Texas Go Math

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 12 Quiz Answer Key.

Texas Go Math Grade 8 Module 12 Quiz Answer Key

Texas Go Math Grade 8 Module 12 Ready to Go On? Answer Key

12.1-12.3 Properties of Translations, Reflections, and Rotations

Use the graph for Exercises 1-2.

Question 1.
Graph the image of triangle ABC after a translation of 6 units to the right and 4 units down. Label the vertices of the image A’, B’, and C’.
Answer:
The transformation is translation 6 units right and 4 unit down.
Apply the algebraic rule on the coordinates of vertices A, B and C to find vertices A’, B’ and C’.
(x, y) → (x+6, y – 4)
A(-4, 5) → A'(2, 1)
B(-4, 3) → B'(2, -1)
C(-1, 3) → C'(5, -1)

Math Grade 8 Answer Key Pdf Module 12 Question 2.
On the same coordinate grid, graph the image of triangle ABC after a reflection across the y-axis. Label the vertices of the images A”, B”, and C”.
Answer:
The transformation is reflection across the x-axis
Apply the algebraic rule on the coordinates of vertices A, B, and C to find vertices A”, B” and C”.
(x, y) → (x, y)
A(-4, 5) → A”(-4, -5)
B(-4, 3) → B”(-4, -3)
C(-1, 3) → C”(-1, -3)

Question 3.
Graph the image of trapezoid HIJK after it is rotated 180° about the origin. Label the vertices of the image. Find the vertices if the trapezoid HIJK is rotated 360°.
Answer:
The transformation is rotation 180° about the origin. Apply the algebraic rule to find points H’, I’. J’ and K’.
The route is (x, y) → (- x, y)
H(0, 4) → H'(0, -4)
I(0, 1) → I'(0, -1)
J(-2, 2) → J'(2, -2)
K(-2, 3) → K'(2, -3)

Question 4.
Vocabulary Translations, reflections, and rotations produce a figure that is ________________ to the original figure.
Answer:
Translations, reflections, and rotations produce a figure that is congruent to the original figure.

12.4 Algebraic Representations of Transformations

Module 12 Quiz Answers Go Math 8th Grade Pdf Question 5.
A triangle has vertices at (2, 3), (-2, 2), and (-3, 5). What are the coordinates of the vertices of the image after the translation (x, y) → (x + 4, y – 3)? Describe the transformation.
Answer:
After the translation of 4 units right and 3 units down, the coordinates of the vertices of the image are:
(x, y) → (x + 4, y – 3)
E(2, 3) → E'(6, 0)
F(-2, 2) → F'(2, -1)
G(-3, 5) → G'(1, 2)

Essential Question

Question 6.
How can you use transformations to solve real-world problems?
Answer:
Transformations are the simple functions in maths that are used when shifting the graphs to the left or right, upward or downward. the new graph is different from the pattern but it has some characteristics that are the same.
Example:
Suppose a country is going through a big economic problem. Let us assume that many people are leaving the country for new opportunities. There will be many people that must remain in the country. So, this is an opportunity for them to start a new business even when the crisis is terrible. If they succeed, when the crisis is over, the business will be prosperous. So they took advantage of a difficult problem transforming it into an opportunity.

Texas Go Math Grade 8 Module 12 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
What would be the orientation of figure L after a translation of 8 units to the right and 3 units up?

Answer:
(C)

Explanation:
After any translation, the orientation of the image of the figure stays the same.
So, the solution is C, because that image has the same orientation as the original figure.

Question 2.
A straw has a diameter of 0.6 cm and a length of 19.5 cm. What is the surface area of the straw to the nearest tenth? Use 3.14 for π.
(A) 5.5 cm²
(B) 11.7 cm²
(C) 36.7 cm²
(D) 37.3 cm²
Answer:(D) 37.3 cm²
Explanation:
Given,
Straw has a diameter of 0.6 cm and a length of 19.5 cm.
Diameter= 0.6 cm
Radius= 0.3 cm
Lenght = 19.5 cm
We know that,
Surface Area = 2πrh + 2πr²
Area = 2×3.14×0.3×19.5 + 2×3.14(0.3)²
Area = 36.738 + 0.5652
Area = 37.3 sq. cm
Thus the correct answer is option A.

Texas Go Math Grade 8 Module 12 Answer Key Question 3.
In what quadrant would the triangle be located after a rotation of 270° clockwise

(A) I
(B) II
(C) III
(D) IV
Answer:
(D) IV

Explanation:
After a rotation of 90° counterclockwise about the origin, the triangle will be in IV quadrant. So, the solution is D.

Question 4.
Which rational number is greater than -3\(\frac{1}{3}\) but less than –\(\frac{4}{5}\)?
(A) -0.4
(B) –\(\frac{9}{7}\)
(C) -0.19
(D) –\(\frac{22}{5}\)
Answer:
(B) –\(\frac{9}{7}\)

Explanation:
Determine the least common multiple of the denominators:
LCM(3, 5, 7, 10, 100) = 2100

Between -3\(\frac{1}{3}\) and –\(\frac{4}{5}\) is the number –\(\frac{9}{7}\).
So, the solution is B.

Question 5.
Which of the following is not true of a trapezoid that has been reflected across the x-axis?
(A) The new trapezoid is the same size as the original trapezoid.
(B) The new trapezoid is the same shape as the original trapezoid.
(C) The new trapezoid is in the same orientation as the original trapezoid.
(D) The x-coordinates of the new trapezoid are the same as the x-coordinates of the original trapezoid.
Answer:
(C) The new trapezoid is in the same orientation as the original trapezoid.

Explanation:
When you reflect a trapezoid across the x-axis, y-coordinates change and x-coordinates are the same as the x-coordinates of the original trapezoid. The new trapezoid has the same size and shape as the original trapezoid.
So, the answer is C, because trapezoids don’t have the same orientation.

Go Math Grade 8 Answer Key Module 12 Assessment Answer Key Question 6.
A triangle with coordinates (6, 4), (2, -1), and (-3, 5) is translated 4 units left and rotated 180° about the origin. What are the coordinates of its image?
(A) (2, 4), (-2, -1), (-7, 5)
(B) (4, 6), (-1, 2), (5, -3)
(C) (4, -2), (-1, 2), (5, 7)
(D) (-2, -4), (2, 1), (7, -5)
Answer:
(D) (-2, -4), (2, 1), (7, -5)

Explanation:
Apply the rule (x, y) → (x – 4, y) for the translation of 4 units left.
(6, 4) → (2, 4)
(2, -1) → (-2, -1)
(-3, 5) → (-7, 5)
Apply the rule (x, y) → (-x, -y) for the 180° rotation about the origin
(2, 4) → (-2, -4)
(-2, -1) → (2, 1)
(-7, 5) → (7, -5)
The coordinates of the image are (- 2, -4), (2, 1), (7, -5)
So, the correct answer is D.

Gridded Response

Question 7.
Solve the equation 3y + 17 = -2y + 25 for y.

Answer:
Given equation 3y + 17 = -2y + 25
3y + 2y + 17 = 25
5y + 17 = 25
5y = 25 – 17
5y = 8
y = 8/5
y = 1.6

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 11 Answer Key Equations and Inequalities with the Variable on Both Sides.

Texas Go Math Grade 8 Module 11 Answer Key Equations and Inequalities with the Variable on Both Sides

Texas Go Math Grade 8 Module 11 Are You Ready? Answer Key

Find the LCD.

Question 1.
8, 12 ________
Answer:
(List the multiples of each number)8 : 8, 16, 24, 32, …..
(Choose the least multiple the lists have in common) 12 : 12, 24, 36, 48. …
LCD(8, 12) = 24

Texas Go Math Grade 8 Pdf Module 11 Answer Key Question 2.
9, 12 ________
Answer:
List the multiples of each number
9 : 9, 18, 27, 36, …….
12 : 12, 24, 36, 48, 56, …
Choose the least multiple the lists have in common
LCD : 36

Question 3.
15, 20 _______
Answer:
List the multiples of each number
15 : 15, 30, 45, 60,…
20 : 20, 40, 60, 80,…
Choose the least multiple the lists have in common
LCD : 60

Question 4.
8, 10 ________
Answer:
List the multiples of each number
8: 8, 16, 24, 32, 40, 48,…
10: 10, 20, 30, 40, 50,…
Choose the least multiple the lists have in common
LCD : 40

Find the product.

Question 5.
0.683 × 100 ________
Answer:
Count the zeros in 100 : 2 zeros
0.683 × 100
Move the decimal point 2 places to the right
68.3

Texas Go Math Grade 8 Answer Key Pdf Module 11 Question 6.
9.15 × 1,000 ________
Answer:
Count the zeros in 1000 : 3 zeros
9.15 × 1000
Move the decimal point 3 places to the right
9150

Question 7.
0.005 × 100 ________
Answer:
Count the zeros in 100 : 2 zeros
0.005 × 100
Move the decimal point 2 places to the right
0.5

Question 8.
1,000 × 1,000 ________
Answer:
Count the zeros in 1000 : 3 zeros
1000 × 1000
Move the decimal point 3 places to the right
1,000,000

Write an algebraic equation for the sentence.

Question 9.
The difference between three times a number and 7 is 14. __________
Answer:
Given,
The difference between three times a number and 7 is 14
Represent the unknown with a variable Times means multiplication
The difference between three times x and 7 is 14
Difference means subtraction
3x – 7 is 14
Place the equal sign
3x – 7 = 14

Question 10.
The quotient of five times a number and 7 is no more than 10. __________
Answer:
The quotient of five times x and 7 is no more than 10.
Represent the unknown with a variable.
(5x) ÷ 7 ≤ 10 (Times means multiplication.)

Module 11 Test Answers Math Inequality Vocabulary Question 11.
14 less than 3 times a number is 5 more than half of the number. ___________
Answer:
Given
14 less than 3 times a number is 5 more than half of the number.
Represent the unknown with a variable Times means multiplication
14 less than 3 times x is 5 more than \(\frac{1}{2}\)x
less than means subtraction
3x – 14 is 5 more than \(\frac{1}{2}\)x
more than means addition
3x – 14 is 5 + \(\frac{1}{2}\)x
Place the equal sign
3x – 14 = 5 + \(\frac{1}{2}\)x

Texas Go Math Grade 8 Module 11 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the bubble map. You may put more than one word in each oval.

Understand Vocabulary

Complete the sentences using the review words.

Question 1.
A value of the variable that makes an equation true is a ___________.
Answer:
A value of the variable that makes an equation true is a “solution“.

Question 2.
The set of all whole numbers and their opposites are ____________.
Answer:
The set of all whole numbers and their opposites are “integers“.

Texas Go Math Grade 8 Answer Key Equations and Inequalities Answer Key Question 3.
An _______________ is an expression that contain s at least one variable.
Answer:
An “algebraic expression” is an expression that contains at least one variable.

Active Reading
Layered Book Before beginning the module, create a layered book to help you learn the concepts in this module. At the top of the first flap, write the title of the book, “Equations and Inequalities with Variables on Both Sides.”Then label each flap with one of the lesson titles ¡n this module. As you study each lesson, write important ideas, such as vocabulary and formulas, under the appropriate flap.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 11.4 Answer Key Inequalities with Rational Numbers.

Texas Go Math Grade 8 Lesson 11.4 Answer Key Inequalities with Rational Numbers

Example 1

Write an inequality to represent the relationship “Twice a number plus four is greater than two-thirds of the number”. Then solve your inequality.
STEP 1: Write an inequality.

Your Turn

Go Math Grade 8 Answer Key Lesson 11.4 Question 1.
Write an inequality to represent the relationship “Three-fourths of a number is greater than five less than the number”. Then solve your inequality.
Answer:
Consider the number = x
¾ × x ≥ 5 – x
3x/4 ≥ 5-x
3x ≥ 4(5 – x)
3x ≥ 20 – 4x
3x + 4x ≥ 20
7x ≥ 20
x ≥ 2.85
¾ × 2.85 ≥ 5 – 2.85
2.13 < 2.15
2.13 is less than 2.15.
Therefore the three-fourths of a number is not greater than five less than the number.

Example 2

Two water tanks hold 28.62 gallons and 31.2 gallons of water. The larger tank is leaking at a rate of 0.12 gallons per hour. The smaller tank is leaking at a rate of 0.08 gallons per hour. After how many hours will there be less water in the larger tank than in the smaller tank?
STEP 1: Write an inequality. Let h represent the number of hours.
amount in larger tank < amount in smaller tank
31.2 – 0.12h < 28.62 – 0.08h

STEP 2: Multiply both sides of the inequality by 10² = 100.
100(31.2) – 100(0.12h) < 100(28.62) – 100(0.08h)
3,120 – 12h < 2,862 – 8h

STEP 3: Use inverse operations to solve the inequality.

So, after 64.5 hours, there will be less water in the larger tank.

