than

Question 2.
Write an expression using math vocabulary for each expression below.
a. 5h – 18
Answer:
Possible answers: the product of 5 and b minus 18, 18 less than 5 times b

b. \(\frac{n}{2}\)
Answer:
Possible answers: the quotient of n and 2, n split into 2 equal groups

c. a + (d – 6)
Answer:
Possible answers: a plus the quantity d minus 6, a increased by the difference of d and 6

d. 10 + 2b
Answer:
Possible answers: 10 plus twice b, the total of 10 and the product of 2 and b

Eureka Math Grade 6 Module 4 Lesson 15 Exit Ticket Answer Key

Question 1.
Write two word expressions for each problem using different math vocabulary for each expression.
a. 5d – 10
Answer:
Possible answers: the product of 5 and d minus 10, 10 less than 5 times d

b. \(\frac{a}{b+2}\)
Answer:
Possible answers: the quotient of a and the quantity of b plus 2, a divided by the sum of b and 2

Question 2.
List five different math vocabulary words that could be used to describe each given expression.
a. 3(d – 2) + 10
Answer:
Possible answers: difference, subtract, product, times, quantity, add, sum

b. \(\frac{a b}{c}\)
Answer:
Possible answers: quotient, divide, split, product, multiply, times, per, each

Eureka Math Grade 6 Module 4 Lesson 15 Opening Exercise Answer Key

Question 1.
Complete the graphic organizer with mathematical words that indicate each operation. Some words may indicate more than one operation.

Answer:

Engage NY Eureka Math 6th Grade Module 4 Lesson 14 Answer Key

Eureka Math Grade 6 Module 4 Lesson 14 Example Answer Key

Example 1.
Fill in the three remaining squares so that all the squares contain equivalent expressions.
Answer:

Example 2.
Fill in a blank copy of the four boxes using the words dividend and divisor so that it is set up for any example.
Answer:

Eureka Math Grade 6 Module 4 Lesson 14 Exercise Answer Key

Exercises

Complete the missing spaces in each rectangle set.

Set A

1. 5 ÷ p
Answer:
5 divided by p, \(\frac{5}{p}\),

2. The quotient of g and h
Answer:
g ÷ h, \(\frac{g}{h}\),

3.
Answer:
23 divided by w, 23 ÷ w, \(\frac{23}{w}\)

4. \(\frac{y}{x+8}\)
Answer:
y divided by the sum of x and 8, y ÷ (x + 8),

5. 7 divided by the quantity a minus 6
Answer:
7 ÷ (a – 6), \(\frac{7}{a-6}\),

6.
Answer:
The sum of m and 11 dividend by 3, (m + 11) ÷ 3, \(\frac{m+11}{3}\)

7. (f + 2) ÷ g
Answer:
The sum of f and 2 dividend by g, \(\frac{f+2}{g}\),

8. \(\frac{c-9}{d+3}\)
Answer:
The quotient of c minus 9 and d plus 3, (c – 9) ÷ (d + 3),

Set B

1. h ÷ 11
Answer:

2. The quotient of m and n
Answer:

3.
Answer:
The quotient of j and 5, j ÷ 5, \(\frac{j}{5}\)

4. \(\frac{h}{m-4}\)
Answer:

5. f divided by the quantity g minus 11
Answer:

6.
Answer:
The sum of a and 5 dividend by 18, (a + 5) ÷ 18, \(\frac{a+5}{18}\)

7. (y – 3) ÷ x
Answer:

8. \(\frac{g+5}{h-11}\)
Answer:
The quotient of g plus 5 divided by the quantity h minus 11,

Set C

1. 6 ÷ k
Answer:

2. The quotient of j and k
Answer:

3.
Answer:
a dividend by 10, a ÷ 10, \(\frac{a}{10}\)

4. \(\frac{15}{f-2}\)
Answer:

5. 13 divided by the sum of h and 1
Answer:

6.
Answer:
The sum of c plus 18 dividend by 3, (c + 18) ÷ 3, \(\frac{c+18}{3}\)

7. (h – 2) ÷ m
Answer:

8. \(\frac{4-m}{n+11}\)
Answer:
The quantity of 4 minus m divided by the sum of n and 11,

Eureka Math Grade 6 Module 4 Lesson 14 Problem Set Answer Key

Complete the missing spaces in each rectangle set.

Answer:
The quotient of h and 16, , \(\frac{h}{16}\)
m divided by the quantity b minus 33, , m ÷ (b – 33)

Answer:

The sum of y and 13 dividend by 2, (y + 13) ÷ 2, \(\frac{y+13}{2}\)

Eureka Math Grade 6 Module 4 Lesson 14 Exit Ticket Answer Key

Question 1.
Write the division expression in words and as a fraction.
(g + 12)÷h
Answer:
g+12
The sum of g and 12 divided by h, \(\frac{g+12}{13}\)

Question 2.
Write the following division expression using the division symbol and as a fraction: f divided by the quantity h minus 3.
Answer:
f ÷ (h – 3) and \(\frac{f}{h-3}\)

Eureka Math Grade 6 Module 4 Lesson 14 Long Division Algorithm Answer Key

Long Division Algorithm

Progression of Exercises

Question 1.
3,282 ÷ 6
Answer:
547

Question 2.
2,712 ÷ 3
Answer:
904

Question 3.
15,036 ÷ 7
Answer:
2,148

Question 4.
1,788 ÷ 8
Answer:
223.5

Question 5.
5,736 ÷ 12
Answer:
478

Question 6.
35,472 ÷ 16
Answer:
2,217

Question 7.
13,384÷28
Answer:
478

Question 8.
31,317÷39
Answer:
803

Question 9.
1,113÷42
Answer:
26.5

Question 10.
4,082 ÷ 52
Answer:
78.5

Engage NY Eureka Math Grade 6 Module 4 Lesson 34 Answer Key

Eureka Math Grade 6 Module 4 Lesson 34 Example Answer Key

Example 1:

Answer:

Example 2:

Kelly works for Quick Oil Change. If customers have to wait longer than 20 minutes for the oil change, the company does not charge for the service. The fastest oil change that Kelly has ever done took 6 minutes. Show the possible customer wait times in which the company charges the customer
Answer:

6 ≤ x ≤ 20

Example 3:

Gurnaz has been mowing lawns to save money for a concert. Gurnaz will need to work for at least six hours to save enough money, but he must work fewer than 16 hours this week. Write an inequality to represent this situation, and then graph the solution.

Answer:

6 ≤ x ≤ 16

Eureka Math Grade 6 Module 4 Lesson 34 Exercise Answer Key

Exercises 1 – 5:

Write an inequality to represent each situation. Then, graph the solution.

Exercise 1.
Blayton is at most 2 meters above sea level.

Answer:

b ≤ 2 where b is Blayton’s position in relationship to sea level in meters.

Exercise 2.
Edith must read for a minimum of 20 minutes.

Answer:

E ≤ 20, where E is the number of minutes Edith reads.

Exercise 3.
Travis milks his cows each morning. He has never gotten fewer than 3 gallons of milk; however, he always gets fewer than 9 gallons of milk.

Answer:

3 ≤ x < 9, where x represents the gallons of milk.

Exercise 4.
Rita can make 8 cakes for a bakery each day. So far, she has orders for more than 32 cakes. Right now, Rita needs more than four days to make all 32 cakes.

