Eureka Math

Engage NY Eureka Math 7th Grade Module 3 Lesson 17 Answer Key

Eureka Math Grade 7 Module 3 Lesson 17 Exploratory Challenge Answer Key

Exploratory Challenge
To find the formula for the area of a circle, cut a circle into 16 equal pieces.

Arrange the triangular wedges by alternating the “triangle” directions and sliding them together to make a “parallelogram.” Cut the triangle on the left side in half on the given line, and slide the outside half of the triangle to the other end of the parallelogram in order to create an approximate “rectangle.”

The circumference is 2πr, where the radius is r. Therefore, half of the circumference is πr.

What is the area of the “rectangle” using the side lengths above?
Answer:
The area of the “rectangle” is base times height, and, in this case, A = πr ∙ r.

Are the areas of the “rectangle” and the circle the same?
Answer:
Yes, since we just rearranged pieces of the circle to make the “rectangle,” the area of the “rectangle” and the area of the circle are approximately equal. Note that the more sections we cut the circle into, the closer the approximation.

If the area of the rectangular shape and the circle are the same, what is the area of the circle?
Answer:
The area of a circle is written as A = πr ∙ r, or A = πr².

Eureka Math Grade 7 Module 3 Lesson 17 Example Answer Key

Example 1.
Use the shaded square centimeter units to approximate the area of the circle.

Question 1.
What is the radius of the circle?
Answer:
10 cm

Question 2.
What would be a quicker method for determining the area of the circle other than counting all of the squares in the entire circle?
Answer:
Count \(\frac{1}{4}\) of the squares needed; then, multiply that by four in order to determine the area of the entire circle.

Question 3.
Using the diagram, how many squares were used to cover one-fourth of the circle?
Answer:
The area of one-fourth of the circle is approximately 79 cm².

Question 4.
What is the area of the entire circle?
Answer:
A ≈ 4 ∙ 79 cm²
A ≈ 316 cm²

Example 2.
A sprinkler rotates in a circular pattern and sprays water over a distance of 12 feet. What is the area of the circular region covered by the sprinkler? Express your answer to the nearest square foot.
Draw a diagram to assist you in solving the problem. What does the distance of 12 feet represent in this problem?
Answer:

The radius is 12 feet.

What information is needed to solve the problem?
Answer:
The formula to find the area of a circle is A = πr². If the radius is 12 ft., then A = π ∙ (12 ft.)² = 144π ft², or approximately 452 ft².

Example 3.
Suzanne is making a circular table out of a square piece of wood. The radius of the circle that she is cutting is 3 feet. How much waste will she have for this project? Express your answer to the nearest square foot.
Draw a diagram to assist you in solving the problem. What does the distance of 3 feet represent in this problem?
Answer:
The radius of the circle is 3 feet.

Question 1.
What information is needed to solve the problem?
Answer:
The area of the circle and the area of the square are needed so that we can subtract the area of the circle from the area of the square to determine the amount of waste.

Question 2.
What information do we need to determine the area of the square and the circle?
Answer:
Circle: just radius because A = πr² Square: one side length

Question 3.
How will we determine the waste?
Answer:
The waste is the area left over from the square after cutting out the circular region. The area of the circle is
A = π ∙ (3 ft.)² = 9π ft² ≈ 28.26 ft². The area of the square is found by first finding the diameter of the circle, which is the same as the side of the square. The diameter is d = 2r; so, d = 2 ∙ 3 ft. or 6 ft. The area of a square is found by multiplying the length and width, so A = 6 ft. ∙ 6 ft. = 36 ft². The solution is the difference between the area of the square and the area of the circle, so 36 ft²-28.26 ft² ≈ 7.74 ft².

Question 4.
Does your solution answer the problem as stated?
Answer:
Yes, the amount of waste is 7.74 ft².

Eureka Math Grade 7 Module 3 Lesson 17 Exercise Answer Key

Exercises 1–3
Solve the problem below individually. Explain your solution.

Exercise 1.
Find the radius of a circle if its circumference is 37.68 inches. Use π ≈ 3.14.
Answer:
If C = 2πr, then 37.68 = 2πr. Solving the equation for r,
37.68 = 2πr
(\(\frac{1}{2 \pi}\))37.68 = (\(\frac{1}{2 \pi}\))2πr
\(\frac{1}{6.28}\) (37.68) ≈ r
6 ≈ r
The radius of the circle is approximately 6 in.

Exercise 2.
Determine the area of the rectangle below. Name two ways that can be used to find the area of the rectangle.

Answer:
The area of the rectangle is 24 cm². The area can be found by counting the square units inside the rectangle or by multiplying the length (6 cm) by the width (4 cm).

Exercise 3.
Find the length of a rectangle if the area is 27 cm² and the width is 3 cm.
Answer:
If the area of the rectangle is Area = length ∙ width, then
27 cm² = l ∙ 3 cm
\(\frac{1}{3}\) ∙ 27 cm² = \(\frac{1}{3}\) ∙ l ∙ 3 cm
9 cm = l

Exercises 4–6

Exercise 4.
A circle has a radius of 2 cm.
a. Find the exact area of the circular region.
Answer:
A = π ∙ (2 cm)² = 4π cm²

b. Find the approximate area using 3.14 to approximate π.
Answer:
A = 4 cm² ∙ π ≈ 4 cm² ∙ 3.14 ≈ 12.56 cm²

Exercise 5.
A circle has a radius of 7 cm.
a. Find the exact area of the circular region.
Answer:
A = π ∙ (7 cm)² = 49π cm²

b. Find the approximate area using \(\frac{22}{7}\) to approximate π.
Answer:
A = 49 ∙ π cm² ≈ (49 ∙ \(\frac{22}{7}\) ) cm² ≈ 154 cm²

c. What is the circumference of the circle?
Answer:
C = 2π ∙ 7 cm = 14π cm ≈ 43.96 cm

Exercise 6.
Joan determined that the area of the circle below is 400π cm². Melinda says that Joan’s solution is incorrect; she believes that the area is 100π cm². Who is correct and why?
Answer:

Melinda is correct. Joan found the area by multiplying π by the square of 20 cm (which is the diameter) to get a result of 400π cm², which is incorrect. Melinda found that the radius was 10 cm (half of the diameter). Melinda multiplied π by the square of the radius to get a result of 100π cm².

Eureka Math Grade 7 Module 3 Lesson 17 Problem Set Answer Key

Question 1.
The following circles are not drawn to scale. Find the area of each circle. (Use \(\frac{22}{7}\) as an approximation for π.)

Answer:

Question 2.
A circle has a diameter of 20 inches.
a. Find the exact area, and find an approximate area using π ≈ 3.14.
Answer:
If the diameter is 20 in., then the radius is 10 in. If A = πr², then A = π ∙ (10 in.)² or 100π in².
A ≈ (100 ∙ 3.14) in² ≈ 314 in².

b. What is the circumference of the circle using π ≈ 3.14 ?
Answer:
If the diameter is 20 in., then the circumference is C = πd or C ≈ 3.14 ∙ 20 in. ≈ 62.8 in.

Question 3.
A circle has a diameter of 11 inches.
a. Find the exact area and an approximate area using π ≈ 3.14.
Answer:
If the diameter is 11 in., then the radius is \(\frac{11}{2}\) in. If A = πr², then A = π ∙ (\(\frac{11}{2}\) in.)²or \(\frac{121}{4}\) π in².
A ≈ (\(\frac{121}{4}\) ∙ 3.14) in² ≈ 94.985 in²

b. What is the circumference of the circle using π ≈ 3.14?
Answer:
If the diameter is 11 inches, then the circumference is C = πd or C ≈ 3.14 ∙ 11 in. ≈ 34.54 in.

Question 4.
Using the figure below, find the area of the circle.

Answer:
In this circle, the diameter is the same as the length of the side of the square. The diameter is 10 cm; so, the radius is 5 cm. A = πr², so A = π(5 cm)² = 25π cm².

Question 5.
A path bounds a circular lawn at a park. If the inner edge of the path is 132 ft. around, approximate the amount of area of the lawn inside the circular path. Use π ≈ \(\frac{22}{7}\) .
Answer:
The length of the path is the same as the circumference. Find the radius from the circumference; then, find the area.
C = 2πr
132 ft. ≈ 2 ∙ \(\frac{22}{7}\) ∙ r
132 ft. ≈ \(\frac{44}{7}\) r
\(\frac{7}{44}\) ∙ 132 ft. ≈ \(\frac{7}{44}\) ∙ \(\frac{44}{7}\) r
21 ft. ≈ r
A ≈ \(\frac{22}{7}\) ∙ (21 ft.)²
A ≈ 1386 ft²

Question 6.
The area of a circle is 36π cm². Find its circumference.
Answer:
Find the radius from the area of the circle; then, use it to find the circumference.
A = πr²
36π cm² = πr²
\(\frac{1}{\pi}\) ∙ 36π cm² = \(\frac{1}{\pi}\) ∙ πr²
36 cm² = r²
6 cm = r
C = 2πr
C = 2π ∙ 6 cm
C = 12π cm

Question 7.
Find the ratio of the area of two circles with radii 3 cm and 4 cm.
Answer:
The area of the circle with radius 3 cm is 9π cm². The area of the circle with the radius 4 cm is 16π cm². The ratio of the area of the two circles is 9π:16π or 9:16.

Question 8.
If one circle has a diameter of 10 cm and a second circle has a diameter of 20 cm, what is the ratio of the area of the larger circle to the area of the smaller circle?
Answer:
The area of the circle with the diameter of 10 cm has a radius of 5 cm. The area of the circle with the diameter of 10 cm is π ∙ (5 cm)², or 25π cm². The area of the circle with the diameter of 20 cm has a radius of 10 cm. The area of the circle with the diameter of 20 cm is π ∙ (10 cm)² or 100π cm². The ratio of the diameters is 20 to 10 or 2:1, while the ratio of the areas is 100π to 25π or 4:1.

Question 9.
Describe a rectangle whose perimeter is 132 ft. and whose area is less than 1 ft². Is it possible to find a circle whose circumference is 132 ft. and whose area is less than 1 ft²? If not, provide an example or write a sentence explaining why no such circle exists.
Answer:
A rectangle that has a perimeter of 132 ft. can have a length of 65.995 ft. and a width of 0 .005 ft. The area of such a rectangle is 0.329975 ft², which is less than 1 ft². No, because a circle that has a circumference of 132 ft. has a radius of approximately 21 ft.
A = πr² = π(21)² = 1387.96 ≠ 1

Question 10.
If the diameter of a circle is double the diameter of a second circle, what is the ratio of the area of the first circle to the area of the second?
Answer:
If I choose a diameter of 24 cm for the first circle, then the diameter of the second circle is 12 cm. The first circle has a radius of 12 cm and an area of 144π cm². The second circle has a radius of 6 cm and an area of 36π cm². The ratio of the area of the first circle to the second is 144π to 36π , which is a 4 to 1 ratio. The ratio of the diameters is 2, while the ratio of the areas is the square of 2, or 4.

Eureka Math Grade 7 Module 3 Lesson 17 Exit Ticket Answer Key

Complete each statement using the words or algebraic expressions listed in the word bank below.

Answer:
1. The length of the height of the rectangular region approximates the length of the radius of the circle.
2. The base of the rectangle approximates the length of one-half of the circumference of the circle.

Question 3.
The circumference of the circle is _______________________.
Answer:
The circumference of the circle is 2πr.

Question 4.
The _________________ of the ___________________ is 2r.
Answer:
The diameter of the circle is 2r.

Question 5.
The ratio of the circumference to the diameter is ______.
Answer:
The ratio of the circumference to the diameter is π.

Question 6.
Area (circle) = Area of (_____________) = \(\frac{1}{2}\) ∙ circumference ∙ r = \(\frac{1}{2}\) (2πr) ∙ r = π ∙ r ∙ r = _____________.
Answer:
Area (circle) = Area of (rectangle) = \(\frac{1}{2}\) ∙ circumference ∙ r = \(\frac{1}{2}\) (2πr) ∙ r = π ∙ r ∙ r = πr².

Engage NY Eureka Math 8th Grade Module 1 Lesson 9 Answer Key

Eureka Math Grade 8 Module 1 Lesson 9 Exercise Answer Key

Are the following numbers written in scientific notation? If not, state the reason.

Exercise 1.
1.908×10¹⁷
Ans:
yes

Exercise 2.
0.325×10^-2
Answer:
no, d<1

Exercise 3.
7.99×10³²
Answer:
yes

Exercise 4.
4.0701 + 10⁷
Answer:
no, it must be a product

Exercise 5.
18.432×5⁸
Answer:
no, d>10 and it is ×5 instead of ×10

Exercise 6.
8×10^-11
Answer:
yes

Exercises 7–9 (10 minutes)
Have students complete Exercises 7–9 independently.

Use the table below to complete Exercises 7 and 8.
The table below shows the debt of the three most populous states and the three least populous states.

Exercise 7.
a. What is the sum of the debts for the three most populous states? Express your answer in scientific notation.
Answer:
(4.07×(10)¹¹)+(3.37×10¹¹)+(2.76×10¹¹)=(4.07+3.37+2.76)×10¹¹
=10.2×10¹¹
=(1.02×10)×10¹¹
=1.02×10¹²

b. What is the sum of the debt for the three least populous states? Express your answer in scientific notation.
Ans:
(4×10⁹)+(4×10⁹)+(2×10⁹)=(4+4+2)×10⁹
=10×10⁹
=(1×10)×10⁹
=1×10¹⁰

c. How much larger is the combined debt of the three most populous states than that of the three least populous states? Express your answer in scientific notation.
Answer:
(1.02×10¹²)-(1×10¹⁰)=(1.02×10²×10¹⁰)-(1×10¹⁰)
=(102×10¹⁰)-(1×10¹⁰)
=(102-1)×10¹⁰
=101×10¹⁰
=(1.01×10²)×10¹⁰
=1.01×10¹²

Exercise 8.
a. What is the sum of the population of the three most populous states? Express your answer in scientific notation.
Answer:
(3.8×10⁷)+(1.9×10⁷)+(2.6×10⁷)=(3.8+1.9+2.6)×10⁷
=8.3×10⁷

b. What is the sum of the population of the three least populous states? Express your answer in scientific notation.
Answer:
(6.9×10⁵)+(6.26×10⁵)+(5.76×10⁵)=(6.9+6.26+5.76)×10⁵
=18.92×10⁵
=(1.892×10)×10⁵
=1.892×10⁶

c. Approximately how many times greater is the total population of California, New York, and Texas compared to the total population of North Dakota, Vermont, and Wyoming?
Answer:
\(\frac{8.3 \times 10^{7}}{1.892 \times 10^{6}}\) = \(\frac{8.3}{1.892}\) × \(\frac{10^{7}}{10^{6}}\)
≈4.39×10
=43.9
The combined population of California, New York, and Texas is about 43.9 times greater than the combined population of North Dakota, Vermont, and Wyoming.

Exercise 9.
All planets revolve around the sun in elliptical orbits. Uranus’s furthest distance from the sun is approximately 3.004×10⁹ km, and its closest distance is approximately 2.749×10⁹ km. Using this information, what is the average distance of Uranus from the sun?
Answer:
average distance = \(\frac{\left(3.004 \times 10^{9}\right)+\left(2.749 \times 10^{9}\right)}{2}\)
= \(\frac{(3.004+2.749) \times 10^{9}}{2}\)
= \(\frac{5.753 \times 10^{9}}{2}\)
=2.8765×10⁹
On average, Uranus is 2.8765×10⁹ km from the sun.

Eureka Math Grade 8 Module 1 Lesson 9 Problem Set Answer Key

Students practice working with numbers written in scientific notation.

Question 1.
Write the number 68,127,000,000,000,000 in scientific notation. Which of the two representations of this number do you prefer? Explain.
Answer:
68 127 000 000 000 000=6.8127×10¹⁶
Most likely, students will say that they like the scientific notation better because it allows them to write less. However, they should also take note of the fact that counting the number of zeros in 68,127,000,000,000,000 is a nightmare. A strong reason for using scientific notation is to circumvent this difficulty: right away, the exponent 16 shows that this is a 17-digit number.

Question 2.
Here are the masses of the so-called inner planets of the solar system.

What is the average mass of all four inner planets? Write your answer in scientific notation.
Answer:

The average mass of the inner planets is 2.9531925×10²⁴ kg.

Eureka Math Grade 8 Module 1 Lesson 9 Exit Ticket Answer Key

Question 1.
The approximate total surface area of Earth is 5.1×10⁸ km². All the salt water on Earth has an approximate surface area of 352,000,000 km², and all the fresh water on Earth has an approximate surface area of 9×10⁶ km². How much of Earth’s surface is covered by water, including both salt and fresh water? Write your answer in scientific notation.
Answer:
(3.52×10⁸)+(9×10⁶)=(3.52×10²×10⁶)+(9×10⁶)
=(352×10⁶)+(9×10⁶)
=(352+9)×10⁶
=361×10⁶
=3.61×10⁸
The Earth’s surface is covered by 3.61×10⁸ km² of water.