Reflect

Question 2.
In Step 2, why do you multiply both sides by 100 rather than by 10?
Answer:
100(31.2) – 100(0.12h) < 100(28.62) – 100(0.08h)
We can multiply by 100 because on the right side after decimal, there are 2 digits, so you can multiply by 100.
If there is 1 digit in the decimal then multiply by 10.

Your Turn

Question 3.
Bamboo Plant A is 1.2 meters tall and growing at a rate of 0.45 meters per day. Bamboo Plant B is 0.85 meters tall and growing 0.5 meters per day. After how many days will Plant B be taller than Plant A?
Answer:
Given,
Bamboo Plant A = 1.2 meters tall.
Plant A growing per day = 0.45 meters.
Bamboo Plant B = 0.85 meters tall.
Plant B growing per day = 0.5 meters
Let us consider number of days = d
1.2 + 0.45d < 0.85 + 0.5d
1.2 – 0.85 < 0.5d – 0.45d
0.35 < 0.005d
d < 0.35/0.005
d < 70 days
1.2 + 0.45(70) < 0.85 + 0.5(70)
32.7 < 35.85
Therefore after 70 days will Plant B be taller than Plant A.

Example 3

Write a real-world situation that can be modeled by the inequality
11 .25x – 20 ≥ 1 0.75x – 12.5.
Each side of the inequality consists of a variable term with a constant subtracted from it. The left side must exceed or be equal to the right side.

The inequality 11 .25x – 20 ≥ 1 0.75x – 12.5 can represent this situation: Ryan earns $11.25 per hour. His transit cost to and from work is $20 per week. Tony earns $10.75 per hour. His weekly transit cost is $12.50. After how many hours of work in a week do Ryan’s earnings minus transit cost exceed Tony’s earnings minus transit cost?

Your Turn

Lesson 11.4 Go Math Grade 8 Answer Key Pdf Question 4.
Write a real-world problem that can be modeled by the inequality
-10 – \(\frac{1}{4}\)x > 20 – \(\frac{1}{2}\)x.
Answer:
-10 – ¼ x > 20 – 1/2x
-10 – 0.25x > 20 – 0.5x

Texas Go Math Grade 8 Lesson 11.4 Guided Practice Answer Key

Write an inequality to represent each relationship. Then solve your inequality. (Example 1)

Question 1.
Three-fourths of a number is less than six plus the number.
Answer:
Consider the number = x
¾ x = 6 + x
3x/4 = 6 + x
3x = 4(6 + x)
3x = 24 + 4x
3x – 4x = 24
-x = 24
X = – 24
The number = -24
¾ (-24) < 6 +(-24)
-18 = -18
No, three fourth of a number is not less than six plus the number.

Question 2.
One-fifth of a number added to eleven is greater than three-fourths of the number.
Answer:
Consider the number = x
⅕ x + 11 ≥ ¾ + x
x/5 + 11 ≥ 3x/4
x/5 + 5 × 11/5 ≥ 3x/4.
x+5 ×11 /5 ≥ 3x/4
x+55/5 ≥ 3x/4
20 × x+55/5 ≥ 20 x 3x/4
4(x + 55)≥ 5 x 3x
4x + 220 ≥ 15x
4x≥ 15x – 220
-11x ≥ -220
X ≥ 20.
⅕ x 20 + 11 ≥ ¾ x 20
4 + 11 ≥ 15
15 ≥ 15
No, the one-fifth of a number added to eleven is not greater than three-fourths of the number.

Question 3.
Ian wants to promote his band on the Internet. Site A offers website hosting for $4.95 per month with a $49.95 startup fee. Site B offers website hosting for $9.95 per month with no startup fee. Write and solve an inequality to determine how many months Ian could have his website on Site B and still keep his total cost less than on Site A (Example 2)
Answer:
Let m represent the number of months
Site B ≤ Site A
Site B offers website hosting per month with no start-up = $9.95 per month.
Site A offers website hosting per month with start-up fee = $4.95 + $49.95.
$9.95m ≤ $4.95m + $49.95
Multiply both sides with 100
100($9.95m) ≤100($4.95m) + 100($49.95)
$995m ≤ $495m + $4995
$995m – $495m ≤ $4995
$500m ≤ $4995
m ≤ $4995/$500
m ≤ 9.99 months
9.95(9.99)≤ $4.95(9.99) + $49.95.
99.40 ≤ 99.40
Therefore 9.99 months Lan could have his website on site B.

Question 4.
Write a real-world problem that can be modeled by the inequality 10x > 5.5x + 31.5. (Example 3)
Answer:
Swetha and Priya invested the same amount in different investment plans. After some period Swetha earned 10 times of that she invested. Priya earned 31.5 more than 5.5 times of his money.
If Swetha earning is more than Priya earning, Then find the amount they invests first time.
Let x be the money.
Swetha earnings = 10x
Priya earnings = 5.5x + 31.5
Swetha earned more than the Priya.
10x > 5.5x + 31.5.

Essential Question Check-In

Question 5.
How can you use inequalities with rational number coefficients and constants to represent real-world problems?
Answer:
Assume that we have an equation ½ x + 20 ≥ 5/6 x + 12
Ramya gets ½ x per day and she gets a $20 per week and Riya gets 5/6 x per day and she get $12 per week. Find how many days Riya wants to work to reach the Ramya.

Texas Go Math Grade 8 Lesson 11.4 Independent Practice Answer Key

Question 6.
Rugs Emporium installs carpet for $80 plus $9.50 per square yard of carpet. Carpets-4-U charges $120 for installation and $7.50 per square yard of carpet. Write and solve an inequality to find the number of square yards of carpet for which Rugs Emporium charges more than Carpets-4-U.
Answer:
Given,
Rugs Emporium installs carpet for $80 plus $9.50 per square yard of carpet.
Carpets-4-U charges $120 for installation and $7.50 per square yard of carpet.
Consider number of square yards = n
$80 + $9.50n≥ $120 + $7.50n
Multiply both sides with 100
100($80) + 100($9.50n) ≥ 100($120) + 100($7.50n)
$8000 + $950n ≥ $12000 + $750n
$950n – $750n ≥ $12000 – $8000
$200n ≥ $4000
n ≥ $4000/$200
n ≥ 20
Therefore number of square yards = 20.

Math Grade 8 Answer Key Pdf Lesson 11.4 Question 7.
Billie needs to have her refrigerator repaired. She contacts two appliance repair companies and is given the rates shown in the table. Write and solve an inequality to find the number of hours for which Ace’s charge is less than or the same as Acme’s charge.

Answer:
Flat free charges of Acme Repair = 49
Charge per hour of Acme repair = 22.50
Flat free charges of Ace Repair = 0
Charge per hour of Ace repair = 34.75
The inequality equation is
Let us consider the number of hours = h
49 + 22.50h ≤ 0 + 34.75h
49 – 0 ≤ 34.75h – 22.50h
49 ≤ 12.25h
h ≤ 49/12.25
h ≤ 4 hours
49 + 22.50(4) ≤ 0 + 34.75(4).
139 ≤ 139
Therefore, Ace’s charge is the same as Acme’s charge.

Question 8.
Write an inequality with the solution x < 12. The inequality should have the variable on both sides, a decimal coefficient of the variable on the left side, and a decimal anywhere on the right side. One of the decimals should be written in tenths, the other in hundredths.
Answer:
The solution is
x < 12
2x – x < 1.75 + 10.25
2x – 10.25 < x + 1.75
1.5x -10.25 + 0.5x < 1.5 + x +0.25

Question 9.
Write an inequality with the solution x ≥ 4. The inequality should have the variable on both sides, a fractional coefficient of the variable on the left side, and a fraction anywhere on the right side.
Answer:
The inequality solution is
x ≥ 4
x ≥ 1.5 + 2.5
2x – x ≥ 3/2 + 5/2
2x – 5/2 ≥ x + 3/2
x – 5/2 ≥ x + 3/2

Question 10.
Multistep According to the Triangle Inequality, the length of the longest side of a triangle must be less than the sum of the lengths of the other two sides. The two shortest sides of a triangle measure \(\frac{3}{10}\) and x + \(\frac{1}{5}\), and the longest side measures \(\frac{3}{2}\)x.
a. Write an inequality that uses the Triangle Inequality to relate the three sides.
Answer:
The length of the longest side of a triangle must be less than the sum of the lengths of the other two sides.
The longest side of the triangle is \(\frac {3}{2} \) x.
The two shortest sides of a triangle is \(\frac {3}{10} \) and x + \(\frac {1}{5} \).
The inequality is
3/2 × x ≤ 3/10 + x + 1/5.

b. For what values of x is the Triangle Inequality true for this triangle?
Answer:
3/2 × x ≤ 3/10 + x + 1/5.
3x/2 ≤ 0.5 + x
3x ≤ 2(0.5 + x)
3x ≤ 1 + 2x
3x – 2x ≤ 1
x≤ 1
The value of x = 1 is the triangle inequality true for this triangle.

Question 11.
Jose and Maris work for different car dealerships. Jose earns a monthly salary of $3,500 plus a 6% commission on sales. Maris earns a monthly salary of $4,000 plus a 4% commission on sales. Above what value of sales are Jose’s monthly earnings more than those of Maris?
Answer:
Jose earns a monthly salary = $3,500 + 6% commission on sales.
Maris earns a monthly salary = $4,000 + 4% commission on sales.
Maris earns more = $4000 – $3500 = $500.
But Joes earns more = 6% – 4% = 2%.
$3500 + 0.06n ≥ $4000 + 0.04n
0.06n – 0.04n ≥ $4000 – $3500
0.02n ≥ $500
n =≥ 500/0.02
n ≥ 5000
$3500 + 0.06(25000) ≥ $4000 + 0.04(25000)
$5000 ≥ $5000
For the value of n = 25000 values of shares are Jose’s monthly earnings more than or equal to Maris.

Question 12.
Multistep Consider the figures shown.

a. Write an expression for the perimeter of each figure.
Answer:
Perimeter of rectangle = 2(l + w)
length = 0.8n + 0.2
breadth = n
Perimeter of a rectangle = 2(0.8n + 0.2 + n)
= 1.6n + 0.4 + n
Perimeter of a rectangle = 2.6n +0.4
Perimeter of triangle = a + b + c
a = 1.2n + 0.4
b = n + 0.6
c = n
= 1.2n + 0.4 + n +0.6 + n
Perimeter of a triangle = 3.2n + 1

b. For what values of n will the rectangle have a greater perimeter than the triangle?
Answer:
The perimeter of the rectangle ≥ perimeter of a triangle.
2.6n + 0.4 ≥ 3.2n + 1
2.6n – 3.2n ≥ 1 – 0.4
-0.6n≥ 0.6
n = -1
2.6(-1) + 0.4 ≥ 3.2(-1) + 1
-2.6 + 0.4 ≥ -3.2 + 1
-2.2 ≥ -2.2
If n = 1 the rectangle has a perimeter greater than or equal to the perimeter of the triangle.

H.O.T. Focus on Higher Order Thinking

Question 13.
Draw Conclusions The inequality y < \(\frac{2}{3}\)x + 5 represents all points on the coordinate plane that are below the line y = \(\frac{2}{3}\)x + 5.
a. If you substitute the x- and y-values of the point (6,11) into the inequality, do you get a true statement? What does this mean?
Answer:
y < \(\frac {2}{3} \)x + 5 equal to
y < ⅔ x + 5.
x = 6
y = 11
11 < ⅔ (11) + 5
11< 12.33
Yes, the given statement is true.

b. If y = 2\(\frac{1}{2}\), what are the possible values of x? What does this mean?
Answer:

Question 14.
Communicate Mathematical Ideas Describe what you can do when solving an inequality to ensure that the coefficient of the variable term will be positive.
Answer: If the coefficient of the variable term will be positive. Then multiply or divide the variable with the same coefficient.

Lesson 11.4 Go Math Answer Key 8th Grade Question 15.
Make a Conjecture For what values of x is the absolute value of x less than or equal to 1.5? (Hint: Substitute numbers into the inequality |x| ≤ 1.5 and see which numbers make the inequality true.) Choose three other positive numbers and answer the question using those numbers. Make a conjecture about when the absolute value of x is less than or equal to a given positive number a.
Answer:
|x| ≤ 1.5
The absolute values of x are 00.1.5, -1.5.
if x = 00 then
|00| ≤ 1.5
if x = 1.5 then
|1.5| ≤ 1.5
if x = -1.5 then
|-1.5| ≤ 1.5

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 12.4 Answer Key Algebraic Representations of Transformations.

Texas Go Math Grade 8 Lesson 12.4 Answer Key Algebraic Representations of Transformations

Your Turn

Question 1.
A rectangle has vertices at (0, -2), (0, 3), (3, -2), and (3, 3). What are the coordinates of the vertices of the image after the translation (x, y) → (x – 6, y – 3)? Describe the translation.
Answer:
Coordinates of the vertices of the image after the translation are:
A'(-6, -5);
B'(-6, 0);
C'(-3, -5);
D'(-3, 0).
The translation is 6 units to the left and 3 units down.