Answer:

x > 4, where x is the number of days Rito has to bake the cakes.

Exercise 5.
Rita must have all the orders placed right now done In 7 days or fewer. How will this change your inequality and your graph?

Answer:

4 < x ≤ 7
Our inequality will change because there is a range for the number of days Rita has to bake the cakes. The graph has changed because Rita is more limited in the amount of time she has to bake the cakes. Instead of the graph showing any number larger than 4, the graph now has a solid circle at 7 because Rita must be done baking the cakes in a maximum of 7 days.

Possible Extension Exercises 6 – 10:

Exercise 6.
Kasey has been mowing lawns to save up money for a concert. He earns $15 per hour and needs at least $90 to go to the concert. How many hours should he mow?

Answer:

15 x ≥ 90
\(\frac{15 x}{15} \geq \frac{90}{15}\)
x ≥ 6
kasey will need to mow for 6 or more hours.

Exercise 7.
Rachel can make 8 cakes for a bakery each day. So far, she has orders for more than 32 cakes. How many days will it take her to complete the orders?

Answer:

8x > 32
\(\frac{8 x}{8}>\frac{32}{8}\)
x > 4

Exercise 8.
Ranger saves $70 each week. He needs to save at least $2,800 to go on a trip to Europe. How many weeks will he need to save?

Answer:

70x ≥ 2800
\(\frac{70 x}{70} \geq \frac{2800}{70}\)
x ≥ 40

Exercise 9.
Clara has less than $75. She wants to buy 3 pairs of shoes. What price shoes can Clara afford if all the shoes are the same price?

Answer:

3x < 75
\(\frac{3 x}{3}<\frac{75}{3}\)
x < 25
Clara can afford shoes that are greater than $0 and less than $25.

Exercise 10.
A gym charges $25 per month plus $4 extra to swim in the pool for an hour. If a member only has $45 to spend each month, at most how many hours can the member swim?

Answer:

4x + 25 ≤ 45
4x + 25 – 25 ≤ 45 – 25
4x ≤ 20
\(\frac{4 x}{4} \leq \frac{20}{4}\)
x ≤ 5

The member can swim in the pool for 5 hours. However, we also know that the total amount of time the member spends in the pool must be greater than or equal to 0 hours because the member may choose not to swim.
0 ≤ x ≤ 5

Eureka Math Grade 6 Module 4 Lesson 34 Problem Set Answer Key

Write and graph an inequality for each problem.

Question 1.
At least 13

Answer:

x ≥ 13

Question 2.
Less than 7

Answer:

x < 7

Question 3.
Chad will need at least 24 minutes to complete the 5k race. However, he wants to finish in under 30 minutes.

Answer:

24 ≤ x < 30

Question 4.
Eva saves $60 each week. Since she needs to save at least $2,400 to go on a trip to Europe, she will need to save for at least 40 weeks.

Answer:

x ≥ 40

Question 5.
Clara has $100. She wants to buy 4 pairs of the same pants. Due to tax, Clara can afford pants that are less than $25.

Answer:
Clara must spend less than $25, but we also know that Clara will spend more than $0 when she buys pants at the store.
0 < x < 25

Question 6.
A gym charges $30 per month plus $4 extra to swim in the poo1 for an hour. Because a member has just $50 to spend at the gym each month, the member can swim at most 5 hours.

Answer:
The member can swim in the pool for 5 hours. However, we also know that the total amount of time the member spends in the pool must be greater than or equal to 0 hours because the member may choose not to swim.
0 ≤ x ≤ 5

Eureka Math Grade 6 Module 4 Lesson 34 Exit Ticket Answer Key

For each question, write an inequality. Then, graph your solution.

Question 1.
Keisha needs to make at least 28 costumes for the school play. Since she can make 4 costumes each week, Keisha plans to work on the costumes for at least 7 weeks.

Answer:
x ≥ 7
Keisha should plan to work on the costumes for 7 or more weeks.

Question 2.
If Keisha has to have the costumes complete in 10 weeks or fewer, how will our solution change?

Answer:
Keisha had 7 or more weeks in problem 1. It will still take her at least 7 weeks, but she cannot have more than 10 weeks.
7 ≤ x ≤ 10

Engage NY Eureka Math 6th Grade Module 4 Lesson 10 Answer Key

Eureka Math Grade 6 Module 4 Lesson 10 Example Answer Key

Example 1.
Write each expression using the fewest number of symbols and characters. Use math terms to describe the expressions and parts of the expressions.

a 6 × b
Answer:
6b; the 6 is the coefficient and a factor and the b is the variable and a factor. We can call 6b the product,
and we can also call it a term.

b. 4 ∙ 3 ∙ h
Answer:
12h; the 12 is the coefficient and a factor and the h b the variable and a factor. We can call 12h the
produce and we can also call it a term.

c. 2 × 2 × 2 × a × b
Answer:
8ab; 8 is the coefficient and a factor, a and b are both varIables and factor, and 8ab is the product and also a term.

d. 5 × m × 3 × p
Answer:
15mp; 15 is the coefficient and factor, m and p are the variables and factors, 15mp is the product and also a term.

e. 1 × g × w
Answer:
1gw or gw; g and w are the variables and factors, 1 is the coefficient and factor if it is included, and gw is the product and also o term.

Example 2.
To expand multiplication expressions, we will rewrite the expressions by Including the “. “ back into the expressions.
a. 5g
Answer:
5 ∙ g

b. 7abc
Answer:
7 ∙ a ∙ b ∙ c

c. 12g
Answer:
12 ∙ g or 2 ∙ 2 ∙ 3 ∙ g

d. 3h ∙ 8
Answer:
3 ∙ h ∙ 8

e. 7g ∙ 9h
Answer:
7 ∙ g ∙ 9 ∙ h or 7 ∙ g ∙ 3 ∙ 3 ∙ h

Example 3.
a. Find the product of 4f ∙ 7g.
Answer:
It may be easier to see how we will use the fewest number of symbols and characters by expanding the expression first.
4 ∙ f ∙ 7 ∙ g
Now we can multiply the numbers and then multiply the variables.
4 ∙ 7 ∙ f ∙ g
28fg

b. Multiply 3de ∙ 9yz.
Answer:
Let’s start again by expanding the expression. Then, we can rewrite the expression by multiplying the numbers and then multiplying the variables.
3 ∙ d ∙ e ∙ 9 ∙ y ∙ z
3 ∙ 9 ∙ d ∙ e ∙ y ∙ z
27 deyz

c. Double the product of 6y and 3bc.
Answer:
6 ∙ y ∙ 3 ∙ b ∙ c
6 ∙ 3 ∙ b ∙ c ∙ y
18bcy
What does it mean to double something?
It means to multiply by 2.
2 ∙ 18bcy
36 bcy