Question 2.
How much of Earth’s surface is covered by land? Write your answer in scientific notation.
Answer:
(5.1×10⁸)-(3.61×10⁸)=(5.1-3.61)×10⁸
=1.49×10⁸
The Earth’s surface is covered by 1.49×10⁸ km² of land.

Question 3.
Approximately how many times greater is the amount of Earth’s surface that is covered by water compared to the amount of Earth’s surface that is covered by land?
Answer:
\(\frac{3.61 \times 10^{8}}{1.49 \times 10^{8}}\)≈2.4
About 2.4 times more of the Earth’s surface is covered by water than by land.

Engage NY Eureka Math 8th Grade Module 1 Lesson 8 Answer Key

Eureka Math Grade 8 Module 1 Lesson 8 Example Answer Key

Example 1.
In 1723, the population of New York City was approximately 7,248. By 1870, almost 150 years later, the population had grown to 942,292. We want to determine approximately how many times greater the population was in 1870 compared to 1723.
The word approximately in the question lets us know that we do not need to find a precise answer, so we approximate both populations as powers of 10.
→ Population in 1723: 7248<9999<10000=10⁴
→ Population in 1870: 942 292<999 999<1 000 000=10⁶
We want to compare the population in 1870 to the population in 1723:
\(\frac{10^{6}}{10^{4}}\)
Now we can use what we know about the laws of exponents to simplify the expression and answer the question:
\(\frac{10^{6}}{10^{4}}\) =10².
Therefore, there were approximately 100 times more people in New York City in 1870 compared to 1723.

Example 2.
Let’s compare the population of New York City to the population of New York State. Specifically, let’s find out how many times greater the population of New York State is compared to that of New York City.
The population of New York City is 8,336,697. Let’s round this number to the nearest million; this gives us 8,000,000. Written as single-digit integer times a power of 10:
8 000 000=8×10⁶.
The population of New York State is 19,570,261. Rounding to the nearest million gives us 20,000,000. Written as a single-digit integer times a power of 10:
20 000 000=2×10⁷.
To estimate the difference in size we compare state population to city population:
\(\frac{2 \times 10^{7}}{8 \times 10^{6}}\)
Now we simplify the expression to find the answer:
\(\frac{2 \times 10^{7}}{8 \times 10^{6}}\) = \(\frac{2}{8}\) × \(\frac{10^{7}}{10^{6}}\)By the product formula
= \(\frac{1}{4}\)×10 By equivalent fractions and the first law of exponents
=0.25×10
=2.5
Therefore, the population of the state is 2.5 times that of the city.

Example 3.
There are about 9 billion devices connected to the Internet. If a wireless router can support 300 devices, about how many wireless routers are necessary to connect all 9 billion devices wirelessly?
Because 9 billion is a very large number, we should express it as a single-digit integer times a power of 10.
9 000 000 000=9×10⁹
The laws of exponents tells us that our calculations will be easier if we also express 300 as a single-digit integer times a power of 10, even though 300 is much smaller.
300=3×10²
We want to know how many wireless routers are necessary to support 9 billion devices, so we must divide
\(\frac{9 \times 10^{9}}{3 \times 10^{2}}\)
Now, we can simplify the expression to find the answer:
\(\frac{9 \times 10^{9}}{3 \times 10^{2}}\) = \(\frac{9}{3}\) × \(\frac{10^{9}}{10^{2}}\) By the product formula
=3×10⁷ By equivalent fractions and the first law of exponents
=30 000 000
About 30 million routers are necessary to connect all devices wirelessly.

Example 4.
The average American household spends about $40,000 each year. If there are about 1×〖10〗⁸ households, what is the total amount of money spent by American households in one year?
Let’s express $40,000 as a single-digit integer times a power of 10.
40000=4×〖10〗⁴
The question asks us how much money all American households spend in one year, which means that we need to multiply the amount spent by one household by the total number of households:
(4×10⁴ )(1×10⁸ )=(4×1)(10⁴×10⁸ ) By repeated use of associative and commutative properties
=4×10¹² By the first law of exponents
Therefore, American households spend about $4,000,000,000,000 each year altogether!

Eureka Math Grade 8 Module 1 Lesson 8 Exercise Answer Key

Exercise 1.
The Federal Reserve states that the average household in January of 2013 had $7,122 in credit card debt. About how many times greater is the U.S. national debt, which is $16,755,133,009,522? Rewrite each number to the nearest power of 10 that exceeds it, and then compare.
Answer:
Household debt=7122<9999<10000=10⁴.
U.S.debt =16 755 133 009 522<99 999 999 999 999<100 000 000 000 000=10¹⁴.
\(\frac{10^{14}}{10^{4}}\) = 10^14-4=10¹⁰. The U.S. national debt is 10¹⁰ times greater than the average household’s credit card debt.

Exercise 2.
There are about 3,000,000 students attending school, kindergarten through Grade 12, in New York. Express the number of students as a single-digit integer times a power of 10.
Answer:
3 000 000=3×10⁶

The average number of students attending a middle school in New York is 8×〖10〗². How many times greater is the overall number of K–12 students compared to the average number of middle school students?
Answer:
\(\frac{3 \times 10^{6}}{8 \times 10^{2}}\)= \(\frac{3}{8}\)×\(\frac{10^{6}}{10^{2}}\)
= \(\frac{3}{8}\)×\(\frac{10^{6}}{10^{2}}\)
=0.375×〖10〗⁴
=3750
There are about 3,750 times more students in K–12 compared to the number of students in middle school.

Exercise 3.
A conservative estimate of the number of stars in the universe is 6×10²². The average human can see about 3,000 stars at night with his naked eye. About how many times more stars are there in the universe compared to the stars a human can actually see?
Answer:
\(\frac{6 \times 10^{22}}{3 \times 10^{3}}\) = \(\frac{6}{3}\) ×\(\frac{10^{22}}{10^{3}}\)= 2×10^22-3 = 2×10¹⁹
There are about 2×10¹⁹ times more stars in the universe compared to the number we can actually see.

Exercise 4.
The estimated world population in 2011 was 7×10⁹. Of the total population, 682 million of those people were left-handed. Approximately what percentage of the world population is left-handed according to the 2011 estimation?
Answer:
682 000 000≈700 000 000=7×10⁸
\(\frac{7 \times 10^{8}}{7 \times 10^{9}}\)=\(\frac{7}{7}\)×\(\frac{10^{8}}{10^{9}}\)
= 1×\(\frac{1}{10}\)
=\(\frac{1}{10}\)
About one-tenth of the population is left-handed, which is equal to 10%.

Exercise 5
The average person takes about 30,000 breaths per day. Express this number as a single-digit integer times a power of 10.
Answer:
30000=3×10⁴

If the average American lives about 80 years (or about 30,000 days), how many total breaths will a person take in her lifetime?
Answer:
(3×10⁴ )×(3×10⁴ )=9×10⁸
The average American takes about 900,000,000 breaths in a lifetime.

Eureka Math Grade 8 Module 1 Lesson 8 Problem Set Answer Key

Students practice estimating size of quantities and performing operations on numbers written in the form of a single-digit integer times a power of 10.

Question 1.
The Atlantic Ocean region contains approximately 2×10¹⁶ gallons of water. Lake Ontario has approximately 8,000,000,000,000 gallons of water. How many Lake Ontarios would it take to fill the Atlantic Ocean region in terms of gallons of water?
Answer:
8 000 000 000 000=8×10¹²
\(\frac{2 \times 10^{16}}{8 \times 10^{12}}\)=\(\frac{2}{8}\)×\(\frac{10^{16}}{10^{12}}\)
=\(\frac{1}{4}\)×10⁴
=0.25×10⁴
=2500
2,500 Lake Ontario’s would be needed to fill the Atlantic Ocean region.

Question 2.
U.S. national forests cover approximately 300,000 square miles. Conservationists want the total square footage of forests to be 300,000〗² square miles. When Ivanna used her phone to do the calculation, her screen showed the following:

a. What does the answer on her screen mean? Explain how you know.
Answer:
The answer means 9×10¹⁰. This is because:
(300 000)²=(3×10⁵)²
=3²×(10⁵ )²
=9×10¹⁰

b. Given that the U.S. has approximately 4 million square miles of land, is this a reasonable goal for conservationists? Explain.
Answer:
4 000 000=4×10⁶. It is unreasonable for conservationists to think the current square mileage of forests could increase that much because that number is greater than the number that represents the total number of square miles in the U.S,
9×10¹⁰>4×10⁶.

Question 3.
The average American is responsible for about 20,000 kilograms of carbon emission pollution each year. Express this number as a single-digit integer times a power of 10.
Answer:
20 000=2×10⁴

Question 4.
The United Kingdom is responsible for about 1× 10⁴ kilograms of carbon emission pollution each year. Which country is responsible for greater carbon emission pollution each year? By how much?
Answer:
2× 10⁴>1×10⁴
America is responsible for greater carbon emission pollution each year. America produces twice the amount of the U.K. pollution.

Eureka Math Grade 8 Module 1 Lesson 8 Exit Ticket Answer Key

Most English-speaking countries use the short-scale naming system, in which a trillion is expressed as 1,000,000,000,000. Some other countries use the long-scale naming system, in which a trillion is expressed as 1,000,000,000,000,000,000,000. Express each number as a single-digit integer times a power of ten. How many times greater is the long-scale naming system than the short-scale?
Answer:
1 000 000 000 000=10¹²
1 000 000 000 000 000 000 000= 10²¹
\(\frac{10^{21}}{10^{12}}\) = 10⁹. The long-scale is about 10⁹ times greater than the short-scale.

Eureka Math Grade 8 Module 1 Lesson 8 Sprint Answer Key

Applying Properties of Exponents to Generate Equivalent Expressions—Round 1
Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. When appropriate, express answers without parentheses or as equal to 1. All letters denote numbers.

Answer:

Applying Properties of Exponents to Generate Equivalent Expressions—Round 2

Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. When appropriate, express answers without parentheses or as equal to 1. All letters denote numbers.

Answer:

Engage NY Eureka Math 8th Grade Module 1 Lesson 7 Answer Key

Eureka Math Grade 8 Module 1 Lesson 7 Exercise Answer Key

Exercise 1.
Let M=993,456,789,098,765. Find the smallest power of 10 that will exceed M.
Answer:
M=993 456 789 098 765 < 999 999 999 999 999 < 1 000 000 000 000 000=10¹⁵. Because M has 15 digits, 10¹⁵ will exceed it.

Exercise 2.
Let M=78,491\(\frac{899}{987}\). Find the smallest power of 10 that will exceed M.
Answer:
M=78491\(\frac{899}{987}\) < 78492<99999<100 000=10⁵.
Therefore, 10⁵ will exceed M.

Exercise 3.
Let M be a positive integer. Explain how to find the smallest power of 10 that exceeds it.
Answer:
If M is a positive integer, then the power of 10 that exceeds it will be equal to the number of digits in M. For example, if M were a 10-digit number, then 10¹⁰ would exceed M. If M is a positive number, but not an integer, then the power of 10 that would exceed it would be the same power of 10 that would exceed the integer to the right of M on a number line. For example, if M=5678.9, the integer to the right of M is 5,679. Then based on the first explanation, 10⁴ exceeds both this integer and M; this is because M=5678.9<5679<10 000=10⁴.

Exercise 4.
The chance of you having the same DNA as another person (other than an identical twin) is approximately 1 in 10 trillion (one trillion is a 1 followed by 12 zeros). Given the fraction, express this very small number using a negative power of 10.
\(\frac{1}{10000000000000}\)
Answer:
\(\frac{1}{10000000000000}\) = \(\frac{1}{10^{13}}\)
= 10^-13

Exercise 5.
The chance of winning a big lottery prize is about 10^-8, and the chance of being struck by lightning in the U.S. in any given year is about 0.000 001. Which do you have a greater chance of experiencing? Explain.
Answer:
0.000 001=10^-6
There is a greater chance of experiencing a lightning strike. On a number line, 10^-8 is to the left of 10^-6. Both numbers are less than one (one signifies 100% probability of occurring). Therefore, the probability of the event that is greater is 10^-6—that is, getting struck by lightning.

Exercise 6.
There are about 100 million smartphones in the U.S. Your teacher has one smartphone. What share of U.S. smartphones does your teacher have? Express your answer using a negative power of 10.
Answer:
\(\frac{1}{100000000}\)=\(\frac{1}{10^{8}}\)=10^-8

Eureka Math Grade 8 Module 1 Lesson 7 Problem Set Answer Key

Question 1.
What is the smallest power of 10 that would exceed 987,654,321,098,765,432?
Answer:
987 654 321 098 765 432<999 999 999 999 999 999<1 000 000 000 000 000 000=10¹⁸

Question 2.
What is the smallest power of 10 that would exceed 999,999,999,991?
Answer:
999 999 999 991<999 999 999 999<1 000 000 000 000=10¹²

Question 3.
Which number is equivalent to 0.000 000 1: 10⁷or 10^-7? How do you know?
Answer:
0.000 000 1=10^-7. Negative powers of 10 denote numbers greater than zero but less than 1. Also, the decimal 0.000 000 1 is equal to the fraction \(\frac{1}{10^{7}}\) which is equivalent to 10^-7.

Question 4.
Sarah said that 0.000 01 is bigger than 0.001 because the first number has more digits to the right of the decimal point. Is Sarah correct? Explain your thinking using negative powers of 10 and the number line.
Answer:
0.000 01= \(\frac{1}{100000}\) = 10^-5 and 0.001= \(\frac{1}{1000}\) =10^-3. On a number line, 10^-5 is closer to zero than 10^-3; therefore, 10^-5 is the smaller number, and Sarah is incorrect.

Question 5.
Order the following numbers from least to greatest:

Answer:
10^-99<10^-17<10^-5<10⁵<10^-14<10³⁰

Eureka Math Grade 8 Module 1 Lesson 7 Exit Ticket Answer Key

Question 1.
Let M=118,526.65902. Find the smallest power of 10 that will exceed M.
Answer:
Since M=118,526.65902<118,527<1,000,000<10⁶, then 10⁶will exceed M.

Question 2.
Scott said that 0.09 was a bigger number than 0.1. Use powers of 10 to show that he is wrong.
Answer:
We can rewrite 0.09 as \(\frac{9}{10^{2}}\) = 9×10^-2 and rewrite 0.1 as \(\frac{1}{10^{1}}\) =1 ×10^-1. Because 0.09 has a smaller power of 10, 0.09 is closer to zero and is smaller than 0.1.

Engage NY Eureka Math 8th Grade Module 1 Lesson 6 Answer Key

Eureka Math Grade 8 Module 1 Lesson 6 Exercise Answer Key

Exercise 1.
Show that (C) is implied by equation (5) of Lesson 4 when m>0, and explain why (C) continues to hold even when
m=0.
Answer:
Equation (5) says for any numbers x, y, (y≠0) and any positive integer n, the following holds: (\(\frac{x}{y}\))ⁿ=\(\frac{x^{n}}{y^{n}}\). So,
(\(\frac{1}{x}\))^m = \(\frac{1^{m}}{x^{m}}\) By (\(\frac{x}{y}\))ⁿ= \(\frac{x^{n}}{y^{n}}\) for positive integer n and nonzero y (5)
= \(\frac{1}{x^{m}}\) Because 1^m =1

If m= 0, then the left side is
(\(\frac{1}{x}\))^m =(\(\frac{1}{x}\))⁰
=1 By definition of x⁰,
and the right side is
\(\frac{1}{x^{m}}\) = \(\frac{1}{x^{0}}\)
= \(\frac{1}{1}\) By definition of x⁰
=1.

Exercise 2.
Show that (B) is in fact a special case of (11) by rewriting it as (x^m)^-1 = x^(-1m) for any whole number m, so that if b=m (where m is a whole number) and a=-1, (11) becomes (B).
Answer:
(B) says x^-m = \(\frac{1}{x^{m}}\).
The left side of (B), x^-m is equal to x^(-1)m.
The right side of (B), \(\frac{1}{x^{m}}\), is equal to (x^m)^-1 by the definition of (x^m)^-1 in Lesson 5.
Therefore, (B) says exactly that (x^m)^-1 = x^(-1)m.

Exercise 3.
Show that (C) is a special case of (11) by rewriting (C) as (x^-1)^m = x^m(-1) for any whole number m. Thus, (C) is the special case of (11) when b=-1 and a=m, where m is a whole number.
Answer:
(C) says (\(\frac{1}{x}\))^m = \(\frac{1}{x^{m}}\) for any whole number m.
The left side of (C) is equal to
(\(\frac{1}{x}\))^m =(x^-1)^mBy definition of x^-1 ,
and the right side of (C) is equal to
\(\frac{1}{x^{m}}\) =x^-m By definition of x^-m, and the latter is equal to x^m(-1). Therefore, (C) says (x^-1)^m) =x^m(-1) for any whole number m.

