## enVision Math Common Core Grade K Answer Key | enVision Math Common Core Grade K Answers

Help kindergarten students to become proficient in maths by downloading enVision Math Common Core Grade 2 Answer Key. This answer key contains the answers for all the questions and all the chapters. It helps to enhance your math skills and score good marks in the exam. Practising envision math answers grade K develops problem-solving ability among elementary school students right from a young age.

## enVision Math Common Core Grade K Answers | enVision Math Common Core Grade K Textbook Answer Key

Preschool students can get enVision Math Common Core K Grade Solution Key in the following sections. It has the answers with detailed explanations for topics such as numbers 0 to 5, compare numbers 0 to 5, numbers 6 to 10, compare numbers 0 to 10, classify and count data, understand addition, understand subtraction, more addition and subtraction.

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## enVision Math Common Core 8th Grade Answers Key Topic 1 Real Numbers

enVision STEM Project

Did You Know?
Natural resources are materials that occur in nature, such as water, fossil fuels, wood, and minerals. Natural resources not only meet basic human needs, but also support industry and economy.

Minerals are used in the manufacturing of all types of common objects, including cell phones, computers, light bulbs, and medicines.

Water, oil, and forests are some of the natural resources that are in danger of someday being depleted.
70% of available fresh water is used in agriculture…

…and 10% for human consumption.

Each person in the United States needs over 48,000 pounds of minerals each year.

About 18 million acres of forest are lost to deforestation each year.

Solar power, wind power and other renewable energy sources are helping to lessen the dependency on oil and fossil fuels.

Fossil fuels are expected to supply almost 80% of world energy use through 2040.

Your Task: Going, Going, Gone?

Natural resource depletion is an important issue facing the world. Suppose a natural resource is being depleted at the rate of 1.333% per year. If there were 300 million tons of this resource in 2005, and there are no new discoveries, how much will be left in the year 2045? You and your classmates will explore the depletion of this resource over time.

Answer:
It is given that
A natural resource is being depleted at the rate of 1.333% per year and there were 300 million tons of this resource in 2005, and there are no new discoveries
So,
The number of resources left in 2045 = The number of resources present in 2005 – The decrease of the number of resources from 2005 to 2045
So,
The number of resources left in 2045 = 300 million – 1.33% of 300 million × (2045 – 2005)
The number of resources left in 2045 = 300 million – 159.6 million
The number of resources left in 2045 = 140.4 million
Hence, from the above,
We can conclude that the number of resources left in 2045 is: 140.4 million

### Topic 1 GET READY!

Review What You Know!

Vocabulary

Choose the best term from the box. Write it on the blank.

Question 1.
A(n) ____ is a decimal that ends in repeating zeros.
Answer:
A “Terminating decimal” is a decimal that ends in repeating zeros
Hence, from the above,
We can conclude that the best term from the box for this definition is “Terminating decimal”

Question 2.
A(n) ____ is a decimal in which a digit or digits repeat
Answer:
A “Repeating decimal” is a decimal in which a digit or digits repeat
Hence, from the above,
We can conclude that the best term from the box for this definition is “Repeating decimal”

Question 3.
A(n) ____ is either a counting number, the opposite of a counting number, or zero
Answer:
An “Integer” is either a counting number, the opposite of a counting number, or zero
Hence, from the above,
We can conclude that the best term from the box for this definition is “Integer”

Question 4.
A(n) ___ is a number that can be used to describe a part of a whole, a part of a set, a location on a number line, or a division of whole numbers.
Answer:
A “Fraction” is a number that can be used to describe a part of a whole, a part of a set, a location on a number line, or a division of whole numbers.
Hence, from the above,
We can conclude that the best term from the box for this definition is “Fraction”

Terminating and Repeating Decimals

Determine whether each decimal is terminating or repeating.

Question 5.
5.692
Answer:
The given decimal is: 5.692
We know that,
A terminating decimal is usually defined as a decimal number that contains a finite number of digits after the decimal point
Hence, from the above,
We can conclude that the given decimal is a terminating decimal

Question 6.
-0.222222…
Answer:
The given decimal is -0.222222….
We know that,
A repeating decimal or recurring decimal is the decimal representation of a number whose digits are periodic and the infinitely repeated portion is not zero
Hence, from the above,
We can conclude that the given decimal is a repeating decimal

Question 7.
7.0001
Answer:
The given decimal is: 7.0001
We know that,
A terminating decimal is usually defined as a decimal number that contains a finite number of digits after the decimal point
Hence, from the above,
We can conclude that the given decimal is a terminating decimal

Question 8.
7.2$$\overline{8}$$
Answer:
The given decimal is: 7.2$$\overline{8}$$
We know that,
A repeating decimal or recurring decimal is the decimal representation of a number whose digits are periodic and the infinitely repeated portion is not zero
Hence, from the above,
We can conclude that the given decimal is a repeating decimal

Question 9.
1.$$\overline{178}$$
Answer:
The given decimal is: 1.$$\overline{178}$$
We know that,
A repeating decimal or recurring decimal is the decimal representation of a number whose digits are periodic and the infinitely repeated portion is not zero
Hence, from the above,
We can conclude that the given decimal is a repeating decimal

Question 10.
-4.03479
Answer:
The given decimal is: -4.03479
We know that,
A terminating decimal is usually defined as a decimal number that contains a finite number of digits after the decimal point
Hence, from the above,
We can conclude that the given decimal is a terminating decimal

Multiplying Integers

Find each product.

Question 11.
2.2
Answer:
The given expression is: 2 × 2
Hence,
2 × 2 = 4

Question 12.
-5. (-5)
Answer:
The given expression is: -5 × -5
We know that,
– × – = +
Hence,
-5 × -5 = 25

Question 13.
7.7
Answer:
The given expression is: 7 × 7
Hence,
7 × 7 = 49

Question 14.
-6 ∙ (-6) ∙ (-6)
Answer:
The given expression is: -6 × -6 × -6
We know that,
– × – × –
= + × –
= –
Hence,
-6 × -6 × -6
= 36 × -6
= -216

Question 15.
10 ∙ 10 ∙ 10
Answer:
The given expression is: 10 × 10 × 10
Hence,
10 × 10 × 10 = 1,000

Question 16.
-9 ∙ (-9) ∙ (-9)
Answer:
The given expression is: -9 × -9 × -9
We know that,
– × – × –
= + × –
= –
Hence,
-9 × -9 × -9
= 81 × -9
= -729

Simplifying Expressions
Simplify each expression.

Question 17.
(4 ∙ 10) + (5 ∙ 100)
Answer:
The given expression is:
(4 × 10) + (5 × 100)
So,
(4 × 10) + (5 × 100)
= 40 + 500
= 540
Hence, from the above,
We can conclude that the value of the expression is: 540

Question 18.
(2100) + (7.10)
Answer:
The given expression is:
2100 + (7 × 10)
So,
2100 + (7 × 10)
= 2100 + 70
= 2170
Hence, from the above,
We can conclude that the value of the given expression is: 2170

Question 19.
(6 · 100) – (1 · 10)
Answer:
The given expression is:
(6 × 100) – (1 × 10)
So,
(6 × 100) – (1 × 10)
= 600 – 10
= 590
Hence, from the above,
We can conclude that the value of the given expression is: 590

Question 20.
(9 ∙ 1,000) + (4 ∙ 10)
Answer:
The given expression is:
(9 × 1,000) + (4 × 10)
So,
(9 × 1,000) + (4 × 10)
= 9,000 + 40
= 9,040
Hence, from the above,
We can conclude that the value of the given expression is: 9,040

Question 21.
(3 · 1,000) – (2 ∙ 100)
Answer:
The given expression is:
(3 × 1,000) – (2 × 100)
So,
(3 × 1,000) – (2 × 100)
= 3,000 – 200
= 2,800
Hence, from the above,
We can conclude that the value of the given expression is: 2,800

Question 22.
(2 ∙ 10) + (7 · 100)
Answer:
The given expression is:
(2 × 10) + (7 × 100)
So,
(2 × 10) + (7 × 100)
= 20 + 700
= 720
Hence, from the above,
we can conclude that the value of the given expression is: 720

Language Development

Fill in the word map with new terms, definitions, and supporting examples or illustrations.

### Topic 1 PICK A PROJECT

PROJECT 1A
Who is your favorite poet, and why?
PROJECT: WRITE A POEM

PROJECT 1B
If you moved to a tiny house, what would you bring with you?
PROJECT: DESIGN A TINY HOUSE

PROJECT 1C
If you could travel anywhere in space, where would you go?
PROJECT: PLAN A TOUR OF THE MILKY WAY

PROJECT 1D
Why do you think people tell stories around a campfire?
PROJECT: TELL A FOLK STORY

Lesson 1.1 Rational Numbers as Decimals

Solve & Discuss It!

Jaylon has a wrench labeled 0.1875 inches and bolts labeled in fractions of an inch. Which size bolt will fit best with the wrench? Explain.

Answer:
It is given that
Jaylon has a wrench labeled 0.1875 inches and bolts labeled in fractions of an inch.
Now,
We know that,
The bolt will be fit in a wrench only when
The size of the bolt (inches) = The size of the wrench (inches)
Now,
The representation of the bolts in the decimal numbers is:
$$\frac{3}{8}$$ = 0.375 inches
$$\frac{1}{8}$$ = 0.046 inches
$$\frac{3}{16}$$ = 0.1875 inches
$$\frac{1}{4}$$ = 0.25 inches
Hence, from the above,
We can conclude that the bolt which has the size $$\frac{3}{16}$$ inches will fit best with the wrench

Reasoning
How can you write these numbers in the same form?
Answer:
The representation of the sizes of bolts in the decimal form is by using the properties of place values
We know that,
A terminating decimal can be written as a fraction by using properties of place value.
Example:
3.75 = three and seventy-five hundredths or $$\frac{375}{100}$$, which is equal to the improper
fraction

Focus on math practices
Reasoning Why is it useful to write a rational number as a fraction or as a decimal?
Answer:
Rational numbers are whole numbers, fractions, and decimals – the numbers we use in our daily lives. They can be written as a ratio of two integers. … The definition says that a number is rational if you can write it in the form $$\frac{a}{b}$$ where a and b are integers, and b is not zero.

? Essential Question
How can you write repeating decimals as fractions?
Answer:
Because repeating decimals are rational numbers, you can write them in fraction form.
STEP 1
Assign a variable to represent the repeating decimal.
STEP 2
Write an equation: variable = decimal.
STEP 3
Multiply each side of the equation by 10d, where d is the number of repeating digits in the repeating decimal.
STEP 4
Subtract equivalent expressions of the variable and the repeating decimal from each side of the equation.
STEP 5
Solve for the variable. Write an equivalent fraction so that the numerator and denominator are integers, if necessary.

Try It!

In another baseball division, one team had a winning percentage of 0.444…. What fraction of their games did this team win?
The team won their games.

Answer:
It is given that
In another baseball division, one team had a winning percentage of 0.444…
Since only 1 number is repeated,
The repeating decimal can be written as 0.$$\overline{4}$$
Now,
According to the steps of converting a repeating decimal to a fraction (Essential Question),
Step 1:
Let x = 0.$$\overline{4}$$
Step 2:
Multiply with 10 on both sides since only 1 number is repeating
So,
10x = 10 (0.$$\overline{4}$$)
10x = 4.$$\overline{4}$$
Step 3:
10x – x = 4.$$\overline{4}$$ – 0.$$\overline{4}$$
9x = 4
x = $$\frac{4}{9}$$
Hence, from the above,
We can conclude that the team won $$\frac{4}{9}$$ of their games

Convince Me!
How do you know what power of ten to multiply by in the second step at the right?
Answer:
Let x be the repeating portion.
Multiply this equation by a power of 10 to move the repeating digits immediately to the left of the decimal point (in other words, to eliminate any zeros preceding the repeating digits).

Try It!
Write the repeating decimal 0.63333… as a fraction.
Answer:
The given repeating decimal is: 0.6333333
Now,
According to the steps of converting a repeating decimal to a fraction (Essential Question),
Step 1:
Let x = 0.6$$\overline{3}$$
Step 2:
Multiply with 10 on both sides since only 1 number is repeating
So,
10x = 10 (0.6$$\overline{3}$$)
10x = 6.$$\overline{3}$$
Step 3:
10x – x = 6.$$\overline{3}$$ – 0.6$$\overline{3}$$
9x = 6.33 – 0.63
9x = 5.7
Divide by 9 into both sides
So,
$$\frac{9}{9}$$x = $$\frac{5.7}{9}$$
x = $$\frac{57}{90}$$
Hence, from the above,
We can conclude that the value of the given repeating decimal in the fraction form is: $$\frac{57}{90}$$

Try It!
Write the repeating decimal 4.1363636… as a fraction.
Answer:
The given repeating decimal is 4.1363636…..
Now,
According to the steps of converting a repeating decimal to a fraction (Essential Question),
Step 1:
Let x = 4.1$$\overline{36}$$
Step 2:
Multiply with 100 on both sides since 2 numbers are repeating
So,
100x = 100 (4.1$$\overline{36}$$)
100x = 41.$$\overline{36}$$
Step 3:
100x – x = 41.$$\overline{36}$$ – 4.1$$\overline{36}$$
99x = 413.636 – 4.136
99x = 409.5
Divide by 99 into both sides
So,
$$\frac{99}{99}$$x = $$\frac{409.5}{99}$$
x = $$\frac{4095}{990}$$
Hence, from the above,
We can conclude that the value of the given repeating decimal in the fraction form is: $$\frac{4095}{990}$$

KEY CONCEPT

Because repeating decimals are rational numbers, you can write them in fraction form.
STEP 1
Assign a variable to represent the repeating decimal.
STEP 2
Write an equation: variable = decimal.
STEP 3
Multiply each side of the equation by 10d, where d is the number of repeating digits in the repeating decimal.
STEP 4
Subtract equivalent expressions of the variable and the repeating decimal from each side of the equation.
STEP 5
Solve for the variable. Write an equivalent fraction so that the numerator and denominator are integers, if necessary.

Do You Understand?

Question 1.
? Essential Question
How can you write repeating decimals as fractions?
Answer:
Because repeating decimals are rational numbers, you can write them in fraction form.
STEP 1
Assign a variable to represent the repeating decimal.
STEP 2
Write an equation: variable = decimal.
STEP 3
Multiply each side of the equation by 10d, where d is the number of repeating digits in the repeating decimal.
STEP 4
Subtract equivalent expressions of the variable and the repeating decimal from each side of the equation.
STEP 5
Solve for the variable. Write an equivalent fraction so that the numerator and denominator are integers, if necessary.

Question 2.
Use Structure Why do you multiply by a power of 10 when writing a repeating decimal as a rational number?
Answer:
The idea is to multiply by some number (10, 100, 1000, etc.) so that when we subtract the original number from the multiple, the repeating part cancels out.

Question 3.
Be Precise How do you decide by which power of 10 to multiply an equation when writing a decimal with repeating digits as a fraction?
Answer:
The idea is to multiply by some number (10, 100, 1000, etc.) so that when we subtract the original number from the multiple, the repeating part cancels out.

Do You Know How?

Question 4.
A survey reported that 63.63% of moviegoers prefer action films. This percent represents a repeating decimal. Write it as a fraction.
Answer:
It is given that
A survey reported that 63.63% of moviegoers prefer action films. This percent represents a repeating decimal
So,
The given repeating decimal is 63.6363…..%
Now,
According to the steps of converting a repeating decimal to a fraction (Essential Question),
Step 1:
Let x = 63.$$\overline{63}$$
Step 2:
Multiply with 100 on both sides since 2 numbers are repeating
So,
100x = 100 (63.$$\overline{63}$$)
100x = 6363.$$\overline{63}$$
Step 3:
100x – x = 6363.$$\overline{63}$$ – 63.$$\overline{63}$$
99x = 6,300
Divide by 99 into both sides
So,
$$\frac{99}{99}$$x = $$\frac{6,300}{99}$$
x = $$\frac{6,300}{99}$$
Hence, from the above,
We can conclude that the value of the given repeating decimal in the fraction form is: $$\frac{6,300}{99}$$

Question 5.
A student estimates the weight of astronauts on the Moon by multiplying their weight by the decimal 0.16666… What fraction can be used for the same estimation?

Answer:
It is given that
A student estimates the weight of astronauts on the Moon by multiplying their weight by the decimal 0.16666…
Now,

Hence, from the above,
We can conclude that the fraction that can be used for the same estimation is: $$\frac{1}{6}$$

Question 6.
Write 2.3181818… as a mixed number.
Answer:
The given repeating decimal is 2.3181818…
Now,
According to the steps of converting a repeating decimal to a fraction (Essential Question),
Step 1:
Let x = 2.3$$\overline{18}$$
Step 2:
Multiply with 100 on both sides since 2 numbers are repeating
So,
100x = 100 (2.3$$\overline{18}$$)
100x = 23.$$\overline{18}$$
Step 3:
100x – x = 23.$$\overline{18}$$ – 2.3$$\overline{18}$$
99x = 231.818 – 2.318
99x = 229.5
Divide by 99 into both sides
So,
$$\frac{99}{99}$$x = $$\frac{229.5}{99}$$
x = $$\frac{2295}{990}$$
x = $$\frac{51}{22}$$
So,
The representation of the above fraction in the mixed form is: 2$$\frac{7}{22}$$
Hence, from the above,
We can conclude that the value of the given repeating decimal in the mixed fraction form is: 2$$\frac{7}{22}$$

Practice & Problem Solving

Leveled Practice In 7 and 8, write the decimal as a fraction or mixed number.

Question 7.
Write the number 0.21212121… as a fraction.
Let x =
100x =
100x – x =
99x =
x =
So 0.2121… is equal to
Answer:
The given repeating decimal is: 0.212121….
Now,

Hence, from the above,
We can conclude that the value of the repeating decimal in the fraction form is: $$\frac{21}{99}$$

Question 8.
Write 3.7 as a mixed number.
Let x =
10x =
9x =
x =
So 3.$$\overline{7}$$ is equal to
Answer:
The given repeating decimal is: 3.$$\overline{7}$$
Now,

Hence, from the above,
We can conclude that the representation of the given repeating decimal in the mixed fraction form is: 3$$\frac{7}{9}$$

Question 9.
Write the number shown on the scale as a fraction.

Answer
The number that is shown on the scale is: 0.233333
Now,
From the: above number,
We can observe that the number is a repeating decimal
Now,

So,
The simplified form of $$\frac{2.1}{9}$$ is: $$\frac{7}{30}$$
Hence, from the above,
We can conclude that the representation of the number that is shown on the scale as a fraction is: $$\frac{7}{30}$$

Question 10.
Tomas asked 15 students whether summer break should be longer. He used his calculator to divide the number of students who said yes by the total number of students. His calculator showed the result as 0.9333…
a. Write this number as a fraction.
Answer:
The given repeating decimal is 0.93333…
Now,

So,
The simplified form of $$\frac{8.4}{9}$$ is: $$\frac{14}{15}$$
Hence, from the above,
We can conclude that the representation of the repeating number in the form of the fraction is: $$\frac{14}{15}$$