Eureka Math Grade 6 Module 4 Lesson 10 Problem Set Answer Key

Question 1.
Rewrite the expression in standard form (use the fewest number of symbols and characters possible).
a. 5 ∙ y
Answer:
5y

b. 7 ∙ d ∙ e
Answer:
7de

c. 5 ∙ 2 ∙ 2 ∙ y ∙ z
Answer:
20yz

d. 3 ∙ 3 ∙ 2 ∙ 5 ∙ d
Answer:
90d

Question 2.
Write the following expressions in expanded form.
a. 3g
Answer:
3 ∙ g

b. 11mp
Answer:
11 ∙ m ∙ p

C. 20yz
Answer:
20 ∙ y ∙ z or 2 ∙ 2 ∙ 5 ∙ y ∙ z

d. 15abc
Answer:
15 ∙ a ∙ b ∙ c or 3 ∙ 5 ∙ a ∙ b ∙ c

Question 3.
Find the product.
a. 5d ∙ 7g
Answer:
35dg

b. 12ab ∙ 3cd
Answer:
36abcd

Eureka Math Grade 6 Module 4 Lesson 10 Exit Ticket Answer Key

Question 1.
Rewrite the expression in standard form (use the fewest number of symbols and characters possible).
a. 5g ∙ 7h
Answer:
35gh

b. 3 ∙ 4 ∙ 5 ∙ m ∙ n
Answer:
60mn

Question 2.
Name the parts of the expression. Then, write it in expanded form.
a. 14b
Answer:
14 ∙ b or 2 ∙ 7 ∙ b
14 is the coefficient, b is the variable, and 14b is a term and the product of 14 × b.

b. 30jk
Answer:
30 ∙ j ∙ k or 2 ∙ 3 ∙ 5 ∙ j ∙ k
30 is the coefficient, i and k are the variables, and 30jk is a term and the product of 30 ∙ j ∙ k.

Engage NY Eureka Math 6th Grade Module 4 Lesson 11 Answer Key

Eureka Math Grade 6 Module 4 Lesson 11 Example Answer Key

a. Use the model to answer the following questions.

How many fives are in the model?
Answer:
5

How many threes are In the model?
Answer:
2

What does the expression represent in words?
Answer:
The sum of two groups of five and two groups of three

What expression could we write to represent the model?
Answer:
2 × 5 + 2 × 3

b. Use the new model and the previous model to answer the next set of questions.

How many fives are in the model?
Answer:
2

How many threes are in the model?
Answer:
2

What does the expression represent in words?
Answer:
Two groups of the sum of five and three

What expression could we write to represent the model?
Answer:
(5 + 3) + (5 + 3) or 2(5 + 3)

Is the model in part (a) equivalent to the model in part (b)?
Answer:
Yes, because both expressions have two 5’s and two 3’s. Therefore, 2 × 5 + 2 × 3 = 2(5 + 3).

d. What relationship do we see happening on either side of the equal sign?
Answer:
On the left-hand side, 2 is being multiplied by 5 and then by 3 before adding the products together. On the right-hand side, the 5 and 3 are added first and then multiplied by 2.

e. In Grade 5 and in Module 2 of this year, you have used similar reasoning to solve problems. What Is the name of the property that is used to say that 2(5 + 3) is the same as 2 × 5 + 2 × 3?
Answer:
The name of the property is the distributive property.

Example 2.
Now we will take a look at an example with variables. Discuss the questions with your partner.

What does the model represent in words?
Answer:
a plus a plus b plus b, two a’s plus two b’s, two times a plus two times b

What does 2a mean?
Answer:
2a means that there are 2 a’s or 2 × a.

How many a’s are in the model?
Answer:
2

How many b’s are in the model?
Answer:
2

What expression could we write to represent the model?
Answer:
2a + 2b

How many a’s are in the expression?
Answer:
2

How many b’s are in the expression?
Answer:
2

What expression could we write to represent the model?
Answer:
(a + b) + (a + b) = 2(a + b)

Are the two expressions equivalent?
Answer:
Yes. Both models include 2 a’s and 2 b’s. Therefore, 2a + 2b = 2(a + b).

Example 3.

Use GCF and the distributive property to write equivalent expressions.
1. 3f + 3g = __________
Answer:
3(f + g)

What is the question asking us to do?
Answer:
We need to rewrite the expression as an equivalent expression in factored form, which means the expression is written as the product of factors. The number outside of the parentheses is the GCF.

How would Problem 1 look if we expanded each term?
Answer:
3 ∙ f + 3 ∙ g

What is the GCF in Problem 1?
Answer:
3

How can we use the GCF to rewrite this expression?
Answer:
3 goes on the outside, and f + g will go inside the parentheses. 3(f + g)

2. 6x + 9y = __________
Answer:
3(2x + 3y)

What is the question asking us to do?
Answer:
We need to rewrite the expression as an equivalent expression in factored form, which means the expression is written as the product of factors. The number outside of the parentheses is the GCF.

How would Problem 2 look if we expanded each term?
Answer:
2 ∙ 3 ∙ x + 3 ∙ 3 ∙ y

What is the GCF in Problem 2?
Answer:
The GCF is 3.

How can we use the GCF to rewrite this expression?
Answer:
I will factor out the 3 from both terms and place it in front of the parentheses. I will place what is left in the terms inside the parentheses: 3(2x + 3y).

3. 3c + 11c = _________
Answer:
c(3 + 11)

Is there a greatest common factor in Problem 3?
Answer:
Yes. When I expand, I can see that each term has a common factor c.
3 ∙ c + 11 ∙ c

Rewrite the expression using the distributive property.
Answer:
c(3 + 11)

4. 24b + 8 = _________
Answer:
8(3b + 1)

Explain how you used GCF and the distributive property to rewrite the expression in Problem 4.
Answer:
I first expanded each term. I know that 8 goes into 24, so I used it in the expansion.
2 ∙ 2 ∙ 2 ∙ 3 ∙ b + 2 ∙ 2 ∙ 2
I determined that 2 ∙ 2 ∙ 2, or 8, is the common factor. So, on the outside of the parentheses I wrote 8, and on the inside I wrote the leftover factor, 3b + 1 ∙ 8(3b + 1)

Why is there a 1 in the parentheses?
Answer:
When I factor out a number, lam leaving behind the other factor that multiplies to make the original number. In this case, when I factor out an 8 from 8, I am left with a 1 because 8 × 1 = 8.

How is this related to the first two examples?
Answer:
In the first two examples, we saw that we could rewrite the expressions by thinking about groups.
We can either think of 24b + 8 as 8 groups of 3b and 8 groups of 1 or as 8 groups of the sum of 3b + 1. This shows that 8(3b) + 8(1) = 8(3b + 1)is the some as 24b + 8.

Eureka Math Grade 6 Module 4 Lesson 11 Exercise Answer Key

Exercise 1.
Apply the distributive property to write equivalent expressions.
a. 7x + 7y
Answer:
7(x + y)

b. 15g + 20h
Answer:
5(3g + 4h)

c. 18m + 42n
Answer:
6(3m + 7n)

d. 30a + 39b
Answer:
3(10a + 13b)

e. 11f + 15f
Answer:
f(11 + 15)

f. 18h + 13h
Answer:
h(18 + 13)

g. 55m + 11
Answer:
11(5m + 1)

h. 7 + 56y
Answer:
7(1 + 8y)

2.
Evaluate each of the expressions below.
a. 6x + 21 y and 3(2x + 7y)                     x = 3 and y = 4
Answer:
6(3) + 21(4)                                             3(23 + 74)
18 + 84                                                    3(6 + 28)
102                                                           3(34)
102                                                           102

b. 5g + 7g and g(5 + 7)                          g = 6
Answer:
5(6) + 7(6)                                              6(5 + 7)
30 + 42                                                   6(12)
72                                                            72

c. 14x + 2 and 2(7x + 1)                          x = 10
Answer:
14(10) + 2                                               2(7.10 + 1)
140 + 2                                                   2(70 + 1)
142                                                          2(71)
142                                                          142

d. Explain any patterns that you notice in the results to parts (a) – c).
Answer:
Both expressions in parts (a) – (c) evaluated to the same number when the indicated value was substituted for the variable. This shows that the two expressions are equivalent for the given values.

e. What would happen if other values were given for the variables?
Answer:
Because the two expressions in each part are equivalent, they evaluate to the same number, no matter what value is chosen for the variable.