Exercise 4.
Proof of Case (iii): Show that when a<0 and b≥0, (x^b )^a=x^ab is still valid. Let a=-c for some positive integer c. Show that the left and right sides of (x^b )^a=x^ab are equal.
The left side is
(x^b )^a=(x^b )^-c
= \(\frac{1}{\left(x^{b}\right)^{c}}\) By x^-m = \(\frac{1}{x^{m}}\) for any whole number m (B)
= \(\frac{1}{x^{c b}}\) . By (x^m)ⁿ=x^mn for all whole numbers m and n (A)
The right side is
x^ab = x^(-c)b
=x^-(cb)
= \(\frac{1}{x^{c b}}\) . By x^-m = \(\frac{1}{x^{m}}\) for any whole number m (B)
So, the two sides are equal.

Eureka Math Grade 8 Module 1 Lesson 6 Problem Set Answer Key

Question 1.
You sent a photo of you and your family on vacation to seven Facebook friends. If each of them sends it to five of their friends, and each of those friends sends it to five of their friends, and those friends send it to five more, how many people (not counting yourself) will see your photo? No friend received the photo twice. Express your answer in exponential notation.

Answer:

The total number of people who viewed the photo is (5⁰+5¹+5²+5³ )×7.

Question 2.
Show directly, without using (11), that (1.27^-36 )⁸⁵= 1.27^-36∙85.
Answer:
(1.27^-36 )⁸⁵= (\(\frac{1}{1.27^{36}}\))⁸⁵ By definition
= \(\frac{1}{\left(1.27^{36}\right)^{85}}\)By (\(\frac{1}{x}\))^m =\(\frac{1}{x^{m}}\) for any whole number m (C)
= \(\frac{1}{1.27^{36 \cdot 85}}\)By (x^m)ⁿ=x^mn for whole numbers m and n (7)
= 1.27^-36∙85 By x^-m = \(\frac{1}{x^{m}}\) for any whole number m (B)

Question 3.
Show directly that (\(\frac{2}{13}\))^-127∙(\(\frac{2}{13}\))^-56=(\(\frac{2}{13}\))^-183.
Answer:

Question 4.
Prove for any nonzero number x, x^-127∙x^-56=x^-183.
Answer:
x^-127∙x^-56 =\(\frac{1}{x^{127}}\) ∙\(\frac{1}{x^{56}}\) By definition
= \(\frac{1}{x^{127} \cdot x^{56}}\) By the product formula for complex fractions
=\(\frac{1}{x^{127+56}}\) By x^m ∙xⁿ=x^m+n for whole numbers m and n (6)
= \(\frac{1}{x^{183}}\)
= x^-183 By x^-m = \(\frac{1}{x^{m}}\) for any whole number m (B)

Question 5.
Prove for any nonzero number x, x^-m ∙x^-n=x^-m-n for positive integers m and n.
Answer:
x^-m∙x^-n= \(\frac{1}{x^{m}}\)∙\(\frac{1}{x^{n}}\) By definition
= \(\frac{1}{x^{m} \cdot x^{n}}\) By the product formula for complex fractions
= \(\frac{1}{x^{m+n}}\) By x^m ∙xⁿ=x^m+n for whole numbers m and n (6)
=x^-(m+n) By x^-m = \(\frac{1}{x^{m}}\) for any whole number m (B)
=x^-m-n

Question 6.
Which of the preceding four problems did you find easiest to do? Explain.
Answer:
Students will likely say that x^-m ∙x^-n=x^-m-n (Problem 5) was the easiest problem to do. It requires the least amount of writing because the symbols are easier to write than decimal or fraction numbers.

Question 7.
Use the properties of exponents to write an equivalent expression that is a product of distinct primes, each raised to an integer power.
Answer:

Eureka Math Grade 8 Module 1 Lesson 6 Exit Ticket Answer Key

Question 1.
Show directly that for any nonzero integer x, x^-5∙x^-7 = x^-12.
Answer:
x^-5∙x^-7 =\(\frac{1}{x^{5}}\) ∙\(\frac{1}{x^{7}}\) By x^-m = \(\frac{1}{x^{m}}\) for any whole number m (B)
=\(\frac{1}{x^{5} \cdot x^{7}}\) By the product formula for complex fractions
=\(\frac{1}{x^{5+7}}\) By x^m ∙xⁿ=x^m+n for whole numbers m and n (6)
=\(\frac{1}{x^{12}}\)
= x^-12 By x^-m = \(\frac{1}{x^{m}}\) for any whole number m (B)

Question 2.
Show directly that for any nonzero integer x, (x^-2 )^-3=x⁶.
Answer:
(x^-2)^-3 = \(\frac{1}{\left(x^{-2}\right)^{3}}\) By x^-m = \(\frac{1}{x^{m}}\) for any whole number m (B)
= \(\frac{1}{x^{-(2 \cdot 3)}}\) By case (ii) of (11)
=\(\frac{1}{x^{-6}}\)
= x⁶ By x^-m = \(\frac{1}{x^{m}}\) for any whole number m (B)

Engage NY Eureka Math 8th Grade Module 1 Lesson 5 Answer Key

Eureka Math Grade 8 Module 1 Lesson 5 Exercise Answer Key

Exercise 1.
Verify the general statement x^-b=\(\frac{1}{x^{b}}\) for x=3 and b=-5.
Answer:
If b were a positive integer, then we have what the definition states. However, b is a negative integer, specifically
b=-5, so the general statement in this case reads
3^-(-5)=\(\frac{1}{3^{-5}}\).
The right side of this equation is

Since the left side is also 3⁵, both sides are equal.
3^-(-5)=\(\frac{1}{3^{-5}}\)=3⁵

Exercise 2.
What is the value of (3×10^-2)?
Answer:
(3×10^-2) = 3 × \(\frac{1}{10^{2}}\) = \(\frac{3}{10^{2}}\) =0.03

Exercise 3.
What is the value of (3×10^-5)?
Answer:
(3×10^-5) = 3×\(\frac{1}{10^{5}}\) = \(\frac{3}{10^{5}}\) =0.00003

Exercise 4.
Write the complete expanded form of the decimal 4.728 in exponential notation.
Answer:
4.728=(4×10⁰)+(7×10^-1)+(2×10^-2)+(8×10^-3)

For Exercises 5–10, write an equivalent expression, in exponential notation, to the one given, and simplify as much as possible.

Exercise 5.
5^-3=
Answer:
\(\frac{1}{5^{3}}\)

Exercise 6.
\(\frac{1}{8^{9}}\) =
Answer:
8^-9

Exercise 7.
3∙2^-4=
Answer:
3∙\(\frac{1}{2^{4}}\) =\(\frac{3}{2^{4}}\)

Exercise 8.
Let x be a nonzero number.
x^-3=
Answer:
\(\frac{1}{x^{3}}\)

Exercise 9.
Let x be a nonzero number.
\(\frac{1}{x^{9}}\) =x^-9

Exercise 10.
Let x,y be two nonzero numbers.
xy^-4 =
Answer:
x∙\(\frac{1}{y^{4}}\) = \(\frac{x}{y^{4}}\)

Exercise 11.
\(\frac{19^{2}}{19^{5}}\) =
Answer:
19^2-5

Exercise 12.
\(\frac{17^{16}}{17^{-3}}\) =
Answer:
17¹⁶×\(\frac{1}{17^{-3}}\) =17¹⁶×17³= 17¹⁶⁺³

Exercise 13.
If we let b=-1 in (11), a be any integer, and y be any nonzero number, what do we get?
Answer:
(y^-1)^a=y^-a

Exercise 14.
Show directly that (\(\frac{7}{5}\))^-4=\(\frac{7^{-4}}{5^{-4}}\).
Answer:
(\(\frac{7}{5}\))^-4=(7∙\(\frac{1}{5}\))^-4 By the product formula
=(7∙5^-1 )^-4 By definition
=7^-4∙(5^-1 )^-4 By (xy)^a=x^a y^a (12)
=7^-4∙5⁴ By (x^b )^a=x^ab (11)
=7^-4∙\(\frac{1}{5^-4}\) By x^-b=\(\frac{1}{x^{b}}\)(9)
=\(\frac{7^{-4}}{5^{-4}}\) By product formula

Eureka Math Grade 8 Module 1 Lesson 5 Problem Set Answer Key

Question 1.
Compute: 3³ ×3² ×3¹ ×3⁰×3^-1 ×3^-2=
Answer:
3³ =27
Compute: 5² ×5¹ 0×5⁸ ×5⁰×5^-10 ×5^-8 =5² =25
Compute for a nonzero number, a: a^m ×aⁿ ×a^l ×a^-n ×a^-m ×a^-l ×a⁰=
Answer:
a⁰=1

Question 2.
Without using (10), show directly that (17.6^-1 )⁸ = 17.6^-8 .
Answer:
(17.6^-1)⁸ =(\(\frac{1}{17.6}\))⁸ By definition
= \(\frac{1^{8}}{17.6^{8}}\) By (\(\frac{x}{y}\))ⁿ = \(\frac{x^{n}}{y^{n}}\) (5)
= \(\frac{1}{17.6^{8}}\)
= 17.6^-8 By definition

Question 3.
Without using (10), show (prove) that for any whole number n and any nonzero number y, (y^-1 )ⁿ =y^-n .
Answer:
(y^-1 )ⁿ =(\(\frac{1}{y}\))ⁿ By definition
=\(\frac{1^{n}}{y^{n}}\) By (\(\frac{x}{y}\))ⁿ = \(\frac{x^{n}}{y^{n}}\)(5)
= \(\frac{1}{y^{n}}\)
= y^-n By definition

Question 4.
Without using (13), show directly that \(\frac{2.8^{-5}}{2.8^{7}}\) = 2.8^-12 .
Answer:
\(\frac{2.8^{-5}}{2.8^{7}}\) = 2.8^-5 × \(\frac{1}{2.8^{7}}\) By the product formula for complex fractions
= \(\frac{1}{2.8^{5}}\) × \(\frac{1}{2.8^{7}}\) By definition
= \(\frac{1}{2.8^{5} \times 2.8^{7}}\) By the product formula for complex fractions
= \(\frac{1}{2.8^{5+7}}\) By x^a∙x^b =x^a+b (10)
= \(\frac{1}{2.8^{12}}\)
= 2.8^-12 By definition

Eureka Math Grade 8 Module 1 Lesson 5 Exit Ticket Answer Key

Write each expression in a simpler form that is equivalent to the given expression.

Question 1.
76543^-4=
Answer:
\(\frac{1}{76543^{4}}\)

Question 2.
Let f be a nonzero number. f^-4=
Answer:
\(\frac{1}{f^{4}}\)

Question 3.
671×28796^-1 =
Answer:
671×\(\frac{1}{28796}\)=\(\frac{671}{28796}\)

Question 4.
Let a, b be numbers (b≠0). ab^-1 =
Answer:
a∙\(\frac{1}{b}\)=\(\frac{a}{b}\)

Question 5.
Let g be a nonzero number. \(\frac{1}{g^{-1}}\) =
Answer:
g

Engage NY Eureka Math 8th Grade Module 1 Lesson 4 Answer Key

Eureka Math Grade 8 Module 1 Lesson 4 Exercise Answer Key

Exercise 1.
List all possible cases of whole numbers m and n for identity (1). More precisely, when m>0 and n>0, we already know that (1) is correct. What are the other possible cases of m and n for which (1) is yet to be verified?
Answer:
Case (A): m>0 and n=0
Case (B): m=0 and n>0
Case (C): m=n=0

Model how to check the validity of a statement using Case (A) with equation (1) as part of Exercise 2. Have students work independently or in pairs to check the validity of (1) in Case (B) and Case (C) to complete Exercise 2. Next, have students check the validity of equations (2) and (3) using Cases (A)–(C) for Exercises 3 and 4.

Exercise 2.
Check that equation (1) is correct for each of the cases listed in Exercise 1.
Answer:
Case (A): x^m∙x⁰=x^m? Yes, because x^m∙x⁰=x^m∙1=x^m.
Case (B): x⁰∙xⁿ=xⁿ? Yes, because x⁰∙xⁿ=1∙xⁿ=xⁿ.
Case (C): x⁰∙x⁰=x⁰? Yes, because x⁰∙x⁰=1∙1=x⁰.

Exercise 3.
Do the same with equation (2) by checking it case-by-case.
Answer:
Case (A): (x^m)⁰=x^0×m? Yes, because x^m is a number, and a number raised to a zero power is 1. 1=x⁰=x^0×m.
S0, the left side is 1. The right side is also 1 because x^0×m=x⁰=1.
Case (B): (x⁰)ⁿ=x^n×0? Yes, because, by definition x⁰=1 and 1ⁿ=1, the left side is equal to 1. The right side is equal to x⁰=1, so both sides are equal.
Case (C): (x⁰)⁰=x^0×0? Yes, because, by definition of the zeroth power of x, both sides are equal to 1.

Exercise 4.
Do the same with equation (3) by checking it case-by-case.
Answer:
Case (A): (xy)⁰=x⁰ y⁰? Yes, because the left side is 1 by the definiti0n of the zeroth power, while the right side is
1×1=1.
Case (B): Since n>0, we already know that (3) is valid.
Case (C): This is the same as Case (A), which we have already shown to be valid.

Exercise 5.
Write the expanded form of 8,374 using exponential notation.
Answer:
8374=(8×10³)+(3×10²)+(7×10¹)+(4×10⁰)

Exercise 6.
Write the expanded form of 6,985,062 using exponential notation.
Answer:
6 985 062
= (6×10⁶)+(9×10⁵)+(8×10⁴)+(5×10³)+(0×10²)+(6×10¹) + (2×10⁰)

Eureka Math Grade 8 Module 1 Lesson 4 Problem Set Answer Key

Let x,y be numbers (x,y≠0). Simplify each of the following expressions.

Question 1.
\(\frac{y^{12}}{y^{12}}\) = y^12-12
Answer:
=y⁰
=1

Question 2.
9¹⁵\(\frac{1}{9^{15}}\) = \(\frac{9^{15}}{9^{15}}\)
Answer:
=9^15-15
=9⁰
=1

Question 3.
(7(123456.789)⁴)⁰=
Answer:
=7⁰ (123456.789)^4×0
=7⁰ (123456.789)⁰
=1

Question 4.
2²∙\(\frac{1}{2^{5}}\)∙2⁵∙\(\frac{1}{2^{2}}\) =\(\frac{2^{2}}{2^{2}}\)∙\(\frac{2^{5}}{2^{5}}\)
Answer:
=2^2-2∙2^5-5
=2⁰∙2⁰
=1

Question 5.

Answer:
= \(\frac{x^{41}}{x^{41}}\)∙\(\frac{y^{15}}{y^{15}}\)
= x^41-41∙y^15-15
=x⁰∙y⁰
= 1

Eureka Math Grade 8 Module 1 Lesson 4 Exit Ticket Answer Key

Question 1.
Simplify the following expression as much as posiible.
Answer:
\(\frac{4^{10}}{4^{10}}\).7⁰ = 4^10-10.1 = 4⁰.1 = 1.1 = 1

Question 2.
Let a and b be two numbers. Use the distributive law and then the definition of zeroth power to show that the numbers (a⁰+b⁰) a⁰ and (a⁰+b⁰) b⁰ are equal.
Answer:
(a⁰+b⁰) a⁰=a⁰∙a⁰+b⁰∙a⁰
=a⁰⁺⁰+a⁰ b⁰
=a⁰+a⁰ b⁰
=1+1∙1
=1+1
=2
(a⁰+b⁰) b⁰=a⁰∙b⁰+b⁰∙b⁰
=a⁰ b⁰+b⁰⁺⁰
=a⁰ b⁰+b⁰
=1∙1+1
=1+1
=2
Since both numbers are equal to 2, they are equal.

Eureka Math Grade 8 Module 1 Lesson 4 Sprint Answer Key

Applying Properties of Exponents to Generate Equivalent Expressions—Round 1
Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. All letters denote numbers.

Answer:

Applying Properties of Exponents to Generate Equivalent Expressions—Round 2

Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. All letters denote numbers.

Answer:

Engage NY Eureka Math 8th Grade Module 1 Lesson 3 Answer Key

Eureka Math Grade 8 Module 1 Lesson 3 Example Answer Key

Examples 1–2
Work through Examples 1 and 2 in the same manner. (Supplement with additional examples if needed.) Have students calculate the resulting exponent; however, emphasis should be placed on the step leading to the resulting exponent, which is the product of the exponents.
Example 1.
(7² )⁶=
Answer:

Example 2.
(1.3)³ )¹⁰=
Answer:
(1.3×1.3×1.3)¹⁰

Eureka Math Grade 8 Module 1 Lesson 3 Exercise Answer Key

Exercise 1.
(15³)⁹=
Answer:
(15)^9×3

Exercise 2.
((-2)⁵ )⁸=
Answer:
(-2)^8×5

Exercise 3.
(3.4¹⁷)⁴=
Answer:
3.4^4×17

Exercise 4.
Let s be a number.
Answer:
(s¹⁷ )⁴=
Answer:
s^4×17

Exercise 5.
Sarah wrote (3⁵ )⁷=3¹². Correct her mistake. Write an exponential expression using a base of 3 and exponents of 5, 7, and 12 that would make her answer correct.
Answer:
Correct way: (3⁵ )⁷=3³⁵; Rewritten Problem: 3⁵×3⁷=3⁵⁺⁷=3¹².