Go Math 8th Grade Answer Key Algebraic Representations of Transformations Question 2.
Triangle ABC has vertices A(-2, 6), 8(0, 5), and C(3, -1). Find the vertices of triangle A’B’C’ after a reflection across the x-axis.
Answer:
The vertices of triangle A’B’C’ after a reflection across the x-axis are:
A'(-2, -6);
B'(0, -5);
C'(3, 1).

Reflect

Question 3.
Communicate Mathematical Ideas How would you find the vertices of an image if a figure were rotated 270° clockwise? Explain.
Answer:
If a figure is rotated 270° clockwise, it’s the same like the figure is rotated 90° counterclockwise. Multiply each y- coordinate by 1, then switch the x- and y-coordinates: (x, y) will be (-y, x).

Your Turn

Question 4.
A triangle has vertices at j(-2, -4), K(1, 5), and L(2, 2). What are the coordinates of the vertices of the image after the triangle is rotated 90° counterclockwise?
Answer:
Step 1: Apply the rule to find the vertices of the image. Multiply each y-coordinate by 1, then switch x- and y coordinates. (x, y) will be (-y, x).
For J(-2, -4) the image is J'(4, -2);
For K(1, 5) the image is K'(-5, 1);
For L(2, 2) the image is L'(-2, 2).

Step 2: Graph the triangle and its image.

Texas Go Math Grade 8 Lesson 12.4 Guided Practice Answer Key

Question 1.
Triangle XYZ has vertices X(-3, -2), Y(-1, 0), and Z(1, -6). Find the vertices of triangle X’Y’Z’ after a translation of 6 units to the right. Then graph the triangle and its image. (Example 1)

Answer:
Step 1: Apply the rule to find the vertices of the image. Add 6 to the x-coordinate (x, y) will be (x + 6, y).
For X(-3, -2) the image is X'(3, -2);
For Y(-1, 0) the image is Y'(5, 0);
For Z(1, -6) the image is Z'(7, -6).

Step 2: Graph the triangle and its image.

Texas Go Math Grade 8 Pdf Algebraic Rule for Reflection Question 2.
Describe what happens to the x- and y-coordinates after a point is reflected across the x-axis. (Example 2)
Answer:
After the point is reflected across the x-axis, the x coordinate does not change and the y coordinate is multiplying by 1.
(x, y) will be (x, -y)

Question 3.
Use the rule (x, y) → (y, -x) to graph the image of the triangle at right. Then describe the transformation. (Example 3)
Answer:
Step 1 Apply the rule to find the vertices of the image.
A(-4, 2) → A'(2, 4)
B(2, 3) → B'(3, -2)
C(-3, 4) → C'(4, 3)
This means that transformation is rotation 90° clockwise.

Step 2: Graph the triangle and its image

Essential Question Check-In

Question 4.
How do the x- and y-coordinates change when a figure is translated right a units and down b units?

Answer:
The x-coordinates increase a units, and y-coordinates decrease b units.

Texas Go Math Grade 8 Lesson 12.4 Independent Practice Answer Key

Write an algebraic rule to describe each transformation. Then describe the transformation.

Question 5.

Answer:
The point (x, y) changes to (x – 2, y – 5). This is an algebraic rule.
The transformation is a translation of 2 units left and 5 units down.

Question 6.

Answer:
The point (x, y) changes to (-x, -y). This is an algebraic rule.
The transformation is a rotation for 180°.

Lesson 12.4 Go Math 8th Grade Pdf Algebraic Representation of Translation Question 7.
Triangle XYZ has vertices X(6, -2.3), Y(7.5, 5), and Z(8, 4). When translated, X’ has coordinates (2.8, -1.3). Write a rule to describe this transformation. Then find the coordinates of Y’ and Z’.
Answer:
X(6, -2.3) → X'(2.8, -1.3)
Find how changes coordinate x.
6 – a = 2.8
a = 6 – 2.8
a = 3.2
And how changes coordinate y.
-2.3 + b = – 1.3
b = 1.3 – (-2.3)
b = -1.2 + 2.3
b = 1
The rule is (x, y) → (x – a, y + b) where a = 3.2 and b = 1.
Apply the rule to find points Y’ and Z’.
Y(7.5, 5) → Y'(7.5 – 3.2, 5 + 1) So, Y’ has coordinates (4.3, 6).
Z(8, 4) → Z'(8 – 3.2, 4 + 1). So, Z’ has coordinates (4.8, 5).

Question 8.
Point L has coordinates (3, -5). The coordinates of point L’ after a reflection are (-3, -5). Without graphing, tell which axis point L was reflected across. Explain your answer.
Answer:
The reflection is across the y-axis because coordinate x of point L changed sign.

Question 9.
Use the rule (x, y) → (x – 2, y – 4) to graph the image of the rectangle. Then describe the transformation.

Answer:
Apply the rule (x, y) → (x – 2, y – 4) to find the vertices of the image A’B’C’D’.
A(1, 1) → A'(-1, -3)
B(4, 1) → B'(2, -3)
C(4, -2) → C'(2, -6)
D(1, -2) → D'(-1, -6)
The transformation is a translation 2 units left and 4 units down.
Graph the rectangle A’B’C’D’.

Question 10.
Parallelogram ABCD has vertices A(-2, -5\(\frac{1}{2}\)), 6(-4, -5\(\frac{1}{2}\)), C(-3, -2), and D(- 1, -2). Find the vertices of parallelogram A’B’C’D’ after a translation of 2\(\frac{1}{2}\) units down.
Answer:
The transformation is a translation of 2\(\frac{1}{2}\) down.
This means that y coordinates change.
Subtract 2\(\frac{1}{2}\) from the y coordinate.

Texas Go Math Grade 8 Algebraic Rule for Transformations Question 11.
Alexandra drew the logo shown on half-inch graph paper. Write a rule that describes the translation Alexandra used to create the shadow on the letter A.

Answer:
The translation is 1 unit right and 0.5 unit down.
So, add 1 to the x coordinate and subtract 0.5 from the y coordinate.
The rule is:
(x, y) → (x + 1, y – 0.5)
One unit is one-half inch.
So, the rule also can be to add 0.5 inch (1 unit) to the x coordinate and subtract 0.25 inch (0.5 unit) from the y coordinate.
(x, y) → (x + 0.5 in, y – 0.25 in)

Question 12.
Kite KLMN has vertices at K(1, 3), L(2, 4), M(3, 3), and N(2, 0). After the kite is rotated, K’ has coordinates (-3, 1). Describe the rotation, and include a rule in your description. Then find the coordinates of L’, M’, and N’.
Answer:
K(1, 3) → K'( 3, 1)
From the point K’ we can see that the rule is:
(x, y) → (y, x)
and this is a rule for the rotation 90° counterclockwise.
Apply it to all points.
L(2, 4) → L'(-4, 2)
M(3, 3) → M'(-3, 3)
N(2, 0) → N'(0, 2)

H.O.T. Focus on Higher Order Thinking

Question 13.
Make a Conjecture Graph the triangle with vertices (-3, 4), (3, 4), and (-5, -5). Use the transformation (y, x) to graph its image.

a. Which vertex of the image has the same coordinates as a vertex of the original figure? Explain why this is true.
Answer:

The point C’ has the same coordinates as a vertex C, because the transformation is:
(x, y) → (y, x) and vertex C has coordinates where x = y. If we replace them, we will get the same point C (-5, -5)

b. What is the equation of a line through the origin and this point?
Answer:
First find coefficient k, if:

The equation of a line through two points (in this case O and C’) is:
y – y₁ = k(x – x₁)
y – 0 = 1 (x – 0)
y = x

c. Describe the transformation of the triangle.
Answer:
This transformation is a reflection across the line y = x because the triangle ABC and A’B’C’ are the same distance from the line y = x, which is a line of reflection.

Go Math Answer Key 8th Grade Algebraic Rules for Transformations Question 14.
Critical Thinking Mitchell says the point (0, 0) does not change when reflected across the x- or y-axis or when rotated about the origin. Do you agree with Mitchell? Explain why or why not.
Answer:
Yes, applying the algebraic rules for the reflection and the rotation on the point O(0, 0), we will get the same point O.
For example reflection across the x-axis:
(x, y) → (x, -y)
(0, 0) → (0, 0)
This also applies to other transformations.

Question 15.
Analyze Relationships Triangle ABC with vertices A(-2, -2), B(-3, 1), and C(1, 1) is translated by (x, y) → (x – 1, y + 3). Then the image, triangle A’B’C’, is translated by (x, y) → (x + 4, y – 1), resulting in A”B”C”.
a. Find the coordinates for the vertices of triangle A”B”C”.
Answer:
First find the coordinates for the vertices B’ and C’.
Apply the rule (x, y) → (x – 1, y + 3).
A(- 2, 2) → A'(-3, 1)
B(-3, 1) → B'(-4, 4)
C(1, 1) → C'(0, 4)
Then apply the rule (x, y) → (x + 4, y – 1) on the coordinates of vertices A’, B’, and C’ to find the coordinates of vertices A”, B” and C”.
A'(- 3, 1) → A”(1, 0)
B'(-4, 4) → B”(0, 3)
C'(0, 4) → C”(4, 3)

b. Write a rule for one translation that maps triangle ABC to triangle A”B”C”.
Answer:
Subtract 1 from the coordinate x and then add 4.
Add 3 to the coordinate y, then subtract 1.
(x, y) → (x – 1 + 4, y + 3 – 1) (Simplify)
(x, y) → (x + 3, y + 2)

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 9 Quiz Answer Key.

Texas Go Math Grade 8 Module 9 Quiz Answer Key

Texas Go Math Grade 8 Module 9 Ready to Go On? Answer Key

9.1 Volume of Cylinders

Find the volume of each cylinder. Round your answers to the nearest tenth If necessary. Use 3.14 for π.

Question 1.

Answer:
Radius of base 6 ft
Height of cylinder = 8 ft
Volume of cylinder = πr²h
Volume = 3.14 × 6² × 8
Volume = 3.14 × 36 × 8
Volume = 904.32 ft³

Go Math Module 9 Answer Key Quiz Answer Question 2.
A can of juice has a radius of 4 inches and a height of 7 inches. What is the 8ft volume of the can?
Answer:
Radius if cyLindrical can = 4 in
Height of cylindrical can = 7 in
Volume of cylinder = πr²h
Volume = 3.14 × 4² × 7
Volume = 351.68 in³
Volume ≈ 351.7 in³

9.2 Volume of Cones

Find the volume of each cone. Round your answers to the nearest tenth if necessary. Use 3.14 for π.

Question 3.

Answer:
Radius of base of cone = 6 cm
Height of cone = 15 cm
Volume of cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × 6² × 15
Volume = 565.2 cm³

Question 4.

Answer:
Radius of base of cone = 12 in
Height of cone = 20 in
Volume of cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × 12² × 20
Volume = 3014.4 cm³

9.3 Volume of Spheres

Find the volume of each sphere. Round your answers to the nearest tenth if necessary. Use 3.14 for π.

Question 5.

Answer:
Radius of sphere = 3 ft
Volume of sphere = \(\frac{4}{3}\)πr³
Volume = \(\frac{4}{3}\) × 3.14 × 3³
Volume = 113.04 ft³
Volume ≈ 113 ft³

Module 9 Volume Module Quiz Answer Key Question 6.

Answer:
Diameter of a sphere = 13 cm
Radius of the sphere = \(\frac{13}{2}\) = 6.5 cm
Volume of the sphere = \(\frac{4}{3}\)πr³
Volume of the sphere = \(\frac{4}{3}\) × 3.14 × 6.5³
Volume of the sphere = 1149.7633 cm³
Volume ≈ 1149.8 cm³

Essential Question

Question 7.
What measurements do you need to know to find the volume of a cylinder? a cone? a sphere?
Answer:
Sphere: For finding the volume of the sphere, the radius is to be measured.
Volume of sphere = \(\frac{4}{3}\)πr³

Cylinder : To calculate the volume of cylinder, we need to find out the base radius of base of the cylinder along with the height of the cylinder.
Volume of cylinder = πr²h

Cone : To calculate the volume of cone, we need to measure the base radius of the base of the cone along with the height of the cone.
Volume of cone = \(\frac{1}{3}\)πr²h

Texas Go Math Grade 8 Module 9 Mixed Review Texas Test Prep Answer Key

Selected Response

Volume of Cylinders and Cones Mini Quiz Answers Question 1.
The bed of a long-bed pickup truck measures 4 feet by 8 feet. Which is the length of the longest thin metal bar that will lie flat in the bed?
(A) 11 ft 3 in.
(B) 10 ft. 0 in.
(C) 8 ft 11 in.
(D) 8 ft 9 in.
Answer:
(D) 8 ft 9 in.

Explanation:
The length of the longest thin metal bar that will lie flat in the bed is equal to the length of the bed’s hypotenuse. Let a = 4 and b = 8. Using the Pythagorean Theorem, we have:
a² + b² = c²
4² + 8² = c²
16 + 64 = c²
80 = c²
Rounding the length of the hypotenuse to the nearest tenth of a foot, we get:
c ≈ 8.9ft.
Therefore, the length of the longest thin metal bar that will lie flat in the bed is 8ft 9 in.