Closing

How can use you use your knowledge of GCF and the distributive property to write equivalent expressions?
Answer:
We can use our knowledge of GCF and the distributive property to change expressions from standard form to factored form.

Find the missing value that makes the two expressions equivalent.
4x + 12y                ___(x + 3y)
35x + 50y              ___(7x + 10y)
18x + 9y                ___(2x + y)
32x + 8y                ___(4x + y)
100x + 700y          ___(x + 7y)
Answer:
4x + 12y                 4 (x + 3y)
35x + 50y               5(7x + 10y)
18x + 9y                 9(2x + y)
32x + 8y                 8(4x + y)
100x + 700y           100(x + 7y)

Explain how you determine the missing number.
Answer:
I would expand each term and determine the greatest common factor. The greatest common factor is the number that is placed on the blank line.

Eureka Math Grade 6 Module 4 Lesson 11 Problem Set Answer Key

Question 1.
Use models to prove that 3(a + b) is equivalent to 3a + 3b.
Answer:

Question 2.
Use greatest common factor and the distributive property to write equivalent expressions in factored form for the following expressions.
a. 4d + 12e
Answer:
4(d + 3e) or 4(1d + 3e)

b. 18x + 30y
Answer:
6(3x + 5y)

c. 21a + 28y
Answer:
7(3a + 4y)

d. 24f + 56g
Answer:
8(3f + 7g)

Eureka Math Grade 6 Module 4 Lesson 11 Exit Ticket Answer Key

Use greatest common factor and the distributive property to write equivalent expressions in factored form.

Question 1.
2x + 8y
Answer:
2(x + 4y)

Question 2.
13ab + 15 ab
Answer:
ab(13 + 15)

Question 3.
20g + 24h
Answer:
4(5g + 6h)

Eureka Math Grade 6 Module 4 Lesson 11 Greatest Common Factor Answer Key

Greatest Common Factor – Round 1
Directions: Determine the greatest common factor of each pair of numbers.

Question 1.
GCF of 10 and 50
Answer:
10

Question 2.
GCF of 5 and 35
Answer:
5

Question 3.
GCF of 3 and 12
Answer:
3

Question 4.
GCF of 8 and 20
Answer:
4

Question 5.
GCF of 15 and 35
Answer:
5

Question 6.
GCF of 10 and 75
Answer:
5

Question 7.
GCF of 9 and 30
Answer:
3

Question 8.
GCF of 15 and 33
Answer:
3

Question 9.
GCF of 12 and 28
Answer:
4

Question 10.
GCF of 16 and 40
Answer:
8

Question 11.
GCF of 24 and 32
Answer:8
Question 12.
GCF of 35 and 49
Answer:
7

Question 13.
GCF of 45 and 60
Answer:
15

Question 14.
GCF of 48 and 72
Answer:
24

Question 15.
GCF of 50 and 42
Answer:
2

Question 16.
GCF of 45 and 72
Answer:
9

Question 17.
GCF of 28 and 48
Answer:
4

Question 18.
GCF of 44 and 77
Answer:
11

Question 19.
GCF of 39 and 66
Answer:
3

Question 20.
GCF of 64 and 88
Answer:
8

Question 21.
GCF of 42 and 56
Answer:
14

Question 22.
GCF of 28 and 42
Answer:
14

Question 23.
GCF of 13 and 91
Answer:
13

Question 24.
GCF of 16 and 84
Answer:
4

Question 25.
GCF of 36 and 99
Answer:
9

Question 26.
GCF of 39 and 65
Answer:
13

Question 27.
GCF of 27 and 87
Answer:
3

Question 28.
GCF of 28 and 70
Answer:
14

Question 29.
GCF of 29 and 91
Answer:
13

Question 30.
GCF of 34 and 51
Answer:
17

Greatest Common Factor – Round 2
Directions: Determine the greatest common factor of each pair of numbers.

Question 1.
GCF of 20 and 80
Answer:
20

Question 2.
GCF of 10 and 70
Answer:
10

Question 3.
GCF of 9 and 36
Answer:
9

Question 4.
GCF of 12 and 24
Answer:
12

Question 5.
GCF of 15 and 45
Answer:
15

Question 6.
GCF of 10 and 95
Answer:
5

Question 7.
GCF of 9 and 45
Answer:
9

Question 8.
GCF of 18 and 33
Answer:
3

Question 9.
GCF of 12 and 32
Answer:
4

Question 10.
GCF of 16 and 56
Answer:
8

Question 11.
GCF of 40 and 7
Answer:
8

Question 12.
GCF of 35 and 63
Answer:
7

Question 13.
GCF of 30 and 75
Answer:
15

Question 14.
GCF of 42 and 72
Answer:
6

Question 15.
GCF of 30 and 28
Answer:
2

Question 16.
GCF of 33 and 99
Answer:
33

Question 17.
GCF of 38 and 76
Answer:
38

Question 18.
GCF of 26 and 65
Answer:
13

Question 19.
GCF of 39 and 48
Answer:
3

Question 20.
GCF of 72 and 88
Answer:
8

Question 21.
GCF of 21 and 56
Answer:
7

Question 22.
GCF of 28 and 52
Answer:
4

Question 23.
GCF of 51 and 68
Answer:
17

Question 24.
GCF of 48 and 84
Answer:
12

Question 25.
GCF of 21 and 63
Answer:
21

Question 26.
GCF of 64 and 80
Answer:
16

Question 27.
GCF of 36 and 90
Answer:
18

Question 28.
GCF of 28 and 98
Answer:
14

Question 29.
GCF of 39 and 91
Answer:
13

Question 30.
GCF of 38 and 95
Answer:
19

Engage NY Eureka Math 6th Grade Module 4 Lesson 12 Answer Key

Eureka Math Grade 6 Module 4 Lesson 12 Example Answer Key

Example 1.
Write an expression that is equivalent to 2(a + b).
Answer:
→ In this example, we have been given the factored form of the expression.
→ To answer this question, we can create a model to represent 2(a + b).
→ Let’s start by creating a model to represent (a + b).

Create a model to represent (a + b).
Answer:

The expression 2(a + b) tells us that we have 2 of the (a + b)’s. Create a model that shows 2 groups of (a + b).
Answer:

How many a’s and how many b’s do you see in the diagram?
Answer:
There are 2 a’s and 2 b’s.

How would the model look if we grouped together the a’s and then grouped together the b’s?
Answer:

What expression could we write to represent the new diagram?
Answer:
2a + 2b

What conclusion can we draw from the models about equivalent expressions?
Answer:
2(a + b) = 2a + 2b

Let a = 3 and b = 4.
Answer:
2(a+b)                      2a + 2b
2(3 + 4)                    2(3) + 2(4)
2(7)                           6 + 8
14                              14

What happens when we double (a + b)?
Answer:
We double a, and we double b.