Exercise 6.
A number y satisfies y²⁴-256=0. What equation does the number x=y⁴ satisfy?
Answer:
Since x=y⁴, then (x)⁶=(y⁴ )⁶. Therefore, x=y⁴ would satisfy the equation x⁶-256=0.

Exercises 7–13 (10 minutes)
Have students complete Exercises 17–12 independently and then check their answers.

Exercise 7.
(11×4)⁹=
Answer:
11^9×1×4^9×1

Exercise 8.
(3²×7⁴ )⁵=
Ans:
3^5×2×7^5×4

Exercise 9.
Let a, b, and c be numbers.
(3² a⁴ )⁵=
Answer:
3^5×2 a^5×4

Exercise 10.
Let x be a number.
(5x)⁷=
Ans:
5^7×1 ∙x^7×1

Exercise 11.
Let x and y be numbers.
(5xy² )⁷=
Ans:
5^7×1 ∙x^7×1∙y^7×2

Exercise 12.
Let a, b, and c be numbers.
(a² bc³ )⁴=
Ans:
a^4×2 ∙b^4×1∙c^4×3

Exercise 13.
Let x and y be numbers, y≠0, and let n be a positive integer. How is (\(\frac{x}{y}\))ⁿ related to xⁿ and yⁿ?
Answer:
(\(\frac{x}{y}\))ⁿ=\(\frac{x^{n}}{y^{n}}\)
Because

Eureka Math Grade 8 Module 1 Lesson 3 Problem Set Answer Key

Question 1.
Show (prove) in detail why (2∙3∙7)⁴=2⁴ 3⁴ 7⁴.
Answer:
(2∙3∙7)⁴=(2∙3∙7)(2∙3∙7)(2∙3∙7)(2∙3∙7)
=(2∙2∙2∙2)(3∙3∙3∙3)(7∙7∙7∙7)
By repeated use of the commutative and associative properties
=2⁴ 3⁴ 7⁴ By definition

Question 2.
Show (prove) in detail why (xyz)⁴=x⁴ y⁴ z⁴ for any numbers x,y,z.
Answer:
The left side of the equation (xyz)⁴ means (xyz) (xyz) (xyz) (xyz). Using the commutative and associative properties of multiplication, we can write (xyz) (xyz) (xyz) (xyz) as (xxxx)(yyyy) (zzzz), which in turn can be written as x⁴ y⁴ z⁴, which is what the right side of the equation states.

Question 3.
Show (prove) in detail why (xyz)ⁿ=xⁿ yⁿ zⁿ for any numbers x, y, and z and for any positive integer n.
Ans:
Beginning with the left side of the equation, (xyz)ⁿ means . Using the commutative and associative properties of multiplication, can be rewritten as and, finally, xⁿ yⁿ zⁿ, which is what the right side of the equation states. We can also prove this equality by a different method, as follows. Beginning with the right side xⁿ yⁿ zⁿ means which by the commutative property of multiplication can be rewritten as . Using exponential notation, can be rewritten as (xyz)ⁿ, which is what the left side of the equation states.

Eureka Math Grade 8 Module 1 Lesson 3 Exit Ticket Answer Key

Write each expression as a base raised to a power or as the product of bases raised to powers that is equivalent to the given expression.

Question 1.
(9³ )⁶=
Ans:
(9³ )⁶=9^6×3=9¹⁸

Question 2.
(113²×37×51⁴ )³=
Ans:
(113²×37×51⁴ )³=((113²×37)×51⁴ )³ By associative law
= (113²×37)³×(51⁴ )³ Because (xy)ⁿ=xⁿ yⁿ for all numbers x, y
= (113²)³×37³×(51⁴)³ Because (xy)ⁿ=xⁿ yⁿ for all numbers x, y
= 113⁶×37³×51¹² Because (x^m)ⁿ=x^mn for all numbers x

Question 3.
Let x,y,z be numbers. (x² yz⁴ )³=
Answer:
(x²yz⁴ )³=((x²×y)×z⁴ )³ By associative law
= (x²×y)³×(z⁴ )³ Because (xy)ⁿ=xⁿ yⁿ for all numbers x, y
=(x² )³×y³×(z⁴ )³ Because (xy)ⁿ=xⁿ yⁿ for all numbers x, y
=x⁶×y³×z¹² Because (x^m)ⁿ=x^mn for all numbers x
= x⁶ y³ z¹²

Question 4.
Let x,y,z be numbers and let m,n,p,q be positive integers. (x^m yⁿ z^p)^q=
Ans:
(x^m yⁿ z^p )^q=((x^m×yⁿ )×z^p)^q By associative law
=(x^m×yⁿ )^q×(z^p )^q Because (xy)ⁿ=xⁿ yⁿ for all numbers x, y
=(x^m )^q×(yⁿ )^q×(z^p)^q Because (xy)ⁿ=xⁿ yⁿ for all numbers x, y
= x^mp×y^nq×z^pq Because (x^m )ⁿ=x^mn for all numbers x
=x^mqy^nqz^pq

Question 5.
\(\frac{4^{8}}{5^{8}}\) =
Answer:
\(\frac{4^{8}}{5^{8}}\) = \(\left(\frac{4}{5}\right)^{8}\)

Engage NY Eureka Math 8th Grade Module 1 Lesson 2 Answer Key

Eureka Math Grade 8 Module 1 Lesson 2 Example Answer Key

Examples 1–2
Work through Examples 1 and 2 in the manner just shown. (Supplement with additional examples if needed.)
It is preferable to write the answers as an addition of exponents to emphasize the use of the identity. That step should not be left out. That is, 5²×5⁴=5⁶ does not have the same instructional value as 5²×5⁴=5²⁺⁴.

Example 1.
5²×5⁴=
Answer:
5²⁺⁴

Example 2.
(-\(\frac{2}{3}\))⁴×(-\(\frac{2}{3}\))⁵=(-\(\frac{2}{3}\))⁴⁺⁵
→ What is the analog of x^m∙xⁿ=x^m+n in the context of repeated addition of a number x?
Allow time for a brief discussion.
→ If we add m copies of x and then add to it another n copies of x, we end up adding m+n copies of x. By the distributive law:
mx+nx=(m+n)x .
This is further confirmation of what we observed at the beginning of Lesson 1: The exponent m+n in x^m+n in the context of repeated multiplication corresponds exactly to the m+n in (m+n)x in the context of repeated addition.

Examples 3–4.
Work through Examples 3 and 4 in the manner shown. (Supplement with additional examples if needed.)
It is preferable to write the answers as a subtraction of exponents to emphasize the use of the identity.

Example 3.

Answer:
(\(\frac{3}{5}\))^8-6

Example 4.
\(\frac{4^{5}}{4^{2}}\) =
Answer:
4^5-2

Eureka Math Grade 8 Module 1 Lesson 2 Exercise Answer Key

Exercise 1.
14²³×14⁸=
Answer:
14²³⁺⁸

Exercise 2.
(-72)¹⁰×(-72)¹³=
Answer:
(-72)¹⁰⁺¹³

Exercise 3.
5⁹⁴×5⁷⁸=
Answer:
5⁹⁴⁺⁷⁸

Exercise 4.
(-3)⁹×(-3)⁵=
Answer:
(-3)⁹⁺⁵

Exercise 5.
Let a be a number.
a²³∙a⁸=
Answer:
a²³⁺⁸

Exercise 6.
Let f be a number.
f¹⁰∙f¹³=
Answer:
f¹⁰⁺¹³

Exercise 7.
Let b be a number.
b⁹⁴∙b⁷⁸=
Answer:
b⁹⁴⁺⁷⁸

Exercise 8.
Let x be a positive integer. If (-3)⁹×(-3)^x=(-3)¹⁴, what is x?
Answer:
x=5

In Exercises 9–16, students need to think about how to rewrite some factors so the bases are the same. Specifically, 2⁴×8²=2⁴×2⁶=2⁴⁺⁶ and 3⁷×9=3⁷×3²=3⁷⁺². Make clear that these expressions can only be combined into a single base because the bases are the same. Also included is a non-example, 5⁴×2¹¹, that cannot be combined into a single base using this identity. Exercises 17–20 offer further applications of the identity.

What would happen if there were more terms with the same base? Write an equivalent expression for each problem.

Exercise 9.
9⁴×9⁶×9¹³=
Answer:
9⁴⁺⁶⁺¹³

Exercise 10.
2³×2⁵×2⁷×2⁹=
Answer:
2^3+5+7+9

Can the following expressions be written in simpler form? If so, write an equivalent expression. If not, explain why not.

Exercise 11.
6⁵×4⁹×4³×6¹⁴=
Answer:
4⁹⁺³×6⁵⁺¹⁴

Exercise 12.
(-4)²∙17⁵∙(-4)³∙17⁷=
Answer:
(-4)²⁺³∙17⁵⁺⁷

Exercise 13.
15²∙7²∙15∙7⁴=
Answer:
15²⁺¹∙7²⁺⁴

Exercise 14.
2⁴×8²=2⁴×2⁶=
Answer:
2⁴⁺⁶

Exercise 15.
3⁷×9=3⁷×3²=
Answer:
3⁷⁺²

Exercise 16.
5⁴ ×2¹¹=
Answer:
Cannot be simplified. Bases are different and cannot be rewritten in the same base.

Exercise 17.
Let x be a number. Rewrite the expression in a simpler form.
(2x³ )(17x⁷ )=
Answer:
34x¹⁰

Exercise 18.
Let a and b be numbers. Use the distributive law to rewrite the expression in a simpler form.
a(a+b)=
Answer:
a²+ab

Exercise 19.
Let a and b be numbers. Use the distributive law to rewrite the expression in a simpler form.
b(a+b)=
Answer:
ab+b²

Exercise 20.
Let a and b be numbers. Use the distributive law to rewrite the expression in a simpler form.
(a+b)(a+b)=
Answer:
a²+ab+ba+b²=a²+2ab+b²

Exercise 21.
\(\frac{7^{9}}{7^{6}}\) =
Answer:
7^9-6

Exercise 22.
\(\frac{(-5)^{16}}{(-5)^{7}}\) =
Answer:
(-5)^16-7

Exercise 23.

Answer:
(\(\frac{8}{5}\))^9-2

Exercise 24.
\(\frac{13^{5}}{13^{4}}\)=
Answer:
13^5-4

Exercise 25.
Let a, b be nonzero numbers. What is the following number?

Answer:
(\(\frac{a}{b}\))^9-2

Exercise 26.
Let x be a nonzero number. What is the following number?
\(\frac{x^{5}}{x^{4}}\) =
Answer:
x^5-4

Can the following expressions be written in simpler forms? If yes, write an equivalent expression for each problem. If not, explain why not.

Exercise 27.
\(\frac{2^{7}}{4^{2}}\) = \(\frac{2^{7}}{2^{4}}\) =
Answer:
2^7-4

Exercise 28.
\(\frac{3^{23}}{27}\)= \(\frac{3^{23}}{3^{3}}\) =
Answer:
3^23-3

Exercise 29.
\(\frac{3^{5} \cdot 2^{8}}{3^{2} \cdot 2^{3}}\)=
Answer:
3^5-2∙2^8-3

Exercise 30.

Answer:
(-2)^7-5∙95^5-4

Exercise 31.
Let x be a number. Write each expression in a simpler form.
a. \(\frac{5}{x^{3}}\)(3x⁸ )=
Answer:
15x⁵

b. \(\frac{5}{x^{3}}\)(-4x⁶ )=
Answer:
-20x³

c. \(\frac{5}{x^{3}}\)(11x⁴ )=
Answer:
55x

Exercise 32.
Anne used an online calculator to multiply 2 000 000 000×2 000 000 000 000. The answer showed up on the calculator as 4e+21, as shown below. Is the answer on the calculator correct? How do you know?

Answer:
2 000 000 000×2 000 000 000 000=4 000 000 000 000 000 000 000.
The answer must mean 4 followed by 21 zeros. That means that the answer on the calculator is correct.
This problem is hinting at scientific notation (i.e., (2× 10⁹)(2×10¹²)=4×10⁹⁺¹²). Accept any reasonable explanation of the answer.

Eureka Math Grade 8 Module 1 Lesson 2 Problem Set Answer Key

To ensure success with Problems 1 and 2, students should complete at least bounces 1–4 with support in class. Consider working on Problem 1 as a class activity and assigning Problem 2 for homework.
Students may benefit from a simple drawing of the scenario. It will help them see why the factor of 2 is necessary when calculating the distance traveled for each bounce. Make sure to leave the total distance traveled in the format shown so that students can see the pattern that is developing. Simplifying at any step will make it difficult to write the general statement for n number of bounces.

Question 1.
A certain ball is dropped from a height of x feet. It always bounces up to \(\frac{2}{3}\) x feet. Suppose the ball is dropped from 10 feet and is stopped exactly when it touches the ground after the 30^th bounce. What is the total distance traveled by the ball? Express your answer in exponential notation.

Answer:

Question 2.
If the same ball is dropped from 10 feet and is stopped exactly at the highest point after the 25^th bounce, what is the total distance traveled by the ball? Use what you learned from the last problem.
Answer:
Based on the last problem, we know that each bounce causes the ball to travel 2(\(\frac{2}{3}\))ⁿ 10 feet. If the ball is stopped at the highest point of the 25^th bounce, then the distance traveled on that last bounce is just (\(\frac{2}{3}\))²⁵ 10 feet because it does not make the return trip to the ground. Therefore, the total distance traveled by the ball in feet in this situation is
10+2(\(\frac{2}{3}\))10+2(\(\frac{2}{3}\))² 10+2(\(\frac{2}{3}\))³ 10+2(\(\frac{2}{3}\))⁴ 10+…..+2(\(\frac{2}{3}\))²³10+2(\(\frac{2}{3}\))²⁴10.

Question 3.
Let a and b be numbers and b≠0, and let m and n be positive integers. Write each expression using the fewest number of bases possible.

Answer:

Question 4.
Let the dimensions of a rectangle be (4×(871209)⁵+ 3×49 762 105) ft. by (7×(871 209)³-(49 762 105)⁴) ft. Determine the area of the rectangle. (Hint: You do not need to expand all the powers.)
Answer:
Area=(4×(871 209)⁵+3×49 762 105) ft.(7×(871 209)³-(49 762 105)⁴ ) ft.
=(28×(871 209)⁸-4×(871 209)⁵ (49 762 105)⁴+21×(871 209)³ (49 762 105)-3×(49 762 105)⁵ ) sq.ft.

Question 5.
A rectangular area of land is being sold off in smaller pieces. The total area of the land is 2¹⁵ square miles. The pieces being sold are 8³ square miles in size. How many smaller pieces of land can be sold at the stated size? Compute the actual number of pieces.
Answer:
8³=2⁹
\(\frac{2^{15}}{2^{9}}\) =2^15-9=2⁶=64
64 pieces of land can be sold.

Eureka Math Grade 8 Module 1 Lesson 2 Exit Ticket Answer Key

Note to Teacher: Accept both forms of the answer; in other words, accept an answer that shows the exponents as a sum or difference as well as an answer where the numbers are actually added or subtracted.

Write each expression using the fewest number of bases possible.

Question 1.
Let a and b be positive integers. 23^a × 23^b=
Answer:
23^a × 23^b = 23^a+b

Question 2.
5³×25=
Answer:
5³×25=5³×5²
= 5³⁺²
=5⁵

Question 3.
Let x and y be positive integers and x>y. \(\frac{11^{x}}{11^{y}}\) =
Answer:
\(\frac{11^{x}}{11^{y}}\) = 11^x-y

Question 4.
\(\frac{2^{13}}{8}\) =
Answer:
\(\frac{2^{13}}{2^{3}}\) = 2^13-3 = 2¹⁰

Engage NY Eureka Math 8th Grade Module 1 Lesson 1 Answer Key

Eureka Math Grade 8 Module 1 Lesson 1 Example Answer Key

Example 1-5
Work through Examples 1–5 as a group, and supplement with additional examples if needed.

Example 1.
5×5×5×5×5×5=
Answer:
5⁶

Example 2.
\(\frac{9}{7}\)×\(\frac{9}{7}\)×\(\frac{9}{7}\)×\(\frac{9}{7}\)=
Answer:
(\(\frac{9}{7}\))⁴

Example 3.
(-\(\frac{4}{11}\))³=
Answer:
(-\(\frac{4}{11}\))×(-\(\frac{4}{11}\))×(-\(\frac{4}{11}\))

Example 4.
(-2)⁶=
Answer:
(-2)×(-2)×(-2)×(-2)×(-2)×(-2)

Example 5.
3.8⁴=3.8×3.8×3.8×3.8

Eureka Math Grade 8 Module 1 Lesson 1 Exercise Answer Key

Exercise 1.