b. How many students said that summer break should be longer?
Answer:
It is given that
Tomas asked 15 students whether summer break should be longer. He used his calculator to divide the number of students who said yes by the total number of students.
So,
The number of students that said summer break should be longer = $$\frac{The number of students that said yes that summer break is longer}{The total number of students} Now, From part (a), The fraction form of his calculated result from part (a) is: 14 / 15 Hence, from the above, We can conclude that the number of students that said the summer break should be longer is: 14 students Question 11. Write 0.\overline{87}$$ as a fraction. Answer: The given repeating decimal is 0.$$\overline{87}$$ Now, Hence, from the above, We can conclude that the representation of the repeating decimal in the fraction form in the simplest form is: $$\frac{29}{33}$$ Question 12. Write 0.$$\overline{8}$$ as a fraction. Answer: The given repeating decimal is 0.$$\overline{8}$$ Now, Hence, from the above, We can conclude that the representation of the given repeating decimal in the fraction form is: $$\frac{8}{9}$$ Question 13. Write 1.$$\overline{48}$$ as a mixed number. Answer: The given repeating decimal number is 1.$$\overline{48}$$ Now, Hence, from the above, We can conclude that the representation of the given repeating decimal in the mixed fraction form is: 1$$\frac{16}{33}$$ Question 14. Write 0.$$\overline{6}$$ as a fraction. Answer: The given repeating decimal is 0.$$\overline{6}$$ Now, Hence, from the above, We can conclude that the representation of the given repeating decimal in the fraction form is: $$\frac{2}{3}$$ Question 15. A manufacturer determines that the cost of making a computer component is 2.161616. Write the cost as a fraction and as a mixed number. Answer: It is given that A manufacturer determines that the cost of making a computer component is 2.161616 So, The given repeating decimal is 2.161616… Now, Hence, from the above, We can conclude that The cost of a computer component in a fraction form is: $$\frac{214}{99}$$ The cost of a computer component in a mixed fraction form is: 2$$\frac{16}{99}$$ Question 16. Reasoning When writing a repeating decimal as a fraction, does the number of repeating digits you use matter? Explain. Answer: No. Even if the number of different digits in the cycle is 1 or 1 million, the method of finding the fraction is the same Question 17. Higher Order Thinking When writing a repeating decimal as a fraction, why does the fraction always have only 9s or 9s and 0s as digits in the denominator? Answer: When writing a repeating decimal as a fraction, the fraction always has only 9s or 9s and 0s as digits in the denominator because we are talking here about a geometric series and they are decimals, so the right side i.e., after the decimal point, the digits are in tenths, hundredths and so on Assessment Practice Question 18. Which decimal is equivalent to $$\frac{188}{11}$$? A. 17.$$\overline{09}$$ B. 17.0$$\overline{09}$$ C. 17.$$\overline{1709}$$ D. 17.$$\overline{1709}$$0 Answer: The given fraction is: $$\frac{188}{11}$$ So, The representation of the given fraction in the decimal form is: we know that, $$\frac{1}{11}$$ = 0.090909….. So, $$\frac{188}{11}$$ = 17.090909…. = 17.$$\overline{09}$$ Hence, from the above, We can conclude that option A matches the representation of the repeating decimal for the given fraction Question 19. Choose the repeating decimal that is equal to the fraction on the left. Answer: Follow the process that is mentioned below to solve the given repeating decimals in the fraction form Now, STEP 1 Assign a variable to represent the repeating decimal. STEP 2 Write an equation: variable = decimal. STEP 3 Multiply each side of the equation by 10d, where d is the number of repeating digits in the repeating decimal. STEP 4 Subtract equivalent expressions of the variable and the repeating decimal from each side of the equation. STEP 5 Solve for the variable. Write an equivalent fraction so that the numerator and denominator are integers, if necessary. Hence, ### Lesson 1.2 Understand Irrational Numbers Explain It! Sofia wrote a decimal as a fraction. Her classmate Nora says that her method and answer are not correct. Sofia disagrees and says that this is the method she learned. A. Construct Arguments Is Nora or Sofia correct? Explain your reasoning. Answer: The given number is 0.121121112111112….. Now, From the given number, We can observe that the given decimal is not a repeating decimal because there are other numbers other than the repeating numbers in the given decimal or a terminating decimal because the decimal is not finite So, Since the given decimal is not a repeating decimal, The method that we used to convert the repeating decimal into a fraction is not applicable Hence, from the above, We can conclude that Nora is correct B. Use Structure What is a nonterminating decimal number that can not be written as a fraction. Answer: A non-terminating, non-repeating decimal is a decimal number that continues endlessly, with no group of digits repeating endlessly. Decimals of this type cannot be represented as fractions, and as a result, are irrational numbers Focus on math practices Construct Arguments is 0.12112111211112… a rational number? Explain. Answer: 0.12112111211112… can’t be represented in the form of $$\frac{p}{q}$$ and it has non terminating non-repeating decimal expansion Hence, from the above, We can conclude that 0.12112111211112… is not a rational number ? Essential Question How is an irrational number different from a rational number? Answer: Numbers that can be expressed in $$\frac{a}{b}$$ or fraction form are rational numbers where a is an integer and b is a non-zero integer and the irrational numbers are the numbers that cannot be written in $$\frac{a}{b}$$ form Try It! Classify each number as rational or irrational. Answer: The given numbers are: Now, We know that, Numbers that can be expressed in $$\frac{a}{b}$$ or fraction form are rational numbers where a is an integer and b is a non-zero integer and the irrational numbers are the numbers that cannot be written in $$\frac{a}{b}$$ form Hence, The representation of the rational and irrational numbers are: Convince Me! Construct Arguments Jen classifies the number 4.567 as irrational because it does not repeat. Is Jen correct? Explain. Answer: The given decimal is: 4.567 We know that, A rational number is a number that can be written in the form of $$\frac{a}{b}$$ A terminating decimal has the finite number of digits without repeating and it is also a rational number So, We can observe that we can write 4.567 as a rational number Hence, from the above, We can conclude that Jen is not correct Try It! Classify each number as rational or irrational and explain. A) $$\frac{2}{3}$$ B) $$\sqrt{25}$$ C) -0.7$$\overline{5}$$ D) $$\sqrt{2}$$ E) 7,548,123 Answer: The given numbers are: A) $$\frac{2}{3}$$ B) $$\sqrt{25}$$ C) -0.7$$\overline{5}$$ D) $$\sqrt{2}$$ E) 7,548,123 Now, We know that, A “Rational number” is a number that can be written in the form of $$\frac{a}{b}$$ Ex: Terminating decimals, perfect squares, etc An “Irrational number” is a number that can not be written in the form of $$\frac{a}{b}$$ Ex: Non-terminating decimals So, From the given numbers, Rational numbers ——> A, B, E Irrational numbers ——> C, D KEY CONCEPT Numbers that are not rational are called irrational numbers. Do You Understand? Question 1. ? Essential Question How is an irrational number different from a rational number? Answer: Numbers that can be expressed in $$\frac{a}{b}$$ or fraction form are rational numbers where a is an integer and b is a non-zero integer and the irrational numbers are the numbers that cannot be written in $$\frac{a}{b}$$ form Question 2. Reasoning How can you tell whether a square root of a whole number is rational or irrational? Answer: If the square root of an integer is itself an integer (Ex: √4 = 2), then by definition it is rational – If the square root is not an integer (Ex: √2 = 1.41414…), then it must be irrational. Put another way the only integers for which the square root of an integer can be rational is if is a perfect square – that is where x is an integer Question 3. Construct Arguments Could a number ever be both rational and irrational? Explain. Answer: No. A rational number is a number that can be expressed as the quotient of two integers. An irrational number is a number that cannot be expressed as a quotient of two integers. So if a number is either rational or irrational, it cannot also be the other. Do You Know How? Question 4. Is the number 65.4349224… rational or irrational? Explain. Answer: The given number is 65.4349224… From the given number, We can observe that the given number is a non-repeating and non-terminating decimal number Hence, from the above, We can conclude that the given number is an irrational number Question 5. Is the number $$\sqrt{2,500}$$ rational or irrational? Explain. Answer: The given number is: $$\sqrt{2,500}$$ We know that, A perfect square number is a rational number So, $$\sqrt{2,500}$$ = 50 Hence, from the above, We can conclude that the given number is a rational number Question 6. Classify each number as rational or irrational. Answer: We know that, A “Rational number” is a number that can be written in the form of $$\frac{a}{b}$$ Ex: Perfect squares An “Irrational number” is a number that can not be written in the form of $$\frac{a}{b}$$ Ex: Non-terminating decimal numbers Hence, The representation of the rational and irrational numbers from the given numbers are: Practice & Problem Solving Question 7. Is 5.787787778… a rational or irrational number? Explain. Answer: The given number is 5.787787778… From the given number, We can observe that the given number is a non-repeating and non-terminating decimal number Hence, from the above, We can conclude that the given number is an irrational number Question 8. Is $$\sqrt{42}$$ rational or irrational? Explain. Answer: The given number is $$\sqrt{42}$$ From the given number, We can observe that the given number is not a perfect square Hence, from the above, We can conclude that the given number is an irrational number Question 9. A teacher places seven cards, lettered A-G, on a table. Which cards show irrational numbers? Answer: The given cards are: We know that, An “Irrational number” is a number that can not be written in the form of $$\frac{a}{b}$$ Hence, from the above, We can conclude that from the given cards, The irrational numbers are: A) π B) 8.25635…, C) 6.$$\overline{31}$$ Question 10. Circle the irrational number in the list below. A) 7.$$\overline{27}$$ B) $$\frac{5}{9}$$ C) $$\sqrt{15}$$ D) $$\sqrt{196}$$ Answer: The given numbers are: A) 7.$$\overline{27}$$ B) $$\frac{5}{9}$$ C) $$\sqrt{15}$$ D) $$\sqrt{196}$$ Now, We know that, A “Rational number” is a number that can be written in the form of $$\frac{a}{b}$$ Ex: Perfect squares An “Irrational number” is a number that can not be written in the form of $$\frac{a}{b}$$ Ex: Non-terminating decimal numbers Hence, from the above, We can conclude that From the given numbers, The irrational numbers are A) and C) Question 11. Lisa writes the following list of numbers. 5.737737773…, 26, $$\sqrt{45}$$, –$$\frac{3}{2}$$, 0, 9 Answer: The given numbers are: A) 5.7377377737… B) 26 C) $$\sqrt{45}$$ D) –$$\frac{3}{2}$$ E) 0 F) 9 Now, We know that, A “Rational number” is a number that can be written in the form of $$\frac{a}{b}$$ Ex: Perfect squares An “Irrational number” is a number that can not be written in the form of $$\frac{a}{b}$$ Ex: Non-terminating decimal numbers a. Which numbers are rational? Answer: From the given numbers, The rational numbers are: B, D, E, and F b. Which numbers are irrational? Answer: From the given numbers, The irrational numbers are: A and C Question 12. Construct Arguments Deena says that 9.565565556… is a rational number because it has a repeating pattern. Do you agree? Explain. Answer: The given number is 9.565565556… From the given number, We can observe that the number is a non-repeating and a non-terminating decimal So, The given number is an irrational number Hence, from the above, We can conclude that we don’t have to agree with Deena Question 13. Is $$\sqrt{1,815}$$ rational? Explain. Answer: The given number is: $$\sqrt{1,815}$$ We know that, A “Rational number” is a number that can be written in the form of $$\frac{a}{b}$$ Ex: Perfect squares Now, From the given square root, We can observe that it won’t form a perfect square Hence, from the above, We can conclude that the given number is an irrational number Question 14. Is the decimal form of $$\frac{13}{3}$$ Explain. Answer: The given number is: $$\frac{13}{3}$$ We know that, $$\frac{13}{3}$$ = 4.3333….. We know that, An “Irrational number” is a number that can not be written in the form of $$\frac{a}{b}$$ Hence, from the above, We can conclude that the decimal form of $$\frac{13}{3}$$ is an irrational number Question 15. Write the side length of the square rug as a square root. Is the side length a rational or irrational number? Explain. Answer: The given figure is: From the given figure, We can observe that the given figure depicts the shape of a square Now, Let the side length of a square be x We know that, Area = (Side length)² x² = 100 x = $$\sqrt{100}$$ We know that, A “Perfect square” is a rational number Hence, from the above, We can conclude that a side length is a rational number Question 16. Reasoning The numbers 2.888… and 2.999… are both rational numbers. What is an irrational number that is between the two rational numbers? Answer: A rational number is a number which is can be represented as the quotient of two numbers without having any remainder i.e., having remainder 0. For example 2.45, 2, 3 etc. An irrational number has a non-zero remainder and has a nonterminating quotient. Hence, The numbers between 2.888… and 2.999… are 2.8889………, 2.8890…….., 2.8891…… etc Question 17. Higher Order Thinking You are given the expressions $$\sqrt{76+n}$$ and $$\sqrt{2 n+26}$$. What is the smallest value of n that will make each number rational? Answer: The given expressions are: $$\sqrt{76+n}$$ and $$\sqrt{2 n+26}$$ Now, To find the smallest value of n so that each expression will be a rational number, $$\sqrt{76+n}$$ = $$\sqrt{2 n+26}$$ Squaring on both sides So, 76 + n = 2n + 26 2n – n = 76 – 26 n = 50 Hence, from the above, We can conclude that the smallest value of n so that the given expressions will become a rational number is: 5 Assessment Practice Question 18. Which numbers are rational? I. 1.1111111… II. 1.567 III. 1.101101110… A. II and III B. III only C. II only D. I and II E I only F. None of the above Answer: The given numbers are: I. 1.1111111… II. 1.567 III. 1.101101110… We know that, A “Rational number” is a number that can be written in the form of $$\frac{a}{b}$$ So, From the given numbers, 1 and 2 are the rational numbers Hence, from the above, we can conclude that option D matches with the given situation Question 19. Determine whether the following numbers are rational or irrational. Answer: The representation of the given numbers as rational and irrational numbers is: ### Lesson 1.3 Compare and Order Real Numbers Solve & Discuss It! Courtney and Malik are buying a rug to fit in a 50-square-foot space. Which rug should they purchase? Explain. Answer: It is given that Courtney and Malik are buying a rug to fit in a 50-square-foot space Now, The given figure is: Now, From the above figure, We can observe that The rugs are in different shapes i.e., a square, a circle, and a rectangle Now, The area of a square rug = The side length of a square rug × The side length of a square rug = 7 × 7 = 49 ft² Now, The area of a circular rug = π × $$\frac{Diameter of a circular rug²}{4}$$ = 3.14 × $$\frac{8 ×8}{4}$$ = 3.14 ×16 = 50.24 ft² Now, The area of a rectangular rug = Length × Width = 6 × 8$$\frac{1}{2}$$ = 6 × $$\frac{17}{2}$$ = $$\frac{6 ×17}{2}$$ = 51 ft² Now, When we compare the area of the rugs, The area of the square rug is less than 50 ft² Hence, from the above, We can conclude that Courtney and Malik should buy the square rug Focus on math practices Make Sense and Persevere How did you decide which rug Courtney and Malik should purchase? Answer: It is given that Courtney and Malik are buying a rug to fit in a 50 ft² space So, To fit in a 50 ft² space, The area of any type of rug should be less than 50 ft² Now, From the above problem, We can observe that The area of the square rug is the only area that is less than 50 ft² Hence, from the above, We can conclude that Courtney and Mali should purchase the rugs based on the areas of the rugs ?Essential Question How can you compare and order rational and irrational numbers? Answer: In the given numbers, one of them is rational while other one is irrational. To make the comparison, let us first make the given irrational number into rational number and then carry out the comparison. So, let us square both the given numbers Try It! Between which two whole numbers is $$\sqrt{12}$$? Answer: The given number is: $$\sqrt{12}$$ Now, Hence, from the above, We can conclude that $$\sqrt{12}$$ is in between 3 and 4 Convince Me! Which of the two numbers is a better estimate for $$\sqrt{12}$$? Explain. Answer: The given number is: $$\sqrt{12}$$ Now, We know that, 12 will be 3² and 4² So, $$\sqrt{12}$$ will be between 3 and 4 Now, When we observe the numbers between 3 and 4 The value of $$\frac{12}$$ will be near to 3.4 Hence, from the above, We can conclude that The two numbers that are better estimate for $$\sqrt{12}$$ is: 3 and 4 Try It! Compare and order the following numbers: $$\sqrt{11}$$, 2$$\frac{1}{4}$$, -2.5, 3.$$\overline{6}$$, -3.97621 … Answer: The given numbers are: $$\sqrt{11}$$, 2$$\frac{1}{4}$$, -2.5, 3.$$\overline{6}$$, -3.97621 … So, $$\sqrt{11}$$ ≅ 3.3 2$$\frac{1}{4}$$ = 2.25 3.$$\overline{6}$$ = 3.666…… So, The order of the numbers from the least to the greatest is: -3.97621…….. < -2.5 < 2.25 < $$\sqrt{11}$$ < 3.$$\overline{6}$$ KEY CONCEPT To compare rational and irrational numbers, you must first find rational approximations of the irrational numbers. You can approximate irrational numbers using perfect squares or by rounding. Do You Understand? Question 1. ? Essential Question How can you compare and order rational and irrational numbers? Answer: In the given numbers, one of them is rational while other one is irrational. To make the comparison, let us first make the given irrational number into rational number and then carry out the comparison. So, let us square both the given numbers Question 2. Reasoning The “leech” is a technical term for the slanted edge of a sail. Is the length of the leech shown closer to 5 meters or 6 meters? Explain. Answer: It is given that The “leech” is a technical term for the slanted edge of a sail Now, The given figure is: Now, From the given figure, We can observe that The length of the leech is: $$\sqrt{30}$$ meters Now, We know that, 5² < 30 < 6² 5 < $$\sqrt{30}$$ < 6 Now, We know that, $$\sqrt{30}$$ ≅ 5.4 So, $$\sqrt{30}$$ is close to 5 Hence, from the above, We can conclude that The length of the leech shown above is close to 5 meters Question 3. Construct Arguments which is a better approximation of $$\sqrt{20}$$, 4.5 or 4.47? Explain. Answer: The given number is: $$\sqrt{20}$$ Now, We know that, 4² < 20 < 5² 4 < $$\sqrt{20}$$ < 5 Now, We know that, 4.5² = 20.25 So, The value of $$\sqrt{20}$$ is close to 4.4 Hence, from the above, We can conclude that The better approximation of $$\sqrt{20}$$ is: 4.4 Do You Know How? Question 4. Approximate $$\sqrt{39}$$ to the nearest whole number. Answer: The given number is: $$\sqrt{39}$$ Now, We know that, 6² < 39 < 7² 6 < $$\sqrt{39}$$ < 7 Now, We know that, 6.5² = 42.25 So, The value of $$\sqrt{39}$$ is close to 6.2 So, The value of $$\sqrt{39}$$ is closes to 6 that is the nearest whole number Hence, from the above, We can conclude that The better approximation of $$\sqrt{39}$$ that is the closest to the whole number is: 6 Question 5. Approximate $$\sqrt{18}$$ to the nearest tenth and plot the number on a number line. Answer: The given number is: $$\sqrt{18}$$ Now, We know that, 4² < 18 < 5² 4 < $$\sqrt{18}$$ < 5 Now, We know that, 4.5² = 20.25 4.2² = 17.69 So, $$\sqrt{18}$$ is close to 4.2 Hence, from the above, We can conclude that The representation of the approximate value of $$\sqrt{18}$$ on the given number line is: The approximate value of $$\sqrt{18}$$ is: 4.2 Question 6. Compare 5.7145… and $$\sqrt{29}$$. Show your work. Answer: The given numbers are: 5.7145…… and $$\sqrt{29}$$ Now, We know that, 5² < 29 < 6² So, 5 < $$\sqrt{29}$$ < 6 Now, We know that, 5.5² = 30.25 5.3² = 28.09 So, The approximate value of $$\sqrt{29}$$ is: 5.3 Now, When we compare the given numbers, We can observ ethat 5.7145….. > 5.3 Hence, from the above, We can conclude that The order of the given numbers is: 5.7145……….. > $$\sqrt{29}$$ Question 7. Compare and order the following numbers 5.2, -5.$$\overline{6}$$, 3$$\frac{9}{10}$$, $$\sqrt{21}$$ Answer: The given numbers are: 5.2, -5.$$\overline{6}$$, 3$$\frac{9}{10}$$, $$\sqrt{21}$$ Now, We know that, -5.$$\overline{6}$$ = -5.666666….. 3$$\frac{9}{10}$$ = 3.9 $$\sqrt{21}$$ ≅ 4.58 So, The order of the given numbers from the least to the greatest is: -5.$$\overline{6}$$ < 3$$\frac{9}{10}$$ < $$\sqrt{21}$$ < 5.2 Hence, from the above, We can conclude that The order of the given numbers from the least to the greatest is: -5.$$\overline{6}$$ < 3$$\frac{9}{10}$$ < $$\sqrt{21}$$ < 5.2 Practice & Problem Solving Question 8. Leveled Practice Find the rational approximation of $$\sqrt{15}$$. a. Approximate using perfect squares. < 15 < = < $$\sqrt{15}$$ < < $$\sqrt{15}$$ < Answer: The given number is: $$\sqrt{15}$$ Now, We know that, By using the approximation using the perfect squares, Hence, from the above, We can conclude that The approximate numbers that are between $$\sqrt{15}$$ are: 3 and 4 b. Locate and plot $$\sqrt{15}$$ on a number line. Find a better approximation using decimals. Answer: From Part (a), We know that, The approximate numbers that are between $$\sqrt{15}$$ are: 3 and 4 Now, We know that, 3.5 ² = 12.25 Now, So, The approximate number that is the closest to $$\sqrt{15}$$ is: 3.8 Hence, from the above, We can conclude that The representation of the approximation of $$\sqrt{15}$$ in the given number line is: The approximate number that is close to $$\sqrt{15}$$ is: 3.8 Question 9. Compare – 1.96312… and –$$\sqrt{5}$$. Show your work. Answer: The given numbers are: -1.96312…… and –$$\sqrt{5}$$ Now, We know that, 2² < 5 < 3² 2 < $$\sqrt{5}$$ < 3 Now, We know that, 2.5² = 6.25 2.2² = 4.84 So, The approximate value of –$$\sqrt{5}$$ is: -2.2 So, The order of the given numbers from the least to the greatest is: -1.96312……. > –$$\sqrt{5}$$ Hence, from the above, We can conclude that The order of the given numbers from the least to the greatest is: -1.96312……. > –$$\sqrt{5}$$ Question 10. Does $$\frac{1}{6}$$, -3, $$\sqrt{7}$$, –$$\frac{6}{5}$$, or 4.5 come first when the numbers are listed from least to greatest? Explain. Answer: The given numbers are: $$\frac{1}{6}$$, -3, $$\sqrt{7}$$, –$$\frac{6}{5}$$, and 4.5 Now, $$\frac{1}{6}$$ = 0.166 $$\sqrt{7}$$ = 2.64 –$$\frac{6}{5}$$ = -1.2 So, The order of the given numbers from the least to the greatest is: -3 < –$$\frac{6}{5}$$ < $$\frac{1}{6}$$ < $$\sqrt{7}$$ < 4.5 Hence, from the above, We can conclude that “-3” will come first when the given numbers will be arranged from the least to the greatest Question 11. A museum director wants to hang the painting on a wall. To the nearest foot, how tall does the wall need to be? Answer: It is given that A museum director wants to hang the painting on a wall Now, The given figure is: Now, From the given figure, We can obsere that The painting on a wall is about $$\sqrt{90}$$ ft Now, We know that, 9² < 90 < 10² 9 < $$\sqrt{90}$$ < 10 Now, We know that, 9.5² = 90.25 So, The approximate value of $$\sqrt{90}$$ is: 9.4 ft Hence, from the above, We can conclude that The height of the wall needed to hang a painting is about 9.4 ft Question 12. Dina has several small clay pots. She wants to display them in order of height, from shortest to tallest. What will be the order of the pots? Answer: It is given that Dina has several small clay pots. She wants to display them in order of height, from shortest to tallest Now, The given heights are: $$\sqrt{8}$$, 2$$\frac{1}{3}$$, $$\sqrt{5}$$, and 2.5 Now, We now that, $$\sqrt{8}$$ ≅ 2.82 in. 2$$\frac{1}{3}$$ = 2.033 in. $$\sqrt{5}$$ ≅ 2.23 in. So, The order of the heights from the shortest to the tallest is: 2$$\frac{1}{3}$$ in. < $$\sqrt{5}$$ in. < 2.5 in. < $$\sqrt{8}$$ in. Hence, from the above, We can conclude that The order of the pots is: 2$$\frac{1}{3}$$ in. < $$\sqrt{5}$$ in. < 2.5 in. < $$\sqrt{8}$$ in. Question 13. Rosie is comparing $$\sqrt{7}$$ and 3.44444…. She says that $$\sqrt{7}$$ > 3.44444… because $$\sqrt{7}$$ = 3.5. a. What is the correct comparison? Answer: It is given that Rosie is comparing $$\sqrt{7}$$ and 3.44444…. She says that $$\sqrt{7}$$ > 3.44444… because $$\sqrt{7}$$ = 3.5. Now, We know that, 2² < 7 < 3² 2 < $$\sqrt{7}$$ < 3 Now, We know that, 2.6² = 6.76 2.7² = 7.29 So, The approximate value of $$\sqrt{7}$$ is: 2.6 So, The order of the given numbers is: $$\sqrt{7}$$ < 3.44444….. Hence, from the above, We can conclude that The correct comparison of the given numbers is: $$\sqrt{7}$$ < 3.44444….. b. Critique Reasoning What mistake did Rosie likely make? Answer: The given numbers are: $$\sqrt{7}$$ and 3.44444….. Now, It is also given that $$\sqrt{7}$$ = 3.5 But, 3.5 = $$\frac{7}{2}$$ Hence, from the above, We can conclude that The mistake Rosie likely make is: She considered $$\sqrt{7}$$ = $$\frac{7}{2}$$ Question 14. Model with Math Approximate – √23 to the nearest tenth. Draw the point on the number line. Answer: The given number is: –$$\sqrt{23}$$ Now, We know that, 4² < 23 < 5² 4 < $$\sqrt{23}$$ < 5 Now, We know that, 4.5² = 20.25 4.7² = 22.09 So, The approximate value of –$$\sqrt{23}$$ is: -4.7 Hence, The representtaion of the approximate value of –$$\sqrt{23}$$ on the given number line is: Question 15. Higher Order Thinking The length of a rectangle is twice the width. The area of the rectangle is 90 square units. Note that you can divide the rectangle into two squares. a. Which irrational number represents the length of each side of the squares? Answer: It is given that The length of a rectangle is twice the width. The area of the rectangle is 90 square units. Note that you can divide the rectangle into two squares. Now, The given figure is: Now, According to the given information, The area of each square = $$\frac{90}{2}$$ = 45 square units Now, We know that, The area of a square = Side² So, Side of a squre = $$\sqrt{The area of a square}$$ So, The side length of each square = $$\sqrt{45}$$ units Hence, from the above, We can conclude that The irrational number that represents the length of each side of the squares is: $$\sqrt{45}$$ units b. Estimate the length and width of the rectangle. Answer: It is given that The length of a rectangle is twice the width. The area of the rectangle is 90 square units. Note that you can divide the rectangle into two squares. Now, The given figure is: Now, Let the width of the rectangle be x units So, The length of the rectangle = 2 (Width) = 2x units Now, We know that, The length of a rectangle = Length × Width So, According to the given information, 90 = 2x × x 90 = 2x² x² = $$\frac{90}{2}$$ x = $$\sqrt{45}$$ Hence, from the above, We can conclude that The length of the rectangle is: 2$$\sqrt{45}$$ units The width of the rectangle is: $$\sqrt{45}$$ units Assessment Practice Question 16. Which list shows the numbers in order from least to greatest? A. -4, –$$\frac{9}{4}$$, $$\frac{1}{2}$$, 3.7, $$\sqrt{5}$$ B. -4, –$$\frac{9}{4}$$, $$\frac{1}{2}$$, $$\sqrt{5}$$, 3.7 C. –$$\frac{9}{4}$$, $$\frac{1}{2}$$, 3.7, $$\sqrt{5}$$, -4 D. –$$\frac{9}{4}$$, -4, $$\frac{1}{2}$$, 3.7, $$\sqrt{5}$$ Answer: The given list of numbers are: -4, –$$\frac{9}{4}$$, $$\frac{1}{2}$$, 3.7, $$\sqrt{5}$$ Now, We know that, –$$\frac{9}{4}$$ = -2.25 $$\frac{1}{2}$$ = 0.5 $$\sqrt{5}$$ ≅ 2.23 So, The order of the given list of numbers from the least to the greatest is: -4 < –$$\frac{9}{4}$$ < $$\frac{1}{2}$$ < $$\sqrt{5}$$ < 3.7 Hence, from the above, We can conclude that The list that shows the numbers from the least to the greatest is: Question 17. The area of a square poster is 31 square inches. Find the length of one side of the poster. Explain. PART A To the nearest whole inch Answer: It is given that The area of a square poster is 31 square inches Now, We know that, The area of a square = Side² Side = $$\sqrt{Area of a square}$$ So, According to the given information, The length of one side of the poster = $$\sqrt{31}$$ inches Now, We know that, 5² < 31 < 6² 5 < $$\sqrt{31}$$ < 6 Now, We know that, 5.5² = 30.25 5.6² = 31.36 So, The approximate value of $$\sqrt{31}$$ is: 5.5 Hence, from the above, We can conclude that The length of one side of the poster to the nearest whole inch is: 6 inches PART B To the nearest tenth of an inch Answer: It is given that The area of a square poster is 31 square inches Now, We know that, The area of a square = Side² Side = $$\sqrt{Area of a square}$$ So, According to the given information, The length of one side of the poster = $$\sqrt{31}$$ inches Now, We know that, 5² < 31 < 6² 5 < $$\sqrt{31}$$ < 6 Now, We know that, 5.5² = 30.25 5.6² = 31.36 So, The approximate value of $$\sqrt{31}$$ is: 5.5 Hence, from the above, We can conclude that The length of one side of the poster to the nearest tenth of an inch is: 5.5 inches ### Lesson 1.4 Evaluate Square Roots and Cube Roots Solve & Discuss It! ACTIVITY Matt and his dad are building a tree house. They buy enough flooring material to cover an area of 36 square feet. What are all possible dimensions of the floor? Answer: It is given that Matt and his dad are building a tree house. They buy enough flooring material to cover an area of 36 square feet Now, To find the length and width of the floor, find the multiples of 36 So, The multiples of 36 are: 36 = 1 × 36, 2 × 18, 3 × 12, 4 × 9, 9 ×4, 12 × 3, 18 × 2, 36 × 1 So, All the possible dimensions of the floor are: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 9 ×4, 12 × 3, 18 × 2, 36 × 1 Hence, from the above, We can conclude that All the possible dimensions of the floor are: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 9 ×4, 12 × 3, 18 × 2, 36 × 1 Look for Relationships Can different floor dimensions result in the same area? Answer: From the above problem, We can observe that All the possible dimensions of the floor are: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 9 ×4, 12 × 3, 18 × 2, 36 × 1 Now, When we find the area by using all the different dimensions of the floor, We can observe that the area of the floor is the same Hence, from the above We can conclude that The different floor dimensions result in the same area Focus on math practices Reasoning Why is there only one set of dimensions for a square floor when there are more sets for a rectangular floor? Are all the dimensions reasonable? Explain. Answer: From the above problem, We can observe that All the possible dimensions of the floor are: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 9 ×4, 12 × 3, 18 × 2, 36 × 1 Now, We know that, A square has the same side lengths A square has the same parallel side lengths Hence, The square floor has only one set of dimensions whereas the rectangular floor has more sets and all the dimensions will be reasonable ? Essential Question How do you evaluate cube roots and square roots? Answer: Let the number be: p Now, The square of a number is: p² The cube of a number is: p³ The square root of a number is: $$\sqrt{p}$$ The cube root of a number is: $$\sqrt[3]{p}$$ Try It! A cube-shaped art sculpture has a volume of 64 cubic feet. What is the length of each edge of the cube? The length of each edge is feet. Answer: It is given that A cube-shaped art sculpture has a volume of 64 cubic feet. Now, We know that, A cube has all the same side lengths Now, Let the side length of a cube be: s So, The volume of a cube (V) = s³ So, Side = $$\sqrt[3]{V}$$ Now, So, Hence, from the above, We can conclude that The length of each edge of the cube is: 4 feet Convince Me! How can you find the cube root of 64? Answer: Let the number be: p Now, The cube root of a number is: $$\sqrt[3]{p}$$ So, The cube root of 64 = $$\sqrt[3]{64}$$ = $$\sqrt[3]{4 × 4 × 4}$$ = 4 Hence, from the above, We can conclude that The cube root of 64 is: 4 Try It! Evaluate. a. $$\sqrt[3]{27}$$ Answer: The given number is: $$\sqrt[3]{27}$$ Now, Hence, from the above, We can conclude that The cube root of the given number is: 3 b. $$\sqrt{25}$$ Answer: The given number is: $$\sqrt{25}$$ Now, Hence, from the above, We can conclude that The square root of the given number is: 5 c. $$\sqrt{81}$$ Answer: The given number is: $$\sqrt{81}$$ Now, Hence, from the above, We can conclude that The square root of the given number is: 9 d. $$\sqrt[3]{1}$$ Answer: The given number is: $$\sqrt[3]{1}$$ Now, Hence, from the above, We can conclude that The cube root of the given number is: 1 Try It! Emily wants to buy a tablecloth to cover a square card table. She knows the tabletop has an area of 9 square feet. What are the minimum dimensions of the tablecloth Emily needs? Emily should buy a tablecloth that measures at least feet by feet. Answer: It is given that Emily wants to buy a tablecloth to cover a square card table. She knows the tabletop has an area of 9 square feet. Now, We know that, The area of a square = Side² So, According to the given information, The area of a square table = Side² Side² = 9 Now, So, Hence, from the above, We can conclude that The minimum dimensions of the table cloth Emily needs is: 3 feet × 3 feet KEY CONCEPT The cube root of a number is a number whose cube is equal to that number. Cubing a number and taking the cube root of the number are inverse operations. The square root of a number is a number whose square is equal to that number. Squaring a number and taking the square root of the number are inverse operations. Do You Understand? Question 1. ? Essential Question How do you evaluate cube roots and square roots? Answer: Let the number be: p Now, The square of a number is: p² The cube of a number is: p³ The square root of a number is: $$\sqrt{p}$$ The cube root of a number is: $$\sqrt[3]{p}$$ Question 2. Generalize A certain number is both a perfect square and a perfect cube. Will its square root and its cube root always be different numbers? Explain. Answer: We know that, A perfect square is a number whose square root is an integer; and a perfect cube is a number whose cube root is an integer. A number that is a perfect square and perfect cube will not always have different numbers as its square root and cube root. Question 3. Critique Reasoning A cube-shaped box has a volume of 27 cubic inches. Bethany says each side of the cube measures 9 inches because 9 × 3 = 27. Is Bethany correct? Explain your reasoning. Answer: It is given that A cube-shaped box has a volume of 27 cubic inches. Bethany says each side of the cube measures 9 inches because 9 × 3 = 27 Now, We know that, A “Cube” has the equal side lengths Now, We know that, The volume of a cube (V) = Side³ So, Side = $$\sqrt[3]{V}$$ Now, For the volume of 27 cubic inches, Side = $$\sqrt[3]{27}$$ = $$\sqrt[3]{3 × 3 ×3}$$ = 3 inches So, Each side of the cube measures 3 inches Hence, from the above, We can conclude that Bethany is not correct Do You Know How? Question 4. A cube has a volume of 8 cubic inches. What is the length of each edge of the cube? Answer: It is given that A cube has a volume of 8 cubic inches Now, We know that, A “Cube” has the equal side lengths Now, We know that, The volume of a cube (V) = Side³ So, Side = $$\sqrt[3]{V}$$ Now, For the volume of 8 cubic inches, Side = $$\sqrt[3]{8}$$ = $$\sqrt[3]{2 × 2 ×2}$$ = 2 inches Hence, from the above, We can conclude that The length of each edge of the given cube is: 2 inches Question 5. Below is a model of the infield of a baseball stadium. How long is each side of the infield? Answer: It is given that Below is a model of the infield of a baseball stadium Now, The given figure is: Now, From the given figure, We can observe that The infield is in the form of a square Now, We know that, The area of a square = Side² So, Each Side of the infield =$$\sqrt{The area of the infield}$$ = $$\sqrt{81}$$ Now, Hence, from the above, We can conclude that The length of each side of the infield is: 9 inches Question 6. Julio cubes a number and then takes the cube root of the result. He ends up with 20. What number did Julio start with? Answer: It is given that Julio cubes a number and then takes the cube root of the result. He ends up with 20 Now, Let us say that the number is x. So The first step that Julio do is to cube the number so that the number will become: x³ Then, The next he did was take cube root of the number, so that the result will become: $$\sqrt[3]{x³}$$ Now, By solving the above expression, $$\sqrt[3]{x³}$$ = x Now, It is given that The end result is 20 So, x = 20 So, Julio started and ended with the same number which is 20. Hence, from the above, We can conclude that The number did Julio start with is: 20 Practice & Problem Solving Leveled Practice In 7 and 8, evaluate the cube root or square root. Question 7. Relate the volume of the cube to the length Answer: The given figure is: Now, We know that, The volume of a cube (V) = Side³ So, Side = $$\sqrt[3]{V}$$ So, Side = $$\sqrt[3]{8}$$ Now, Hence, from the above, We can conclude that The length of each edge of the cube is: 2 cm Question 8. Relate the area of the square to the length of each edge. Answer: The given figure is: Now, We know that, The area of a square (V) = Side² So, Side = $$\sqrt{V}$$ So, Side = $$\sqrt{16}$$ Now, Hence, from the above, We can conclude that The length of each side of the square is: 4 cm Question 9. Would you classify the number 169 as a perfect square, a perfect cube, both, or neither? Explain. Answer: The given number is: 169 Now, We know that, A perfect cube is a number that can be expressed as the product of three equal integers A perfect square is a number that can be expressed as the product of two equal integers Now, 169 can be written as: 169 = 13 × 13 Hence, from the above, We can conclude that 169 would be classified as a perfect square Question 10. The volume of a cube is 512 cubic inches. What is the length of each side of the cube? Answer: It is given that The volume of a cube is 512 cubic inches Now, We know that, The volume of a cube (V) = Side³ So, Side = $$\sqrt[3]{V}$$ So, Side = $$\sqrt[3]{512}$$ So, Hence, from the above, We can conclude that The length of each edge of the cube is: 8 inches Question 11. A square technology chip has an area of 25 square cm. How long is each side of the chip? Answer: It is given that A square technology chip has an area of 25 square cm. Now, We know that, The area of a square = Side² So, Side = $$\sqrt{The area of a square}$$ So, Side = $$\sqrt{25}$$ So, Hence, from the above, We can conclude that The length of each side of the chip is: 5 cm Question 12. Would you classify the number 200 as a perfect square, a perfect cube, both, or neither? Explain. Answer: The given number is: 200 Now, We know that, A perfect cube is a number that can be expressed as the product of three equal integers A perfect square is a number that can be expressed as the product of two equal integers Now, 200 can be written as: 200 = 100 × 2 = 10 × 10 × 2 Hence, from the above, We can conclude that 200 would not be classified neither as a perfect square nor a perfect cube Question 13. A company is making building blocks. What is the length of each side of the block? Answer: It is given that A company is making building blocks Now, The given figure is: Now, From the given figure, We can observe that The building blocks is in the form of a cube Now, We know that, The volume of a cube (V) = Side³ So, Side = $$\sqrt[3]{V}$$ So, Side = $$\sqrt[3]{1}$$ Now, Hence, from the above, We can conclude that The length of each side of the block is: 1 ft Question 14. Mrs. Drew wants to build a square sandbox with an area of 121 square feet. What is the total length of wood Mrs. Drew needs to make the sides of the sandbox? Answer: It is given that Mrs. Drew wants to build a square sandbox with an area of 121 square feet Now, We know that, The area of a square (A) = Side² So, Side = $$\sqrt{A}$$ So, Side = $$\sqrt{121}$$ Now, So, The side of the sandbox is: 11 feet Now, To find the total length of wood Mrs. Drew needs to make the sides of the sandbox = 4 × (The length of the side of the sandbox) = 4 × 11 = 44 feet Hence, from the above, We can conclude that The total length of wood Mrs. Drew needs to make the sides of the sandbox is: 44 feet Question 15. Construct Arguments Diego says that if you cube the number 4 and then take the cube root of the result, you end up with 8. Is Diego correct? Explain. Answer: It is given that Diego says that if you cube the number 4 and then take the cube root of the result, you end up with 8 Now, According to the given information, Step 1: 4³ = 4 × 4 × 4 = 64 Step 2: $$\sqrt[3]{64}$$ = $$\sqrt[3]{4 × 4 ×4}$$ = 4 But, It is given that The end result is 8 and we got 4 Hence,f rom the above, We can conclude that Diego is not correct Question 16. Higher Order Thinking Talia is packing a moving box. She has a square-framed poster with an area of 9 square feet. The cube-shaped box has a volume of 30 cubic feet. Will the poster lie flat in the box? Explain. Answer: It is given that Talia is packing a moving box. She has a square-framed poster with an area of 9 square feet. The cube-shaped box has a volume of 30 cubic feet Now, The given figure is: Now, To make the square-framed poster fit into a cube-shaped box, The side of square-framed poster < The side of each edge of the cube-shaped box Now, We know that, The area of a square = Side² The volume of a cube = Side³ So, $$\sqrt{9}$$ = 3 feet $$\sqrt[3]{30}$$ = 3.10 feet So, 3 < 3.10 So, The side of square-framed poster < The side of each edge of the cube-shaped box Hence, from the above, We can conclude that The poster lie flat in the box Assessment Practice Question 17. Which expression has the greatest value? A. $$\sqrt{49}$$ . 2 B. $$\sqrt{49}$$ – $$\sqrt{16}$$ C. $$\sqrt{25}$$ + $$\sqrt{16}$$ D. $$\sqrt{25}$$.3 Answer: The given expressions are: a. The given expression is: $$\sqrt{49}$$ . 2 Now, We know that, $$\sqrt{49}$$ = 7 So, The value of the given expression is: 14 b. The given expression is: $$\sqrt{49}$$ – $$\sqrt{16}$$ Now, We know that, $$\sqrt{49}$$ = 7 $$\sqrt{16}$$ = 4 So, The value of the given expression is: 3 c. The given expression is: $$\sqrt{25}$$ + $$\sqrt{16}$$ Now, We know that, $$\sqrt{25}$$ = 5 $$\sqrt{16}$$ = 4 So, The value of the given expression is: 9 d. The given expression is: $$\sqrt{25}$$.3 Now, We know that, $$\sqrt{25}$$ = 5 So, The value of the given expression is: 15 Hence, from the above, We can conclude that The expression that has the greatest value is: Question 18. A toy has various shaped objects that a child can push through matching holes. The area of the square hole is 8 square cm. The volume of a cube-shaped block is 64 cubic cm PART A Which edge length can you find? Explain. Answer: It is given that A toy has various shaped objects that a child can push through matching holes. The area of the square hole is 8 square cm. The volume of a cube-shaped block is 64 cubic cm Now, We know that, The area of a square (A) = Side² The volume of a cube (V) = Side³ So, $$\sqrt{8}$$ = 2.82 cm $$\sqrt[3]{64}$$ = 4 cm Hence, from the above, We can conclude that The value of the edge lengths you found are: The side of a square-shaped hole is: 2.82 cm The side of a cube-shaped block is: 4 cm PART B Will the block fit in the square hole? Explain. Answer: Now, From Part A, We can observe that The side of a square-shaped hole is: 2.82 cm The side of a cube-shaped block is: 4 cm Now, For the block to fit in the square hole, The side of the block < The side of the hole But, 4 cm > 2.82 cm Hence, from the above, We can conclude that The block will not fit in the square hole ### Lesson 1.5 Solve Equations Using Square Roots and Cube Roots Solve & Discuss It! Janine can use up to 150 one-inch blocks to build a solid, cube-shaped model. What are the dimensions of the possible models that she can build? How many blocks would Janine use for each model? Explain. Answer: It is given that Janine can use up to 150 one-inch blocks to build a solid, cube-shaped model Now, The given figure is: Now, We know that, The volume of a cube = Length × Width × Height Now, To find the dimensions of the possible models that Janine can model, We have to find the multiples of 150 in terms of three So, 150 = 25 × 6 150 = 5 × 5 × 6 So, The total number of blocks Janine would use for each model = The sum of the above three multiples of 150 = 5 + 5 + 6 = 16 blocks Hence, from the above, We can conclude that The dimensions of the possible model that Janine can build is: 5 × 5 × 6 The total number of blocks Janine would use for each model is: 16 blocks Look for Relationships How are the dimensions of a solid related to its volume? Answer: The volume, V , of any rectangular solid is the product of the length, width, and height. We could also write the formula for volume of a rectangular solid in terms of the area of the base. The area of the base, B , is equal to length × Width. Focus on math practices Reasoning Janine wants to build a model using $$\frac{1}{2}$$-inch cubes. How many $$\frac{1}{2}$$-inch cubes would she use to build a solid, cube-shaped model with side lengths of 4 inches? Show your work. Answer: It is given that Janine wants to build a model using $$\frac{1}{2}$$-inch cubes and a cube-shaped model with side lengths of 4 inches Now, According to the given information, The number of $$\frac{1}{2}$$-inch cubes would Janine used to build a solid = $$\frac{1}{2}$$ × 8 blocks Hence, from the above, We can conclude that The number of $$\frac{1}{2}$$-inch cubes would she use to build a solid, cube-shaped model with side lengths of 4 inches is: 8 blocks ? Essential Question How can you solve equations with squares and cubes? Answer: The steps to solve equations with squares are: Step 1: Divide all terms by a (the coefficient of x2). Step 2: Move the number term (c/a) to the right side of the equation. Step 3: Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation. The steps to solve equations with cubes are: Step 1: Divide all terms by a (the coefficient of x³). Step 2: Move the number term ($$\frac{d}{a}$$) to the right side of the equation. Step 3: Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation. Try It! What is the side length, s, of the square below? Each side of the square measures meters. Answer: The given figure is: Now, From the given figure, We can observe that It is a square Now, We know that, The area of a square = Side² Now, So, Hence, from the above, We can conclude that Each side of the square measures 10 meters Convince Me! Why are there two possible solutions to the equation s2 = 100? Explain why only one of the solutions is valid in this situation. Answer: The given equation is: s² Now, We know that, The square of a positive number or a negative number is always positive and the square root of a number must always be positive So, s² = 100 s = $$\sqrt{100}$$ s = ± 10 Now, We know that, The side of any figure will always be positive Hence, from the above, We can conclude that Only one of the solutions is valid in this situation because of the property of sides of the geometrical figures Try It! Solve x3 = 64. Answer: The given equation is: x³ = 64 So, x = $$\sqrt[3]{64}$$ x = $$\sqrt[3]{4 × 4 × 4}$$ x = 4 Hence, from the above, We can conclude that The value of x for the given equation is: 4 Try It! a. Solve a3 = 11. Answer: The given equation is: a³ = 11 So, a = $$\sqrt[3]{11}$$ Now, We know that, The cube of a number will always be positive Hence, from the above, We can conclude that The possible solution for the given equation is: $$\sqrt[3]{11}$$ b. Solve c2 = 27. Answer: The given equation is: c² = 27 So, c = ±$$\sqrt{27}$$ Hence, from the above, We can conclude that The possible solutions for the given equation is: $$\sqrt{27}$$, –$$\sqrt{27}$$ KEY CONCEPT You can use square roots to solve equations involving squares. x2 = a $$\sqrt{x^{2}}$$ = $$\sqrt{a}$$ x = + $$\sqrt{a}$$, –$$\sqrt{a}$$ You can use cube roots to solve equations involving cubes. x2 = b Vx3 = xb x3 = b $$\sqrt[3]{x^{3}}$$ = $$\sqrt[3]{b}$$ x = $$\sqrt[3]{b}$$ Do You Understand? Question 1. ? Essential Question How can you solve equations with squares and cubes? Answer: The steps to solve equations with squares are: Step 1: Divide all terms by a (the coefficient of x2). Step 2: Move the number term (c/a) to the right side of the equation. Step 3: Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation. The steps to solve equations with cubes are: Step 1: Divide all terms by a (the coefficient of x³). Step 2: Move the number term ($$\frac{d}{a}$$) to the right side of the equation. Step 3: Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation. Question 2. Be Precise Suri solved the equation x2 = 49 and found that x = 7. What error did Suri make? Answer: It is given that Suri solved the equation x2 = 49 and found that x = 7. Now, The given equation is: x² = 49 Now, We know that, The square of a number is always positive but the square root of a number can either be positive or negative So, x = $$\sqrt{49}$$ x = ±7 x = 7, -7 Hence, from the above, We can conclude that The error did Suri made is that he did not consider the negative square root of 49 Question 3. Construct Arguments There is an error in the work shown below. Explain the error and provide a correct solution. x3 = 125 $$\sqrt[3]{x^{3}}$$ = $$\sqrt[3]{125}$$ x = 5 and x = -5 Answer: The given equation is: x³ = 125 Now, We know that, The cube root of a number will always be positive So, $$\sqrt[3]{x^{3}}$$ = $$\sqrt[3]{125}$$ x = $$\sqrt[3]{x^{3}}$$ = $$\sqrt[3]{5 × 5 × 5}$$ x = 5 Hence, from the above, We can conclude that The error is not considering that the cube root of a number will always be positive The correct solution is: x = 5 Question 4. Why are the solutions to x2 = 17 irrational? Answer: The given equation is: x² = 17 So, x = ±$$\sqrt{17}$$ Now, We know that, The perfect square will be an integer and we know that an integer is a rational number Now, When we look at 17, It is not a perfect square Hence, from the above, We can conclude that The solutions of the given equation are irrational Do You Know How? Question 5. If a cube has a volume of 27 cubic cm, what is the length of each edge? Use the volume formula, V = s3, and show your work. Answer: It is given that A cube has a volume of 27 cubic cm Now, We know that, The volume of a cube (V) = Side³ So, Side = $$\sqrt[3]{V}$$ Now, According to the given information, Side = $$\sqrt[3]{27}$$ Side = $$\sqrt[3]{3 × 3 × 3}$$ Side = 3 cm Hence, from the above, We can conclude that The length of each edge of the give cube is: 3 cm Question 6. Darius is building a square launch pad for a rocket project. If the area of the launch pad is 121 square cm, what is its side length? Use the area formula, A = s2, and show your work. Answer: It is given that Darius is building a square launch pad for a rocket project. If the area of the launch pad is 121 square cm, Now, We know that, The area of a square (A) = Side² So, Side = $$\sqrt{A}$$ Now, According to the given information, Side = $$\sqrt{121}$$ Side = $$\sqrt{11 × 11}$$ Side = 11 cm Hence, from the above, We can conclude that The side length of the launch pad is: 1 cm Question 7. Solve the equation x3 = -215. Answer: The given equation is: x³ = -215 Now, We know that, The square root won’t accept negative values but the cube root will accept both positive and negative values So, x = $$\sqrt[3]{-215}$$ Now, Hence, from the above, We can conclude that The solution for the given equation is: -5.99 Practice & Problem Solving Leveled Practice In 8 and 9, solve. Question 8. z2 = 1 Answer: The given equation is: z² = 1 Now, Hence, from the above, We can conclude that Question 9. a3 = 216 Answer: The given equation is: a³ = 216 Now, Hence, from the above, We can conclude that The solution for the given equation is: 6 Question 10. Solve v2 = 47. Answer: The give equation is: v² = 47 Now, $$\sqrt{v²}$$ = $$\sqrt{47}$$ v = ±$$\sqrt{47}$$ Now, Hence, from the above, We can conclude that The solutions for the given equation are: 6.85, and -6.85 Question 11. The area of a square photo is 9 square inches. How long is each side of the photo? Answer: It is given that The area of a square photo is 9 square inches Now, We know that, The area of a square (A) = Side² (s) So, s² = 9 $$\sqrt{s²}$$ = $$\sqrt{9}$$ s = 3 inches [Because the length of the side will never be negative] Hence, from the above, We can conclude that The length of each side of the photo is: 3 inches Question 12. Solve the equation y2 = 81. Answer: The given equation is: y² = 81 Now, $$\sqrt{y²}$$ = ±$$\sqrt{81}$$ y = ±9 Hence, from the above, We can conclude that The solutions for the given equation are: 9, and -9 Question 13. Solve the equation w3 = 1,000. Answer: The given equation is: w³ = 1,000 Now, $$\sqrt[3]{w³}$$ = $$\sqrt[3]{1,000}$$ w = $$\sqrt[3]{10 × 10 × 10}$$ w = 10 Hence, from the above, We can conclude that The solution for the given equation is: 10 Question 14. The area of a square garden is shown. How long is each side of the garden? Answer: It is given that The area of a square garden is shown. Now, The given figure is: Now, From the given figure, We can observe that The area of the square garden is: 121 ft² Now, We know that, The area of a square (A) = s² So, According to the given information, s² = 121 $$\sqrt{s²}$$ = $$\sqrt{121}$$ s = 11 ft [Since the length of the side will never be negative] Hence, from the above, We can conclude that The length of each side of the garden is: 11 ft Question 15. Solve b2 = 77. Answer: The given equation is: b² = 77 Now, $$\sqrt{b²}$$ = ±$$\sqrt{77}$$ Now, Hence, from the above, We can conclude that The solutions for the givene quation are: 8.77, and -8.77 Question 16. Find the value of c in the equation c3 = 1,728. Answer: The given equation is: c³ = 1,728 Now, $$\sqrt[3]{c³}$$ = $$\sqrt[3]{1,728}$$ c = 12 Hence, from the above, We can conclude that The value of c for the given equation is: 12 Question 17. Solve the equation v3 = 12. Answer: The given equation is: v³ = 12 Now, $$\sqrt[3]{v³}$$ = $$\sqrt[3]{12}$$ Now, Hence, from the above, We can conclude that The solution for the given equation is: 2.28 Question 18. Higher Order Thinking Explain why $$\sqrt[3]{-\frac{8}{27}}$$ is –$$\frac{2}{3}$$ Answer: The given equation is: $$\sqrt[3]{-\frac{8}{27}}$$ Now, We know that, $$\sqrt[3]{\frac{a}{b}}$$ = $$\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$$ So, $$\sqrt[3]{-\frac{8}{27}}$$ = –$$\frac{\sqrt[8]{a}}{\sqrt[3]{27}}$$ = –$$\frac{\sqrt[2 × 2 ×2]{a}}{\sqrt[3]{3 ×3 × 3}}$$ = –$$\frac{2}{3}$$ Hence, from the above, We can conclude that $$\sqrt[3]{-\frac{8}{27}}$$ is –$$\frac{2}{3}$$ due to the below property of Exponents: $$\sqrt[3]{\frac{a}{b}}$$ = $$\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$$ Question 19. Critique Reasoning Manolo says that the solution of the equation g2 = 36 is g = 6 because 6 × 6 = 36. Is Manolo’s reasoning complete? Explain. Answer: It is given that Manolo says that the solution of the equation g2 = 36 is g = 6 because 6 × 6 = 36 Now, The given equation is: g² = 36 Now, We know that, The square of a number will always be positive but the square root of a number will either be positive or negative So, $$\sqrt{g²}$$ = ±$$\sqrt{36}$$ g = ±6 So, The solutions for the given equation are: 6, -6 Hence, from the above, We can conclude that Manolo’s reasoning is not complete Question 20. Evaluate $$\sqrt[3]{-512}$$. a. Write your answer as an integer. Answer: The given equation is: $$\sqrt[3]{-512}$$ Now, $$\sqrt[3]{-512}$$ = $$\sqrt[3]{(-8) × (-8) × (-8)}$$ = -8 Hence, from the above, We can conclude that The value of $$\sqrt[3]{-512}$$ as an integer is: -8 b. Explain how you can check that your result is correct. Answer: The given equation is: $$\sqrt[3]{-512}$$ Now, From part (a), We get the value of the given equation is: 8 Now, $$\sqrt[3]{(-8) × (-8) × (-8)}$$ = $$\sqrt[3]{64 × (-8)}$$ = $$\sqrt[3]{-512}$$ Hence, from the above, We can conclude that The result is correct because the givene quation and the result are the same Question 21. Yael has a square-shaped garage with 228 square feet of floor space. She plans to build an addition that will increase the floor space by 50%. What will be the length, to the nearest tenth, of one side of the new garage? Answer: It is given that Yael has a square-shaped garage with 228 square feet of floor space. She plans to build an addition that will increase the floor space by 50%. Now, According to the given information, 50% of 228 = $$\frac{50}{100}$$ × 228 = $$\frac{50 × 228}{100}$$ = 114 square feet So, The area of the new garage = 228 + 114 = 342 square feet Now, We know that, The area of a square (A) = Side (s)² So, s² = 342 $$\sqrt{s²}$$ = $$\sqrt{342}$$ Now, Hence, from the above, We can conclude that The length of one side of the new garage is: 18.5 feet Assessment Practice Question 22. The Traverses are adding a new room to their house. The room will be a cube with a volume of 6,859 cubic feet. They are going to put in hardwood floors, which costs 10 per square foot. How much will the hardwood floors cost? Answer: It is given that The Traverses are adding a new room to their house. The room will be a cube with a volume of 6,859 cubic feet. They are going to put in hardwood floors, which costs 10 per square foot Now, We know that, The volume of a cube (V) = s³ So, According to the given information, s³ = 6,859 $$\sqrt[3]{s³}$$ = $$\sqrt[3]{6,859}$$ s = 19 feet So, The length of each edge of the new room is: 19 feet Now, To find the total cost of hardwood floors, find the perimeter of the room and multiply the result with the cost per square foot Now, We know that, The perimeter of a cube = 6s So, The perimeter of the cube (p) = 6 × 19 = 114 feet Now, The total cost of hardwood floors = 114 × 10 = 1,140 Hence, from the above, We can conclude that The total cost of hardwood floors is: 1,140 Question 23. While packing for their cross-country move, the Chen family uses a crate that has the shape of a cube. PART A If the crate has the volume V = 64 cubic feet, what is the length of one edge? It is given that While packing for their cross-country move, the Chen family uses a crate that has the shape of a cube and the crate has the volume V = 64 cubic feet Now, We know that, The volume of a cube (V) = s³ So, According to the given information, s³ = 64 $$\sqrt[3]{s³}$$ = $$\sqrt[3]{64}$$ s = 4 feet Hence, from the above, We can conclude that The length of each edge of the crate is: 4 feet PART B The Chens want to pack a large, framed painting. If the framed painting has the shape of a square with an area of 12 square feet, will the painting fit flat against a side of the crate? Explain. Answer: It is given that The Chens want to pack a large, framed painting. If the framed painting has the shape of a square with an area of 12 square feet Now, We know that, The area of a square (A) = s² So, According to the given information, s² = 12 Now, $$\sqrt{s²}$$ = $$\sqrt{12}$$ Now, So, According to the given information, The side of the crate > The side of the painting 4 > 2.28 Hence, from the above, We can conclude that The painting will fit flat against a side of the crate ### Topic 1 MID-TOPIC CHECKPOINT Question 1. Vocabulary How can you show that a number is a rational number? Lesson 1.2 Answer: A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers Question 2. Which shows 0.2$$\overline{3}$$ as a fraction? Lesson 1.1 A. $$\frac{2}{33}$$ B. $$\frac{7}{33}$$ C. $$\frac{23}{99}$$ D. $$\frac{7}{30}$$ Answer: The given expression is: 0.2$$\overline{3}$$ Hence, from the above, We can conclude that The options that show 0.2$$\overline{3}[/latex as a fraction is: Question 3. Approximate [latex]\sqrt{8}$$ to the nearest hundredth. Show your work. Lesson 1.3 Answer: The given expression is: $$\sqrt{8}$$ Now, Hence, from the above, We can conclude that The approximate value of the given expression to the nearest hundredth is: 2.82 Question 4. Solve the equation m2 = 14. Lesson 1.5 Answer: The given equation is: m² = 14 Now, $$\sqrt{m²}$$ = ±$$\sqrt{14}$$ Now, Hence, from the above, We can conclude that The solutions for the given equation are: 3.74, -3.74 Question 5. A fish tank is in the shape of a cube. Its volume is 125 ft3. What is the area of one face of the tank? Lessons 1.4 and 1.5 Answer: It is given that A fish tank is in the shape of a cube. Its volume is 125 ft3. Now, We know that, The volume of a cube (V) = s³ So, s³= 125 Now, $$\sqrt[3]{s³}$$ = $$\sqrt[3]{125}$$ s = $$\sqrt[3]{5 × 5 × 5}$$ s = 5 ft Now, We know that, The surface area of a cube = 4s² So, The area of one face of the tank = 4 × 5² = 4 × 25 = 100 ft² Hence, from the above, We can conclude that The area of one face of the tank is: 100 ft² Question 6. Write 1.$$\overline{12}$$ as a mixed number. Show your work. Lesson 1.1 Answer: The given expression is: 1.$$\overline{12}$$ Now, The representation of the given expression in the form of a fraction is: $$\frac{28}{25}$$ Now, The representation of $$\frac{28}{25}$$ into a mixed number is: 1$$\frac{3}{25}$$ Hence, from the above, We can conclude that The representation of 1.$$\overline{12}$$ as a mixed number is: 1$$\frac{3}{25}$$ How well did you do on the mid-topic checkpoint? Fill in the stars. ### Topic 1 MID-TOPIC PERFORMANCE TASK Six members of the math club are forming two teams for a contest. The teams will be determined by having each student draw a number from a box. PART A The table shows the results of the draw. The students who drew rational numbers will form the team called the Tigers. The students who drew irrational numbers will form the team called the Lions. List the members of each team. Answer: It is given that Six members of the math club are forming two teams for a contest. The teams will be determined by having each student draw a number from a box. Now, The given table is: Now, We know that, The numbers that can be written in the form of $$\frac{p}{q}$$ are “Rational numbers” The numbers that can not be written in the form of $$\frac{p}{q}$$ are “Irrational numbers” Now, From the given table, The list of rational numbers are: 6.$$\overline{34}$$, $$\sqrt{36}$$, 6.3$$\overline{4}$$ The list of Irrational numbers are: $$\sqrt{38}$$, 6.343443444….,, $$\sqrt{34}$$ Hence, from the above, We can conclude that The students present in the Tigers Team are: 6.$$\overline{34}$$, $$\sqrt{36}$$, 6.3$$\overline{4}$$ The students present in the Lions Team are: $$\sqrt{38}$$, 6.343443444….,, $$\sqrt{34}$$ PART B The student on each team who drew the greatest number will be the captain of that team. Who will be the captain of the Tigers? Show your work. Answer: From Part A, We can observe that The students present in the Tigers Team are: 6.$$\overline{34}$$, $$\sqrt{36}$$, 6.3$$\overline{4}$$ Now, From the above list of numbers, We can observe that 6.$$\overline{34}$$ is the greater number Hence, from the above, We can conclude that The captain of the Tigers Team is: Anya PART C Who will be the captain of the Lions? Show your work. Answer: The students present in the Lions Team are: $$\sqrt{38}$$, 6.343443444….,, $$\sqrt{34}$$ Now, From the above list of numbers, We can observe that 6.343443444…., is the greater number Hence, from the above, We can conclude that The captain of the Lions Team is: Ryan ### Lesson 1.6 Use Properties of Integer Exponents Solve & Discuss It! One band’s streaming video concert to benefit a global charity costs 1.00 to view. The first day, the concert got 2,187 views. The second day, it got about three times as many views. On the third day, it got 3 times as many views as on the second day. If the trend continues, how much money will the band raise on Day 7? Answer: It is given that One band’s streaming video concert to benefit a global charity costs 1.00 to view. The first day, the concert got 2,187 views. The second day, it got about three times as many views. On the third day, it got 3 times as many views as on the second day So, According to the given information, The amount of money got on the first day of concert = (The total number of views) × 1 = 2,187 × 1 = 2,187 The amount of money got on the second day of concert = (The amount of money got on the first day of concert) × 3 = 2,187 × 3 = 6,561 The amount of money got on the third day of concert = (The amount of money got on the second day of concert) × 3 = 6,561 × 3 = 19,683 The amount of money got on the fourth day of concert = (The amount of money got on the third day of concert) × 3 = 19,683 × 3 = 59,049 The amount of money got on the fifth day of concert = (The amount of money got on the fourth day of concert) × 3 = 59,049 × 3 = 1,77,147 The amount of money got on the sixth day of concert = (The amount of money got on the fifth day of concert) × 3 = 1,77,147 × 3 = 5,31,441 The amount of money got on the seventh day of concert = (The amount of money got on the sixth day of concert) × 3 = 5,31,441 × 3 = 15,94,323 Hence, from the above, We can conclude that The amount of money the Band raise on Day 7 is: 15,94,323 Focus on math practices Use Structure Use prime factorization to write an expression equivalent to the amount of money raised by the band on the last day of the week. Answer: From the above, We can observe that The amount of money got on the last day of the wee is: 15,94,323 Now, By using the Prime factorisation method, Hence, from the above, We can conclude that The prime factorisation of 15,94,323 is: 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 ? Essential Question How do properties of integer exponents help you write equivalent expressions? Answer: The properties of integer exponents can be used to write equivalent expressions by combining numeric or algebraic expressions that have a common base, distributing exponents to products and quotients, and simplifying powers of powers. Try It! The local zoo welcomed a newborn African elephant that weighed 34 kg. It is expected that at adulthood, the newborn elephant will weigh approximately 34 times as much as its birth weight. What expression represents the expected adult weight of the newborn elephant? Answer: It is given that The local zoo welcomed a newborn African elephant that weighed 34 kg. It is expected that at adulthood, the newborn elephant will weigh approximately 34 times as much as its birth weight Now, According to the given information, The expected adult weight of the newborn elephant = (The actual weight of the newborn elephant) × (The number of times the expected weight will be as much as its birth weight) = 34 × 34 Now, According to the “Product of Powers Property”, When multiplying two powers with the same bases, add the exponents So, The expected adult weight of the newborn elephant = 34 + 4 = 38 kg Hence, from the above, We can conclude that The expected adult weight of the newborn elephant is: 38 kg Convince Me! Explain why the Product of Powers Property makes mathematical sense. Answer: The Power of a Product rule states that a term raised to a power is equal to the product of its factors raised to the same power Try It! Write equivalent expressions using the properties of exponents. a. (73)2 Answer: The given expression is: (7³)² Now, We know that, The “Power of Powers Property” states that to find the power of a power, multiply the exponents So, (7³)² = 73 × 2 = 76 Hence, from the above, We can conclude that The value of the given expression is: 76 b. (45)3 Answer: The given expression is: (45)3 Now, We know that, The “Power of Powers Property” states that to find the power of a power, multiply the exponents So, (45)3 = 43 × 5 = 415 Hence, from the above, We can conclude that The value of the given expression is: 415 c. 94 × 84 Answer: The given expression is: 94 × 84 Now, We know that, By using the “Power of Products Property”, when multiplying two exponential expressions with the same exponent and different powers, multiply the bases and keep the exponents the same So, 94 × 84 = (9 × 8)4 = 724 Hence, from the above, We can conclude that The value of the given expression is: 724 d. 89 ÷ 83 Answer: The given expression is: 89 ÷ 83 Now, We know that, The “Quotient of powers Property” states that when dividing two exponential expressions with the same base, subtract the exponents So, 89 ÷ 83 = 89 – 3 = 86 Hence, from the above, We can conclude that The value of the given expression is: 86 KEY CONCEPT Use these properties when simplifying expressions with exponents (when a, m, and n ≠ 0). Do You Understand? Question 1. Essential Question How do properties of integer exponents help you write equivalent expressions? Answer: The properties of integer exponents can be used to write equivalent expressions by combining numeric or algebraic expressions that have a common base, distributing exponents to products and quotients, and simplifying powers of powers. Question 2. Look for Relationships If you are writing an equivalent expression for 23 × 24, how many times would you write 2 as a factor? Answer: It is given that you are writing an equivalent expression for 23 × 24 Now, The given expression is: 23 × 24 Now, We know that, The “Power of Powers Property” states that to find the power of a power, multiply the exponents So, 23 × 24 = 23 + 4 = 27 = 2 × 2 × 2 × 2 × 2 × 2 × 2 Hence, from the above, We can conclude that We would write 2 as a factor 7 times Question 3. Construct Arguments Kristen wrote 58 as an expression equivalent to (52)4. Her math partner writes 56. Who is correct? Answer: It is given that Kristen wrote 58 as an expression equivalent to (52)4. Her math partner writes 56 Now, The given expression is: 58 Now, We know that, (am)n = amn Now, According to the above Property, 58 = 54 × 2 = (54)2 = (52)4 Hence, from the above, We can conclude that Kristen is correct Question 4. Critique Reasoning Tyler says that an equivalent expression for 23 × 53 is 109. Is he correct? Explain. Answer: It is given that Tyler says that an equivalent expression for 23 × 53 is 109 Now, The given expression is: 23 × 53 Now, We know that, am × bm = (a × b)m So, 23 × 53 = (2 × 5)3 = 103 So, The equivalent expression for 23 × 53 is: 103 Hence, from the above, We can conclude that Tyler is not correct Do You Know How? Question 5. Write an equivalent expression for 712 × 74. Answer: The given expression is: 712 × 74 Now, We know that, am × an = am + n So, 712 × 74 = (7)12 + 4 = 716 Hence, from the above, We can conclude that The equivalent expression for the given expression is: 716 Question 6. Write an equivalent expression for (82)4. Answer: The given expression is: (82)4 Now, We know that, (am)n = amn So, (82)4 = 82 × 4 = 88 Hence, from the above, We can conclude that The equivalent expression for the given expression is: 88 Question 7. A billboard has the given dimensions. Using exponents, write two equivalent expressions for the area of the rectangle. Answer: It is given that A billboard has the given dimensions. Now, The given figure is: Now, From the given figure, We can observe that The billboard is in the form of a rectangle. Now, From the given figure, We can observe that The length of the billboard is: 7² ft The width of the billboard is: 4² ft Now, We know that, We know that, am × bm = (a × b)m So, The area of the rectangle (A) = 7² × 4² = (7 × 4)² = 28² ft² Hence, from the above, We can conclude that The two equivalent expressions for the area of the rectangle is: 7² × 4² and 28² ft² Question 8. Write an equivalent expression for 189 – 184. Answer: The given expression is: 189 – 184 Now, 189 – 184 = 184 (185 – 1) Hence, from the above, We can conclude that The equivalent expression for the given expression is: 184 (185 – 1) Practice & Problem Solving Leveled Practice In 9-12, use the properties of exponents to write an equivalent expression for each given expression. Question 9. 28 × 24 Answer: The given expression is: 28 × 24 Now, Hence, from the above, We can conclude that The equivalent expression for the given expression is: 212 Question 10. $$\frac{8^{7}}{8^{3}}$$ Answer: The given expression is: $$\frac{87}{83}$$ Now, We know that, $$\frac{am}{an}$$ = am – n So, Hence, from the above, We can conclude that The equivalent expression for the given expression is: 84 Question 11. (34)5 Answer: The given expression is: (34)5 Now, We know that, (am)n = amn Now, Hence, from the above, We can conclude that The equivalent expression for thegiven expression is: 320 Question 12. 39 × 29 Answer: The given expression is: 39 × 29 Now, We know that, am × bm = (a × b)m Now, Hence, from the above, We can conclude that The equivalent expression for the given expression is: (3 × 2)9 Question 13. a. How do you multiply powers that have the same base? Answer: If the two exponential expressions have the same base and different exponents and both are multiplying each other, then add the exponents by keeping the bases same b. How do you divide powers that have the same base? Answer: If the two exponential expressions have the same base and different exponents and both are dividing each other, then subtract the exponents by keeping the bases same c. How do you find the power of a power? Answer: When an exponential expression contains the power of a power, we will multiply both the powers by keeping the base constant d. How do you multiply powers with different bases but the same exponent? Answer: If the two exponential expressions have the same exponent and different bases and both are multiplying each other, then multiply the bases by keeping the exponents same Question 14. Which expressions are equivalent to 211? Select all that apply. $$\frac{2^{23}}{2^{12}}$$ 27 ∙ 24 $$\frac{2^{9}}{2^{2}}$$ 22 ∙ 29 Answer: The given expression is: 211 Hence, from the above, We can conclude that The expressions that are equivalent to 211 are: In 15-18, use the properties of exponents to write an equivalent expression for each given expression. Question 15. (44)3 Answer: The given expression is: (44)3 Now, We know that, (am)n = am × n So, (44)3 = 44 × 3 = 412 Hence, from the above, We can conclude that (44)3 = 412 Question 16. $$\frac{3^{12}}{3^{3}}$$ Answer: The given expression is: $$\frac{3^{12}}{3^{3}}$$ Now, We know that, $$\frac{am}{an}$$ = am – n So, $$\frac{3^{12}}{3^{3}}$$ = 312 – 3 = 39 Hence, from the above, We can conclude that $$\frac{3^{12}}{3^{3}}$$ = 39 Question 17. 45 × 42 Answer: The given expression is: 45 × 42 Now, We know that, am × an = am + n So, 45 × 42 = 45 + 2 = 47 Hence, from the above, We can conclude that 45 × 42 = 47 Question 18. 64 × 24 Answer: The given expression is: 64 × 24 Now, We know that, am × bm = (a × b)m So, 64 × 24 = (6 × 2)4 = 124 Hence, from the above, We can conclude that 64 × 24 = 124 Question 19. Critique Reasoning Alberto incorrectly stated that $$\frac{5^{7}}{5^{4}}$$ = 13. What was Alberto’s error? Explain your reasoning and find the correct answer. Answer: It is given that Alberto incorrectly stated that $$\frac{5^{7}}{5^{4}}$$ = 13 Now, The given expression is: $$\frac{5^{7}}{5^{4}}$$ Now, We know that, $$\frac{am}{an}$$ = am – n> So, $$\frac{5^{7}}{5^{4}}$$ = 5 7 – 4 = 53 So, From the above, We can observe that Alberto applied the “Quotient of Powers Property” incorrectly Hence, from the above, We can conclude that Alberto’s error is: Alberto applied the “Quotient of Powers Property” incorrectly Question 20. Is the expression 8 × 85 equivalent to (8 × 8)5? Explain. Answer: Question 21. Is the expression (32)-3 equivalent to (33)-2? Explain. Answer: Question 22. Is the expression 32 ∙ 3-3 equivalent to 33 ∙ 3-2? Explain. Answer: Question 23. Model with Math What is the width of the rectangle written as an exponential expression? Answer: Question 24. Simplify the expression $$\left(\left(\frac{1}{2}\right)^{3}\right)^{3}$$. Answer: Question 25. Higher Order Thinking Use a property of exponents to write (3b)5 as a product of powers. Answer: Assessment Practice Question 26. Select all the expressions equivalent to 45 ∙ 410 45 + 410 43 ∙ 45 43 ∙ 412 43 + 412 418 – 43 415 Question 27. Your teacher asks the class to evaluate the expression (23)1. Your classmate gives an incorrect answer of 16. PART A Evaluate the expression. PART B What was the likely error? A. Your classmate divided the exponents. B. Your classmate multiplied the exponents. C. Your classmate added the exponents. D. Your classmate subtracted the exponents. Answer: ### Lesson 1.7 More Properties of Integer Exponents ACTIVITY Explore It! Calvin and Mike do sit-ups when they work out. They start with 64 sit-ups for the first set and do half as many each subsequent set. Look for Relationships Determine whether the relationship shown for Set 1 is also true for Sets 2-5. A. What representation can you use to show the relationship between the set number and the number of sit-ups? B. What conclusion can you make about the relationship between the number of sit-ups in each set? Focus on math practices Use Structure How could you determine the number of sit-up sets Calvin and Mike do? ? Essential Question What do the Zero Exponent and Negative Exponent Properties mean? Try It! Evaluate a. (-7)0 b. (43)0 c. 10 d. (0.50) Convince Me! Why is 2(70) = 2? Try It! Write each expression using positive exponents. a. 8-2 b. 2-4 c. 3-5 Answer: Try It! Write each expression using positive exponents. a. $$\frac{1}{5^{-3}}$$ b. $$\frac{1}{2^{-6}}$$ KEY CONCEPT Use these additional properties when simplifying or generating equivalent expressions with exponents (when a ≠ 0 and n ≠ 0). Zero Exponent Property a0 = 1 Negative Exponent Property a-n = $$\frac{1}{a^{n}}$$ Do You Understand? Question 1. Essential Question What do the Zero Exponent and Negative Exponent Properties mean? Question 2. Reasoning In the expression 9-12, what does the negative exponent mean? Question 3. Reasoning in the expression 3(20), what is the order of operations? Explain how you would evaluate the expression. Do You Know How? Question 4. Simplify 1,999,9990. Answer: Question 5. a. Write 7-6 using a positive exponent. b. Rewrite $$\frac{1}{10^{-3}}$$ using a positive exponent. Answer: Question 6. Evaluate 27xoy-2 for x = 4 and y = 3. Answer: Practice & Problem Solving Leveled Practice In 7-8, complete each table to find the value of a nonzero number raised to the power of 0. Question 7. Answer: Question 8. Answer: Question 9. Given: (-3.2)0 a. Simplify the given expression. b. Write two expressions equivalent to the given expression. Explain why the three expressions are equivalent. Answer: Question 10. Simplify each expression for x = 6. a. 12x0(x-4) b. 14(x-2) In 11 and 12, compare the values using >,<, or =. Question 11. 3-2 1 Answer: Question 12. $$\left(\frac{1}{4}\right)^{0}$$ 1 In 13 and 14, rewrite each expression using a positive exponent. Question 13. 9-4 Answer: Question 14. $$\frac{1}{2^{-6}}$$ Answer: Question 15. Given: 9y0 a. Simplify the expression for y = 3. b. Construct Arguments Will the value of the given expression vary depending on y? Explain. Answer: Question 16. Simplify each expression for x = 4. a. -5x-4 b. 7x-3 Answer: Question 17. Evaluate each pair of expressions. a. (-3)-8 and -3-8 b. (-3)-9 and -3-9 Answer: Question 18. Be Precise To win a math game, Lamar has to pick a card with an expression that has a value greater than 1. The card Lamar chooses reads $$\left(\frac{1}{2}\right)^{-4}$$. Does Lamar win the game? Explain. Answer: Question 19. Simplify the expression. Assume that x is nonzero. Your answer should have only positive exponents. x-10 ∙ x6 Answer: Question 20. Higher Order Thinking a. Is the value of the expression $$\left(\frac{1}{4^{-3}}\right)^{-2}$$ greater than 1, equal to 1, or less than 1? b. If the value of the expression is greater than 1, show how you can change one sign to make the value less than 1. If the value is less than 1, show how you can change one sign to make the value greater than 1. If the value is equal to 1, show how you can make one change to make the value not equal to 1. Assessment Practice Question 21. Which expressions are equal to 5-3? Select all that apply. 125 125-1 53 $$\frac{1}{5^{3}}$$ $$\frac{1}{125}$$ Answer: Question 22. Which expressions have a value less than 1 when x = 4? Select all that apply. $$\left(\frac{3}{x^{2}}\right)^{0}$$ $$\frac{x^{0}}{3^{2}}$$ $$\frac{1}{6^{-x}}$$ $$\frac{1}{x^{-3}}$$ 3x-4 ### Lesson 1.8 Use Powers of 10 to Estimate Quantities ACTIVITY Explain It! Keegan and Jeff did some research and found that there are approximately 7,492,000,000,000,000,000 grains of sand on Earth. Jeff says that it is about 7 × 1015 grains of sand. Keegan says that this is about 7 × 1018 grains of sand. A. How might Jeff have determined his estimate? How might Keegan have determined his estimate? B. Whose estimate, Jeff’s or Keegan’s, is more logical? Explain. Focus on math practices Be Precise Do you think the two estimates are close in value? Explain your reasoning. ? Essential Question when would you use a power of 10 to estimate a quantity? Try It! Light travels 299,792,458 meters per second. Sound travels at 332 meters per second. Use a power of 10 to compare the speed of light to the speed of sound. 299,792,458 rounded to the greatest place value is There are zeros in the rounded number. The estimated speed of light is × 10 meters per second. 3 × 10 > 3 × 10, so the speed of light is faster than the speed of sound. 322 rounded to the greatest place value is There are zeros in the rounded number. The estimated speed of sound is × 10 meters per second. Convince Me! Country A has a population of 1,238,682,005 and Country B has a population of 1,106,487,394. How would you compare these populations? Try It! There are approximately 1,020,000,000 cars in the world. The number of cars in the United States is approximately 239,800,000. Compare the number of cars in the world to that in the United States. KEY CONCEPT You can estimate a very large or very small number by rounding the number to its greatest place value, and then writing that number as a single digit times a power of 10. Do You Understand? Question 1. ? Essential Question when would you use powers of 10 to estimate a quantity? Answer: Question 2. Construct Arguments Kim writes an estimate for the number 0.00436 as 4 × 103. Explain why this cannot be correct. Answer: Question 3. Be Precise Raquel estimated 304,900,000,000 as 3 × 108. What error did she make? Answer: Do You Know How? Question 4. Use a single digit times a power of 10 to estimate the height of Mt. Everest to the nearest ten thousand feet. Answer: Question 5. A scientist records the mass of a proton as 0.0000000000000000000000016726231 gram. Use a single digit times a power of 10 to estimate the mass. Answer: Question 6. The tanks at the Georgia Aquarium hold approximately 8.4 × 106 gallons of water. The tanks at the Audubon Aquarium of the Americas hold about 400,000 gallons of water. Use a single digit times a power of 10 to estimate how many times greater the amount of water is at the Georgia Aquarium. Answer: Practice & Problem Solving Leveled Practice in 7-9, use powers of 10 to estimate quantities. Question 7. A city has a population of 2,549,786 people. Estimate this population to the nearest million. Express your answer as the product of a single digit and a power of 10. Rounded to the nearest million, the population is about Written as the product of a single digit and a power of ten, this number is Answer: Question 8. Use a single digit times a power of 10 to estimate the number 0.00002468. Rounded to the nearest hundred thousandth, the number is about Written as a single digit times a power of ten, the estimate is Answer: Question 9. The approximate circumferences of Earth and Saturn are shown. How many times greater is the circumference of Saturn than the circumference of Earth? The circumference of Saturn is Saturn’s circumference is about times greater than the circumference of Earth. Answer: Question 10. Estimate 0.037854921 to the nearest hundredth. Express your answer as a single digit times a power of ten. Answer: Question 11. Compare the numbers 6 × 10-6 and 2 × 10-8. a. Which number has the greater value? b. Which number has the lesser value? c. How many times greater is the greater number? Answer: Question 12. Taylor made 43,785 last year. Use a single digit times a power of ten to express this value rounded to the nearest ten thousand. Answer: Question 13. The length of plant cell A is 8 × 10-5 meter. The length of plant cell B is 0.000004 meter. How many times greater is plant cell A’s length than plant cell B’s length? Answer: Question 14. Critique Reasoning The diameter of one species of bacteria is shown. Bonnie approximates this measure as 3 × 10-11 meter. Is she correct? Explain. Answer: Question 15. The populations of Cities A and B are 2.6 × 105 and 1,560,000, respectively. The population of City C is twice the population of City B. The population of City C is how many times the population of City A? Answer: Assessment Practice Question 16. Earth is approximately 5 × 109 years old. For which of these ages could this be an approximation? A. 4,762,100,000 years B. 48,000,000,000 years C. 4.45 × 109 years D. 4.249999999 × 109 years Answer: Question 17. PART A Express 0.000000298 as a single digit times a power of ten rounded to the nearest ten millionth. PART B Explain how negative powers of 10 can be helpful when writing and comparing small numbers. Answer: ### Lesson 1.9 Understand Scientific Notation ACTIVITY Solve & Discuss It! Scientists often write very large or very small numbers using exponents. How might a scientist write the number shown using exponents? Use Structure How can you use your knowledge of powers of 10 to rewrite the number? Focus on math practices Look for Relationships What does the exponent in 1015 tell you about the value of the number? ? Essential Question What is scientific notation and why is it used? Try It! The height of Angel Falls, the tallest waterfall in the world, is 3,212 feet. How do you write this number in scientific notation? Convince Me! Why do very large numbers have positive exponents when written in scientific notation? Explain. Try It! A common mechanical pencil lead measures about 0.005 meter in diameter. How can you express this measurement using scientific notation ? Try It! Write the numbers in standard form. a. 9.225 × 1018 b. 6.3 × 10-8 Answer: KEY CONCEPT Scientific notation is a way to write very large numbers or very small numbers. Scientists use scientific notation as a more efficient and convenient way of writing such numbers. A number in scientific notation is the product of two factors. The first factor must be greater than or equal to 1 and less than 10. The second factor is a power of 10. To write a number in scientific notation in standard form, multiply the decimal number by the power of 10. Do You Understand? Question 1. ?Essential Question What is scientific notation and why is it used? Answer: Question 2. Critique Reasoning Taylor states that 2,800,000 in scientific notation is 2.8 × 10-6 because the number has six places to the right of the 2. Is Taylor’s reasoning correct? Answer: Question 3. Construct Arguments Sam will write 0.000032 in scientific notation. Sam thinks that the exponent of 10 will be positive. Do you agree? Construct an argument to support your response. Answer: Do You Know How? Question 4. Express 586,400,000 in scientific notation. Answer: Question 5. The genetic information of almost every living thing is stored in a tiny strand called DNA. Human DNA is 3.4 × 10-8 meter long. Write the length in standard form. Answer: Question 6. The largest virus known to man is the Megavirus, which measures 0.00000044 meter across. Express this number in scientific notation. Answer: Question 7. How would you write the number displayed on the calculator screen in standard form? Answer: Practice & Problem Solving Leveled Practice In 8 and 9, write the numbers in the correct format. Question 8. The Sun is 1.5 × 108 kilometers from Earth. 1.5 × 108 is written as in standard form. Answer: Question 9. Brenna wants an easier way to write 0.0000000000000000587. 0.0000000000000000587 is written as × 10 in scientific notation. Answer: Question 10. Is 23 × 10-8 written in scientific notation? Justify your response. Answer: Question 11. Is 8.6 × 107 written in scientific notation? Justify your response. Answer: Question 12. Simone evaluates an expression using her calculator. The calculator display is shown at the right. Express the number in standard form. Answer: Question 13. Express the number 0.00001038 in scientific notation. Answer: Question 14. Express the number 80,000 in scientific notation. Answer: Question 15. Peter evaluates an expression using his calculator. The calculator display is shown at the right. Express the number in standard form. Answer: Question 16. a. What should you do first to write 5.871 × 10-7 in standard form? b. Express the number in standard form Answer: Question 17. Express 2.58 × 10-2 in standard form. Answer: Question 18. At a certain point, the Grand Canyon is approximately 1,600,000 centimeters across. Express this number in scientific notation. Answer: Question 19. The length of a bacterial cell is 5.2 × 10-6 meter. Express the length of the cell in standard form. Answer: Question 20. Higher Order Thinking Express the distance 4,300,000 meters using scientific notation in meters, and then in millimeters. Answer: Assessment Practice Question 21. Which of the following numbers are written in scientific notation? 12 × 106 12 6.89 × 106 6.89 0.4 4 × 10-1 Answer: Question 22. Jeana’s calculator display shows the number to the right. PART A Express this number in scientific notation. PART B Express this number in standard form. ### 3-Act Mathematical Modeling: Hard-Working Organs 3-ACT MATH ACT 1 Question 1. After watching the video, what is the first question that comes to mind? Answer: Question 2. Write the Main Question you will answer. Answer: Question 3. Construct Arguments Predict an answer to this Main Question. Explain your prediction. Answer: Question 4. On the number line below, write a number that is too small to be the answer. Write a number that is too large. Answer: Question 5. Plot your prediction on the same number line. АСТ 2 Question 6. What information in this situation would be helpful to know? How would you use that information? Answer: Question 7. Use Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically. Answer: Question 8. Model with Math Represent the situation using mathematics. Use your representation to answer the Main Question. Answer: Question 9. What is your answer to the Main Question? Is it greater or less than your prediction? Explain why. Answer: ACT 3 Question 10. Write the answer you saw in the video. Answer; Question 11. Reasoning Does your answer match the answer in the video? If not, what are some reasons that would explain the difference? Answer: Question 12. Make Sense and Persevere Would you change your model now that you know the answer? Explain. Answer: ACT 3 Reflect Question 13. Model with Math Explain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question? Answer: Question 14. Generalize What pattern did you notice in your calculations? How did that pattern help you solve the problem? Answer: SEQUEL Question 15. Use Structure How many times does a heart beat in a lifetime? Use your solution to the Main Question to help you solve. Answer: ### Lesson 1.10 Operations with Numbers in Scientific Notation Solve & Discuss It! The homecoming committee wants to fly an aerial banner over the football game. The banner is 1,280 inches long and 780 inches tall. How many different ways can the area of the banner be expressed? Focus on math practices Be Precise Which of the solutions is easiest to manipulate? ? Essential Question How does using scientific notation help when computing with very large or very small numbers? Try It! The planet Venus is on average 2.5 × 107 kilometers from Earth. The planet Mars is on average 2.25 × 108 kilometers from Earth. When Venus, Earth, and Mars are aligned, what is the average distance from Venus to Mars? Answer: Convince Me! In Example 1 and the Try it, why did you move the decimal point to get the final answer? Try It! There are 1 × 1014 good bacteria in the human body. There are 2.6 x 1018 good bacteria among the spectators in a soccer stadium. About how many spectators are in the stadium? Express your answer in scientific notation. KEY CONCEPT Operations with very large or very small numbers can be carried out more efficiently using scientific notation. The properties of exponents apply when carrying out operations. Do You Understand? Question 1. ? Essential Question How does using scientific notation help when computing with very small or very large numbers? Question 2. Use Structure When multiplying and dividing two numbers in scientific notation, why do you sometimes have to rewrite one factor? Answer: Question 3. Use Structure For the sum of (5.2 × 104) and (6.95 × 104) in scientific notation, why will the power of 10 be 105? Answer: Do You Know How? Question 4. A bacteriologist estimates that there are 5.2 × 104 bacteria growing in each of 20 petri dishes. About how many bacteria in total are growing in the petri dishes? Express your answer in scientific notation. Answer: Question 5. The distance from Earth to the Moon is approximately 1.2 × 109 feet. The Apollo 11 spacecraft was approximately 360 feet long. About how many spacecraft of that length would fit end to end from Earth to the Moon? Express your answer in scientific notation. Answer: Question 6. The mass of Mars is 6.42 × 1023 kilograms. The mass of Mercury is 3.3 × 1023 kilograms. a. What is the combined mass of Mars and Mercury expressed in scientific notation? b. What is the difference in the mass of the two planets expressed in scientific notation? Answer: Practice & Problem Solving Leveled Practice In 7 and 8, perform the operation and express your answer in scientific notation. Question 7. (7 × 10-6)(7 × 10-6) Answer: Question 8. (3.76 × 105) + (7.44 × 105) Answer: Question 9. What is the value of n in the equation 1.9 × 107 = (1 × 105)(1.9 × 10n)? Answer: Question 10. Find (5.3 × 103) – (8 × 102). Express your answer in scientific notation. Answer: Question 11. What is the mass of 30,000 molecules? Express your answer in scientific notation. Answer: Question 12. Critique Reasoning Your friend says that the product of 4.8 × 108 and 2 × 10-3 is 9.6 × 10-5. Is this answer correct? Explain. Answer: Question 13. Find $$\frac{7.2 \times 10^{-8}}{3 \times 10^{-2}}$$. Write your answer in scientific white notation. Answer: Question 14. A certain star is 4.3 × 102 light years from Earth. One light year is about 5.9 × 1012 miles. How far from Earth (in miles) is the star? Express your answer in scientific notation. Question 15. The total consumption of fruit juice in a particular country in 2006 was about 2.28 × 109 gallons. The population of that country that year was 3 × 108. What was the average number of gallons consumed per person in the country in 2006? Answer: Question 16. The greatest distance between the Sun and Jupiter is about 8.166 × 108 kilometers. The greatest distance between the Sun and Saturn is about 1.515 × 109 kilometers. What is the difference between these two distances? Answer: Question 17. What was the approximate number of pounds of garbage produced per person in the country in one year? Express your answer in scientific notation. Answer: Question 18. Higher Order Thinking a. What is the value of n in the equation 1.5 × 1012 = (5 × 105)(3 × 10n)? b. Explain why the exponent on the left side of the equation is not equal to the sum of the exponents on the right side. Assessment Practice Question 19. Find (2.2 × 105) ÷ (4.4 × 10-3). When you regroup the factors, what do you notice about the quotient of the decimal factors? How does this affect the exponent of the quotient? Answer: Question 20. Which expression has the least value? A. (4.7 × 104) + (8 × 104) B. (7.08 × 103) + (2.21 × 103) C. (5.43 × 108) – (2.33 × 108) D. (9.35 × 106) – (6.7 × 106) Answer: ### Topic 1 Review ? Topic Essential Question What are real numbers? How are real numbers used to solve problems? Vocabulary Review Use Vocabulary in Writing Use vocabulary words to explain how to find the length of each side of a square garden with an area of 196 square inches. Concepts and Skills Review LESSON 1.1 Rational Numbers as Decimals Quick Review You can write repeating decimals in fraction form by writing two equations. You multiply each side of one equation by a power of 10. Then you subtract the equations to eliminate the repeating decimal. Practice Write each number as a fraction or a mixed number. Question 1. 