Example 2.
Write an expression that Is equivalent to double (3x + 4y).

How can we rewrite double (3x + 4y)?
Answer:
Double is the same as multiplying by two.
2(3x + 4y) or 6x + 8y

Is this expression In factored form, expanded form, or neither?
Answer:
The first expression is in factored form, and the second expression is in expanded form.

Let’s start this problem the same way that we started the first example. What should we do?
Answer:
We can make a model of 3x + 4y.

Are there terms that we can combine in this example?
Answer:

Yes. There are 6 x’s and 8y’s.
So, the model is showing 6x + 8y.

What is an equivalent expression that we can use to represent 2(3x + 4y)?
Answer:
2(3x + 4y) = 6x + 8y
This is the same as 2(3x) + 2(4y).

Summarize how you would solve this question without the model.
Answer:
When there is a number outside the parentheses, I would multiply it by all the terms on the inside of the parentheses.

Example 3.
Write an expression in expanded form that is equivalent to the model below.

What factored expression is represented in the model?
Answer:
y(4x + 5)

How can we rewrite this expression in expanded form?
Answer:
y(4x) + y(5)
4xy + 5y

Example 4.
Write an expression in expanded form that is equivalent to 3(7d + 4e).
Answer:
We will multiply 3 × 7d and 3 × 4e.
We would get 21d + 12e. So, 3(7d+ 4e) = 21d + 12e.

Eureka Math Grade 6 Module 4 Lesson 12 Exercise Answer Key

Exercises
Create a model for each expression below. Then, write another equivalent expression using the distributive property.

Exercise 1.
3(x + y)
Answer:

3x + 3y

Exercise 2.
4(2h + g)
Answer:

8h + 4g

Apply the distributive property to write equivalent expressions in expanded form.

Exercise 3.
8(h + 3)
Answer:
8k + 24

Exercise 4.
3(2h + 7)
Answer:
6k + 21

Exercise 5.
5(3x + 9y)
Answer:
15x + 45y

Exercise 6.
4(11k + 3g)
Answer:
44h + 12g

Exercise 7.

Answer:
7jk + 12jm

Exercise 8.
a(9b + 13)
Answer:
9ab + 13a

Eureka Math Grade 6 Module 4 Lesson 12 Problem Set Answer Key

Question 1.
Use the distributive property to write the following expressions in expanded form.
a. 4(x + y)
Answer:
4x + 4y

b. 8(a + 3b)
Answer:
8a + 24h

c. 3(2x + 11y)
Answer:
6x + 33y

d. 9(7a + 6b)
Answer:
63a + 54b

e. c(3a + b)
Answer:
3ac + bc

f. y(2x + 11z)
Answer:
2xy + 11yz

Question 2.
Create a model to show that 2(2x + 3y) = 4x + 6y.
Answer:

Eureka Math Grade 6 Module 4 Lesson 12 Exit Ticket Answer Key

Use the distributive property to write the following expressions in expanded form.
Question 1.
2(b + c)
2h + 2c

Question 2.
5(7h + 3m)
Answer:
35h + 15m

Question 3.
e(f + g)
Answer:
ef + eg

Eureka Math Grade 6 Module 4 Lesson 12 Opening Exercise Answer Key

a. Create a model to show 2 × 5.
Answer:

b. Create a model to show 2 × b, or 2b
Answer:

Engage NY Eureka Math 6th Grade Module 4 Lesson 13 Answer Key

Eureka Math Grade 6 Module 4 Lesson 13 Example Answer Key

Example 1.
Write an expression showing 1 ÷ 2 without the use of the division symbol.
Answer:

What can we determine from the model?
Answer:
1 ÷ 2 is the same as \(\frac{1}{2}\).

Example 2.
Write an expression showing a ÷ 2 without the use of the division symbol.
Answer:

What can we determine from the model?
Answer:
a ÷ 2 is the same as \(\frac{a}{2}\).

When we write division expressions using the division symbol, we represent
Answer:
dividend ÷ divisor.

How would this look when we write division expressions using a fraction?
Answer:

Example 3
a. Write an expression showing a ÷ b without the use of the division symbol.
Answer:
\(\frac{a}{b}\)

b. Write an expression for g divided by the quantity h plus 3.
Answer:
\(\frac{g}{h+3}\)

c. Write an expression for the quotient of the quantity m reduced by 3 and 5.
Answer:
\(\frac{m-3}{5}\)

Eureka Math Grade 6 Module 4 Lesson 13 Exercise Answer Key

Write each expression two ways: using the division symbol and as a fraction.

a. 12 divided by 4
Answer:
12 ÷ 4 and \(\frac{12}{4}\)

b. 3 divided by 5
Answer:
3÷ 5 and \(\frac{3}{5}\)

c. a divided by 4
Answer:
a ÷ 4 and \(\frac{a}{4}\)

d. The quotient of 6 and m
Answer:
6 ÷ m and \(\frac{6}{m}\)

e. Seven divided by the quantity x plus y
Answer:
7 ÷ (x + y) and \(\frac{7}{x+y}\)

f. y divided by the quantity x minus 11
Answer:
y ÷ (x – 11) and \(\frac{y}{x-11}\)

g. The sum of the quantity h and 3 divided by 4
Answer:
(h + 3) ÷ 4 and \(\frac{h+3}{4}\)

h. The quotient of the quantity k minus 10 and m
Answer:
(k – 10) ÷ m and \(\frac{k-10}{m}\)

Eureka Math Grade 6 Module 4 Lesson 13 Problem Set Answer Key

Question 1.
Rewrite the expressions using the division symbol and as a fraction.

a. Three divided by 4
Answer:
3 ÷ 4 and \(\frac{3}{4}\)

b. The quotient of m and 11
Answer:
m ÷ 11 and \(\frac{m}{11}\)

c. 4 divided by the sum of h and 7
Answer:
4 ÷ (h + 7) and \(\frac{4}{h+7}\)

d. The quantity x minus 3 divided by y
Answer:
(x – 3) ÷ y and \(\frac{x-3}{y}\)

Question 2.
Draw a model to show that x ÷ 3 s the same as \(\frac{x}{3}\).
Answer:

Eureka Math Grade 6 Module 4 Lesson 13 Exit Ticket Answer Key

Rewrite the expressions using the division symbol and as a fraction.

Question 1.
The quotient of m and 7
Answer:
m ÷ 7 and \(\frac{m}{7}\)

Question 2.
Five divided by the sum of a and b
Answer:
5 ÷ (a + b) and \(\frac{5}{a+b}\)

Question 3.
The quotient of k decreased by 4 and 9
Answer:
(k – 4) ÷ 9 and \(\frac{k-4}{9}\)

Engage NY Eureka Math 6th Grade Module 4 Lesson 9 Answer Key

Eureka Math Grade 6 Module 4 Lesson 9 Example Answer Key

Example 1.
Create a bar diagram to show 3 plus 5.
Answer:

How would this look it you were asked to show 5 plus 3?
Answer:
There would be 5 tiles and then 3 tiles.

Are these two expressions equivalent?
Answer:
Yes. Both 3 + 5 and 5 + 3 have a sum of 8.

Example 2.