Answer:
4⁷

Exercise 2.

Answer:

Exercise 3.

Answer:
(-11.63)³⁴

Exercise 4.

Answer:
15 times

Exercise 5.

=(-5)¹⁰

Exercise 6.

Answer:
(\(\frac{7}{2}\))²¹

Exercise 7.

Answer:
(-13)⁶

Exercise 8.

Answer:
(-\(\frac{1}{14}\))¹⁰

Exercise 9.

Answer:
x¹⁸⁵

Exercise 10.

Answer:
n times

Exercise 11.
Will these products be positive or negative? How do you know?

Answer:
This product will be positive. Students may state that they computed the product and it was positive. If they say that, let them show their work. Students may say that the answer is positive because the exponent is positive; however, this would not be acceptable in view of the next example.

This product will be negative. Students may state that they computed the product and it was negative. If so, ask them to show their work. Based on the discussion of the last problem, you may need to point out that a positive exponent does not always result in a positive product.

The two problems in Exercise 12 force the students to think beyond the computation level. If students struggle, revisit the previous two problems, and have them discuss in small groups what an even number of negative factors yields and what an odd number of negative factors yields.

Exercise 12.
Is it necessary to do all of the calculations to determine the sign of the product? Why or why not?

Answer:
Students should state that an odd number of negative factors yields a negative product.

Students should state that an even number of negative factors yields a positive product.

Exercise 13.
Fill in the blanks indicating whether the number is positive or negative.
If n is a positive even number, then (-55)ⁿ is _____
Answer:
positive.
If n is a positive odd number, then (-72.4)ⁿ is __
Answer:
negative.

Exercise 14.
Josie says that Is she correct? How do you know?
Answer:
Students should state that Josie is not correct for the following two reasons: (1) They just stated that an even number of factors yields a positive product, and this conflicts with the answer Josie provided, and (2) the notation is used incorrectly because, as is, the answer is the negative of 15⁶, instead of the product of 6 copies of -15. The base is (-15). Recalling the discussion at the beginning of the lesson, when the base is negative it should be written clearly by using parentheses. Have students write the answer correctly.

Eureka Math Grade 8 Module 1 Lesson 1 Problem Set Answer Key

Question 1.
Use what you know about exponential notation to complete the expressions below.

Answer:
(-5)¹⁷

Answer:
19 times

Answer:
45 times

Answer:
6⁴

=4.3¹³

Answer:
(-1.1)⁹

(\(\frac{2}{3}\))¹⁹

(-\(\frac{11}{5}\))^x
Answer:
x times

Answer:
15 times

Answer:
a^m

Question 2.
Write an expression with (-1) as its base that will produce a positive product, and explain why your answer is valid.
Answer:
Accept any answer with (-1) to an exponent that is even.

Question 3.
Write an expression with (-1) as its base that will produce a negative product, and explain why your answer is valid.
Answer:
Accept any answer with (-1) to an exponent that is odd.

Question 4.
Rewrite each number in exponential notation using 2 as the base.
Answer:
8=
Answer:
2³

16=
Answer:
2⁴

32 =
Answer:
2⁵

64=
Answer:
2⁶

128=
Answer:
2⁷

256=
Answer:
2⁸

Question 5.
Tim wrote 16 as (-2)⁴. Is he correct? Explain.
Answer:
Tim is correct that 16=(-2)⁴. (-2)(-2)(-2)(-2)=(4)(4)=16.

Question 6.
Could -2 be used as a base to rewrite 32? 64? Why or why not?
Answer:
A base of -2 cannot be used to rewrite 32 because (-2)⁵=-32. A base of -2 can be used to rewrite 64 because (-2)⁶=64. If the exponent, n, is even, (-2)ⁿ will be positive. If the exponent, n, is odd, (-2)ⁿ cannot be a positive number.

Eureka Math Grade 8 Module 1 Lesson 1 Exit Ticket Answer Key

Question 1.
a. Express the following in exponential notation:

Answer:
(-13)³⁵

b. Will the product be positive or negative? Explain.
Answer:
The product will be negative. The expanded form shows 34 negative factors plus one more negative factor. Any even number of negative factors yields a positive product. The remaining 35th negative factor negates the resulting product.

Question 2.
Fill in the blank:

Answer:
4 times

Question 3.
Arnie wrote:

Is Arnie correct in his notation? Why or why not?
Answer:
Arnie is not correct. The base, -3.1, should be in parentheses to prevent ambiguity. At present the notation is not correct.

Engage NY Eureka Math 7th Grade Module 3 End of Module Assessment Answer Key

Eureka Math Grade 7 Module 3 End of Module Assessment Task Answer Key

Question 1.
Gloria says the two expressions \(\frac{1}{4}\) (12x+24)-9x and -6(x+1) are equivalent. Is she correct? Explain how you know.
Answer:
No, Gloria is not correct.
The standard form of \(\frac{1}{4}\) (12x+24)-9x is -6x + 6 and the standard form of -6(x + 1) is -6x – 6. -6x + 6 is not equivalent to -6x – 6
\(\frac{1}{4}\) (12x+24)-9x
\(\frac{1}{4}\)(12x) + \(\frac{1}{4}\)(24) – 9x
3x + 6 – 9x
3x – 9x + 6
-6x + 6
-6(x + 1)
(-6)(x) + (-6)(1)
-6x – 6

Question 2.
A grocery store has advertised a sale on ice cream. Each carton of any flavor of ice cream costs $3.79.
a. If Millie buys one carton of strawberry ice cream and one carton of chocolate ice cream, write an algebraic expression that represents the total cost of buying the ice cream.
Answer:
3.79(s + c)

b. Write an equivalent expression for your answer in part (a).
Answer:
3.79s + 3.79c

c. Explain how the expressions are equivalent.
Answer:
Part b is the same expression as part a with the distributive property applied and in standard form.

Question 3.
A new park was designed to contain two circular gardens. Garden A has a diameter of 50 m, and garden B has a diameter of 70 m.
a. If the gardener wants to outline the gardens in edging, how many meters will be needed to outline the smaller garden? (Write in terms of π.)
Answer:
C = 2πr r = \(\frac{1}{2} \cdot 50\) = 25
C = 2π(25)
C = 50πm
The smaller garden will need 50πm of ending

b. How much more edging will be needed for the larger garden than the smaller one? (Write in terms of π.)
Answer:
C = 2πr r = \(\frac{1}{2} \cdot 70\) = 35
C = 2π(35)
C = 70πm
Larger garden – smaller garden
70πm – 50πm
The larger garden will need 50πm of ending

c. The gardener wishes to put down weed block fabric on the two gardens before the plants are planted in the ground. How much fabric will be needed to cover the area of both gardens? (Write in terms of π.)
Answer:
A_larger + A_smaller
πr² + πr²
π(35)² + π(25)²
1225π + 625π
1850πm² of fabric will be needed to cover the area of both gardens.

Question 4.
A play court on the school playground is shaped like a square joined by a semicircle. The perimeter around the entire play court is 182.8 ft., and 62.8 ft. of the total perimeter comes from the semicircle.

a. What is the radius of the semicircle? Use 3.14 for π.
Answer:
\(\frac{1}{2}\)C = 62.8
\(\frac{1}{2}\)(2πr) = 62.8
πr = 62.8
62.8÷ 3.14 = 20

r = 20
The radius of the semi-circle is 20ft.

b. The school wants to cover the play court with sports court flooring. Using 3.14 for π, how many square feet of flooring does the school need to purchase to cover the play court?
Answer:
Are square + Area semicircle
s.s + \(\frac{1}{2}\)(πr²)
40.40 + \(\frac{1}{2}\)(3.14 (20)²)
The school needs to purchase enough flooring to cover 2,228 ft²
1600 + \(\frac{1}{2}\)(3.14(400))
1600 + 628
2228

Question 5.
Marcus drew two adjacent angles.

a. If ∠ABC has a measure one-third of ∠CBD, then what is the degree measurement of ∠CBD?
Answer:

Let m be the measure of ∠CBD in degrees
∠ABC + ∠CBD = 180
\(\frac{1}{3}\)m + m = 180
1\(\frac{1}{3}\)m = 180
(\(\frac{3}{4}\)m)\(\frac{4}{3}\)m = 180(\(\frac{3}{4}\)m)
m = 135

b. If the measure of ∠CBD is 9(8x+11) degrees, then what is the value of x?
Answer:
135 = 9(8x+11)
135 = 72x + 99
135 – 99 = 72x + 99 – 99
36 = 72x
36.\(\frac{1}{72}\) = 72x.\(\frac{1}{72}\)m
x = \(\frac{1}{72}\)m

135 = 9(8x+11)
135 × \(\frac{1}{9}\) = 9(8x + 11) × \(\frac{1}{9}\)
15 – 11 = 8x + 11 – 11
4 = 8x
4.\(\frac{1}{8}\) = 8x.\(\frac{1}{8}\)
\(\frac{1}{2}\) = X

Question 6.
The dimensions of an above-ground, rectangular pool are 25 feet long, 18 feet wide, and 6 feet deep.
a. How much water is needed to fill the pool?
Answer:
V = l.w.h

V = 25ft. ∙8ft∙6ft
V = 2,700ft³

b. If there are 7.48 gallons in 1 cubic foot, how many gallons are needed to fill the pool?
Answer:

To fill pool, 20,196 gallons are needed

c. Assume there was a hole in the pool, and 3,366 gallons of water leaked from the pool. How many feet did the water level drop?
Answer:

The water level dropped one foot.

d. After the leak was repaired, it was necessary to lay a thin layer of concrete to protect the sides of the pool. Calculate the area to be covered to complete the job.
Answer:

base : 25.18
lateral faces: 2(6∙18) and 2(6∙25)
(25∙18) + 2(6∙18) + 2(6∙25)
450 + 216 + 300
966
The surface area that needs to be covered is 966ft²

Question 7.
Gary is learning about mosaics in art class. His teacher passes out small square tiles and encourages the students to cut up the tiles in various angles. Gary’s first cut tile looks like this:

Answer:
a. Write an equation relating ∠TIL with ∠LIE.
Answer:
3m + (m – 10) = 90

b. Solve for m.
Answer:
3m + (m – 10) = 90
3m + m – 10 = 90
4m – 10 = 90
4m – 10 + 10 = 90 + 10
4m = 100
4m∙\(\frac{1}{4}\) = 100∙\(\frac{1}{4}\)
m = 25

c. What is the measure of ∠TIL?
Answer:
3m
3(25) = 75
The measure of ∠TIL is 75°

d. What is the measure of ∠LIE?
Answer:
m – 10
25 – 10
The measure of ∠LIE is 15°

Engage NY Eureka Math 7th Grade Module 3 Lesson 9 Answer Key

Eureka Math Grade 7 Module 3 Lesson 9 Example Answer Key

Example 1.
Fred and Sam are a team in the local 138.2 mile bike-run-athon. Fred will compete in the bike race, and Sam will compete in the run. Fred bikes at an average speed of 8 miles per hour and Sam runs at an average speed of 4 miles per hour. The bike race begins at 6:00 a.m., followed by the run. Sam predicts he will finish the run at 2:33 a.m. the next morning.
a. How many hours will it take them to complete the entire bike-run-athon?
Answer:
From 6:00 a.m. to 2:00 a.m. the following day is 20 hours.
33 minutes in hours is \(\frac{33}{60}\)=\(\frac{11}{20}\)=0.55, or 0.55 hours.
Therefore, the total time it will take to complete the entire bike-run-athon is 20.55 hours.

b. If t is how long it takes Fred to complete the bike race, in hours, write an expression to find Fred’s total distance.
Answer:
d=rt
d=8t
The expression of Fred’s total distance is 8t.

c. Write an expression, in terms of t to express Sam’s time.
Answer:
Since t is Fred’s time and 20.55 is the total time, then Sam’s time would be the difference between the total time and Fred’s time. The expression would be 20.55-t.

d. Write an expression, in terms of t, that represents Sam’s total distance.
Answer:
d=rt
d=4(20.55-t)
The expressions 4(20.55-t) or 82.2-4t is Sam’s total distance.

e. Write and solve an equation using the total distance both Fred and Sam will travel.
Answer:
8t+4(20.55-t)=138.2
8t+82.2-4t=138.2
8t-4t+82.2=138.2
4t+82.2=138.2
4t+82.2-82.2=138.2-82.2
4t+0=56
(\(\frac{1}{4}\))(4t)=(\(\frac{1}{4}\))(56)
t=14
Fred’s time: 14 hours
Sam’s time: 20.55-t=20.55-14=6.55
6.55 hours

f. How far will Fred bike, and how much time will it take him to complete his leg of the race?
Answer:
8(14)=112
Fred will bike 112 miles and will complete the bike race in 14 hours.

g. How far will Sam run, and how much time will it take him to complete his leg of the race?
Answer:
4(20.55-t)
4(20.55-14)
4(6.55)
26.2
Sam will run 26.2 miles, and it will take him 6.55 hours.

Answer:

→ How do you find the distance traveled?
→ Multiply the rate of speed by the amount of time.
→ Model how to organize the problem in a distance, rate, and time chart.

Answer:

→ Explain how to write the equation to have only integers and no decimals. Write the equation.
→ Since the decimal terminates in the tenths place, if we multiply every term by 10, the equation would result with only integer coefficients. The equation would be 40t+822=1382.

Example 2.
Shelby is seven times as old as Bonnie. If in 5 years, the sum of Bonnie’s and Shelby’s ages is 98, find Bonnie’s present age. Use an algebraic approach.

Answer:

x+5+7x+5=98
8x+10=98
8x+10-10=98-10
8x=88
(\(\frac{1}{8}\))(8x)=(\(\frac{1}{8}\))(88)
x=11
Bonnie’s present age is 11 years old.

→ The first step we must take is to write expressions that represent the present ages of both Bonnie and Shelby. The second step is to write expressions for future time or past time, using the present age expressions. How would the expression change if the time were in the past and not in the future?
→ If the time were in the past, then the expression would be the difference between the present age and the amount of time in the past.

Eureka Math Grade 7 Module 3 Lesson 9 Opening Exercise Answer Key

Heather practices soccer and piano. Each day she practices piano for 2 hours. After 5 days, she practiced both piano and soccer for a total of 20 hours. Assuming that she practiced soccer the same amount of time each day, how many hours per day, h, did Heather practice soccer?
Answer:
h: hours per day that soccer was practiced
5(h+2)=20
5h+10=20
5h+10-10=20-10
5h=10
(\(\frac{1}{5}\))(5h)=(\(\frac{1}{5}\))(10)
h=2
Heather practiced soccer for 2 hours each day.

Over 5 days, Jake practices piano for a total of 2 hours. Jake practices soccer for the same amount of time each day. If he practiced piano and soccer for a total of 20 hours, how many hours, h, per day did Jake practice soccer?
Answer:
h: hours per day that soccer was practiced
5h+2=20
5h+2-2=20-2
5h=18
(\(\frac{1}{5}\))(5h)=(18)(\(\frac{1}{5}\))
h=3.6
Jake practiced soccer 3.6 hours each day.

Eureka Math Grade 7 Module 3 Lesson 9 Problem Set Answer Key

Question 1.
A company buys a digital scanner for $12,000. The value of the scanner is 12,000(1-\(\frac{n}{5}\)) after n years. The company has budgeted to replace the scanner when the trade-in value is $2,400. After how many years should the company plan to replace the machine in order to receive this trade-in value?
Answer:
12,000(1-\(\frac{n}{5}\))=2,400
12,000-2,400n=2,400
-2,400n+12,000-12,000=2,400-12,000
-2,400n=-9,600
n=4
They will replace the scanner after 4 years.

Question 2.
Michael is 17 years older than John. In 4 years, the sum of their ages will be 49. Find Michael’s present age.
Answer:
x represents Michael’s age now in years.

x+4+x-17+4=49
x+4+x-13=49
2x-9=49
2x-9+9=49+9
2x=58
(\(\frac{1}{2}\))(2x)=(\(\frac{1}{2}\))(58)
x=29
Michael’s present age is 29 years old.

Question 3.
Brady rode his bike 70 miles in 4 hours. He rode at an average speed of 17 mph for t hours and at an average rate of speed of 22 mph for the rest of the time. How long did Brady ride at the slower speed? Use the variable t to represent the time, in hours, Brady rode at 17 mph.
Answer:

The total distance he rode: 17t+22(4-t)
The total distance equals 70 miles:
17t+22(4-t)=70
17t+88-22t=70
-5t+88=70
-5t+88-88=70-88
-5t=-18
t=3.6
Brady rode at 17 mph for 3.6 hours.