0.$$\overline{7}$$ Answer: Question 2. 0.0$$\overline{4}$$ Answer: Question 3. 4.$$\overline{45}$$ Question 4. 2.191919…. Answer: LESSON 1.2 Understand Irrational Numbers Quick Review An irrational number is a number that cannot be written in the form $$\frac{a}{b}$$, where a and b are integers and b ≠ 0. Rational and irrational numbers together make up the real number system. Practice Question 1. Determine which numbers are irrational. Select all that apply. $$\sqrt{36}$$ $$\sqrt{23}$$ -4.232323…. 0.151551555…. 0.3$$\overline{5}$$ π Question 2. Classify -0.$$\overline{25}$$ as rational or irrational. Explain. Answer: LESSON 1.3 Compare and Order Real Numbers Quick Review To compare and order real numbers, it helps to first write each number in decimal form. Practice Question 1. Between which two whole numbers does $$\sqrt{89}$$ lie? $$\sqrt{89}$$ is between and Answer: Question 2. Compare and order the following numbers. Locate each number on a number line. 2.$$\overline{3}$$, $$\sqrt{8}$$, 2.5, 2$$\frac{1}{4}$$ Answer: LESSON 1.4 Evaluate Square Roots and Cube Roots Quick Review Remember that a perfect square is the square of an integer. A square root of a number is a number that when multiplied by itself is equal to the original number. Similarly, a perfect cube is the cube of an integer. A cube root of a number is a number that when cubed is equal to the original number. Practice Classify each number as a perfect square, a perfect cube, both, or neither. Question 1. 27 Answer: Question 2. 100 Answer: Question 3. 64 Answer: Question 4. 24 Answer: Question 5. A gift box is a cube with a volume of 512 cubic inches. What is the length of each edge of the box? Answer: LESSON 1.5 Solve Equations Using Square Roots and Cube Roots Quick Review You can use square roots to solve equations involving squares. You can use cube roots to solve equations involving cubes. Equations with square roots often have two solutions. Look at the context to see whether both solutions are valid. Practice Solve for x. Question 1. x3 = 64 Answer: Question 2. x2 = 49 Answer: Question 3. x3 = 25 Answer: Question 4. x2 = 125 Answer: Question 5. A container has a cube shape. It has a volume of 216 cubic inches. What are the dimensions of one face of the container? LESSON 1.6 Use Properties of Integer Exponents Quick Review These properties can help you write equivalent expressions that contain exponents. Product of Powers Property am.an = am+n Power of Powers Property (am)n = amn Power of Products Property an ∙ bn = (a ∙ b)n Quotient of Powers Property am ÷ an = am-n, when a ≠ 0 Practice Use the properties of exponents to write an equivalent expression for each given expression. Question 1. 64 ∙ 63 Answer: Question 2. (36)-2 Answer: Question 3. 73 ∙ 23 Answer: Question 4. 410 ÷ 44 Answer: LESSON 1.7 More Properties of Integer Exponents Quick Review The Zero Exponent Property states that any nonzero number raised to the power of 0 is equal to 1. The Negative Exponent Property states that for any nonzero rational number a and integer n, a-n = $$\frac{1}{a^{n}}$$ Practice Write each expression using positive exponents. Question 1. 9-4 Answer: Question 2. $$\frac{1}{3^{-5}}$$ Answer: Evaluate each expression for x = 2 and y = 5 Question 3. -4x-2 + 3y0 Answer: Question 4. 2x0y-2 Answer: LESSON 1.8 Use Powers of 10 to Estimate Quantities Quick Review You can estimate very large and very small quantities by writing the number as a single digit times a power of 10. Practice Question 1. In the year 2013 the population of California was about 38,332,521 people. Write the estimated population as a single digit times a power of 10. Answer: Question 2. The wavelength of green light is about 0.00000051 meter. What is this estimated wavelength as a single digit times a power of 10? Answer: Question 3. The land area of Connecticut is about 12,549,000,000 square meters. The land area of Rhode Island is about 2,707,000,000 square meters. How many times greater is the land area of Connecticut than the land area of Rhode Island? Answer: LESSON 1.9 Understand Scientific Notation Quick Review A number in scientific notation is written as a product of two factors, one greater than or equal to 1 and less than 10, and the other a power of 10. Practice Question 1. Write 803,000,000 in scientific notation. Answer: Question 2. Write 0.0000000068 in scientific notation. Answer: Question 3. Write 1.359 × 105 in standard form. Answer: Question 4. The radius of a hydrogen atom is 0.000000000025 meter. How would you express this radius in scientific notation? LESSON 1.10 Operations with Numbers in Scientific Notation Quick Review When multiplying and dividing numbers in scientific notation, multiply or divide the first factors. Then multiply or divide the powers of 10. When adding and subtracting numbers in scientific notation, first write the numbers with the same power of 10. Then add or subtract the first factors, and keep the same power of 10. If the decimal part of the result is not greater than or equal to 1 and less than 10, move the decimal point and adjust the exponent. Practice Perform each operation. Express your answers in scientific notation. Question 1. (2.8 × 104) × (4 × 105) Answer: Question 2. (6 × 109) ÷ (2.4 × 103) Question 3. (4.1 × 104) + (5.6 × 106) Answer: Question 4. The population of Town A is 1.26 × 105 people. The population of Town B is 2.8 × 104 people. How many times greater is the population of Town A than the population of Town B? Answer: ### Topic 1 Fluency Practice Crisscrossed Solve each equation. Write your answers in the cross-number puzzle below. Each digit, negative sign, and decimal point of your answer goes in its own box. Across A -377 = x – 1,000 B x3 = 1,000 C x3 = -8 D x + 7 = -209 F x + 19 = -9 J 14 + x = -9 L m – 2.02 = -0.58 M -3.09 + x = -0.7 N -2.49 = -5 + x Q x – 3.5 = -3.1 T q – 0.63 = 1.16 V 8.3 + x = 12.1 Down A y – 11 = 49 B x + 8 = 20 C z3 = -1,331 D 11 + x = 3 E x – 14 -7.96 F 14 + x = -19 G d + 200 = 95 H x2 = 144 K -12 = t – 15.95 P 0.3 + x = 11 R x – 3 = -21 S – 7 = -70 + y #### enVision Math Common Core Grade 8 Answer Key ## enVision Math Common Core Grade 8 Answer Key Topic 8 Solve Problems Involving Surface Area And Volume ## enVision Math Common Core 8th Grade Answers Key Topic 8 Solve Problems Involving Surface Area And Volume Topic 8 Essential Question How can you find volumes and surface areas of three-dimensional figures? Answer: The “Surface area” is the sum of the areas of all faces (or surfaces) on a 3D shape. Ex: A cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces We know that, The volume of a three-dimensional figure = Cross-sectional area × length 3-ACT MATH Measure Up Have you ever heard of the terms griffin beaker, Erlenmeyer flask, or graduated cylinder? Maybe you’ve used them in your science class. Each piece of equipment in a chemistry lab has a specific purpose, so containers come in many shapes. It’s sometimes necessary to pour a solution from one container to another. Think about this during the 3-Act Mathematical Modeling lesson. Topic 8 ënVision STEM Project Did You Know? The production of packaging is a huge industry employing over five million people with annual sales of more than 400 billion dollars. Packaging materials protect and deliver food and products to consumers. A plastic bottle takes 450–1,000 years to biodegrade. Seabirds are dying of starvation with stomachs full of plastic and Styrofoam. Polystyrene foam lasts forever! Eco-friendly packaging materials are being made from mushrooms and bamboo. There is even a drink bottle made from recyclable paper. New technology results in packaging materials that are both affordable and biodegradable. Environmentally friendly companies are producing sustainable packaging. In addition to using recyclable materials, they reduce the water, natural resources, and energy needed for production. They minimize waste when designing products. Your Task: Wrap it Up! Engineers consider several factors when designing product packaging. These factors include cost efficiency and eco-friendly design so that materials are disposable, recyclable, biodegradable, and not wasted. Suppose you are an engineer working for Liquid Assets, an environmentally friendly company that designs, builds, and packages water purifiers. You and your classmates will use your knowledge of volume and surface area to determine an environmentally sound way to package the purifiers. ### Topic 8 GET READY! Review What You Know! Vocabulary Choose the best term from the box to complete each definition. base diameter radius three-dimensional two-dimensional Question 1. The __________ is the distance from the center to the edge of a circle. Answer: We know that, The “Radius” is the distance from the center to the edge of a circle Hence, from the above, We can conclude that the best term to complete the given definition is: Radius Question 2. A shape that has length, width, and height is ___________. Answer: We know that, A shape that has a length, width, and height is known as “Three-dimensional” Hence, from the above, We can conclude that the best term to complete the given definition is: Three dimensional Question 3. Any side of a cube can be considered a __________. Answer: We know that, Any side of a cube can be considered a “Base” Hence, from the above, We can conclude that the best term to complete the given definition is: Base Question 4. A shape that has length and width, but not height, is ___________. Answer: We know that, A shape that has length and width, but not height is known as “Two-dimensional” Hence, from the above, We can conclude that the best term to complete the given definition is: Two-dimensional Question 5. The _____________ of a circle is a line segment that passes through its center and has endpoints on the circle. Answer: We know that, The “Diameter” of a circle is a line segment that passed through its center and has endpoints on the circle Hence, from the above, We can conclude that the best term to complete the given definition is: Diameter Multiplying with Decimals Find the product. Question 6. 14 ∙ 3.5 = _______ Answer: The given expression is: 14 × 3.5 So, 14 × 3.5 = 49.0 Hence, from the above, We can conclude that the value for the given expression is: 49 Question 7. 9 ∙ 3.14 = _________ Answer: The given expression is: 9 × 3.14 So, 9 × 3.14 = 28.26 Hence, from the above, We can conclude that the value for the given expression is: 28.26 Question 8. 4.2 ∙ 10.5 = _________ Answer: The given expression is: 4.2 × 10.5 So, 4.2 × 10.5 = 44.1 Hence, from the above, We can conclude that the value for the given expression is: 44.1 Areas of Circles Find the area of each circle. Use 3.14 for π. Question 9. A = ________ Answer: The given circle is: From the given circle, The radius is: 8 cm Now, We know that, The area of the circle = πr² So, The area of the given circle = 3.14 × 8² = 200.96 cm² Hence, from the above, We can conclude that the area for the given circle is: 200.96 cm² Question 10. A = _________ Answer: The given circle is: From the given circle, The diameter is: 12 cm Now, We know that, Radius = $$\frac{Diameter}{2}$$ Radius = $$\frac{12}{2}$$ Radius = 6 cm Now, We know that, The area of the circle = πr² So, The area of the given circle = 3.14 × 6² = 113.04 cm² Hence, from the above, We can conclude that the area of the given circle is: 113.04 cm² Use the Pythagorean Theorem Find the missing side length of the triangle. Question 11. x = _________ Answer: The given triangle is: Now, Fro the given figure, We can observe that the triangle is a right triangle Now, We know that, According to the Pythagorean Theorem, c² = a² + b² Where, c is the length of the hypotenuse a and b are the side lengths So, 13² = 12²+ x² x²= 169 – 144 x² = 25 x = $$\sqrt{25}$$ x = 5 in. Hence, from the above, We can conclude that the missing side length of the given triangle is: 5 in. Question 12. x = __________ Answer: The given triangle is: Now, Fro the given figure, We can observe that the triangle is a right triangle Now, We know that, According to the Pythagorean Theorem, c² = a² + b² Where, c is the length of the hypotenuse a and b are the side lengths So, 30² = 24²+ x² x²= 900 – 576 x² = 324 x = $$\sqrt{324}$$ x = 18 in. Hence, from the above, We can conclude that the missing side length of the given triangle is: 18 in. Language Development Complete the word web. Write keywords, ideas, examples, or illustrations that connect to each new vocabulary term. ### Topic 8 Pick A Project PROJECT 8A What makes a concert rock? PROJECT: DESIGN PROPS OR STAGE STRUCTURES PROJECT 8B What is the most interesting museum you have visited? PROJECT: MAKE A MODEL OF A MUSEUM PROJECT 8C Where around the United States can you find quarries? PROJECT: POUR AND MEASURE SAND PROJECT 8D If you were cast in a play, would it be a comedy or a drama? Why? PROJECT: WRITE A SKIT ### Lesson 8.1 Find Surface Area of Three-Dimensional Figures Explore It! Andrea is designing the packaging for a tube-shaped container. I can… find the surface areas of cylinders, cones, and spheres. A. Model with Math What two-dimensional shape represents the top and bottom of the container? What two-dimensional shape represents the tube? Draw a net of the tube-shaped container. Answer: It is given that Andrea is designing the packaging for a tube-shaped container. Now, The given arrangement for tube shaped container is: Now, From the given arrangement, We can observe that The two-dimensional shape that represents the top and bottom of the container is: Circle The two-dimensional shape that represents the tube is: Rectangle Hence, The representation of the tube-shaped container is: B. Look for Relationships The circular top and bottom fit perfectly on the ends of the container. How are the measures of the circles and the rectangle related? Answer: The representation of the tube-shaped container is: Now, From the given figure, We can observe that the tube-shaped container is made up of 2 circles and 1 rectangle So, The total surface area of the tube-shaped container is the sum of the areas of 2 circles and 1 rectangle Now, We know that, The area of a circle = πr² The area of a rectangle = Height× Base Where, The base of the rectangle is a circle So, The circumference of the circle = 2πr Hence, The surface area of the tube-shaped container = 2πr² + 2πrh Focus on math practices Model with Math How can you check whether the net that you drew accurately represents the tube-shaped container? Answer: The representation of the tube-shaped container is: Now, When the top and bottom of the container correctly fit the tube, That is the representation that the net you drew accurately represents the tube-shaped container Essential Question How are the areas of polygons used to find the surface area formulas for three-dimensional figures? Answer: We know that, A three-dimensional figure is a combination of some two-dimensional figures Ex: We can make a “Cuboid” from the combination of “Rectangles” We can make a “Sphere” from the combination of ‘Circles” So, The total surface area of a three-dimensional figure can be given as the sum of all the areas of the two-dimensional figures that are used to make the three-dimensional figure Try it What is the surface area of a cylinder with a height of 9.5 inches and a radius of 2.5 inches? The surface area of the cylinder is __________ square inches. Answer: It is given that The height of a cylinder is: 9.5 inches The radius of a cylinder is: 2.5 inches Now, We know that, The surface area of the cylinder (S.A) = 2πr² + 2πrh = 2 × 3.14 × (2.5)² + 2 × 3.14 × 9.5 × 2.5 = 39.25 + 149.15 = 188.4 square inches Hence, from the above, We can conclude that the surface area of the given cylinder is: 188.4 square inches Convince Me! How can you find the surface area of a cylinder if you only know its height and the circumference of its base? S.A. = 2(πr2) + (2πr)h = 2π(________2) + 2π(_______)(________) = _______π + _______π = _______π Answer: It is given that We know only the height of the cylinder and the circumference of its base Now, We know that, The surface area of a cylinder (S.A) = 2πr² + 2πrh So, S.A = 2π (r²) + 2π (r) (h) S.A = 2π + π S.A = 3π Hence, from the above, We can conclude that the surface area of a cylinder with only its height and the circumference of its base is: 3π Try It! a. What is the surface area of a cone with a radius of 7 feet and a slant height of 9 feet? Use $$\frac{22}{7}$$ for π. Answer: It is given that The radius of the cone is (r): 7 feet The slant height of the cone is (l): 9 feet Now, We know that, The surface area of the cone (S.A) = πr² + πrl So, S.A = $$\frac{22}{7}$$ × 7² + $$\frac{22}{7}$$ × 7 × 9 = 154 + 198 = 352 square feet Hence, from the above, We can conclude that the surface area of the given cone is: 352 square feet b. What is the surface area of a sphere with a diameter of 2.7 inches? Use 3.14 for π. Answer: It is given that, The diameter of a sphere is: 2.7 inches Now, We know that, Radius = $$\frac{Diameter}{2}$$ So, Radius of the sphere = $$\frac{2.7}{2}$$ So, The radius of the sphere (r) = 1.35 inches Now, We know that, The surface area of the sphere (S.A) = 4πr² So, S.A = 4 × 3.14 × (1.35)² = 22.89 square inches Hence, from the above, We can conclude that the surface area of the given sphere is: 22.89 square inches KEY CONCEPT Formulas for finding the area of polygons can be used to find the surface areas of cylinders, cones, and spheres. Do You Understand? Question 1. Essential Question How are the areas of polygons used to find the surface area formulas for three-dimensional figures? Answer: We know that, A three-dimensional figure is a combination of some two-dimensional figures Ex: We can make a “Cuboid” from the combination of “Rectangles” We can make a “Sphere” from the combination of ‘Circles” So, The total surface area of a three-dimensional figure can be given as the sum of all the areas of the two-dimensional figures that are used to make the three-dimensional figure Question 2. Reasoning Why is the length of the base of the rectangle the same as the circumference of the circles in the net of a cylinder? Answer: The representation of the cylinder is: Now, If you look at the net, the curved surface of the cylinder is rectangular in shape. The length of the rectangle is the same as the circumference of the circle. Since the length of the rectangle wraps around the circle rim, it is the same length as the circumference of the circle. Question 3. Construct Arguments Aaron says that all cones with a base circumference of 8 inches will have the same surface area. Is Aaron correct? Explain. Answer: It is given that Aaron says that all cones with a base circumference of 8 inches will have the same surface area Now, We know that, The surface area of a cone = πr² + πrl Now, From the above formula, We can conclude that the surface area of the cone does not depend only on the circumference of the base but also we need the side length of the cone part as well So, All cones with a base circumference of 8 inches will not have the same surface area. Hence, from the above, We can conclude that Aaron is not correct Do You Know How? Question 4. What is the surface area of the cylinder? Use 3.14 for π, and round to the nearest tenth. Answer: The given figure is: From the given figure, The diameter of the cylinder is: 2 mm The height of the cylinder is: 10 mm So, Radius of the cylinder = $$\frac{Diameter of the cylinder}{2}$$ = $$\frac{2}{2}$$ = 1 mm Now, We know that, The surface area of the cylinder (S.A) = 2πr² + 2πrh So, S.A = 2 × 3.14 × 1² + 2 × 3.14 × 1 × 10 = 6.28 + 62.8 = 69.08 mm² Hence, from the above, We can conclude that the surface area of the given cylinder is: 69.08 mm² Question 5. What is the surface area of the cone to the nearest tenth? Use 3.14 for π. Answer: The given figure is: From the given figure, The slant height of the cone is (l): 4 ft The radius of the cone is: 3 ft Now, We know that, The surface area of the cone (S.A) = πr² + πrl So, S.A = 3.14 × 3² + 3.14 × 3 × 4 = 28.26 + 37.68 = 65.94 ft² Hence, from the above, We can conclude that the surface area of the given cone is: 65.94 ft² Question 6. What is the surface area of the sphere in terms of π? Answer: The given figure is: From the given figure, The diameter of the sphere is: 2 cm Now, We know that, The radius of the sphere (r) = $$\frac{Diameter of the sphere}{2}$$ = $$\frac{2}{2}$$ = 1 cm Now, We know that, The surface area of the sphere (S.A) = 4πr² So, S.A = 4 × 3.14 × 1² = 12.56 cm² Hence, from the above, We can conclude that the surface area of the given sphere is: 12.56 cm² Practice & Problem Solving Leveled Practice In 7-8, find the surface area. Question 7. What is the surface area of the cylinder? Use 3.14 for π, and round to the nearest tenth. S.A. = 2(πr2) + (2πr)h = 2π(________2) + 2 π(_______)(________) = 2 π(________) + 2 π(_________) = _______π + _______π = _______π ≈ ________ cm2 Answer: The given figure is: From the given figure, The radius of the cylinder is: 3 cm The height of the cylinder is: 5 cm Now, We know that, The surface area of the cylinder (S.A) = 2πr² + 2πrh So, S.A = 2 × 3.14 × 3² + 2 × 3.14 × 3 × 5 = 56.52 + 94.2 = 150.72 cm² Hence, from the above, We can conclude that the surface area of the given cylinder is: 150.72 cm² Question 8. What is the surface area of the cone? Use 3.14 for π. S.A. = πr2 + πlr = π(________2) + π(_______)(________) = ________ π + _________π = _______π ≈ ________ cm2 Answer: The given figure is: From the given figure, The radius of the cone (r) is: 7 cm The slant height of the cone (l) is: 13 cm Now, We know that, The surface area of the cone (S.A) = πr² + πrl So, S.A = 3.14 × 7² + 3.14 × 7 × 13 = 153.86 + 285.74 = 439.6 cm² Hence, from the above, We can conclude that the surface area of the given cone is: 439.6 cm² Question 9. Construct Arguments Sasha incorrectly claimed that the surface area of the cylinder is about 76.9 square inches. Explain her likely error and find the correct surface area of the cylinder. Answer The given figure is: : Now, From the given figure, The diameter of the cylinder is: 7 in. The height of the cylinder is: 19 in. Now, We know that, Radius = $$\frac{Diameter}{2}$$ So, Radius of the circle = $$\frac{7}{2}$$ = 3.5 in. Now, We know that, The surface area of the cylinder (S.A) = 2πr² + 2πrh So, S.A = 2 × 3.14 × (3.5)² + 2 × 3.14 × 3.5 × 19 = 76.93 + 417.62 = 494.55 in.² Hence, from the above, We can conclude that The correct surface area of the given cylinder is: 494.55 in.² The mistake made by Sasha is: She adds only the area of the top and bottom but not the area of the rectangle and the area of the top and bottom of the cylinder Question 10. A theme park has a ride that is located in half a sphere. The ride goes around the widest part of the sphere, which has a circumference of 514.96 yards. What is the surface area of the sphere? Estimate to the nearest hundredth using 3.14 for π. Answer: It is given that A theme park has a ride that is located in half a sphere. The ride goes around the widest part of the sphere, which has a circumference of 514.96 yards Now, The given sphere is: Now, We know that, Circumference = 2πr So, Circumference of the theme park = 514.96 yd 2πr = 514.96 r = $$\frac{514.96}{2 × 3.14}$$ = 2.38 yd Now, We know that, The surface area of the sphere (S.A) = 4πr² So, S.A = 4 × 3.14 × (2.38)² = 71.14 yd² Hence, from the above, We can conclude that the surface area of the given sphere is: 71.14 yd² Question 11. Find the amount of wrapping paper you need to wrap a gift in the cylindrical box shown. You need to cover the top, the bottom, and all the way around the box. Use 3.14 for π, and round to the nearest tenth. Answer: The given figure is: Now, From the given figure, The radius of the cylinder is 9 in. The height of the cylinder is: 8 in Now, We know that, The surface area of the cylinder (S.A) = 2πr² + 2πrh So, S.A = 2 × 3.14 × 9² + 2 × 3.14 × 9 × 8 = 508.68 + 452.16 = 960.84 in.² Hence, from the above, We can conclude that the amount of wrapping paper you need to wrap a gift in the cylindrical box is: 960.84 in.² Question 12. Donna paints ornaments for a school play. Each ornament is made up of two identical cones, as shown. How many bottles of paint does she need to paint 70 ornaments? Answer: It is given that Donna paints ornaments for a school play. Each ornament is made up of two identical cones, as shown Now, From the given figure, We can observe that The radius of a cone is: 4.1 cm The slant height of a cone is: 8.9 cm Now, We know that, The surface area of a cone (S.A) = πr² + πrl So, S.A = 3.14 × (4.1)² + 3.14 × 4.1 × 8.9 = 52.78 + 114.57 = 167.35 cm² So, The surface area of the second cone (S.A) = 167.35 cm² So, The S.A of 70 ornaments = 70 × (167.35 × 2) = 23,429 cm² So, The number of bottles of paint she needed to paint 70 ornaments = $$\frac{23,429}{2,000}$$ = 11.7 ≅ 12 bottles Hence, from the above, We can conclude that she need 12 bottles of paint to paint 70 ornaments Question 13. Higher-Order Thinking a. What is the surface area of the cone? Use 3.14 for π, and round to the nearest whole number. Answer: The given figure is: Now, From the given figure, The diameter of the cone is: 6 cm The slant height of the cone is: 12 cm Now, We know that, Radius = $$\frac{Diameter}{2}$$ Radius = $$\frac{6}{2}$$ Radius = 3 cm Now, We know that, The surface area of the cone (S.A) = πr² + πrl So, S.A = 3.14 × 3² + 3.14 × 3 × 12 = 28.26 + 113.04 = 141.3 cm² Hence, from the above, We can conclude that the surface area of the given cone is: 141.3 cm² b. Reasoning Suppose the diameter and the slant height of the cone are cut in half. How does this affect the surface area of the cone? Explain. Answer: It is given that the diameter and the slant height of the cone are cut in half So, S.A = π($$\frac{d}{2}$$)² + π ($$\frac{d}{2}$$) ($$\frac{l}{2}$$) = 3.14 × $$\frac{6²}{4}$$ + 3.14 × 3 × 6 = 28.26 + 56.52 = 84.78 cm² Now, From part (a), The S.A of the cone is: 141.3 cm² So, The ratio of S.A of the cones obtained from part (a) and part (b) respectively = $$\frac{141.3}{84.78}$$ = 1.666 Hence, from the above, We can conclude that the S.A of the cone we obtained in part (b) is 1.666 times of the S.A of the cone we obtained in part (b) Assessment Practice Question 14. What is the surface area of the sphere? Use 3.14 for π, and round to the nearest tenth. A 254.5 cm2 B. 56.55 cm2 C. 1,017.4 cm2 D. 4,071.5 cm2 Answer: The given figure is: From the given figure, The radius of the sphere is: 9 cm Now, We know that, The surface area of the sphere (S.A) = 4πr² So, S.A = 4 × 3.14 × 9² = 1,017.36 cm² ≈ 1,017.4 cm² Hence, from the above, We can conclude that the S.A of the given sphere is: 1,017.4 cm² Question 15. What is the approximate surface area of the cone, in square inches? Use 3.14 for π, and round to the nearest whole number. Answer: The given figure is: Now, From the given figure, The slant height of the cone (l) is 40 in. The diameter of the cone (d) is 40 in. Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{40}{2}$$ r = 20 in. Now, We know that, The surface area of the cone (S.A) = πr² + πrl So, S.A = 3.14 × 20² + 3.14 × 20 × 40 = 1,256 + 2,512 = 3,768 in.² Hence, from the above, We can conclude that the surface area of the given cone is: 3,768 in² ### Lesson 8.2 Find Volume of Cylinders Explain It! Jenna and Ricardo are buying a new fish tank for the growing population of zebrafish in their science lab. Jenna says the tanks hold the same amount of water because they have the same dimensions. Ricardo says that he can fill the bottom of the rectangular tank with more cubes, so it can hold more water. I can… use what I know about finding volumes of rectangular prisms to find the volume of a cylinder. A. Look for Relationships How are the shapes of the two fish tanks alike? How are they different? Answer: When we observe the fish tanks of Jenna and Ricardo, We can observe that a. The heights of the two fish tanks are the same b. The number of cubes that can be filled at the bottom is different c. The number of cubes filled in Jenna’s fish tank is less than that of the number of cubes filled in Ricardo’s fish tank d. The amount of water that can hold in Jenna’s fish tank is less than the amount of water that can hold in Ricardo’s fish tank B. Critique Arguments Who do you think is correct, Ricardo or Jenna? Explain. Answer: It is given that Jenna and Ricardo are buying a new fish tank for the growing population of zebrafish in their science lab. Jenna says the tanks hold the same amount of water because they have the same dimensions. Ricardo says that he can fill the bottom of the rectangular tank with more cubes, so it can hold more water. Now, To find which fish tank holds more water, find the volume of the two fish tanks Now, For Jenna’s fish tank: The fish tank is in the form of a cylinder Now, We know that, The volume of the cylinder (V) = πr²h = $$\frac{πd²h}{4}$$ So, V = $$\frac{3.14 × 24 × 24 × 48}{4}$$ = $$\frac{86,814.72}{4}$$ = 21,073.68 in³ For Ricardo’s fish tank: The fish tank is in the form of a rectangular prism Now, We know that, The volume of a rectangular prism (V) = Length × Width × Height So, V = 24 × 24 × 48 = 27,648 in³ Hence, from the above, We can conclude that Ricardo is correct on the basis of volumes of their fish tanks Focus on math practices Use Structure How can you use what you know about areas of two-dimensional figures and volumes of prisms to compare the volumes of the fish tanks? Answer: We know that, Volume = Area × Length (or) Height (or) Depth Where, “Area” is the area of two-dimensional figures like rectangles, circles, etc. Essential Question How is the volume of a cylinder related to the volume of a rectangular prism? Answer: Rectangular prisms and cylinders are somewhat similar because they both have two bases and a height. The formula for the volume of a rectangular solid is V=Bh can also be used to find the volume of a cylinder Where, “B” in the rectangular prism is the area of the rectangle “B” in the cylinder is the area of the circle Try It! The area of the base of the cylinder is 78.5 in.2. What is the volume of the cylinder? V = Bh = _______ ∙ _______ = _______ The volume of the cylinder is ________ cubic inches. Answer: It is given that The area of the base of the cylinder is 78.5 in.2 Now, The given figure is: Now, We know that, The volume of a cylinder (V) = Bh Where, “B” is defined as the area of the circle So, V = 78.5 × 11 = 863.5 in.³ Hence, from the above, We can conclude that the volume of the given cylinder is: 863.5 cubic inches Convince Me! Why can you use the formula V = Bh to find the volume of a cylinder? Answer: The representation of the cylinder is: Now, From the given figure, We know that, The two circles that are in the top and bottom positions are congruent So, The area for both the circles is also the same Now, We know that, Volume = Area × Height Now, We know that, Area of the circle = πr² Hence, The volume of the cylinder (V) = πr²h Try It! Lin is building a cylindrical planter with a base diameter of 15 inches. She has 5,000 cubic inches of soil to fill her planter. What is the height of the largest planter Lin can build? Use 3.14 for π, and round to the nearest inch. Answer: It is given that Lin is building a cylindrical planter with a base diameter of 15 inches. She has 5,000 cubic inches of soil to fill her planter. So, From the given information, The volume of the cylindrical planter = 5,000 cubic inches The diameter of the cylindrical planter = 15 inches Now, We know that, The volume of a cylinder (V) = πr²h = $$\frac{πd²h}{4}$$ So, 5,000 = $$\frac{3.14 × 15 × 15 × h}{4}$$ 5,000 = 176.625h h = $$\frac{5,000}{176.625}$$ h = 28.3 inches Hence, from the above, We can conclude that the height of the largest planter Lin can build is: 28.3 inches KEY CONCEPT The formula for the volume of a cylinder is the same as the formula for the volume of a prism. The formula for the volume of a cylinder is V= Bh, where B is the area of the circular base and h is the height of the cylinder. Do You Understand? Question 1. Essential Question How is the volume of a cylinder related to the volume of a rectangular prism? Answer: Rectangular prisms and cylinders are somewhat similar because they both have two bases and a height. The formula for the volume of a rectangular solid is V=Bh can also be used to find the volume of a cylinder Where, “B” in the rectangular prism is the area of the rectangle “B” in the cylinder is the area of the circle Question 2. Use Structure What two measurements do you need to know to find the volume of a cylinder? Answer: We know that, The volume of a cylinder (V) = πr²h Now, From the given formula, We can observe that π is a constant Hence, from the above, We can conclude that the two measurements that needed to be known are: a. Radius of the cylinder b. The height of the cylinder Question 3. Reasoning Cylinder A has a greater radius than Cylinder B. Does Cylinder A necessarily have a greater volume than Cylinder B? Explain. Answer: It is given that Cylinder A has a greater radius than Cylinder B Now, We know that, The volume of a cylinder (V) = πr²h Now, Fro the above, We can observe that Volume (V) ∝ Radius² So, When we increase the value of the radius, the value of the volume will automatically increase Hence, from the above, We can conclude that cylinder A has a greater volume than Cylinder B Do You Know How? Question 4. What is the volume of the cylinder? Express your answer in terms of π. Answer: The given figure is: Now, We know that, The volume of a cylinder (V) = Bh Where, B = πr² So, V = 4π × 10 V = 40π mm³ Hence, from the above, We can conclude that the volume of the given cylinder in terms of π is: 40π mm³ Question 5. What is the approximate height of the cylinder? Use 3.14 for π, and if necessary, round to the nearest tenth. Answer: The given figure is: Now, From the given figure, We can observe that The volume of a cylinder (V) = 314 ft³ The radius of a cylinder = 10 ft Now, We know that, The volume of a cylinder (V) = πr²h So, 314 = 3.14 × 10 × 10 × h h = $$\frac{314}{3.14 × 10 × 10}$$ h = 1 ft Hence, from the above, We can conclude that the height of the given cylinder is: 1 ft Question 6. What is the volume of the cylinder? Use 3.14 for π, and if necessary, round to the nearest tenth. Answer: The given figure is: Now, From the given figure, We can observe that The height of a cylinder = 4 cm The circumference of a circle = 22.4 cm Now, We know that, The circumference of a circle = 2πr So, 2πr = 22.4 r = $$\frac{22.4}{2 × 3.14}$$ r = 3.56 cm Now, We know that, The volume of a cylinder (V) = πr²h So, V = 3.14 × 3.56 × 3.56 × 4 = 159.2 cm³ Hence, from the above, We can conclude that the volume of the given cylinder is: 159.2 cm³ Practice & Problem Solving Question 7. Leveled Practice What is the volume of a cylinder with a radius of 5 centimeters and height of 2.5 centimeters? Use 3.14 for π. V = π ________ 2 ∙ ________ = π ______ ∙ ________ = _______ π The volume of the cylinder is about _______ cubic centimeters. Answer: It is given that The radius of a cylinder (r) = 5 cm The height of a cylinder (h) = 2.5 cm Now, We know that, The volume of a cylinder (V) = πr²h So, V = 3.14 × 5 × 5 × 2.5 = 196.25 cm Hence, from the above, We can conclude that the volume of the given cylinder is: 196.25 cm³ (or) 62.5π cm³ Question 8. Find the volume of each cylinder in terms of. Which cylinder has the greater volume? Cylinder A: Area of Base = 61 ft2, height = 10 ft Cylinder B: Circumference = 6π ft, height = 6 ft Answer: The given data is: Cylinder A: Area of Base = 61 ft2, height = 10 ft Cylinder B: Circumference = 6π ft, height = 6 ft Now, We know that, The volume of a cylinder (V) = Bh Where, B is the area of the circle Now, For Cylinder A: V = 61 × 10 = 610 ft³ For Cylinder B: We know that, Circumference = 2πr So, 2πr = 6π r = $$\frac{6π}{2π}$$ = 3 ft So, V = 3.14 × 3 × 3 × 6 = 169.56 ft³ Hence, from the above, We can conclude that Cylinder A has the greatest volume when we compare the volumes of Cylinder A and Cylinder B Question 9. The volume of a cylinder is 2251 cubic inches, and the height of the cylinder is 1 inch. What is the radius of the cylinder? Answer: It is given that The volume of a cylinder is 2251 cubic inches, and the height of the cylinder is 1 inch Now, We know that, The volume of a cylinder (V) = πr²h So, 2251 = 3.14 × r² × 1 r² = $$\frac{2251}{3.14}$$ r² = 716.87 in² r = 26.77 in Hence, from the above, We can conclude that the radius of the given cylinder is: 26.77 in. Question 10. A company is designing a new cylindrical water bottle. The volume of the bottle is 103 cubic centimeters. What is the radius of the water bottle? Estimate using 3.14 for π, and round to the nearest hundredth. Answer: It is given that A company is designing a new cylindrical water bottle. The volume of the bottle is 103 cubic centimeters Now, The given figure is: Now, We know that, The volume of a cylinder (V) = πr²h So, 103 = 3.14 × r² × 8.1 r² = $$\frac{103}{3.14 × 8.1}$$ r² = 4.05cm² r = 2.01 cm Hence, from the above, We can conclude that the radius of the given cylinder is: 2.01 cm Question 11. Use the figure at the right. a. Find the volume of the cylinder in terms of π. Answer: The given figure is: Now, From the given figure, We can observe that The radius of the cylinder (r) = 4 in. The height of the cylinder (h) = 3 in Now, We know that, The volume of a cylinder (V) = πr²h So, V = π × 4 × 4 × 3 V = 48π in.³ Hence, from the above, We can conclude that the volume of the given cylinder in terms of π is: 48π in.³ b. Is the volume of a cylinder, which has the same radius but twice the height, greater or less than the original cylinder? Explain. Answer: From part (a), The radius of the cylinder (r) = 4 in. The height of the cylinder (h) = 3 in So, Now, For part (b), The radius of the cylinder (r) = 4 in. The height of the cylinder (h) = 6 in. So, V = π × 4 × 4 × 6 = 96π in.³ Hence, from the above, We can conclude that the volume of the cylinder we obtained in part (b) is greater than the volume of the cylinder we obtained in part (a) Question 12. Reasoning A rectangular piece of cardboard with dimensions 6 inches by 8 inches is used to make the curved side of a cylinder-shaped container. Using this cardboard, what is the greatest volume the cylinder can hold? Explain. Answer: It is given that A rectangular piece of cardboard with dimensions 6 inches by 8 inches is used to make the curved side of a cylinder-shaped container. Now, Let the height of the cylinder be: 6 inches (or) 8 inches Now, We know that, The volume of a rectangular prism (V) = Length × Width × Height So, For h = 6 inches: V = 6 × 8 × 6 = 288 cubic inches For h = 8 inches: V = 6 × 8 × 8 = 384 cubic inches Hence, from the above, We can conclude that the greatest volume the given cylinder can hold is: 384 cubic inches Question 13. The cylinder shown has a volume of 885 cubic inches. a. What is the radius of the cylinder? Use 3.14 for π. Answer: It is given that The cylinder shown has a volume of 885 cubic inches. Now, The given figure is: Now, From the given figure, We can observe that The height of a cylinder (h) = 11.7 in. Now, We know that, The volume of a cylinder (V) = πr²h So, 885 = 3.14 × r² × 11.7 r² = $$\frac{885}{3.14 × 11.7}$$ r² = 24.08 r = 4.9 in. Hence, from the above, We can conclude that the radius of the given cylinder is: 4.9 in. b. Reasoning If the height of the cylinder is changed, but the volume stays the same, then how will the radius change? Explain. Answer: We know that, The volume of a cylinder (V) = πr²h Now, It is given that Volume —–> Constant Height —– > Changed Now, From the given formula, If the volume is constant, then h ∝ $$\frac{1}{r²}$$ Hence, from the above relation, We can conclude that If the value of height increases, then the value of radius decreases If the value of height decreases, then the value of radius increases Question 14. Toy rubber balls are packaged in a cylinder that holds 3 balls. Find the volume of the cylinder. Use 3.14 for π, and round to the nearest tenth. Answer: It is given that Toy rubber balls are packaged in a cylinder that holds 3 balls Now, The given figure is: Now, From the given figure, We can observe that The height of the cylinder (h) = 20.7 cm The diameter of the cylinder (d) = 6.9 cm Now, We know that, The volume of a cylinder (V) = πr²h = $$\frac{πd²h}{4}$$ So, V = $$\frac{3.14 × 20.7 × 20.7 × 6.9}{4}$$ = 2,321 cm³ Hence, from the above, We can conclude that the volume of the given cylinder is: 2,321 cm³ Question 15. Higher-Order Thinking An insulated collar is made to cover a pipe. Find the volume of the material used to make the collar. Let r = 3 inches, R= 5 inches, and h = 21 inches. Use 3.14 for π, and round to the nearest hundredth. Answer: It is given that An insulated collar is made to cover a pipe. Now, The given data is: r = 3 inches, R= 5 inches, and h = 21 inches Now, The radius of insulated collar = R – r = 5 – 3 = 2 inches Now, The volume of a cylinder (V) = πr²h So, The volume of the material that is used to make the collar (V) = 3.14 × 2 × 2 × 21 = 263.76 cubic inches Hence, from the above, We can conclude that the volume of the material that is used to make the collar is: 263.76 cubic inches Assessment Practice Question 16. The volume of a cylinder is 1,029π cubic centimeters. The height of the cylinder is 21 centimeters. What is the radius, to the nearest centimeter, of the cylinder? Answer: It is given that The volume of a cylinder is 1,0291 cubic centimeters. The height of the cylinder is 21 centimeters. Now, We know that, The volume of a cylinder (V) = πr²h So, 1,029π = πr² × 21 r² = $$\frac{1,029}{21}$$ r² = 49 r = 7 cm Hence, from the above, We can conclude that the radius of the given cylinder is: 7 cm Question 17. The diameter of a cylinder is 7 yards. The height is 12 yards. What is the volume, in terms of π and to the nearest cubic yard, of the cylinder? Answer: It is given that The diameter of a cylinder is 7 yards. The height is 12 yards Now, We know that, The volume of a cylinder (V) = πr²h = $$\frac{πd²h}{4}$$ So, V = $$\frac{3.14 × 7 × 12 × 7}{4}$$ = 1,846.32 yards³ (or) 588π yards³ Hence, from the above, We can conclude that the volume of the given cylinder is: 1,846.32 cubic yards (or) 588π cubic yards Question 18. A cylinder is shown. What statements about the cylinder are true? ☐ The radius of the cylinder is 2 ft. ☐ The diameter of the cylinder is 4 yd. ☐ The height of the cylinder is 8 in. ☐ The volume of the cylinder is 32 in.2. ☐ The volume of the cylinder is 32π in.3. Answer: The given figure is: Now, From the given figure, We can observe that The diameter of the cylinder = 4 in. The height of the cylinder = 8 in. So, The radius of the cylinder = $$\frac{Diameter}{2}$$ = 2 in. Now, We know that, The volume of a cylinder (V) = πr²h So, V = 3.14 × 2 × 2 × 8 = 32π in.³ = 100.48 in.³ Hence, from the above, We can conclude that the statements that are true about the given cylinder are: ### Topic 8 MID-TOPIC CHECKPOINT Question 1. Vocabulary Select all the statements that describe surface area and volume. Lessons 8-1 and 8-2 ☐ Surface area is the sum of the areas of all the surfaces of a figure. ☐ Volume is the distance around a figure. ☐ The surface area is a three-dimensional measure. ☐ Volume is the amount of space a figure occupies. ☐ Volume is a three-dimensional measure. Answer: The above statements that describe the surface area and volume are: In 2-4, use the figure at the right. Sallie packed a cone-shaped cup inside of a cylindrical package. Question 2. The cone-shaped cup is made out of paper. How much paper was used to make the cup, excluding the opening at the top of the cup? Use 3.14 for π, and round to the nearest tenth. Lesson 8-1 Answer: It is given that Sallie packed a cone-shaped cup inside of a cylindrical package and the cone-shaped cup is made out of paper Now, The given figure is: Now, From the given figure, We can observe that The slant height of cone (l) is: 33 cm The diameter of the cone (d) is: 20 cm So, The radius of cone (r) = $$\frac{Diameter}{2}$$ = 10 cm Now, We know that, The surface area of a cone (S.A) = πr² + πrl So, S.A = 3.14 × 10 × 10 + 3.14 × 10 × 33 = 314 + 1,036.2 = 1,350.2 cm² Hence, from the above, We can conclude that We have to use 1,350.2 cm² of paper was used to make the cup, excluding the opening at the top of the cup Question 3. The cylindrical package is made out of cardboard. In terms of π, how much cardboard was used to make the package? Lesson 8-1 Answer: It is given that the cylindrical package is made out of cardboard Now, The given figure is: Now, From the given figure, We can observe that The height of the cylinder (h) is: 33 cm The diameter of the cylinder (d) is: 20 cm So, Radius = $$\frac{Diameter}{2}$$ = $$\frac{20}{2}$$ = 10 cm Now, We know that, The surface area of a cylinder (S.A) = 2πr² + 2πrh So, S.A = 2π × 10² + 2π × 10 × 33 = 200π + 660π = 860π cm² Hence, from the above, We can conclude that 860π cm² of cardboard was used to make the package Question 4. How much space does the package occupy in terms of π? Lesson 8-1 Answer: The given figure is: Now, From the above, We can observe that the given figure is a cylinder Now, From the given figure, We can observe that The height of the cylinder (h) is: 33 cm The diameter of the cylinder (d) is: 20 cm So, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{20}{2}$$ r = 10 cm Now, We know that, The volume of a cylinder (V) = πr²h So, V = π × 10² × 33 = 3,300π cm³ Hence, from the above, We can conclude that 3,300π cm³ of space does the package occupy in terms of π Question 5. What is the surface area of the sphere in terms of π? Lesson 8-1 Answer: The given figure is: Now, From the given figure, We can observe that The radius of the sphere (r) is: 3 ft Now, We know that, The surface area of a sphere (S.A) = 4πr² So, S.A = 4π × 3² = 36π ft² Hence, from the above, We can conclude that the surface area of the given sphere is: 36π ft² Question 6. The volume of the cylinder is 400π cm3. What is the height of the cylinder? Lesson 8-2 A. 5 cm B. 16 cm C. 25 cm D. 80 cm Answer: It is given that The volume of the cylinder is 400π cm³ Now, The given figure is: Now, From the given figure, We can observe that The diameter of the cylinder (d) is: 10 cm So, Radius = $$\frac{Diameter}{2}$$ = $$\frac{10}{2}$$ = 5 cm Now, We know that, The volume of a cylinder (V) = πr²h So, 400π = π × 5² × h h = $$\frac{400}{25}$$ h = 16 cm Hence, from the above, We can conclude that the volume of the given cylinder is: 16 cm ### Topic 8 MID-TOPIC PERFORMANCE TASK Melissa designed a sculpture in which a cylinder-shaped section was removed from a cube. PART A Before painting the surface of the sculpture, Melissa wants to sand the surface where the cylinder section was removed. What is the surface area of the section she will sand? Use 3.14 for π. Explain how you found the surface area. Answer: It is given that Melissa designed a sculpture in which a cylinder-shaped section was removed from a cube. Now, The given figure is: Now, From the given figure, The diameter of the cylinder (d) is: 5 cm The height of the cylinder (d) is: 10 cm So, The radius of the cylinder (r) = $$\frac{Diameter of the cylinder}{2}$$ r = 2.5 cm Now, We know that, We know that, The surface area of a cylinder (S.A) = 2πr² + 2πrh So, S.A = 2 × 3.14 × 2.5 × 2.5 + 2 × 3.14 × 2.5 × 10 = 39.25 + 157 = 196.25 cm² Hence, from the above, We can conclude that 196.25 cm² is the surface area of the section Melissa will sand PART B Melissa has a can of spray paint that covers about 6,500 square centimeters. Can Melissa apply two coats of paint to the entire sculpture? Explain. Use 3.14 for π. Answer: The given figure is: Now, From the above, We can observe that The scripture is the combination of the cube and the cylinder Now, We know that, The surface area of a cube (S.A) = 6a² Where, a is the side of the cube So, S.A of the cube = 6 × 10² = 600 cm² Now, From part (a), The S.A of the cylinder = 196.25 cm² So, The S.A of the scripture = 600 + 196.25 = 796.25 cm² So, 6,500 > 796.25 × 2 Hence, from the above, We can conclude that Melissa can apply two coats of paint to the entire sculpture PART C What is the volume of the sculpture? Use 3.14 for π. Answer: The given figure is: Now, From the above, We can observe that The scripture is the combination of the cube and the cylinder Now, The diameter of the cylinder (d) is: 5 cm The height of the cylinder (h) is: 10 cm So, The radius of the cylinder (r) is: 2.5 cm Now, We know that, The volume of the cylinder (V) = πr²h The volume of the cube (V) = a³ Where, a is the side of the cube So, The volume of the scripture (V) = πr²h + a³ So, V = 3.14 × 2.5 × 2.5 × 10 + 10³ = 196.25 + 1,000 = 1,196.25 cm³ Hence, from the above, We can conclude that the volume of the scripture is: 1,196.25 cm³ ### Lesson 8.3 Find Volume of Cones Solve & Discuss It! A landscape architect uses molds for casting rectangular pyramids and rectangular prisms to make garden statues. He plans to place each finished pyramid on top of a prism. If one batch of concrete mix makes one prism or three pyramids, how does the volume of one pyramid compare to the volume of one prism? Explain. I can… find the volume of cones. Answer: It is given that A landscape architect uses molds for casting rectangular pyramids and rectangular prisms to make garden statues. He plans to place each finished pyramid on top of a prism and one batch of concrete mix makes one prism or three pyramids Now, We know that, The prism is in the shape of the cylinder The pyramid is in the shape of the cone Now, We know that, The volume of a prism (V) = πr²h The volume of a pyramid (V) = $$\frac{1}{3}$$πr²h So, The volume of a prism = 3 × The volume of a pyramid Hence, from the above, We can conclude that The volume of a prism is 3 times the volume of a pyramid Look for Relationships What do you notice about the dimensions of the bases of the pyramid and prism? How are the heights of the two solids related? Answer: The given figure is: Now, From the given figure, We can observe that a. The dimensions of the bases of the prism and the pyramid are the same b. The heights of the pyramid and the prism are the same Focus on math practices Make Sense and Persevere If the architect mixes 10 batches of concrete, how many sculptures combining 1 prism and 1 pyramid could he make? Explain. Answer: It is given that one batch of concrete mix makes one prism or three pyramids So, The ratio of the mix of prism and pyramid in 1 batch is: 1 : 3 Hence, For 10 batches of the mix of concrete, We can make sculptures combining 1 prism and 1pyramid are: 10 prisms and 30 pyramids Essential Question How is the volume of a cone related to the volume of a cylinder? Answer: The volumes of a cone and a cylinder are related in the same way as the volumes of a pyramid and a prism is related. If the heights of a cone and a cylinder are equal, then the volume of the cylinder is 3 times as much as the volume of a cone Try It! Find the volume of the cone. Use 3.14 for π. The volume of the cone is about ________ cubic inches. V = ____ πr²h ≈ _______(3.14) (______)2 (4) = ________ (3.14) (______)(4) = _________ Answer: The given figure is: From the above figure, We can observe that The radius of the cone (r) is 1.5 in. The height of the cone (h) is 4 in. Now, We know that, The volume of the cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × (1.5)² × 4 = 9.42 in.³ Hence, from the above, We can conclude that the volume of the given cone is: 9.42 in.³ Convince Me! If you know the volume of a cone, how can you find the volume of a cylinder that has the same height and radius as the cone? Answer: We know that, If the cylinder and the cone has the same height and radius, then The volume of cone = $$\frac{1}{3}$$ × The volume of cylinder Try It! Find the volume of each cone. a. Use $$\frac{22}{7}$$ for π. Express the answer as a fraction. Answer: The given figure is: Now, From the given figure, We can observe that The slant height (l) is: 5 mm The height of the cone (h) is: 3 mm Now, We know that, l² = r²+ h² So, 5² = r² + 3² r² = 25 – 9 r² = 16 r = 4 mm Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 4² × 3 = $$\frac{1,056}{21}$$ mm³ Hence, from the above, We can conclude that the volume of the given cone in the fraction form is: $$\frac{1,056}{66}$$ mm³ b. Express the volume in terms of π. Answer: The given figure is: From the given figure, We can observe that The height of the cone (h) is: 21 ft The circumference of the circle (C) is: 16π ft Now, We know that, The circumference of the circle (C) = 2πr So, 2πr = 16π r = 8 ft Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × π × 8² × 21 = 448π ft³ Hence, from the above, We can conclude that the volume of the given cone in terms of π is: 448π ft³ KEY CONCEPT The volume of a cone is $$\frac{1}{3}$$ the volume of a cylinder with the same base and height. The formula for the volume of a cone is V = $$\frac{1}{3}$$Bh, where B is the area of the base and his the height of the cone. Do You Understand? Question 1. Essential Question How is the volume of a cone related to the volume of a cylinder? Answer: The volumes of a cone and a cylinder are related in the same way as the volumes of a pyramid and a prism is related. If the heights of a cone and a cylinder are equal, then the volume of the cylinder is 3 times as much as the volume of a cone Question 2. Use Structure What dimensions do you need to find the volume of a cone? Answer: To find the volume of a cone, The dimensions we need are: a. The radius of the cone b. The height of the cone Question 3. Look for Relationships If you know a cone’s radius and slant height, what must you do before you can find its volume? Answer: If you know a cone’s radius (r) and slant height(l), then The volume of a cone (V) is given as: V = $$\frac{1}{3}$$πr² × $$\sqrt{l² – r²}$$ Do You Know How? Question 4. Wanda found a cone-shaped seashell on the beach. The shell has a height of 63 millimeters and a base radius of 8 millimeters. What is the volume of the seashell? Estimate 63 mm using $$\frac{22}{7}$$ for π. Answer: It is given that Wanda found a cone-shaped seashell on the beach. The shell has a height of 63 millimeters and a base radius of 8 millimeters Now, The given figure is: Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 8² × 63 = 4,224 mm³ Hence, from the above, We can conclude that the volume of the seashell is: 4,224 mm³ Question 5. What is the volume of the cone? Estimate using 3.14 for π, and round to the nearest tenth. Answer: The given figure is: Now, From the given figure, We can observe that The height of the cone (h) is: 40mm The slant height of the cone (l) is: 41 mm Now, We know that, l² = r² + h² So, 41² = r²+ 40² r² = 81 r = 9 mm Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × 9² × 40 = 3,391.2 mm³ Hence, from the above, We can conclude that the volume of the given cone is: 3,391.2 mm³ Question 6. What is the volume of the cone in terms of π if the circumference of the base is 1.4π feet? Answer: The given figure is: Now, From the given figure, We can observe that The height of the cone (h) is: 2.7 ft Now, We know that, Circumference (C) = 2πr So, 2πr = 1.4π r = 0.7 ft Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × π × (0.7)² × 2.7 = 0.441π ft³ Hence, from the above, We can conclude that the volume of the given cone is: 0.441π ft³ Practice & Problem Solving Multimedia Leveled Practice In 7 and 8, find the volumes of the cones. Question 7. What is the volume of the cone? Write your answer in terms of π. Answer: From the given figure, We can observe that The radius of the cone (r) is: 3 cm The height of the cone (h) is: 4 cm Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × π × 3² × 4 = 12π cm³ Hence, from the above, We can conclude that the volume of the given cone is: 12π cm³ Question 8. What is the volume of the cone to the nearest hundredth? Use 3.14 for π. Answer: From the given figure, We can observe that The radius of the cone (r) is: 16 units The height of the cone (h) is: 36 units Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × 16² × 36 = 9,646.08 units³ Hence, from the above, We can conclude that the volume of the given cone is: 9,646.08 units³ Question 9. If a cone-shaped hole is 3 feet deep and the circumference of the base of the hole is 44 feet, what is the volume of the hole? Use $$\frac{22}{7}$$ for π. Answer: It is given that A cone-shaped hole is 3 feet deep and the circumference of the base of the hole is 44 feet Now, We know that, Circumference (C) = 2πr So, 2πr = 44 r = $$\frac{44}{2π}$$ r = 7 feet Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 7² × 3 = 154 feet³ Hence, from the above, We can conclude that the volume of the given cone-shaped hole is: 154 feet³ Question 10. The volume of the cone is 462 cubic yards. What is the radius of the cone? Use $$\frac{22}{7}$$ for π. Answer: The given figure is: Now, From the given figure, We can observe that The height of the cone (h) is: 9 yd Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, 462 = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × r² × 9 r² = $$\frac{462}{9.428}$$ r = 7 yd Hence, from the above, We can conclude that the radius of the given cone is: 7 yd Question 11. A city engineer determines that 5,500 cubic meters of sand will be needed to combat erosion at the city’s beach. Does the city have enough sand to combat erosion? Use $$\frac{22}{7}$$ for π. Explain. Answer: It is given that A city engineer determines that 5,500 cubic meters of sand will be needed to combat erosion at the city’s beach. Now, The given figure is: Now, From the given figure, We can observe that The slant height of the cone(l) is: 37 m The height of the cone (h) is: 35 m Now, We know that, l² = r²+ h² 37² = r² + 35² r² = 37² – 35² r = 12 m Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 12² × 35 = 5,280 m³ So, 5,500 > 5,280 Hence, from the above, We can conclude that the city has enough sand to combat erosion Question 12. A water tank is shaped like the cone shown. a. How much water can the tank hold? Use 3.14 for π, and round to the nearest tenth. Answer: The given figure is: Now, From the given figure, We can observe that The slant height of the cone (l) is: 61 ft The height of the cone (h) is: 60 ft Now, We know that, l² = r²+ h² 61² = r² + 60² r² = 61² – 60² r = 11 ft Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × 11² × 60 = 7,598.8 ft³ Hence, from the above, We can conclude that the amount of water that the tank can hold is: 7,598.8 ft³ b. If water is drained from the tank to fill smaller tanks that each hold 500 cubic feet of water, how many smaller tanks can be filled? Answer: It is given that the water is drained from the tank to fill smaller tanks that each holds 500 cubic feet of water Now, From part (a), We know that, The amount of water that the tank can hold is: 7,598.8 ft³ Now, The number of smaller tanks that can be filled = $$\frac{The amount of water that the tank can hold}{The amount of water that the smaller tank can hold}$$ = $$\frac{7,598.8}{500}$$ ≅ 15 Hence, from the above, We can conclude that the number of smaller tanks that can be filled is: 15 Question 13. An ice cream cone is filled exactly level with the top of a cone. The cone has a 9-centimeter depth and a base with a circumference of 91 centimeters. How much ice cream is in the cone in terms of π? Answer: It is given that An ice cream cone is filled exactly level with the top of a cone. The cone has a 9-centimeter depth and a base with a circumference of 91 centimeters Now, We know that, Circumference (C) = 2πr So, 2πr = 91 r = $$\frac{91}{2π}$$ r = 14.49 cm Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × π × (14.49)² × 9 = 629.88π cm³ Hence, from the above, We can conclude that the amount of ice cream that is in the cone in terms of π is: 629.88π cm³ Question 14. In the scale model of a park, small green cones represent trees. What is the volume of one green cone? Use $$\frac{22}{7}$$ for π. Answer: The given green cone is: Now, From the given green cone, We can observe that The slant height of the cone (l) is: 65 mm The height of the cone (h) is: 63 mm Now, We know that, l² = r²+ h² 65² = r² + 603² r² = 65² – 63² r = 16 mm Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 16² × 63 = 16,896 mm³ Hence, from the above, We can conclude that the volume of one green cone is: 16,896 mm³ Question 15. Reasoning Compare the volumes of two cones. One has a radius of 5 feet and a slant height of 13 feet. The other one has a height of 5 feet and a slant height of 13 feet. a. Which cone has the greater volume? Answer: The given data is: Cone 1: Radius: 5 feet Slant height: 13 feet Cone 2: Height: 5 feet Slant height: 13 feet Now, We know that, l² = r²+ h² So, For cone 1: 13² = h² + 5² h² = 13² – 5² h = 12 feet For cone 2: 13² = r² + 5² r² = 13² – 5² r = 12 feet Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, For Cone 1: V = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 5² × 12 = 314.28 feet³ For Cone 2: V = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 12² × 5 = 754.28 feet³ Hence, from the above, We can conclude that Cone 2 has the greater volume b. What is the volume of the larger cone in terms of π? Answer: From part (a), We can observe that, Cone 2 has a greater volume Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × π × 12² × 5 = 240π feet³ Question 16. An artist makes a cone-shaped sculpture for an art exhibit. If the sculpture is 7 feet tall and has a base with a circumference of 24.492 feet, what is the volume of the sculpture? Use 3.14 for π, and round to the nearest hundredth. Answer: It is given that An artist makes a cone-shaped sculpture for an art exhibit and the sculpture is 7 feet tall and has a base with a circumference of 24.492 feet Now, We know that, Circumference (C) = 2πr So, 2πr = 24.492 r = $$\frac{24.492}{2π}$$ r = 3.9 feet Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × 3.9² × 7 = 111.43 feet³ Hence, from the above, We can conclude that the volume of the given cone-shaped sculpture is: 111.43 feet³ Question 17. Higher-Order Thinking A cone has a radius of 3 and a height of 11. a. Suppose the radius is increased by 4 times its original measure. How many times greater is the volume of the larger cone than the smaller cone? Answer: It is given that A cone has a radius of 3 and a height of 11 Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × 3² × 11 = 103.62 units³ Now, If the radius is increased by 4 times of its original measure, then V = $$\frac{1}{3}$$ × 3.14 × (3 × 4)² × 11 = 1,657.92 units³ Now, The number of times the volume of the larger cone than the smaller cone = $$\frac{1,657.92}{103.62}$$ = 16 Hence, from the above, We can conclude that the volume of the larger cone is 16 times greater than the volume of the smaller cone b. How would the volume of the cone change if the radius were divided by four? Answer: From part (a), We know that, The volume of the cone with a radius of 3 and a height of 11 units is: 103.62 units³ Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × (3 ÷ 4)² × 11 = 6.476 units³ Now, The ratio of the volume of the changed cone to the volume of the original cone = $$\frac{6.476}{103.62}$$ = 0.06 Hence, from the above, We can conclude that the volume of the cone will be 0.06 times of the original cone Assessment Practice Question 18. List the cones described below in order from least volume to greatest volume. • Cone 1: radius 6 cm and height 12 cm • Cone 2: radius 12 cm and height 6 cm • Cone 3: radius 9 cm and height 8 cm A. Cone 2, Cone 3, Cone 1 B. Cone 1, Cone 3, Cone 2 C. Cone 2, Cone 1, Cone 3 D. Cone 1, Cone 2, Cone 3 Answer: The given data is: a. Cone 1: radius 6 cm and height 12 cm b. Cone 2: radius 12 cm and height 6 cm c. Cone 3: radius 9 cm and height 8 cm Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, For Cone 1: V = $$\frac{1}{3}$$ × 3.14 × 6² × 12 = 452.16 cm³ For Cone 2: V = $$\frac{1}{3}$$ × 3.14 × 12² × 6 = 904.32 cm³ For Cone 3: V = $$\frac{1}{3}$$ × 3.14 × 9² × 8 = 678.24 cm³ Hence, from the above, We can conclude that the order of the volumes from the least to the greatest is: Cone 1 < Cone 3 < Cone 2 Question 19. What is the volume, in cubic inches, of a cone that has a radius of 8 inches and a height of 12 inches? Use 3.14 for π, and round to the nearest hundredth. Answer: It is given that A cone has a radius of 8 inches and a height of 12 inches Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × 8² × 12 = 803.84 inches³ Hence, from the above, We can conclude that the volume of the given cone is: 803.84 inches³ ### Lesson 8.4 Find Volume of Spheres Explore It! Marshall uses the beaker to fill the bowl with water. I can… find the volume of a sphere and use it to solve problems. A. Draw and label three-dimensional figures to represent the beaker and the bowl. Answer: The representation of the three-dimensional figures that represent the beaker and bowl are: B. Marshall has to fill the beaker twice to completely fill the bowl with water. How can you use an equation to represent the volume of the bowl? Answer: The given figures are: Now, It is given that Marshall has to fill the beaker twice to completely fill the bowl with water Now, From part (a), We can observe that the beaker is in the form of a cone Now, We know that, The volume of a cone = $$\frac{1}{3}$$πr²h Now, According to the given situation, The volume of the bowl(V’) = 2 × The volume of the beaker(V) V’ = 2 × $$\frac{1}{3}$$πr²h Now, From the given figures, We can observe that The height of a bowl is 2 times its radius So, V’ = $$\frac{2}{3}$$πr² (2r) V’ = $$\frac{4}{3}$$πr³ Hence, from the above, We can conclude that the volume of the bowl is: $$\frac{4}{3}$$πr³ Focus on math practices Reasoning How is the volume of a sphere and the volume of a cone related? What must be true about the radius and height measurements for these relationships to be valid? Answer: We know that, The relationship between the volume of the sphere and the volume of the cone is: The volume of the sphere = 2 × The volume of the cone Now, For the above relationship to be valid, a. The heights of the sphere and the cone must be the same b. The radius of the sphere and the cone must be the same Essential Question How is the volume of a sphere related to the volume of a cone? Answer: The volume of a sphere is twice the volume of a cone that has the same circular base and height. i.e., The volume of the sphere = 2 × The volume of the cone Try It! What is the volume of a ball with a diameter of 6 centimeters? Use 3.14 for π. V = $$\frac{4}{3}$$ πr3 = $$\frac{4}{3}$$ π _______3 ≈ ________ ∙ ________ = __________ The volume of the ball is about _______ cm3 Answer: It is given that The diameter of a ball is: 6 cm Now, We know that, The ball is in the form of a sphere Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{6}{2}$$ r = 3 cm Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × 3.14 × 3³ = 113.04 cm³ Hence, from the above, We can conclude that the volume of the given ball is about 113.04 cm³ Convince Me! How is the volume of a sphere related to the volume of a cone that has the same circular base and height? Answer: The relationship between the volume of a cone and the volume of a sphere that has the same circular base and height is: The volume of the sphere = 2 × The volume of the cone Try It! What is the volume of the composite figure shown? Use 3.14 for π. Answer: The given composite figure is: Now, We know that, A composite figure is made up of 2 or more two-dimensional figures Now, The given composite figure is made up of Hemisphere and Cone Now, We know that, The volume of the given composite figure = The volume of the Hemisphere + The volume of the cone Now, From the given composite figure, We can observe that Radius (r) = 2 in. Height (h) = 6 in. Now, We know that, The volume of the Hemisphere (V) = $$\frac{2}{3}$$πr³ So, V = $$\frac{2}{3}$$ × 3.14 × 2³ = 16.74 in.³ Now, We know that, The volume of the cone (V’) = $$\frac{1}{3}$$πr²h So, V’ = $$\frac{1}{3}$$ × 3.14 × 2² × 6 = 25.12 in.³ So, The volume of the given composite figure = 16.74 + 25.12 = 41.84 in.³ Hence, from the above, We can conclude that the volume of the given composite figure is: 41.84 in.³ KEY CONCEPT The volume of a sphere is twice the volume of a cone that has the same circular base and height. The formula for the volume of a sphere with radius r is V = $$\frac{4}{3}$$ πr3. Do You Understand? Question 1. Essential Question How is the volume of a sphere related to the volume of a cone? Answer: The volume of a sphere is twice the volume of a cone that has the same circular base and height. i.e., The volume of the sphere = 2 × The volume of the cone Question 2. Critique Reasoning Kristy incorrectly says that the volume of the sphere below is 144π cubic units. What mistake might Kristy have made? Answer: The given sphere is: Now, It is given that Kristy incorrectly says that the volume of the sphere below is 144π cubic units Now, From the given sphere, We can observe that Radius (r) = 6 units Now, We know that, The volume of a spher (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × π × 6³ = 288π units³ Hence, from the above, We can conclude that the mistake done by Kristy is: Considering the sphere as the hemisphere and calculated the volume of the hemisphere instead of the sphere Question 3. Generalize Mehnaj has a set of blocks that are all the same height. The cone-shaped block has a volume of 125 cubic inches. The sphere-shaped block has a volume of 250 cubic inches. What do you know about the radius of the base of the cone-shaped block? Explain. Answer: It is given that Mehnaj has a set of blocks that are all the same height. The cone-shaped block has a volume of 125 cubic inches. The sphere-shaped block has a volume of 250 cubic inches So, From the given situation, We can observe that a. The heights of the two cone-shaped blocks are the same b. The volumes of the two cone-shaped blocks are different Now, We know that, The volume of a cone = $$\frac{1}{3}$$πr²h So, V1 = $$\frac{1}{3}$$πr1²h1 V2 = $$\frac{1}{3}$$πr2²h2 So, V1/V2 = r1²/ r2² r1²/ r2² = $$\frac{125}{250}$$ r1² / r2² = 0.5 r1 / r2 = 0.707 Hence, from the above, We can conclude that the radius of the first cone-based block is 0.707 times the radius of the second cone-shaped block Do You Know How? Question 4. Clarissa has a decorative bulb in the shape of a sphere. If it has a radius of 3 inches, what is its volume? Use 3.14 for π. Answer: It is given that Clarissa has a decorative bulb in the shape of a sphere and it has a radius of 3 inches Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × 3.14 × 3³ = 113.04 inches³ Hence, from the above, We can conclude that the volume of the given decorative bulb is: 113.04 inches³ Question 5. A sphere has a surface area of about 803.84 square centimeters. What is the volume of the sphere? Use 3.14 for π and round to the nearest whole number. Answer: It is given that A sphere has a surface area of about 803.84 square centimeters Now, We know that, The surface area of a sphere (S.A) = 4πr² So, 4πr² = 803.84 r² = $$\frac{803.84}{4π}$$ r² = 64 r = 8 cm Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × 3.14 × 8³ = 2,143.57 cm³ ≈2,144 cm³ Hence, from the above, We can conclude that the volume of the given sphere is about 2,144 cm³ Question 6. A water pipe is a cylinder 30 inches long, with a radius of 1 inch. At one end of the cylinder, there is a hemisphere. What is the volume of the water pipe? Explain. Answer: It is given that A water pipe is a cylinder 30 inches long, with a radius of 1 inch. At one end of the cylinder, there is a hemisphere. So, The volume of the water pipe = The volume of the cylinder + The volume of the hemisphere Now, From the given water pipe, We can observe that Radius (r) = 1 in. Height (h) = 30 in. Now, We know that, The volume of the Hemisphere (V) = $$\frac{2}{3}$$πr³ So, V = $$\frac{2}{3}$$ × 3.14 × 1³ = 2.09 in.³ Now, We know that, The volume of the cylinder (V’) = πr²h So, V’ = 3.14 × 1² × 30 = 94.2 in.³ So, The volume of the given water pipe = 2.09 + 94.2 = 96.29 in.³ Hence, from the above, We can conclude that the volume of the given water pipe is: 96.29 in.³ Practice & Problem Solving Question 7. Leveled Practice What is the amount of air, in cubic centimeters, needed to fill the stability ball? Use 3.14 for π, and round to the nearest whole number. Use the formula The volume of the stability ball is approximately __________ cubic centimeters. Answer: From the given figure, We can observe that The stability ball is in the form of a sphere Now, The diameter of the stability ball (d) is: 55 cm Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{55}{2}$$ r = 27.5 cm Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × 3.14 × (27.5)³ = 87,069.53 cm³ ≈87,070 cm³ Hence, from the above, We can conclude that the volume of the stability ball is approximately 87,070 cm³ Question 8. The spherical balloon has a 22-inch. diameter when it is fully inflated. Half of the air is let out of the balloon. Assume that the balloon remains a sphere. Keep all answers in terms of π. a. Find the volume of the fully-inflated balloon. Answer: It is given that The spherical balloon has a 22-inch. diameter when it is fully inflated. Half of the air is let out of the balloon. Now, From the given situation, The diameter of the fully inflated balloon (d) is: 22 inches Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{22}{2}$$ r = 11 cm Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × π × 11³ = 1,774.6π cm³ Hence, from the above, We can conclude that the volume of the fully inflated balloon is: 1,774.6π cm³ b. Find the volume of the half-inflated balloon. Answer: It is given that The spherical balloon has a 22-inch. diameter when it is fully inflated. Half of the air is let out of the balloon. So, The diameter of half-inflated balloon (d) = $$\frac{The diameter of fully inflated balloon}{2}$$ d = $$\frac{22}{2}$$ d = 11 cm Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{11}{2}$$ r = 5.5 cm Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × π × (5.5)³ = 221.8π cm³ Hence, from the above, We can conclude that the volume of the half-inflated balloon is: 221.8π cm³ c. What is the radius of the half-inflated balloon? Round to the nearest tenth. Answer: From part (b), We can observe that The diameter of half-inflated balloon (d) = $$\frac{The diameter of fully inflated balloon}{2}$$ d = $$\frac{22}{2}$$ d = 11 cm Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{11}{2}$$ r = 5.5 cm Hence, from the above, We can conclude that the radius of the half-inflated balloon is: 5.5 cm Question 9. Find the volume of the figure. Use 3.14 for π, and round to the nearest whole number. Answer: The given figure is: Now, From the given figure, We can observe that The figure is a combination of a hemisphere and a cone So, From the given figure, Diameter (d) =14 cm Height (h) = 17 cm Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{14}{2}$$ r = 7 cm Now, We know that, The volume of the figure (V) = The volume of a hemisphere + The volume of a cone Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × 7² × 17 = 871.87 cm³ Now, We know that, The volume of a hemisphere (V’) = $$\frac{2}{3}$$πr³ So, V’ = $$\frac{2}{3}$$ × 3.14 × 7³ = 718.01 cm³ So, The volume of the given figure = 871.87 + 718.01 = 1,589.88 cm³ ≈1,590 cm³ Hence, from the above, We can conclude that the volume of the given figure is: 1,590 cm³ Question 10. The surface area of a sphere is about 2,826 square millimeters. What is the volume of the sphere? Use 3.14 for π, and round to the nearest whole number. Answer: It is given that The surface area of a sphere is about 2,826 square millimeters Now, We know that, The surface area of a sphere (S.A) = 4πr² So, 4πr² = 2,826 r² = $$\frac{2,826}{4π}$$ r² = 225 r = 15 mm Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × 3.14 × 15³ = 14,130 mm³ Hence, from the above, We can conclude that the volume of the given sphere is: 14,130 mm³ Question 11. A sphere has a volume of 1,837.35 cubic centimeters. What is the radius of the sphere? Use 3.14 for π, and round to the nearest tenth. Answer: It is given that A sphere has a volume of 1,837.35 cubic centimeters Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, 1,837.35 = $$\frac{4}{3}$$πr³ πr³ = $$\frac{1,837.35 × 3}{4}$$ πr³ = 1,378.01 r³ = $$\frac{1,378.01}{π}$$ r³ = 438.85 r = 0.333 cm Hence, from the above, We can conclude that the radius of the given sphere is: 0.333 cm Question 12. Find the volume of the solid. Use 3.14 for π, and round to the nearest whole number. Answer: The given figure is: From the above, We can observe that the given figure is a composite figure Now, From the given figure, We can observe that Radius (r) = 4 m Height (h) = 17 m Now, We know that, The volume of the given figure = The volume of a hemisphere + The volume of a cylinder Now, We know that, The volume of the Hemisphere (V) = $$\frac{2}{3}$$πr³ So, V = $$\frac{2}{3}$$ × 3.14 × 4³ = 133.97 m³ Now, We know that, The volume of the cylinder (V’) = πr²h So, V’ = 3.14 × 4² × 17 = 854.08 m³ So, The volume of the given figure = 133.9 + 854.08 = 987.98 m³ Hence, from the above, We can conclude that the volume of the given figure is: 987.98 m³ Question 13. Your friend says that the volume of a sphere with a diameter of 3.4 meters is 164.55 cubic meters. What mistake might your friend have made? Find the correct volume. Use 3.14 for π and round to the nearest hundredth. Answer: It is given that Your friend says that the volume of a sphere with a diameter of 3.4 meters is 164.55 cubic meters Now, From the given information, Diameter (d) = 3.4 m Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{3.4}{2}$$ r = 1.7 m Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × 3.14 × (1.7)³ = 20.56 m³ Hence, from the above, We can conclude that The mistake made by your friend is: Consideration of diameter as radius and find the volume of the given sphere Question 14. A solid figure has a cone and hemisphere hollowed out of it. What is the volume of the remaining part of the solid? Use 3.14 for π, and round to the nearest whole number. Answer: It is given that A solid figure has a cone and hemisphere hollowed out of it So, The volume of the remaining part of the solid = | The volume of a hemisphere – The volume of a cone | Now, From the given solid, We can observe that Radius (r) = 6 in. Height = 23 in. Now, We know that, The volume of a hemisphere (V) = $$\frac{2}{3}$$πr³ So, V = $$\frac{2}{3}$$π × 6³ = 144π in.³ Now, We know that, The volume of a cone (V’) = $$\frac{1}{3}$$πr²h So, V’ = $$\frac{1}{3}$$π × 6² × 23 = 276π in.³ So, The volume of the remaining part of the solid = |144π – 276π| = 132π = 132 × 3.14 = 414.48 in.³ ≈415 in.³ Hence, from the above, We can conclude that the volume of the remaining part of the given solid is about 415 in.³ Question 15. Higher-Order Thinking A student was asked to find the volume of a solid where the inner cylinder is hollow. She incorrectly said the volume is 2,034.72 cubic inches. Answer: It is given that A student was asked to find the volume of a solid where the inner cylinder is hollow. She incorrectly said the volume is 2,034.72 cubic inches. a. Find the volume of the solid. Use 3.14 for π. Round to the nearest whole number. Answer: The given solid is: Now, From the given solid, We can observe that it is a combination of a cone and a cylinder So, The volume of the given solid = The volume of a cylinder + The volume of a cone Now, From the given solid, We can observe that The radius of the cylinder is 3 in. The height of the cylinder is 15 in. The diameter of the cone is 12 in. The slant height of the cone is 9 in. Now, We know that, The volume of a cylinder (V) = πr²h So, V = 3.14 × 3² × 15 = 423.9 in.³ Now, The volume of a cone: We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{12}{2}$$ r = 6 in. Now, We know that, l² = r² + h² So, 9² = 6² + h² h² = 9² – 6² h² = 45 h = 6.70 in. Now, We know that, The volume of a cone (V’) = $$\frac{1}{3}$$πr²h So, V’ = $$\frac{1}{3}$$ × 3.14 × 6² × 6.70 = 252.45 in.³ So, The volume of the given solid = 423.9 + 252.45 = 676.3 in.³ ≈676 in.³ Hence, from the above, We can conclude that the volume of the given solid is about 676 in.³ b. What mistake might the student have made? Answer: Assessment Practice Question 16. The spherical boulder is 20 feet in diameter and weighs almost 8 tons. Find its volume. Use 3.14 for π. Round to the nearest cubic foot. Answer: It is given that The spherical boulder is 20 feet in diameter and weighs almost 8 tons. Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{20}{2}$$ r = 10 feet So, V = $$\frac{4}{3}$$ × 3.14 × 10³ = 4,186.6 feet³ ≈ 4,187 feet³ Hence, from the above, We can conclude that the volume of the given spherical boulder is: 4,187 feet³ Question 17. A bowl is in the shape of a hemisphere (half a sphere) with a diameter of 13 inches. Find the volume of the bowl. Use 3.14 for π, and round to the nearest cubic inch. Answer: It is given that A bowl is in the shape of a hemisphere (half a sphere) with a diameter of 13 inches. Find the volume of the bowl Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{13}{2}$$ r = 6.5 inches Now, We know that, The volume of a hemisphere (V) = $$\frac{2}{3}$$πr³ So, V = $$\frac{2}{3}$$ × 3.14 × (6.5)³ = 574.88 inches³ ≈ 575 inches³ Hence, from the above, We can conclude that the volume of the given bowl is about 575 inches³ 3-ACT MATH 3-Act Mathematical Modeling: Measure Up ACT 1 1. After watching the video, what is the first question that comes to mind? Answer: Question 2. Write the Main Question you will answer. Answer: Question 3. Construct Arguments Predict an answer to this Main Question. Explain your prediction. Answer: Question 4. On the number line below, write a number that is too small to be the answer. Write a number that is too large. Answer: Question 5. Plot your prediction on the same number line. Answer: ACT 2 Question 6. What information in this situation would be helpful to know? How would you use that information? Answer: Question 7. Use Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically. Answer: Question 8. Model with Math Represent the situation using mathematics. Use your representation to answer the Main Question. Answer: Question 9. What is your answer to the Main Question? Is it higher or lower than your prediction? Explain why. Answer: ACT 3 Question 10. Write the answer you saw in the video. Answer: Question 11. Reasoning Does your answer match the answer in the video? If not, what are some reasons that would explain the difference? Answer: Question 12. Make Sense and Persevere Would you change your model now that you know the answer? Explain. Answer: ACT 3 Extension Reflect Question 13. Model with Math Explain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question? Answer: Question 14. Make Sense and Persevere When did you struggle most while solving the problem? How did you overcome that obstacle? Answer: SEQUEL Question 15. Generalize Suppose you have a graduated cylinder half the height of the one in the video. How wide does the cylinder need to be to hold the liquid in the flask? Answer: ### Topic 8 REVIEW Topic Essential Question How can you find volumes and surface areas of three-dimensional figures? Answer: The “Surface area” is the sum of the areas of all faces (or surfaces) on a 3D shape. Ex: A cuboid has 6 rectangular faces. To find the surface area of a cuboid, add the areas of all 6 faces We know that, The volume of a three-dimensional figure = Cross-sectional area × length Vocabulary Review Complete each definition and then provide an example of each vocabulary word. Vocabulary composite figure cone cylinder sphere Answer: Use Vocabulary in Writing Draw a composite figure that includes any two of the following: a cylinder, a cone, a sphere, and a hemisphere. Label each part of your drawing. Then describe each part of your composite figure. Use vocabulary terms in your description. Concepts and Skills Review Lesson 8.1 Find Surface Area of Three-Dimensional Figures Quick Review Surface area is the total area of the surfaces of a three-dimensional figure. The chart gives formulas for finding the surface area of a cylinder, a cone, and a sphere. Example What is the surface area of the cylinder? Use 3.14 for π. Answer: Practice Question 1. What is the surface area of the cone? Use 3.14 for π. Answer: The given figure is: From the given figure, We can observe that The slant height of the cone (l) is: 13 m The radius of the cone (r) is: 5 m Now, We know that, The surface area of a cone (S.A) = πr² + πrl So, S.A = 3.14 × 5² + 3.14 × 5 × 13 = 78.5 + 204.1 = 282.6 m² Hence, from the above, We can conclude that the surface area of the given cone is: 282.6 m² Question 2. What is the surface area of the sphere in terms of π? Answer: The given figure is: From the given figure, We can observe that The diameter of the sphere (d) is: 10 cm Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{10}{2}$$ r = 5 cm Now, We know that, The surface area of a sphere (S.A) = 4πr² So, S.A = 4 × 3.14 × 5² = 314 cm² Hence, from the above, We can conclude that the surface area of the given sphere is: 314 cm² Question 3. What is the surface area of the cylinder in terms of π? Answer: The given figure is: Now, From the given figure, We can observe that, The diameter of the cylinder (d) is 12 in. The height of the cylinder (h) is 15 in. Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{12}{2}$$ r = 6 in. Now, We know that, The surface area of a cylinder (S.A) = 2πr² + 2πrh So, S.A = 2 × 3.14 × 6² + 2 × 3.14 × 6 × 15 = 226.08 + 565.2 = 791.28 in.³ Hence, from the above, We can conclude that the surface area of the given cylinder is: 791.28 in.³ Lesson 8.2 Find Volume of Cylinders Quick Review The volume of a cylinder is equal to the area of its base times its height. V = area of base · height, or V = πr²h Example What is the volume of the cylinder? Use 3.14 for π. Answer: Practice Question 1. What is the volume of the cylinder in terms of π? Answer: The given figure is: From the given figure, We can observe that The diameter of the cylinder (d) is: 2 m The height of the cylinder (h) is: 6 m Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{2}{2}$$ r = 1 m Now, We know that, The volume of a cylinder (V) = πr²h So, V = 3.14 × 1² × 6 = 18.84 m³ Hence, from the above, We can conclude that the volume of the given cylinder is: 18.84 m³ Question 2. The volume of the cylinder is 141.3 cubic centimeters. What is the radius of the cylinder? Use 3.14 for π. Answer: It is given that The volume of the cylinder is 141.3 cubic centimeters Now, The given figure is: Now, From the given figure, We can observe that The height of the cylinder (h) is: 5 cm Now, We know that, The volume of a cylinder (V) = πr²h So, 141.3 = 3.14 × r² × 5 r² = $$\frac{141.3}{3.14 × 5}$$ r²= 9 r = 3 cm Hence, from the above, We can conclude that the radius of the given cylinder is: 3 cm Lesson 8.3 Find Volume of Cones Quick Review To find the volume of a cone, use the formula V = $$\frac{1}{3}$$πr2h. Example What is the volume of the cone? Use 3.14 for π. Answer: Practice Question 1. What is the volume of the cone in terms of π? Answer: The given figure is: Now, From the given figure, We can observe that The radius of the cone (r) is 3 in. The height of the cone (h) is 8 in. Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × 3² × 8 = 75.36 in.³ Hence, from the above, We can conclude that the volume of the given cone is: 75.36 in.³ Question 2. What is the volume of the cone? Use 3.14 for π. Answer: The given figure is: Now, From the given figure, We can observe that The radius of the cone (r) is: 4 cm The slant height of the cone (h) is: 5 cm Now, We know that, l² = r² + h² 5² = 4² + h² h² = 5² – 4² h² = 9 h = 3 cm Now, We know that, The volume of a cone (V) = $$\frac{1}{3}$$πr²h So, V = $$\frac{1}{3}$$ × 3.14 × 4² × 3 = 50.24 cm³ Hence, from the above, We can conclude that the volume of the given cone is: 50.24 cm³ Lesson 8.4 Find Volume of Spheres Quick Review To find the volume of a sphere, use the formula V = $$\frac{4}{3}$$πr3 Example Find the volume of the composite figure. Use 3.14 for π. Answer: First, find the volume of the sphere. Use 3.14 for π. V = $$\frac{4}{3}$$πr3 = $$\frac{4}{3}$$π(3.5)3 → Substitute 3.5 for r. = 57.17π ≈ 179.5 cm3 Divide by 2 to find the volume of the hemisphere: 179.5 ÷ 2 ≈ 89.75 cubic centimeters. Then, find the volume of the cone. Use 3.14 for π. V = $$\frac{1}{3}$$πr2 h = $$\frac{1}{3}$$π(3.5)2(14) → Substitute 3.5 for r and 14 for h. = 57.171 ≈ 179.5 cm3 The volume of the composite figure is approximately 89.75 + 179.5 ≈ 269.25 cubic centimeters. Practice Question 1. What is the volume of the sphere? Use $$\frac{22}{7}$$ for π. Answer: The given figure is: Now, From the given figure, We can observe that The radius of the sphere (r) is: 14 cm Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × $$\frac{22}{7}$$ × 14³ = 11,498.6 cm³ Hence, from the above, We can conclude that the volume of the given sphere is: 11,498.6 cm³ Question 2. The surface area of a sphere is 1,017.36 square inches. What is the volume of the sphere? Use $$\frac{22}{7}$$ for π. Answer: It is given that The surface area of a sphere is 1,017.36 square inches. Now, We know that, The surface area of a sphere (S.A) = 4πr² So, 4πr² = 1,017.36 r² = $$\frac{1,017.36}{4π}$$ r² = 80.92 r = 8.99 inches Now, We know that, The volume of a sphere (V) = $$\frac{4}{3}$$πr³ So, V = $$\frac{4}{3}$$ × $$\frac{22}{7}$$ × (8.99)³ = 3,044.68 inches³ Hence, from the above, We can conclude that the volume of the given sphere is: 3,044.68 inches³ Question 3. What is the volume of the composite figure? Use $$\frac{22}{7}$$ for π. Answer: The given figure is: Now, From the above, We can observe that the given figure is a combination of a hemisphere and a cone So, The volume of the given figure = The volume of a hemisphere + The volume of a cone Now, From the given figure, We can observe that Diameter (d) = 4 cm Height (h) = 10 cm Now, We know that, Radius (r) = $$\frac{Diameter}{2}$$ r = $$\frac{4}{2}$$ r = 2 cm Now, We know that, The volume of a hemisphere (V) = $$\frac{2}{3}$$πr³ So, V = $$\frac{2}{3}$$ × $$\frac{22}{7}$$ × 2³ = 16.76 cm³ Now, We know that, The volume of a cone (V’) = $$\frac{1}{3}$$πr²h So, V’ = $$\frac{1}{3}$$ × $$\frac{22}{7}$$ × 2² × 10 = 41.90 cm³ So, The volume of the given figure = 41.90 + 16.76 = 58.66 cm³ Hence, from the above, We can conclude that the volume of the given figure is: 58.66 cm³ ### Topic 8 Fluency Practice Hidden Clue For each ordered pair, solve the equation to find the unknown coordinate. Then locate and label the corresponding point on the graph. Draw line segments to connect the points in alphabetical order. Use the completed picture to help answer the riddle below. What do squares, triangles, pentagons, and octagons have in common? Answer: Step 1: For each ordered pair, solve the equation to find the unknown coordinate. Step 2: Then locate and label the corresponding point on the graph. Step 3: Draw line segments to connect the points in alphabetical order. Step 4: Use the completed picture to help answer the riddle below. All the squares, triangles, pentagons, and octagons have in common the sum of all the total angles i.e., 360° #### enVision Math Common Core Grade 8 Answer Key ## enVision Math Common Core Grade 7 Answer Key Topic 5 Solve Problems Using Equations and Inequalities ## enVision Math Common Core 7th Grade Answers Key Topic 5 Solve Problems Using Equations and Inequalities Review What You Know! Vocabulary Choose the best term from the box to complete each definition. Question 1. A statement that contains the symbols <, >, ≤, or ≥ is called a(n) Answer: Inequality, Explanation: An open sentence that contains the symbol < , ≤ , > , or ≥ is called an inequality. Inequalities can be solved the same way as equations. Question 2. Properties that state that performing the same operation on both sides of an equation will keep the equation true are called Answer: Properties of equation, Explanation: The first four properties of equality–those that deal with operations–allow us to add, subtract, multiply and divide variables. They also formally express the idea that when we perform the same operations on both sides of an equation, the two sides are still equivalent. In other words, that we can perform the same operation on both sides without changing the values of the variables these are called as properties of equation. Question 3. Addition and subtraction have a(n) ___ because they can “undo” each other. Answer: Inverse relationship, Explanation: An inverse operation are two operations that undo each other, e.g. addition and subtraction or multiplication and division. Can perform the same inverse operation on each side of an equivalent equation without changing the equality. Question 4. Terms that have the same variable are called ___ Answer: Liketerms, Explanation: Terms that have the same variable are called like terms as like terms are terms that have the same variables and powers example ax + bx here there are two terms ax, bx both have x a same variable x and power also same. Properties of Equality Use properties to solve each equation for x. Question 5. x + 9.8 = 14.2 Answer: x = 4.4, Explanation: Given x + 9.8 = 14.2, after subtracting 9.8 both sides we get x + 9.8 – 9.8 = 14.2 – 9.8, x + 0 = 4.4, therefore x = 4.4. Question 6. 14x = 91 Answer: x = 6 remainder 7, Explanation: Given 14x = 91, dividing both sides by 14 we get 14x ÷14 = 91 ÷ 14, x = 91 ÷14, 6 14)91( 84 7 therefore x = 6 with remainder 7. Question 7. $$\frac{1}{3}$$x = 24 Answer: x = 72, Explanation: Given $$\frac{1}{3}$$x = 24, multiplying both sides by 3 we get $$\frac{1}{3}$$x X 3 = 24 X 3, therefore x= 24 X 3= 72. Like Terms Combine like terms in each expression. Question 8. $$\frac{1}{4}$$k + $$\frac{1}{4}$$m – $$\frac{2}{3}$$k + $$\frac{5}{9}$$m Answer: $$\frac{1}{4}$$k + $$\frac{1}{4}$$m – $$\frac{2}{3}$$k + $$\frac{5}{9}$$m = – $$\frac{5}{12}$$k + $$\frac{29}{36}$$m, Explanation: Given $$\frac{1}{4}$$k + $$\frac{1}{4}$$m – $$\frac{2}{3}$$k + $$\frac{5}{9}$$m, we combine like terms in expression as ($$\frac{1}{4}$$k – latex]\frac{2}{3}k) +
($$\frac{1}{4}$$m + $$\frac{5}{9}$$m) =
($$\frac{1}{4}$$ – $$\frac{2}{3}$$)k +
($$\frac{1}{4}$$ + $$\frac{5}{9}$$)m,
before subtracting or adding we make common
denominators for the both terms so
$$\frac{1 X 3 – 2 X 4}{12}$$k +
$$\frac{1 X 9 + 5 X 4}{36}$$m =
$$\frac{3 – 8}{12}$$k + $$\frac{ 9+ 20}{36}$$m = $$\frac{- 5}{12}$$k + $$\frac{29}{36}$$m =
–$$\frac{5}{12}$$k + $$\frac{29}{36}$$m.