Question 2.
Using 3.14 for π, what is the volume of the cylinder below to the nearest tenth?

(A) 102 cubic yards
(B) 347.6 cubic yards
(C) 1,091.6 cubic yards
(D) 4,366.4 cubic yards
Answer:
(C) 1,091.6 cubic yards

Explanation:
Base diameter 11.4 yd
Base radius r = \(\frac{11.4}{2}\)
Base radius r = 5.7 yd
Height = 10.7 yd
Volume of a cylinder = πr²h
V = 3.14 × 5.7² × 10.7
V = 1091.59902 yd³
V ≈ 1091.6 yd³

Question 3.
Rhett made mini waffle cones for a birthday party. Each waffle cone was 3.5 inches high and had a radius of 0.8 inches. What is the volume of each cone to the nearest hundredth?
(A) 1.70 cubic inches
(B) 2.24 cubic inches
(C) 2.34 cubic inches
(D) 8.79 cubic inches
Answer:
(C) 2.34 cubic inches

Explanation:
Base radius r = 0.8 in
Height = 3.5 in
Volume of a cone = \(\frac{1}{3}\)πr²h
V = \(\frac{1}{3}\) × 3.14 × 0.8² × 3.5
V = 2.3445 in³
V ≈ 2.34 in³

Volume of Cylinders and Cones Mini Quiz Answer Key Question 4.
Using 3.14 for π, what is the volume to the nearest tenth of a cone that has a height of 17 meters and a base with a radius of 6 meters?
(A) 204 cubic meters
(B) 640.6 cubic meters
(C) 2,562.2 cubic meters
(D) 10,249 cubic meters
Answer:
(B) 640.6 cubic meters

Explanation:
Find the volume of a cone that has a height of 17 meters and a base with a radius of 6 meters.
Use the formula for the volume of a cone: \(\frac{1}{3}\)πr²h
Base radius r = 6 m
Height = 17 m
Volume of a cone = \(\frac{1}{3}\)πr²h
V = \(\frac{1}{3}\) × 3.14 × 6² × 17
V = 640.56 m³
V ≈ 640.6 m³

Question 5.
Using 3.14 for π, what is the volume of the sphere to the nearest tenth?

(A) 4180 cubic centimeters
(B) 5572.5 cubic centimeters
(C) 33,434.7 cubic centimeters
(D) 44,579.6 cubic centimeters
Answer:
(B) 5572.5 cubic centimeters

Explanation:
Diameter of the sphere = 22 cm
Radius r = \(\frac{22}{2}\) cm
Radius r = 11 cm
Volume of sphere = \(\frac{4}{3}\)πr³
Volume = \(\frac{4}{3}\) × 3.14 × 11³
Volume = 5572.4533 cm³
Volume ≈ 5572.5 cm³

Gridded Response

Volume of Spheres Mini Quiz Answer Key Module 9 Test Answers Question 6.
A diagram of a deodorant container is shown. It is made up of a cylinder and half of a sphere. What is the volume of the whole container to the nearest tenth cubic centimeter? Use 3.14 for π.

Answer:
a) Radius of a cylinder as well as the half sphere = 1.6 cm
Height = 6.2 cm
Volume of the half sphere = \(\frac{2}{3}\)πr³
Volume of the half sphere = \(\frac{2}{3}\) × 3.14 × 1.6³
Volume of the half sphere = 8.574 cm³

b) Volume of the cylinder = πr²h
Volume of the cylinder = 3.14 × 1.6² × 6.2
Volume of the cylinder = 49.838 cm³

c) Volume of the whole figure = Volume of the half sphere + Volume of the cylinder
Volume of the whole figure = 8.5 74 + 49.838
Volume of the whole figure = 58.4 12 cm³
Volume of the whole figure ≈ 58.4 cm³

a) Volume of the half sphere ≈ 8.574 cm³
b) Volume of the cylinder ≈ 49.838 cm³
c) Volume of the whole figure ≈ 58.4 cm³

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 11.3 Answer Key Inequalities with the Variable on Both Sides.

Texas Go Math Grade 8 Lesson 11.3 Answer Key Inequalities with the Variable on Both Sides

Texas Go Math Grade 8 Lesson 11.3 Explore Activity Answer Key

Modeling a Real-World Situation with an inequality

Many real-world situations can be modeled by inequalities. Some phrases that indicate an inequality are “less than 7” “greater than7” no more than,” and “at least.”

Super-Clean house cleaning company charges a fee of $384 to power wash a house plus $2 per linear foot. Power Bright charges $6 per linear foot, but no flat fee. Write an inequality that can be solved to find the number of linear feet a house must have to make the total cost charged by Super-Clean less than the cost charged by Power Bright.
A. Translate from words into symbols.
Let! represent the number of ______________ of the house.

Answer:
Less than 7 symbol = <7.
Greater than 7 symbol = >7
No more than symbol = >/ 7
At least 7 = > or = 7
Given that,
Super-Clean house cleaning company charges a fee to power wash a house = $384 + $2 per linear foot.
Power Bright charges $6 per linear foot, but no flat fee.
Let us consider the number of linear feet of houses = x
$384 + $2x < $6x

B. Write the inequality.
Answer:
Super-Clean house cleaning company charges a fee to power wash a house = $384 + $2 per linear foot.
Power Bright charges $6 per linear foot, but no flat fee.
Let us consider the number of linear feet of houses = x
$384 + $2x < $6x
$384 < $6x – 2x
$384 < $4x
$384/$4 < x
$96 < x
The total cost charged by Super-Clean is $96 less than the cost charged by Power Bright.

Reflect

Question 1.
How did you decide which inequality symbol to use?
Answer:
There are different types of the symbols They are less than/ greater than,
Equal to, not equal to, greater than or equal, less than or equal.
You will decide the inequality by comparing the numbers and the range vales that satisfy the conditions of the given variable.

Your Turn

Question 2.
The temperature in Amarillo is 74°F and is increasing at a rate of 2°F per hour. In Houston, it is 68F and increasing 4°F per hour. Write and solve an inequality to find how long it will take for the temperature in Houston to exceed the temperature in Amarillo.
Answer:
Given that,
The temperature in Amarillo = 74°F it is increasing at a rate of 2°F per hour.
The temperature in the Houston = 68 F it increases 4°F per hour.
Let us consider temperature time = x
68 + 4x > 74 + 2x
4x – 2x > 74 – 68
2x > 6
x > 3
The temperature in Houston takes 3°F to exceed the temperature in Amarillo.

Question 3.
Write a real-world situation that can be modeled by the inequality 46h > 84+ 25h.
Answer:
The inequality 46h > 84+ 25h. Represents Roopa has a salary per month is 84 and Karthik has a salary of 25h per month. And Geetha has a salary per month is 46h. Both the Roopa and Karthik have a salary has more than Geetha. Find how many months the Roopa and Karthik will work to reach the Geetha salary.

Texas Go Math Grade 8 Lesson 11.3 Guided Practice Answer Key

Question 1.
The Doily Record charges a fee of $525 plus $75 per week to run an ad. The Chronicle charges $150 per week. (Explore Activity and Example 1)
a. Write an inequality that can be solved to find the number of weeks an ad must run to make the total cost of running an ad in The Daily Record less than the cost in The Chronicle.
Answer:
The Doily Record charges to run an ad = $525 + $75 per week.
The Chronicle charges $150 per week.
Let us consider x = number of weeks.
$525 + $75x < $150x
From the above inequality equation, we can find the number of weeks an ad must run to make the total cost of running an ad in the daily record less than the cost in the chronicle.

b. Solve your inequality.
Answer:
Let us consider x = the number of weeks.
$525 + $75x < $150x
$525 < $150x – $75x
$525 < $75x
$525/$75 < x
7 < x.
Therefore 7 weeks an ad must run to make the total cost of running an ad in the daily record less than the cost in the chronicle.

Question 2.
The inventory report at Jacob’s Office Supplies shows that there are 150 packages of pencils and 120 packages of markers. If the store sells 7 packages of pencils each day and 5 packages of markers each day, write and solve an inequality to find in how many days the number of packages of pencils will be fewer than the number of packages of markers. (Example 1)
Answer:
Given that,
There are 150 packages of pencils and 120 packages of markers.
In the store, they sell 7 packages of pencils and 5 packages of markers.
Let us consider the number of days = x.
The inequality equation is
150 – 7x < 120 – 5x.
150 -120 < -5x + 7x
30 < 2x
30/2 < x
15 < x
Therefore, the number of days is 15.
15 packages of pencils will be fewer than the number of packages of markers.

Question 3.
Write a real-world situation that can be modeled by the inequality 834 + 14s > 978 – 10s. Then solve the inequality. (Example 2)
Answer:
The inequality 834 + 14s > 978 – 10s can represent that Roshan earns 834 per month and he gets 14 per week.
Diya earns 978 each month and she uses 10 per each week to her uses. After how many the Diya reached the Roshan.

Essential Question Check-In

Question 4.
How can you use inequalities to represent real-world problems?
Answer:
Using the inequalities is the best solution to determine the problems.
It is simple to determine the number of products that should be produced to maximize the profit.
For example, you have an inequality equation that is
20x > 30 + 15x.
Here the 20x represents the selling price of the apple
30 + 15x represents the charges of each apple.
The selling price is greater than the charges of the apple.
Find the number of apples must sell to make a profit.

Texas Go Math Grade 8 Lesson 11.3 Independent Practice Answer Key

Question 5.
The school band is selling pizzas for $7 each to raise money for new uniforms. The supplier charges $100 plus $4 per pizza.
a. Write an inequality that can be solved to find the number of pizzas the band members must sell to make a profit.
Answer:
Given that,
The school band selling pizzas = $7 per each.
The supplier charges = $100 + $4 per pizza.
The in-equality equation is
Consider the number of pizzas = x
$100 + $4x = $7x
The equation shows the number of pizzas the band members must sell to make a profit.

b. Solve your inequality.
Answer:
The in-equality equation is
Consider the number of pizzas = x
$100 + $4x = $7x
$100 = $7x – $4x
$100 = $3x
x = $100/$3
x = 33.3
Therefore the 33.3 number of pizzas the band members must sell to make a profit.

Go Math Grade 8 Lesson 11.3 Inequalities Answer Key Question 6.
Nadia was offered a job selling ads for a magazine. She must agree to one of the following payment options:
Choice A: $110 for each ad she sells
Choice B: A weekly salary of $150, plus $85 for each ad she sells
Write and solve an inequality to determine the number of ads Nadia would need to sell per week in order for Choice A to be the better choice.
Answer:
Given that choice A = $100 per each ad
Choice B = $150 per week + $85 per each ad.
Let us consider a number of ads = x
$100x < $150 + $85x
$100x – $85 < $150
$15x < $150
x < 150/15.
x < 10
Nadia would need 10 ads to sell per week in order for Choice A to be the better choice.

Question 7.
Represent Real-World Problems Write a real-world situation that can be modeled by the inequality 20x > 30 + 15x.
Answer:
Ravi has 30 biscuits and Rosy has 15x biscuits. Both Ravi and Rosy have more than 20x.

Write an inequality to represent each relationship. Then solve your inequality.

Question 8.
Ten less than five times a number is less than six times the number decreased by eight.
Answer:
The inequality equation is
Let us consider the number = x
10 -5x < 6x -8
10 – 8 < 6x – 5x
2 < x.

Question 9.
The sum of a number and twenty is greater than four times the number decreased by one.
Answer:
The inequality equation is
Let x be the number
$x + 20 > $4x – 1
20 + 1 > $4x – $x
21 > $3x
x > 21/3
x > $7.

Go Math Grade 8 Lesson 11.3 Answer Key Question 10.
Bob’s Bagels offers pre-paid cards and has the specials shown. Carol has a $50 card she uses to buy coffee and a bagel each week. Diego has a $60 card he uses to buy tea and a breakfast sandwich each week. Write and solve an inequality to find the number of weeks in which the balance on Carol’s card will be greater than the balance on Diego’s card.

Answer:
Given that,
Bob’s Bagels offers pre-paid cards.
Carol has a card = $50.
she uses the card to buy coffee and a bagel each week.
Diego has a card = $60.
He uses the card to buy tea and a breakfast sandwich each week.
The inequality equality equation is
Let us consider the number of weeks = x
$50 – $3x < $60 – $5x.
-3x + 5x < $60 – $50.
2x < 10
x < 5
Therefore the number of weeks is 5.

H.O.T. Focus on Higher-on-Order Thinking

Question 11.
Critique Reasoning Ahmad and Cameron solved the same inequality but got different answers. One of the solutions is incorrect. Find and correct the error.

Answer:
By solving the two equations the answer is x > -3.
x > -3 = -3 < x.
So, Ahmad’s solution is the correct answer.
Cameron’s solution is incorrect.
The error is x < -3. In the last step of the solution.