Question 4.
Caitlan went to the store to buy school clothes. She had a store credit from a previous return in the amount of $39.58. If she bought 4 of the same style shirt in different colors and spent a total of $52.22 after the store credit was taken off her total, what was the price of each shirt she bought? Write and solve an equation with integer coefficients.
Answer:
t: the price of one shirt
4t-39.58=52.22
4t-39.58+39.58=52.22+39.58
4t+0=91.80
(\(\frac{1}{4}\))(4t)=(\(\frac{1}{4}\))(91.80)
t=22.95
The price of one shirt was $22.95.

Question 5.
A young boy is growing at a rate of 3.5 cm per month. He is currently 90 cm tall. At that rate, in how many months will the boy grow to a height of 132 cm?
Answer:
Let m represent the number of months.
3.5m+90=132
3.5m+90-90=132-90
3.5m=42
(\(\frac{1}{3.5}\))(3.5m)=(\(\frac{1}{3.5}\))(42)
m=12
The boy will grow to be 132 cm tall 12 months from now.

Question 6.
The sum of a number, \(\frac{1}{6}\) of that number, 2 \(\frac{1}{2}\) of that number, and 7 is 12 \(\frac{1}{2}\). Find the number.
Answer:
Let n represent the given number.
n+\(\frac{1}{6}\) n+(2 \(\frac{1}{2}\))n+7=12 \(\frac{1}{2}\)
n(1+\(\frac{1}{6}\)+\(\frac{5}{2}\))+7=12 \(\frac{1}{2}\)
n(\(\frac{6}{6}\)+\(\frac{1}{6}\)+\(\frac{15}{6}\))+7=12 \(\frac{1}{2}\)
n(\(\frac{22}{6}\))+7=12 \(\frac{1}{2}\)
1\(\frac{1}{3}\) n+7-7=12 \(\frac{1}{2}\)-7
1\(\frac{1}{3}\) n+0=5 \(\frac{1}{2}\)
1\(\frac{1}{3}\) n=5 \(\frac{1}{2}\)
\(\frac{3}{11}\)∙1\(\frac{1}{3}\) n=\(\frac{3}{11}\)∙1\(\frac{1}{2}\)
1n=\(\frac{3}{2}\)
n=1 \(\frac{1}{2}\)
The number is 1 \(\frac{1}{2}\).

Question 7.
The sum of two numbers is 33 and their difference is 2. Find the numbers.
Answer:
Let x represent the first number, then 33-x represents the other number since their sum is 33.
x-(33-x)=2
x+(-(33-x))=2
x+(-33)+x=2
2x+(-33)=2
2x+(-33)+33=2+33
2x+0=35
2x=35
\(\frac{1}{2}\)∙2x=\(\frac{1}{2}\)∙35
1x=\(\frac{35}{2}\)
x=17 \(\frac{1}{2}\)
33-x=33-(17 \(\frac{1}{2}\))=15 \(\frac{1}{2}\)
{17 \(\frac{1}{2}\),15 \(\frac{1}{2}\)}

Question 8.
Aiden refills three token machines in an arcade. He puts twice the number of tokens in machine A as in machine B, and in machine C, he puts \(\frac{3}{4}\) of what he put in machine A. The three machines took a total of 18,324 tokens. How many did each machine take?
Answer:
Let A represent the number of tokens in machine A. Then \(\frac{1}{2}\) A represents the number of tokens in machine B, and \(\frac{3}{4}\) A represents the number of tokens in machine C.
A+\(\frac{1}{2}\) A+\(\frac{3}{4}\) A=18,324
\(\frac{9}{4}\) A=18,324
A=8,144
Machine A took 8,144 tokens, machine B took 4,072 tokens, and machine C took 6,108 tokens.

Question 9.
Paulie ordered 250 pens and 250 pencils to sell for a theatre club fundraiser. The pens cost 11 cents more than the pencils. If Paulie’s total order costs $42.50, find the cost of each pen and pencil.
Answer:
Let l represent the cost of a pencil in dollars. Then, the cost of a pen in dollars is l+0.11.
250(l+l+0.11)=42.5
250(2l+0.11)=42.5
500l+27.5=42.5
500l+27.5+(-27.5)=42.5+(-27.5)
500l+0=15
500l=15
\(\frac{500l}{500}\)=\(\frac{15}{500}\)
l=0.03
A pencil costs $0.03, and a pen costs $0.14.

Question 10.
A family left their house in two cars at the same time. One car traveled an average of 7 miles per hour faster than the other. When the first car arrived at the destination after 5 \(\frac{1}{2}\) hours of driving, both cars had driven a total of 599.5 miles. If the second car continues at the same average speed, how much time, to the nearest minute, will it take before the second car arrives?
Answer:
Let r represent the speed in miles per hour of the faster car, then r-7 represents the speed in miles per hour of the slower car.
5 \(\frac{1}{2}\) (r)+5 \(\frac{1}{2}\) (r-7)=599.5
5 \(\frac{1}{2}\) (r+r-7)=599.5
5 \(\frac{1}{2}\) (2r-7)=599.5
1\(\frac{1}{2}\) (2r-7)=599.5
\(\frac{2}{11}\)∙\(\frac{11}{2}\) (2r-7)=\(\frac{2}{11}\)∙599.5
1∙(2r-7)=\(\frac{1199}{11}\)
2r-7=109
2r-7+7=109+7
2r+0=116
2r=116
\(\frac{1}{2}\)∙2r=\(\frac{1}{2}\)∙116
1r=58
r=58
The average speed of the faster car is 58 miles per hour, so the average speed of the slower car is 51 miles per hour.
distance=rate∙time
d=51∙5 \(\frac{1}{2}\)
d=51∙\(\frac{11}{2}\)
d=280.5
The slower car traveled 280.5 miles in 5 \(\frac{1}{2}\) hours.
d=58∙5 \(\frac{1}{2}\)
d=58∙\(\frac{11}{2}\)
d=319
OR
599.5-280.5 = 319
The faster car traveled 319 miles in 5 \(\frac{1}{2}\) hours.
The slower car traveled 280.5 miles in 5 \(\frac{1}{2}\) hours. The remainder of their trip is 38.5 miles because
319-280.5=38.5.
distance=rate∙time
38.5=51 (t)
\(\frac{1}{51}\) (38.5)=\(\frac{1}{51}\) (51)(t)
\(\frac{38.5}{51}\)=1t
\(\frac{77}{102}\)=t
This time is in hours. To convert to minutes, multiply by 60 minutes per hour.
\(\frac{77}{102}\)∙60=\(\frac{77}{51}\)∙30=\(\frac{2310}{51}\)≈45
The slower car will arrive approximately 45 minutes after the first.

Question 11.
Emily counts the triangles and parallelograms in an art piece and determines that altogether, there are 42 triangles and parallelograms. If there are 150 total sides, how many triangles and parallelograms are there?
Answer:
If t represents the number of triangles that Emily counted, then 42-t represents the number of parallelograms that she counted.
3t+4(42-t)=150
3t+4(42+(-t))=150
3t+4(42)+4(-t)=150
3t+168+(-4t)=150
3t+(-4t)+168=150
-t+168=150
-t+168-168=150-168
-t+0=-18
-t=-18
-1∙(-t)=-1∙(-18)
1t=18
t=18
There are 18 triangles and 24 parallelograms.

Note to the Teacher: Problems 12 and 13 are more difficult and may not be suitable to assign to all students to solve independently.

Question 12.
Stefan is three years younger than his sister Katie. The sum of Stefan’s age 3 years ago and \(\frac{2}{3}\) of Katie’s age at that time is 12. How old is Katie now?
Answer:
If s represents Stefan’s age in years, then s+3 represents Katie’s current age, s-3 represents Stefan’s age 3 years ago, and s also represents Katie’s age 3 years ago.
(s-3)+(\(\frac{2}{3}\))s=12
s+(-3)+\(\frac{2}{3}\) s=12
s+\(\frac{2}{3}\) s+(-3)=12
\(\frac{3}{3}\) s+\(\frac{2}{3}\) s+(-3)=12
\(\frac{5}{3}\) s+(-3)=12
\(\frac{5}{3}\) s+(-3)+3=12+3
\(\frac{5}{3}\) s+0=15
\(\frac{5}{3}\) s=15
\(\frac{3}{5}\)∙\(\frac{5}{3}\) s=\(\frac{3}{5}\)∙15
1s=3∙3
s=9
Stefan’s current age is 9 years, so Katie is currently 12 years old.

Question 13.
Lucas bought a certain weight of oats for his horse at a unit price of $0.20 per pound. The total cost of the oats left him with $1. He wanted to buy the same weight of enriched oats instead, but at $0.30 per pound, he would have been $2 short of the total amount due. How much money did Lucas have to buy oats?
Answer:
The difference in the costs is $3.00 for the same weight in feed.
Let w represent the weight in pounds of feed.
0.3w-0.2w=3
0.1w=3
\(\frac{1}{10}\) w=3
10∙\(\frac{1}{10}\) w=10∙3
1w=30
w=30
Lucas bought 30 pounds of oats.
Cost=unit price×weight
Cost=($0.20 per pound)∙(30 pounds)
Cost=$6.00
Lucas paid $6 for 30 pounds of oats. Lucas had $1 left after his purchase, so he started with $7.

Eureka Math Grade 7 Module 3 Lesson 9 Exit Ticket Answer Key

Question 1.
Brand A scooter has a top speed that goes 2 miles per hour faster than Brand B. If after 3 hours, Brand A scooter traveled 24 miles at its top speed, at what rate did Brand B scooter travel at its top speed if it traveled the same distance? Write an equation to determine the solution. Identify the if-then moves used in your solution.
Answer:
x: speed, in mph, of Brand B scooter
x+2: speed, in mph, of Brand A scooter
d=rt
24=(x+2)(3)
24=3(x+2)

Possible solution 1:
24 = 3(x+2)
8=x+2
8-2=x+2-2
6=x
If-then Moves: Divide both sides by 3.
Subtract 2 from both sides.

Possible solution 2:
24=3(x+2)
24=3x+6
24-6=3x+6-6
18=3x+0
(\(\frac{1}{3}\))(18)=(\(\frac{1}{3}\))(3x)
6=x
If-then Moves: Subtract 6 from both sides.
Multiply both sides by \(\frac{1}{3}\).

Question 2.
At each scooter’s top speed, Brand A scooter goes 2 miles per hour faster than Brand B. If after traveling at its top speed for 3 hours, Brand A scooter traveled 40.2 miles, at what rate did Brand B scooter travel if it traveled the same distance as Brand A? Write an equation to determine the solution and then write an equivalent equation using only integers.
Answer:
x: speed, in mph, of Brand B scooter
x+2: speed, in mph, of Brand A scooter
d=rt
40.2=(x+2)(3)
40.2=3(x+2)
Possible solution 1:
40.2=3(x+2)
13.4=x+2
134=10x+20
134-20=10x+20-20
114=10x
(\(\frac{1}{10}\))(114)=(\(\frac{1}{10}\))(10x)
11.4=x

Possible solution 2:
40.2=3(x+2)
40.2=3x+6
402=30x+60
402-60=30x+60-60
342=30x
(\(\frac{1}{30}\))(342)=(\(\frac{1}{30}\))(30x)
11.4=x

Brand B’s scooter travels at 11.4 miles per hour.

Engage NY Eureka Math 7th Grade Module 3 Lesson 10 Answer Key

Eureka Math Grade 7 Module 3 Lesson 10 Example Answer Key

Example 1.
Estimate the measurement of x. ___
In a complete sentence, describe the angle relationship in the diagram.

Answer:
∠BAC and ∠CAD are angles on a line and their measures have a sum of 180°.

Write an equation for the angle relationship shown in the figure and solve for x. Then, find the measures of ∠BAC and confirm your answers by measuring the angle with a protractor.
x + 132 = 180
x + 132 – 132 = 180 – 132
x = 48
m∠BAC = 48°

Example 2.
In a complete sentence, describe the angle relationship in the diagram.

Answer:
∠AEL and ∠LEB are angles on a line and their measures have a sum of 180°. ∠AEL and ∠KEB are vertical angles and are of equal measurement.

Write an equation for the angle relationship shown in the figure and solve for x and y. Find the measurements of ∠LEB and ∠KEB.
Answer:
y = 144°; m∠KEB = 144° (or vert. ∠s are = )
x + 144 = 180
x + 144 – 144 = 180 – 144
x = 36
m∠LEB = 36°

Example 3.
In a complete sentence, describe the angle relationships in the diagram.

Answer:
∠GKE, ∠EKF, and ∠GKF are angles at a point and their measures have a sum of 360°.

Write an equation for the angle relationship shown in the figure and solve for x. Find the measurement of ∠EKF and confirm your answers by measuring the angle with a protractor.
Answer:
x + 90 + 135 = 360
x + 225 = 360
x + 225 – 225 = 360 – 225
x = 135
m∠EKF = 135°

Example 4.
The following two lines intersect. The ratio of the measurements of the obtuse angle to the acute angle in any adjacent angle pair in this figure is 2:1. In a complete sentence, describe the angle relationships in the diagram.

Answer:
The measurement of an obtuse angle is twice the measurement of an acute angle in the diagram.

Label the diagram with expressions that describe this relationship. Write an equation that models the angle relationship and solve for x. Find the measurements of the acute and obtuse angles.
Answer:
2x + 1x = 180
3x = 180
(\(\frac{1}{3}\))(3x) = (\(\frac{1}{3}\))(180)
x = 60
Acute angle = 60°
Obtuse angle = 2x° = 2(60°) = 120°

Eureka Math Grade 7 Module 3 Lesson 10 Opening Exercise Answer Key

Use the diagram to complete the chart.

Answer:

Eureka Math Grade 7 Module 3 Lesson 10 Exercise Answer Key

Exercise 1.
In a complete sentence, describe the angle relationship in the diagram.

Answer:
∠BAC, ∠CAD, and ∠DAE are angles on a line and their measures have a sum of 180°.

Find the measurements of ∠BAC and ∠DAE.
Answer:
3x + 90 + 2x = 180
5x + 90 = 180
5x + 90 – 90 = 180 – 90
(\(\frac{1}{5}\))(5x) = (\(\frac{1}{5}\))(90)
x = 18
m∠BAC = 3(18°) = 54°
m∠DAE = 2(18°) = 36°

Exercise 2.
In a complete sentence, describe the angle relationships in the diagram.

Answer:
∠JEN and ∠NEM are adjacent angles and, when added together, are the measure of ∠JEM; ∠JEM and ∠KEL are vertical angles and are of equal measurement.

Write an equation for the angle relationship shown in the figure and solve for x.
Answer:
3x + 16 = 85
3x + 16 – 16 = 85 – 16
3x = 69
(\(\frac{1}{3}\))3x = 69(\(\frac{1}{3}\))
x = 23

Exercise 3.
In a complete sentence, describe the angle relationships in the diagram.

Answer:
∠EAH, ∠GAH, ∠GAF, and ∠FAE are angles at a point and their measures sum to 360°.

Find the measurement of ∠GAH.
Answer:
(x + 1) + 59 + 103 + 167 = 360
x + 1 + 59 + 103 + 167 = 360
x = 30
m∠GAH = (30 + 1)° = 31°

Exercise 4.
The ratio of m∠GFH to m∠EFH is 2∶3. In a complete sentence, describe the angle relationships in the diagram.

Answer:
The measurement of ∠GFH is \(\frac{2}{3}\) the measurement of ∠EFH; The measurements of ∠GFH and ∠EFH have a sum of 90°.

Find the measures of ∠GFH and ∠EFH.
Answer:
2x + 3x = 90
5x = 90
(\(\frac{1}{5}\))(5x) = (\(\frac{1}{5}\))(90)
x = 18
m∠GFH = 2(18°) = 36°
m∠EFH = 3(18°) = 54°

Eureka Math Grade 7 Module 3 Lesson 10 Problem Set Answer Key

For each question, use angle relationships to write an equation in order to solve for each variable. Determine the indicated angles. You can check your answers by measuring each angle with a protractor.

Question 1.
In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measurement of ∠DAE.

Answer:
One possible response: ∠CAD, ∠DAE, and ∠FAE are angles on a line and their measures sum to 180°.
90 + x + 65 = 180
x + 155 = 180
x + 155 – 155 = 180 – 155
x = 25
m∠DAE = 25°

Question 2.
In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measurement of ∠QPR.

Answer:
∠QPR, ∠RPS, and ∠SPT are angles on a line and their measures sum to 180°.
f + 154 + f = 180
2f + 154 = 180
2f + 154 – 154 = 180 – 154
2f = 26
(\(\frac{1}{2}\))2f = (\(\frac{1}{2}\))26
f = 13
m∠QPR = 13°

Question 3.
In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measurements of ∠CQD and ∠EQF.

Answer:
∠BQC, ∠CQD, ∠DQE, ∠EQF, and ∠FQG are angles on a line and their measures sum to 180°.
10 + 2x + 103 + 3x + 12 = 180
5x + 125 = 180
5x + 125 – 125 = 180 – 125
5x = 55
(\(\frac{1}{5}\))5x = (\(\frac{1}{5}\))55
x = 11
m∠CQD = 2(11°) = 22°
m∠EQF = 3(11°) = 33°

Question 4.
In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measure of x.