Question 9.
-4b + 2w +(-4b) + 8w
Answer:
-8b + 10w,

Explanation:
Given -4b + 2w +(-4b) + 8w we combine like terms as
– 4b + (-4b) and 2 w + 8w =
– 4b + (-4b) + 2w + 8w =
– 8b + 10w, therefore -4b + 2w +(-4b) + 8w = -8b + 10w.

Question 10.
6 – 5z + 8 – 4z + 1
Answer:
-9z + 15 or 15 – 9z,

Explanation:
Given 6 – 5z + 8 – 4z + 1 we combine like terms as
(6 + 8 + 1) and (-5z – 4z) as
6 + 8 + 1 – 5z – 4z = 15 – 9z,
therefore 6 – 5z + 8 – 4z + 1 = 15 – 9z or -9z + 15.

Inequalities

Question 11.
Write an inequality that represents the situation:
A large box of golf balls has more than 12 balls.
Describe how your inequality represents the situation.
Answer:
12 > x is Inequality,

Explanation:
Given a large box of golf balls has more than 12 balls means
12 > x is inequality, where x = the number of golf balls,
inequality states that the number of golf balls is over 12.

Language Development

Fill in the Venn diagram to compare and contrast equations and inequalities.

In the box below, draw pictures to represent the
terms and phrases in the overlap section of your diagram.
Answer:

PICK A PROJECT

PROJECT 5A

How many different ways could you sort a basket of vegetables?
PROJECT: COMPARING WITH A VENN DIAGRAM

Explanation:
Ways of sorting a basket of vegetables as per
1. A part of a plant used as a food (spinach),
2. Do not contain seeds (potatoes),
3. Distinct in taste – can be sweet (sweet potato), salty,
sore or bitter (bitter gourd),
4. Mostly green in color (cucumber),
5. Supply fiber, vitamins, minerals and trace elements,
6. Involved in vegetative reproduction.

PROJECT 5B
Which character would you be from your favorite play? Why?
PROJECT: WRITE A PLAY

Answer:
Character I would be from my favorite plays is Cinderella,
Always Be Kind No Matter What –
A little kindness goes a long way, and no doubt
Cinderella is a kind soul. Her kindness can make
her seem like a pushover (especially when her evil
stepmother and stepsisters are overloading her with housework!).

Explanation:
Play:

1. Cinderella is a princess.
2. She was born to 2 lovely parents.
3. Her mother dies early, and her father remarries.
4. But the stepmother and Cinderella’s two
stepsisters treated her like a servant.
5. Cinderella meets the Prince once in the forest.
6. She goes to the royal ball ceremony with the
fairy godmother’s help and dances with the prince.
7. The carriage, driver, horse, dress, etc.,
were all magical and would have returned
to normal just after the clock struck midnight.
8. In a hurry to leave the palace before midnight,
Cinderella forgets her slippers.
9. The prince traces her back by those slippers and marries her.
10. They live happily ever after.
At the end Cinderella kind heart wins,
so it’s my favorite play.

PROJECT 5C
If you could live in another country, where would you live, and why?
PROJECT: EXCHANGE SOUVENIRS

Answer:
I would live in Canada,
Many people have the dream of living in another country for at least some time during their life. And with good reason. Their are so many benefits.

Moving overseas can dramatically changes my life for the better.
In fact, there are a whole host of reasons why I decided to move to another country.
Living overseas can offer new opportunities, new lifestyles,
new careers and a new direction, financially better prospects.

Explanation:
I would prefer living in canada because
It’s an absolutely stunning place,
Great job opportunities,
People are family friendly,
The best of modern metropolitan living,
Familiar culture and language,
Free health care,
The Canadian sense of humor,
Canadian crime rates are incredibly low,
Canada is an ‘education superpower’ country,
World leaders in quantum computing,
medical research and space science,
Canada has a wonderfully varied climate,
Canadians breathe some of the cleanest air on this planet,
Great food, Niagara Falls, Cheap living costs,
A country you can be proud of living.

PROJECT 5D
How would you prepare for being on a game show?
PROJECT: SOLVE RANDOMIZED EQUATIONS AND INEQUALITIES

Preparing for being on a game show
1. Studying the format of the show,
2.Taking it seriously and practice,
3. Don’t be guilt – No matter how much I am
losing by keep myself cool and dignity,
4. Stay positive,
5. Enjoy it.

### Lesson 5.1 Write Two-Step Equations

Marley collects golf balls. His neighbor Tucker
collects 3 more than twice as many golf balls as Marley.

A. How can you use a table to represent the number of golf balls in Marley’s collection, m, and the number of golf balls in Tucker’s collection?
B. How can you use an algebraic expression to represent
the number of golf balls in Tucker’s collection?
a.

b. Algebraic expression is t = 2m + 3,

Explanation:
Given Marley collects golf balls. His neighbor Tucker
collects 3 more than twice as many golf balls as Marley so
a. The table to represent the number of golf balls in
Marley’s collection m and the number of golf balls
in Tucker’s collection are if Marley’s has 10 golf balls
then Tucker will have 2 X 10 + 3 = 20 + 3 = 23,
if Marley’s has 12 golf balls then Tucker will have
2 X 12 + 3 = 24 + 3 = 27, if Marley’s has 15 golf balls
then Tucker will have 2 X 15 + 3 = 30 + 3 = 33,
if Marley’s has 18 golf balls
then Tucker will have 2 X 18 + 3 = 36 + 3 = 39 and
if Marley’s has 20 golf balls
then Tucker will have 2 X 20 + 3 = 40 + 3 = 43 golf balls
as shown above in the table.
b. As Marley collects golf balls, His neighbor Tucker
collects 3 more than twice as many golf balls as Marley so
the algebraic expression to represent
the number of golf balls in Tucker’s collection is
t = 2m + 3 here t is to represent Tucker and m
is to represent Marley.

Focus on math practices
Look for Relationships How do the terms of the
expression you wrote in Part B relate to the values in the table?
Answer:
Yes, the terms of the expression I wrote in Part B
relate to the values in the table,

Explanation:
As the values in the table of Marley’s collection of
golf balls and Tucker’s collection of golf balls are related
with the terms of expression as Tucker’s collection
of golf balls are dependent on collection of
Marley’s golf balls as twice plus 3 more so values
of table are m = m and t = 2m + 3,
therefore the terms of expression I wrote in Part B
relate to the values in the table.

Essential Question
How does an equation show the relationship between variables and
other quantities in a situation?
Answer:
Yes, an equation show the relationship between
variables and other quantities in a situation with
equality and inequalities signs,

Explanation:
An equation is distinct because it has an equals sign and
that in itself creates a relationship.
Usually, it’s relating the left side to the right side in
terms of that fact that they’re equal to each other and
unlike inequalities which form more of a relation between two quantities that can be greater than or less than.

Try It!

Cole buys a new laptop for $335. He makes a down payment of$50 and pays the rest in 6 equal monthly payments, p.
What equation represents the relationship between the cost of the laptop and Cole’s payments?
Answer:

Explanation:
Given Cole buys a new laptop for $335. He makes a down payment of$50 and pays the rest in 6 equal monthly payments, p.
The equation cost = $50 + 6 X p represents the relationship between the cost of the laptop and Cole’s payments as shown above. Convince Me! Why are both multiplication and addition used in the equation that represents Cole’s monthly payments? Answer: Cole’s makes a down payment of$50 first and pays
the rest in 6 equal monthly payments, p not all at same time,
So, both multiplication and addition are used in the equation,

Explanation:
As given Cole buys a new laptop for $335. He makes a down payment of$50 first and
pays the rest in 6 equal monthly payments, p.
So first we write multiplication as 6 X p then we add
initial down payment $50 in the equation. Try It! Marcia and Tamara are running a race. Marcia has run 4 kilometers. Tamara has completed of the race and is 2.5 kilometers ahead of Marcia. Write an equation that represents the relationship between the distances each girl has run. Let k represent the total length of the race in kilometers. Answer: Marica = m = 4km, k = Tamara = t = 2. 5km + m = 2.5km + 4km = 6.5km, Explanation: Given Marcia and Tamara are running a race. Marcia has run 4 kilometers. Tamara has completed of the race and is 2.5 kilometers ahead of Marcia. Let k represent the total length of the race in kilometers, m for Marica and t for Tamara and Marica m = 4 km and Tamara and total length of the race is t or k = 2.5km + 4km = 6.5km, Equation that represents the relationship between the distances each girl has run is k = t = 2.5 km + m. Try It! At the mall, Claire buys a hat that is 60% off and socks that are reduced to$5.49.
She spends a total of $9.49. Let x represent the cost of the hat. Which of the following equations correctly represents Claire’s shopping trip? Answer: Equation correctly representing Claire’s shopping trip is 0.4x + 5.49 = 9.49, Explanation: Given at the mall, Claire buys a hat that is 60% off and socks that are reduced to$5.49.
She spends a total of $9.49. Let x represent the cost of the hat means and 60% off so 40%x = 0.4x, Socks =$5.49 and total = $9.49, So the following equation correctly represents Claire’s shopping trip is 0.4x + 5.49 = 9.49. KEY CONCEPT You can write an equation with more than one operation to represent a situation. Do You Understand? Question 1. Essential Question How does an equation show the relationship between variables and other quantities in a situation? Answer: Yes, an equation show the relationship between variables and other quantities in a situation with equality and inequalities signs, Explanation: An equation is distinct because it has an equals sign and that in itself creates a relationship. Usually, it’s relating the left side to the right side in terms of that fact that they’re equal to each other and unlike inequalities which form more of a relation between two quantities that can be greater than or less than. Question 2. Use Structure Do the equations $$\frac{1}{5}$$x + 2 = 6 and $$\frac{1}{5}$$(x + 2) represent the same situation? Explain. Answer: Equations $$\frac{1}{5}$$x + 2 = 6 and $$\frac{1}{5}$$(x + 2) do not represent the same situation, Explanation: Given equations are 1. $$\frac{1}{5}$$x + 2 = 6 and 2. $$\frac{1}{5}$$(x + 2), 1. If we solve $$\frac{1}{5}$$x + 2 = 6 we get $$\frac{1}{5}$$x = 6 -2, $$\frac{1}{5}$$x = 4, x = 4 X 5 = 20 and 2. If we solve $$\frac{1}{5}$$(x + 2) we get $$\frac{1}{5}$$ X x + $$\frac{1}{5}$$ X 2 = $$\frac{1}{5}$$ X x = $$\frac{5}{2}$$, x = $$\frac{5}{2}$$ X 5, therefore x = $$\frac{25}{2}$$ as 20 ≠ $$\frac{25}{2}$$, So equations $$\frac{1}{5}$$x + 2 = 6 and $$\frac{1}{5}$$(x + 2) do not represent the same situation. Question 3. How do you decide which operations to use when writing an equation? Answer: The order of operations is a rule that tells the correct sequence of steps for evaluating a math expression. We can remember the order using PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). Explanation: The order of operations define the priority in which complex equations are solved. The top priority is your parenthesis, then exponents, followed by multiplication and division, and finally addition and subtraction (PEMDAS). Do You Know How? Question 4. Rita started the day with r apps. Then she deleted 5 apps and still had twice as many apps as Cora has. Write an equation that represents the number of apps each girl has. Answer: Equation : r – 5 = 2c, Cora = 36 apps, Rita =77 apps, Explanation: Given Rita started the day with r apps. Then she deleted 5 apps and still had twice as many apps as Cora has. Let us take Cora as c = 36 apps and Rita has twice as Cora after deleting 5 apps means the equation is 2c = r – 5, and Rita has r = 2c + 5 = 2 X 36 + 5 = 72 + 5 = 77 apps, therefore, Equation : r – 5 = 2c, Cora = 36 apps, Rita =77 apps respectively. Question 5. Write a problem that could be represented by the equation 5n – 6 = 19. Answer: Jim’s age is 6 years less than 5 times his younger brother’s age, Explanation: Given to write a problem that could be represented by the equation 5n – 6 = 19, So, lets take Jim’s age at present is 19 years which is 6 years less than 5 times his younger brother’s age. Question 6. Kayleigh babysat for 11 hours this week. That was 5 fewer than $$\frac{2}{3}$$ as many hours as she babysat last week, h. Write an equation to represent the number of hours she babysat each week. Answer: Equation to represent the number of hours she babysat each week is $$\frac{2}{3}$$h – 5 = 11, Explanation: Given Kayleigh babysat for 11 hours this week. That was 5 fewer than $$\frac{2}{3}$$ as many hours as she babysat last week, h. The equation to represent the number of hours she babysat each week is $$\frac{2}{3}$$h – 5 = 11. Practice & Problem Solving Question 7. A farmer ships oranges in wooden crates. Suppose each orange weighs the same amount. The total weight of a crate filled with g oranges is 24.5 pounds. Write an equation that represents the relationship between the weight of the crate and the number of oranges it contains. Answer: Equation that represents the relationship between the weight of the crate and the number of oranges it contains is Explanation: Given a farmer ships oranges in wooden crates. Suppose each orange weighs the same amount as 0.38lb. The total weight of a crate filled with g oranges is 24.5 pounds. So an equation that represents the relationship between the weight of the crate and the number of oranges it contains is 24.5 = 15lb + (0.38lb X g) and 1 pound = 1 lb. Question 8. Jordan wrote the following description: Three fewer than one fourth of x is 12. Write an equation to represent the description. Answer: Equation: $$\frac{1}{4}$$x – 3 = 12, Explanation: Given Jordan wrote the following description: Three fewer than one fourth of x is 12. So the equation to represent the description is $$\frac{1}{4}$$x – 3 = 12. Question 9. At a graduation dinner, an equal number of guests were seated at each of 3 large tables, and 7 late-arriving guests were seated at a smaller table. There were 37 guests in all . If n represents the number of people seated at each of the large tables, what equation represents the situation? Answer: The number of people seated at each of the large tables, represented by an equation is $$\frac{1}{3}$$n + 7 = 37, Explanation: Given at a graduation dinner, an equal number of guests were seated at each of 3 large tables and 7 late-arriving guests were seated at a smaller table. There were 37 guests in all. If n represents the number of people seated at each of the large tables, The equation representing the situation is $$\frac{1}{3}$$n + 7 = 37. Question 10. Last night, 4 friends went out to dinner at a restaurant. They split the bill evenly. Each friend paid$12.75 for
his or her meal and each left the same amount for a tip, t.
The total dinner bill including the tip was $61. What equation could you use to describe the situation? Answer: Equation: 4 X$12.75 + 4t  = $61, Explanation: Given last night, 4 friends went out to dinner at a restaurant. They split the bill evenly. Each friend paid$12.75 for
his or her meal and each left the same amount for a tip, t.
The total dinner bill including the tip was $61. Therefore, equation to describe the situation is 4 X$12.75 + 4t  = $61. Question 11. Mia buys 4$$\frac{1}{5}$$ pounds of plums. The total cost after using a coupon for 55¢ off her entire purchase was$3.23. If c represents the cost of the plums
in dollars per pound, what equation could represent the situation?
Answer:
Equation:  $3.23 = 4$$\frac{1}{5}$$ – .55, Explanation: Given Mia buys 4$$\frac{1}{5}$$ pounds of plums. The total cost after using a coupon for 55¢ off her entire purchase was$3.23. If c represents the cost of the plums
in dollars per pound, the equation representing
the situation is $3.23 = 4$$\frac{1}{5}$$ – .55, as 1 dollar = 100 cents. For 12 and 13, use the equation shown at the right. Question 12. Describe a situation that the equation could represent. Answer: There are 6 groups of children each group there are 15 students and out of which 3 students are absent, so number of students present are g represented by equation g + 3 ÷ 6 = 15, Explanation: Given equation as g + 3 ÷ 6 = 15, let us take the situation as there are 6 groups of children each group there are 15 students and out of which 3 students are absent, so number of students present are g, so that g + 3 = 15 X 6, therefore g + 3 ÷ 6 = 15. Given the Question 13. Reasoning Would the situation you wrote for Problem 12 work if the denominator in the equation were doubled? Explain why or why not. Answer: Yes, Explanation: Reasoning is that if the situation I wrote for Problem 12 will work if the denominator in the equation were doubled because the value of number of students present will change or increase. Question 14. You want to buy a pet iguana. You already have$12 and
plan to save $9 per week. a. Model with Math If w represents the number of weeks until you have enough money to buy the iguana, what equation represents your plan to afford the iguana? b. Explain how you could set up an equation to find the amount of money you should save each week to buy the iguana in 6 weeks. Answer: a. Equation :$48 = $12 +$9w,
b. Equation to find the amount(a) of money I should save
each week to buy the iguana in 6 weeks is $48 =$12 + 6a or $60 by 6, Explanation: Given I want to buy a pet iguana. I already have$12 and
plan to save $9 per week and iguana costs$48,
a. If w represents the number of weeks until
I have enough money to buy the iguana, the equation
representing my plan to afford the iguana is $48 =$12 + $9w, b. If the amount(a) of money I should save each week to buy the iguana in 6 weeks is$48 =  $12 + 6a or a =$60 by 6.

Question 15.
In a certain country, the life expectancy of a
woman born in 1995 was 80.2 years.
Between 1995 and 2005, the life expectancy
increased 0.4 year every 5 years.
a. If L represents the life expectancy of a woman born in 2005,
what equation could you use to represent the situation?
b. Reasoning Could two differences equations be used to
find the value of L? Explain.
Answer:
a. L = 80.2 + 80.2 X 2(0.4),
b. Yes, Equation 1: L = 80.2 + 80.2 X 2(0.4) and
Equation 2: L – 80.2 = 2 X 0.4 X 80.2,

Explanation:
Given in a certain country, the life expectancy of a
woman born in 1995 was 80.2 years.
Between 1995 and 2005, the life expectancy
increased 0.4 year every 5 years.
a. If L represents the life expectancy of a woman born in 2005,
The equation that could be used to represent the situation is
L = 80.2 + 80.2 X 2(0.4), we multiply 80.2 with 2 of 0.4 as
every 5 years it is increased by 0.4 and it means from
1995 and 2005 it will increases twice of 0.4,
b. Yes two equations can be used to find the value of L as
First we calculate L by Equation 1: L = 80.2 + 80.2 X 2(0.4) and
Second we calculate L by Equation 2: L – 80.2 = 2 X 0.4 X 80.2.

Question 16.
Higher Order Thinking Use the equation 5x – 13 = 12
a. Write a description that represents the equation.
b. Of the numbers 1, 2, 3, 4, and 5, which are solutions to the equation?
Answer:
a. Description:
There are few sets of books as x in the library and
each set contains 5 books in that 13 books children
took to home and rest are 12 books in the libray,
So find out how many number of books x were there  in the library,
b. Solution to the equation is 5,

Explanation:
Given the equation 5x – 13 = 12,
a. Wrote the description that represents the equation as
there are few sets of books as x in the library and
each set contains 5 books in that 13 books children
took to home and rest are 12 books in the libray,
So find out how many number of books x were there  in the library,
b. Of the numbers 1, 2, 3, 4, and 5, the solution to the equation
5x – 13 = 12 is 5x = 12 + 13,
5x = 25, therefore x = 25 by 5 = 5, therefore solution to
the equation 5x – 13 = 12 is 5.

Assessment Practice

Question 17.
A garden contains 135 flowers, each of which is either red or yellow.
There are 3 beds of yellow flowers and 3 beds of red flowers.
There are 30 yellow flowers in each yellow flower bed.
PART A
If r represents the number of red flowers in each red flower bed,
what equation could you use to represent the number of red and yellow flowers?

PART B
Write another real-world situation that your equation from Part A could represent.
Answer:
Part A:
Equation to represent the number of red and
yellow flowers is 135 = 3 X 30 + 3r,

Part B:
Real World situation :
There are 135 number of mangoes with fruit seller out of
which 3 baskets mangoes are green in color and 3 baskets of
mangoes are yellow in color and there are total 30 yellow mangoes
in each yellow basket, write an equation to represent the number of
green and yellow mangoes?,

Explanation:
Given a garden contains 135 flowers, each of which is either red or yellow.
There are 3 beds of yellow flowers and 3 beds of red flowers.
There are 30 yellow flowers in each yellow flower bed.
PART A
If r represents the number of red flowers in each red flower bed,
the equation I could use to represent the number of red and
yellow flowers is 135 = 3 X 30 + 3r and
PART B
Wrote another real-world situation that my equation
from Part A could represent as Real World situation :
There are 135 number of mangoes with fruit seller out of
which 3 baskets mangoes are green in color and 3 baskets of
mangoes are yellow in color and there are total 30 yellow mangoes
in each yellow basket, write an equation to represent the number of
green and yellow mangoes? respectively.

### Lesson 5.2 Solve Two-Step Equations

Elizabeth wrote the following clues. What is the relationship between the shapes?

Answer:
1. 4c + t = 2s + t, 2 circles = square
2. 3t + 2c = 2c + s, 3 triangles = square
3. s + 2c = 6t, square = 6 triangles – 2 circles,

Explanation:
Elizabeth wrote the clues as
1. 4 circles + 1 triangle = 2 squares and 1 triangle which means
4 circles = 2 squares, so 1 square is equal to 2 circles.
2. 3 triangles + 2 circles = 2 circles + 1 square,
so 1 square is equal to 3 triangles,
3. 1 square + 2 circles = 6 triangles,
so, 1 square = 6 triangles – 2 circles.

Use Structure
How can you use properties of equality to reason about these equations?
Answer:
1. Addition Property,
2. Addition property,
3. Substitution property,

Explanation:
As we know
1. 4c + t = 2s + t,
Addition Property says for all real numbers x, y, and z ,
if x=y , then x + z = y + z, So 4c + t = 2s + t means 4c = 2s,
2. 3t + 2c = 2c + s means as per Addition Property
3t = s,
3. 1 square + 2 circles = 6 triangles we use
substitution property for circles and triangles as
1 square = 2 circles and 1 square = 3 triangles,
So 1 square + 1 square = 2 X 3 triangles =  2 squares,
2 squares = 2 squares.

Focus on math practices
Look for Relationships Complete the equation with only triangles
using the relationships from the clues shown above.

Answer:
Equation : 4c + 1t = 7t,

Explanation:
As we know 4c = 2s and 1s = 3t means 1s = 2c,
so 3t = 2c therefore 4c = 6t as given 4c +1t = 6t + 1t = 7t
therefore the equation with only triangles
using the relationships from the clues shown above is 4c + 1t = 7t.

Essential Question
How is solving a two-step equation similar to solving a one-step equation?
Answer:
Yes, solving a two-step equation is similar to solving a one-step equation,

Explanation:
Solving a one-step or two-step equation:
In solving an equation is to have only variables on
one side of the equal sign and numbers on the
other side of the equal sign.
The other alike is to have the number in front of the variable
equal to one the variable does not always have to be x.
These equations can use any letter as a variable.

Try It!
Andrew rents bowling shoes for $4. He bowls 2 games. Andrew spent a total of$22. How much was the cost of each game, b?
Complete the bar diagrams, and then solve the problem.

Answer:

Explanation:
Given Andrew rents bowling shoes for $4. He bowls 2 games. Andrew spent a total of$22. How much was the cost of each game, b,
Completed the bar diagrams as shown above,
Total spent = shoe rental + 2 X cost of each game,
$22 =$4 + 2 X b,
$22 –$4 = $4 + 2b –$4,
$18 = 2b, Now each game cost is$18 by 2 = 2b by 2
b = $9, therefore cost of each game is$9.

Convince Me!
What were the two steps you used to solve this equation?
Answer:
Two step equations can be solved in two steps using two different
properties of equality, I used subtraction property and Division property,

Explanation:
Total spent = shoe rental + 2 X cost of each game is given
$22 =$4 + 2 X b,
First we use the subtraction property of equality to
isolate the term containing the variable,
$22 –$4 = $4 + 2b –$4,
$18 = 2b, Secondly we use the division property of equality to isolate the variable or get the variable by itself on one side of the equation as$18 by 2 = 2b by 2
b = $9, therefore cost of each game is$9.

Try It!
Kirsty ran 24 laps in a charity run and then walked
0.2 kilometer to the presentation table.
The total distance Kirsty traveled was 29.6 kilometers.
What was the distance of each lap? Explain how you solved the problem.
Answer:
1.225km is equal to each laps,

Explanation:
Given Kristy total distance traveled = 29.6km
and Kristy ran 24 laps and later walk 0.2km.
It simply implies that
Total distance traveled = distance covered running + distance covered walking,
Since we know that Total distance traveled = 29.6km,
distance covered running = 24 laps,
distance covered walking = 0.2km,
Distance covered running in km = total distance traveled – distance covered walking,
Distance covered running in km = 29.6km – 0.2km = 29.4km.
To now find the distance for each lap.
Since we have: Distance covered running in km = 29.4km.
Distance covered running in lap = 24 laps i.e 24 laps = 29.4km,
1 lap = x Use cross-multiple 24 laps X x =
29.4km × 1 lap x = 29.4km / 24 x = 1.225km,
Therefore 1.225km is equal to each laps.

KEY CONCEPT
The properties of equality can be applied the same way when
solving two-step equations as when solving one-step equations.

Answer:
Yes, The properties of equality can be applied the same way when
solving two-step equations as when solving one-step equations.
Two step equations can be solved in two steps using two different
properties of equality, We used subtraction property and division property,

Explanation:
We have 5x + 27 = 122,
First we use the subtraction property of equality to
isolate the term containing the variable,
5x + 27 – 27 = 122 – 27,
5x = 95,
Secondly we use the division property of equality to
isolate the variable or get the variable by itself on
one side of the equation as 5x by 5 = 95 by 5
x = 19, therefore the properties of equality can be applied
the same way when solving two-step equations as
when solving one-step equations.

Do You Understand?

Question 1.
Essential Question How is solving a two step equation similar to
solving a one-step equation?
Answer:
Yes, solving a two-step equation is similar to solving a one-step equation,

Explanation:
Solving a one-step or two-step equation:
In solving an equation is to have only variables on
one side of the equal sign and numbers on the
other side of the equal sign.
The other alike is to have the number in front of the variable
equal to one the variable does not always have to be x.
These equations can use any letter as a variable.

Question 2.
Use Structure Preston uses the bar diagram below to represent 4x – 3 = 13.
How would you use the bar diagram to solve for x?

Answer:
By using bar diagram we solve x = 4,

Explanation:
A helpful way to solve problem is to use a bar diagram.
A bar is used to represent the whole. So we label it
with the amount of the whole. Finally, we indicate the required
number of parts with a question mark and determine the value,
By seeing bar diagram we have the total + 3 is divided into
4 parts of x now we solve x as 4x – 3 = 13 by adding 3 both sides,
4x – 3 + 3 = 13 + 3,
4x = 16, Now dividing both sides by 4 we get
4x ÷ 4 = 16 ÷ 4,
x = 4.