How can we show a number increased by 2?
Answer:
a + 2 or 2 + a

Can you prove this using a model? If so, draw the model.
Answer:
Yes. I can use a bar diagram.

Example 3.
Write an expression to show the sum of m and k.
Answer:
m + k or k + m

Which property can be used in Examples 1 – 3 to show that both expressions given are equivalent?
Answer:
The commutative property of addition

Example 4
How can we show 10 minus 6?
Draw a bar diagram to model this expression.
Answer:

What expression would represent this model?
Answer:
10 – 6

Could we also use 6 – 10?
Answer:
No. If we started with 6 and tried to take 10 away the models would not match.

Example 5
How can we write an expression to show 3 less than a number?
Start by drawing a diagram to model the subtraction. Are we taking away from the 3 or the unknown number?
Answer:
We are taking 3 away from the unknown number.

What expression would represent this model?
Answer:
The expression is n – 3

Example 6
How would we write an expression to show the number c being subtracted from the sum of a and b?

Start by writing an expression for “the sum of a and b.”
Answer:
a + b or b + a

Now, show c being subtracted from the sum.
Answer:
a + b – c or b + a – c

Example 7
Write an expression to show c minus the sum of a and b.
Answer:
c – (a + b)

Why are parentheses necessary in this example and not the others?
Answer:
Without the parentheses, only a is being taken away from c, where the expression says that a + b should be taken away from c.

Replace the variables with numbers to see if c – (a + b) is the same as c – a + b.

Eureka Math Grade 6 Module 4 Lesson 9 Exercise Answer Key

Exercise 1.
Write an expression to show the sum of 7 and 1. 5.
Answer:
7 + 1.5 or 1.5 + 7

Exercise 2.
Write two expressions to show w increased by 4. Then, draw models to prove that both expressions represent the same thing.
Answer:
w + 4 and 4 + w

Exercise 3.
Write an expression to show the sum of a, b, and c.
Answer:
Answers will vary. Below are possible answers.
a + b + c               b + c + a           c + b + a
a + c + b               b + a + c           c + a + b

Exercise 4.
Write an expression and a model showing 3 less than p.
Answer:
p – 3

Exercise 5.
Write an expression to show the difference of 3 and p.
Answer:
3 – p

Exercise 6.
Write an expression to show 4 less than the sum of g and 5.
Answer:
g + 5 – 4 or 5 + g – 4

Exercise 7.
Write an expression to show 4 decreased by the sum of g and 5.
Answer:
4 – (g + 5) or 4 – (5 + g)

Exercise 8.
Should Exercises 6 and 7 have different expressions? Why or why not?
Answer:
The expressions are different because one includes the word “decreased by”and the other has the words ‘less than.” The words “less than” give the amount that was taken away first, whereas the word “decreased by” gives us a starting amount and then the amount that was taken away.

Eureka Math Grade 6 Module 4 Lesson 9 Problem Set Answer Key

Question 1.
Write two expressions to show a number increased by 11. Then, draw models to prove that both expressions represent the same thing.
Answer:
a + 11 and 11 + a

Question 2.
Write an expression to show the sum of x and y.
Answer:
x + y or y + x

Question 3.
Write an expression to show h decreased by 13.
Answer:
h – 13

Question 4.
Write an expression to show k less than 3. 5.
Answer:
3.5 – k

Question 5.
Write an expression to show the sum of g and h reduced by 11.
Answer:
g + h – 11

Question 6.
Write an expression to show 5 less than y, plus g.
Answer:
y – 5 + g

Question 7.
Write an expression to show 5 less than the sum of y and g.
Answer:
y + g – 5

Eureka Math Grade 6 Module 4 Lesson 9 Exit Ticket Answer Key

Question 1.
Write an expression showing the sum of 8 and a number f.
Answer:
8 + f or f + 8

Question 2.
Write an expression showing 5 less than the number k.
Answer:
k – 5

Question 3.
Write an expression showing the sum of a number h and a number w minus 11.
Answer:
h + w – 11

Engage NY Eureka Math 6th Grade Module 4 Lesson 8 Answer Key

Eureka Math Grade 6 Module 4 Lesson 8 Example Answer Key

Example 1.
Additive Identity Property of Zero
g + 0 = g

Remember a letter in a mathematical expression represents a number. Can we replace g with any number?
Answer:
Yes

Choose a value for g, and replace g with that number in the equation. What do you observe?
Answer:
The value of g does not change when 0 is added to g.

Repeat this process several times, each time choosing a different number for g.

Will all values of result in a true number sentence?
Answer:
Yes

Write the mathematical language for this property below:
Answer:
g + 0 = g, additive identity property of zero. Any number added to zero equals itself.

Example 2.
Multiplicative Identity Property of One
g × 1 = g
Remember a letter in a mathematical expression represents a number. Can we replace g with any number?
Answer:
Yes

Choose a value for g, and replace g with that number in the equation. What do you observe?
Answer:
The value of g does not change when g is multiplied by 1.

Will all values of result in a true number sentence? Experiment with different values before making your claim.
Answer:
Yes

Write the mathematical language for this property below:
Answer:
g × 1 = g, multiplicative identity property of one. Any number multiplied by one equals itself.

Example 3.
Commutative Property of Addition and Multiplication
3 + 4 = 4 + 3
3 × 4 = 4 × 3

Replace the 3’s in these number sentences with the letter
Answer:
a + 4 = 4 + a
a × 4 = 4 × a

Choose a value for a, and replace with that number in each of the equations. What do you observe?
Answer:
The result is a true number sentence.

Will all values of result in a true number sentence? Experiment with different values before making your claim.
Answer:
Yes, any number, even zero, can be used in place of the variable

Now, write the equations again, this time replacing the number 4 with a variable, b.
Answer:
a + b = b + a
a × b = b × a

Will all values of and result in true number sentences for the first two equations? Experiment with different values before making your claim.
Answer:
Yes

Write the mathematical language for this property below:
Answer:
a + b = b + a commutative property of addition. Order does not matter when adding.
a × b = b × a commutative property of multiplication. Order does not matter when multiplying.

Example 4.
3 + 3 + 3 + 3 = 4 × 3
3 ÷ 4 = \(\frac{3}{4}\)

Replace the 3’s in these number sentences with the letter
Answer:
a + a + a + a = 4 × a
a ÷ 4 = \(\frac{a}{4}\)

Choose a value for a, and replace a with that number in each of the equations. What do you observe?
Answer:
The result is a true number sentence.

Will all values of a result in a true number sentence? Experiment with different values before making your claim.
Answer:
Yes, any number, even zero, can be used in place of the variable

Now, write the equations again, this time replacing the number 4 with a variable, b.
Answer:
a + a + a + a = b × a
a ÷ b = \(\frac{a}{b}\), b ≠ 0

Will all values of and result in true number sentences for the equations? Experiment with different values before making your claim.
Answer:
In the equation a + a + a + a = b × a, any value can be substituted for the variable a, but only 4 can be used for b since there are exactly 4 copies of a in the equation.
It is true for all values of and all values of b ≠ 0.