Go Math Lesson 11.3 8th Grade Inequalities Question 12.
Represent Real-World Problems Meena sells apple pies at the farmer’s market. She charges $12 for each pie. It costs her $5 to make each pie, and there is a $35 fee she must pay each week to have a booth at the market.
a. Write and solve an inequality to find the number of pies Meena must sell each week in order to make a profit.
Answer:
Given that,
Meena Charges for each apple pie in the farmers market = $12
The cost of each apple pie = $5.
She pay each week to have a booth in the market = $35
The inequality equation is
$12x – $5x > $35
$7x > $35.
x > 35/7
x > 5
Meena must sell more than 5 pies in each week.

b. What If? Suppose the fee to have a booth at the farmer’s market increases to $40. Will the new fee increase the number of pies Meena will have to sell in order to make a profit? Explain your reasoning.
Answer:
Meena Charges for each apple pie in the farmers market = $12
The cost of each apple pie = $5.
She pays each week to have a booth in the market = $40
The inequality equation is
$12x – $5x > $40
$7x > $40
x > 40/7
x > 5.71
After increasing the booth charges Meena must sell more than 5.71 pies each week.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 9.2 Answer Key Volume of Cones.

Texas Go Math Grade 8 Lesson 9.2 Answer Key Volume of Cones

Texas Go Math Grade 8 Lesson 9.2 Explore Activity Answer Key

Modeling the Volume of a Cone

A cone is a three-dimensional figure that has one vertex and one circular base.

To explore the volume of a cone, Sandi does an experiment with a cone and a cylinder that have congruent bases and heights. She fills the cone with popcorn kernels and then pours the kernels into the cylinder. She repeats this until the cylinder is full.

Sandi finds that it takes 3 cones to fill the volume of the cylinder.
STEP 1: What is the formula for the volume V of a cylinder with base area B and height h? __________
STEP 2: What is the area of the base of the cone? __________
STEP 3: Sandi found that, when the bases and height are the same, __________ times V_cone = V_cylinder.
STEP 4: How does the volume of the cone compare to the volume of the cylinder?
Volume of cone:

Reflect

Question 1.
Use the conclusion from this experiment to write a formula for the volume of a cone in terms of the height and the radius. Explain.
Answer:
From the experiment, we found that:
V_cone = \(\frac{1}{3}\) V_cylinder
Therefore,
V_cone = \(\frac{1}{3}\)πr²h
where r is the radius of the base of the cone and h is the height of the cone.

Lesson 9.2 Volume of Cones Answers Key 8th Grade Question 2.
How are the formulas for the volume of a cone and a pyramid similar?
Answer:
Find the similarities and differences between the pyramid and the cone and their voLumes. Also, notice what type of a base has the pyramid and the cone. Make conclusions based on that.
Volume of Pyramid or Cones = \(\frac{1}{3}\) × Area of Base × height = \(\frac{1}{3}\)Bh
The volume of three cones is equal to the volume of one cylinder with the same base and height.
Similarly, the volume of three pyramids is equal to the volume of one prism with the same base and height.
The difference in the volume of the pyramid and cone is in the base.
The base of a cone is a circle whose formula is B = πr²
The base of a pyramid is a square.
Formulas for the volume of a cone and a pyramid is \(\frac{1}{3}\)Bh

Example 1

Find the volume of each cone. Round your answers to the nearest tenth. Use 3.14 for π.

The volume is about 150.7 ft³.

Reflect

Question 3.
How can you rewrite the formula for the volume of a cone using the diameter d instead of the radius r?
Answer:
The volume of the cone is given as:
V = \(\frac{1}{3}\)πr²h ……….. (1)
On the other hand, the relation between the radius r and the diameter d can be expressed as:
r = \(\frac{d}{2}\) ………….. (2)
Substituting (2) into (1) so that we can rewrite the formula for the volume of a cone using the diameter d instead of the radius r:
V = \(\frac{1}{3}\)πr²h
V = \(\frac{1}{3}\) ∙ π ∙ (\(\frac{d}{2}\))² ∙ h
V = \(\frac{1}{3}\) ∙ π ∙ \(\frac{d^{2}}{4}\) ∙ h
V = \(\frac{1}{12}\) ∙ π ∙ d² ∙ h

Your Turn

Find the volume of each cone. Round your answers to the nearest tenth. Use 3.14 for π.

Question 4.

Answer:
The diameter d of the base of the cone is 15 cm, therefore the radius r is 75 cm. The height h of the cone is 16 cm.
V = \(\frac{1}{3}\)πr²h
V = \(\frac{1}{3}\) ∙ 3.14 ∙ (7.5)² ∙ 16
V = 942
The volume of the cone is 942 cm³

Lesson 9.2 Cone Volume with Diameter Texas Go Math 8th Grade Question 5.

Answer:
Here, radius is 2ft and height is 3 ft.
volume of the cone = \(\frac{1}{3}\) × π × r² × h
= \(\frac{1}{3}\) × 3.14 × 2² × 3
= 12.56 ft³

Example 2

For her geography project, Karen built a clay model of a volcano in the shape of a cone. Her model has a diameter of 12 inches and a height of 8 inches. Find the volume of clay in her model to the nearest tenth. Use 3.14 for π.

STEP 1: Find the radius.
r = \(\frac{12}{2}\) = 6 in.
STEP 2: Find the volume of clay.

The volume of the clay is about 301.4 in³.

Your Turn

Question 6.
The cone of the volcano Paricutin in Mexico had a height of 410 meters and a diameter of 424 meters. Approximate the volume of the cone.
Answer:
Here, height of the volcano = 410 meters and the diameter is 424 meters.
the radius of the volcano is 424/2 = 212 meters
volume of the cone = \(\frac{1}{3}\) × π × r² × h
= \(\frac{1}{3}\) × 3.14 × 212² × 410
= 19286968.53 m³

Texas Go Math Grade 8 Lesson 9.2 Guided Practice Answer Key

Volume of Cone Worksheet Answer Key Go Math Lesson 9.2 8th Grade Question 1.
The area of the base of a cylinder is 45 square inches and its height is 10 inches. A cone has the same area for its base and the same height. What is the volume of the cone? (Explore Activity)

The volume of the cone is ___________ in³.
Answer:
The area of the base of a cylinder, B = 45 in²
Height of the cylinder, h = 10 inch
Volume of the cylinder, V_cylinder = B × h
= 45 × 10 = 450 inch³
Volume of the cone, V_cone = \(\frac{1}{3}\)V_cylinder
= \(\frac{1}{3}\)(450 inch³) = 150 inch³
So, the volume of the cone is
V_cone = 150 in³

Question 2.
A cone and a cylinder have congruent height and bases. The volume of the cone is 18 m³. What is the volume of the cylinder? Explain. (Explore Activity)
Answer:
The volume of the cone is 18 m³. We know that:
V_cone = \(\frac{1}{3}\)V_cylinder
V_cylinder = 3V_cone
V_cylinder = 3 ∙ 18
V_cylinder = 54 m³

Find the volume of each cone. Round your answer to the nearest tenth if necessary. Use 3.14 for π. (Example 1)

Question 3.

Answer:
The diameter of the cone is 6 ft.
so, the radius of the cone is 3 ft
the height of the cone is 7 ft.
the volume of cone = \(\frac{1}{3}\) × π × r² × h
= \(\frac{1}{3}\) × 3.14 × 3² × 7
= 65.94 ft³

Go Math 8th Grade Answer Key Pdf Volume of a Cone Using Diameter Question 4.

Answer:
Here, the radius is 33 inch and the height is 100 inch
volume of cone = \(\frac{1}{3}\) × π × r² × h
= \(\frac{1}{3}\) × 3.14 × 33² × 100
= 113982 in³

Question 5.
Gretchen made a paper cone to hold a gift for a friend. The paper cone was 15 inches high and had a radius of 3 inches. Find the volume of the paper cone to the nearest tenth. Use 3.14 for π. (Example 2)
Answer:
Here, the radius is 3 inch and the height is 15 inch
volume of cone = \(\frac{1}{3}\) × π × r² × h
= \(\frac{1}{3}\) × 3.14 × 3² × 15
= 141.3 in³

Question 6.
A cone-shaped building is commonly used to store sand. What would be the volume of a cone-shaped building with a diameter of 50 meters and a height of 20 meters? Round your answer to the nearest tenth. Use 3.14 for π. (Example 2)
Answer:
The diameter of the cone is 50 meters.
so, the radius of the cone is 25 meters.
the height of the cone is 20 meters.
the volume of cone = \(\frac{1}{3}\) × π × r² × h
= \(\frac{1}{3}\) × 3.14 × 25² × 20
= 13083.33 m³

Essential Question Check-In

Question 7.
How do you find the volume of a cone? For help, use the model in the Explore Activity.
Answer:
V_cone = \(\frac{1}{3}\)V_cylinder
V_cone = \(\frac{1}{3}\)πr²h

Texas Go Math Grade 8 Lesson 9.2 Independent Practice Answer Key

Find the volume of each cone. Round your answers to the nearest tenth if necessary. Use 3.14 for π.

Question 8.

Answer:
Radius r = 7 mm
Height = 8 mm
Volume of cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × (7)² × 8
Volume = 410.29 mm³
Volume ≈ 410.3 mm³

Go Math Grade 8 Lesson 9.2 Answer Key Volume of Cone with Diameter and Height Question 9.

Answer:
Radius r = 2 in
Height = 6 in
Volume of cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × (2)² × 6
Volume = 25.12 in³
Volume ≈ 25.1 in³

Question 10.
A cone has a diameter of centimeters and a height of 11.5 centimeters.
Answer:
Diameter of base = 6 cm
Radius = \(\frac{6}{2}\) cm
Radius r = 3 cm
Height = 11.5 cm
Volume of cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × (3)² × 11.5
Volume = 108.33 cm³
Volume ≈ 108.3 cm³

Question 11.
A cone has a radius of 3 meters and a height of 10 meters.
Answer:
Radius r = 3 m
Height = 10 m
Volume of cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × (3)² × 10
Volume = 94.2 m³

Round your answers to the nearest tenth if necessary. Use 3.14 for π.

Question 12.
Antonio is making mini waffle cones. Each waffle cone is 3 inches high and has a radius of inch. What is the volume of a waffle cone?
Answer:
Radius = \(\frac{3}{4}\) in
Radius r = 0.75 in
Height = 3 in
Volume of each waffle cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × (0.75)² × 3
Volume = 1.76625 in³
Volume ≈ 1.8 in³

Go Math Grade 8 Volume of Cone Worksheet with Answers Pdf Question 13.
A snack bar sells popcorn in cone-shaped containers. One container has a diameter of 8 inches and a height of 10 inches. How many cubic inches of popcorn does the container hold?
Answer:
Diameter of base = 8 in
Radius = \(\frac{8}{2}\) in
Radius r = 4 in
Height = 10 in
Volume of cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × (4)² × 10
Volume = 167.466 in³
Volume ≈ 167.5 in³

Question 14.
A volcanic cone has a diameter of 300 meters and a height of 150 meters. What is the volume of the cone?
Answer:
Diameter of a volcanic cone = 300 m
Radius = \(\frac{300}{2}\) = 150 m
Height = 150 m
Volume of cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × (150)² × 150
Volume = 3532500 m³
Volume of a volcanic cone = 3532500 m³

Question 15.
Multistep Orange traffic cones come in a variety of sizes. Approximate the volume, in cubic inches, of a traffic cone that has a height of 2 feet and a diameter of 10 inches. Use 3.14 for π.
Answer:
Diameter of a traffic cone = 10 in
Radius = \(\frac{10}{2}\) = 5 in
Height = 2 ft
1 foot is equal to 12 inches, so 2 ft = 2 ∙ 12 = 24 in
Volume of the traffic cone = \(\frac{1}{3}\)πr²h
Volume = \(\frac{1}{3}\) × 3.14 × (5)² × 24
Volume = 628 in³
Volume of the traffic cone = 628 in³

Find the missing measure for each cone. Round your answers to the nearest tenth if necessary. Use 3.14 for π.