Question 3.
Clara has solved the problem 6p – 12 = 72 and says that p = 14.
How can you check to see if Clara is correct?
Answer:
By solving using two-step equation we get p = 14,
we check Clara is correct,

Explanation:
Given Clara equation as 6p – 12 = 72 we solve
First by the addition property of equality to
isolate the term containing the variable,
6p – 12 + 12 = 72 + 12,
6p = 84,
Secondly use the division property of equality to
isolate the variable or get the variable by itself on
one side of the equation as
6p by 6 = 84 by 6, we get p = 14, so checked Clara is correct.

Do You Know How?

Question 4.
Clyde is baking, and the recipe requires 1$$\frac{1}{3}$$ cups of flour.
Clyde has 2 cups of flour, but he is doubling the recipe to make twice as much.
How much more flour does Clyde need?

a. Write an equation to represent the problem.
Let c represent the amount of flour Clyde needs.
b. Solve the equation.
Answer:
a. Equation: c = 2 X 1$$\frac{1}{3}$$ – 2,
b. Clyde needs $$\frac{2}{3}$$ cup more,

Explanation:
Given Clyde is baking, and the recipe requires 1$$\frac{1}{3}$$ cups of flour.
Clyde has 2 cups of flour, but he is doubling the recipe to make twice as much.
If c represent the amount of flour Clyde needs the equation is
c = 2 X 1$$\frac{1}{3}$$ – 2,
b. Now solving c = 2 X $$\frac{1 X 3 + 1}{3}$$ – 2,
c = 2 X $$\frac{4}{3}$$ – 2,
c = $$\frac{2 X 4}{3}$$ – 2,
c = $$\frac{8}{3}$$ – 2,
c = $$\frac{8 – 6}{3}$$,
c = $$\frac{2}{3}$$ .

Question 5.
Four times a number, n, added to 3 is 47.
a. Write an equation that you can use to find the number.
b. What is the number represented by n?
Answer:
a. Equation: 4n + 3 = 47,
b. The number represented by n is 11,

Explanation:
Given four times a number n added to 3 is 47 means
a. 4 X n + 3 is equal to 47,
therefore the equation is 4n + 3 = 47,
b. Upon solving 4n + 3 = 47 we get value of n, So
first we subtract 3 both sides as
4n + 3  – 3 = 47 – 3,
4n = 44, now we divide both sides by 4 we get
4n ÷ 4 = 44 ÷ 4, so n = 11, therefore the number
represented by n is 11.

Practice & Problem Solving

Question 6.
Use the bar diagram to help you solve the equation 4x – 12 = 16.

Answer:
By using bar diagram we solve x = 7,

Explanation:
A helpful way to solve problem is to use a bar diagram.
A bar is used to represent the whole. So we label it
with the amount of the whole. Finally, we indicate the required
number of parts with a question mark and determine the value,
By seeing bar diagram we have the total + 12 is divided into
4 parts of x now we solve x as 4x – 12 = 16 by adding 12 both sides,
4x – 12 + 12 = 16 + 12,
4x = 28, Now dividing both sides by 4 we get
4x ÷ 4 = 28 ÷ 4,
x = 7.

Question 7.
Complete the steps to solve the equation.

Answer:

Explanation:
Completed the steps to solve the equation as shown above
first we subtract 2 both sides as
1/5t + 2 – 2 = 17 – 2,
1/5t = 15, now we multiply both sides by 5 we get
1/5t = 15 X 5,  we get t = 75.

Question 8.
Use the bar diagram to write an equation. Then solve for x.

Answer:
Equation is 3x – 5 = 7 and x = 4,

Explanation:
Using the bar diagram we have 3 times x and subtracting 5
we are getting total as 7, so the equation will be
3x – 5 = 7, now solving to get x first we add 5 both sides as
3x – 5 + 5 = 7 + 5,
3x = 12, now we divide both sides by 3,
3x ÷ 3 = 12 ÷ 3, we get x = 4.

Question 9.
While shopping for clothes, Tracy spent $38 less than 3 times what Daniel spent. Write and solve an equation to find how much Daniel spent. Let x represent how much Daniel spent. Answer: Equation for Daniel spent is : 3x – 38 = 10, Daniel spent is$16,

Explanation:
Given while shopping for clothes, Tracy spent $38 less than 3 times what Daniel spent. Let x represent how much Daniel spent. So an equation to find how much Daniel spent is 3x – 38 = 10, now we solve by adding 38 both sides as 3x – 38 + 38 = 10 + 38, 3x = 48, now we divide both sides by 3, 3x ÷ 3 = 48 ÷ 3, we get x = 16, therefore, Daniel spent is$16.

Question 10.
Solve the equation 0.5p – 3.45 = -1.2.
Answer:
p = 4.5,

Explanation:
Given to solve the equation 0.5p – 3.45 = -1.2, first
we add +3.45 both side as
0.5p – 3.45 + 3.45 = -1.2 + 3.45,
0.5p = 2.25, now we divide both sides by 0.5 as
0.5p  ÷ 0.5 = 2.25 ÷ 0.5, we get p = 4.5.

Question 11.
Solve the equation $$\frac{n}{10}$$ + 7 = 10.
Answer:
n = 30,

Explanation:
Given to solve the equation $$\frac{n}{10}$$ + 7 = 10,
first we subtract 7 both sides as
$$\frac{n}{10}$$ + 7 – 7  = 10 – 7,
$$\frac{n}{10}$$ = 3, now we multiply both sides by 10 as
$$\frac{n}{10}$$ X 10 = 3 X 10,
we get n = 30.

Question 12.
A group of 4 friends went to the movies. In addition to their tickets,
they bought a large bag of popcorn to share for $6.25. The total was$44.25.
a. Write and solve an equation to find the cost of one movie ticket, m.
b. Draw a model to represent the equation.
Answer:
a. 4m + 6.25 = 44.25,
The cost of one movie ticket m is $9.5, b. Explanation: Given a group of 4 friends went to the movies. In addition to their tickets, they bought a large bag of popcorn to share for$6.25. The total was $44.25. a. Wrote and solved an equation to find the cost of one movie ticket, m as 4m + 6.25 = 44.25, first we subtract 6.25 both sides as 4m + 6.25 – 6.25 = 44.25 – 6.25, 4m = 38, now we divide both sides by 4 as 4m ÷ 4 = 38 ÷ 4, we get m =$9.5.
b. Drawn a bar diagram model to represent the equation
as shown above.

Question 13.
Oliver incorrectly solved the equation 2x + 4 = 10.
He says the solution is x = 7.
a. What is the correct solution?
b. What mistake might Oliver have made?
Answer:
a. The correct solution is x = 3,
b. Instead of subtracting 4 both sides Oliver added
4 both sides,

Explanation:
Given Oliver incorrectly solved the equation 2x + 4 = 10 as
he says the solution is x = 7,
a. The correct solution is first subtract 4 both sides as
2x + 4 – 4 = 10 – 4,
2x = 6 now divide both sides by 2 as
2x ÷ 2 = 6 ÷ 2, we get x = 3,
b. Mistake Oliver had made is if he says
solution is x = 7 and after solving we are getting
x as 3 means he added more 4 both sides.

Question 14.
Use the equation 4.9x – 1.9 = 27.5.
a. Make Sense and Persevere What two properties of equality
do you need to use to solve the equation?
b. The solution is x = .
Answer:
a. We need two properties of equality as
1. Addition property of equality and
2. Division property of equality,
b. The solution is x =  6,

Explanation:
Given the equation 4.9x – 1.9 = 27.5,
two properties of equality we need to use to solve the equation is
first addition property of equality by adding 1.9 both sides as
4.9x – 1.9 + 1.9 = 27.5 + 1.9,
4.9x = 29.4 now we use division property of equality by
dividing 4.9 both sides as
4.9x ÷ 4.9 =29.4 ÷ 4.9, we get x = 6,
b. As solved in bit a. we get the solution for x as 6.

Question 15.
Higher Order Thinking
At a party, the number of people who ate meatballs was
11 fewer than $$\frac{1}{3}$$ of the total number of people.
Five people ate meatballs

a. Write and solve an equation to find the number of people at the party.
Let x represent the number of people at the party.
b. Write a one-step equation that has the same solution.
Answer:
a. Equation: 5 = $$\frac{1}{3}$$ x – 11,
The number of people at the party are x = 48,
b. One-step equation that has the same solution is
$$\frac{1}{3}$$ x   = 16,

Explanation:
At a party, the number of people who ate meatballs was
11 fewer than $$\frac{1}{3}$$ of the total number of people.
Five people ate meatballs, So equation to find the number of
people at the party.
a. Let x represent the number of people at the party. Therefore
Equation is $$\frac{1}{3}$$ x – 11 = 5,
b. As we have 5 people ate meatballs and 11 are fewer than $$\frac{1}{3}$$
of the total number of people. Total number of people a are equal to 16,
therefore one-step equation that has the same solution is
$$\frac{1}{3}$$ x   = 16.

Assessment Practice

Question 16.
In a week, Tracy earns $12.45 less than twice the amount Kayla earns. Tracy earns$102.45. How much does Kayla earn?
Answer:
Kayla earns = $57.45, Explanation: Given in a week, Tracy earns$12.45 less than twice the amount Kayla earns.
Let Kayla earns x and Tracy earns Earns in week= $102.45, Tracy = 2x – 12.45, 102.45 = 2x – 12.45, 2x = 102.45 + 12.45, 2x = 114.90, x= 114.90/2, x =$57.45, therefore Kayla earns 57.45 dollars.

Question 17.
Solve the equation 2x + 4$$\frac{1}{5}$$ = 9.
Explain the steps and properties you used.
Answer:
x = $$\frac{12}{5}$$,
Properties used are 1. Subtraction property of equation and
2. Division property of equation,

Explanation:
Given the equation as 2x + 4$$\frac{1}{5}$$ = 9,
2x + $$\frac{4 X 5 + 1}{5}$$ = 9,
2x + $$\frac{21}{5}$$ = 9, Now we use subtraction property of equation
both sides we subtract $$\frac{21}{5}$$ as
2x + $$\frac{21}{5}$$ – $$\frac{21}{5}$$ = 9 – $$\frac{21}{5}$$,
2x = $$\frac{45 – 21}{5}$$,
2x = $$\frac{24}{5}$$, Now we use division property of equation so we
divide both sides by 2 as 2x/2 = $$\frac{24}{5}$$/2, we get
x = $$\frac{12}{5}$$.

### Lesson 5.3 Solve Equations Using the Distributive Property

Explain It!
Six friends go jet skiing. The total cost for the adventure is $683.88, including a$12 fee per person to rent flotation vests.
Marcella says they can use the equation 6r + 12 = 683.88
to find the jet ski rental cost, r, per person.
Julia says they need to use the equation 6(r + 12) = 683.88.

A. Construct Arguments Whose equation accurately represents the situation?
Construct an argument to support your response.
B. What error in thinking might explain the inaccurate equation?
Answer:
A. Argument : Julia is right the equation 6(r +12) = 683.33 is true,
B. Error in thinking may be instead of 6 persons X $12 fee per person to rent flotation vests, Marcella says they can use the equation 6r + 12 = 683.88 not 6r + 6 X 12 = 683.88, Explanation: Given Six friends go jet skiing. The total cost for the adventure is$683.88,
including a $12 fee per person to rent flotation vests. Marcella says they can use the equation 6r + 12 = 683.88 to find the jet ski rental cost, r, per person. Julia says they need to use the equation 6(r + 12) = 683.88. A. Argument : Julia is right the equation 6(r +12) = 683.33 is true, because the total cost for the adventure is$683.88,
including a $12 fee per person to rent flotation vests means 6 multiply by r in addition to 6 multiply by$12 is
equal to total cost, but not Marcella as she says they
can use the equation 6r + 12 = 683.88,
B. Error in thinking may be instead of 6 persons X $12 fee per person to rent flotation vests, Marcella says they can use the equation 6r + 12 = 683.88 means she is not considering 6 persons rent flotation vests only for one person she is considering, i.e Marcella is saying in the equation for 1 person rent flotation vests instead of for six persons which makes inaccurate equation. Focus on math practices Use Structure How can you use the correct equation to determine the jet ski rental cost per person? Answer: The correct equation is 6(r +$12) = $683.88 and the jet ski rental cost per person is$101.98,

Explanation:
Given six friends go jet skiing. The total cost for the adventure is $683.88, including a$12 fee per person to rent flotation vests.
We have correct equation as 6(r + $12) =$683.88,
r + 12 = 683.88 ÷ 6,
r = (683.88 ÷ 6) – 12,
So the correct equation to determine the jet ski rental cost per person is
r =  $113.98 –$12 = $101.98, therefore The correct equation is 6(r +$12) = $683.88 and the jet ski rental cost per person is$101.98.

Essential Question
How does the Distributive Property help you solve equations?
Answer:
Distributive property helps in simplifying the problems by
breaking the expressions into addition or subtraction.
The distributive property states that when a factor is multiplied
by the sum or subtract of two numbers, we can multiply each of the two numbers
by that factor and then add or subtract them.

Explanation:
Distributive property is defined as the algebraic property used
to multiply two or more numbers within the parenthesis.
We can say that the distributive property helps in simplifying
the problems by breaking the expressions into addition or subtraction.
It multiplies the number outside parentheses which is equal
to the addition or subtraction of product.
Apart from distributive property, there are two types of properties
known as Commutative and Associative.
This property of multiplication is basically used in addition or subtraction.
To solve equation of distributive property, we need to expand the equation,
find the products and add or subtract.

Try It!

A collector has a box of 32 figurines. The value of each figurine
increased by $2.32 over the past year. The box of figurines is now worth$114.24. What was the original cost, x, of one figurine?
The original cost of one figurine was .

Answer:
The original cost of one figurine was $1.25 Explanation: Given a collector has a box of 32 figurines. The value of each figurine increased by$2.32 over the past year.
The box of figurines is now worth $114.24. The original cost of one figurine was 32(x + 2.32) = 114.24, Applying distributive property as 32 X x + 32 X 2.32 = 114.24, 32x + 74.24 = 114.24, 32x = 114.24 – 74.24, 32x = 40, x = 40 ÷ 32, x= 1.25. Convince Me! Can the equation 32x + 2.32 = 114.24 be used to find the original cost of each figurine in the problem above? Explain. Answer: No, the equation 32x + 2.32 = 114.24 cannot be used to find the original cost of each figurine in the problem above, Explanation: As the given equation is 32x + 2.32 = 114.24 cannot be used to find the original cost of each figurine in the problem above because here in the equation it is not using the value of each figurine increased by$2.32 over the past year,
instead of using for 32 figurines it is using only for 1 figurine
it has to use as 32(x + 2.32) = 114.24 not 32x + 2.32 = 114.24,
therefor, the equation 32x + 2.32 = 114.24 cannot be used to
find the original cost of each figurine in the problem above.

Try It!

Use the Distributive Property to solve each equation.
a. –$$\frac{1}{2}$$(b – 6) = 5,
Answer:
b = -2,

Explanation:
Given equation as –$$\frac{1}{2}$$(b – 6) = 5 using
distributive property as –$$\frac{1}{2}$$ X b + (-$$\frac{1}{2}$$) X (-6) = 5,
–$$\frac{b}{2}$$ + $$\frac{6}{2}$$ = 5,
–$$\frac{b}{2}$$ + 3 = 5,
–$$\frac{b}{2}$$ = 5 – 3,
–$$\frac{b}{2}$$ = 2,
-b = 2 X 2,
– b =  4, therefore b = -4.

b. 0.4(x – 0.45) = 9.2,
Answer:
x = 23.45,

Explanation:
Given equation as 0.4(x-0.45) = 9.2 using distributive property as
0.4 X x – 0.4 X 0.45 = 9.2,
0.4x – 0.18 = 9.2,
0.4x = 9.38,
x = 9.38 ÷ 0.4,
x = 23.45.

c. -4(p – 212) = 44,
Answer:
p = 201,

Explanation:
Given equation as -4(p-212) = 44 using distributive property as
-4 X p – 4 X -212 = 44,
-4p + 848 = 44,
-4p = 44 – 848,
p = -804 ÷ -4,
p = 201.

KEY CONCEPT

When solving equations written in the form p(x + 9) = r,
you can use the Distributive Property to multiply the
two terms in the parentheses by the term outside the parentheses.

Do You Understand?

Question 1.
Essential Question How does the Distributive Property help you solve equations?
Answer:
Distributive property helps in simplifying the problems by
breaking the expressions into addition or subtraction.
The distributive property states that when a factor is multiplied
by the sum or subtract of two numbers,
we can multiply each of the two numbers
by that factor and then add or subtract them,
for equation 6(x + 8.5) = 123 we use distributive property
we get x = 12,

Explanation:
Distributive property is defined as the algebraic property used
to multiply two or more numbers within the parenthesis.
We can say that the distributive property helps in simplifying
the problems by breaking the expressions into addition or subtraction.
It multiplies the number outside parentheses which is equal
to the addition or subtraction of product.
Apart from distributive property, there are two types of properties
known as Commutative and Associative.
This property of multiplication is basically used in addition or subtraction.
To solve equation of distributive property, we need to expand the equation,
find the products and add or subtract for equation 6(x + 8.5) = 123
we use distributive property as 6x + 6 X 8.5 = 123,
6x + 51 = 123,
6x = 123 -51,
6x = 72,
x = 72 ÷ 6 = 12.

Question 2.
Make Sense and Persevere How are the terms in parentheses
affected when multiplied by a negative coefficient when the
Distributive Property is applied?
Answer:
If a number outside the parentheses has a negative sign then
the first and simplest way is to change each positive or negative sign
of the terms that were inside the parentheses.
Negative or minus signs become positive or plus signs.
Similarly, positive or plus signs become negative or minus signs,

Explanation:
The terms in parentheses affected when multiplied by a
negative coefficient when the distributive property is applied is the first and
simplest way is to change each positive or negative sign
of the terms that were inside the parentheses.
Negative or minus signs become positive or plus signs.
Similarly, positive or plus signs become negative or minus signs,
Example : If equation is -6(x – 4) if we apply distributive property
-6 X x  -6 X -4,
-6x + 24, means positive x becomes negative 6x and negative 4
has become positive 24 when multiplied by -6.

Question 3.
Reasoning How can an area model help you set up an equation
for a problem situation?
Answer:
Area of a shape is the space occupied by the shape.
The area of the given shape is the shaded part which means
we can set up an equation for a problem situation given,

Explanation:

As shown above the area model for the equation is
(2x + 5)(x+3) as shown.

Do You Know How?

Question 4.
A family of 7 bought tickets to the circus. Each family member also
bought a souvenir that cost $6. The total amount they spent was$147.
How much did one ticket cost?
Answer:
Equation : 7(t + 6) = $147, The cost of one ticket is$15,

Explanation:
Given a family of 7 bought tickets to the circus. Each family member also
bought a souvenir that cost $6. The total amount they spent was$147.
So the equation is 7(t + 6) = 147 on solving we get cost of one ticket as
7t + 42 = 147,
7t = 147 – 42,
7t = 105,
t = 105/7 = 15,
therefore cost of one ticket is $15. Question 5. David reads the problem: Ally bought a T-shirt and a pair of shorts on sale, which reduced prices by $$\frac{1}{4}$$. The total savings on the two garments was$10.25.
Find the original price for the pair of shorts.

David says that the original price of the shorts was $41. Does his answer seem reasonable? Defend your answer by writing and solving an equation that represents the situation. Answer: No, David answer is not reasonable, the correct equation is $$\frac{1}{4}$$ (18 + s) = 10.25, where s is price of the shorts, Explanation: Given David reads the problem: Ally bought a T-shirt and a pair of shorts on sale, which reduced prices by $$\frac{1}{4}$$. The total savings on the two garments was$10.25.
Find the original price for the pair of shorts.
David says that the original price of the shorts was $41. Now we will check let us take s for price of the shorts, and price of T-shirt is$18,
So the equation is $$\frac{1}{4}$$ (18 + s) = 10.25,
$$\frac{1}{4}$$ X 18 + $$\frac{1}{4}$$ X s = 10.25,
18 + s = 10.25 X 4,
18 + s = 41,
therefore s = 41 – 18 = 23,
So No, David answer is not reasonable the price of shorts is $23 not$41.

Question 6.
Which of the following shows the correct use of the
Distributive Property when solving (33 – x) = 135.2?
A. (33 – x) = 1_-3 • 135.2
B. $$\frac{1}{3}$$ • 33 – $$\frac{1}{3}$$x = $$\frac{1}{3}$$ • 135.2
C. $$\frac{1}{3}$$ • 33 + $$\frac{1}{3}$$x = 135.2
D. $$\frac{1}{3}$$ • 33 – $$\frac{1}{3}$$x = 135.2
Answer:
B. $$\frac{1}{3}$$ • 33 – $$\frac{1}{3}$$x = $$\frac{1}{3}$$ • 135.2,

Explanation:
Given to show the correct use of the
Distributive Property when solving (33 – x) = 135.2? as
the correct answer is B. $$\frac{1}{3}$$ • 33 – $$\frac{1}{3}$$x = $$\frac{1}{3}$$ • 135.2, if we multiply both sides with $$\frac{1}{3}$$
we get the correct equation as (33 – x) = 135.2 because
$$\frac{1}{3}$$(33 – x) = $$\frac{1}{3}$$ X135.2 means,
$$\frac{1}{3}$$ • 33 – $$\frac{1}{3}$$x = $$\frac{1}{3}$$ • 135.2,
therefore the correct answer is B. $$\frac{1}{3}$$ • 33 – $$\frac{1}{3}$$x = $$\frac{1}{3}$$ • 135.2.

Practice & Problem Solving

Leveled Practice For 7-10, use the Distributive Property to solve the equations.

Question 7.
-2(x + 5) = 4

Answer:

Explanation:
Given equation as -2(x + 5) = 4, using Distributive Property
we get -2 X x + (-2 X 5) = 4,
-2x – 10 = 4,
-2x = 14, So x = -14/2 = -7,
therefore x = -7.

Question 8.
3.2 = $$\frac{4}{5}$$(b – 5)

Answer:

Explanation:
Given equation as 3.2 = $$\frac{4}{5}$$(b – 5) using Distributive Property,
we get 3.2 = $$\frac{4}{5}$$  X b + $$\frac{4}{5}$$(-5),
3.2 = $$\frac{4}{5}$$b – 4,
3.2 + 4 = $$\frac{4}{5}$$b,
7.2 = $$\frac{4}{5}$$b,
b = $$\frac{5}{4}$$ X 7.2,
b = 5 X 1.8,
therefore b = 9.

Question 9.
$$\frac{1}{8}$$(p + 24) = 9

Answer:

Explanation:
Given equation as $$\frac{1}{8}$$(p + 24) = 9 using Distributive Property,
$$\frac{1}{8}$$ X p + $$\frac{1}{8}$$ X 24 = 9,
$$\frac{1}{8}$$p + 3 = 9,
$$\frac{1}{8}$$ p = 9 – 3,
$$\frac{1}{8}$$p = 6,
p = 6 X 8 = 48 or we can solve $$\frac{1}{8}$$(p + 24) = 9 as
p + 24 = 9 X 8,
p + 24 =72,
p = 72 – 24 = 48.

Question 10.
$$\frac{2}{3}$$(6a + 9) = 20.4

Answer:

Explanation:
Given equation as $$\frac{2}{3}$$(6a + 9) = 20.4 using Distributive Property,
$$\frac{2}{3}$$ X 6a + $$\frac{2}{3}$$ X 9 = 20.4 9,
4a + 6 = 20.4,
4a = 20.4 – 6,
4a = 14.4,
a = 14.4/4,
a = 3.6.

Question 11.
Use the equation at the right.
a. Make Sense and Persevere If you apply the Distributive Property first to solve the equation,
what operation will you need to use last?

b. If instead you divide first to solve the equation,
what operation would you need to use last?
Answer:
a. We use division operation last, If we apply
the Distributive Property first to solve the equation,
b. We use multiplication last If instead we divide first to solve the equation,

Explanation:
Given equation as 6($$\frac{d}{3}$$ – 5) = 34,
a. If we apply the Distributive Property first to solve the equation,
operation I will need to use last is division as we know
1.. Multiply the term outside of the parentheses by each term
in the parentheses,
2. We combine like terms,
3. we divide to find the final solution,
so 6 X $$\frac{d}{3}$$ – 6 X 5 = 34,
2d – 30 = 34,
2d = 34 + 30,
2d = 64,
d = 64/2 = 32,
b. Now If instead I divide first to solve the equation,
operation I would need to use last is
6 X $$\frac{d}{3}$$ – 6 X 5 = 34,
$$\frac{d}{3}$$ – 5 = 34/6,
$$\frac{d}{3}$$ = 5.66 + 5,
$$\frac{d}{3}$$ = 10.66
d= 10.66 X 3 = 31.98 ≈ 32, we use multiplication.

Question 12.
A family buys 4 airline tickets online. The family buys travel
insurance that costs $19 per ticket. The total cost is$752.
Let x represent the price of one ticket.
a. Write an equation to represent this situation.
b. What is the price of one ticket?
Answer:
a. Equation: 4(x + 19) = 752,
b. The price of one ticket is $169, Explanation: Given a family buys 4 airline tickets online. The family buys travel insurance that costs$19 per ticket. The total cost is $752. Let x represent the price of one ticket. a. As 4 tickets multiplied by each ticket and 4 tickets multiplied by travel insurance of$19 is equal to $752, So, the equation is 4(x + 19) = 752, b. Now solving the equation 4(x + 19) = 752, 4x + 76 = 752, 4x = 752 – 76, 4x = 676, x = 676/4 = 169, therefore the price of one ticket is$169.

Question 13.
A local charity receives $$\frac{1}{3}$$ of funds
raised during a craft fair and a bake sale.
The total amount given to charity was $137.45. How much did the bake sale raise? Answer: The Bake Sale raised funds of$159.75,

Explanation:
Given a local charity receives $$\frac{1}{3}$$ of funds
raised during a craft fair and a bake sale.
The total amount given to charity was $137.45 and Craft Fair raised$252.60, lets take Bake sale raised funds of b,
So $$\frac{1}{3}$$(252.60 + b) = 137.45,
252.60 + b = 137.45 X 3,
252.60 + b = 412.35,
b= 412.35 – 252.60,
b = 159.75,
therefore the Bake Sale raised funds of $159.75. Question 14. The solution shown for the equation is incorrect. a. What is the correct solution? b. What was the likely error? Answer: a. The correct solution is r = 8, b. Instead of +8 it is showing -8, Explanation: a. The given equation is -3(6 – r) = 6, -3 X 6 -3 X -r = 6, -18 + 3r = 6, 3r = 6 + 18, 3r = 24, r = 24/3 = 8, b. The likely error while solving is -3 X -r it is +3r, but instead of 3r it is taken as -3r so it is showing r = -8 but the correct solution is r =8. Question 15. Vita wants to center a towel bar on her door that is 27$$\frac{1}{2}$$ inches wide. She determines that the distance from each end of the towel bar to the end of the door is 9 inches. Write and solve an equation to find the length of the towel bar. Answer: Equation: x + 18 = 27$$\frac{1}{2}$$, The length of the towel bar is 9$$\frac{1}{2}$$ inches, Explanation: Let x inches be the length of the towel bar, The width of the door is 9 inches on the each side of the towel bar so the width of the door is 9 + x + 9 = x + 18 inches. It is given that the width of the door is 27$$\frac{1}{2}$$ inches, S0 27$$\frac{1}{2}$$ inches = x +18, x = 27$$\frac{1}{2}$$ – 18, x = $$\frac{55}{2}$$ – 18, x = $$\frac{55 – 36}{2}$$, x = $$\frac{19}{2}$$ as numerator is greater than denominator, we write mixed fraction as x = 9$$\frac{1}{2}$$ inches. therefore, the length of the towel bar is 9$$\frac{1}{2}$$ inches. Question 16. Higher Order Thinking A cell phone plan is shown at the right. The rates, which include an unlimited data plan, are the same each month for 7 months. The total cost for all 7 months is$180.39.
Let m represent the average number of minutes that exceeds
700 minutes each month.
a. Write an equation to represent the given situation.

b. Solve the equation to determine how many additional minutes,
on average, you use each month.
Answer:
a. Equation : 7(19.70 + 1.97 + m X 0.05) = 180.39,
b. 82 additional minutes, on average, can be used each month,

Explanation:
Given a cell phone plan is shown at the right.
The rates, which include an unlimited data plan, are the same
each month for 7 months. The total cost for all 7 months is $180.39. Let m represent the average number of minutes that exceeds 700 minutes each month. a. Equation is 7(19.70 + 1.97 + m X 0.05) = 180.39, b. Solving 7(19.70 + 1.97 + m X 0.05) = 180.39, 21.67 + 0.05m = 180.39/ 7, 0.05m = 25.77 – 21.67, 0.05m = 4.1, m = 4.1/0.05 = 82, therefore, 82 additional minutes, on average, can be used each month. Assessment Practice Question 17. Fidel earns a fixed amount, m, for each television he sells, and an additional$15 if the buyer gets an extended warranty.
Fidel sells 12 televisions with extended warranties, earning $900. Write an equation to represent the situation. Then solve the equation to find the amount earned for each television sold. Answer: Equation: 12(m +15) = 900, The amount earned for each television sold is$60,

Explanation:
Given, Amount earned by Fidel for each TV is m,
Additional amount for extended warranty is $15, Televisions sold are 12, Amount of televisions sold is$900,
Amount earned by each TV * Number of TV’s +
Additional amount * Number of TV’s = Total amount,
So the equation is 12(m + 15) = 900,
Dividing both sides by 12, we get  m + 15 =  900/12,
m +15 = 75, m =75 – 15 = 60,
therefore, Fidel earns $60 for each television sold. ### TOPIC 5 MID-TOPIC CHECKPOINT Question 1. Vocabulary Explain how to isolate the variable in the equation –$$\frac{2}{3}$$n + 7 = 15. Lesson 5-2 Answer: n = -12, Explanation: To isolate a variable is to perform the following steps: 1. Eliminate any fractions present by multiplying both sides by any denominators present. 2. Get all terms with the variable on the left side of the equation and all terms without the variable on the other side. 3. Combine the terms on the left side, if possible. 4. Factor out the variable on the left side, if necessary. 5. Divide both sides by the coefficient of the variable, leaving the variable isolated. Now the equation given is –$$\frac{2}{3}$$n + 7 = 15, solving –$$\frac{2}{3}$$n = 15 – 7, n = – $$\frac{3{2}$$ X 8, n = -12. Question 2. Jake paid$13.50 for admission to the county fair and
bought 9 tickets to play games.
If he spent a total of $36, what is the cost, c, of one ticket? Write and solve an equation. Lessons 5-1 and 5-2 Answer: Equation: 9c + 13.50 = 36, The cost of one ticket is$2.5,

Explanation:
Given Jake paid $13.50 for admission to the county fair and bought 9 tickets to play games. If he spent a total of$36 and the cost c of one ticket is
Equation 9c + 13.5 = 36,
9c + $13.50 =$36, subtracting 13.5 both sides
9c + 13.50 – 13.5 = 36 -13.5,
9c =  22.5, dividing both sides by 9 we get
9c/9 = 22.5/9,
c = 2.5, therefore cost of one ticket is $2.5. Question 3. Select all the equations that are equivalent to $$\frac{1}{2}$$(4 + 8x) = 17. Lesson 5-3 Answer: The equations that are equivalent to $$\frac{1}{2}$$(4 + 8x) = 17 are 4x = 15 and 4 + 8x = 34, Explanation: Given equation is $$\frac{1}{2}$$(4 + 8x) = 17, multiplying both side by 2 we get 4 + 8x = 34 and now dividing both sides by 2 we get 2 + 4x = 17, 4x = 15, so the equations that are equivalent to $$\frac{1}{2}$$(4 + 8x) = 17 are 4x = 15 and 4 + 8x = 34. Question 4. Clara has 9 pounds of apples. She needs 1$$\frac{1}{4}$$ pounds to make one apple pie. If she sets aside 1.5 pounds of apples to make applesauce, how many pies, p, can she make? Write and solve an equation. Lessons 5-1 and 5-2 Answer: Equation: 1$$\frac{1}{4}$$p + 1.5 = 9, Number of apple pies Clara can make are 6, Explanation: Given Clara has 9 pounds of apples. She needs 1$$\frac{1}{4}$$ pounds to make one apple pie. If she sets aside 1.5 pounds of apples to make applesauce, Number of pie’s p she can make are equation is 1$$\frac{1}{4}$$p + 1.5 = 9, $$\frac{4 X 1 + 1}{4}$$p + 1.5 = 9, $$\frac{5}{4}$$p + 1.5 = 9, $$\frac{5}{4}$$p = 9 – 1.5, $$\frac{5}{4}$$p = 7.5, 5p = 7.5 X 4, 5p = 30, p = 30/5 = 6, therefore number of apple pies Clara can make are 6. Question 5. Solve the equation -4(1.75 + x) = 18. Show your work. Lesson 5-3 Answer: Solving the equation -4(1.75 + x) = 18 we get x = -6.25, Explanation: Given equation is -4(1.75 + x) = 18, dividing both side by -4, -4(1.75 + x)/-4 = 18/-4, 1.75 + x = – 4.5, x = – 4.5 -1.75, x = – 6.25. Question 6. Four friends attend a school play and pay$6.75 per ticket.
Each also buys a Healthy Snack Bag sold by the Theater Club.
If the friends spent a total of $37.00, how much did each Healthy Snack Bag cost, b? Write and solve an equation. Lessons 5-1, 5-2,5-3 Answer: Equation:4(6.75 + b) = 37, Each Healthy Snack Bag cost is$2.5,

Explanation:
Given four friends attend a school play and pay $6.75 per ticket. Each also buys a Healthy Snack Bag sold by the Theater Club. If the friends spent a total of$37.00, Now if each Healthy
Snack Bag cost, b then the is equation 4(6.75 + b) = 37,
dividing both sides by 4 we get 6.75 + b = 37/4,
6.75 + b = 9.25,
b = 9.25 – 6.75,
b = 2.5, therefore each Healthy Snack Bag cost is $2.5. How well did you do on the mid-topic checkpoint? Fill in the stars. ### TOPIC 5 MID-TOPIC PERFORMANCE TASK Marven and three friends are renting a car for a trip. Rental prices are shown in the table. PART A Marven has a coupon that discounts the rental of a full-size car by$25.
They decide to buy insurance for each day. If the cost is $465, how many days, d, will they rent the car? Write and solve an equation. Answer: Equation : 49d + 21d – 25 = 465, Number of days they will rent the car is 7 days, Explanation: Given Marven has a coupon that discounts the rental of a full-size car by$25.
They decide to buy insurance for each day. If the cost is $465, So number of many days, d, will they rent the car is 49d + 21d – 25 = 465, 70d = 465 + 25, 70d = 490, d = 490/70, d= 7, therefore number of days they will rent the car is 7 days. PART B If they still use the coupon, how many days could they rent the small car with insurance if they have$465 to spend?
Answer:
If they still use the coupon, how many days could they rent the small car
with insurance if they have $465 to spend is 8 days, Explanation: If they still use the coupon, Number of days could they rent the small car with insurance if they have$465 to spend is
39d + 21d – 25 = 465,
60d = 465 + 25,
60d = 490,
d = 8.166, minimum 8 days.

PART C
They rent a car with insurance for 5 days but lost their coupon.
If Marven and the three friends spend $75 each, which car did they rent? Write and solve an equation to justify your answer. Answer: Equation: 5(p + 21) = 4 X 75, They rent the small car, Explanation: If they rent a car with insurance for 5 days but lost their coupon. If Marven and the three friends spend$75 each means 4 X 75,
Let price be p for the rent of the car,
5(p + 21)  = 4 X 75,
5p + 105 = 300,
5p = 300 – 105,
5p = 195,
p = 195/5,
p = 39, So car did they rent is small car.

### Lesson 5.4 Solve Inequalities Using Addition or Subtraction

Explain It!

Selena and Martin are waiting at the bus stop. The number lines show the
possible wait times in minutes, t, for Selena and Martin.

A. Construct Arguments Who anticipates a longer wait?
Justify your response with a mathematical explanation.
Answer:
Martin anticipated longer,

Explanation:
Based on inequalities, Martin anticipated a longer waiting time.
This is because he thinks that the bus will arrive greater than 15 minutes.

B. If Selena and Martin both wait 10 minutes for the bus,
whose possible wait time was closer to his or her actual wait time? Explain.
Answer:
Selena is closer,

Explanation:
If both Selena and Martin waited 10 minutes for the bus. Selena is closer to the actual wait time because she anticipated the waiting time for the bus less than 15 minutes before it arrives.

Focus on math practices
Be Precise If Selena and Martin both wait exactly 15 minutes for the bus,
whose possible wait time was closer to his or her actual wait time? Explain.
Answer:
Selena is closer to actual wait time,

Explanation:
If Selena and Martin both wait exactly 15 minutes for the bus.
Selena is closer to the actual wait time.
This is because her possible wait time included 15 minutes.
Her anticipation for the waiting time is exactly the same as the actual waiting time.
Unlike Martin that he anticipated waiting time is greater than 15 mins,
which does not include 15 minutes.

Essential Question
How is solving inequalities with addition and subtraction similar to and
different from solving equations with addition and subtraction?
Answer:
Similarity: Both uses properties of equality or inequality,
Different: An equation has one solution while an inequality has more than one solution,

Explanation:
There is a similarity when solving inequalities with addition and subtraction
from solving equations with addition and subtraction because
both uses properties of equality or inequality.
Inverse relationship between addition and subtraction to
isolate the variable in an equation or inequality. The difference
between the two is the solution for inequality and equation,
There is only one solution for an equation while in an inequality
there can be more than one solution.

Try It!

Kyoko has completed 26 hours of community service.
Her goal is to complete at least 90 hours this semester.
Write and solve an inequality to show how many more hours, h,
Kyoko needs to complete to meet her goal.
Use the number line to graph the solutions.

Answer:

Explanation:
Given Kyoko has completed 26 hours of community service.
Her goal is to complete at least 90 hours this semester.
Wrote and solved an inequality to show how many more hours, h,
Kyoko needs to complete to meet her goal as h + 26 ≤ 90,
Used the number line to graph the solutions as shown above.

Convince Me!
Is there more than one solution to the problem about Kyoko?
Explain. Give one value that is a solution and one value that is not a solution.
Answer:
Yes , there more than one solution to the problem about Kyoko,

Explanation:
Yes , there more than one solution to the problem about Kyoko
because of the indicated range for time.
Kyoko must complete at least 64 hours more to achieve her
goal in the community service.
She cannot go beyond 64 hours because it will be more than
the required number of hours. 65 hours is not a solution to the problem.

Try It!

The speed limit on a road drops down to 15 miles per hour around a curve.
Mr. Gerard slows down by 10 miles per hour as he drives around the curve.
He never drives above the speed limit.
At what speed was Mr. Gerard driving before the curve? Graph the solution.
Answer:
Mr. Gerard was driving less than or equal to 25 miles per hour before the curve,

Explanation:
Given the speed limit on a road drops down to
15 miles per hour around a curve.
Mr. Gerard slows down by 10 miles per hour
as he drives around the curve.
He never drives above the speed limit.
The speed Mr. Gerard was driving before the curve is
x – 10 ≤ 15, inequality for the situation,
x – 10 + 10 ≤ 15 + 10, addition property of inequality,
x ≤ 25, his driving speed before the curve,
therefore, Mr. Gerard was driving less than or equal to
25 miles per hour before the curve.

Try It!
Solve the inequality n – 1$$\frac{3}{4}$$ ≤ –$$\frac{5}{8}$$ Then graph the solution.
Answer:
n ≤ 1$$\frac{1}{8}$$  or n ≤ 1.125,

Explanation:
n – 1$$\frac{3}{4}$$ ≤ –$$\frac{5}{8}$$,
n – 1$$\frac{3}{4}$$ + 1$$\frac{3}{4}$$ ≤ –$$\frac{5}{8}$$ + 1$$\frac{3}{4}$$, addition property of inequality,
n ≤ –$$\frac{5}{8}$$ + $$\frac{7}{4}$$, simplifying
n ≤ $$\frac{-5 + 14}{8}$$,
n ≤ $$\frac{9}{8}$$, as numerator is greater we write in mixed fraction as
n ≤ 1$$\frac{1}{8}$$  or n ≤ 1.125.

KEY CONCEPT
Solving inequalities with addition and subtraction is the same as solving equations with addition and subtraction.
Use the inverse relationship between addition and subtraction to isolate the variable.

Do You Understand?

Question 1.
Essential Question How is solving inequalities with addition and subtraction
similar to and different from solving equations with addition and subtraction?
Answer:
Similarity: Both uses properties of equality or inequality,
Different: An equation has one solution while an inequality has more than one solution,

Explanation:
There is a similarity when solving inequalities with addition and subtraction
from solving equations with addition and subtraction because
both uses properties of equality or inequality.
Inverse relationship between addition and subtraction to
isolate the variable in an equation or inequality. The difference
between the two is the solution for inequality and equation,
There is only one solution for an equation while in an inequality
there can be more than one solution.

Question 2.
Be Precise How do the solutions of the two inequalities differ?
Are any of the solutions the same? Explain.
a. x+ 5 < 8 and x + 5 > 8
b. x + 5 ≤ 8 and x + 5 ≥ 8

Answer:
a. The number 3 is not part of the solution for both inequalities,

b. The number 3 is one the solutions for both inequalities,

Explanation:
a. x + 5 – 5 < 8 – 5, (subtraction property of inequality)
x < 3, (solution to the equation) and
x + 5 – 5 > 8 – 5, (subtraction property of inequality)
x > 3, (solution to the equation),
Both solution uses subtraction property of equality,
However the value of the solutions are different,
One is less than 3 while the other is greater than 3.

b. x + 5 – 5 ≤ 8 – 5, (subtraction property of inequality)
x ≤ 3, (solution to the equation) and
x + 5 – 5 ≥ 8 – 5, (subtraction property of inequality)
x ≥ 3, (solution to the equation),
Both solution uses subtraction property of equality,
Both inequalities have 3 as part of the solution.

Question 3.
Reasoning Write two different inequalities in which one of the solutions is
the same as the solution to x – 23 = 191.
Answer:
The inequalities are x- 23 ≤ 191 and x – 23 < 191,
the first inequality has a similar solution to the indicated equation
because of the inequality symbol,

Explanation:
1. x -23 ≤ 191 (first inequality)
x -23 + 23 ≤ 191 + 23 (addition property)
x ≤ 214 (range of solution to the inequality),
x -23 < 191 (second inequality)
x -23 + 23 < 191 + 23 (addition property)
x < 214 (range of solution to the inequality),

2. The solution to the equation is
x – 23 + 23 = 191 + 23,(addition property of equality)
x = 214 (solution to the equation).

Do You Know How?

Question 4.
Solve each inequality. Then graph the solution.
a. x + 5 > 3

Answer:
x > -2,

Explanation:
x + 5 > 3( inequality),
x + 5 – 5 > 3 – 5, subtract 5 from both sides,
x > -2,
The graph of the solution x > -2 is shown above.

b. x + 5 ≤ 3

Answer:
x ≤ -2,

Explanation:
x + 5 ≤ 3 (inequality),
x + 5 – 5 ≤ 3 – 5, subtract 5 from both sides,
x ≤ -2,
The graph of the solution x ≤ -2 is shown above.

c. x – $$\frac{3}{2}$$ ≤ -3

Answer:
x ≤ –$$\frac{3}{2}$$,

Explanation:
x – $$\frac{3}{2}$$ ≤ -3 (inequality),
x – $$\frac{3}{2}$$ + $$\frac{3}{2}$$ ≤ -3 + $$\frac{3}{2}$$(adding $$\frac{3}{2}$$ both sides),
≤ –$$\frac{3}{2}$$,
The graph of the solution x > -2 is shown above.

Question 5.
Elanor is driving below the speed limit on a highway.
a. Write the inequality to show how much faster Elanor can drive without going over the speed limit.

b. Solve the inequality you wrote. By how much can Elanor increase her speed?
Answer:
a. x + 43.5 < 55,
b. Elanor increase her speed by x < 11.5,

Explanation:
a. Let x be the increase in the speed limit,
The inequality for the given situation where Elanor can
drive without going over the speed limit is x + 43.5 < 55,

b. Now we find the value of x in the given inequality is
x + 43.5 < 55, subtracting 43.5 both sides as
x + 43.5 – 43.5 < 55 – 43.5, we get
x< 11.5, So Elanor increase her speed by x < 11.5.

Practice & Problem Solving

Leveled Practice In 6 and 7, fill in the boxes to solve each inequality. Then graph the solutions.

Leveled Practice In 6 and 7, fill in the boxes to solve each inequality. Then graph the solutions.

Question 6.
x + 5 < 7

Answer:
x < 2,

Explanation:
Given inequality as x + 5 < 7, now we subtract 5 from both sides,
x + 5 – 5 < 7 – 5, we get
x < 2.

Question 7.
x – 4 ≥ 12

Answer:
x  ≥ 16,

Explanation:
Given inequality as x – 4 ≥ 12, now we add 4 on both sides,
x – 4 + 4 ≥ 12 + 4, we get
x ≥ 16.

Question 8.
Solve x + 10 ≥ 14. Then graph the solution.

Answer:
x ≥ 4,

Explanation:
Given inequality as x + 10 ≥ 14, now we subtract 10 from both sides,
x + 10 – 10 ≥ 14 – 10, we get x ≥ 4.

Question 9.
Solve x – 20 ≤ -11. Then graph the solution.

Answer:
x ≤ 9,

Explanation:
Given inequality as x – 20 ≤ – 11, now we add 20 on both sides,
x – 20 + 20 ≥ -11 + 20, we get
x ≤ 9.

Question 10.
The maximum number of students in a classroom is 26.
If there are 16 students signed up for the art class, how many more
students can join the class without exceeding the maximum?
Answer:
There are at most 10 students who can join the art class,

Explanation:
Let x be the number of students that can join the art class,
x + 16 ≤ 26,
x + 16 – 16 ≤ 26 – 16, subtract 16 from both sides,
x ≤ 10, Therefore there are at most 10 students who can join the art class.

Question 11.
Higher Order Thinking The inequality x + c > -2.55 has the
solution x > 4.85 What is the value of c? How do you know?
Answer:
c = -7.4,

Explanation:
x + c = -2.55,
4.85 + c = -2.55, Substitute the value of x,
4.85 + c – 4.85 = -2.55 – 4.85, subtract -4.85 from both sides,
c = -7.4,
The value of c = -7.4, substitute this to the original inequality,
so that it will make the inequality true,
x + c > -2.55,
x + (-7.4) > -2.55, substitute the value of c,
x + (-7.4) + 7.4 > -2.55 + 7.4, add 7.4 on both sides,
x > 4.85 which is true.

Question 12.
Rina is climbing a mountain. She has not yet reached base camp.
Write an inequality to show the remaining distance, d, in feet
she must climb to reach the peak.

Answer:
Inequality: d + 9,695 > 12,358,
Rina must climb d > 2,663 feet to reach the peak,

Explanation:
The inequality to determine the distance is
d + 9,695 > 12,358,
d + 9,695 – 9,695 > 12,358 – 9,695, subtract 9,695 from both sides,
d > 2,663, therefore, Rina must climb d > 2,663 feet to reach the peak.

Question 13.
On a math test, students must solve the inequality x – 5< 11 and then graph the solution. Mason said the solution is x < 6 and graphed the solution as shown below.

a. What error did Mason make?
b. Show the correct solution on the number line.

Answer:
a. Mason did not use the inverse operations to determine the solution,
b.

Explanation:
a. Mason did not use the inverse operations to determine the solution of x,
Mason subtracted 5 to the right side when it should have been added this
is the correct way of using inverse operation,
b. x – 5 < 11,
x – 5 + 5 < 11 + 6, adding 5 on both sides,
x < 16, the graph of solution is shown on number line above.

Question 14.
Model with Math Dani’s neighbors paid her to take care of their bird during their vacation.
Dani spent $4 of her earnings on an afternoon snack and$16 on a new book.
Afterward, she had at least $8 left. Write an inequality to represent how much Dani’s neighbors paid her. Answer: Inequality : x – (4 + 16) ≥ 8, Dani’s neighbors paid her x ≥ 28, Explanation: Given Dani’s neighbors paid her to take care of their bird during their vacation. Dani spent$4 of her earnings on an afternoon snack and
$16 on a new book. Afterward, she had at least$8 left,
Let x be the amount Dani’s neighbors paid her, So the inequality is
x – (4 + 16) ≥ 8, now solving the inequality
x – 20 + 20 ≥ 8 + 20, adding 20 on both sides,
x ≥ 28, therefore, Dani’s neighbors paid her x ≥ 28.

Question 15.
Reasoning The temperature in a greenhouse should be 67°F or higher.
One morning, the heater stopped working.
The temperature dropped 4 degrees before someone fixed the heater.
The temperature was still at least 67°F when the heater started working again.
How can you best describe the temperature in the greenhouse before
the heater stopped working?

Answer:
The temperature in the greenhouse before
the heater stopped working is at least 71°F,

Explanation:
The temperature in a greenhouse should be 67°F or higher.
One morning, the heater stopped working.
The temperature dropped 4 degrees before someone fixed the heater.
The temperature was still at least 67°F when the heater started working again.
Let x be the initial temperature of the heater, so
x – 4 ≥ 67,
x + 4 + 4 ≥ 67 + 4, adding 4 on both sides,
x ≥ 71, therefore, the temperature in the greenhouse before
the heater stopped working is at least 71°F.

Assessment Practice

Question 16.
Ramiro has $21. He wants to buy a skateboard that costs$47.
How much more money does he need to have at least $47? Write an inequality that represents the situation. Solve the inequality and graph your solution. Answer: Inequality : 47 ≤ x + 21, 26 ≤ x or x ≥ 26, Explanation: Given Ramiro has$21. He wants to buy a skateboard that costs $47. Let x be more money does he need to have at least$47,
So the inequality is 47 ≤ x+21,
47 – 21≤  x + 21 – 21, subtract 21 from both sides,
26 ≤ x, The graph 26 ≤ x or x ≥ 26 is shown above.

Question 17.
Kendra has $7.35 in her purse. She needs at least$2.87 more to buy a special bead.
What is the total amount, x, she needs for the bead?
Which inequalities can be used to represent the situation?
Select all that apply.

Answer:
Inequalities that can be used to represent the
situation x = 10.22 are
1. x – 7.35 ≤ 2.87,
2.  x + 7.35  ≥ 2.87,
3. x – 7.35 ≥ 2.87,
4. x ≥ 10.22,
5. x ≤ 10.22,

Explanation:
Given Kendra has $7.35 in her purse. She needs at least$2.87
more to buy a special bead.
The total amount, x, she needs for the bead is x = 7.35 + 2.87,
So x = 10.22, Given
1. x + 7.35 ≤ 2.87, on solving
x + 7.35 – 7.35 ≤ 2.87 – 7.35, subtracting 7.35 from both sides,
x ≤ 4.48 but we got x = 10.22, So
x + 7.35 ≤ 2.87 is incorrect as x is not equal to x = 10.22 and
also not less than 4.48.

2. x – 7.35 ≤ 2.87, on solving
x – 7.35 + 7.35 ≤ 2.87 + 7.35, adding 7.35 from both sides,
x ≤ 10.22 we got x = 10.22, So
x – 7.35 ≤ 2.87 is correct as x is equal to x = 10.22.

3. x + 7.35 ≥ 2.87,on solving
x + 7.35 – 7.35 ≥ 2.87 – 7.35, subtracting 7.35 from both sides,
x ≥ 4.48 but we got x = 10.22, So
x + 7.35 ≥ 2.87 is correct as x = 10.22 whose value is
greater than 4.48.

4. x – 7.35 ≥ 2.87,on solving
x – 7.35 + 7.35 ≥ 2.87 + 7.35, adding 7.35 from both sides,
x ≥ 10.22 but we got x = 10.22, So
x – 7.35 ≥ 2.87 is correct as x value is equal to 10.22.

5. x ≥ 10.22 is correct as x is equal to x = 10.22.

6. x ≤ 2.87 is incorrect as x is not equal to 10.22 and is not
also less than 2.87.

7. x ≤ 10.22 is correct as x is equal to 10.22.

8. x ≤ 4.48 is incorrect as x value is not equal to 10.22 and
also not less than 4.48.
Therefore,

Inequalities that can be used to represent the
situation x = 10.22 are
1. x – 7.35 ≤ 2.87,
2.  x + 7.35  ≥ 2.87,
3. x – 7.35 ≥ 2.87,
4. x ≥ 10.22,
5. x ≤ 10.22,

### Lesson 5.5 Solve Inequalities Using Multiplication or Division

Solve & Discuss It!

Alex and Hope were trying to solve -6x > 24.
Whose inequality shows the solution? Show your work.

Answer:
Hope’s work x < -4 shows the inequality to the solution,

Explanation:
Given Alex and Hope were trying to solve -6x > 24.
Hope shows the solution as x < -4 means x values are less than
-4 which will be -5,-6, -7 and so on, if we substitute x as -5,
we get -6(-5) = 30 which is greater than 24,
therefore, Hope’s work x < -4 shows the inequality to the solution,

Construct Arguments Why does more than one value of x
make the inequality true?
Answer:
Yes, because there may be more than one solutions to make
make the inequality true,

Explanation:
Most of the time, an inequality has more than one or even infinity solutions.
For example the inequality: x>3 . The solutions of this inequality are
“all numbers strictly greater than 3”, therefore more than one value of x
make the inequality true.

Focus on math practices
Be Precise What do you notice about the inequality symbols used
in the original inequality and in the correct solution?
Answer:
The inequality symbol in the solution changed,

Explanation:
The inequality symbol (>) changed in the solution,
The inequality symbol became <, This is because of the
negative integer divided to both sides of the inequality.

Essential Question
How is solving inequalities with multiplication and division similar to and
different from solving equations with multiplication and division?
Answer:
Solving inequalities is very similar to solving equations,
but sometimes we have to reverse the symbol.
we must reverse the inequality symbol when we multiply or
divide both sides of the equation by a negative number.
1) It must be multiplication or division (not addition or subtraction)
2) The number being multiplied or divided must be negative.

Explanation:
Solving inequalities is very similar to solving equations,
but sometimes we have to reverse the symbol.

For example:
Multiplication Properties of Inequality,
c is positive,
a<b, then ac<bc ← -2<3,
a>b, then ac>bc ←-2(4)<3(4),
c is negative,
a<b, then ac>bc ← -2<3,
a>b, then ac<bc ←-2(-4)>3(-4),
Division property of inequality,
c is positive,
a<b, then a/c<b/c ← 2<8,
a>b, than a/c>b/c ←2/4<8/4,
c is negative,
a<b, then a/c>b/c ←2<8,
a>b, then a/c<b/c ← 2/-4>8/-4, or

Solving Inequalities is very similar to solving Equations.
The same general technique applies.
That technique for solving equations is:
Whatever we do to one side of the equation,
we have to do to the other side to preserve the equality,
The technique for solving inequalities is:
Whatever we do to one side of the inequality,
we have to do to the other side to preserve the inequality
the techniques are the same.
The technique means:
If we multiply or divide one side of an equation or inequality
by the same number, we have to multiply or divide the other side
of the equation or inequality by the same number.
The difference between solving equations and solving inequalities is:
If you multiply or divide an inequality by a negative number,
then the inequality reverses.
The following examples will show this rule in action.
The first example will be multiplication.
The second example will be division.
The first example starts with:
7 is greater than 5.
Multiply both sides of that inequality by -5 and we get:
7*-5 = -35 on the left side of the inequality.
5*-5 = -25 on the right side of the inequality.
The result is -35 is smaller than -25.
The inequality started as greater than and
became smaller than because we were multiplying both sides
of the inequality by a negative number.
It’s clear to see that 7 really is greater than 5.
It is also clear to see that -35 is less than -25.
The second example is simply the reverse of the first example and starts with:
-35 is smaller than -25.
Divide both sides of this inequality by -5 and we get:
-35 / -5 = 7 on the left side of the inequality.
-25 / -5 = 5 on the right side of the inequality.
The result is 7 is greater than 5.
The inequality started as less than and became greater than
because you were dividing both sides of the inequality by a negative number.
We needed to do that to preserve the inequality.
It’s clear with these numbers that reversing the inequality is
essential when we are multiplying both sides of the inequality or
dividing both sides of the inequality by a negative number.

Try It!

Solve the inequality $$\frac{d}{7}$$ > 15. Then graph the solution.

Answer:
d> 105,

Explanation:
Solved the inequality $$\frac{d}{7}$$ > 15 as
$$\frac{d}{7}$$ > 15, multiplying both sides by 7,
we get d > 105, Shown the graph of the solution d > 105 above.

Convince Me!
Frances solved the inequality 5g ≥ 35. She says that 7 is
a solution to the inequality. Is Frances correct? Explain.
Answer:
Yes, Frances is correct,

Explanation:
Given Frances solved the inequality 5g ≥ 35.
She says that 7 is a solution to the inequality on solving
5g ≥ 35, dividing both sides by 5 we get
g ≥ 7, means g is greater than 7 and we get g is
equal to 7, therefore, Yes, Frances is correct.