Eureka Math Grade 6 Module 4 Lesson 8 Problem Set Answer Key

Question 1.
State the commutative property of addition using the variables a and b.
Answer:
a + b = b + a

Question 2.
State the commutative property of multiplication using the variables a and b.
Answer:
a × b = b × a

Question 3.
State the additive property of zero using the variable b.
Answer:
b + 0 = b

Question 4.
State the multiplicative identity property of one using the variable b.
Answer:
b × 1 = b

Question 5.
Demonstrate the property listed in the first column by filling in the third column of the table.

Answer:

Question 6.
Why is there no commutative property for subtraction or division? Show examples.
Answer:
Answers will vary. Examples should show reasoning and proof that the commutative property does not work for subtraction and division. An example would be 8 ÷ 2 and 2 ÷ 8. 8 ÷ 2 = 4, but 2 ÷ 8 = \(\frac{1}{4}\).

Eureka Math Grade 6 Module 4 Lesson 8 Exit Ticket Answer Key

Question 1.
State the commutative property of addition, and provide an example using two different numbers.
Answer:
Any two different addends can be chosen, such as 5 + 6 = 6 + 5.

Question 2.
State the commutative property of multiplication, and provide an example using two different numbers.
Answer:
Any two different factors can be chosen, such as 4 × 9 = 9 × 4.

Question 3.
State the additive property of zero, and provide an example using any other number.
Answer:
Any nonzero addend can be chosen, such as 3 + 0 = 3.

Question 4.
State the multiplicative identity property of one, and provide an example using any other number.
Answer:
Any nonzero factor can be chosen, such as 12 × 1 = 12.

Eureka Math Grade 6 Module 4 Lesson 8 Opening Exercise Answer Key

4 + 0 = 4
4 × 1 = 4
4 ÷ 1 = 4
4 × 0 = 0
1 ÷ 4 = \(\frac{1}{4}\)

How many of these statements are true?
Answer:
All of them

How many of those statements would be true if the number was replaced with the number in each of the number sentences?
Answer:
All of them

Would the number sentences be true if we were to replace the number with any other number?

What if we replaced the number 4 with the number 0? Would each of the number sentences be true?
Answer:
No. The first four are true, but the last one, dividing by zero, is not true.

What if we replace the number 4 with a letter g? Please write all 4 expressions below, replacing each 4 with a g.
Answer:
g + 0 = g
g × 1 = g
g ÷ 1 = g
g × 0 = 0
1 ÷ g = \(\frac{1}{g}\)

Are these all true (except for g= 0) when dividing?
Answer:
Yes

Eureka Math Grade 6 Module 4 Lesson 8 Division of Fractions II Answer Key

Division of Fractions II – Round 1

Directions: Determine the quotient of the fractions and simplify.

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

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

Question 3.
\(\frac{6}{10} \div \frac{4}{10}\)
Answer:
\(\frac{6}{4}=\frac{3}{2}=1 \frac{1}{2}\)

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

Question 5.
\(\frac{2}{7} \div \frac{6}{7}\)
Answer:
\(\frac{2}{6}=\frac{1}{3}\)

Question 6.
\(\frac{11}{9} \div \frac{8}{9}\)
Answer:
\(\frac{11}{8}=1 \frac{3}{8}\)

Question 7.
\(\frac{5}{13} \div \frac{10}{13}\)
Answer:
\(\frac{5}{10}=\frac{1}{2}\)

Question 8.
\(\frac{7}{8} \div \frac{13}{16}\)
Answer:
\(\frac{14}{13}=1 \frac{1}{13}\)

Question 9.
\(\frac{3}{5} \div \frac{7}{10}\)
Answer:
\(\frac{6}{7}\)

Question 10.
\(\frac{9}{30} \div \frac{3}{5}\)
Answer:
\(\frac{9}{18}=\frac{1}{2}\)

Question 11.
\(\frac{1}{3} \div \frac{4}{5}\)
Answer:
\(\frac{5}{12}\)

Question 12.
\(\frac{2}{5} \div \frac{3}{4}\)
Answer:
\(\frac{8}{15}\)

Question 13.
\(\frac{3}{4} \div \frac{5}{9}\)
Answer:
\(\frac{27}{20}=1 \frac{7}{20}\)

Question 14.
\(\frac{4}{5} \div \frac{7}{12}\)
Answer:
\(\frac{48}{35}=1 \frac{13}{35}\)

Question 15.
\(\frac{3}{8} \div \frac{5}{2}\)
Answer:
\(\frac{6}{40}=\frac{3}{20}\)

Question 16.
\(3 \frac{1}{8} \div \frac{2}{3}\)
Answer:
\(\frac{75}{16}=4 \frac{11}{16}\)

Question 17.
\(1 \frac{5}{6} \div \frac{1}{2}\)
Answer:
\(\frac{22}{6}=\frac{11}{3}=3 \frac{2}{3}\)

Question 18.
\(\frac{5}{8} \div 2 \frac{3}{4}\)
Answer:
\(\frac{20}{88}=\frac{5}{22}\)

Question 19.
\(\frac{1}{3} \div 1 \frac{4}{5}\)
Answer:
\(\frac{5}{27}\)

Question 20.
\(\frac{3}{4} \div 2 \frac{3}{10}\)
Answer:
\(\frac{30}{92}=\frac{15}{46}\)

Question 21.
\(2 \frac{1}{5} \div 1 \frac{1}{6}\)
Answer:
\(\frac{66}{35}=1 \frac{31}{35}\)

Question 22.
\(2 \frac{4}{9} \div 1 \frac{3}{5}\)
Answer:
\(\frac{110}{72}=\frac{55}{36}=1 \frac{19}{36}\)

Question 23.
\(1 \frac{2}{9} \div 3 \frac{2}{5}\)
Answer:
\(\frac{55}{153}\)

Question 24.
\(2 \frac{2}{3} \div 3\)
Answer:
\(\frac{8}{9}\)

Question 25.
\(1 \frac{3}{4} \div 2 \frac{2}{5}\)
Answer:
\(\frac{35}{48}\)

Question 26.
\(4 \div 1 \frac{2}{9}\)
Answer:
\(\frac{36}{11}=3 \frac{3}{11}\)

Question 27.
\(3 \frac{1}{5} \div 6\)
Answer:
\(\frac{16}{30}=\frac{8}{15}\)

Question 28.
\(2 \frac{5}{6} \div 1 \frac{1}{3}\)
Answer:
\(\frac{51}{24}=2 \frac{3}{24}=2 \frac{1}{8}\)

Question 29.
\(10 \frac{2}{3} \div 8\)
Answer:
\(\frac{32}{24}=\frac{4}{3}=1 \frac{1}{3}\)

Question 30.
\(15 \div 2 \frac{3}{5}\)
Answer:
\(\frac{75}{13}=5 \frac{10}{13}\)

Division of Fractions II – Round 2

Directions: Determine the quotient of the fractions and simplify.