Texas Go Math Grade 8 Pdf Volume of Pyramids and Cones Worksheet Question 16.
radius= _________
height = 6 in.
volume = 100.48 in³
Answer:
Let radius be R.
Height = 6 in
Volume = 100.4 in³
Volume of cone = \(\frac{1}{3}\)πr²h
V = \(\frac{1}{3}\)πr²h
So, we have
\(\sqrt{\frac{3 V}{h \pi}}\) = R
\(\sqrt{\frac{3 \times 100.48}{6 \times 3.14}}\) = R
\(\sqrt{\frac{301.44}{18.84}}\) = R
R = \(\sqrt{16}\)
R = 4 in

Question 17.
diameter = 6cm
height = ____________
volume = 56.52 cm³
Answer:
Let height be h
Diameter 6 cm
Radius = \(\frac{6}{2}\) cm
Radius 3 cm
Volume = 56.52 cm³
Volume of cone = \(\frac{1}{3}\)πr²h
V = \(\frac{1}{3}\)πr²h
So, we have
\(\frac{3 V}{r^{2} \pi}\) = h
\(\frac{3 \times 56.52}{3^{2} \times 3.14}\) = h
\(\frac{169.56}{28.26}\) = h
h = 6 cm

Question 18.
The diameter of a cone-shaped container is 4 inches, and its height Is 6 inches. How much greater is the volume of a cylinder shaped container with the same diameter and height? Round your answer to the nearest hundredth. Use 3.14 for π.
Answer:
The diameter of a cone, d = 4 inch
radius of a cone, r = \(\frac{d}{2}\) = \(\frac{4}{2}\) = 2 inches
height of a cone, h = 6 inches
So, Volume of a cone, V_cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × 3.14 × (2)² × 6
= 25.12 in³
And, volume of a cylinder with same diameter and height, V_cylinder
V = πr²h = 3.14 × (2)² × 6
= 75.36 in³
The volume of the cylinder is 50.24 in³ greater than the volume of cone.
The volume of the cylinder is three times the volume of cone.

H.O.T. Focus on Higher Order Thinking

Question 19.
Alex wants to know the volume of sand in an hourglass. When all the sand is in the bottom, he stands a ruler up beside the hourglass and estimates the height of the cone of sand.

a. What else does he need to measure to find the volume of sand?
Answer:
To find the volume of the sand, he also needs to measure the radius of the base of the hourglass.

b. Make a Conjecture If the volume of sand is increasing at a constant rate, is the height increasing at a constant rate? Explain.
Answer:
The volume of the cone is linearly proportional to the height of the cone. Therefore, if the volume is increasing at a constant rate, the height is also increasing at a constant rate.

Go Math 8th Grade Pdf Lesson 9.2 Volume of Cones Worksheet Answer Key Question 20.
Problem Solving The diameter of a cone is x cm, the height is 18 cm, and the volume is 301.44 cm³. What is x? Use 3.14 for π.
Answer:

We know that the diameter of the circle is twice its radius, therefore
x = 2 ∙ r
x = 2 ∙ 4
x = 8 cm

Question 21.
Analyze Relationships A cone has a radius of 1 foot and a height of 2 feet. How many cones of liquid would it take to fill a cylinder with a diameter of 2 feet and a height of 2 feet? Explain.
Answer:
The diameter of the base of the cylinder is 2 feet, which means that its radius is 1 foot Its height is 2 feet. The volume of this cylinder is:
V_cylinder = πr²h
V_cylinder = 3.14 ∙ 1² ∙ 2
V_cylinder = 6.28
The radius of the base of the cone is 1 foot and the height of the cone is 2 feet The volume of the cone is:
V_cone = \(\frac{1}{3}\)πr²h
V_cone = \(\frac{1}{3}\) ∙ 3.14 ∙ 1² ∙ 2
V_cone = \(\frac{1}{3}\) ∙ 628
V_cone = \(\frac{1}{3}\) ∙ V_cylinder
V_cone ≈ 2.09
It would take 3 cones of liquid to fill the cylinder.

Question 22.
Critique Reasoning Herb knows that the volume of a cone is one-third that of a cylinder with the same base and height. He reasons that a cone with the same height as a given cylinder but 3 times the radius should therefore have the same volume as the cylinder, since \(\frac{1}{3}\) ∙ 3 = 1. Is Herb correct? Explain.
Answer:
The volume of the given cylinder is:
V_cylinder = πr²h
The volume of the cone with the same height h as a given cyLinder but 3 times the radius r is:
V_cone = \(\frac{1}{3}\)π(3r)²h
V_cone = 3πr²h
V_cone = 3V_cylinder
As we can see, Herb is not correct. The volume of the cone is not equal to the volume of the cylinder, but it is three times the volume of the cylinder.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Unit 3 Study Guide Review Answer Key.

Texas Go Math Grade 8 Unit 3 Study Guide Review Answer Key

Texas Go Math Grade 8 Unit 3 Exercises Answer Key

Module 7 Angle Relationships in Parallel Lines and Triangles

Question 1.
If m∠GHA = 76°, find the measures of the given angles. (Lesson 7.1)

m∠EGC = _____________
m∠EGD = _____________
m∠BHF = _____________
m∠HGD = _____________
Answer:
(i) ∠GHA = ∠EGC (corresponding angles) = 106°
(ii) ∠EGC + ∠EGD = 180°
∠EGD = 180 – 106 = 74°
(iii) ∠GHA = ∠BHF (vertical angles) = 106°
(iv) ∠GHA = ∠HGD (alternate interior angles) = 106°

8th Grade Math Unit 3 Study Guide Question 2.
Find the measure of the missing angles. (Lesson 7.2)

m∠JKL = _____________
m∠LKM = _____________
Answer:
From the Triangle Sum Theorem we have:
m∠L + m∠M + m∠LKM = 1800
We substitute the given angle measures and we solve for x.
125° + 40° + m∠LKM = 180°
165° + m∠LKM = 180°
165° – 165° + m∠LKM = 180° – 165°
m∠LKM = 15°
∠JKL and ∠LKM are supplementary angles, therefore
m∠JKL + m∠LKM = 180°
m∠JKL + 150 = 180°
m∠JKL + 15° – 15° = 180° 15°
m∠JKL = 165°

Question 3.
Is the larger triangle similar to the smaller triangle? Explain your answer. (Lesson 7.3)
Answer:
yes; the triangles are similar because all of their angles are congruent both triangles have a right angle and angle B.

Question 4.
Find the value of x and y in the figure. (Lesson 7.3)

Answer:
In similar triangles, corresponding side lengths are proportional.

Also,

Module 8 The Pythagorean Theorem

Find the missing side lengths. Round your answers to the nearest hundredth. (Lesson 8.1)

Question 1.

Answer:
Base = 10 ft
Height 10 ft
Hypotenuse = \(\sqrt{\text { base }^{2}+\text { height }^{2}}\)
Hyp = \(\sqrt{10^{2}+10^{2}}\)
Hyp = \(\sqrt{100+100}\)
Hyp = \(\sqrt{200}\)
Hyp = \(10 \sqrt{2}\) ft

Unit 3 Math Test 8th Grade Answer Key Question 2.

Answer:
We want to find r, the length from a bottom corner to the opposite top corner. First, we find s, the length of the diagonal across the bottom of the box.
w² + l² = s²
10² + 25² = s²
100 + 625 = s²
725 = s²
s ≈ 26.9 cm
We use our expression for s to find r.
h² + s² = r²
10² + 725 = r²
100 + 725 = r²
825 = r²
r ≈ 28.7 cm

Question 3.
Hye Sun has a modern coffee table whose top is a triangle with the following side lengths: 8 feet, 6 feet, and 5 feet. Is Hye Sun’s coffee table top a right triangle? (Lesson 8.2)
Answer:
Formula for the right-angle triangle = a² + b² = c²
a = 6 feet
b = 5 feet
c = 8 feet
= 6² + 5² = 8²
= 36 + 25 = 64
= 61 > 64
The given triangle is not a right-angled triangle.

Unit 3 Study Guide Math Grade 8 Question 4.
Find the length of each side of triangle ABC. If necessary, round your answers to the nearest hundredth. (Lesson 8.3)

Answer:
The Length of the vertical. leg is 6 units.
\(\overline{A C}\) = 6 units
The length of the horizontaL leg is 3 units
\(\overline{A B}\) = 3 units
Let a = 6 and b = 3. Let C represent the Length of the hypotenuse. We use the Pythagorean Theorem to find C.
a² + b² = c²
6² + 3² = c²
36 + 9 = c²
45 = c²
c = \(\sqrt {45}\)
c ≈ 6.71
\(\overline{B C}\) = 6.71 units

Module 9 Volume

Find the volume of each figure. Round your answers to the nearest hundredth. (Lessons 9.1, 9.2, 9.3)

Question 1.

Answer:
Base diameter = 12 mm
Base radius r = \(\frac{12}{2}\)
Base radius r = 6 mm
Height = 4 mm
VoLume of a cylinder = πr²h
V = 3.14 × 6² × 4
V = 3.14 × 36 × 6
V = 452.16 mm³

Question 2.

Answer:
Base diameter 50 in
Base radius r = \(\frac{50}{2}\)
Base radius r = 25 in
Height = 34 in
VoIime of a cone = \(\frac{1}{3}\)πr²h
V = \(\frac{1}{3}\) × 3.14 × 25² × 34
V = 22241.66667 in³
V ≈ 22241.67 in³

8th Grade Unit 3 Study Guide Answer Key Question 3.
Find the volume of a ball with a radius of 1.68 inches.
Answer:
The ball is in the shape of a sphere
The volume of the sphere = V = 4/3πr².sphere
Radius r = 1.68 inches
= 4/3 × π × (1.68) ²
= 4/3 × 3.14 × 2.82
= 5.025 sq. inches.
The volume of the sphere = V = 5.025 sq. inches.

Module 10 Surface Area

Find the lateral and total surface area of each figure. If necessary, round your answers to the nearest hundredth. (Lessons 10.1, 10.2)

Question 1.

Answer:
Given that the shape is the cylinder.
The formula for the total surface area of a cylinder = 2πrh + 2πr².
Diameter = 5ft =5/2
Radius = 2.5ft.
Height = 5ft
= 2(3.14) (2.5) (5) + 2(3.14) (5) ²
= 78.5 + 157
= 235.5 sq. ft.
The total surface area of a cylinder = 235.5 sq. ft.
The lateral surface area of a cylinder = 2πrh
= 2(3.14)(2.5)(5)
= 78.5 sq. ft
The lateral surface area of a cylinder = 78.5 sq. ft.

Question 2.

Answer:
Given that the shape is the rectangular prism
The formula for the surface area of a rectangular prism = 2(lb + bh + lh).
Length = l = 7.5ft
Breadth = b = 4ft
Height = h = 2ft
= 2(7.5 × 4 + 4 × 2 + 7.5 × 2).
= 2(30 + 8 + 15)
= 2(53)
= 106 cu. ft
The surface area of a rectangular prism = 106 cubic ft.
The lateral surface area of the rectangular prism = 2(l + b)h.
= 2(7.5 + 4) × 2.
= 60 × 2
= 120 sq. ft
The lateral surface area of the rectangular prism = 120 sq. ft

Grade 8 Unit 3 Test Math Answer Key Question 3.
A cylindrical water tower has a height of 50 meters and a radius of 20 meters.
Answer:
Given that,
The height of the cylindrical water tank = 50 meters.
The radius of the cylindrical water tank = 20 meters.
The formula for the total surface area of a cylinder = 2πrh + 2πr².
= 2 × 3.14 × 20 × 50 + 2 × 3.14 × (20)².
= 6280 + 2512
= 8792 sq. meter.
The total surface area of a cylinder = 8792 sq. meter
The lateral surface area of a cylinder = 2πrh
= 2(3.14)(20)(50)
= 6280 sq. meter
The lateral surface area of a cylinder = 6280 sq. meters.

Texas Go Math Grade 8 Unit 3 Performance Tasks Answer Key

Question 1.
CAREERS IN MATH Hydrologist A hydrologist needs to estimate the mass of water in an underground aquifer, which is roughly cylindrical in shape. The diameter of the aquifer is 65 meters, and its depth is 8 meters. One cubic meter of water has a mass of about 1000 kilograms.

a. The aquifer is completely filled with water. What is the total mass of the water in the aquifer? Explain how you found your answer. Use 3.14 for π and round your answer to the nearest kilogram.
Answer:
a) Since the aquifer is completely filled with water, we need to find the volume of the aquifer Its diameter is 65 meters, which means that its radius is 32-5 meters.
V = πr²h
V ≈ 3.14 ∙ 32.5² ∙ 8
V ≈ 26, 533m³
Since one cubic meter of water has a mass of about 1000 kilograms, then we have
26. 533 ∙ 1000 = 26, 533,000kg
The aquifer has about 26,533,000 kg of water

b. Another cylindrical aquifer has a diameter of 70 meters and a depth of 9 meters. The mass of the water in it is 2.7 × 10⁷ kilograms. Is the aquifer totally filled with water? Explain your reasoning.
Answer:
The diameter of the other aquifer is 70 meters, which means that its radius is 35 meters.
V = πr²h
V ≈ 3.14 ∙ 35² ∙ 9
V ≈ 34,618.5m³
Since one cubic meter of water has a mass of about 1,000 kilograms, we have
34,618.5 ∙ 1000 = 34, 618, 500 kg
On the other hand, we are told that there are 27 ∙ 10⁷ kg of water in the aquifer, which is more than the aquifer can hold.