Try It!
Solve each inequality. Then graph the solution.

a. 149.76 > -19.2x

Answer:
-7.8 < x or x > -7.8,

Explanation:
Given to solve inequality 149.76 > -19.2x,
149.76/- 19.2 > -19.2x/-19.2, dividing both sides by -19.2 and
reverse the inequality symbol,
-7.8 < x, shown the value of x as -7.8 < x or x > -7.8 on the graph above.

b. -3.25y < -61.75

Answer:
-3.25y < -61.75,

Explanation:
Given to solve inequality -3.25y < -61.75,
-3.25y/- 3.25 > -61.75/-3.25, dividing both sides by -3.25 and
reverse the inequality symbol,
y > 19, shown the value of y as y > 19 on the graph above.

Try It!
Solve each inequality. Then graph the solution.

a. $$\frac{k}{-0.5}$$ < 12

Answer:
k> -6,

Explanation:
Given to solve $$\frac{k}{-0.5}$$ < 12 inequality,
– 0.5$$\frac{k}{-0.5}$$ > – 0.5(12), multiplying both sides by -0.5 and
reverse the inequality symbol,
k > -6, The graph of the solution k > -6 is as shown above.

b. –$$\frac{5}{4}$$h ≥ 25

Answer:
h ≤ -20,

Explanation:
Given to solve –$$\frac{5}{4}$$h ≥ 25 inequality,
-4 (-$$\frac{5}{4}$$h) ≤ -4(25) multiplying both sides by -4 and
reverse the inequality symbol,
5h ≤ -100,
5h/5 ≤ -100/5, divide both sides by 5,
5h ≤ -20,The graph of the solution 5h ≤ -20 is as shown above.

KEY CONCEPT

Solving inequalities with multiplication and division is the
same as solving equations with multiplication and division
when the values are positive. Use the inverse relationship
between multiplication and division to isolate the variable.

When multiplying or dividing by negative values, the inequality symbol is reversed.

Do You Understand?

Question 1.
Essential Question
How is solving inequalities with multiplication and division
similar to and different from solving equations with multiplication and division?
Answer:
Solving inequalities is very similar to solving equations,
but sometimes you have to reverse the symbol.
we must reverse the inequality symbol when we multiply or
divide both sides of the equation by a negative number.
1) It must be multiplication or division (not addition or subtraction)
2) The number being multiplied or divided must be negative.

Explanation:
Solving inequalities is very similar to solving equations,
but sometimes you have to reverse the symbol.

For example:
Multiplication Properties of Inequality,
c is positive,
a<b, then ac<bc ← -2<3,
a>b, then ac>bc ←-2(4)<3(4),
c is negative,
a<b, then ac>bc ← -2<3,
a>b, then ac<bc ←-2(-4)>3(-4),
Division property of inequality,
c is positive,
a<b, then a/c<b/c ← 2<8,
a>b, than a/c>b/c ←2/4<8/4,
c is negative,
a<b, then a/c>b/c ←2<8,
a>b, then a/c<b/c ← 2/-4>8/-4, or

Solving Inequalities is very similar to solving Equations.
The same general technique applies.
That technique for solving equations is:
Whatever we do to one side of the equation,
we have to do to the other side to preserve the equality,
The technique for solving inequalities is:
Whatever we do to one side of the inequality,
we have to do to the other side to preserve the inequality
the techniques are the same.
The technique means:
If you multiply or divide one side of an equation or inequality
by the same number, we have to multiply or divide the other side
of the equation or inequality by the same number.
The difference between solving equations and solving inequalities is:
If we multiply or divide an inequality by a negative number,
then the inequality reverses.
The following examples will show this rule in action.
The first example will be multiplication.
The second example will be division.
The first example starts with:
7 is greater than 5.
Multiply both sides of that inequality by -5 and we get:
7*-5 = -35 on the left side of the inequality.
5*-5 = -25 on the right side of the inequality.
The result is -35 is smaller than -25.
The inequality started as greater than and
became smaller than because we were multiplying both sides
of the inequality by a negative number.
It’s clear to see that 7 really is greater than 5.
It is also clear to see that -35 is less than -25.
The second example is simply the reverse of the first example and starts with:
-35 is smaller than -25.
Divide both sides of this inequality by -5 and we get:
-35 / -5 = 7 on the left side of the inequality.
-25 / -5 = 5 on the right side of the inequality.
The result is 7 is greater than 5.
The inequality started as less than and became greater than
because you were dividing both sides of the inequality by a negative number.
We needed to do that to preserve the inequality.
It’s clear with these numbers that reversing the inequality is
essential when we are multiplying both sides of the inequality or
dividing both sides of the inequality by a negative number.
or
The similarity is the method of solving it when the
values are positive. The difference is the method of
solving it when the values are negative.

Explanation:
The similarities of solving equation and solving inequalities
Both are used in determining the values of variable
Both can be solved either by one step or two steps
Both have same process when the values are positive
The difference between solving equation and solving inequalities
The properties used when solving equations and solving inequalities
The method of solving equation and inequalities differ
when the values are negative.

Question 2.
Construct Arguments Why is -x < 3 equivalent to x > -3?
Provide a convincing argument.
Answer:
The two inequalities are equivalent,

Explanation:
The variable cannot be negative, therefore it is need
to show the variable as positive.
-x < 3,
-1x/-1 > 3/-1, divide both sides by -1 and
reverse the inequality symbol,
x > -3,
The two inequalities are equivalent, this is because
the first inequality has negative variable.
it needs to be changed into a positive variable.
After solving the first inequality, the result is the same
with the second inequality.

Question 3.
If a, b, and care rational numbers and a > b,
is ac > bc always true? Justify your answer.
Answer:
The given inequality is not always true,

Explanation:
The inequality ac > bc is not always true,
This is because when c is a negative integer,
the inequality becomes false.
Example for this is a = 3, b = 2, c = -1,
Substituting to the inequality ac > bc =
3(-1) > 2(-1) = -3 < -2.

Do You Know How?

Question 4.
Solve each inequality. Then graph the solution.
a. 4x > 12

Answer:
x > 3,

Explanation:
Given to solve inequality 4x > 12,
4x/4 > 12/4, divide both sides by 4
x > 3, shown the value of x as x > 3 on the graph above.

b. $$\frac{x}{4}$$ ≤ -12

Answer:
x ≤ -48,

Given to solve inequality $$\frac{x}{4}$$ ≤ -12
4 X $$\frac{x}{4}$$ ≤ 4 X -12, multiplying both sides by 4
x ≤ -48, shown the value of x as x ≤ -48 on the graph above.

c. -4x > 12

Answer:
x < -3,

Explanation:
Given to solve -4x > 12 inequality,
-4x/-4 < 12/-4 dividing both sides by -4 and
reverse the inequality symbol,
x < -3,
x < -3,The graph of the solution x <-3 is as shown above.

Question 5.
Vanna is saving for a trip. The hotel room will be $298.17 for 3 nights, and there will be additional fees. What is her daily cost? a. Write an inequality for the situation. b. Solve the inequality. Then provide a statement that represents the solution of the problem. Answer: a. 3x ≥ 298.17, b. x≥ 99.39 , The stay of Vanna in the hotel plus additional cost fees cost at least$99.39,

Explanation:
Given Vanna is saving for a trip. The hotel room will be $298.17 for 3 nights, and there will be additional fees, Let x be the daily cost of the stay in the hotel room, So a. The inequality for the situation is 3x ≥ 298.17, b. Solving 3x ≥ 298.17 we get daily cost 3x/3 ≥ 298.17/3, dividing both sides by 3 we get x ≥ 99.39, therefore the stay of Vanna in the hotel plus additional cost fees cost at least$99.39.

Practice & Problem Solving

Leveled Practice In 6-9, fill in the boxes to solve the inequality.
Then graph the solution.

Question 6.

Answer:
m ≤ 7,

Explanation:
Given to fill the boxes and to solve the 8m ≤ 56 inequality,
8m/8 ≤ 56/8 dividing both sides by 8,
m ≤ 7,The graph of the solution m ≤ 7 is as shown above.

Question 7.

Answer:
x> 6,

Explanation:
Given to fill the boxes and to solve the –$$\frac{4}{3}$$x < -8 inequality,
-3 X $$\frac{4}{3}$$ x > -3(-8), multiplying both sides by -3 and
reverse the inequality symbol,
4x > 24,
4x/4 > 24/4, dividing both sides by 4 we get
x > 6, so the graph of the solution x > 6 is as shown above.

Question 8.

Answer:
x < -8,

Explanation:
Given to fill the boxes and to solve the -7x > 56 inequality,
– 7x/(-7) < 56/(-8), dividing both sides by -7 and
reverse the inequality symbol,
x < -8, so the graph of the solution x < -8 is as shown above.

Question 9.

Answer:
m ≤ -10,

Explanation:
Given to fill the boxes and to solve the $$\frac{m}{-5}$$ ≥ 2 inequality,
-5 X $$\frac{m}{-5}$$ ≤ -5(2), multiplying both sides by -5 and
reverse the inequality symbol,
m ≤ -10, so the graph of the solution m ≤ -10 is as shown above.

Question 10.
Kyra and five friends shared a bag of fruit snacks.
Each person got no more than 3 fruit snacks.
The inequality x ÷ 6 ≤ 3 represents this situation.
Solve the inequality to find the possible numbers of fruit
snacks that were in the bag.
Answer:
There are at most 18 fruit snacks that were in the bag,

Explanation:
Given Kyra and five friends shared a bag of fruit snacks.
Each person got no more than 3 fruit snacks.
The inequality x ÷ 6 ≤ 3 represents this situation the
possible numbers of fruit snacks that were in the bag are
x ÷ 6 ≤ 3,
6 X (x ÷ 6) ≤ 6 X 3, multiplying both sides by 6,
x ≤ 18, therefore, there are at most 18 fruit snacks that were in the bag.

Question 11.
Over the next 17 months, Eli needs to read more than 102 e-books.
The inequality 17x > 102 represents the number of e-books he
needs to read per month.
Solve the inequality to find the number of e-books Eli needs to read per month.
Answer:
Eli needs to read at least 6 e -books per month,

Explanation:
Given over the next 17 months, Eli needs to read more than 102 e-books.
The inequality 17x > 102 represents the number of e-books he
needs to read per month. Now solving the inequality to
find the number of e-books Eli needs to read per month as
17x > 102,
17x ÷17 > 102 ÷17, dividing both sides by 17,
x > 6, therefore, Eli needs to read at least 6 e -books per month.

Question 12.
Brittney can spend no more than $15 for new fish in her aquarium. a. Let f be the number of fish she can buy. What inequality represents the problem? b. How many fish can Brittney buy? Answer: a. 3f ≤ 15, b. Brittney can buy no more than 5 fishes, Explanation: Given Brittney can spend no more than$15 for new fish in her aquarium.
a. Let f be the number of fish she can buy.
So inequality that represents the problem is 3f ≤ 15,
b. Solving the inequality equation 3f ≤ 15,
3f ÷ 3 ≤ 15 ÷ 3, dividing both sides by 3 we get,
f ≤ 5, So Brittney can buy no more than 5 fishes.

Question 13.
Isaac has a bag of n peanuts. He shares the peanuts with 5 of his friends.
Each person gets at least 18 peanuts. The inequality 18 ≤ n ÷ 6
represents this situation. Graph the solution of this inequality.
Answer:
The number of peanuts Isaac has a bag of at least 108,

Explanation:
Given Isaac has a bag of n peanuts. He shares the peanuts with 5 of his friends.
Each person gets at least 18 peanuts. The inequality 18 ≤ n ÷ 6
represents this situation. So the number of peanuts Isaac have are
6 X 18 ≤ 6 X n ÷ 6, multiplying the inequality both sides by 6,
108 ≤ n or n ≥ 108, so the graph of the solution
108 ≤ n or n ≥ 108 is as shown above.

Question 14.
a. Solve the inequality – 3x < 12.
b. Reasoning
Describe how you know the direction of the inequality sign
without solving the inequality.
Answer:
a. x > -4,
b. The inequality sign changes because of the
negative integer in the variable,

Explanation:
a. Given to solve the inequality -3x < 12,
-3x ÷ -3< 12 ÷ -3, dividing both sides by -3 and
reverse the inequality symbol,
x > -4.
b. The inequality sign can be determined without solving
because there is negative integer in the variable.
With the negative integer, automatically the inequality sign
changes.

Question 15.
Higher Order Thinking Renata and her family go through an
average of more than 15 cans of sparkling water each day.
They buy cases of 24 cans at $3.50 a case. a. Write an inequality for the number of cases they go through in 30 days. b. Solve the inequality in part a. If they buy only full cases, how much do they spend on sparkling water in 30 days? Answer: a. x ≥ 18.75 is the inequality for the number of cases they go through in 30 days, b.$66.50 they spend on sparkling water in 30 days,

Explanation:
Given Renata and her family go through an
average of more than 15 cans of sparkling water each day.
They buy cases of 24 cans at $3.50 a case. a. 1 day = 15 cans and 24 cans =$3.50,
So let x be number of cases
30 multiply 15 ÷ 24 = 18.75 almost 19,
therefore x ≥ 18.75 is the inequality for the
number of cases they go through in 30 days,
b. 19 X 3.5 = $66.50 they spend on sparkling water in 30 days. Question 16. Solve the inequality. Graph the solution on the number line. -6.25x > -38$$\frac{3}{4}$$ Answer: x < 6.2, Explanation: Given to solve the inequality -6.25x > -38$$\frac{3}{4}$$, converting the mixed number to decimal before solving the inequality, -38$$\frac{3}{4}$$ = -38.75, -6.25x > -38.75, -6.25x ÷ -6.25 > -38.75 ÷ -6.25, dividing both sides by -6.25 and reverse the inequality symbol, x < 6.2, the graph of the solution x < 6.2 is shown above. Assessment Practice Question 17. Cynthia plans to build a tree house that is $$\frac{1}{3}$$ the size of Andrew’s tree house. Cynthia plans to make the area of her tree house at least 13 square feet. PART A Write and solve an inequality to find the area of Andrew’s tree house. Let x be the area of Andrew’s tree house. Answer: $$\frac{1}{3}$$x ≥ 13, Explanation: Given Cynthia plans to build a tree house that is $$\frac{1}{3}$$ the size of Andrew’s tree house. Cynthia plans to make the area of her tree house at least 13 square feet. The inequality for the given situation is $$\frac{1}{3}$$x ≥ 13, Now solving 3 X $$\frac{1}{3}$$x ≥ 3 X 13, multiplying both sides by 3 we get x ≥ 39 is the area of Andrew’s tree house. PART B Describe how you know which tree house is larger without solving the inequality. Answer: It is because of the factor that was multiplied to the area of the Andrew’s tree house which is less than 1, Explanation: Since the factor that was multiplied to the area of the Andrew’s tree house which is less than 1, it can be easily be determined that Cynthia’s tree house is smaller than Andrew’s tree house. 3-ACT Math ACT 1 Question 1. After watching the video, what is the first question that comes to mind? Answer: Question 2. Write the Main Question you will answer. Answer: Question 3. Construct Arguments Make a prediction to answer this Main Question. Explain your prediction. Question 4. On the number line below, write a number that is too small to be the answer. Write a number that is too large. Too small Answer: Question 5. Plot your prediction on the same number line. Answer: ACT 2 Question 6. What information in this situation would be helpful to know? How would you use that information? Answer: Question 7. Use Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically Answer: Question 8. Model with Math Represent the situation using mathematics. Use your representation to answer the Main Question. Answer: Question 9. What is your answer to the Main Question? Is it higher or lower than your initial prediction? Explain why. Answer: ACT 3 Question 10. Write the answer you saw in the video. Answer: Question 11. Reasoning Does your answer match the answer in the video? If not, what are some reasons that would explain the difference? Answer: Question 12. Make Sense and Persevere Would you change your model now that you know the answer? Explain. Answer: ACT 3 Extension Reflect Question 13. Model with Math Explain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question? Answer: Question 14. Reasoning If all single tracks were on sale for 10% off, how would your model change? How would the answer to the Main Question change? Answer: SEQUEL Question 15. Make Sense and Persevere Suppose you have a$50 gift card to the same site.
You want to buy an album with 16 tracks for $12.99 and then use the rest of the gift card for single tracks. How many songs can you buy with the gift card? Answer: ### Lesson 5.6 Solve Two-Step Inequalities Solve & Discuss It! Rico and Halima are shopping for craft sticks, glue, and electrical tape for a science project. Together, they have$30 to spend on supplies.
How should they spend their $30 if they need at least 1,000 craft sticks? Focus on math practices Make Sense and Persevere At the store, Rico and Halima find boxes of 500 craft sticks for$7.50.
Which boxes of craft sticks should they buy?
Answer:
Rico and Halima should buy the box with 500 craft sticks,

Explanation:
Rico and Halima should buy the box with 500 craft sticks,
buying 2 boxes of 500 craft sticks will give them a total
of 1000 craft sticks at a lower price which will cost them $15, instead of buying at least 4 boxes of 275 craft sticks. Essential Question How is solving a two-step inequality similar to and different from solving a two-step equation? Answer: Equations and Inequalities – Two-step equations and inequalities – It takes two steps to solve an equation or inequality that has more than one operation: Simplify using the inverse of addition or subtraction. Simplify further by using the inverse of multiplication or division. Explanation: It takes two steps to solve an equation or inequality that has more than one operation: Simplify using the inverse of addition or subtraction. Simplify further by using the inverse of multiplication or division. when you multiply or divide an inequality by a negative number, we must reverse the inequality symbol. Try It! Erin has$52 to spend at the florist. She wants to buy a vase for $11.75 and several roses for$3.50 each. What are the possible numbers of roses Erin can buy?

Answer:
Erwin can buy up to 11 roses,

Explanation:
Given Erin has $52 to spend at the florist. She wants to buy a vase for$11.75 and
several roses for $3.50 each. The possible numbers of roses Erin can buy are 11.75 + 3.50r ≤ 52, 11.75 + 3.50r – 11.75 ≤ 52 – 11.75, subtracting 11.75 from both sides, 3.50r ≤ 40.25, 3.50r ÷ 3.50 ≤ 40.25 ÷ 3.50, dividing both sides by 3.50 we get r ≤ 11.50, therefore Erwin can buy up to 11 roses. Convince Me! What properties did you use to solve the inequality? 300 5-6 Solve Two-Step Inequalities. Answer: Subtraction property, Division property. Explanation: Solving two-step inequalities, Example: x + 8 > 5, Isolated the variable x by subtracting 8 from both sides of the inequality. x + 8 – 8 > 5 – 8 => x > −3, Therefore, x > −3. Try It! The Jazz Band needs to raise at least$600 to travel to an upcoming competition.
The members of the band have already raised $350. If they sell calendars for$8 each,
how many calendars would they need to sell to exceed their goal?
Answer:
Jazz band needs to sell at least 32 calendars to exceed their goal,

Explanation:
Given the Jazz Band needs to raise at least $600 to travel to an upcoming competition. The members of the band have already raised$350. If they sell calendars for $8 each, number of calendars would they need to sell to exceed their goal if x is the number of calendars, 350 + 8x ≥ 600, 350 + 8x – 350 ≥ 600 -350, subtracting 350 from both sides, 8x ≥ 250, 8x ÷ 8 ≥ 250 ÷ 8, dividing both sides by 8, x ≥ 31.25, therefore, Jazz band needs to sell at least 32 calendars to exceed their goal. Try It! Solve the inequality 5 – $$\frac{1}{2}$$x > 30. Answer: x < -50, Explanation: Given the inequality 5 – $$\frac{1}{2}$$x > 30, 5 – $$\frac{1}{2}$$x -5 > 30 – 5, subtract 5 from both the sides, –$$\frac{1}{2}$$x > 25, -2 X – $$\frac{1}{2}$$x > -2 X 25, multiplying both sides by -2 and reverse the inequality symbol as x < -50. KEY CONCEPT Like two-step equations, solving two-step inequalities involves carrying out two different operations—addition or subtraction, and multiplication or division. Unlike two-step equations, which have a single solution, two-step inequalities have multiple solutions. Do You Understand? Question 1. Essential Question How is solving a two-step inequality similar to and different from solving a two-step equation? Answer: The similarity between the two is that they are carrying out two different operations, The difference is that the number of solutions they have, Explanation: The similarity between solving a two-step inequality and two-step equation is that they are carrying out two different operations, The difference between solving a two-step inequality and two-step equation is that two-step equation only have single solution unlike two-step inequalities have multiple solutions. Question 2. Reasoning What is the difference between the number of solutions for a two-step equation and for a two-step inequality? Answer: Two-step equations only have a single solution, Two-step inequalities have multiple solutions, Explanation: Two-step equations there is only a single solution, Two-step inequalities there are multiple solutions, this is because of the inequality symbol, which makes the inequality true for as long as the value of the variable is within the given solution. Question 3. Why are inverse relationships between operations used to solve two-step inequalities? Answer: Inverse relationships between operations are used to isolate the variable on the one side of the inequality, Explanation: Inverse relationships between operations are used to solve two-step inequalities in order to isolate the variable on the one side of the inequality, removing the values together with the variable means that the inverse operation should be used on both sides. Do You Know How? Question 4. Joe ran 3 miles yesterday and wants to run at least 12 miles this week. Write an inequality that can be used to determine the additional number of days Joe must run this week if each run is 3 miles. Then solve the inequality. Answer: The inequality for the given situation is 3 + 3x ≥ 12, Joe need to run at least 3 days to meet his target, Explanation: Given Joe ran 3 miles yesterday and wants to run at least 12 miles this week. Let x be number of days Joe must run this week if each run is 3 miles. The inequality for the given situation is 3 + 3x ≥ 12, 3 + 3x -3 ≥ 12 -3, subtracting 3 both sides, 3x ≥ 9, dividing both sides by 3 we get x ≥ 3, therefore, Joe need to run at least 3 days to meet his target. Question 5. Solve 4 + 6.5x < 36.5. Answer: Solving 4 + 6.5x < 36.5, we get x < 5, Explanation; Given the inequality 4 + 6.5x < 36.5, 4 + 6.5x -4 < 36.5 -4, subtracting 4 from both the sides, 6.5x < 32.5, 6.5x ÷ 6.5 < 32.5 ÷ 6.5, dividing both sides by 6.5, x < 5. Question 6. Tomas has$1,000 to spend on a vacation. His plane ticket costs $348.25. If he stays 5.5 days at his destination, how much can he spend each day? Write an inequality and then solve. Answer: Inequality: 348.25 + 5.5x ≤ 1000, Tomas can spend no more than$118.50 per day,

Explanation:
Given Tomas has $1,000 to spend on a vacation. His plane ticket costs$348.25.
If he stays 5.5 days at his destination let x be the amount,
the inequality for the given situation is 348.25 + 5.5x ≤ 1000,
348.25 + 5.5x – 348.25 ≤ 1000 – 348.25, subtracting 348.25 from both the sides,
5.5x ≤ 651.75,
5.5x ÷ 5.5 ≤ 651.75 ÷ 5.5, dividing both sides by 5.5,
x ≤ 118.50, therefore, Tomas can spend no more than $118.50 per day. Question 7. Solve 12 – $$\frac{3}{5}$$x > 39. Answer: Solving 12 – $$\frac{3}{5}$$x > 39, we get x < -45, Explanation: Given 12 – $$\frac{3}{5}$$x > 39, 12 – $$\frac{3}{5}$$x – 12 > 39 – 12, subtracting 12 from both sides, – $$\frac{3}{5}$$x > 27, – 5 X- $$\frac{3}{5}$$x < – 5 X 27, multiplying both sides by -5 and reverse the inequality symbol, 3x < -135, 3x ÷ 3 < -135 ÷ 3, divide both sides by 3, x < -45. Practice & Problem Solving Leveled Practice For 8 and 9, fill in the boxes to write and solve each inequality. Question 8. Eight less than the product of a number n and $$\frac{1}{5}$$ is no more than 95. Answer: n ≤ 515, Explanation: The inequality is $$\frac{1}{5}$$ x – 8 ≤ 95, $$\frac{1}{5}$$ x – 8 + 8 ≤ 95 +8, adding 8 on both sides, $$\frac{1}{5}$$ x ≤ 103, 5 X $$\frac{1}{5}$$ x ≤ 5 X 103, multiplying both sides by 5, x ≤ 515. Question 9. Seven more than the quotient of a number b and 45 is greater than 5. Answer: b > -90, Explanation: Given $$\frac{b}{45}$$ + 7 > 5, $$\frac{b}{45}$$ + 7 – 7 > 5 – 7, subtracting 7 from both sides $$\frac{b}{45}$$ > -2, 45 X $$\frac{b}{45}$$ > 45 X (-2), multiplying both sides by 45 we get, b > -90. Question 10. Solve the inequalities and compare. a. Solve 2x + 6 < 10. b. Solve -2x + 22 < 18. c. Which is the correct comparison of solutions for 2x + 6 < 10 and -2x + 22 < 18? A. The inequalities have some common solutions. B. The inequalities have one common solution. C. The inequalities have no common solutions. D. The inequalities have the same solutions. Answer: a. x < 2, b. x > 2, c. C. The inequalities have no common solutions, Explanation: a. Solving 2x + 6 < 10, 2x + 6 -6 < 10 -6, subtracting 6 from both sides, 2x < 4, 2x ÷ 2 < 4 ÷ 2, dividing both sides by 2, so x < 2, b. Solving -2x + 22 < 18, -2x +22 -22 < 18 -22, subtracting 22 from both sides, -2x < -4, -2x ÷ -2 > -4 ÷ -2, dividing both sides by -2 and reverse the inequality symbol, so x > 2, c. The correct comparison of solutions for the two given inequalities 2x + 6 < 10 and -2x + 22 < 18 are the inequalities have no common solutions, the first inequality has a solution that the value of x should be less than 2 to make the inequality true, the second inequality has a solution that the value of x should be greater than 2. Question 11. Make Sense and Persevere Talia has a daily budget of$94 for a car rental.
Write and solve an inequality to find the greatest distance Talia can drive
each day while staying within her budget.

Answer:
Tali can drive no more than 320 miles,

Explanation:
Let x be the distance per mile,
The inequality for the given situation is 30 + 0.20x ≤ 94,
solving 30 + 0.20x ≤ 94,
30 + 0.20x -30 ≤ 94 -30, subtracting 30 from both sides,
0.20 x ≤ 64,
0.20 x ÷0.20 ≤ 64 ÷ 0.20,dividing both sides by 0.20
x ≤ 320, therefore, Tali can drive no more than 320 miles.

Question 12.
Model with Math A manager needs to rope off a
rectangular section for a private party. The length of
the section must be 7.6 meters. The manager can use
no more than 28 meters of rope. What inequality could
you use to find the possible width, w, of the roped-off section?
Answer:
Inequality:
2(7.6) + 2(w) ≤ 28,
The width of the roped – off section should be no
more than 6.4 meters,

Explanation:
Given A manager needs to rope off a
rectangular section for a private party. The length of
the section must be 7.6 meters. The manager can use
no more than 28 meters of rope. The inequality to find
the possible width, w, of the roped-off section is 2(7.6) + 2(w) ≤ 28,
15.2 + 2w ≤ 28, multiplying the values,
15.2 + 2w – 15.2 ≤ 28 – 15.2, substracting 15.2 from both sides,
2w ≤ 12.8,
2w ÷ 2 ≤ 12.8 ÷ 2, dividing both sides by 2,
w ≤ 6.4 , therefore, the width of the roped – off section should be no
more than 6.4 meters.

Question 13.
Higher Order Thinking Andrea went to the store to buy a
sweater that was on sale for 40% off the original price.
It was then put on clearance at an additional 25% off the sale price.
She also used a coupon that saved her an additional $5. Andrea did not spend more than$7.60 for the sweater.
What are the possible values for the original price of the sweater?

Answer:
The price of the sweater is no more than $28, Explanation: Given Andrea went to the store to buy a sweater that was on sale for 40% off the original price. It was then put on clearance at an additional 25% off the sale price. She also used a coupon that saved her an additional$5.
Andrea did not spend more than $7.60 for the sweater. Let x be the original price of the sweater so the inequality is 0.75(0.6x) – 5 ≤ 7.6, multiplying the values 0.75(0.6x) – 5 ≤ 7.6, adding 5 both sides, 0.45 x – 5 + 5 ≤ 7.6 + 5, 0.45x ≤ 12.6, 0.45x ÷ 0.45 ≤ 12.6 ÷ 0.45, dividing both sides by 0.45, x ≤ 28, the price of the sweater is no more than$28.

Question 14.
A pool can hold 850 gallons. It now has 598 gallons of
water and is being filled at the rate shown.
How many more minutes, m, can water continue to
flow into the pool before it overflows? Write and solve an inequality.

Answer:
Inequality :
598 + 15.75m ≤ 850,
The water can  continue to flow into the pool before it overflows
no more 16 minutes,

Explanation:
Given A pool can hold 850 gallons. It now has 598 gallons of
water and is being filled at the rate shown. So more minutes, m,
can water continue to flow into the pool before it overflows is
solving the inequality 598 + 15.75m ≤ 850,
598 + 15.75m – 598 ≤ 850 – 598, subtracting 598 from both sides,
15.75m ≤ 252,
15.75m ÷15.75 ≤ 252 ÷ 15.75, dividing both sides by 15.75 we get
m ≤ 16, therefore the water can  continue to flow into the pool
before it overflows no more 16 minutes.

Assessment Practice

Question 15.
Use the rectangle diagram at the right.
PART A
Write and solve an inequality to find the values of x for
which the perimeter of the rectangle is less than 120.

Answer:
Inequality:
2(x + 4) + 2x < 120,
x < 28,

Explanation:
Given length of rectangle x + 4 and breadth as x and
for which the perimeter of the rectangle is less than 120,
The inequality for the given situation is 2(x + 4) + 2x < 120,
2x + 8 +2x < 120, using distributive property,
4x + 8 < 120, combining like terms,
4x + 8 – 8 < 120 – 8, subtracting 8 from both sides,
4x < 112,
4x ÷ 4 < 112 ÷ 4, dividing both sides by 4 we get
x < 24.

PART B
Based on your answer to Part A, are there any values that
can be eliminated from the solution set? Explain.
Answer:
Yes, 28 can be eliminated from the solution set,

Explanation:
There is a value that can be eliminated in the solution set,
28 is not part of the solution set since the solution set is
less than 28, it means any value that is 27 or below.

Question 16.
Kari is building a rectangular garden bed. The length is 6 feet.
She has 20 feet of boards to make the sides.
Write and solve an inequality to find the possible width of her garden bed.
Answer:
2(6 + w) ≤ 20,
The width of Kari garden bed is less than 4,

Explanation:
Given Kari is building a rectangular garden bed. The length is 6 feet.
She has 20 feet of boards to make the sides.
Let w be the possible width of her garden bed.
The inequality is 2(6 + w) ≤ 20,solving for w,
2(6 + w) ≤ 20, multiplying the values,
12 + 2w ≤ 20, subtracting 12 both sides,
12 + 2w -12 ≤ 20 -12, subtracting 12 both sides,
2w ≤ 8,
2w ÷ 2 ≤ 8 ÷ 2, dividing both sides by 2,
w ≤ 4, therefore the width of Kari garden bed is less than 4.

### Lesson 5.7 Solve Multi-Step Inequalities

Explore It

Charlene has 2 flash drives of the same size that
she uses to store pictures and videos. Each drive is
holding the same number of GB of data, d.
She wants to move everything to a memory card that can hold up to 8 GB.

A. Charlene is going to delete 1 GB of data from each flash drive.
How can the total amount of data left on the two flash drives be
represented as an expression?
B. How can the expression you wrote be used to
write an inequality that shows the maximum amount of
data each flash drive can have on it in order to have all
the data transfer to the 8 GB memory card?
Answer:
a. 2(d – 1),
b. 2(d – 1) ≤ 8,

Explanation:
Given Charlene has 2 flash drives of the same size that
she uses to store pictures and videos. Each drive is
holding the same number of GB of data, d.
She wants to move everything to a memory card that can hold up to 8 GB.
a. Charlene is going to delete 1 GB of data from each flash drive.
So the total amount of data left on the two flash drives be
represented as an expression as 2(d – 1).
b. The expression I wrote can be used
for the inequality that shows the maximum amount of
data each flash drive can have on it in order to have all
the data transfer to the 8 GB memory card is 2(d – 1) ≤ 8.

Focus on math practices
Reasoning If each flash drive has 5 GB of memory, can all of
the data be transferred to the memory card? Explain.
Answer:
A flash drive has 5 GB of memory, can have all of its
data be transferred to the memory card,

Explanation:
Using the inequality to determine If each flash drive
has 5 GB of memory can be transferred to the memory card so
2(d – 1) ≤ 8,
2d – 2 ≤ 8, applying distributive property,
2d – 2 + 2 ≤ 8 + 2, adding 2 on both sides,
2d ≤ 10,
2d ÷ 2 ≤ 10 ÷ 2, dividing both sides by 2,
d ≤ 5,
From the solution set, a flash drive can hold no more than
5 GB, therefore, a flash drive has 5 GB of memory,
can have all of its data be transferred to the memory card.

Essential Question
How is solving a multi-step inequality similar to and different
from solving a multi-step equation.
Answer:
Two are similar because of the methods used in solving,
The difference between the two is the number of solution,

Explanation:
Solving a multi-step inequality and solving a multi-step equation is similar
because of the methods used in solving like using the Distributive Property,
combining like terms and use inverse relationships and properties to solve them.
The difference between a multi-step inequality and solving a multi-step equation is
the number of solution. For multi-step equation there is one possible solution,
for multi-step inequality there are many possible solutions.

Try It!

Twice the difference of Felipe’s age, f, and 4 is at least 2.
What are possible values for Felipe’s age? Graph the solution.

Explanation:
Given twice the difference of Felipe’s age, f, and 4 is at least 2,
the possible values for Felipe’s age are inequality 2(f-4) ≥ 2,
2f – 8 ≥ 2, used distributive property,
2f – 8 + 8 ≥ 2 + 8, adding 8 on both sides,
2f ÷ 2 ≥ 10 ÷ 2, dividing both sides by 2,
f ≥ 5, so the graph of the solution f ≥ 5 is as shown above.

Convince Me!
Describe the similarity between the process of solving an inequality with two steps and
solving an inequality with more than two steps.

Try It!
Solve the inequality – 1 – 6(6 + 2x) < 11. Then graph the solution.

Solve the inequality 3(4 – 6) + 2 ≥ 2(-t + 3) + 4. Then graph the solution.

Answer:

Explanation:
Solving the inequality – 1 – 6(6 + 2x) < 11,
-1 -36 -12x < 11, used distributive property,
-37 -12x < 11, combining like terms,
-37 -12x + 37 < 11 + 37, Adding 37 on both sides,
-12x > 48,
-12x ÷ 12 > 48 ÷ 12, dividing both sides by -12 and reverse inequality symbol,
x > 4, so the graph of the solution x > 4 is as shown above.

Solving the inequality 3(4 – 6) + 2 ≥ 2(-t + 3) + 4,
12 – 18 + 2 ≥ -2t + 6 + 4, used distributive property,
-4 ≥ -2t + 10, combining like terms,
-4 – 10 ≥ -2t + 10 – 10, subtracting 10 from both sides,
-14 ≥ -2t,
-14 ÷ -2 ≤ -2t ÷ -2, dividing both sides by -2 and reverse inequality symbol,
7 ≤ t, so the graph of the solution 7 ≤ t is as shown above.

KEY CONCEPT

Solving multi-step inequalities is similar to solving multi-step equations.
You may need to use the Distributive Property, combine like terms, and
use inverse relationships and properties to solve them.
4(y – 4) + 8 ≤ 20
4y – 16 + 8 ≤ 20
4y – 8 ≤ 20
4y – 8 + 8 ≤ 20 + 8
4y ≤ 28

Do You Understand?

Question 1.
Essential Question How is solving a multi step inequality similar to and
different from solving a multi-step equation?
Answer:
Two are similar because of the methods used in solving,
The difference between the two is the number of solution,

Explanation:
Solving a multi-step inequality and solving a multi-step equation is similar
because of the methods used in solving like using the Distributive Property,
combining like terms and use inverse relationships and properties to solve them.
The difference between a multi-step inequality and solving a multi-step equation is
the number of solution. For multi-step equation there is one possible solution,
for multi-step inequality there are many possible solutions.

Question 2.
Be Precise Explain how you would combine like terms and use properties of
operations to solve the inequality 5(2t + 3) – 3t < 16.
Answer:
t < 1/7,

Explanation:
Given to solve the inequality 5(2t + 3) – 3t < 16,
10t + 15 – 3t <16, used distributive property,
7t + 15 < 16, combining like terms,
7t + 15 -15 < 16 – 15, subtracting 15 from both sides of the inequality,
7t < 1,
7t ÷ 7 < 1 ÷7, dividing both sides by 7 we get,
t < 1/7.

Question 3.
Critique Reasoning Gloria’s solution to a multi-step inequality is r> 7.
She states that the graph will have an open dot at 7 and extend with
an arrow to the right indefinitely. Is she correct? Explain.
Answer:
Gloria’s describes the solution correctly,

Explanation:
Gloria is correct because it will open dot as the solution does not
include 7 as the part of the solution, the inequality symbol is greater
than which means that the arrow will extend to the right.

Do You Know How?

Question 4.
Solve the inequality 2(n + 3) – 4 < 6. Then graph the solution.

Answer:

Explanation:
Given 2(n + 3) – 4 < 6,
2n + 6 – 4 < 6, using distributive property,
2n + 2 < 6, combined like terms,
2n + 2 – 2 < 6 – 2, subtracting 2 from both sides,
2n < 4,
2n ÷ 2 < 4 ÷ 2, dividing both sides by 2,
n < 2, so the graph of the solution n < 2 is as shown above.

Question 5.
Solve the inequality -2(x + 3) + 2 ≥ 6. Then graph the solution.

Answer:

Explanation:
Given -2(x + 3) + 2 ≥ 6,
-2x – 6 + 2 ≥ 6, using distributive property,
-2x – 4 ≥ 6, combined like terms,
-2x – 4 + 4 ≥ 6 + 4, adding 4 on both sides,
-2x ≥ 10,
-2x ÷ -2 ≤ 10 ÷ -2, dividing both sides by -2 and reversing inequality symbol,
x ≤ 5, so the graph of the solution x ≤ 5 is as shown above.

Question 6.
Three times the difference of Federico’s age and 4, increased by 7, is greater than 37.
What are possible values of Federico’s age? Graph his possible ages on the number line.

Answer:

Explanation:
Given three times the difference of Federico’s age and 4, increased by 7
is greater than 37. The possible values of Federico’s age are
3(x – 4) + 7 > 37,
3x – 12 + 7 > 37, using distributive property,
3x – 5 > 37, combined like terms,
3x – 5 + 5 > 37 + 5, adding 5 on both sides,
3x > 42,
3x ÷ 3 > 42 ÷ 3, dividing both sides by 3,
x > 14, so there Federico’s age should be greater than 14,
The graph of the possible ages of Federico’s is as shown above.

Practice & Problem Solving

Question 7.
Use the inequality 18 < -3(4x – 2).
a. Solve the inequality for x.
b. Which graph shows the solution to the inequality?

Answer:

Explanation:
a. Given to solve the inequality 18 < -3(4x -2),
18 < -12x + 6, using distributive property,
18 – 6 < -12x + 6 – 6, subtracting 6 from both sides,
12 < -12x,
12 ÷ -12 > -12x ÷ -12, dividing both sides by -12 and reversing inequality symbol,
-1 > x,
b. Graph A shows the solution to the inequality 18 < -3(4x -2) as -1 > x which
is shown above.

Question 8.
Michelle says that the solution to the inequality 2(4y – 3) > -22 is y> -3.5.
Her work is shown.
2(4y – 3) > -22
8y > -28
y > -3.5
a. What was Michelle’s mistake?
b. What is the solution to the inequality?
Answer:
a. Michelle did not use the inverse relationship in order to isolate the variable,
b. The solution is y > -2,

Explanation:
a. After using distributive property, Michelle did not use the inverse relationship
in order to isolate the variable, she thought that subtracting 6 on the right side of inequality
will remove the 6 as well on left side of the inequality,
b. Solving the inequality 2(4y – 3) > -22,
8y – 6 > -22, using distributive property,
8y – 6 + 6 > -22 + 6, adding 6 on both sides,
8y > -16,
8y ÷ 8 > -16 ÷ 8, dividing both sides by 8 we get
y > -2.

Question 9.
Model with Math The length of a picture frame is 7 inches more than the width.
For what values of x is the perimeter of the picture frame greater than 154 inches?

Answer:
The width of the picture frame is greater than 35 inches,

Explanation:
Given the length of a picture frame is 7 inches more than the width.
So the values of x for which the perimeter of the picture frame will be
greater than 154 inches is
2(x + 7) + 2 (x) > 154,
2x + 14 + 2x > 154, using distributive property,
4x + 14 > 154, combined like terms,
4x + 14 – 14 > 154 – 14, subtracting 14 on both sides,
4x > 140,
4x ÷ 4 > 140 ÷ 4, dividing both sides by 4 we get
x > 35, therefore the width of the picture frame is greater than 35 inches.

Question 10.
Critique Reasoning Sierra says that she can simplify the left side of
the inequality 2(-3 + 5) + 2 ≥ -4(x – 2) – 3 by combining the terms
within the parentheses, but that she can’t do the same on the right side.
Is Sierra correct? Explain.
Answer:
Sierra is correct about combining the terms,

Explanation:
Sierra is correct as the terms inside the parentheses are both constant,
Performing the operation inside the parentheses can combine
the two constants. She is also correct that on the right side of inequality,
the term are not the same, therefore it cannot be combined.

Question 11.
a. Solve the inequality 30 ≥ 6($$\frac{2}{3}$$z + $$\frac{1}{3}$$).
b. Solve the inequality 15.6 < 2.7(z – 1) – 0.6.
c. Are there any values of z that solve both inequalities?
Use a number line to support your answer.
Answer:
a. 7 ≥ z,
b. 7< z,
c.

Explanation:
a. Solving the inequality 30 ≥ 6($$\frac{2}{3}$$z + $$\frac{1}{3}$$),
30 ≥ 4z + 2, using distributive property,
30 – 2 ≥ 4z + 2 – 2, subtracting 2 on both sides,
28 ≥ 4z,
28 ÷ 4 ≥ 4z ÷ 4, dividing both sides by 4 we get 7 ≥ z.

b. Solving the inequality 15.6 < 2.7(z – 1) – 0.6,
15.6 < 2.7z – 2.7 – 0.6, using distributive property,
15.6 < 2.7z – 3.3, combined like terms,
15.6 + 3.3 < 2.7z – 3.3 + 3.3, adding 3.3 on both sides,
18.9 < 2.7z,
18.9 ÷ 2.7 < 2.7z ÷ 2.7, dividing both sides by 2.7,
we get 7 < z.

c. There are no values that could solve both inequalities,
This is because, the first inequality has the solution set of any
value no more than 7 while the second inequality has the solution
set of any value greater than 7. The solution of first inequality, includes
7 while the solution for the second inequality does not include 7.

Question 12.
Mr. Lin baked banana bread for a bake sale to raise money for the math team.
He said that he added a spoonful of walnuts for each of the students in
his three classes, and that he added more than 250 walnuts. He used the
inequality 16W + 24W + 10w > 250 to represent the situation, where w represents
the number of walnuts in each spoonful. How many walnuts could be in each spoonful?

Answer:
Each spoonful have greater than 5 walnuts,

Explanation:
Given Mr. Lin baked banana bread for a bake sale to raise money for the math team.
He said that he added a spoonful of walnuts for each of the students in
his three classes and that he added more than 250 walnuts. He used the
inequality 16W + 24W + 10w > 250 to represent the situation, where w represents
the number of walnuts in each spoonful. So number of many walnuts could be in
each spoonful is writing inequality 16W + 24W + 10w > 250 and solving as
50w > 250, combining like terms,
50w ÷ 50 > 250 ÷ 50, dividing both sides by 50,
we get w >5, therefore, each spoonful have greater than 5 walnuts.

Question 13.
Use both the Addition and Multiplication Properties of Inequality
to solve the inequality.
Graph the solutions on a number line. 2(3y – 5) < -16

Answer:

Explanation:
Solving the inequality 2(3y – 5) < -16,
6y – 10 < -16, using distributive property,
6y – 10 + 10 < -16 + 10, adding 10 on both sides,
6y < -6,
6y ÷ 6 < -6 ÷ 6, dividing both sides by 6,
we get y < -1.

Question 14.
Higher Order Thinking Solve each of the given inequalities for z.
Which of the inequalities has 5 as a solution?

Answer:
The inequality that has a solution of 5 is Inequality 1,

Explanation:
Given to solve the inequality 4(2.8z +1.75) > -26.6,
11.2z + 7 > -26.6, using distributive property,
11.2z + 7 – 7 > -26.6 – 7, subtracting 7 on both sides,
11.2z > -33.6,
11.2z ÷ 11.2 > -33.6 ÷ 11.2, dividing both sides by 11.2,
z > -3,
Given to solve the inequality 2(1.9z +1.5) ≤ 18.2,
3.8z + 3 ≤ 18.2, using distributive property,
3.8z + 3 – 3 ≤ 18.2 – 3 subtracting 7 on both sides,
3.8z ≤ 15.2,
3.8z ÷ 3.8 ≤ 15.2 ÷ 3.8, dividing both sides by 3.8,
z ≤ 4,
So the inequality that has a solution of 5 is Inequality 1.

Assessment Practice

Question 15.
The school band needs $500 to buy new hats. They already have$200.
They are selling bumper stickers for $1.50 each. How many bumper stickers do they need to sell to have at least$500? Write and solve an inequality that represents the situation.
Answer:
200 bumper stickers do they need to sell,
Inequality : 1.50b + 200 = 500,

Explanation:
Given the school band needs $500 to buy new hats. They already have$200.
They are selling bumper stickers for $1.50 each. Let b be the number of bumper stickers do they need to sell to have at least$500? The inequality that represents the situation is
1.50b + 200 ≤ 500, solving,
1.50b + 200 – 200 ≤ 500 – 200, subtracting 200 both sides,
1.50b ≤ 300,
1.50b ÷1.50 ≤ 300 ÷ 1.50, dividing both sides by 1.50,
b ≤ 200, therefore 200 bumper stickers do they need to sell and
Inequality : 1.50b + 200 = 500.

### Topic 5 REVIEW

Topic Essential Question

How can you solve real-world and mathematical problems with numerical and
algebraic equations and inequalities?
Answer:
Through the use of equations and inequalities , we will be able to determine
the possible solutions to the problem,

Explanation:
Real world and mathematical problems can be solved through the use
of numerical and algebraic equations and inequalities.
In a certain problem there are unknown that needs to be solved.
Therefore, using equations and inequalities in order to determine
the unknown it will help to easily solve the problem.
Through the use of equations and inequalities we will be
able to determine the possible solutions to the problem.

Vocabulary Review

Complete each definition and then provide an example of each vocabulary word used.

Question 1.
You when you divide both sides of the equation 3n = 12 by 3.
Answer:
Isolate the variable,
When we divide both sides of the equation 3n = 12 by 3, we isolate
the variable,

Explanation:
3n = 12,
3n ÷ 3 = 12 ÷ 3, Dividing both sides by 3 we get n = 4.

Question 2.
A statement that contains >, <, ≥, ≤, or ≠ to compare two expressions is a(n)
Answer:
Inequality,

Explanation:
Inequality is a statement that contains >, <, ≥, ≤, or ≠ to compare two expressions.

Question 3.
You can use the to remove parentheses in
the process of solving the equation – 10(x + 5) = 40.
Answer:
Distributive Property,

Explanation:
In the the process of solving the equation – 10(x + 5) = 40,
we use distributive property to remove parentheses as
-10x -50 = 40.

Use Vocabulary in Writing
Write an equation or inequality to represent the following situation:
17 is at least 5 more than 3 times x. Explain how you wrote your
equation or inequality. Use vocabulary from Topic 5 in your explanation.
Answer:
Equation or Inequality : 17 > 3x + 5,

Explanation:
Given the situation as 17 is at least 5 more than 3 times x,
So wrote the equation or inequality as 17 > 5 + 3x means
17 is greater than 5 and 3 times of x(3x), Vocabulary used is
equation or inequality.

Concepts and Skills Review

LESSONS 5-1 AND 5-2
Write Two-Step Equations | Solve Two-Step Equations

Quick Review
Equations can be used to represent situations. Two-step equations have two
different operations.
The properties of equality can be applied the same way when solving
two-step equations as when solving one-step equations.

Practice

Question 1.
The total number of students in the seventh grade is 9 more than 4 times
as many students as are in the art class. There are 101 students in the seventh grade.
Write and solve an equation to find the number of students in the art class.
Let x represent the number of students in the art class.
Answer:
Equation : 4x + 9 = 101,
There are 23 students in the art class,

Explanation:
Given the total number of students in the seventh grade is 9 more than 4 times
as many students as are in the art class. There are 101 students in the seventh grade.
Let x represent the number of students in the art class.
So the equation to find the number of students in the art class is  4x + 9 = 101, on solving
4x + 9 – 9 = 101 – 9, subtracting 9 from both sides,
4x ÷ 4 = 92 ÷ 4, Dividing both sides by 4 we get
x = 23, therefore there are 23 students in the art class,

Question 2.
List the steps to solve the following equation: 5x – 6 = 44.
Then solve for x.
Answer:
x = 10,

Explanation:
The steps to solve the following equation : 5x – 6 = 44 is
5x – 6 + 6 = 44 + 6, adding 6 on both sides,
5x = 50,
5x ÷ 5 = 50 ÷ 5, dividing both sides by 5 we get x = 10.

Question 3.
Solve for the given variable.
a. 4y + 3 = 19
b. $$\frac{1}{2}$$n – 3 = 5
Answer:
a. y = 4,
b. n = 16,

Explanation:
Solving:
a. 4y + 3 = 19,
4y + 3 – 3 = 19 – 3, subtracting 3 on both sides,
4y = 16,
4y ÷ 4 = 16 ÷ 4, dividing both sides by 4 we get y = 4.

b. $$\frac{1}{2}$$n – 3 = 5,
$$\frac{1}{2}$$n – 3  + 3 = 5 + 3, adding 3 on both sides,
$$\frac{1}{2}$$n = 8,
2 X $$\frac{1}{2}$$n = 2 X 8, multiplying both sides by 2 we get
n = 16.

LESSON 5.3 Solve Equations Using the Distributive Property

Quick Review
Use the Distributive Property to solve problems of the form p(x + q) = r.

Practice

Question 1.
There are 450 seats in the lower level of a concert hall with
b balcony seats in the upper level. So far, 170 tickets have been sold,
which is $$\frac{1}{5}$$ of the total number of seats in the concert hall.
How many tickets sold are balcony seats?
Answer:
There are 400 balcony seats in the concert hall,

Explanation:
Given there are 450 seats in the lower level of a concert hall with
b balcony seats in the upper level. So far, 170 tickets have been sold,
which is $$\frac{1}{5}$$ of the total number of seats in the concert hall.
Number of tickets sold which are balcony seats are
Equation: $$\frac{1}{5}$$(b + 450) = 170,
$$\frac{1}{5}$$b + 90 = 170, Using distributive property,
$$\frac{1}{5}$$b + 90 – 90 = 170 – 90, subtracting 90 from both sides,
$$\frac{1}{5}$$b = 80,
5 X $$\frac{1}{5}$$b = 5 X 80 , multiplying both sides by 5 we get
b = 400, therefore there are 400 balcony seats in the concert hall.

Question 2.
Solve the equation -4(8 + y) = 90.
Answer:
y = -30.5,

Explanation:
Given to solve -4(8 + y) = 90,
-32 – 4y = 90, Using distributive property,
-32 – 4y + 32 = 90 + 32, adding 32 on both sides,
-4y = 122,
-4y ÷ -4 = 122 ÷ -4, dividing both sides by -4 we get y = -30.5.

LESSON 5-4 Solve Inequalities Using Addition or Subtraction

Quick Review
When you add or subtract the same number on both sides of an inequality,
the relationship between the sides stays the same.
Solutions to inequalities can be graphed on number lines.

Practice

Question 1.
Carson’s wheelbarrow can hold 345 pounds. If he has 121 pounds
of rock in the wheelbarrow, what number of pounds, p, can he put in
the wheelbarrow without going over the weight limit?
Answer:
The weight that can put in the wheelbarrow is no more than 224 pounds,

Explanation:
Given Carson’s wheelbarrow can hold 345 pounds.
If he has 121 pounds of rock in the wheelbarrow,
So number of pounds p can he put in the wheelbarrow without
going over the weight limit is  121 + p ≤ 345,
121 + p – 121 ≤ 345 – 121, subtracting 121 from both sides we get
p ≤ 224, therefore the weight that can put in the wheelbarrow is
no more than 224 pounds.

Question 2.
Solve x – 19 < 81. Then graph the solution.
Answer:

Explanation:
Solving x – 19 < 81 as
x – 19 + 19 < 81 + 19, adding 19 on both sides,
x < 100 and graph of the solution x < 100 is as shown above.

LESSON 5-5 Solve Inequalities Using Multiplication or Division

Quick Review
When you multiply or divide both sides of an inequality by the same positive number,
the inequality remains true. When you multiply or divide both sides of an inequality by
the same negative number, you need to reverse the inequality symbol,
but the inequality remains true.

Practice

Question 1.
Travis has 3 months to save money for a trip. An airplane ticket costs more than $300. If he saves the same amount of money, a, each month, how much does he need to save each month to pay for the ticket? Answer: Inequality : 3a > 300, Travis should save more than$100 each month to pay for the ticket,

Explanation:
Given Travis has 3 months to save money for a trip.
An airplane ticket costs more than $300. If he saves the same amount of money a each month, So he need to save each month to pay for the ticket is 3a > 300, Now on simplifying 3a ÷ 3 = 300 ÷ 3, Dividing both sides by 3 we get a >100, therefore, Travis should save more than$100 each month to pay for the ticket.

Question 2.
Solve –$$\frac{1}{8}$$y ≤ 34. Then graph the solution.
Answer:

Explanation:
Solving –$$\frac{1}{8}$$y ≤ 34,
-8 X –$$\frac{1}{8}$$y ≤ -8 X 34, Multiplying both sides by -8 and
reverse the inequality symbol,
y ≥ -272 and graph of the solution y ≥ -272 is as shown above.

LESSON 5-6 Solve Two-Step Inequalities

Quick Review
Inverse relationships and properties can be used to isolate the variable and
solve two step inequalities in the form px + q < r or px + q > r in the same way
that they are used to solve two-step equations.

Practice

Question 1.
The school band gets $5 for each T-shirt they sell at a fundraiser. They have a goal of raising$150. If $45 has been raised so far, how many more T-shirts do they have to sell to reach or exceed the goal? Answer: The school band needs to sell at least 21 T-shirts, Explanation: Given the school band gets$5 for each T-shirt they sell at a fundraiser.
They have a goal of raising $150. If$45 has been raised so far,
more T-shirts they do have to sell to reach or exceed the goal are
45 + 5x ≥ 150,
45 + 5x – 45 ≥ 150 – 45, subtracting 45 from both sides,
5x ≥ 105,
5x ÷ 5 ≥ 105 ÷ 5, dividing both sides by 5 we get x ≥ 21,
therefore the school band needs to sell at least 21 T-shirts.

Question 2.
Solve the inequality –8 –$$\frac{1}{3}$$n ≤ -25.
Answer:
n ≥ 51,

Explanation:
Solving the inequality –8 –$$\frac{1}{3}$$n ≤ -25,
–8 –$$\frac{1}{3}$$n + 8 ≤ -25 + 8, adding 8 on both sides,
–$$\frac{1}{3}$$n ≤ -17,
-3 X –$$\frac{1}{3}$$n ≥ – 3 X -17, Multiplying both sides by -3 and
reverse the inequality symbol we get n ≥ 51.

LESSON 5-7 Solve Multi-Step Inequalities

Quick Review
Solving a multi-step inequality is similar to solving a multi-step equation.
All of the rules and properties for solving one- and
two step inequalities apply to solving multi-step inequalities.

Practice

Question 1.
Solve 1.9(2.3n + 6) + 10.45 > 43.7. Then graph the solution.

Answer:

Explanation:
Given to solve 1.9(2.3n + 6) + 10.45 > 43.7,
4.37n + 11.4 + 10.45 > 43.7, using distributive property,
4.37n + 21.85 > 43.7, combining like terms,
4.37n + 21.85 – 21.85 > 43.7 – 21.85, subtracting 21.85 from both sides,
4.37n > 21.85,
4.37n ÷ 4.37 > 21.85 ÷ 4.37, dividing both sides by 4.37 we get
n > 5 and graph of the solution n > 5
is as shown above.

Question 2.
Solve 4(-2n + 2.5) – 8 ≤ 50. Then graph the solution.

Answer:

Explanation:
Given to solve 4(-2n + 2.5) – 8 ≤ 50,
-8n + 10 – 8 ≤ 50, using distributive property,
-8n + 2 ≤ 50, combining like terms,
-8n +2 -2 ≤ 50 – 2, subtracting -2 from both sides,
-8n ≤ 48,
-8n ÷ -8 ≥ 48 ÷ -8, dividing both sides by -8 and reverse
the inequality symbol we get n ≥ -6 and
graph of the solution n ≥ -6
is as shown above.

### Topic 5 Fluency Practice

Crisscrossed

Solve each problem. Write your answers in the cross-number puzzle below.
Each digit, decimal point, dollar sign, and percent symbol of your answer goes in its own box.
Round money amounts to the nearest cent as needed.