Answer:
All of the angles in the diagram are angles at a point and their measures sum to 360°.
4(x + 71) = 360
4x + 284 = 360
4x + 284 – 284 = 360 – 284
4x = 76
(\(\frac{1}{4}\))4x = (\(\frac{1}{4}\))76
x = 19

Question 5.
In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measures of x and y.

Answer:
∠CKE, ∠EKD, and ∠DKB are angles on a line and their measures sum to 180°. Since ∠FKA and ∠AKE form a straight angle and the measurement of ∠FKA is 90°, ∠AKE is 90°, making ∠CKE and ∠AKC form a right angle and their measures have a sum of 90°.
x + 25 + 90 = 180
x + 115 = 180
x + 115 – 115 = 180 – 115
x = 65

(65) + y = 90
65 – 65 + y = 90 – 65
y = 25

Question 6.
In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measures of x and y.

Answer:
∠EAG and ∠FAK are vertical angles and are of equal measurement. ∠EAG and ∠GAD form a right angle and their measures have a sum of 90°.
2x + 24 = 90
2x + 24 – 24 = 90 – 24
2x = 66
(\(\frac{1}{2}\))2x = (\(\frac{1}{2}\))66
x = 33
3y = 66
(\(\frac{1}{3}\))3y = (\(\frac{1}{3}\))66
y = 22

Question 7.
In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measures of ∠CAD and ∠DAE.

Answer:
∠CAD and ∠DAE form a right angle and their measures have a sum of 90°.
(\(\frac{3}{2}\) x + 20) + 2x = 90
\(\frac{7}{2}\) x + 20 = 90
\(\frac{7}{2}\) x + 20 – 20 = 90 – 20
\(\frac{7}{2}\) x = 70
(\(\frac{2}{7}\)) \(\frac{7}{2}\) x = 70(\(\frac{2}{7}\))
x = 20
m∠CAD = \(\frac{3}{2}\) (20°) + 20° = 50°
m∠DAE = 2(20°) = 40°

Question 8.
In a complete sentence, describe the relevant angle relationships in the following diagram. Find the measure of ∠CQG.

Answer:
∠DQE and ∠CQF are vertical angles and are of equal measurement. ∠CQG and ∠GQF are adjacent angles and their measures sum to the measure of ∠CQF.
3x + 56 = 155
3x + 56 – 56 = 155 – 56
3x = 99
(\(\frac{1}{3}\))3x = (\(\frac{1}{3}\))99
x = 33
m∠CQG = 3(33°) = 99°

Question 9.
The ratio of the measures of a pair of adjacent angles on a line is 4:5.
a. Find the measures of the two angles.
Answer:
∠1 = 4x, ∠2 = 5x
4x + 5x = 180
9x = 180
(\(\frac{1}{9}\))9x = (\(\frac{1}{9}\))180
x = 20
∠1 = 4(20°) = 80°
∠2 = 5(20°) = 100°

b. Draw a diagram to scale of these adjacent angles. Indicate the measurements of each angle.
Answer:

Question 10.
The ratio of the measures of three adjacent angles on a line is 3:4:5.
a. Find the measures of the three angles.
Answer:
∠1 = 3x, ∠2 = 4x, ∠3 = 5x
3x + 4x + 5x = 180
12x = 180
(\(\frac{1}{12}\))12x = (\(\frac{1}{12}\))180
x = 15
∠1 = 3(15°) = 45°
∠2 = 4(15°) = 60°
∠3 = 5(15°) = 75°

b. Draw a diagram to scale of these adjacent angles. Indicate the measurements of each angle.
Answer:

Eureka Math Grade 7 Module 3 Lesson 10 Exit Ticket Answer Key

In a complete sentence, describe the relevant angle relationships in the following diagram. That is, describe the angle relationships you could use to determine the value of x.
Answer:
∠KAE and ∠EAF are adjacent angles whose measurements are equal to ∠KAF; ∠KAF and ∠JAG are vertical angles and are of equal measurement.

Use the angle relationships described above to write an equation to solve for x. Then, determine the measurements of ∠JAH and ∠HAG.

Answer:
5x + 3x = 90 + 30
8x = 120
(\(\frac{1}{8}\))(8x) = (\(\frac{1}{8}\))(120)
x = 15
m∠JAH = 3(15°) = 45°
m∠HAG = 5(15°) = 75°

Engage NY Eureka Math 7th Grade Module 3 Lesson 8 Answer Key

Eureka Math Grade 7 Module 3 Lesson 8 Example Answer Key

Example 1.
Julia, Keller, and Israel are volunteer firefighters. On Saturday, the volunteer fire department held its annual coin drop fundraiser at a streetlight. After one hour, Keller had collected $42.50 more than Julia, and Israel had collected $15 less than Keller. The three firefighters collected $125.95 in total. How much did each person collect?
Find the solution using a tape diagram.
Answer:

3 units + 42.50+27.50=125.95
3 units + 70=125.95
3 units =55.95
1 unit =18.65

42.50+27.50=70
125.95-70=55.95
55.95÷3=18.65

Julia collected $18.65. Keller collected $61.15. Israel collected $46.15.

What were the operations we used to get our answer?
Answer:
First, we added 42.50 and 27.50 to get 70. Next, we subtracted 70 from 125.95. Finally, we divided 55.95 by 3 to get 18.65.

The amount of money Julia collected is j dollars. Write an expression to represent the amount of money Keller collected in dollars.
Answer:
j+42.50

Using the expressions for Julia and Keller, write an expression to represent the amount of money Israel collected in dollars.
Answer:
j+42.50-15
or
j+27.50

Using the expressions written above, write an equation in terms of j that can be used to find the amount each person collected.
Answer:
j+(j+42.50)+(j+27.50)=125.95

Solve the equation written above to determine the amount of money each person collected, and describe any if-then moves used.
Answer:
j+(j+42.50)+(j+27.50)=125.95
3j+70=125.95 Any order, any grouping
(3j+70)-70=125.95-70 If-then move: Subtract 70 from both sides (to make a 0).
3j+0=55.95 Any grouping, additive inverse
3j=55.95 Additive identity
(\(\frac{1}{3}\))(3j)=(55.95)(\(\frac{1}{3}\)) If-then move: Multiply both sides by \(\frac{1}{3}\) (to make a 1).
(\(\frac{1}{3}\)∙3)j=18.65 Associative property
1∙j=18.65 Multiplicative inverse
j=18.65 Multiplicative identity
If Julia collected $18.65, then Keller collected $18.65+$42.50=$61.15, and Israel collected
$61.15-$15=$46.15.

Example 2.
You are designing a rectangular pet pen for your new baby puppy. You have 30 feet of fence barrier. You decide that you would like the length to be 6\(\frac{1}{3}\) feet longer than the width.
Draw and label a diagram to represent the pet pen. Write expressions to represent the width and length of the pet pen.

Answer:
Width of the pet pen: x ft.
Then, (x+6 \(\frac{1}{3}\)) ft. represents the length of the pet pen.
Find the dimensions of the pet pen.
Answer:
Arithmetic
(30-6\(\frac{1}{3}\)-6 \(\frac{1}{3}\))÷4
17 \(\frac{1}{3}\)÷4
4 \(\frac{1}{3}\)
The width is 4 \(\frac{1}{3}\) ft.
The length is 4 \(\frac{1}{3}\) ft.+6 \(\frac{1}{3}\) ft.=10 \(\frac{2}{3}\) ft.
Algebraic
x+(x+6 \(\frac{1}{3}\))+x+(x+6 \(\frac{1}{3}\))=30
4x+12 \(\frac{2}{3}\)=30
4x+12 \(\frac{2}{3}\)-12 \(\frac{2}{3}\)=30-12 \(\frac{2}{3}\)
If-then move: Subtract 12 \(\frac{2}{3}\) from both sides.
4x=17 \(\frac{1}{3}\)
(\(\frac{1}{4}\))(4x)=(17 \(\frac{1}{3}\))(\(\frac{1}{4}\))
If-then move: Multiply both sides by \(\frac{1}{4}\)
x=4 \(\frac{1}{3}\)

If the perimeter of the pet pen is 30 ft. and the length of the pet pen is 6 \(\frac{1}{3}\) ft. longer than the width, then the width would be 4 \(\frac{1}{3}\) ft., and the length would be 4 \(\frac{1}{3}\) ft.+6 \(\frac{1}{3}\) ft.=10 \(\frac{2}{3}\) ft.

Example 3.
Nancy’s morning routine involves getting dressed, eating breakfast, making her bed, and driving to work. Nancy spends \(\frac{1}{3}\) of the total time in the morning getting dressed, 10 minutes eating breakfast, 5 minutes making her bed, and the remaining time driving to work. If Nancy spends 35 \(\frac{1}{2}\) minutes getting dressed, eating breakfast, and making her bed, how long is her drive to work?
Write and solve this problem using an equation. Identify the if-then moves used when solving the equation.
Answer:
Total time of routine: x minutes
\(\frac{1}{3}\) x+10+5=35 \(\frac{1}{2}\)
\(\frac{1}{3}\) x+15=35 \(\frac{1}{2}\)
\(\frac{1}{3}\) x+15-15=35 \(\frac{1}{2}\)-15
If-then move: Subtract 15 from both sides.
\(\frac{1}{3}\) x+0=20 \(\frac{1}{2}\)
3(\(\frac{1}{3}\) x)=3(20 \(\frac{1}{2}\))
If-then move: Multiply both sides by 3.
x=61 \(\frac{1}{2}\)
If-then move: Subtract 15 from both sides.
61 \(\frac{1}{2}\)-35 \(\frac{1}{2}\)=26
It takes Nancy 26 minutes to drive to work.

Is your answer reasonable? Explain.
Answer:
Yes, the answer is reasonable because some of the morning activities take 35 \(\frac{1}{2}\) minutes, so the total amount of time for everything will be more than 35 \(\frac{1}{2}\) minutes. Also, when checking the total time for all of the morning routine, the total sum is equal to total time found. However, to find the time for driving to work, a specific activity in the morning, it is necessary to find the difference from the total time and all the other activities.

Example 4.
The total number of participants who went on the seventh-grade field trip to the Natural Science Museum consisted of all of the seventh-grade students and 7 adult chaperones. Two-thirds of the total participants rode a large bus, and the rest rode a smaller bus. If 54 students rode the large bus, how many students went on the field trip?
Answer:
Arithmetic Approach:

Total on both buses: (54÷2)×3=81
Total number of students: 81-7=74; 74 students went on the field trip.
Algebraic Approach: Challenge students to build the equation and solve it on their own first. Then, go through the steps with them, pointing out how we are “making zeros” and “making ones.” Point out that, in this problem, it is advantageous to make a 1 first. (This example is an equation of the form p(x+q)=r.)
Number of students: s
Total number of participants: s+7
\(\frac{2}{3}\) (s+7)=54
\(\frac{3}{2}\) (\(\frac{2}{3}\) (s+7))=\(\frac{3}{2}\) (54)
If-then move: Multiply both sides by \(\frac{3}{2}\) (to make a 1).
(\(\frac{3}{2}\)∙\(\frac{2}{3}\))(s+7)=81
If-then move: Subtract 7 from both sides (to make a 0).
1(s+7)=81
s+7=81
(s+7)-7=81-7
s+0=74
s=74
74 students went on the field trip.

Eureka Math Grade 7 Module 3 Lesson 8 Opening Exercise Answer Key

Recall and summarize the if-then moves.
Answer:
If a number is added or subtracted to both sides of a true equation, then the resulting equation is also true:
If a=b, then a+c=b+c.
If a=b, then a-c=b-c.
If a number is multiplied or divided to each side of a true equation, then the resulting equation is also true:
If a=b, then ac=bc.
If a=b and c≠0, then a÷c=b÷c.

Write 3+5=8 in as many true equations as you can using the if-then moves. Identify which if-then move you used.
Answer:
Answers will vary, but some examples are as follows:
If 3+5=8, then 3+5+4=8+4. Add 4 to both sides.
If 3+5=8, then 3+5-4=8-4. Subtract 4 from both sides.
If 3+5=8, then 4(3+5)=4(8). Multiply both sides by 4.
If 3+5=8, then (3+5)÷4=8÷4. Divide both sides by 4.

Eureka Math Grade 7 Module 3 Lesson 8 Problem Set Answer Key

Write and solve an equation for each problem.

Question 1.
The perimeter of a rectangle is 30 inches. If its length is three times its width, find the dimensions.
Answer:
The width of the rectangle: w inches
The length of the rectangle: 3w inches
Perimeter=2(length+width)

2(w+3w)=30
2(4w)=30
8w=30
(\(\frac{1}{8}\))(8w)=(\(\frac{1}{8}\))(30)
w=3 \(\frac{3}{4}\)
OR
2(w+3w)=30
(w+3w)=15
4w=15
w=3 \(\frac{3}{4}\)

The width is 3 \(\frac{3}{4}\) inches.
The length is (3)(3\(\frac{3}{4}\) in.)=(3)(\(\frac{15}{4}\) in.)=11 \(\frac{1}{4}\) in.

Question 2.
A cell phone company has a basic monthly plan of $40 plus $0.45 for any minutes used over 700. Before receiving his statement, John saw he was charged a total of $48.10. Write and solve an equation to determine how many minutes he must have used during the month. Write an equation without decimals.
Answer:
The number of minutes over 700: m minutes
40+0.45m=48.10
0.45m+40-40=48.10-40
0.45m=8.10
(\(\frac{1}{0.45}\))(0.45m)=8.10(\(\frac{1}{0.45}\))
m=18
4000+45m=4810
45m+4000-4000=4810-4000
45m=810
(\(\frac{1}{45}\))(45m)=810(\(\frac{1}{45}\))
m=18
John used 18 minutes over 700 for the month. He used a total of 718 minutes.

Question 3.
A volleyball coach plans her daily practices to include 10 minutes of stretching, \(\frac{2}{3}\) of the entire practice scrimmaging, and the remaining practice time working on drills of specific skills. On Wednesday, the coach planned 100 minutes of stretching and scrimmaging. How long, in hours, is the entire practice?
Answer:
The duration of the entire practice: x hours
\(\frac{2}{3}\) x+\(\frac{10}{60}\)=\(\frac{100}{60}\)
\(\frac{2}{3}\) x+\(\frac{1}{6}\)=\(\frac{5}{3}\)
\(\frac{2}{3}\) x+\(\frac{1}{6}\)–\(\frac{1}{6}\)=\(\frac{5}{3}\)–\(\frac{1}{6}\)
\(\frac{2}{3}\) x=\(\frac{9}{6}\)
(\(\frac{3}{2}\))(\(\frac{2}{3}\) x)=\(\frac{3}{2}\) (\(\frac{9}{6}\))
x=\(\frac{27}{12}\)=2 \(\frac{1}{4}\)
The entire practice is a length of 2\(\frac{1}{4}\) hours, or 2.25 hours.

Question 4.
The sum of two consecutive even numbers is 54. Find the numbers.
Answer:
First consecutive even integer: x
Second consecutive even integer: x+2
x+(x+2)=54
2x+2=54
2x+2-2=54-2
2x+0=52
(\(\frac{1}{2}\))(2x)=(\(\frac{1}{2}\))(52)
x=26
The consecutive even integers are 26 and 28.

Question 5.
Justin has $7.50 more than Eva, and Emma has $12 less than Justin. Together, they have a total of $63.00. How much money does each person have?
Answer:
The amount of money Eva has: x dollars
The amount of money Justin has: (x+7.50) dollars
The amount of money Emma has: ((x+7.50)-12) dollars, or (x-4.50) dollars
x+(x+7.50)+(x-4.50)=63
3x+3=63
3x+3-3=63-3
3x+0=60
(\(\frac{1}{3}\))3x=(\(\frac{1}{3}\))60
x=20
If the total amount of money all three people have is $63, then Eva has $20, Justin has $27.50, and Emma has $15.50.

Question 6.
Barry’s mountain bike weighs 6 pounds more than Andy’s. If their bikes weigh 42 pounds altogether, how much does Barry’s bike weigh? Identify the if-then moves in your solution.
Answer:
If we let a represent the weight in pounds of Andy’s bike, then a+6 represents the weight in pounds of Barry’s bike.
a+(a+6)=42
(a+a)+6=42
2a+6=42
2a+6-6=42-6
If 2a+6=42, then 2a+6-6=42-6.
2a+0=36
2a=36
\(\frac{1}{2}\)∙2a=\(\frac{1}{2}\)∙36
If 2a = 36, then \(\frac{1}{2}\)∙2a = \(\frac{1}{2}\)∙36.
1∙a=18
a=18

Barry’s Bike: a+6
(18)+6=24
Barry’s bike weighs 24 pounds.