Texas Go Math Grade 8 Answer Key Pdf Unit 3 Study Guide Question 2.
From his home, Myles walked his dog north 5 blocks, east 2 blocks, and then stopped at a drinking fountain. He then walked north 3 more blocks and east 4 more blocks, It started to rain so he cut through a field and walked straight home.
a. Draw a diagram of his path.
Answer:

b. How many blocks did Myles walk in all? How much longer was his walk before it started to rain than his walk home?
Answer:
Before it started to rain, Myles walked 14 blocks (black route).
5 + 2 + 3 + 4 = 14
To find how many blocks he walks after it started to rain, we let a = 8 blocks and b = 6 blocks. Let C represent the Length of the hypotenuse (the red route, the road straight home). We use the Pythagorean Theorem to find C.
a² + b² = c²
8² + 6² = c²
64 + 36 = c²
100 = c²
c = \(\sqrt {100}\)
c = 10 blocks
Myles walked 24 blocks in total (14 from home to the destination and 10 back home). The walk before it started to rain was 4 blocks longer than the walk home.

Texas Go Math Grade 8 Unit 3 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
Which of the following angle pairs formed by a transversal that intersects two parallel lines are not congruent?
(A) alternate interior angles
(B) adjacent angles
(C) corresponding angles
(D) alternate exterior angles
Answer:
(B) adjacent angles

Study Guide 8th Grade Math Unit 3 Answer Key Question 2.
The measures of the three angles of a triangle are given by 3x + 1, 2x – 3, and 9x. What is the measure of the smallest angle?
(A) 13°
(B) 23°
(C) 29°
(D) 40°
Answer:
(B) 23°

Explanation:
The angles of the given triangle = {(3x + 1), (2x-3), 9x}
As we know the sum of all three angles of a triangle is 180°
Thus the sum of the given angles should be also 180°
Therefore,
(3x + 1) + (2x – 3)9x = 180°
3x + 2x + 9x + 1 – 3 = 180
14x – 2 = 180
14x = 182
x = \(\frac{182}{14}\)
x = 13°.
Thus the angles are;
3x + 1 = 3(13) + 1 = 40°
2x – 3 = 2(13) – 3 = 23°
9x = 9(13) = 117°
Since 23° < 40° < 117°
Therefore the smallest angle is 23°.

Question 3.
Using 3.14 for π, what is the volume of the cylinder?

(A) 200 cubic yards
(B) 628 cubic yards
(C) 1256 cubic yards
(D) 2512 cubic yards
Answer:
(B) 628 cubic yards

Explanation:
Diameter of base 10 yd
Radius = \(\frac{10}{2}\)
Radius r = 5 yd
Height = 8yd
Volume of cylinder = πr²h
Volume = 3.14 × (5)² × 8
Volume 628 yd³

Unit 3 Study Guide Answer Key Math 8th Grade Question 4.
Which of the following is not true?
(A) \(\sqrt {36}\) + 2 > \(\sqrt {16}\) + 5
(B) 5π > 17
(C) \(\sqrt {10}\) + 1 < \(\frac{9}{2}\)
(D) 5 – \(\sqrt {35}\) < 0 Answer: (A) \(\sqrt {36}\) + 2 > \(\sqrt {16}\) + 5

Explanation:
(A) \(\sqrt {36}\) + 2 > \(\sqrt {16}\) + 5
6 + 2 > 4 + 5
8 > 9
This statement is false as 8 is not greater than 9.

(B) 5π > 17
5 ∙ 3.14 < 17
15.7 < 17
This statement is true as 17 is greater than 15.7.

(C) \(\sqrt {10}\) + 1 < \(\frac{9}{2}\)
3.16 + 1 < 4.5
4.16 < 4.5
This statement is true as 4.5 is greater than 4.16.

(D) 5 – \(\sqrt {35}\) < 0
5 – 5.92 < 0
5 – 5.92 < 0
-0.92 < 0
This statement is true as 0 is greater than -0.92.

Question 5.
A pole is 65 feet tall. A support wire is attached to the top of the pole and secured to the ground 33 feet from the pole’s base. Find the approximate length of the wire.
(A) 32 feet
(B) 73 feet
(C) 56 feet
(D) 60 feet
Answer:
(B) 73 feet

Explanation:
Let a = 65 feet and b = 33 feet Let C represent the length of the hypotenuse, the length of the wire. We use the Pythagorean Theorem to find C.
a² + b² = c²
65² + 33² = c²
4225 + 1089 = c²
5314 = c²
c = \(\sqrt {5314}\)
c ≈ 72.897
c ≈ 73 feet

Question 6.
Using 3.14 for π, what is the volume of the sphere 7 cm to the nearest tenth?

(A) 205.1 cm³
(B) 1077 cm³
(C) 179.5 cm³
(D) 4308.1 cm³
Answer:
(C) 179.5 cm³

Explanation:
Diameter of base = 7 cm
Radius = \(\frac{7}{2}\) cm
Radius r = 3.5 cm
Volume of sphere = \(\frac{4}{3}\)πr³
Volume = \(\frac{4}{3}\) × 3.14 × (3.5)³
Volume = 179.5033 cm³
Volume ≈ 179.5cm³

Question 7.
Which set of lengths are not the side lengths of a right triangle?
(A) 28, 45, 53
(B) 13, 84, 85
(C) 36, 77, 85
(D) 16, 61, 65
Answer:
(D) 16, 61, 65

Explanation:
Check if the lengths in Option A form a right triangle.
Let a = 28, b = 45 and c = 53. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
28² + 45² = 53²
784 + 2025 = 2809
2809 = 2809
True
Since 28² + 45² = 53², the triangle is a right triangle.

Check if the lengths in Option B form a right triangle.
Let a. = 13, b = 84 and c = 85. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
13² + 84² = 85²
169 + 7056 = 7225
7225 = 7225
True
Since 13² + 84² = 85², the triangle is a right triangle.

Check if the Lengths in Option C form a right triangLe.
Let a = 36, b = 77 and c = 85. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
36² + 77² = 85²
1296 + 5929 = 7225
7225 = 7225
True
Since 36² + 77² = 85², the triangle is a right triangle.

Check if the Lengths in Option D form a right triangle.
Let a = 16, b = 61 and c = 65. Using the converse of the Pythagorean Theorem, we have:
a² + b² = c²
16² + 61² = 65²
256 + 3721 = 4225
3977 = 4225
False
Since 16² + 61² = 65², the triangle is not a right triangle.

Unit 3 Review/Test 8th Grade Math Answer Key Question 8.
Which statement describes the solution of a system of linear equations for two lines with different slopes and different y-intercepts?
(A) one non-zero solution
(B) infinitely many solutions
(C) no solution
(D) solution of 0
Answer:
(A) one non-zero solution

Question 9.
What is the side length of a cube that has a surface area of 486 square inches?
(A) 7 inches
(B) 8 inches
(C) 9 inches
(D) 10 inches
Answer:
(C) 9 inches

Explanation:
Volume of the given cube 729 in²
Volume of a cube of side 7 in 7³ = 343
Volume of a cube of side 8 in 8³ = 512
Volume of a cube of side 9 in 9³ = 729
Volume of a cube of side 10 in = 10³ = 1000
A cube of side 9 in will have volume of 729 in³. Thus option C is correct.

Question 10.
A box in the shape of a triangular prism is shown below. What is the surface area of the box?

(A) 96 in²
(B) 102 in²
(C) 108 in²
(D) 116 in²
Answer:
Given that the box is in the shape of a triangular prism.
The formula of the surface area of the triangular prism = bh + (b1 + b2 + b3) l.
b = 3 inches.
h = 4 inches.
b1 = 4 inches.
b2 = 3 inches.
b3 = 5 inches.
l = 8 inches.
= 3 × 4 + (4 + 3 + 5)8
= 108 sq. in.
The surface area of the triangular prism = 108 sq. in.
Option B is the correct answer

Gridded Response

Question 11.
What is the value of h in the triangle below?

Answer:
In similar triangles the corresponding sides and lengths are proportional.
h/8 = 9/10
h/8 = 0.9
h = 0.9 × 8
h = 7.2
The value of h in the triangle is 7.2

Grade 8 Math Unit 3 Assessment Answer Key Question 12.
Chandra is working on an art project using a cone that has a height of 11 inches and a base radius of 2 inches. Using 3.14 for π, what is the volume of the cone to the nearest tenth of a cubic inch?

Answer:
Given that the shape is a cone.
The formula for the volume of the cone = 1/3hπr².
Height = h = 11 inches.
Radius = r = 2 inches.
= ⅓ × 11 × 3.14 × (2) ².
= 45.59 sq. inches.
The volume of the cone = 45.59 sq. inches.

Hot Tips! Read graphs and diagrams carefully. Look at the labels for important information.

Question 13.
The net of a cylinder is shown below.

Using 3.14 for π, what is the lateral surface area of the cylinder in square meters?

Answer:

Texas Go Math Grade 8 Unit 3 Vocabulary Preview Answer Key

Use the puzzle to preview key vocabulary from this unit. Unscramble the circled letters to answer the riddle at the bottom of the page.

Across
1. The angle formed by two sides of a triangle (2 words) (Lesson 7-2)
5. A three-dimensional figure that has two congruent circular bases. (Lesson 9-1)
6. A three-dimensional figure with all points the same distance from the center. (Lesson 9-3)

Down
2. The line that intersects two or more lines. (Lesson 7-1)
3. The side opposite the right angle ¡n a right triangle. (Lesson 8-1)
4. Figures with the same shape but not necessarily the same size. (Lesson 7-3)
5. A three-dimensional figure that has one vertex and one circular base. (Lesson 9-2)

Question.
What do you call an angle that is adorable?
Answer:
____ ____ ____ ____ ____ ____ ____ ____ ____ ____!

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 11 Quiz Answer Key.

Texas Go Math Grade 8 Module 11 Quiz Answer Key

Texas Go Math Grade 8 Module 11 Ready to Go On? Answer Key

11.1 Equations with the Variable on Both Sides

Solve.

Question 1.
4a – 4 = 8 + a ____________
Answer:
4a – 4 = 8 + a
(Subtract a from both sides.) 4a – 4 – a = 8 + a – a
(Add 4 to both sides.) 3a – 4 = 8
(Divide both sides by 3.) 3a = 12
a = 4
The statement is true. There is one solution.

Grade 8 Math Module 11 Answer Key Question 2.
4x + 5 = x + 8 ___________
Answer:
4x + 5 = x + 8
(Subtract x from both sides) 4x + 5 – x = x + 8 – x
(Subtract 5 from both sides.) 3x + 5 = 8
(Divide both sides by 3.) 3x = 3
x = 1
The statement is true. There is one solution.

Question 3.
Hue is arranging chairs. She can form 6 rows of a given length with 3 chairs left over, or 8 rows of that same length if she gets 11 more chairs. Write and solve an equation to find how many chairs are in that row length.
Answer:
(Write an equation) 6x + 3 = 8x – 11
(Subtract 3 from both sides.) 6x + 3 – 3 = 8x – 11 – 3
(Subtract 8x from both sides.) 6x = 8x – 14
(Divide by -2.)  – 2x = -14
x = 7
There are 7 chairs in each row.

11.2 Equations with Rational Numbers

Solve.

Question 4.
\(\frac{2}{3}\)s – \(\frac{2}{3}\) = \(\frac{s}{6}\) + \(\frac{4}{3}\) ____________
Answer:
(Multiply both sides by the LCM(3, 6) = 6) \(\frac{2}{3}\)s – \(\frac{2}{3}\) = \(\frac{s}{6}\) + \(\frac{4}{3}\)
6 ∙ \(\frac{2}{3}\)s – 6 ∙ \(\frac{2}{3}\) = 6 ∙ \(\frac{s}{6}\) + 6 ∙ \(\frac{4}{3}\)
4s – 4 = n + 8
(Add 4 to both sides) 4s – 4 + 4 = s + 8 + 4
4s = s + 12
(Subtract n from both sides) 4s – s = s + 12 – s
3s = 12
(Divide both sides by 3) \(\frac{3s}{3}\) = \(\frac{12}{3}\)
s = 4

Math Quiz for Grade 8 Module 11 Test Answers Question 5.
1.5d + 3.25 = 1 + 2.25d _____________
Answer:
1.5d + 3.25 = 1 + 2.25d
(Subtract 3.25 from both sides) 1.5d + 3.25 – 3.25 = 1 + 2.25d – 3.25
1.5d = 2.25d – 2.25
(Subtract 2.25d from both sides.) 1.5d – 2.25d = 2.25d – 2.25 – 2.25d
-0.75d = -2.25
(Divide both sides by -0.75) \(\frac{-0.75d}{-0.75}\) = \(\frac{-2.25}{-0.75}\)
d = 3

Question 6.
Happy Paws charges $19.00 plus $1.50 per hour to keep a dog during the day. Woof Watchers charges $15.00 plus $2.75 per hour. Write and solve an equation to find how many hours the total cost of the services is equal.
Answer:
Happy Paws charge for x hours
1.5x + 19
Woof Watchers charge for x hours
2.75x + 15
Put two expressions as equal
2.75x + 15 = 1.5x + 19
Subtract 1.5 from both sides
2.75x + 15 – 1.5x = 1.5x + 19 – 1.5x
1.25x + 15 = 19
Subtract 15 from both sides
1.25x + 15 – 15 = 19 – 15
1.25x = 4
Divide both sides by 1.25
x = \(\frac{4}{1.25}\) = 3.2
The total cost of the services is equal after $3.2 hrs.