ACROSS
A. Antonia buys 0.75 yard of fabric at $12.00 per yard. If she pays 5% sales tax, what is the total cost of the fabric? B. Five friends plan to split a restaurant bill evenly. The total cost of the meal is$89.75, and they want to leave a 20% tip.
What amount should each friend pay?
E. Kaylie buys a sweater on sale for $40.11. If the discount is 20% off and she pays$1.91 in sales tax, what is the original price of the sweater?
F. Randy buys a pair of shoes that were originally priced at $147. He receives a 35% discount and pays 8.5% sales tax. How much does Randy pay? G. A basketball player makes 8 of 22 shots in Game 1, 6 of 15 shots in Game 2, and 10 of 23 shots in Game 3. What percent of the shots did the player make in the three games? Answer: Across: A. The total cost of the fabric is$9.45,
B. The amount should each friend pay is $21.54, E. The original price of the sweater is$47.75,
F. Randy paid $103.67, G. Percent of the shots did the player make in the three games is 40%, Explanation: Across: A. Given Antonia buys 0.75 yard of fabric at$12.00 per yard.
If she pays 5% sales tax, so 0.75 X $12 =$9,
Now 5% on $9 is 5 X 9 ÷100 =$0.45,
therefore the total cost of the fabric is $9 +$0.45 = $9.45. B. Given five friends plan to split a restaurant bill evenly. The total cost of the meal is$89.75 and they want to leave a 20% tip.
So amount should each friend pay must be
89.75 X 20 ÷100 = $17.95 is the tip, the total amount is$89.75 + $17.95 =$107.7 which is paid by
5 friends, therefore the amount each friend must pay is
$107.7 ÷ 5 =$21.54.
E. Given Kaylie buys a sweater on sale for $40.11. If the discount is 20% off and she pays$1.91 in sales tax,
Let x be the original price of the sweater so
Original Price = (Paid Price – Tax) + 20% discount,
x = (40.11 -1.91) + 20% discount,
x  = $38.20 +( (20 X x) ÷ 100), x – (20x ÷ 100)=$38.20,
100x – 20x = $38.20 X 100, 80x =$3,820.
x  = $3,820 ÷ 80, x =$47.75.
F. Given Randy buys a pair of shoes that were originally priced at $147. He receives a 35% discount and pays 8.5% sales tax. Let Randy paid p, Randy receives 35% discount means ($147 – ($147 X 35% discount)) = ($147 – $51.45) =$95.55,
on $95.55 Randy pays 8.5% sales tax so 95.55 X 8.5 ÷ 100 =$8.12175,
Therefore Randy paid p = $95.55 +$8.12175 = $103.67175 ≈$103.67.
G. Given a basketball player makes 8 of 22 shots in Game 1,
6 of 15 shots in Game 2 and 10 of 23 shots in Game 3,
In Game 1 percent of the shots did the player make is 8 ÷ 22 = 0.36 = 36%,
In Game 2 percent of the shots did the player make is 6 ÷ 15 = 0.40 = 40%,
In Game 3 percent of the shots did the player make is 10 ÷ 23 = 0.43 = 43%,
Therefore percent of the shots did the player make in the three games is
(36% + 40% + 43%) ÷ 3 = 39.33% ≈ 40%.

DOWN
A. Jack buys a tablet that costs $99 and a memory card that costs$15.
He has a coupon for a 15% discount. What is the amount of the discount on the two items?
B. Tara buys two pairs of socks for $4.99 each and three T-shirts for$11.45 each.
If she pays 6% sales tax, what is the total amount of her purchase?
C. Sunil receives a 20% discount on a concert ticket that costs $75. If Sunil pays$3.30 in sales tax on the discounted ticket, what is the sales tax rate?
D. Dylan works for 4 hours and is paid $17.50 per hour. He must pay 15% in income taxes. What amount does he earn after taxes? E. Miles earns a 6% commission on each vehicle he sells. Today he sold a truck for$18,500 and a car for $9,600. What is the total amount of his commission for these vehicles? Answer: A. The amount of the discount on the two items is$17.10,
B. The total amount of Tara’s purchase is $47.00, C. The sales tax rate is 5.5%, D. Dylan earn after taxes is$59.50,
E. The total amount of Miles commission for these vehicles is $1,686, Explanation: A. Given Jack buys a tablet that costs$99 and a memory card that costs $15. He has a coupon for a 15% discount. So tablet costs$99 – ($99 X 15 ÷ 100) =$99 –  $14.85 =$84.15 and Memory card costs $15 – ($15 X 15 ÷ 100) =
$15 –$2.25 = $12. 75, Therefore the amount of the discount on the two items is discount amount on tablet + discount amount on memory card =$ 14.85 +  $2.25 =$17.10.
B. Given Tara buys two pairs of socks for $4.99 each and three T-shirts for$11.45 each.
If she pays 6% sales tax, To find the total amount of her purchase first we will calculate
cost for socks + cost for T-shirts and then add sales tax too, so
2($4.99) + 3(11.45) =$9.98 + $34.35 =$44.33,
Now sales tax = $44.33 X 6 ÷ 100 =$2.6598, therefore $44.33 +$2.6598 = $46.9898 ≈$47.00.
C. Given Sunil receives a 20% discount on a concert ticket that costs $75. If Sunil pays$3.30 in sales tax on the discounted ticket,
The discounted ticket price is ($75 –$75 X 20 ÷ 100) = $75 –$15 = $60, let the sales rate tax is r, so$3.30 = r% of 60, therefore r = (3.30 ÷ 60) X 100= 0.55 X 100 = 5.5%.
D. Given Dylan works for 4 hours and is paid $17.50 per hour. He must pay 15% in income taxes. So amount does he earn after taxes would be first he gets$17.50 X 4 = $70, So amount paid to income tax will be$70 X 15 ÷ 100 = $10.5, Therefore Dylan earn after taxes is$70 – $10.5 =$59.50.
E. Given Miles earns a 6% commission on each vehicle he sells.
Today he sold a truck for $18,500 and a car for$9,600.
So first commission on truck sold is $18,500 X 6 ÷ 100 =$1,110 and
commission on car sold is $9,600 X 6 ÷ 100 =$576,
therefore the total amount of his commission for these vehicles is
$1,110 +$576 = \$1,686.

## enVision Math Common Core 8th Grade Answers Key Topic 6 Congruence And Similarity

Topic Essential Question
How can you show that two figures are either congruent or similar to one another?
Answer:
When two line segments have the same length, we can say that they are congruent. When two figures have the same shape and size, we can say that the two figures are congruent. These two triangles are congruent. We can also say that their side lengths are the same and that their angle measures are the same

3-ACT MATH

Tricks of the Trade
All kinds of objects in nature have symmetry: beehives, pine cones, butterflies, and snowflakes, to name a few. If you look closely, you will start to see patterns left and right. Think about this during the 3-Act Mathematical Modeling lesson.

Topic 6 ënVision STEM Project

Did You Know?
Trees provide wood for cooking and heating for half of the world’s population.

As trees grow, carbon dioxide is removed from the atmosphere for photosynthesis. Forests are called “carbon sinks” because one acre of forest absorbs six tons of carbon dioxide and puts out four tons of oxygen.

Trees provide lumber for buildings, tools, and furniture. Other products include rubber, sponges, cork, paper, chocolate, nuts, and fruit.
About 30% of the land is covered by forests.

Forests are now being managed to preserve wildlife and old-growth forests, protect biodiversity, safeguard watersheds, and develop recreation, as well as extract timber.

Forests also need to be managed to prevent raging wildfires, invasive species, overgrazing, and disease.

Your Task: Forest Health
The proper management of forests is a growing science. You and your classmates will learn about forest health indicators and use what you know about similar triangles and ratios to gather and interpret data in order to assess the health of a forest.

### Topic 6 GET READY!

Review What You Know!

Vocabulary
Choose the best term from the box to complete each definition.
adjacent angles
complementary angles
supplementary angles
vertical angles

Question 1.
_________ have a sum of 90°.
Answer:
We know that,
The “Complementary angles” have a sum of 90°
Hence, from the above,
We can conclude that the best term to complete the given definition is: Complementary angles

Question 2.
_________ share the same ray.
Answer:
We know that,
The “Adjacent Angles” share the same ray
Hence, from the above,
We can conclude that the best term to complete the given definition is: Adjacent angles

Question 3.
_________ are pairs of opposite angles made by intersecting lines.
Answer:
We know that,
The “Vertical Angles” are pairs of opposite angles made by intersecting lines
Hence, from the above,
We can conclude that the best term to complete the given definition is: Vertical angles

Question 4.
__________ have a sum of 180°
Answer:
We know that,
The “Supplementary Angles” have a sum of 180°
Hence, from the above,
We can conclude that the best term to complete the given definition is: Supplementary angles

Multiplying Real Numbers

Simplify the expression.
Question 5.
5 × 2 = ________
Answer:
The given expression is:
5 × 2
Hence, from the above,
We can conclude that
5 × 2 = 10

Question 6.
6 × $$\frac{1}{2}$$ = ________
Answer:
The given expression is:
6 × $$\frac{1}{2}$$
So,
6 × $$\frac{1}{2}$$
= $$\frac{6}{2}$$
= 3
Hence, from the above,
We can conclude that
6 × $$\frac{1}{2}$$ = 3

Question 7.
12 × $$\frac{1}{3}$$ = ________
Answer:
The given expression is:
12 × $$\frac{1}{3}$$
So,
12 × $$\frac{1}{3}$$
= $$\frac{12}{3}$$
= 4
Hence, from the above,
We can conclude that
12 × $$\frac{1}{3}$$ = 4

Identifying Points on a Coordinate Plane

Name the location of the point.

Answer:
The given coordinate plane is:

From the given coordinate plane,
We can observe that

Question 8.
point W
Answer:
From the given coordinate plane,
We can observe that point W lies where both the x-axis and y-axis are negative
Hence, from the above,
We can conclude that point W lies in the third quadrant

Question 9.
point X
Answer:
From the given coordinate plane,
We can observe that point X lies where the x-axis is negative and the y-axis is positive
Hence, from the above,
We can conclude that point X lies in the second quadrant

Question 10.
point Y
Answer:
From the given coordinate plane,
We can observe that point Y lies where both the x-axis and y-axis are positive
Hence, from the above,
We can conclude that point Y lies in the first quadrant

Question 11.
point Z
Answer:
From the given coordinate plane,
We can observe that point Z lies where the x-axis is positive and the y-axis is negative
Hence, from the above,
We can conclude that point Z lies in the fourth quadrant

Supplementary Angles

The angles are supplementary. Find the missing angle measure.
Question 12.

Answer:
The given figure is:

We know that,
The sum of the supplementary angles is: 180°
Now,
Let the missing angle measure be: x°
So,
x° + 130° = 180°
x° = 180° – 130°
x° = 50°
Hence, from the above,
We can conclude that the missing angle measure is: 50°

Question 13.

Answer:
The given figure is:

We know that,
The sum of the supplementary angles is: 180°
Now,
Let the missing angle measure be: x°
So,
x° + 139° = 180°
x° = 180° – 139°
x° = 41°
Hence, from the above,
We can conclude that the missing angle measure is: 41°

Language Development
Complete the graphic organizer with an illustration for each transformation. Write either congruent or similar to make the given statement true.

Answer:

### Topic 6 PICK A PROJECT

PROJECT 6A
How might an artist use mathematics?
PROJECT: WRITE A BIOGRAPHY

PROJECT 6B
What geometric shapes do you see around you?
PROJECT: RECORD A VIDEO ABOUT SIMILAR FIGURES

PROJECT 6C
What different types of bridges have you crossed?
PROJECT: BUILD A MODEL OF A TRUSS BRIDGE

PROJECT 6D
What shapes tessellate?
PROJECT: DESIGN A TESSELLATION

### Lesson 6.1 Analyze Translations

Solve & Discuss It!
Ashanti draws a trapezoid on the coordinate plane and labels it in Figure 1. Then she draws Figure 2. How can she determine whether the figures have the same side lengths and the same angle measures?

I can… translate two-dimensional figures.
Answer:
It is given that
Ashanti draws a trapezoid on the coordinate plane and labels it in Figure 1. Then she draws Figure 2.
The given figure is:

From the figure,
We can observe that the first figure was translated or moved to the right by some units and became a second figure
We know that,
A “Translation” is a transformation that moves every point in a figure the same distance in the same direction
We know that,
The pre-image and image in the “Translation” are similar
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal.
Hence, from the above,
We can conclude that by using the “Translation” property, she can determine whether the figures have the same side lengths and the same angle measures

Focus on math practices
Be Precise How do you know that the method you described shows whether the side lengths and angle measures are equal? Explain.
Answer:
We know that,
A “Translation” is a transformation that moves every point in a figure the same distance in the same direction
We know that,
The pre-image and image in the “Translation” are similar
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal.

Essential Question
How does a translation affect the properties of a two-dimensional figure?
Answer:
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.

Try It!

The clients also want the small table below the window moved 5 feet to the right. Where should the architect place the small table? Draw the new location of the table on the plan.
Answer:
It is given that
The clients also want the small table below the window moved 5 feet to the right
Now,
The original plan is:

From the above plan,
We can observe that the window is at most right
So,
When we move the table,
The new location of the table will be below the window
Hence,
The location of the table on the new plan is:

Convince Me!
An equilateral triangle with side lengths 5 inches is translated 3 units down and 2 units right. Describe the shape and dimensions of the translated figure.
Answer:
It is given that
An equilateral triangle with side lengths 5 inches is translated 3 units down and 2 units right.
We know that,
In an equilateral triangle, all the side lengths are equal
Now,
Let the dimensions of an equilateral triangle be (x1, y1), (x2, y2), and (x3, y3)
So,
The dimensions of the translated figure are: (x1 + 2, y1 – 3), (x2 + 2, y2 – 3), and (x3 + 2, y3 – 3)
Now,
Let us suppose the dimensions of an equilateral to be:
(1, 1), (1, 6), and (5, 3)
So,
The dimensions of the translated equilateral triangle are:
(3, -2), (3, 3), and (7, 0)
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way
So,
The shape of the translated figure remains the same as the original figure
Hence,
The representation and shape of the original and translated equilateral triangles are:

Try It!

Triangle ABC is translated 5 units right and 1 unit down. Graph and label the image A’ B’C’. If m∠A = 30° in ΔABC, then what is m∠A in ΔA’B’C’?

Answer:
The given figure is:

From the given figure,
The dimensions of ΔABC are:
A (-2, 3), B (-3, -4), and C (-5, -1)
Now,
Let the dimensions of an equilateral triangle be (x1, y1), (x2, y2), and (x3, y3)
So,
The dimensions of the translated figure are: (x1 + 5, y1 – 1), (x2 + 5, y2 – 1), and (x3 + 5, y3 – 1)
So,
The dimensions of the translated triangle (A’B’C’) are:
A (3, 2), B (2, -5), and C (0, -2)
We know that,
A translated figure has the same shape as the original shape i.e, the lengths and the angles of the translated figure are also the same as the original figure
So,
m∠A in Triangle ABC = m∠A in Triangle A’B’C’ = 30°
Hence,
The representation of the original and translated triangles are:

KEY CONCEPT

A translation, or slide, is a transformation that moves Preimage to every point of a figure the same distance and the same direction.

Do You Understand?
Question 1.
Essential Question How does a translation affect the properties of a two-dimensional figure?
Answer:
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.

Question 2.
Construct Arguments Triangle L’M’N’ is the image of triangle LMN after a translation. How are the side lengths and angle measures of the triangles related? Explain.
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
Hence, from the above,
We can conclude that
The side lengths and the angle measures of Triangle L’M’N’ is the same as The side lengths and the angle measures of Triangle LMN

Question 3.
Generalize Sanjay determined that one vertex of a figure was mapped to its image by translating the point 2 units left and 7 units down. What is the rule that maps the other vertices of the figure to their images?
Answer:
It is given that
Sanjay determined that one vertex of a figure was mapped to its image by translating the point 2 units left and 7 units down
Now,
Now,
Let the dimensions of any vertex of the given be (x, y)
So,
The dimensions of any vertex of the translated figure are: (x – 2, y + 7)
Hence, from the above,
We can conclude that
The rule that maps the other vertices of the figure to their images is: (x – 2, y – 7)

Do You Know How?
In 4-6, use the coordinate plane.

Question 4.
Which figure is a translation of Figure A? Explain
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
So,
When we observe the coordinate plane,
We can see that A and C have the same shape i.e., have the same length and the same angle
Hence, from the above,
We can conclude that
Figure C is a translation of Figure A

Question 5.
Graph the translation of Figure A 3 units right and 4 units up.
Answer:
From the given coordinate plane,
The dimensions of Figure A are:
(-2, -1), (-4, -1), (-4, -2), (-3, -2), (-3, -3), (-4, -3), (-4, -4), and (-2, -4)
So,
The rule that maps the vertices of the given figure to their images is: (x + 3, y + 4)
So,
The vertices of the image are:
(1, 3), (-1, 3), (-1, 2), (0, 2), (0, 1), (-1, 1), (-1, 0), and (1, 0)
Hence,
The representation of Figure A and its translated figure is:

Question 6.
Describe the translation needed to move Figure B to the same position as the image from Item 5.
Answer:
From the given coordinate plane,
The dimensions of figure B are:
(-2, 4), (-4, 4), (-4, 3), (-3, 3), (-3, 2), (-4, 2), (-4, 1), and (-2, 1)
Hence,
The translation needed to move figure B to the same position as the image from Item 5 is:
The X-axis: Translate 3 units right
The Y-axis: Translate 1 unit down

Practice & Problem Solving

Question 7.
Graph G’R’A’M’, the image of GRAM after a translation 11 units right and 2 units up.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The dimensions of GRAM are:
G (-8, 2), R (-8, 6), A (-4, 6), and M (-5, 2)
So,
After a translation of 11 units right and 2 units up,
The dimensions of G’R’A’M’ are:
G’ (-8 + 11, 2 + 2), R’ (-8 + 11, 6 + 2), A’ (-4 + 11, 6 + 2), and M’ (-5 + 11, 2 + 2)
So,
G’ (3, 4), R’ (3, 8), A’ (7, 8), and M’ (6, 4)
Hence,
The representation of GRAM and its image G’R’A’M’ is:

Question 8.
∆A’ B’ C’ is a translation of ∆ABC. Describe the translation.

Answer:
The given coordinate plane is:

From the coordinate plane,
The dimensions of ΔABC are:
A (1, -5), B (-2, -2), and C (0, 0)
Now,
The dimensions of ΔA’B’C’ are:
A’ (0, -1), B’ (-3, 2), and C’ (-1, 4)
So,
By observing the dimensions of ΔABC and ΔA’B’C’,
The translation between ΔABC and ΔA’B’C is:
The x-axis: Translate 1 unit left
The y-axis: Translate 4 units up

Question 9.
Which triangle is the image of ∆DEF after a translation? Describe the translation.

Answer:
The given figure is:

We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
So,
From the given figure,
We can observe that
ΔMNO has the same shape as ΔDEF
So,
The image of ΔDEF is: ΔMNO
Now,
The dimensions of ΔDEF are:
D (-8, -10), F (-6, -10), and E (-8, -4)
Now,
The dimensions of ΔMNO are:
M (2, -10), O (4, -10), and N (2, -4)
So,
By observing the dimensions of ΔDEF and ΔMNo,
The translation between ΔDEF and ΔMNO is:
The x-axis: Translate 10 units right
The y-axis: No Translation required

Question 10.
The vertices of figure QRST are translated 3 units left and 11 units down to form figure Q’R’S’T’. Explain the similarities and differences between the two figures.
Answer:
It is given that
The vertices of figure QRST are translated 3 units left and 11 units down to form figure Q’R’S’T’.
Hence,
The similarities between figure QRST and figure Q’R’S’T’ are:
a. The two figures have the same side length
b. The two figures have the same angle measure
c. The two figures have the same shape
The differences between figure QRST and figure Q’R’S’T’ are:
a. Different dimensions of the vertices of QRST and Q’R’S’T’
b. Different positions of figure QRST and figure Q’R’S’T’

Question 11.
Graph the image of the given triangle after a translation of 3 units right and 2 units up.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The dimensions of the given triangle are:
(-3, 4), (-5, -1), and (-8, 3)
Now,
After a translation 3 units right and 2 units up,
The dimensions of the given triangle are:
(-3 + 3, 4 + 2), (-5 + 3, -1 + 2), and (-8 + 3, 3 + 2)
(0, 6), (-2, 1), and (-5, 5)
Hence,
The representation of the given triangle and its image is:

Question 12.
Quadrilateral P’Q’R’ S’ is the image of quadrilateral PQRS after a translation.

Answer:
It is given that
Quadrilateral P’Q’R’ S’ is the image of quadrilateral PQRS after a translation.
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.

a. If the length of side PQ is about 2.8 units, what is the length of side P’ Q’?
Answer:
It is given that
The length of the side PQ is about 2.8 units
Hence, from the above,
We can conclude that the length of side P’Q’ is also about 2.8 units

b. If m∠R = 75°, what is m∠R’?
Answer:
It is given that
m∠R = 75°
Hence, from the above,
We can conclude that
m∠R’ = 75°

Question 13.
Higher-Order Thinking A farmer has a plot of land shaped like the figure in the graph. There is another identical plot of land 120 yards east and 100 yards north of the original plot.

a. Draw the image after the given translation.
Answer:
It is given that
A farmer has a plot of land shaped like the figure in the graph. There is another identical plot of land 120 yards east and 100 yards north of the original plot.
Now,
The given plot of land is:

Now,
From the given plot of land,
The dimensions of land are:
(0, 0), (0, 300), (300, 0), and (300, 300)
Now,
After the translation of 120 units right and 100 units up,
The dimensions of the identical plot are:
(0 + 120, 0 + 100), (0 + 120, 300 + 100), (300 + 120, 0 + 100), and (300 + 120, 300 + 100)
(120, 100), (120, 400), (420, 100), and (420, 400)
Hence,
The representation of the plot and its image after the translation is:

b. Find the combined area of the 2 plots in square yards.
Answer:
From part (a),
We can observe that the shape of the plot of land and its identical is like a square
We know that,
The square has the equal side lengths
Now,
From part (a),
We can observe that the side length of the plot of land and its identical is: 300 yards
Now,
We know that,
The area of square = Side²
So,
The area of the plot of land = 300²
= 90,000 yard²
The area of the identical plot of land = 300²
= 90,000 yard²
So,
The combined area of the 2 plots = 90,000 + 90,000
= 1,80,000 yard²
Hence, from the above,
We can conclude that the combined area of the 2 plots is: 1,80,000 yard²

Assessment Practice
Question 14.
What is true about the preimage of a figure and its image created by a translation? Select all that apply.
☐ Each point in the image moves the same distance and direction from the preimage.
☐ Each point in the image has the same x-coordinate as the corresponding point in the preimage.
☐ Each point in the image has the same y-coordinate as the corresponding point in the preimage.
☐ The preimage and the image are the same size.
☐ The preimage and the image are the same shape.
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
Hence,
The statements that are true about the preimage of a figure and its image created by a translation are:

Question 15.
The vertices of parallelogram QUAD are Q(-7, -7), U(-6, -4), A(-2,-4), and D(-3, -7).

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The dimensions of the parallelogram QUAD are:
Q (-7, -7), U (-6, -4), A (-2, -4), and D (-3, -7)
Now,
After the translation of 11 units right and 9 units up,
The dimensions of the parallelogram QUAD are:
Q’ (-7 + 11, -7 + 9), U’ (-6 + 11, -4 + 9), A’ (-2 + 11, -4 + 9), and D’ (-3 + 11, -7 + 9)
Q’ (4, 2), U’ (5, 5), A’ (9, 5), and D’ (8, 2)

PART A
Graph and label the image of QUAD after a translation 11 units right and 9 units up.
Answer:
The representation of the parallelogram QUAd and its image is:

PART B
If m∠U = 110°, what is m∠ U’?
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
Hence, from the above,
We can conclude that
m∠U’ = 110°

PART C
If the length of side UA is 4 units, what is the length of side U’ A’?
Answer:
We know that,
When you translate something in geometry, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, it remains a segment, and its length doesn’t change. Similarly, if you translate an angle, the measure of the angle doesn’t change.
Hence, from the above,
We can conclude that
The length of the side U’ A’ is: 4 units

### Lesson 6.2 Analyze Reflections

Solve & Discuss It!
Dale draws a triangle on grid paper and labels it in Figure 1. Then using his pencil as a guide, he draws another triangle directly on the opposite side of the pencil so that the vertical side is now one square to the right of the pencil instead of one square to the left of the pencil. He labels this triangle in Figure 2. How are the figures the same? How are they different?

I can… reflect two-dimensional figures.
Answer:
Dale draws a triangle on grid paper and labels it in Figure 1. Then using his pencil as a guide, he draws another triangle directly on the opposite side of the pencil so that the vertical side is now one square to the right of the pencil instead of one square to the left of the pencil. He labels this triangle in Figure 2.
Hence,
The representation of Figure 1 and Figure 2 is:

So,
From the above representation,
The similarities between Figure 1 and Figure 2 are:
a. Both the figures have the same side lengths
b. Both the figures have the same angle measures
c. Both the figures have the same size
d. Both the figures have the same distance
The differences between Figure 1 and Figure 2 are:
a. The positions of both figures are different
b. The orientations of both figures are different
c. The directions of both the figures are different

Look for Relationships
What do you notice about the size, shape, and direction of the two figures?
Answer:
The similarities between Figure 1 and Figure 2 are:
a. Both the figures have the same side lengths
b. Both the figures have the same angle measures
c. Both the figures have the same size
d. Both the figures have the same distance
The differences between Figure 1 and Figure 2 are:
a. The positions of both figures are different
b. The orientations of both figures are different
c. The directions of both the figures are different

Focus on math practices
Reasoning Dale draws a line in place of his pencil and folds the grid paper along the line. How do the triangles align when the grid paper is folded? Explain.
Answer:
It is given that
Dale draws a line in place of his pencil and folds the grid paper along the line
Now,
The representation of Figure 1 and Figure 2 are:

Now,
From the given figure,
We can observe that the two triangles will stack on each other when the grid paper is folded i.e.,
a. The vertical side of figure 2 is on top of the vertical side of figure 1
b. The base of figure 2 is on top of the base of figure 1
c. The hypotenuse of figure 2 is on top of the hypotenuse of figure 1

Essential Question
How does a reflection affect the properties of a two-dimensional figure?
Answer:
When the reflection takes place along the x-axis,
The values of x will remain constant and the values of y will have a sign change
Ex:
When (x, y) and (-x, y) reflects along the x-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)

Try It!

While updating the design, the architect accidentally clicked on the chair and reflected it across the centerline. Draw the new location of the chair on the plan.

Answer:
It is given that
While updating the design, the architect accidentally clicked on the chair and reflected it across the centerline.
Hence,
The representation of the new location of the chair is:

Convince Me!
How do the preimage and image compare after a reflection?
Answer:
A reflection is a transformation that turns a figure into its mirror image by flipping it over a line. The line of reflection is the line that a figure is reflected over. If a point is on the line of reflection then the image is the same as the preimage. Otherwise,
the image is not the same as the preimage. Images are always congruent to preimages

Try It!

Quadrilateral KLMN has vertices at K(2, 6), L(3, 8), M(5, 4), and N(3, 2). It is reflected across the y-axis, resulting in quadrilateral K’L’M’N’. What are the coordinates of point N’?
Answer:
It is given that
Quadrilateral KLMN has vertices at K(2, 6), L(3, 8), M(5, 4), and N(3, 2). It is reflected across the y-axis, resulting in quadrilateral K’L’M’N’
Now,
We know that,
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
So,
The reflection of N (3, 2) i.e., the coordinates of N’ is: (-3, 2)
Hence, from the above,
We can conclude that the coordinates of N’ are: (-3, 2)

Try It!

Polygon ABCDE is reflected across the line x = -2. Graph and label the image A’B’C’D’E’. Is m∠A= M∠A? Explain.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of polygon ABCDE are:
A (-4, 4), B (-3, 3), C (-3, 2), D (-5, 1), and E (-5, 3)
It is given that polygon ABCDE is reflected along x = -2 i.e., alone the x-axis
We know that,
When (x, y) and (-x, y) reflects along the y-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)
So,
The vertices of the reflection of polygon ABCDE i..e, A’B’C’D’E’ are:
A’ (-4, -4), B’ (-3, -3), C’ (-3, -2), D’ (-5, -1), and E’ (-5, -3)
We know that,
In reflection,
The side lengths and the angle measures in the image and the preimage are the same
Hence,
The representation of polygon ABCDE and its reflection polygon A’B’C’D’E’ is:

Hence,
m ∠A = M ∠A

KEY CONCEPT

A reflection, or flip, is a transformation that flips a figure across a line of reflection. The preimage and image are the same distance from the line of reflection but on opposite sides. They have the same size and shape but different orientations.

Do You Understand?
Question 1.
Essential Question How does a reflection affect the properties of a two-dimensional figure?
Answer:
When the reflection takes place along the x-axis,
The values of x will remain constant and the values of y will have a sign change
Ex:
When (x, y) and (-x, y) reflects along the x-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)

Question 2.
Generalize What do you notice about the corresponding coordinates of the preimage and image after a reflection across the x-axis?
Answer:
When the reflection takes place along the x-axis,
The values of x will remain constant and the values of y will have a sign change
Ex:
When (x, y) and (-x, y) reflects along the x-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)

Question 3.
Construct Arguments Jorge said the y-values would stay the same when you reflect a preimage across the line y = 5 since the y-values stay the same when you reflect a preimage across the y-axis. Is Jorge correct? Explain.
Answer:
It is given that
Jorge said the y-values would stay the same when you reflect a preimage across the line y = 5 since the y-values stay the same when you reflect a preimage across the y-axis.
Now,
We know that,
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
Hence, from the above,
We can conclude that Jorge is correct

Do You Know How?
Question 4.
Is AX’ Y’ Z’ a reflection of AXYZ across line g?

Answer:
The given figure is:

Now,
From the given figure,
We can observe that ΔXYZ is reflected across the line g i.e., y-axis
We know that,
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
So,
From the figure,
We can observe that
The negative x-coordinates of the vertices of ΔXYZ became the positive x -coordinates for Δ X’Y’Z’
Hence, from the above,
We can conclude that ΔX’Y’Z’ is the reflection of ΔXYZ across the line g

Use the coordinate grid below for 5 and 6.

Question 5.
Describe the reflection of figure EFGH.
Answer:
The given coordinate plane is:

From the given coordinate plane,
We can observe that
The reflection of figure EFGH is: Figure E’F’G’H’
Now,
We can observe that
The reflection of the figure EFGH takes place across the x-axis
So,
The x-coordinates of the vertices of the reflection of the figure EFGH will be constant and only y-coordinates will have a change in value
Hence,
The figure EFGH will flip i.e., top becomes down and vice-versa to form a reflection i.e., figure E’F’G’H’

Question 6.
Draw the image that would result from a reflection of figure E’F’G’H across the line x = -1.
Answer:
The given coordinate plane is:

Now,
From the reflection of EFGH i.e., figure E’F’G’H’,
We can observe that the vertices of the figure E’F’G’H’ are:
E’ (-8, 7), F’ (-5, 5), G’ (-4, 6), and H’ (-2, 6)
Now,
To form the image of the figure E’F’G’H’,
We need to reflect the figure E’F’G’H’ across the y-axis
When the reflection takes place along the x-axis,
The values of x will remain constant and the values of y will have a sign change
Ex:
When (x, y) and (-x, y) reflects along the x-axis,
The reflection of (x, y) will become (x, -y)
The reflection of (-x, y) will become (-x, -y)
So,
The vertices for the image of the figure E’F’G’H’ are:
E’ (-8, -7), F’ (-5, -5), G’ (-4, -6), and H’ (-2, -6)
Hence,
The representation of the figure E’F’G’H’ and its image is:

Practice & Problem Solving

Question 7.
Leveled Practice Trapezoid ABCD is shown. Draw the reflection of trapezoid ABCD across the y-axis.

Plot the points and draw trapezoid A’ B’C’D’.
Answer:
From the given trapezoid ABCD,
The vertices are:
A (2, 8), B (6, 8), C (8, 3), and D (1, 3)
It is given that
Draw the reflection of trapezoid ABCD across the y-axis i.e., y is constant
Now,
We know that,
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
So,
The vertices for the reflection of trapezoid ABCD are:
A’ (-2, 8), B (-6, 8), C’ (-8, 3), and D (-1, 3)
So,
The points of the preimage and image are:

Hence,
The representation of trapezoid A’B’C’D’ is:

Question 8.
Reasoning is triangle A’ B’C’a reflection of triangle ABC across the line? Explain.

Answer:
The given figure is:

Now,
From the given figure,
We can observe that the reflection of ΔABC takes place across the y-axis
Now,
When the reflection takes place along the y-axis,
The values of y will remain constant and the values of x will have a sign change
Ex:
When (x, y) and (x, -y) reflects along the y-axis,
The reflection of (x, y) will become (-x, y)
The reflection of (x, -y) will become (-x, -y)
So,
From the vertices of ΔABC,
The x-coordinates are negative and the y-coordinates are positive
So,
For the reflection of ΔABC along the y-axis,
The x-coordinates will have to become positive and the y-coordinates will be positive as in the vertices of ΔABC
Hence, from the above,
We can conclude that ΔA’B’C’ is the reflection of ΔABC across the given line

Question 9.
Your friend gives you the graph of quadrilateral ABCD and its image, quadrilateral A’B’C’D’. What reflection produces this image?

Answer:
It is given that
Your friend gives you the graph of quadrilateral ABCD and its image, quadrilateral A’B’C’D’
Now,
The quadrilateral ABCD and its image quadrilateral A’B’C’D’ is:

Now,
From the given figure,
We can observe that
The quadrilateral ABCD and its image will have the same y-coordinates but the x-coordinates are different
We know that,
When the reflection takes place across the y-axis,
The y-coordinates are the same for the image and the preimage
The x-coordiantes have sign changes with the same values for the image and the preimage
Hence, from the above,
We can conclude that the reflection across the y-axis produces the image quadrilateral A’B’C’D’

Question 10.
Construct Arguments Your friend incorrectly says that the reflection of ∆EFG to its image ∆E’ F’G’ is a reflection across the y-axis.

a. What is your friend’s mistake?
Answer:
The given coordinate plane is:

From the above figure,
We can observe that
For the vertices of the given figure,
The x-coordinates remain the same whereas the y-coordinates are different
We know that,
When the reflection takes place across the x-axis,
The x-coordinates of the preimage and the image are the same
The y-coordinates of the preimage and the image will have sign change with the same values
Hence, from the above,
We can conclude that the reflection of ΔEFG takes pace across the x-axis instead of across the y-axis

b. What is the correct description of the reflection?
Answer:
From the given figure,
We can observe that
For the vertices of the given figure,
The x-coordinates remain the same whereas the y-coordinates are different
We know that,
When the reflection takes place across the x-axis,
The x-coordinates of the preimage and the image are the same
The y-coordinates of the preimage and the image will have sign change with the same values
Hence, from the above,
We can conclude that the reflection of ΔEFG to its image ΔE’F’G’ takes pace across the x-axis

Question 11.
Make Sense and Persevere The vertices of ∆ABC are A(-5, 5), B(-2,5), and C(-2, 3). If ∆ ABC is reflected across the line y = -1, find the coordinates of the vertex C’
Answer:
It is given that
The vertices of ∆ABC are A(-5, 5), B(-2,5), and C(-2, 3) and ∆ ABC is reflected across the line y = -1
We know that,
When the reflection takes place across the y-axis,
The y-coordinates of the preimage and the image are the same
The x-coordinates of the preimage and the image will have sign change with the same values
So,
The reflection of the vertex C (-2, 3) is: (2, 3)
Hence, from the above,
We can conclude that the coordinates of the vertex C’ i.e., the reflection of the vertex C is: (2, 3)

Question 12.
Higher Order Thinking What reflection of the parallelogram ABCD results in image A’B’C’D?

Answer:
The given figure is:

From the given figure,
We can observe that the parallelogram ABCD and its reflection A’B’C’D’ are parallel to the x-axis
Now,
We can observe that,
For both the parallelogram ABCD and its reflection A’B’C’D’,
The x-coordinates are changing but the y-coordinates remain the same
Now,
We know that,
When the reflection takes place across the y-axis,
The y-coordinates of the preimage and the image are the same
The x-coordinates of the preimage and the image will have sign change with the same values
Hence, from the above,
We can conclude that the reflection across the y-axis of the parallelogram ABCD results in image A’B’C’D

Assessment Practice
Question 13.
∆JAR has vertices J(4, 5), A6, 4), and R(5,2). What graph shows ∆JAR and its image after a reflection across the line y = 1?
PART A

Answer:
It is given that
∆JAR has vertices J(4, 5), A6, 4), and R(5,2)
Now,
We know that,
The reflection across the y-axis is:
The y -coordinates are the same
The x-coordinates will have sign changes with the same value
So,
The reflection of the vertices of ΔJAR is:
J’ (-4, 5), A’ (-6, 4), and R (-5, 2)
Hence,f rom the above,
We can conclude that option D matches with the given vertices of the reflection i.e., ΔJ’A’R’

PART B
The measure of ∠A = 90°. What is m∠A’?
Answer:
We know that,
In Reflection,
The side lengths and the angle measures of the preimage and the image are the same
Hence, from the above,
We can conclude that
∠A = ∠A’ = 90°

### Lesson 6.3 Analyze Rotations

Explain It!
Maria boards a car at the bottom of the Ferris wheel. She rides to the top, where the car stops. Maria tells her friend that she completed $$\frac{1}{4}$$ of the turn before the car stopped.

I can… rotate a two-dimensional figure.

A. Do you agree with Maria? Explain.
Answer:
It is given that
Maria boards a car at the bottom of the Ferris wheel. She rides to the top, where the car stops. Maria tells her friend that she completed $$\frac{1}{4}$$ of the turn before the car stopped.
Now,
We know that,
From the given figure,
The Ferris wheel looks like a circle
We know that
The total angle measure of a circle (Ferris wheel) is: 360°
But,
It is given that
Maria tells her friend that she completed $$\frac{1}{4}$$ of the turn before the car stopped
But, from the figure,
We can observe that she completed the $$\frac{1}{2}$$ of the ride
Hence, from the above,
We can conclude that we can’t agree with Maria

B. How could you use angle measures to describe the change in position of the car?
Answer:
We know that,
From the given figure,
The Ferris wheel looks like a circle
We know that
The total angle measure of a circle (Ferris wheel) is: 360°
So,
The starting position of the car in terms of angle measure is given as 0°
The $$\frac{1}{2}$$ of the position of the car in terms of angle measure = $$\frac{360°}{2}$$
= 180°
Hence, from the above,
We can conclude that
In terms of angle measures,
The starting position of the car is: 0°
The $$\frac{1}{2}$$ of the car is: 180°

Focus on math practices
Construct Arguments How can you describe Maria’s change in position when her car returns to the position at which she began the ride?
Answer:
We know that,
The starting position of the car is: 0°
The $$\frac{1}{2}$$ of the car is: 180°
Now,
If Maria’s car returns to the position at which she began the ride, then
Maria has completed the Ferris wheel (or) Maria returned back from her previous $$\frac{1}{2}$$ of the position of the car
Hence,
In terms of angle measures,
The change in Maria’s position when her car returns to the position at which she began the ride is: 360° (or) -180°
Here,
-180° represents that Maria returned back from her previous $$\frac{1}{2}$$ of the position of the car

Essential Question
How does rotation affect the properties of a two-dimensional figure?
Answer:
When you rotate a two-dimensional figure, you are just moving it.
Ex:
If you rotate a rectangle, then it will remain a rectangle, just moved wherever you move it. This is similarly the same with an angle and aside length, the measure of the angle and the side length won’t change.

Try It!

The architect continues to rotate the umbrella in a counterclockwise direction until it is in its original position. What is the angle of this rotation?

Answer:
It is given that
The architect continues to rotate the umbrella in a counterclockwise direction until it is in its original position.
Now,
We know that,
If any figure rotates until it comes to its original position again, then that means the figure completed a full cycle (or) revolution and the angle of the full cycle is: 360°
Hence, from the above,
We can conclude that the architect continues to rotate the umbrella for 360° in a counterclockwise direction until it is in its original position

Convince Me!
How does an image compare to its preimage after a -45° rotation?
Answer:

Try It!

The coordinates of the vertices of quadrilateral HIJK are H(1,4), I(3, 2), J(-1,-4), and K(-3, -2). If quadrilateral HIJK is rotated 270° about the origin, what are the vertices of the resulting image, quadrilateral H’I’I’K?
Answer:
It is given that
The coordinates of the vertices of quadrilateral HIJK are H(1,4), I(3, 2), J(-1,-4), and K(-3, -2)
Now,.
We know that,
The change in the x and y-coordinates for the given angle of rotation is:

Now,
It is given that to rotate the quadrilateral HIJK 270° about the origin to form the quadrilateral H’I’J’K’
So,
The vertices of the quadrilateral H’I’J’K’ are:
H’ (4, -1), I’ (2, -3), J’ (-4, 1), K’ (-2, 3)
Hence,
The representation of the quadrilateral HIJK and its image H’I’J’K’ is:

Try It!

Describe the rotation that maps ∆FGH to ∆FG’H’.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔFGH are:
F (-5, -2), G (-1, -2), and H (-1, 1)
The vertices of ΔF’G’H’ are:
F’ (5, 2), G’ (1, 2), and H’ (1, -1)
Now,
To find the rotation that maps ΔFGH through ΔF’G’H’, the following steps are:
Step 1:
Draw rays from the origin through point G and point G’
Step 2:
Measure the angle formed by the rays
So,
The representation of ΔFGH and its image ΔF’G’H’ with its angle of rotation is:

Hence, from the above,
We can conclude that the angle of rotation that maps ΔFGH through ΔF’G’H’ is: 180°

KEY CONCEPT

A rotation is a transformation that turns a figure about a fixed point called the center of rotation. The angle of rotation is the number of degrees the figure is rotated. The x- and y-coordinates change in predictable ways when rotated.

Do You Understand?
Question 1.
Essential Question How does rotation affect the properties of a two-dimensional figure?
Answer:
When you rotate a two-dimensional figure, you are just moving it.
Ex:
If you rotate a rectangle, then it will remain a rectangle, just moved wherever you move it. This is similarly the same with an angle and aside length, the measure of the angle and the side length won’t change.

Question 2.
Reasoning If a preimage is rotated 360 degrees about the origin how can you describe its image?
Answer:
We know that,
To complete the full cycle i.e., to a preimage to return to its original position, the angle measure is: 360°
Hence, from the above,
We can conclude that
If a preimage is rotated 360° about the origin, then the image and the preimage are the same

Question 3.
Construct Arguments In Example 3, side AB is parallel to side DC. How are side A’ B’ and side D’ C’ related? Explain.

Answer:
In Example 3,
The given two-dimensional figure is a parallelogram
We know that,
In a parallelogram, the opposite sides have the same side lengths
So,
Now,
It is given that
In parallelogram ABCD,
AB is parallel to CD
So,
In parallelogram A’B’C’D’,
A’B’ is parallel to C’D’ since these two sides are just the images of the sides AB and CD
Hence, from the above,
We can conclude that A’B’ is parallel to C’D’

Do You Know How?
Question 4.
The coordinates of the vertices of rectangle ABCD are A(3,-2), B(3, 2), C(-3, 2), and D(-3,-2).
a. Rectangle ABCD is rotated 90° about the origin. What are the coordinates of the vertices of rectangle A’B’C’D’?
Answer:
It is given that
The coordinates of the vertices of rectangle ABCD are A(3,-2), B(3, 2), C(-3, 2), and D(-3,-2).
Now,
We know that,
The change in the x and y-coordinates for the given angle of rotation is:

Now,
It is given that to rotate the rectangle ABCD 90° about the origin to form the rectangle A’B’C’D’
So,
The vertices of the rectangle A’B’C’D’ are:
A’ (2, 3), B’ (-2, 3), C’ (-2, -3), and D’ (2, -3)
Hence,
The representation of the rectangle ABCD and its image rectangle A’B’C’D’ is:

b. What are the measures of the angles of A’B’C’D’?
Answer:
From part (a),
We can observe that all the angles of the rectangle A’B’C’D’ are the same
Hence, from the above,
We can conclude that the angle measures of the rectangle A’B’C’D’ are:
∠A = 90°, ∠B = 90°, ∠C = 90°,and ∠D = 90°

Question 5.
Describe the counterclockwise rotation that maps ∆QRS to ∆Q’R’S.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔQRS are:
Q (6, 2), R (4, 9), and S (2, 7)
The vertices of ΔQ’R’S’ are:
Q’ (2, -6), R’ (9, -4), and S’ (7, -2)
Now,
To find the rotation that maps ΔQRS through ΔQ’R’S’, the following steps are:
Step 1:
Draw rays from the origin through point Q and point Q’
Step 2:
Measure the angle formed by the rays
So,
The representation of ΔQRS and its image ΔQ’R’S’ with its angle of rotation is:

Hence, from the above,
We can conclude that the angle of rotation that maps ΔQRS through ΔQ’R’S’ is: 90°

Practice & Problem Solving

Question 6.
What is the angle of rotation about the origin that maps △ΡQR to △Ρ’ Ο’ R’?

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔPQR are:
P (3, 3), Q (5, 3), and R (5, 7)
The vertices of ΔP’Q’R’ are:
P’ (-3, 3), Q’ (-3, 5), and S’ (-7, 5)
Now,
To find the rotation that maps ΔPQR through ΔP’Q’R’, the following steps are:
Step 1:
Draw rays from the origin through point P and point P’
Step 2:
Measure the angle formed by the rays
So,
The representation of ΔPQR and its image ΔP’Q’R” with its angle of rotation is:

Hence, from the above,
We can conclude that the angle of rotation that maps ΔPQR through ΔP’Q’R’ is: 90°

Question 7.
Is △X’ Y’Z’ a rotation of △XYZ? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔXYZ are:
X (-2, 3), Y (-5, 4), and Z (-2, 7)
The vertices of ΔX’Y’Z’ are:
X’ (3, 2), Y’ (4, 5), and Z’ (7, 2)
Now,
To find the angle of rotation,
Compare the x and y-coordinates of ΔXYZ and ΔX’Y’Z’
So,
(x, y) before rotation ——-> (y, -x) after rotation
We know that,

Hence, from the above,
We can conclude that ΔX’Y’Z’ is a rotation of ΔXYZ

Question 8.
△PQR is rotated 270° about the origin. Graph and label the coordinates of P’, Q’, and R’.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔPQR are:
P (2, 3), Q (4, 6), and R (2, 7)
It is given that
Rotate ΔPQR 270° at the origin
We know that,

So,
The vertices of ΔP’Q’R’ when the angle of rotation is 270° are:
P’ (3, -2), Q’ (6, -4), and R’ (7, -2)
Hence,
The representation of ΔPQR and its image ΔP’Q’R’ is:

Question 9.
Is △P’ Q’R’a 270° rotation of △PQR about the origin? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔPQR are:
P (3, 3), Q (7, 4), and R (3, 5)
The vertices of ΔP’Q’R’ are:
P’ (-3, 3), Q’ (-4, 7), and R’ (-5, 3)
Now,
To find the angle of rotation,
Compare the x and y-coordinates of ΔPQR and ΔP’Q’R’
So,
(x, y) before rotation ——-> (-y, x) after rotation
We know that,

Hence, from the above,
We can conclude that ΔP’Q’R’ is a rotation of ΔPQR

Question 10.
Reasoning Explain why any rotation can be described by an angle between 0° and 360°.
Answer:
If you rotate an object 360°, it’s like the object never moved because the object would still be in the same spot as if you didn’t move it.
Hence,
Any rotation can be described by an angle of 0° to 360°

Question 11.
Rotate rectangle KLMN 270° about the origin.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of rectangle KLMN are:
K (-3, 2), L (-5, 2), M (-5, 4), and N (-3, 4)
It is given that
Rotate rectangle KLMN 270° at the origin
We know that,

So,
The vertices of rectangle K’L’M’N’ when the angle of rotation is 270° are:
K’ (2, 3), L’ (2, 5), M’ (4, 5), and N’ (4, 3)
Hence,
The representation of rectangle KLMN and its image rectangle K’L’M’N’ is:

Question 12.
Higher-Order Thinking An architect is designing a new windmill with four sails. In her sketch, the sails’ center of rotation is the origin, (0, 0), and the tip of one of the sails, point Q, has coordinates (2, -3). She wants to make another sketch that shows the windmill after the sails have rotated 270° about their center of rotation. What would be the coordinates of?
Answer:
It is given that
An architect is designing a new windmill with four sails. In her sketch, the sails’ center of rotation is the origin, (0, 0), and the tip of one of the sails, point Q, has coordinates (2, -3). She wants to make another sketch that shows the windmill after the sails have rotated 270° about their center of rotation
So,
We have to rotate point Q 270° about the origin
We know that,

So,
When we rotate any point 270° about the origin,
(x, y) before rotation ——–> (y, -x)
So,
The representation of point Q after representation is: (-3, -2)
Hence,
The representation of point Q and its image Q’ is:

Assessment Practice
Question 13.
A rotation about the origin maps △TRI to △T’ R’I’.
PART A
Which graph shows an angle you could measure to find the angle of rotation about the origin?

Answer:
It is given that
A rotation about the origin maps △TRI to △T’ R’I’
Now,
We know that,
We can find the angle of rotation only when the vertex maps with its image but not with any other images
Hence, from the above,
We can conclude that option A matches the given situation

PART B
What is the angle of rotation about the origin?
A. 90°
B. 180°
C. 270°
D. 360°
Answer:
From part (a),
When we observe option A,
The vertices of ΔTRI are:
T (-3, 3), R (-5, 3), and I (-4, 5)
The vertices of ΔT’R’I’ are:
T’ (3, -3), R’ (5, -3), and I’ (4, -5)
We know that,

So,
When we compare the vertices of ΔTRI and ΔT’R’I’,
We can observe that
(x, y) before rotation ———–> (-x, -y) after rotation
Hence, from the above,
We can conclude that the angle of rotation about the origin is: 180°

### Lesson 6.4 Compose Transformations

Solve & Discuss It!
How can you map Figure A onto Figure B?

I can… describe and perform a sequence of transformations.
Answer:
The given figure is:

Now,
The steps to obtain Figure B from Figure A are:
Step 1:
Reflect Figure A across the x-axis
Step 2:
Reflect the figure that we obtained in Step 1 across the y-axis
Hence,
The representation of the mapping of Figure A and Figure B using the above steps is:

Focus on math practices
Look for Relationships Is there another transformation or sequence of transformations that will map Figure A to Figure B?
Answer:
Yes, there are another sequence of transformations that will map Figure A to figure B
Now,
The steps to obtain Figure B from Figure A are:
Step 1:
Reflect Figure A across the y-axis
Step 2:
Reflect the figure that we obtained in Step 1 across the x-axis
Hence,
The representation of the mapping of Figure A and Figure B using the above steps is:

Essential Question
How can you use a sequence of transformations to map a preimage to its image?
Answer:
Mathematical transformations involve changing an image in some prescribed manner. There are four main types of transformations They are:
A) Translation B) Rotation C) Reflection D) Dilation

Try It!

Ava decided to move the cabinet to the opposite wall. What sequence of transformations moves the cabinet to its new position?

Answer:
It is given that
Ava decided to move the cabinet to the opposite wall
So,
From the figure,
We can observe that
To move the cabinet wall to the opposite wall, the following sequences of transformations have to be followed:
Step 1:
Translate the cabinet 8 units down
Step 2:
Rotate the cabinet 360° counterclockwise
Hence, from the above,
We can conclude that
The new position of the cabinet is:

Convince Me!
Ava decides that she would like the chairs to be placed directly across from the couch. What is a sequence of transformations that she can use to move the chairs to their new positions?

Try It!

What is another sequence of transformations that maps △ABC onto △A” B” C”?

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (-5, 5), B (-3, 3), and C (-6, 1)
Now,
Another sequence of transformations that maps ΔABC onto ΔA”B” C” is:
Step 1:
Draw ΔABC and make its reflection across the y-axis and name it ΔA’B’C’
Step 2:
Translate ΔA’B’C’ 2 units to the right and 6 units down
Step 3:
Reflect the figure we obtained in step 2 across the y-axis
Hence, from the above,
We can conclude that
The representation of another sequence of transformations that maps ΔABC onto ΔA”B” C” is:

KEY CONCEPT
You can use a sequence of two or more transformations to map a preimage to its image.
You can map △ABC onto △Α” Β” C” by translation of 3 units right followed by a 90° clockwise rotation about the origin.

Do You Understand?
Question 1.
Essential Question How can you use a sequence of transformations to map a preimage to its image?
Answer:
Mathematical transformations involve changing an image in some prescribed manner. There are four main types of transformations They are:
A) Translation B) Rotation C) Reflection D) Dilation

Question 2.
Make Sense and Persevere A preimage is rotated 180° about the origin and then rotated 180° about the origin again. Compare the preimage and image.
Answer:
It is given that
A preimage is rotated 180° about the origin and then rotated 180° about the origin again
Now,
When the preimage is rotated 180° and again 180°
The image will be the same as the preimage
Hence, from the above,
We can conclude that
When a preimage is rotated 180° about the origin and then rotated 180° about the origin again, the preimage and the image will be the same

Question 3.
Reasoning A figure ABC, with vertices A(2, 1), B(7, 4), and C(2, 7), is rotated 90° clockwise about the origin, and then reflected across the x-axis. Describe another sequence that would result in the same image.
Answer:
It is given that
A figure ABC, with vertices A(2, 1), B(7, 4), and C(2, 7), is rotated 90° clockwise about the origin, and then reflected across the x-axis
So,
The steps for another sequence of transformations that would result in the same image as the given situation is:
Step 1:
Draw the given vertices of Triangle ABC
Step 2:
Rotate Triangle ABC 90° counterclockwise
Step 3:
Reflect the image that we obtained in step 2 across the x-axis so that we will get the same image as in the given situation
Hence,
The representation of another sequence of transformations is:

Do You Know How?
In 4-6, use the diagram below.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of Figure WXYZ are:
W (2, 4), X (5, 4), Y (5, 2), and Z (2, 2)
The vertices of Figure W’X’Y’Z’ are:
W’ (-4, -4), X’ (-4, -1), Y’ (-2, -1), and Z’ (-2, -4)

Question 4.
Describe a sequence of transformations that maps rectangle WXYZ onto rectangle W’X’Y’Z’.
Answer:
The steps that we have to follow to obtain the given sequence of transformations that maps rectangle WXYZ onto rectangle W’X’Y’Z’ are:
Step 1:
Draw rectangle WXYZ
Step 2:
Reflect rectangle WXYZ across the x-axis
Step 3:
Rotate the image we obtained in step 2 90° counterclockwise
Step 4:
Translate the image we obtained in step-3 6 units left so that we can obtain rectangle W’X’Y’Z’
Hence,
The representation of the sequence of transformations for the given situation is:

Question 5.
Describe another way that you could map rectangle WXYZ onto W’X’Y’Z’.
Answer:
The steps for another way of transformations that maps rectangle WXYZ onto W’X’Y’Z’ are:
Step 1:
Draw Rectangle WXYZ
Step 2:
Rotate rectangle WXYZ 90° counterclockwise
Step 3:
Rotate the image we obtained in step-2 180° counterclockwise
Step 4:
Translate the image we obtained in step 3 6 units down so that we can obtain rectangle W’X’Y’Z’
Hence,
The representation of another sequence of transformations is:

Question 6.
Draw the image of rectangle WXYZ after a reflection across the line y = 1 and a translation 1 unit right. Label the image W” X” Y” Z”.
Answer:
The vertices of rectangle WXYZ are:
W (2, 4), X (5, 4), Y (5, 2), and Z (2, 2)
After the reflection across the line y = 1,
The vertices of rectangle WXYZ are:
W’ (2, -5), X’ (5, -5), Y’ (5, -3), and Z’ (2, -3)
After the translation of 1 unit right,
The vertices of rectangle WXYZ are:
W” (3, -5), X” (6, -5), Y” (6, -3), and Z” (3, -3)
Hence,
The representation of rectangle WXYZ and its image W”X” Y”Z” after the above sequence of transformations is:

Practice & Problem Solving

Question 7.
Leveled Practice Describe a sequence of transformations that maps △QRS onto △TUV.

A translation ________ units left and ________ units down, followed by a _________ across the ________.
Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔQRS are:
Q (3, 4), R (7, 4), and S (6, 9)
The vertices of ΔTUV are:
T (0, 0), U (-4, 0), and V (-3, 5)
Now,
In order to obtain the vertices of ΔTUV, the following transformations we have to follow are:
Step 1:
Translate the vertices of ΔQRS 3 units left and 4 units down
Step 2:
Reflect the image we obtained in step 1 across the x-axis in order to get the vertices of ΔTUV
Hence, from the above,
We can conclude that
A translation of 3 units left and 4 units down, followed by a reflection across the x-axis

Question 8.
Model with Math
A family moves a table, shown as rectangle EFGH, by translating it 3 units left and 3 units down followed by a 90° rotation about the origin. Graph E’ F’G’H’ to show the new location of the table.