Question 7.
Trevor and Marissa together have 26 T-shirts to sell. If Marissa has 6 fewer T-shirts than Trevor, find how many T-shirts Trevor has. Identify the if-then moves in your solution.
Answer:
Let t represent the number of T-shirts that Trevor has, and let t-6 represent the number of T-shirts that Marissa has.
t+(t-6)=26
(t+t)+(-6)=26
2t+(-6)=26
2t+(-6)+6=26+6
If-then move: Addition property of equality

2t+0=32
2t=32
\(\frac{1}{2}\)∙2t=\(\frac{1}{2}\)∙32
If-then move: Multiplication property of equality
1∙t=16
t=16
Trevor has 16 T-shirts to sell, and Marissa has 10 T-shirts to sell.

Question 8.
A number is 1\(\frac{1}{7}\) of another number. The difference of the numbers is 18. (Assume that you are subtracting the smaller number from the larger number.) Find the numbers.
Answer:
If we let n represent a number, then 1\(\frac{1}{7}\) n represents the other number.
n-(1\(\frac{1}{7}\) n)=18
7\(\frac{1}{7}\) n-1\(\frac{1}{7}\) n=18
6\(\frac{1}{7}\) n=18
\(\frac{7}{6}\)∙6\(\frac{1}{7}\) n=\(\frac{7}{6}\)∙18
1n=7∙3
n=21
The numbers are 21 and 3.

Question 9.
A number is 6 greater than \(\frac{1}{2}\) another number. If the sum of the numbers is 21, find the numbers.
Answer:
If we let n represent a number, then \(\frac{1}{2}\) n+6 represents the first number.
n+(\(\frac{1}{2}\) n+6)=21
(n+\(\frac{1}{2}\) n)+6=21
(\(\frac{2}{2}\) n+\(\frac{1}{2}\) n)+6=21
\(\frac{3}{2}\) n+6=21
\(\frac{3}{2}\) n+6-6=21-6
\(\frac{3}{2}\) n+0=15
\(\frac{3}{2}\) n=15
\(\frac{2}{3}\)∙\(\frac{3}{2}\) n=\(\frac{2}{3}\)∙15
1n=2∙5
n=10
Since the numbers sum to 21, they are 10 and 11.

Question 10.
Kevin is currently twice as old as his brother. If Kevin was 8 years old 2 years ago, how old is Kevin’s brother now?
Answer:
If we let b represent Kevin’s brother’s age in years, then Kevin’s age in years is 2b.
2b-2=8
2b-2+2=8+2
2b=10
(\(\frac{1}{2}\))(2b)=(\(\frac{1}{2}\))(10)
b=5
Kevin’s brother is currently 5 years old.

Question 11.
The sum of two consecutive odd numbers is 156. What are the numbers?
Answer:
If we let n represent one odd number, then n+2 represents the next consecutive odd number.
n+(n+2)=156
2n+2-2=156-2
2n=154
(\(\frac{1}{2}\))(2n)=(\(\frac{1}{2}\))(154)
n=77
The two numbers are 77 and 79.

Question 12.
If n represents an odd integer, write expressions in terms of n that represent the next three consecutive odd integers. If the four consecutive odd integers have a sum of 56, find the numbers.
Answer:
If we let n represent an odd integer, then n+2, n+4, and n+6 represent the next three consecutive odd integers.
n+(n+2)+(n+4)+(n+6)=56
4n+12=56
4n+12-12=56-12
4n=44
n=11
The numbers are 11, 13, 15, and 17.

Question 13.
The cost of admission to a history museum is $3.25 per person over the age of 3; kids 3 and under get in for free. If the total cost of admission for the Warrick family, including their two 6-month old twins, is $19.50, find how many family members are over 3 years old.
Answer:
If we let w represent the number of Warrick family members, then w-2 represents the number of family members over the age of 3 years.
3.25(w-2)=19.5
3.25w-6.5=19.5
3.25w-6.5+6.5=19.5+6.5
3.25w=26
w=8
w-2=6
There are 6 members of the Warrick family over the age of 3 years.

Question 14.
Six times the sum of three consecutive odd integers is -18. Find the integers.
Answer:
If we let n represent the first odd integer, then n+2 and n+4 represent the next two consecutive odd integers.
6(n+(n+2)+(n+4))=-18
6(3n+6)=-18
18n+36=-18
18n+36-36=-18-36
18n=-54
n=-3
n+2=-1
n+4=1
The integers are -3, -1, and 1.

Question 15.
I am thinking of a number. If you multiply my number by 4, add -4 to the product, and then take \(\frac{1}{3}\) of the sum, the result is -6. Find my number.
Answer:
Let n represent the given number.
\(\frac{1}{3}\) (4n+(-4))=-6
\(\frac{4}{3}\) n-\(\frac{4}{3}\)=-6
\(\frac{4}{3}\) n-\(\frac{4}{3}\)+\(\frac{4}{3}\)=-6+\(\frac{4}{3}\)
\(\frac{4}{3}\) n=\(\frac{-14}{3}\)
n=-3 \(\frac{1}{2}\)

Question 16.
A vending machine has twice as many quarters in it as dollar bills. If the quarters and dollar bills have a combined value of $96.00, how many quarters are in the machine?
Answer:
If we let d represent the number of dollar bills in the machine, then 2d represents the number of quarters in the machine.
2d∙(\(\frac{1}{4}\))+1d∙(1)=96
\(\frac{1}{2}\) d+1d=96
1 \(\frac{1}{2}\) d=96
\(\frac{3}{2}\) d=96
\(\frac{2}{3}\) (\(\frac{3}{2}\) d)=\(\frac{2}{3}\) (96)
d=64
2d=128
There are 128 quarters in the machine.

Eureka Math Grade 7 Module 3 Lesson 8 Exit Ticket Answer Key

Mrs. Canale’s class is selling frozen pizzas to earn money for a field trip. For every pizza sold, the class makes $5.35. They have already earned $182.90, but they need $750. How many more pizzas must they sell to earn $750? Solve this problem first by using an arithmetic approach, then by using an algebraic approach. Compare the calculations you made using each approach.
Answer:
Arithmetic Approach:
Amount of money needed: 750-182.90=567.10
Number of pizzas needed: 567.10÷5.35=106
If the class wants to earn a total of $750, then they must sell 106 more pizzas.

Algebraic Approach:
Let x represent the number of additional pizzas they need to sell.
5.35x+182.90=750
5.35x+182.90-182.90=750-182.90
5.35x+0=567.10
(\(\frac{1}{5.35}\))(5.35x)=(\(\frac{1}{5.35}\))(567.10)
x=106
OR
5.35x+182.90=750
100(5.35x+182.90)=100(750)
535x+18290=75000
535x+18290-18290=75000-18290
(\(\frac{1}{5.35}\))(535x)=(\(\frac{1}{5.35}\))(56710)
x=106
If the class wants to earn $750, then they must sell 106 more pizzas.
Both approaches subtract 182.90 from 750 to get 567.10. Dividing by 5.35 is the same as multiplying by \(\frac{1}{5.35}\). Both result in 106 more pizzas that the class needs to sell.

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

Eureka Math Grade 7 Module 3 Lesson 7 Example Answer Key

Example
The ages of three sisters are consecutive integers. The sum of their ages is 45. Calculate their ages.
a. Use a tape diagram to find their ages.
Answer:

45-3=42
42÷3=14

Youngest sister: 14 years old
2^nd sister: 15 years old
Oldest sister: 16 years old

b. If the youngest sister is x years old, describe the ages of the other two sisters in terms of x, write an expression for the sum of their ages in terms of x, and use that expression to write an equation that can be used to find their ages.
Answer:
Youngest sister: x years old
2nd sister: (x+1) years old
Oldest sister: (x+2) years old
Sum of their ages: x+(x+1)+(x+2)
Equation: x+(x+1)+(x+2)=45

c. Determine if your answer from part (a) is a solution to the equation you wrote in part (b).
Answer:
x+(x+1)+(x+2)=45
14+(14+1)+(14+2)=45
45=45
True
→ Let x be an integer; write an algebraic expression that represents one more than that integer.
→ x+1
→ Write an algebraic expression that represents two more than that integer.
→ x+2
Discuss how the unknown unit in a tape diagram represents the unknown integer, represented by x. Consecutive integers begin with the unknown unit; then, every consecutive integer thereafter increases by 1 unit.

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

Your brother is going away to college, so you no longer have to share a bedroom. You decide to redecorate a wall by hanging two new posters on the wall. The wall is 14 feet wide, and each poster is four feet wide. You want to place the posters on the wall so that the distance from the edge of each poster to the nearest edge of the wall is the same as the distance between the posters, as shown in the diagram below. Determine that distance.

Answer:
14-4-4=6
6÷3=2
The distance is 2 feet

Your parents are redecorating the dining room and want to place two rectangular wall sconce lights that are 25 inches wide along a 10 \(\frac{2}{3}\)-foot wall so that the distance between the lights and the distances from each light to the nearest edge of the wall are all the same. Design the wall and determine the distance.

Answer:
25 in. = \(\frac{25}{12}\) ft. = 2 \(\frac{1}{12}\) ft.
(10 \(\frac{2}{3}\) ft. – 2 \(\frac{1}{12}\) ft. – 2 \(\frac{1}{12}\) ft.) ÷ 3
(10 \(\frac{8}{12}\) ft. – 2 \(\frac{1}{12}\) ft. – 2 \(\frac{1}{12}\) ft.) ÷ 3
(6 \(\frac{6}{12}\) ft.) ÷ 3
6 \(\frac{1}{2}\) ft. ÷ 3
\(\frac{13}{2}\) ft. ÷ 3
\(\frac{13}{2}\) ft. × \(\frac{1}{3}\)
\(\frac{13}{6}\) ft.
2 \(\frac{1}{6}\) ft.
OR
10 \(\frac{2}{3}\) ft.
10 ft.×12 \(\frac{in}{ft.}\) + \(\frac{2}{3}\) ft. × 12 \(\frac{in}{ft}\)
120 in. + 8 in.
128 in.
(128 in. – 25 in. – 25 in.) ÷ 3
78 in. ÷ 3
26 in.

Let the distance between a light and the nearest edge of a wall be x ft. Write an expression in terms of x for the total length of the wall. Then, use the expression and the length of the wall given in the problem to write an equation that can be used to find that distance.
Answer:
3x+2 \(\frac{1}{12}\)+2 \(\frac{1}{12}\)
3x+2 \(\frac{1}{12}\)+2 \(\frac{1}{12}\)=10 \(\frac{2}{3}\)

Now write an equation where y stands for the number of inches: Let the distance between a light and the nearest edge of a wall be y inches. Write an expression in terms of y for the total length of the wall. Then, use the expression and the length of the wall to write an equation that can be used to find that distance (in inches).
Answer:
2 \(\frac{1}{12}\) feet =25 inches; therefore, the expression is 3y+25+25.
10 \(\frac{2}{3}\) feet =128 inches; therefore, the equation is 3y+25+25=128.

What value(s) of y makes the second equation true: 24, 25, or 26?
Answer:
y=24
3y+25+25=128
3(24)+25+25=128
72+25+25=128
122=128
False
y=25
3y+25+25=128
3(25)+25+25=128
75+25+25=128
125=128
False
y=26
3y+25+25=128
3(26)+25+25=128
78+25+25=128
128=128
True

Since substituting 26 for y results in a true equation, the distance between the light and the nearest edge of the wall should be 26 in.

Eureka Math Grade 7 Module 3 Lesson 7 Exercise Answer Key

Sophia pays a $19.99 membership fee for an online music store.
a. If she also buys two songs from a new album at a price of $0.99 each, what is the total cost?
Answer:
$21.97

b. If Sophia purchases n songs for $0.99 each, write an expression for the total cost.
Answer:
0.99n+19.99

c. Sophia’s friend has saved $118 but is not sure how many songs she can afford if she buys the membership and some songs. Use the expression in part (b) to write an equation that can be used to determine how many songs Sophia’s friend can buy.
Answer:
0.99n+19.99=118

d. Using the equation written in part (c), can Sophia’s friend buy 101, 100, or 99 songs?
Answer:
n=99
0.99n+19.99=118
0.99(99)+19.99=118
98.01+19.99=118
118=118
True
n=100
0.99n+19.99=118
0.99(100)+19.99=118
99+19.99=118
118.99=118
False
n=101
0.99n+19.99=118
0.99(101)+19.99=118
99.99+19.99=118
119.98=118
False

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

Question 1.
Check whether the given value is a solution to the equation.
a. 4n-3=-2n+9 n=2
Answer:
4(2)-3=-2(2)+9
8-3=-4+9
5=5
True

b. 9m-19=3m+1 m=\(\frac{10}{3}\)
Answer:
9(\(\frac{10}{3}\))-19=3(\(\frac{10}{3}\))+1
\(\frac{90}{3}\)-19=30/3+1
30-19=10+1
11=11
True

c. 3(y+8)=2y-6 y=30
Answer:
3(30+8)=2(30)-6
3(38)=60-6
114=54
False

Question 2.
Tell whether each number is a solution to the problem modeled by the following equation.
Mystery Number: Five more than -8 times a number is 29. What is the number?
Let the mystery number be represented by n.
The equation is 5+(-8)n=29.
a. Is 3 a solution to the equation? Why or why not?
Answer:
No, because 5-24≠29.

b. Is -4 a solution to the equation? Why or why not?
Answer:
No, because 5+32≠29.

c. Is -3 a solution to the equation? Why or why not?
Answer:
Yes, because 5+24=29.

d. What is the mystery number?
Andrew:
-3 because 5 more than -8 times -3 is 29.

Question 3.
The sum of three consecutive integers is 36.
a. Find the smallest integer using a tape diagram.
Answer:

36-3=33
33÷3=11
The smallest integer is 11.

b. Let n represent the smallest integer. Write an equation that can be used to find the smallest integer.
Answer:
Smallest integer: n
2^nd integer: (n+1)
3^rd integer: (n+2)
Sum of the three consecutive integers: n+(n+1)+(n+2)
Equation: n+(n+1)+(n+2)=36.

c. Determine if each value of n below is a solution to the equation in part (b).
n=12.5
Answer:
No, it is not an integer and does not make a true equation.
n=12
Answer:
No, it does not make a true equation.
n=11
Answer:
Yes, it makes a true equation.

Question 4.
Andrew is trying to create a number puzzle for his younger sister to solve. He challenges his sister to find the mystery number. “When 4 is subtracted from half of a number, the result is 5.” The equation to represent the mystery number is \(\frac{1}{2}\) m-4=5. Andrew’s sister tries to guess the mystery number.

a. Her first guess is 30. Is she correct? Why or why not?
Answer:
No, it does not make a true equation.
\(\frac{1}{2}\) (30)-4=5
15-4=5
11=5
False

b. Her second guess is 2. Is she correct? Why or why not?
Answer:
No, it does not make a true equation.
\(\frac{1}{2}\) (2)-4=5
1-4=5
-3=5
False

c. Her final guess is 4 \(\frac{1}{2}\). Is she correct? Why or why not?
Answer:
No, it does not make a true equation.
\(\frac{1}{2}\) (4 \(\frac{1}{2}\))-4=5
2 \(\frac{1}{4}\)-4=5
-1 \(\frac{3}{4}\)=5
False

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

Question 1.
Check whether the given value of x is a solution to the equation. Justify your answer.
a. \(\frac{1}{3}\) (x+4)=20 x=48
Answer:
(\(\frac{1}{3}\) (48+4)=20
\(\frac{1}{3}\) (52)=20
False, 48 is NOT a solution to \(\frac{1}{3}\) (x+4)=20.
17 \(\frac{1}{3}\)=20

b. 3x-1=5x+10 x=-5 \(\frac{1}{2}\)
(3(-5 \(\frac{1}{2}\))-1=5(-5 \(\frac{1}{2}\))+10
–\(\frac{33}{2}\)-1=-\(\frac{55}{2}\)+10
True, -5 \(\frac{1}{2}\) is a solution to 3x-1=5x+10.
–\(\frac{35}{2}\)=-\(\frac{35}{2}\)

Question 2.
The total cost of four pens and seven mechanical pencils is $13.25. The cost of each pencil is 75 cents.
a. Using an arithmetic approach, find the cost of a pen.
Answer:
(13.25-7(0.75))÷4
(13.25-5.25)÷4
8÷4
2
The cost of a pen is $2.

b. Let the cost of a pen be p dollars. Write an expression for the total cost of four pens and seven mechanical pencils in terms of p.
Answer:
4p+7(0.75) or 4p+5.25

c. Write an equation that could be used to find the cost of a pen.
Answer:
4p+7(0.75)=13.25 or 4p+5.25=13.25

d. Determine a value for p for which the equation you wrote in part (b) is true.
Answer:
4p+5.25=13.25
4(2)+5.25=13.25
8+5.25=13.25
True, when p=2, the equation is true.
13.25=13.25

e. Determine a value for p for which the equation you wrote in part (b) is false.
Answer:
Any value other than 2 will make the equation false.