11.3 inequalities with the Variable on Both Sides

Question 7.
Write an inequality to represent the relationship “Two less than 2 times a number is greater than the number plus 64″. Then solve your inequality.
Answer:
Let us consider the number = x
2 – 2x ≥ x + 60
-2x -x ≥ 60 – 2
-3x ≥ 58
x ≥ 58/-3
x ≥ -19.33
2 – 2(-19.33) ≥ -19.33 + 60
40.66 ≥ 40.67.

11.4 Inequalities with Rational Numbers

Question 8.
One prepaid cell phone company charges $0.028 per minute and a $3 monthly fee. Another company charges $0.034 per minute with no monthly fee. For what number of minutes per month are the charges for the first company cheaper?
Answer:
Let us consider the number of charges = x
The first company charges = $0.028 and $3 for the monthly fee.
The secondary company charges = $0.034 and no monthly fee.
The inequality equation is
$0.029 + $3 ≤ $0.034x
Multiply both sides by 1000
28x + 2000 ≤ 34x
2000 ≤ 34x + 28x
2000 ≤ 62x
x ≥ 2000/62
X ≥ 32.2
32.2 minutes per month are the charges for the first company cheaper.

Essential Question

Grade 8 Module 11 Answer Key Math Quiz Answers Question 9.
How can you use equations with the variable on both sides to solve real-world problems?
Answer:
I could join a game club for $15 and rent games for $2 each, or I could pay $3 for each video rental, which is the better deal?

Texas Go Math Grade 8 Module 11 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
Two cars are traveling in the same direction. The first car is going 40 mi/h and the second car is going 55 mi/h. The first car left 3 hours before the second car. Which equation could you solve to find how many hours it will take for the second car to catch up to the first car?
(A) 55t + 3 = 40t
(B) 55t + 165 = 40t
(C) 40t + 3 = 55t
(D) 40t + 120 = 55t
Answer:
(D) 40t + 120 = 55t

Explanation:
We write an expression representing the distance that car A has traveled. Let’s denote the time by t.
3 * 40 + 40t = 120 + 40t
We write an expression representing the distance that car B has traveled. Let’s denote the time by t
55t
We write an equation that can be solved to find the time it will take for the second car to catch up to the first car.
40t + 120 = 55t

Question 2.
A wide-screen television display measures approximately 15 inches high and 27 inches wide. A television is advertised by giving the approximate length of the diagonal of its screen. How should the television be advertised?
(A) 36 inches
(B) 31 inches
(C) 30 inches
(D) 21 inches
Answer:
The wide-screen television display is in the shape of a rectangle
The formula for the length of the diagonal of a rectangle = c² = a² + b²
The length of the rectangle = a = 15
Wide of the rectangle = b = 27
c² = 15² + 27²
c² = 225 + 729
c = sq. root (954)
c = 31 sq. inches.
Option B is the correct answer.

Grade 8 Math Module 11 Answer Key Pdf Question 3.
Shawn’s Rentals charges $27.50 per hour to rent a surfboard and a wet suit. Darla’s Surf Shop charges $23.25 per hour to rent a surfboard plus $17 extra for a wetsuit. For what total number of hours are the charges for Shawn’s Rentals the same as the charges for Darla’s Surf Shop?
(A) 3
(B) 4
(C) 5
(D) 6
Answer:
(B) 4

Explanation:
Shawn’s rentaL charge after x hrs
27.5x
Darlas surf shop charge after x hr
23.25x + 17
Equate the two expressions
23.25x + 17 = 27.5x
Subtract 23.25x from both sides
23.25x + 17 – 23.25x = 27.5x – 23.25x
4.25x = 17
Divide both sides by 4.25
x = \(\frac{17}{4.25}\) = 4
The charge would be equal after $4 hrs.

Question 4.
Which of the following is an irrational number?
(A) -8
(B) 4.63
(C) \(\sqrt{11}\)
(D) \(\frac{1}{3}\)
Answer:
(C) \(\sqrt{11}\)

Explanation:
\(\sqrt{11}\) is irrational.

Question 5.
Carlos has at least as many action figures in his collection as Josh. Carlos has 5 complete sets plus 4 individual figures. Josh has 3 complete sets plus 14 individual figures. Which inequality represents how many action figures can be in a complete set?
(A) x ≥ 7
(B) x ≤ 7
(C) x ≥ 5
(D) x ≤ 5
Answer:
Let us consider the action figures in a complete set = x
The Charles has 5 complete sets + 4 individual figures
Josh has 3 complete sets + 14 individual figures
The inequality equation is
5x + 4 ≤ 3x + 14
5x – 3x ≤ 14 – 4
2x ≤ 10
x ≤ 10/2
x ≤ 5
Option D is the correct answer.

Module 11 Math Quiz for Grade 8 with Answers Question 6.
Which inequality represents the solution to 1.25x + 2.5 < 2.75x – 6.5?
(A) x > 6
(B) x < 6
(C) x > 2.25
(D) x < 2.25
Answer:
1.25x + 2.5 < 2.75x – 6.5
1.25x – 2.75x < -6.5 – 2.5
-1.50x < -9
1.5x < 9
x < 9/1.5
x < 6
Option B is the correct answer.

Gridded Response

Question 7.
If both figures have the same perimeter, what is the perimeter of each figure?

Answer:
Given that,
The perimeter of the rectangle = perimeter of the triangle
2(l + w) = a + b + c
l = x + 5
w = x
a = x + 7
b = x + 4
c = x + 11
2(x + 5 + x) = x + 7 + x + 4 + x + 11
4x + 10 = 3x + 22
22 – 10 = 4x – 3x
x = 12.
3(12) + 22 = 4(12) + 10
58 = 58
The perimeter of the rectangle is 58 sq. inches.
The perimeter of the triangle is 58 sq. inches.

Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Module 10 Quiz Answer Key.

Texas Go Math Grade 8 Module 10 Quiz Answer Key

Texas Go Math Grade 8 Module 10 Ready to Go On? Answer Key

10.1 Surface Area of Prisms

Find the lateral and total surface area of each prism. Round your answers to the nearest tenth if necessary.

Question 1.

Answer:
Given that the prism is a rectangular prism.
Lateral surface area of a rectangular prism = 2(l + w)h.
Length = l = 18cm
Width = w = 10cm
Height = h = 6cm.
=2(18 + 10)6.
=2(28)6.
= 336 square units.

Surface Area Quiz Answer Key Grade 8 Math Question 2.

Answer:
Given that the prism is the triangular prism.
The formula for the lateral surface of the triangular prism = perimeter of the base x height of the prism = (a + b + c)h.
H = 8yd
A = 25yd
B = 10yd
C = 6yd
= (25 + 10 + 6)8
= (41)8
= 328yd.
The lateral surface area of the triangular prism = 328 yd²
328 rounded to the nearest ten is 330. Because in the 336 the one’s place is greater than the 5. Thus replace 0 by 0 and tens place by 3.

Question 3.

Answer:
Given that the prism is the triangular prism.
The formula for the lateral surface of the triangular prism = perimeter of the base x height of the prism = (a + b + c)h.
H = 15m
A = 17m
B = 13m
C = 16m
= (17 + 13 + 16)15
= (46)15
= 690m.
The lateral surface of the triangular prism = 690m²
690 rounded to the nearest ten is 690.

Question 4.

Answer:
Given that the prism is a rectangular prism.
Lateral surface area of a rectangular prism = 2(l + w)h.
Length = l = 14.5mm
Width = w = 10mm
Height = h = 4.2mm
=2(14.5 + 10)4.2.
=2(24.5)4.2.
= 205.8mm²
205.8 rounded to the nearest ten is 205. Because in the 205.8 the decimal place is greater than the 5. So replace the 8 by 0

10.2 Surface Area of Cylinders

Find the lateral and total surface area of each cylinder. Round your answers to the nearest tenth if necessary. Use 3.14 for π.

Question 5.

Answer:
Formula for the lateral surface area of the cylinder is 2πRH.
Diameter = 18yd
Radius = 18/2 = 9
H = height = 14yd
R = radius = 9yd
LA = 2 x 3.14 x 9 x 14.
lateral surface area of the cylinder = 791.28yd.
791.28 is rounded to the nearest tenth is 791.20. Because in the decimal the one’s place is greater than 5 it is replaced by 0.

Grade 8 Math Module 10 Answer Key Question 6.

Answer:
The formula for the lateral surface area of the cylinder is 2πRH.
H = height = 9ft
R = radius = 7ft
LA = 2 x 3.14 x 7 x 9.
the lateral surface area of the cylinder= 395.64ft
395.64 is rounded to the nearest tenth is 1582.60. Because in the decimal the one’s place is less than 5 so it is replaced by 0.

Essential Question

Question 7.
How can finding surface area help you solve packaging problems?
Answer: The packaging boxes are in the shape of a cuboid and it has 6 faces. Its label has length, width, and height. The surface area of the prism is SA = 2lw + 2lh + 2hw.

Texas Go Math Grade 8 Module 10 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
Grain is stored in cylindrical structures called silos. Which is the best estimate for the volume of a silo with a diameter of 12.3 feet and a height of 25 feet?
(A) 450 cubic feet
(B) 900 cubic feet
(C) 2970 cubic feet
(D) 10,800 cubic feet
Answer:
The formula for the volume of the cylinder = 2πr2h.
H = 25feet
Diameter = 12.3 feet
Radius = d/2 = 12.3/2 =6.15.
= 2 x 3.14 x (12.3)2 x 25
= 2969.06625 = 2970 cubic feet.
Option C is the correct answer.

Question 2.
What is the side length of a cube that has a total surface area of 384 square inches?
(A) 6 inches
(B) 7 inches
(C) 8 inches
(D) 9 inches
Answer:
Given that the total surface area of the cube = 384 inches
The formula for the total surface area = 6a²
= 6(6)² = 216 inches.
= 6(7)² = 294 inches.
= 6(8)² = 384 inches.
= 6(9)² = 486 inches.
Option C is the correct answer.

Question 3.
A block of cheese is shown in the shape of the triangular prism below.

What is the total surface area of the block of cheese?
(A) 120 in²
(B) 494 in²
(C) 510 in²
(D) 640 in²
Answer:
There are two triangles in the triangular prism
Area of the one triangle = ½ x b x h
= ½ x 17 x 15 = 127.5
Area of the two triangles = ½ x b x h
= 1/2 x 8 x 13 = 52
The formula for the total surface area of the triangle = area of one triangle + area of two triangles.
= 179.5 square inches.

Question 4.
Using 3.14 for π, what is the lateral surface area of the cylinder to the nearest unit?

(A) 122 yd²
(B) 204 yd²
(C) 383 yd²
(D) 587 yd²
Answer:
The formula for the lateral surface area of the cylinder is 2πRH.
Diameter = 11.4yd
Radius = 11.4/2 = 5.7yd
H = height = 10.7yd
R = radius = 5.7yd
LA = 2 x 3.14 x 5.7 x 10.7.
the lateral surface area of the cylinder= 383.01yd²
383.01 is rounded to the nearest tenth of 383. Because in the decimal the one’s place is less than 5 so it is replaced by 0.
Option C is the correct answer.

Module 10 Review Quiz Grade 8 Math Question 5.
The net of a cylinder is shown.
What is the total surface area of the cylinder represented by the net?

(A) 1607.7 m²
(B) 1856 m²
(C) 4220.2 m²
(D) 5827.8 m²
Answer:
The formula for the surface area of the cylinder = 2πr(r + h).
H = 42m
Diameter = 32m
Radius = 32/2 = 16m
= 2 x 3.14 x 16(16 + 42)
= 100.48(58)
= 5827.8
Option D is the correct answer.

Question 6.
A cylinder has a circumference of 12π cm and a height that is half the radius. What is the total surface area of the cylinder?
(A) 367π cm²
(B) 72π cm²
(C) 108π cm²
(D) 144π cm²
Answer:
The formula for the surface area of the cylinder = 2πr(r + h).
Circumference = 12π/π
= 12 is the diameter.
Radius = d/2 = 12/2 = 6
R = 6
H = half the radius = 6/2 = 3.
= 2 x π x 6(6 + 3)
= 6.28 x 54
= 108πcm²
Option C is the correct answer.

Gridded Response

Question 7.
A rectangular prism has a length of 4 feet, a width of 1 yard, and a height of 3 inches. What is the total surface area of the prism to the nearest tenth of a square foot?

Answer:
The formula for the surface area of the rectangular prism = 2(lb + bh + lh).
L = 4 feet.
1 feet 12 inches
4 feets = 12 x 4 = 48 inches.
Width = breadth = B = 1 yard.
1 yard = 36 inches.
H = 3 inches.
= 2(48 x 36 + 36 x 3 + 48 x 3)
= 2(1728 + 108 + 144)
= 3960 inches = 27.5 square foot
27.5 nearest tenth to square foot is 27.5