Answer:
It is given that
A family moves a table, shown as rectangle EFGH, by translating it 3 units left and 3 units down followed by a 90° rotation about the origin
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of rectangle EFGH are:
E (3, 3), F (8, 3), G (8, 7), and H (3, 7)
Now,
To obtain rectangle E’F’G’H’,
The following series of transformations are:
Step 1:
By translating 3 units left and 3 units down,
The vertices of rectangle EFGH will become:
E (3 – 3, 3 – 3), F (8 – 3, 3 – 3), G (8 – 3, 7 – 3), and H (3 – 3, 7 – 3)
E (0, 0), F (5, 0), G (5, 4), and H (0, 4)
Step 2:
Rotate the vertices we obtain in step 1 90° counterclockwise about the origin
We know that,
When we rotate a point 90° counterclockwise about the origin,
(x, y) before rotation ——> (-y, x) after rotation
So,
The vertices of rectangle E’F’G’H’ are:
E’ (0, 0), F’ (0, 5), G’ (-4, 5), and H’ (-4, 0)
Hence,
The representation of the new location of the table is:

Question 9.
Describe a sequence of transformations that maps quadrilateral ABCD to quadrilateral HIJK.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of quadrilateral ABCD are:
A (3, 1), B (4, 1), C (4, 3), and D (3, 3)
The vertices of quadrilateral HIJK are:
H (-3, 0), I (-2, 0), J (-2, -2), and K (-3, -2)
Now,
The series of transformations that maps quadrilateral ABCD onto quadrilateral HIJK are:
Step 1:
Draw quadrilateral ABCD
Step 2:
Reflect quadrilateral ABCD across the x-axis
So,
The vertices of quadrilateral ABCD are:
A (3, -1), B (4, -1), C (4, -3), and D (3, -3)
Step 3:
Translate 6 units left and 1 unit up
So,
The vertices that we obtained in step 2 will become (The vertices of quadrilateral HIJK):
H (-3, 0), I (-2, 0), J (-2, -2), and K (-3, -2)
Hence,
The representation of the series of transformations that map quadrilateral ABCD onto quadrilateral HIJK are:

Question 10.
Map △QRS to △Q’R’ S’with a reflection across the y-axis followed by a translation 6 units down.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔQRS are:
Q (-3, 5), R (-2, 4), and S (-5, 3)
Now,
The series of transformations that maps ΔQRS to ΔQ’R’S’ as given above are:
Step 1:
Reflect the vertices of ΔQRS along the y-axis
So,
Q (3, 5), R (2, 4), and S (5, 3)
Step 2:
Translate the vertices that we obtain in step-1 6 units down so that we can obtain the vertices of ΔQ’R’S’
So,
Q’ (3, 5 – 6), R’ (2, 4 – 6), and S’ (5, 3 – 6)
Q’ (3, -1), R’ (2, -2), and S’ (5, -3)
Hence,
The representation of the series of transformations for the given situation is:

Question 11.
Higher-Order Thinking A student says that he was rearranging furniture at home and he used a glide reflection to move a table with legs from one side of the room to the other. Will a glide reflection result in a functioning table? Explain.
Answer:
It is given that
A student says that he was rearranging furniture at home and he used a glide reflection to move a table with legs from one side of the room to the other.
We know that,
A glide reflection is a  sequence of translation and reflection
Now,
From the given situation,
We can observe that the table is moving from one room to the other
So,
The “Translation” occurs
Now,
After moving he will rearrange the table in the room
So,
A “Reflection” may take place
Hence, from the above,
We can conclude that a glide reflection result in a functioning table

Assessment Practice
Question 12.
PART A
Which sequence of transformations maps rectangle ABCD onto rectangle A’ B’C’D?

A. translation 6 units down, reflection across the x-axis
B. reflection across the x-axis, translation 6 units right
C. reflection across the x-axis, translation 6 units left
D. translation 6 units left, reflection across the y-axis
Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of rectangle ABCD are:
A (2, -2), B (4, -2), C (4, -3), and D (2, -3)
The vertices of rectangle A’B’C’D’ are:
A’ (-4, 2), B’ (-2, 2), C’ (-2, 3), and D’ (-4, 3)
So,
The sequence of transformations that maps rectangle ABCD to rectangle A’B’C’D’ is:

Hence, from the above,
We can conclude that option C matches the given situation

PART B
Describe a sequence of transformations that maps A’B’C’ D’ onto ABCD.
Answer:
The sequence of transformations that maps A’B’C’D’ onto ABCD is:
Step 1:
Draw the rectangle ABCD
So,
The vertices of rectangle ABCD are:
A (2, -2), B (4, -2), C (4, -3), and D (2, -3)
Step 2:
Reflect rectangle ABCD across the x-axis
So,
The vertices of rectangle ABCD are:
A (2, 2), B (4, 2), C (4, 3), and D (2, 3)
Step 3:
Translate the vertices that we obtained in step 2 6 units left so that we can obtain rectangle A’B’C’D’
So,
A’ (2 – 6, 2), B’ (4 – 6, 2), C’ (4 – 6, 3), and D’ (2 – 6, 3)
So,
A’ (-4, 2), B’ (-2, 2), C’ (-2, 3), and D’ (-4, 3)

Question 13.
PART A
Which figures are the image of Figure A after a reflection across the x-axis and a translation of 4 units right?

A. Figure B
B. Figure C
C. Figure D
D. Figure E
Answer:
We know that,
A “Reflection” is called a “Flip” but the reflection does not affect the shape and length of the figure
So,
From the given figures,
When we observe Figures A and B,
We can say that Figure B is a reflection of A because the shape and length is the same
But,
when we observe the other figures,
We have different shapes and lengths between the preimage and the image
Hence, from the above,
We can conclude that
Figure B is the image of Figure A after a reflection across the x-axis and a translation of 4 units right

PART B
Which figure can be transformed into Figure G after a rotation 90° about the origin, then a translation 13 units right and 4 units down?
A. Figure B
B. Figure D
C. Figure E
D. Figure F
Answer:
From the given figures,
We have to obtain
The coordinates of Figure G are:
(6, -6), (9, -6), (9, -9), (7, -9), and (6, -5)
We know that,
When a point rotates 90° about the origin,
(x, y) before rotation ——-> (-y, x) after rotation
Now,
For the translation of 13 units right and 4 units down,
The vertices will be like: (x + 13, y – 4)
Hence, from the above,
We can conclude that option A matches the given situation

3-ACT MATH

3-Act Mathematical Modeling: Tricks of the Trade

ACT 1
Question 1.
After watching the video, what is the first question that comes to mind?
Answer:

Question 2.
Write the Main Question you will answer.

Answer:

Question 3.
Make a prediction to answer this Main Question.
Answer:

Question 4.
Construct Arguments Explain how you arrived at your prediction.
Answer:

ACT 2
Question 5.
What information in this situation would be helpful to know? How would you use that information?

Answer:

Question 6.
Use Appropriate Tools What tools can you use to solve the problem? Explain how you would use them strategically.
Answer:

Question 7.
Model with Math Represent the situation using mathematics. Use your representation to answer the Main Question.
Answer:

Question 8.
What is your answer to the Main Question? Does it differ from your prediction? Explain.

Answer:

ACT 3
Question 9.
Write the answer you saw in the video.

Answer:

Question 10.
Reasoning Does your answer match the answer in the video? If not, what are some reasons that would explain the difference?
Answer:

Question 11.
Make Sense and Persevere Would you change your model now that you know the answer? Explain.
Answer:

ACT 3 Extension
Reflect
Question 12.
Model with Math Explain how you used a mathematical model to represent the situation. How did the model help you answer the Main Question?

Answer:

Question 13.
Make Sense and Persevere When did you struggle most while solving the problem? How did you overcome that obstacle?
Answer:

SEQUEL
Question 14.
Be Precise Find another optical illusion online involving shapes that look different but are the same. Explain how you know the shapes are the same.

Answer:

### Lesson 6.5 Understand Congruent Figures

Solve & Discuss It!
Simone plays a video game in which she moves shapes into empty spaces. After several rounds, her next move must fit the blue piece into the dashed space. How can Simone move the blue piece to fit in the space?

I can… use a sequence of translations, reflections, and rotations to show that figures are congruent.
Answer:
It is given that
Simone plays a video game in which she moves shapes into empty spaces. After several rounds, her next move must fit the blue piece into the dashed space.
Now,
From the given figure,
We can observe that
The blue piece and the dashed  piece are the reflections of each other
Hence,
Simone move the blue piece to fit in the space by the translation followed by reflection

Reasoning
How can you use what you know about sequences of transformations to move the piece?
Answer:
From the given figure,
We can observe that
The blue piece should be moved to the place of the dashed space and it will be possible only due to the “Translation”
But,
We can observe that it does not fit into the dashed space.
So,
Reflect the blue piece that it can fit into the dashed piece
Hence, from the above,
We can conclude that the sequence of transformations we can use to move the transformation are:
a. Translation b. Reflection

Focus on math practices
Construct Arguments How do you know that the piece that fits into the space is the same as the original blue shape? Explain.
Answer:
We know that,
In the reflection,
a. The shape of the preimage and the image are the same
b. The length of the preimage and the image are the same
c. The orientation of the image and the preimage are different
Hence, from the above,
We can conclude that
Due to the properties of the reflection,
We know that the piece that fits into space is the same as the original shape

Essential Question
How does a sequence of translations, reflections, and rotations result in congruent figures?
Answer:
If we copy one figure on tracing paper and move the paper so the copy covers the other figure exactly, then that suggests they are congruent. We can prove that two figures are congruent by describing a sequence of translations, rotations, and reflections that move one figure onto the other so they match up exactly.

Try It!

How can you determine whether the orange and blue rectangles are congruent?

Answer:
The given coordinate plane is:

From the given coordinate plane,
We can observe that
The blue rectangle has a length of 6 units and a width of 5 units
The orange rectangle has a length of 5 units and a width of 6 units
Now,
We can say that
We can obtain the orange rectangle by rotating the blue rectangle
We know that,
Rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figure is not changed, then the figures are congruent
Hence, from the above,
We can conclude that the orange and blue rectangles are congruent

Convince Me!
Quadrilateral PQRS is congruent to quadrilateral P’ Q’R’S. What do you know about how these figures relate?
Answer:
It is given that
Quadrilateral PQRS is congruent to quadrilateral P’ Q’R’S
We know that,
When the two figures are congruent,
a. The shapes of the two figures are the same
b. The lengths of the two figures are the same
c. The angle measures of the two figures are the same
Hence,
In Quadrilateral PQRS and Quadrilateral P’Q’R’S’, the two figures are said to be congruent when
a. PQ = P’Q’ and RS = R’S’
b. ∠P = ∠P’, ∠Q = ∠Q’, ∠R = ∠R’, and ∠S = ∠S’

Try It!

Are the figures congruent? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of Figure 1 are:
(1, 5), (3, 7), (3, 5), and (2, 3)
The vertices of Figure 2 are:
(6, 3), (8, 3), (10, 2), and (8, 1)
Now,
Find out whether Translation, Reflection, and Rotation is possible between 2 figures or not
So,
The representation of Figure 1 and Figure 2 are:

Now,
From the given figures,
We can observe that none of the transformations is possible
Hence, from the above,
We can conclude that the two figures are not congruent

KEY CONCEPT

Two-dimensional figures are congruent if there is a sequence of translations, reflections, and rotations that maps one figure onto the other.

Do You Understand?
Question 1.
Essential Question How does a sequence of translations, reflections, and rotations result in congruent figures?
Answer:
If we copy one figure on tracing paper and move the paper so the copy covers the other figure exactly, then that suggests they are congruent. We can prove that two figures are congruent by describing a sequence of translations, rotations, and reflections that move one figure onto the other so they match up exactly.

Question 2.
Reasoning Does a sequence of transformations have to include a translation, a reflection, and a rotation to result in congruent figures? Explain.
Answer:
Rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figure is not changed, then the figures are congruent

Question 3.
Construct Arguments Is there a sequence of reflections, rotations, and translations that makes the preimage and image not only congruent, but identical in orientation? Explain.
Answer:
We know that,
In the reflection,
The orientation of the preimage and the image will differ
Hence, from the above,
We can conclude that
There is a sequence of reflections, rotations, and translations that makes the preimage and image only congruent but not identical in orientation

Do You Know How?
Question 4.
A rectangle with an area of 25 square centimeters is rotated and reflected in the coordinate plane. What will be the area of the resulting image? Explain.
Answer:
It is given that
A rectangle with an area of 25 square centimeters is rotated and reflected in the coordinate plane.
We know that,
In a sequence of transformations like Translation, Rotation, and Reflection,
The shapes and side lengths of the image and the preimage are the same
Since the side lengths are the same, the area will also be the same
Hence, from the above,
We can conclude that the area of the image will also be 25 square centimeters

In 5 and 6, use the coordinate grid below.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (1, 4), B (2, 2), and C (5, 2)
The vertices of ΔDEF are:
D (9, 9), E (8, 7), and F (5, 7)
The vertices of ΔGHI are:
G (6, 6), H (8, 5), and I (8, 1)

Question 5.
Is △ABC ≅ △DEF? Explain.
Answer:
To find whether ΔABC is congruent to ΔDEF or not,
Step 1:
Reflect Triangle ABC across the x-axis
Step 2:
Translate the image that we obtained in step-1 10 units right and 5 units up
So,
The representation of step 1 and step 2 is:

Hence, from the above,
We can observe that the vertices we obtained in step 2 are the same as ΔDEF
Hence,
ΔABC is congruent to ΔDEF

Question 6.
Is △ABC ≅ △GHI? Explain.
Answer:
The representation of △ABC and △GHI is:

From the above,
We can observe that △ABC and △GHI do not have the same size
We know that,
In order to be 2 figures congruent,
a. The sizes of the figures would be the same
b. The shapes of the figures would be the same
c. The side lengths of the figures should be the same
Hence, from the above,
We can conclude that
△ABC is not congruent to △GHI

Practice & Problem Solving

Question 7.
△Q’R’ S’ is the image of △QRS after a reflection across the y-axis and a translation 6 units down. Is the image the same size and shape as the preimage?

△QRS and △Q’R’S’ _________ the same size and shape.
Answer:
It is given that
△Q’R’ S’ is the image of △QRS after a reflection across the y-axis and a translation 6 units down
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔQRS are:
Q (-3, 5), R (-2, 4), and S (-5, 3)
The vertices of ΔQ’R’S’ are:
Q’ (3, -1), R’ (2, -2), and S’ (5, -3)
Now,
Step 1:
After a reflection of ΔQRS across the y-axis,
The vertices of ΔQRS will be:
Q (3, 5), R (2, 4), and S (5, 3)
Step 2:
After a translation of 6 units down,
The vertices that we obtained in Step 1 are:
Q’ (3, 5 – 6), R’ (2, 4 – 6), and S’ (5, 3 – 6)
Q’ (3, -1), R’ (2, -2), and S’ (5, -3)
So,
The vertices of ΔQ’R’S’ we obtained from the coordinate plane and the vertices of ΔQ’R’S’ we obtained after the sequence of transformations are the same
Hence, from the above,
We can conclude that △QRS and △Q’R’S’ have the same size and shape and the image (ΔQ’R’S’) is the same size and shape as the preimage (ΔQRS)

Question 8.
Is △DEF ≅ △D’ E’F’? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔDEF are:
D (5, 5), E (6, 3), and F (2, 4)
The vertices of ΔD’E’F’ are:
D’ (-2, -1), E’ (-3, 1), and F’ (1, 0)
So,
The representtaion of the sequence of transformations of ΔDEF to show it is congruent to ΔD’E’F’ is:

Hence, from the above,
We can conclude that ΔDEF is congruent to ΔD’E’F’

Question 9.
Construct Arguments Describe a way to show that quadrilateral ABCD is congruent to quadrilateral A’B’C’D’.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of quadrilateral ABCD are:
A (3, 5), B (5, 5), C (5, 4), and D (3, 4)
The vertices of quadriateral A’B’C’D’ are:
A’ (-3, 0), B’ (-5, 0), C’ (-5, -1), and D’ (-3, -1)
So,
The representtaion of the transformation of sequences that shows quadrilateral ABCd is congruent to quadrilateral A’B’C’D’ is:

Hence, from the above,
We can conclude that quadrilateral ABCD is congruent to quadrilateral A’B’C’D’

Question 10.
You are making two triangular flags for a project and need the flags to be the same shape and size. △XYZ and △X’Y’Z’ are the flags you have drawn. Are the flags the same shape and size? Explain.

Answer:
It is given that
You are making two triangular flags for a project and need the flags to be the same shape and size. △XYZ and △X’Y’Z’ are the flags you have drawn.
Now,
The given coordinate plane is:

From the given coordiate plane,
The vertices of ΔXYZ are:
X (5, 6), Y (6, 1), and Z (2, 2)
he vertices of ΔX’Y’Z’ are:
X’ (-2, 0), Y’ (-3, -5), and Z’ (1, -4)
So,
The representtaion of the sequence of transformation to find whether two flags have the same size and the same shape or not is:

Question 11.
Which two triangles are congruent? Describe the sequence of transformations that maps one figure onto the other.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (-3, 2), B (-2, 7), and C (-7, 4)
The vertices of ΔQRS are:
Q (4, 7), R (2, 2), and S (7, 2)
The vertices of ΔXYZ are:
X (-2, -7), Y (-2, -3), and Z (-7, -5)
The vertices of ΔDEF are:
D (7, -5), E (2, -2), and F (2, -7)
Now,
We know that,
The two figures are said to be congruent when
a. The 2 figures have the same size i.e., the same length and the same angle measure
b. The 2 figures have the same shape
So,
The figures that are congruent to each other are represented as:

So,
From the above,
We can say that ΔQRS and ΔDEf are congruent
Hence,
The sequence of Transformations to show ΔQRS and ΔDEF are congruent is:

Question 12.
Is △LMN ≅ △XYZ? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔLMN are:
L (7, 9), M (9, 5), and N (6, 5)
The vertices of ΔXYZ are:
X (2, 2), Y (5, 4), and Z (5, 1)
Now,
We know that,
The two figures are said to be congruent when
a. The 2 figures have the same size i.e., the same length and the same angle measure
b. The 2 figures have the same shape
So,
The sequence of transformations to find whether ΔLMN is congruent to ΔXYZ is:

Hence, from the above,
We can conclude that ΔLMN is not congruent to ΔXYZ since they don’t have the same side lengths

Question 13.
Higher-Order Thinking A student was asked to describe a sequence of transformations that maps △DEF onto △D’ E’F’, given that △DEF ≅ △D’ E’F’. She incorrectly said the sequence of transformations that maps △DEF onto △D’ E’F’ is a reflection across the x-axis, followed by a translation of 6 units right and 4 units up. What mistake did the student likely make?

Answer:
It is given that
A student was asked to describe a sequence of transformations that maps △DEF onto △D’ E’F’, given that △DEF ≅ △D’ E’F’. She incorrectly said the sequence of transformations that maps △DEF onto △D’ E’F’ is a reflection across the x-axis, followed by a translation of 6 units right and 4 units up
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔDEF are:
D (4, 5), E (5, 1), and F (1, 2)
The vertices of ΔD’E’F’ are:
D’ (-2, -1), E’ (-1, 3), and F’ (-5, 2)
Now,
The correct sequence of Transformations to show ΔDEF and ΔD’E’F’ are congruent is:

Hence, from the above,
We can conclude that the mistake done by a student is the interchange of the sequence of Transformations
Hence,
The correct sequence of Transformations is:
Translation of 6 units left and 4 units down followed by the reflection across the x-axis

Assessment Practice
Question 14.
PART A
How can you determine whether △DEF ≅ △D’ E’ F?

A. Determine whether a sequence of rotations maps △DEF onto △D’E’F’.
B. Determine whether a sequence of transformations maps △DEF onto △D’ E’F’.
C. Determine whether a sequence of translations maps △DEF onto △D’ E’F’.
D. Determine whether a sequence of reflections maps △DEF onto △D’ E’F’.
Answer:
We know that,
If we want to find whether the given figures are congruent or not, then
We have to determine whether a sequence of transformations maps ΔDEF onto ΔD’E’F’
Hence, from the above,
We can conclude that option A matches the given situation

PART B
Is △DEF ≅ △D’ E’ F? Explain.
Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔDEF are:
D (5, 5), E (6, 3), and F (2, 4)
The vertices of ΔD’E’F’ are:
D’ (-2, -1), E’ (-3, 1), and F’ (1, 0)
So,
The representation of the sequence of transformations to find out whether ΔDEf and ΔD’E’F’ are congruent or not is:

Hence, from the above,
We can conclude that ΔDEF is congruent to ΔD’E’F’

### Topic 6 MID-TOPIC CHECKPOINT

Question 1.
Vocabulary Describe three transformations where the image and preimage have the same size and shape. Lesson 6-1, Lesson 6-2, and Lesson 6-3
Answer:
We know that,
There are four types of transformations. They are:
a. Translation
b. Reflection
c. Rotation
d. Translation
Now,
Some transformations keep the pre-image and image congruent. Congruent means that they are the same size and shape or that they have the same measurements. They make not have the same orientation
Hence,
Three of the four transformations that preserve the size and shape of the pre-image are: Translation, Rotation, and Reflections

For 2-6, use the figures below.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of the quadrilateral MNPQ are:
M (1, 2), N (2, 4), P (4, 5), and Q (3, 3)
The vertices of the quadrilateral RSTU are:
R (1, -2), S (2, -4), T (4, -5), and U (3, -3)

Question 2.
What are the coordinates of each point after quadrilateral RSTU is rotated 90° about the origin? Lesson 6-3
Answer:
We know that,
When any point is rotated 90° about the origin,
(x, y) before rotation ——–> (-y, x) after rotation
Hence,
After 90° rotation about the origin,
The coordinates of each point of the quadrilateral RSTU will become:
R (2, 1), S (4, 2), T (5, 4), and U (3, 3)

Question 3.
What are the coordinates of each point after quadrilateral MNPQ is translated 2 units right and 5 units down? Lesson 6-1
Answer:
We know that,
When the translation occurs,
(x, y) before Translation ——-> (x + h, y + k) after Translation
Where,
h is the translation on the x-axis
k is the translation on the y-axis
We will take the positive value of h when the translation occurs on the right side of the x-axis
We will take the negative value of h when the translation occurs on the left side of the x-axis
We will take the positive value of k when the translation occurs on the top side of the y-axis
We will take the negative value of k when the translation occurs on the down side of the y-axis
Hence,
After the Translation of 2 units right and 5 units down,
The coordinates of each point of the quadrilateral MNPQ will become:
M (1 + 2, 2 – 5), N (2 + 2, 4 – 5), P (4 + 2, 5 – 5), and Q (3 + 2, 3 – 5)
M (3, -3), N (4, -1), P (6, 0), and Q (5, -2)

Question 4.
What are the coordinates of each point after quadrilateral MNPQ is reflected across the x-axis and then translated into 3 units left? Lessons 6-2 and 6-4
Answer:
We know that,
The vertices of the quadrilateral MNPQ are:
M (1, 2), N (2, 4), P (4, 5), and Q (3, 3)
So,
After the vertices of the quadrilateral, MNPQ reflected across the x-axis,
The vertices of the quadrilateral MNPQ will become:
M (1, -2), N (2, -4), P (4, -5), and Q (3, -3)
Now,
After the Translation of 3 units left,
The vertices of the quadrilateral MNPQ will become:
M (1 – 3, -2), N (2 – 3, -4), P (4 – 3, -5), and Q (3 – 3, -3)
M (-2, -2), N (-1, -4), P (1, -5), and Q (0, -3)
Hence, from the above,
We can conclude that the coordinates of each point after quadrilateral MNPQ is reflected across the x-axis and then translated into 3 units left are:
M (-2, -2), N (-1, -4), P (1, -5), and Q (0, -3)

Question 5.
Which series of transformations maps quadrilateral MNPQ onto quadrilateral RSTU? Lesson 6-4
A. reflection across the x-axis, translation 4 units down
B. reflection across the y-axis, translation 4 units down
C. rotation 180° about the origin, and then reflection across the x-axis
D. rotation 180° about the origin, and then reflection across the y-axis
Answer:
The representation of the series of transformations that maps the quadrilateral MNPQ onto the quadrilateral RSTU is:

Hence, from the above,
We can conclude that option A matches the given situation

Question 6.
Is quadrilateral MNPQ congruent to quadrilateral RSTU? Explain. Lesson 6-5
Answer:
We know that,
The 2 figures are said to be congruent only when:
a. The shapes of the 2 figures are the same
b. The sizes of the 2 figures are the same (Side lengths, and Angle measures)
Now,

From the above,
We can observe that the side lengths are not the same
Hence, from the above,
We can conclude that the quadrilateral MNPQ is not congruent to the quadrilateral RSTU

### Topic 6 MID-TOPIC PERFORMANCE TASK

A tessellation is a design in a plane that uses one or more congruent figures, with no overlaps and no gaps, to cover the entire plane. A tessellation of an equilateral triangle is shown.

PART A
Explain how the tessellation of an equilateral triangle is formed using reflections.
Answer:
We know that,
When you cut a shape out of paper, then flip it over, the flipped shape looks like a mirror image of the original shape. So a tessellation made with this technique is called a “Reflection tessellation”
Now,
See if the figure will fit together with no gaps. The answer is yes, the figures will tessellate because it is made up of two shapes that do tessellate

PART B
Explain how the tessellation of an equilateral triangle is formed using rotations.
Answer:
We know that
A “Rotational tessellation” is a pattern where the repeating shapes fit together by rotating 90 degrees
Now,
See if the figure will fit together with no gaps. The answer is yes, the figures will tessellate because it is made up of two shapes that do tessellate

PART C
Which of the regular polygon(s) below can be tessellated using a series of transformations?

Answer:
We know that,
Equilateral triangles, squares, and regular hexagons are the only regular polygons that will tessellate. Therefore, there are only three regular tessellations.
Hence, from the above,
We can conclude that square and pentagon can be tessellated by using a series of transformations

### Lesson 6.6 Describe Dilations

Solve & Discuss It!
A landscape architect designs a small splash pad represented by △ABC. Then she decides to make the splash pad larger as shown by △ADE. How are the splash pad designs alike? How are they different?

I can… dilate two-dimensional figures
Answer:
It is given that
A landscape architect designs a small splash pad represented by △ABC. Then she decides to make the splash pad larger as shown by △ADE.
Now,
The given landscape architect is:

Now,
From the given landscape architect,
We can observe that the difference in the splash designs
Hence, from the above landscape architect,
We can conclude that
The splash designs are alike in:
a. Shape b. Angle measures c. Orientation
The splash designs are different in:
a. Side lengths

Look for Relationships
How can you use what you know about scale drawings to compare and contrast the designs?
Answer:
A dilation is a transformation that results in an image with the same shape, angle measures, and orientation as the preimage, but different side lengths.
We know that,
When the scale factor is greater than 1, the dilation is a reduction.
When the scale factor is between 0 and 1, the dilation is an enlargement.

Focus on math practices
Reasoning Paul wants to make two square picnic tables. One table will have side lengths that are $$\frac{1}{2}$$ of the lengths of the second table. How do the tablets compare? Explain.
Answer:
It is given that
Paul wants to make two square picnic tables. One table will have side lengths that are $$\frac{1}{2}$$ of the lengths of the second table
So,
The size of the tables are in the ratio of $$\frac{1}{2}$$ : 1
So,
The size of the tables are in the ratio of 1:2
Hence, from the above,
We can conclude that the side lengths of one table are 2 times the side lengths of the second table

Essential Question
What is the relationship between a preimage and its image after a dilation?
Answer:
After dilation, the pre-image and image have the same shape but not the same size.
In terms of Sides:
In dilation, the sides of the pre-image and the corresponding sides of the image are proportional.

Try It!

F’G’ H’I’ is the image of FGHI after a dilation with center at the origin. What is the scale factor?

The ratio of a side length in FGHI to a corresponding side length in F’GH’I’is: $$\frac{}{}$$
The scale factor is __________.
Answer:
It is given that
F’G’ H’I’ is the image of FGHI after a dilation with center at the origin.
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of FGHI are:
F (1, 1), G (1, 2), H (2, 2), and I (2, 1)
The vertices of F’G’H’I’ are:
F’ (5, 5), G’ (5, 10), H’ (10, 10), and I’ (10, 5)
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
The ratio of a side length in FGHI to a corresponding side length in F’GH’I’is:
$$\frac{Side length of F’G’}{Side length of FG}$$ (or) $$\frac{Side length of H’I’}{Side length of HI}$$
= $$\frac{0 + 5}{0 + 1}$$ (or) $$\frac{0 + 5}{0 + 1}$$
= 5
So,
The scale factor is: 5
Hence, from the above,
We can conclude that
The ratio of a side length in FGHI to a corresponding side length in F’GH’I’is: $$\frac{5}{1}$$
The scale factor is 5.

Convince Me!
Quadrilateral WXYZ is the image of quadrilateral FGHI after a dilation with center at the origin and a scale factor of 3.5. What are the coordinates of the vertices of quadrilateral WXYZ?
Answer:
It is given that
Quadrilateral WXYZ is the image of quadrilateral FGHI after a dilation with center at the origin and a scale factor of 3.5
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of FGHI are:
F (1, 1), G (1, 2), H (2, 2), and I (2, 1)
It is given that
The scale factor is: 3.5
So,
The vertices of the quadrilateral WXYZ are:
W ( 1 × 3.5, 1 ×3.5), X (1 × 3.5, 2 × 3.5), Y (2 × 3.5, 2 × 3.5), and Z (2 × 3.5, 1 × 3.5)
So,
W (3.5, 3.5), X (3.5, 7), Y (7, 7), and Z (7, 3.5)
Hence, from the above,
We can conclude that
The coordinates of the vertices of quadrilateral WXYZ are:
W (3.5, 3.5), X (3.5, 7), Y (7, 7), and Z (7, 3.5)

Try It!

A dilation maps point L(3, 6) to its image L’ (2, 4). Complete the dilation of figure LMN and label the image L’M’N’. What is the scale factor? What is the length of side M’N’?

Answer:
It is given that
A dilation maps point L(3, 6) to its image L’ (2, 4). Complete the dilation of figure LMN and label the image L’M’N’
Now,
The given coordinate plane is:

From the coordinate plane,
The vertices of ΔLMN are:
L (3, 6), M (3, 3), and N (6, 3)
It is given that the image of L is: L’ (2, 4)
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
The scale factor = $$\frac{Side length of L’}{Side length of L}$$
= $$\frac{4 – 2}{6 – 3}$$
= $$\frac{2}{3}$$
So,
The scale factor is: $$\frac{2}{3}$$
Now,
The coordinates of M’ = The coordinates of M × $$\frac{2}{3}$$
= (3, 3) × $$\frac{2}{3}$$
= (3 × $$\frac{2}{3}$$, 3 × $$\frac{2}{3}$$)
= (2, 2)
The coordinates of N’ = The coordinates of N × $$\frac{2}{3}$$
= (6, 3) × $$\frac{2}{3}$$
= (6 × $$\frac{2}{3}$$, 3 × $$\frac{2}{3}$$)
= (4, 2)
Now,
Compare M’ and N’ with (x 1, y 1), (x 2, y 2 )
We know that,
The distance between 2 points = √(x 2 – x 1 ) 2 + (y 2 – y 1 ) 2
So,
The side length of M’N’ = √(4 – 2 ) 2 + (2 – 2 )
= 2 units
Hence, from the above,
We can conclude that
The scale factor is: $$\frac{2}{3}$$
The side length of M’N’ is: 2 units

KEY CONCEPT

A dilation is a transformation that results in an image with the same shape, angle measures, and orientation as the preimage, but different side lengths.
When the scale factor is greater than 1, the dilation is an enlargement.

When the scale factor is between 0 and 1, the dilation is a reduction.

Do You Understand?
Question 1.
Essential Question What is the relationship between a preimage and its image after a dilation?
Answer:
After dilation, the pre-image and image have the same shape but not the same size.
In terms of Sides:
In dilation, the sides of the pre-image and the corresponding sides of the image are proportional.

Question 2.
Generalize When will a dilation be a reduction? When will it be an enlargement?
Answer:
When the scale factor is greater than 1, the dilation is an enlargement.

When the scale factor is between 0 and 1, the dilation is a reduction.

Question 3.
Reasoning Flora draws a rectangle with points at (12, 12), (15, 12), (15,9), and (12, 9). She dilates the figure with center at the origin and a scale factor of $$\frac{3}{4}$$. what is the measure of each angle in the image? Explain.
Answer:
It is given that
Flora draws a rectangle with points at (12, 12), (15, 12), (15,9), and (12, 9). She dilates the figure with center at the origin and a scale factor of $$\frac{3}{4}$$
So,
The vertices for the image of the given rectangle = (x × $$\frac{3}{4}$$, y × $$\frac{3}{4}$$)
So,
The vertices for the image of the given rectangle wil be:
( 12 × $$\frac{3}{4}$$, 12 × $$\frac{3}{4}$$), (15 × $$\frac{3}{4}$$, 12 × $$\frac{3}{4}$$), (15 × $$\frac{3}{4}$$, 9 × $$\frac{3}{4}$$), and (12 × $$\frac{3}{4}$$, 9 × $$\frac{3}{4}$$)
= (9, 9), (11.25, 9), (11.25, 6.75), and (9, 6.75)
Now,
We know that,
In dilation,
a. The preimage and image are the same in shape, orientation, and angle measures
b. the preimage and the image are different in size and the side lengths
So,
The representation of the angle measures in the preimage and image of the given rectangle is:

Hence, from the above,
We can conclude that the measure of each angle in the image is: 90°

Do You Know How?
In 4-6, use the coordinate grid below.

Answer:
The given coordinate plane is:

From the coordinate plane,
The vertices of Figure 1 are:
(4, 4), (6, 8), and (8, 4)
The vertices of Figure 2 are:
(2, 2), (3, 4), and (4, 2)
The vertices of Figure 3 are:
(1, 1), (1.5, 2), and (2, 1)

Question 4.
Figure 3 is the image of Figure 1 after a dilation with a center at the origin. What is the scale factor? Explain.
Answer:
It is given that
Figure 3 is the image of Figure 1 after a dilation with a center at the origin
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = $$\frac{4}{1}$$ (or) $$\frac{6}{1.5}$$ (or) $$\frac{8}{2}$$
= 4
Hence, from the above,
We can conclude that the scale factor is “4” so that Figure 3 is the image of Figure 1

Question 5.
What are the coordinates of the image of Figure 2 after a dilation with center at the origin and a scale factor of 3?
Answer:
We know that,
The vertices of Figure 2 are:
(2, 2), (3, 4), and (4, 2)
So,
With a scale factor of 3,
The vertices of Figure 2 will become:
(2 × 3, 2 × 3), (3 × 3, 4 × 3), and (4 × 3, 2 × 3)
= (6, 6), (9, 12), and (12, 6)
Hence, from the above,
We can conclude that the coordinates of the image of Figure 2 after a dilation at the origin and a scale factor of 3 are:
(6, 6), (9, 12), and (12, 6)

Question 6.
Which figures represent a dilation with a scale factor of $$\frac{1}{2}$$?
Answer:
We know that,
The vertices of Figure 1 are:
(4, 4), (6, 8), and (8, 4)
The vertices of Figure 2 are:
(2, 2), (3, 4), and (4, 2)
The vertices of Figure 3 are:
(1, 1), (1.5, 2), and (2, 1)
Now,
If we consider Figure 2 as the preimage and Figure 1 as the image, then
Scale factor = $$\frac{2}{4}$$ (or) $$\frac{3}{6}$$ (or) $$\frac{4}{8}$$
= $$\frac{1}{2}$$
If we consider Figure 3 as the preimage and Figure 2 as the image, then
Scale factor = $$\frac{1}{2}$$ (or) $$\frac{1.5}{3}$$ (or) $$\frac{2}{4}$$
= $$\frac{1}{2}$$
Hence, from the above,
We can conclude that (Figure 2, Figure 1) and (Figure 3, Figure 2) represent a dilation with a scale factor of $$\frac{1}{2}$$

Practice & Problem Solving

Question 7.
Leveled Practice Draw the image of △DEF after a dilation with center (0, 0) and scale factor of 2.

Find the coordinates of each point in the original figure.
D(____, (____) E(____, (____) F(____, (____)
Multiply each coordinate by _______.
Find the coordinates of each point in the image:
D'(____, (____) E'(____, (____) F'(____, (____)
Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of ΔDEF are:
D (0, 0), E (2, 0), and F (0, 2)
Now,
With a scale factor of 2,
Multiply each coordinate with 2
The coordinates of ΔDEF will become:
D (0 × 2, 0 × 2), E (2 × 2, 0 × 2), and F (0 × 2, 2 × 2)
D (0, 0), E (4, 0), and F (0, 4)
So,
The coordinates of the points for the image of ΔDEF are:
D’ (0, 0), E’ (4, 0), and F’ (0, 4)
Hence,
The representation of ΔDEF and its image ΔD’E’F’ is:

Question 8.
Find the scale factor for the dilation shown.

Answer:
The given coordinate plane is:

Now,
From the given coordinate plane,
The vertices of Figure DEFG are:
D (0, 0), E (5, 0), F (5, 6), and G (0, 6)
The vertices of Figure D’E’F’G’ are:
D’ (0, 0), E’ (15, 0), G’ (15, 18), and H’ (0, 18)
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = $$\frac{15}{5}$$ (or) $$\frac{18}{6}$$
= 3
Hence, from the above,
We can conclude that
The scale factor for the given dilation is: 3

Question 9.
Critique Reasoning For the dilation with center (0, 0) shown on the graph, your friend says the scale factor is $$\frac{5}{2}$$. What is the correct scale factor? What mistake did your friend likely make?

Answer:
It is given that
For the dilation with center (0, 0) shown on the graph, your friend says the scale factor is $$\frac{5}{2}$$
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (0, 5), B (-5, 0), and C (5, -5)
The vertices of ΔA’B’C’ are:
A’ (0, 2), B’ (-2, 0), C’ (2, -2)
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = $$\frac{2}{5}$$ (or) $$\frac{-2}{-5}$$
= $$\frac{2}{5}$$
Hence, from the above,
We can conclude that
The correct scale factor is: $$\frac{2}{5}$$
The mistake done by your friend is:
The consideration of the scale factor as $$\frac{Length of the preimage}{length of the image}$$ instead of $$\frac{Length of the image}{Length of the preimage}$$

Question 10.
The smaller figure is the image of dilation of the larger figure. The origin is the center of dilation. Tell whether the dilation is an enlargement or a reduction. Then find the scale factor of the dilation.

Answer:
It is given that
The smaller figure is the image of dilation of the larger figure. The origin is the center of dilation
Now,
The given coordinate plane is:

From the given coordinate plane,
The coordinates of each point of the preimage are:
(3, 6), (15, 6), (15, 18), and (3, 18)
The coordinates of each point of the image are:
(1, 2), (5, 2), (5, 6), and (1, 6)
Now,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = $$\frac{1}{3}$$ (or) $$\frac{2}{6}$$ (or) $$\frac{5}{15}$$ (or) $$\frac{6}{18}$$
= $$\frac{1}{3}$$
Now,
Since the scale factor is less than 1, the dilation is a reduction
Hence, from the above,
We can conclude that
The given dilation is a reduction
The scale factor is: $$\frac{1}{3}$$

Question 11.
Higher-Order Thinking Q’R’S’T’ is the image of QRST after a dilation with center at the origin.

Answer:
It is given that
Q’R’S’T’ is the image of QRST after a dilation with center at the origin
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of the parallelogram QRST are:
Q (4, 4), R (16, 4), S (20, 16), and T (8, 16)
The vertices of the parallelogram Q’R’S’T’ are:
Q’ (1, 1), R’ (4, 1), S’ (5, 4), and T’ (2, 4)

a. Find the scale factor.
Answer:
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = $$\frac{4}{1}$$ (or) $$\frac{16}{4}$$ (or) $$\frac{20}{5}$$ (or) $$\frac{8}{2}$$
= 4
Hence, from the above,
We can conclude that the scale factor is: 4

b. Find the area of each parallelogram. What is the relationship between the areas?
Answer:
We know that,
Area of the parallelogram = Base × Height
Now,
the representation of the side lengths of the parallelogram QRST and its dilation Q’R’S’T’ is:

So,
The area of the parallelogram QRST = 12.6 × 12
= 151.2 square units
The area of the parallelogram Q’R’S’T’ = $$\frac{The area of the parallelogram QRST}{4}$$
= 37.8 square units
Hence, from the above,
We can conclude that
The area of the parallelogram QRST is: 151.2 sq. units
The area of the parallelogram Q’R’S’T’ is: 37.8 sq. units
The relationship between the two areas is:
The area of the parallelogram Q’R’S’T’ = $$\frac{The area of the parallelogram QRST}{4}$$

Assessment Practice
Question 12.
Triangle PQR is the image of △JKL after a dilation. Is the dilation an enlargement or a reduction? Explain.

A. An enlargement, because the image is larger than the original figure
B. An enlargement, because the image is smaller than the original figure
C. A reduction, because the image is smaller than the original figure
D. A reduction, because the image is larger than the original figure
Answer:
It is given that
Triangle PQR is the image of △JKL after a dilation
Now,
From the given figure,
We can observe that ΔPQR is larger than ΔJKL
So,
We can say that the dilation is an enlargement
Hence, from the above,
We can conclude that option A matches with the given situation

Question 13.
Rectangle QUAD has coordinates Q(0, 0), U(0, 3), A(6, 3), and D(6, 0). Q’U’ A’D’ is the image of QUAD after a dilation with center (0, 0) and a scale factor of 6. What are the coordinates of point D’? Explain.
Answer:
It is given that
Rectangle QUAD has coordinates Q(0, 0), 4(0, 3), A6, 3), and D(6, 0). Q’U’ A’D’ is the image of QUAD after a dilation with center (0, 0) and a scale factor of 6.
So,
With a scale factor of 6,
The vertices of the rectangle QUAD will become:
Q (0 × 6, 0 × 6), U (0 × 6, 3 × 6), A (6 × 6, 3 × 6) and D (6 × 6, 0 × 6)
= Q (0, 0), U (0, 18), A (36, 18), and D (6, 0)
So,
After a dilation,
The vertices of the rectangle QUAD will become the vertices of the rectangle Q’U’A’D’
So,
The vertices of the rectangle Q’U’A’D’ are:
Q’ (0, 0), U’ (0, 18), A’ (36, 18), and D’ (36, 0)
Hence, from the above,
We can conclude that the coordinates of point D’ are: (36, 0)

### Lesson 6.7 Understand Similar Figures

Solve & Discuss It!
Andrew draws the two figures shown on a coordinate plane. How are the two figures alike? How are they different? How do you know?

I can… use a sequence of transformations, including dilations, to show that figures are similar.
Answer:
It is given that
Andrew draws the two figures shown on a coordinate plane
Now,
The given figures are:

From the above,
We can observe that
ΔABC is the preimage
ΔA’B’C’ is the image
We can also observe that
The image is smaller than the image since the transformation we used when drawing the image is a “Dilation”
Hence,
Since the transformation we used is a “Dilation”,
The two figures are similar in terms of:
a. Shape b. Size c. Orientation d. Angle measures
The two figures are different in terms of:
a. Side lengths

Look for Relationships
Is △ABC a preimage of △A’B’C’? How do you know?
Answer:
The given figures are:

From the given figures,
We can clearly observe that the transformation called “Dilation” takes place
Since the dilation takes place,
We can conclude that ΔABC is a preimage of ΔA’B’C’

Focus on math practices
Reasoning How can you use the coordinates of the vertices of the triangles to identify the transformation that maps △ABC to △A’B’C’? Explain.
Answer:
The given figures are:

Now,
Fro the given figures,
We know that a transformation called a “Dilation” takes place
So,
We know that,
The “Scale factor” is the ratio of a length in the image to the corresponding length in the preimage
So,
Scale factor = [altex]\frac{The x-coordinate (or) y-coordinate that maps A’ or B’ or C’}{The x-coordinate (or) y-coordinate that maps A or B or C}[/latex}
Hence, from the above,
We can conclude that
By using the “Scale factor”, the coordinates of the vertices of the triangles can be used to identify the transformation that maps △ABC to △A’B’C’

Essential Question
How are similar figures related by a sequence of transformations?
Answer:
Two figures are similar if and only if one figure can be obtained from the other by a single transformation, or a sequence of transformations, including translations, reflections, rotations, and/or dilations. Similarity transformations preserve shape, but not necessarily size, making the figures “similar”

Try It!

Is △ΑΒC similar to △A’ B’ C?
The triangles _________ similar.

Answer:
The given figures are:

Now,
If we want to find the 2 triangles are similar or not, find the scale factor and find whether they have the same shape or not
If the scale factor for 2 triangles is the same, then those triangles are similar
Now,
We know that,
The scale factor is defined as the distance from this center point to a point on the preimage and also the distance from the center point to a point on the image
Now,
For AB and A’B’,
Scale factor = $$\frac{27}{12}$$
= $$\frac{9}{4}$$
For BC and B’C’,
Scale factor = $$\frac{27}{12}$$
= $$\frac{9}{4}$$
For AC and A’C’,
Scale factor = $$\frac{18}{8}$$
= $$\frac{9}{4}$$
Since the scale factor is equal for all the corresponding sides
We can conclude that ΔABC is similar to ΔA’B’C’

Convince Me!
What sequence of transformations shows that △ABC is similar to △A’ B’C’?
Answer:
The sequence of Transformations that shows ΔABC is similar to ΔA’B’C’ is:
a. Rotation b. Dilation c. Translation

Try It!

a. Graph the image of JKL after a reflection across the line x = 1 followed by dilation with a scale factor of $$\frac{1}{2}$$ and center of dilation point J’.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔJKL are:
J (-2, -4), K (-4, 0), and L (-2, 1)
The vertices of ΔPQR are:
P (-4, 2), Q (4, 6), and R (6, 2)
Now,
After the reflection of ΔJKL across the line x = 1,
J (-1, 4), K (-3, 0), and L (-1, -1)
So,
With a scale factor of $$\frac{1}{2}$$,
The vertices of ΔJKL will become:
J (-1, 4) × $$\frac{1}{2}$$, K (-3, 0) × $$\frac{1}{2}$$, and L (-1, -1) × $$\frac{1}{2}$$
So,
J’ (-0.5, 2), K’ (-1.5, 0), and L’ (-0.5, -0.5)
Hence,
The representation of the image of JKL after a reflection across the line x = 1 followed by dilation with a scale factor of $$\frac{1}{2}$$ and center of dilation point J’ is:

b. Is △JKL similar to △PQR?
Answer:
We know that,
The vertices of ΔJKL are:
J (-2, -4), K (-4, 0), and L (-2, 1)
The vertices of ΔPQR are:
P (-4, 2), Q (4, 6), and R (6, 2)
Now,
Step 1:
Rotate the vertices of ΔJKL 270° about the origin
We know that,
(x, y) before 270° rotation —-> (y, -x) after 270° rotation
So,
The vertices of ΔJKL will be:
J (-4, 2), K (0, 4), and L (1, 2)
Step 2:
Dilate the vertices of ΔJKL we obtained in step 1 by a scale factor of 2
So,
The vertices of ΔJKL will be:
J (-8, 4), K (0, 8), and L (2, 4)
Step 3:
Translate the vertices of ΔJKL we obtained in step 2 by 4 units right and 2 units down
So,
The vertices of ΔJKL will become:
P (-4, 2), Q (4, 6), and R (6, 2)
So,
The representation of the sequence of transformations is:

Hence, from the above,
We can conclude that ΔJKL is similar to ΔPQR

KEY CONCEPT

Two-dimensional figures are similar if there is a sequence of rotations, reflections, translations, and dilations that maps one figure onto the other.

Do You Understand?
Question 1.
Essential Question How are similar figures related by a sequence of transformations?
Answer:
Two figures are similar if and only if one figure can be obtained from the other by a single transformation, or a sequence of transformations, including translations, reflections, rotations, and/or dilations. Similarity transformations preserve the shape, but not necessarily size, making the figures “similar”

Question 2.
Be Precise How do the angle measures and side lengths compare in similar figures?
Answer:
Matching sides of two or more polygons are called corresponding sides, and matching angles are called corresponding angles. If two figures are similar, then the measures of the corresponding angles are equal and the ratios of the lengths of the corresponding sides are proportional.

Question 3.
Generalize Does a given translation, reflection, or rotation, followed by a given dilation, always map a figure to the same image as that same dilation followed by that same translation, reflection, or rotation? Explain.
Answer:
We know that,
The images formed by reflection, translation, and rotation have the same shape and size whereas dilation has the same shape but a different size
In the same way,
The image formed by dilation has the same shape but a different size whereas the image formed by reflection, translation, and rotation has the same shape and size as the image formed in dilation
Hence, from the above,
We can conclude that
The translation, reflection, or rotation, followed by a given dilation, always map a figure to the same image as that same dilation followed by that same translation, reflection, or rotation

Do You Know How?
Question 4.
Is trapezoid ABCD – trapezoid EFGH? Explain.

Answer:
The given figures are:

From the given figures,
We can observe that
The corresponding angles i.e., the opposite angles are equal
Now,
The ratio of the side lengths of the corresponding figures = $$\frac{32}{16}$$ (or) $$\frac{28}{14}$$ (or) $$\frac{20}{10}$$
= 2
So,
The ratio of the side lengths of the corresponding figures are the same
Now,
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal. This common ratio is called the scale factor
Hence, from the above,
We can conclude that
Trapezoid ABCD – Trapezoid EFGH

Use the graph for 5 and 6.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔABC are:
A (-4, -2), B (-1, -2), and C (-3, -3)
The vertices of ΔDEF are:
D (9, -5), E (6, -8), and F (1, -5)

Question 5.
△ABC is dilated by a factor of 2 with a center of dilation at point C, reflected across the x-axis, and translated 3 units up. Graph the resulting similar figure.
Answer:
We know that,
The vertices of ΔABC are:
A (-4, -2), B (-1, -2), and C (-3, -3)
So,
Step 1:
After dilation by a factor of 2:
Multiply all the coordinates of each point by 2
So,
The vertices of ΔABC are:
A (-8, -4), B (-2, -4), and C (-6, -6)
Now,
With a dilation at point C,
Add all the coordinates of each point with -3
So,
The vertices of ΔABC are:
A (-11, -7), B (-5, -7), and C (-9, -9)
Step 2:
When the vertices of ΔABC that we obtained in step 1 reflects across the x-axis,
The vertices of ΔABC will be:
A (-11, 7), B (-5, 7), and C (-9, 9)
Step 3:
When the vertices of ΔABC we obtained in step 2 translated by 3 units up,
Add all the y-coordinates of each point with 3
So,
The vertices of ΔABC will be:
A (-11, 10), B (-5, 10), and C (-9, 12)
Hence,
The representation of all the steps is:

Question 6.
Is △ABC similar to △DEF? Explain.
Answer:
We know that,
The vertices of ΔABC are:
A (-4, -2), B (-1, -2), and C (-3, -3)
The vertices of ΔDEF are:
D (9, -5), E (6, -8), and F (1, -5)
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal
So,
The representation of ΔABC and ΔDEF are:

From the above,
We can observe that the ratio of the corresponding side lengths are not in the same proportion
Hence, from the above,
We can conclude that ΔABC is not similar to ΔDEF

Practice & Problem Solving

Question 7.
Leveled Practice RSTU and VXYZ are quadrilaterals. Given RSTU ~ VXYZ, describe a sequence of transformations that maps RSTU to VXYZ.

• reflection across the ________
• translation ______ unit(s) left and _______ unit(s) up
• dilation with center (0,0) and a scale factor of ________
Answer:
The given coordinate plane is:

From the coordinate plane,
The vertices of the quadrilateral VXYZ are:
V (-2, -1), X (-3, 0), Y (-2, 2), and Z (-1, 0)
The vertices of the quadrilateral RSTU are:
R (3, 0), S (6, 3), T (3, 9), and U (0, 3)
Now,
Step 1:
Reflect the vertices of the quadrilateral VXYZ across the y-axis
So,
The vertices of the quadrilateral VXYZ will be:
V (2, -1), X (3, 0), Y (2, 2), and Z (1, 0)
Step 2:
Dilate the vertices of the quadrilateral VXYZ with a scale factor of 3
So,
Multiply all the coordinates of all the vertices of the quadrilateral VXYZ by 3
So,
The vertices of the quadrilateral VXYZ will be:
V (6, -3), X (9, 0), Y (6, 6), and Z (3, 0)
Step 3:
Translate the vertices of the quadrilateral VXYZ 3 units left and 3 units up
So,
The vertices of the quadrilateral VXYZ will become:
R (3, 0), S (6, 3), T (3, 9), and U (0, 3)
Hence,
The sequence of Transformations that map the quadrilateral VXYZ onto the quadrilateral RSTU is:

Question 8.
Reasoning Is △MNO similar to △PQO? Explain.

Answer:
The given coordinate plane is:

From the coordinate plane,
The vertices of ΔMNO are:
M (6, 6 ), N (0, 6), and O (0, 0)
The vertices of ΔPQO are:
P (-12, -9), Q (0, -9), and O (0, 0)
We know that,
Two figures are said to be similar if they are the same shape. In more mathematical language, two figures are similar if their corresponding angles are congruent, and the ratios of the lengths of their corresponding sides are equal
So,
The representation of ΔMNO and ΔPQO are:

From the above,
We can observe that
a. The ratio of the corresponding side lengths are not in the same proportion
b. The angle measures are not the same
Hence, from the above,
We can conclude that ΔMNO is not similar to ΔPQO

Question 9.
△PQR is dilated by a scale factor of 2 with a center of dilation (0, 0) and rotated 180° about the origin. Graph the resulting similar △XYZ.

Answer:
It is given that
△PQR is dilated by a scale factor of 2 with a center of dilation (0, 0) and rotated 180° about the origin.
Now,
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔPQR are:
P (2, 2), Q (4, 2), and R (3, 4)
Now,
Step 1:
Dilation of ΔPQR by a scale factor of 2:
Multiply all the coordinates of each point of ΔPQR by 2
So,
The vertices of ΔPQR will be:
P (4, 4), Q (8, 4), and R (6, 8)
Step 2:
Rotation of 180° counterclockwise about the origin:
We know that,
(x, y) before rotating 180° ——- > (-x, -y) after rotating 180°
So,
The vertices of ΔPQR that we obtained in step 1 will become:
X (-4, -4), Y (-8, -4), and Z (-6, -8)
Hence,
The graph of the resulting ΔXYZ is:

Question 10.
Describe a sequence of transformations that shows that quadrilateral RSTU is similar to quadrilateral VXYZ.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of the quadrilateral VXYZ are:
V (-5, -5), X (-5, -1), Y (-1, -1), and Z (-1, -5)
The vertices of the quadrilateral RSTU are:
R (-16, -14), S (-16, -6), T (-8, -6), and U (-8, -14)
Now,
Step 1:
Translate the quadrilateral VXYZ 3 units left and 2 units down
So,
The vertices of the quadrilateral VXYZ will be:
V (-8, -7), X (-8, -3), Y (-4, -3), and Z (-4, -7)
Step 2:
Dilate the vertices of the quadrilateral VXYZ we obtained in step 1 by a scale factor of 2
So,
The vertices of the quadrilateral VXYZ will be:
R (-16, -14), S (-16, -6), T (-8, -6), and U (-8, -14)
Hence,
The sequence of Transformations that shows the quadrilateral VXYZ is similar to the quadrilateral RSTU is:

Question 11.
Construct Arguments Is △PQR similar to △XYZ? Explain.

Answer:
The given coordinate plane is:

From the given coordinate plane,
The vertices of ΔXYZ are:
X (4, 4), Y (4, 8), and Z (8, 8)
The vertices of ΔPQR are:
P (2, -2), Q (4, -2), and R (4, -4)
Now,
Step 1:
Rotate ΔXYZ 270° about the origin
We know that,
(x, Y) before 270° rotation —— > (y, -x) after 270° rotation
So,
The vertices of ΔXYZ will become:
X ( 4, -4), Y (8, -4), and Z (8, -8)
Step 2:
Dilate ΔXYZ with a scale factor of $$\frac{1}{2}$$
So,
The vertices of ΔXYZ that we obtained in step 1 will become:
P (2, -2), Q (4, -2), and R (4, -4)
So,
After the sequence of Rotation and Dilation of ΔXYZ,
We obtained the vertices of ΔPQR
Hence, from the above,
We can conclude that ΔPQR is similar to ΔXYZ

Question 12.
Higher-Order Thinking Given △JKL ~ △XYZ, find two possible coordinates for missing point Y. For each coordinate chosen, describe a sequence of transformations, including a dilation, that will map △JKL to △XYZ.

Answer:
It is given that
Given △JKL ~ △XYZ
Now,
The given coordinate plane is:

From the coordinate plane,
The vertices of ΔJKL are:
J (2, 8), K (6, 2), and L (2, 2)
The vertices of ΔXYZ are:
X (-2, 5), Y (x, y), and Z (-2, 2)
Now,
Step 1:
Reflect ΔJKL across the y-axis
So,
The vertices of ΔJKL will be:
J (-2, 8), K (-6, 2), and L (-2, 2)
Step 2:
Dilate the vertices of ΔJKL that we obtained in step by a scale factor of $$\frac{1}{2}$$
So,
The vertices of ΔJKL will be:
(-1, 4), K (-3, 1), and L (-1, 1)
Step 3:
Translate the vertices of ΔJKL that we obtained in step 2 by 1 unit left and 1 unit up
So,
The vertices of ΔJKL will become:
X (-2, 5), Y (-4, 2), and Z (-2, 2)
Hence, from the above,
We can conclude that the coordinates for the missing point Y are: (-4, 2)

Assessment Practice
Question 13.
Rajesh is making pennants in preparation for a school soccer game. He wants the pennants to be similar triangles. Which of these triangles could he use for the pennants?

A. △QRS and △TVW
B. △QRS and △XYZ
C. △TVW and △JKL
D. △TVW and △XYZ
Answer:
It is given that
Rajesh is making pennants in preparation for a school soccer game. He wants the pennants to be similar triangles.
Now,
If we want to find which of the triangles are similar, find the scale factor and find whether they have the same shape or not
If the scale factor for 2 triangles is the same, then those triangles are similar
Now,
We know that,
The scale factor is defined as the distance from this center point to a point on the preimage and also the distance from the center point to a point on the image
Now,
For ΔQRS and ΔTVW,
Scale factor = $$\frac{3}{1}$$ (or) $$\frac{3}{1.5}$$
= 3 (or) 2
For ΔQRS and ΔXYZ,
Scale factor = $$\frac{3}{1.5}$$ (or) $$\frac{3}{1.5}$$
= 2
Hence from the above,
We can conclude that ΔQRS and ΔXYZ are similar triangles and he can use these triangles for the pennants

Question 14.
Determine whether the following pairs of triangles are similar or not similar.