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

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

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

Question 4.
\(\frac{3}{8} \div \frac{5}{8}\)
Answer:
\(\frac{3}{5}\)

Question 5.
\(\frac{1}{4} \div \frac{3}{12}\)
Answer:
\(\frac{3}{3}\) = 1

Question 6.
\(\frac{1}{4} \div \frac{3}{12}\)
Answer:
\(\frac{14}{3}=4 \frac{2}{3}\)

Question 7.
\(\frac{8}{15} \div \frac{4}{5}\)
Answer:
\(\frac{8}{12}=\frac{2}{3}\)

Question 8.
\(\frac{5}{6} \div \frac{5}{12}\)
Answer:
\(\frac{10}{5}\) = 2

Question 9.
\(\frac{3}{5} \div \frac{7}{9}\)
Answer:
\(\frac{27}{35}\)

Question 10.
\(\frac{3}{10} \div \frac{3}{9}\)
Answer:
\(\frac{27}{30}=\frac{9}{10}\)

Question 11.
\(\frac{3}{4} \div \frac{7}{9}\)
Answer:
\(\frac{27}{28}\)

Question 12.
\(\frac{7}{10} \div \frac{3}{8}\)
Answer:
\(\frac{56}{30}=\frac{28}{15}=1 \frac{13}{15}\)

Question 13.
\(4 \div \frac{4}{9}\)
Answer:
\(\frac{36}{4}\) = 9

Question 14.
\(\frac{5}{8} \div 7\)
Answer:
\(\frac{5}{56}\)

Question 15.
\(9 \div \frac{2}{3}\)
Answer:
\(\frac{27}{2}=13 \frac{1}{2}\)

Question 16.
\(\frac{5}{8} \div 1 \frac{3}{4}\)
Answer:
\(\frac{20}{56}=\frac{5}{14}\)

Question 17.
\(\frac{1}{4} \div 2 \frac{2}{5}\)
Answer:
\(\frac{5}{48}\)

Question 18.
\(2 \frac{3}{5} \div \frac{3}{8}\)
Answer:
\(\frac{104}{15}=6 \frac{14}{15}\)

Question 19.
\(1 \frac{3}{5} \div \frac{2}{9}\)
Answer:
\(\frac{72}{10}=7 \frac{2}{10}=7 \frac{1}{5}\)

Question 20.
\(4 \div 2 \frac{3}{8}\)
Answer:
\(\frac{32}{19}=1 \frac{13}{19}\)

Question 21.
\(1 \frac{1}{2} \div 5\)
Answer:
\(\frac{3}{10}\)

Question 22.
\(3 \frac{1}{3} \div 1 \frac{3}{4}\)
Answer:
\(\frac{40}{21}=1 \frac{19}{21}\)

Question 23.
\(2 \frac{2}{5} \div 1 \frac{1}{4}\)
Answer:
\(\frac{48}{25}=1 \frac{23}{25}\)

Question 24.
\(3 \frac{1}{2} \div 2 \frac{2}{3}\)
Answer:
\(\frac{21}{16}=1 \frac{5}{16}\)

Question 25.
\(1 \frac{4}{5} \div 2 \frac{3}{4}\)
Answer:
\(\frac{36}{55}\)

Question 26.
\(3 \frac{1}{6} \div 1 \frac{3}{5}\)
Answer:
\(\frac{95}{48}=1 \frac{47}{48}\)

Question 27.
\(3 \frac{3}{5} \div 2 \frac{1}{8}\)
Answer:
\(\frac{144}{85}=1 \frac{59}{85}\)

Question 28.
\(5 \div 1 \frac{1}{6}\)
Answer:
\(\frac{30}{7}=4 \frac{2}{7}\)

Question 29.
\(3 \frac{3}{4} \div 5 \frac{1}{2}\)
Answer:
\(\frac{30}{44}=\frac{15}{22}\)

Question 30.
\(4 \frac{2}{3} \div 5 \frac{1}{4}\)
Answer:
\(\frac{56}{63}=\frac{8}{9}\)

Engage NY Eureka Math 6th Grade Module 4 Lesson 7 Answer Key

Eureka Math Grade 6 Module 4 Lesson 7 Example Answer Key

Example 1.

What is the length of one side of this square?
Answer:
3 units

What is the formula for the area of a square?
Answer:
A = s²

What is the square’s area as a multiplication expression?
Answer:
3 units × 3 units

What is the square’s area?
Answer:
9 square units

We can count the units. However, look at this other square. Its side length is 23 That is just too many tiny units to draw. What expression can we build to find this square’s area?
Answer:
23 × 23 cm

What is the area of the square? Use a calculator if you need to.
Answer:
529 cm²

Example 2.

What does the letter represent in this blue rectangle?
Answer:
b = 8

With a partner, answer the following question: Given that the second rectangle is divided into four equal parts, what number does the x represent?
Answer:
x = 8

How did you arrive at this answer?
Answer:
We reasoned that each width of the 4 congruent rectangles must be the same. Two 4 lengths equals

What is the total length of the second rectangle? Tell a partner how you know.
Answer:
The length consists of 4 segments that each has a length of 4 cm. 4 × 4 cm = 16 cm.

If the two large rectangles have equal lengths and widths, find the area of each rectangle.
Answer:
8 cm × 16 cm = 128 cm²

Discuss with your partner how the formulas for the area of squares and rectangles can be used to evaluate area for a particular figure.
Answer:

Example 3.

What does the l represent in the first diagram?
Answer:
The length of the rectangular prism

What does the w represent in the first diagram?
Answer:
The width of the rectangular prism

What does the h represent in the first diagram?
Answer:
The height of the rectangular prism

Since we know the formula to find the volume is V = l × w ×h, what number can we substitute for the l in the formula? Why?
Answer:
6, because the length of the second right rectangular prism is 6 cm.

What other number can we substitute for the l?
Answer:
No other number can replace the l. Only one number can replace one letter.

What number can we substitute for the in the formula? Why?
Answer:
2, because the width of the second right rectangular prism is 2 cm.

What number can we substitute for the h in the formula?
Answer:
8 because the height of the second right rectangular prism is 8 cm.

Determine the volume of the second right rectangular prism by replacing the letters in the formula with their appropriate numbers.
Answer:
V = l × w × h; V = 6 cm × 2 cm × 8 cm = 96 cm³

Eureka Math Grade 6 Module 4 Lesson 7 Exercise Answer Key

Exercise 1.
Complete the table below for both squares. Note: These drawings are not to scale.


Answer:

Exercise 2.


Answer:

Exercise 3.
Complete the table for both figures. Using a calculator is appropriate.


Answer:

Eureka Math Grade 6 Module 4 Lesson 7 Problem Set Answer Key

Question 1.
Replace the side length of this square with 4 in., and find the area.

Answer:
The student should draw a square, label the side 4 in., and calculate the area to be 16 in².

Question 2.
Complete the table for each of the given figures.


Answer:

Question 3.
Find the perimeter of each quadrilateral in Problems 1 and 2.
Answer:
p = 16 in. p = 118 p = 35 yd.

Question 4.
Using the formula V = l × w × h, find the volume of a right rectangular prism when the length of the prism is 45 cm, the width is 12 cm, and the height is 10 cm.
Answer:
V = l × w × h; V = 45 cm × 12 cm × 10 cm = 5,400 cm³

Eureka Math Grade 6 Module 4 Lesson 7 Exit Ticket Answer Key

Question 1.
In the drawing below, what do the letters and represent?

Answer:
Length and width of the rectangle

Question 2.
What does the expression l + w + l + w represent?
Answer:
Perimeter of the rectangle, or the sum of the sides of the rectangle

Question 3.
What does the expression l ∙ w represent?
Answer:
Area of the rectangle

Question 4.
The rectangle below is congruent to the rectangle shown in Problem 1. Use this information to evaluate the expressions from Problems 2 and 3.

Answer:
l = 5 and w = 2 p = 14 units A = 10 units²