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Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 3 Review Test to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 3 Review Test Answer Key

Concepts and Skills

Solve each linear equation. Show your work.

Question 1.
2(x – 5) – 8 = 20
Answer:
Given,
2(x – 5) – 8 = 20
2x – 10 – 8 = 20
2x – 18 = 20
2x = 20 + 18
2x = 38
x = 38/2
x = 19

Question 2.
2x – (5 – x) = \(\frac{5}{2}\)
Answer:
Given,
2x – (5 – x) = \(\frac{5}{2}\)
2x – 5 + x = \(\frac{5}{2}\)
3x – 5 = \(\frac{5}{2}\)
3x = \(\frac{5}{2}\) + 5

Question 3.
\(\frac{1}{4}\)(x + 2) – 2 = 0.5
Answer:
Given,
\(\frac{1}{4}\)(x + 2) – 2 = 0.5
\(\frac{1}{4}\)x + \(\frac{1}{2}\) – 2 – \(\frac{1}{2}\) = 0
\(\frac{1}{4}\)x – 2 = 0
\(\frac{1}{4}\)x = 2
x = 2 × 4
x = 8

Question 4.
4x – \(\frac{5-2x}{5}\) = \(\frac{3}{5}\)
Answer:
4x – \(\frac{5-2x}{5}\) = \(\frac{3}{5}\)
20x – (5 – 2x) = 3
20x – 5 + 2x = 3
22x – 5 = 3
22x = 3 + 5
22x = 8
x = 8/22 or 4/11

Write each repeating decimal as a fraction. Show your work.

Question 5.
\(0 . \overline{2}\)
Answer: \(\frac{2}{9}\)

Question 6.
\(0.9 \overline{3}\)
Answer:
\(\frac{14}{15}\)

Question 7.
\(0.2 \overline{6}\)
Answer: \(\frac{4}{15}\)

Question 8.
\(0.3 \overline{16}\)
Answer: \(\frac{313}{990}\)

Tell whether each equation has one solution, no solution, or an infinite number of solutions. Show your work.

Question 9.
2x + 4 = -2(\(\frac{1}{2}\) – x)
Answer:
Given,
2x + 4 = -2(\(\frac{1}{2}\) – x)
2x + 4 = -1 + 2x
2x – 2x + 4 = -1
4 = -1
No solution

Question 10.
6y + (16 – 2y) = 4(4 + y)
Answer:
Given,
6y + (16 – 2y) = 4(4 + y)
6y + 16 – 2y = 16 + 4y
4y + 16 = 16 + 4y
It has solution or Identity

Question 11.
4x + 5 = 2x – 7
Answer:
Given,
4x + 5 = 2x – 7
4x – 2x = -7 – 5
2x = -12
x = -6
One solutions

Question 12.
2x + 5 = -4(-\(\frac{5}{4}\) – \(\frac{1}{2}\)x)
Answer:
Given,
2x + 5 = -4(-\(\frac{5}{4}\) – \(\frac{1}{2}\)x)
2x + 5 = 5 + 2x
Identity

Find the value of y when x = 4.

Question 13.
4x – 2 = y + 5
Answer:
Given,
4x – 2 = y + 5
4(4) – 2 = y + 5
16 – 2 = y + 5
14 – 5 = y
y = 9

Question 14.
x – 4y = 2
Answer:
Given,
x – 4y = 2
4 – 4y = 2
4 – 2 = 4y
2 = 4y
y = 2/4
y = 1/2 or 0.5

Question 15.
y – x = \(\frac{1}{3}\)(x + 14)
Answer:
Given,
y – x = \(\frac{1}{3}\)(x + 14)
y – 4 = \(\frac{1}{3}\)(4 + 14)
y – 4 = \(\frac{18}{3}\)
y = \(\frac{18}{3}\) + 4
y = 6 + 4
y = 10

Question 16.
\(\frac{3x-7}{y}\) = \(\frac{1}{3}\)
Answer:
Given,
\(\frac{3x-7}{y}\) = \(\frac{1}{3}\)
x = 4
\(\frac{3(4)-7}{y}\) = \(\frac{1}{3}\)
\(\frac{5}{y}\) = \(\frac{1}{3}\)
5 × 3 = 1 × y
y = 15

Question 17.
\(\frac{1}{7}\)(3x + y) = x
Answer:
Given,
\(\frac{1}{7}\)(3x + y) = x
x = 4
\(\frac{1}{7}\)(3(4) + y) = 4
12 + y = 4 – \(\frac{1}{7}\)
y = 4 – \(\frac{1}{7}\) – 12
y = -8 – \(\frac{1}{7}\)
y = -8.14

Question 18.
\(\frac{3y+1}{4}\) = 2x
Answer:
Given,
\(\frac{3y+1}{4}\) = 2x
x = 4
\(\frac{3y+1}{4}\) = 2 × 4
\(\frac{3y+1}{4}\) = 8
3y + 1 = 8 × 4
3y + 1 = 32
3y = 31
y = \(\frac{31}{3}\)

Express x in terms of y. Find the value of x when y = -2.

Question 19.
4x + y = 2(x + 3y)
Answer:
Given,
4x + y = 2(x + 3y)
4x + y = 2x + 6y
4x – 2x = 6y – y
2x =  5y
y = -2
2x = 5(-2)
2x = -10
x = -10/2
x = -5

Question 20.
3(x – 2y) = 4x + 5y
Answer:
Given,
3(x – 2y) = 4x + 5y
3x – 6y = 4x + 5y
3x – 4x = 5y + 6y
-x = 11y
y = -2
x = -11(-2)
x = 22

Question 21.
\(\frac{1}{3}\)x + \(\frac{5}{6}\)y = 2
Answer:
Given,
\(\frac{1}{3}\)x + \(\frac{5}{6}\)y = 2
y = -2
\(\frac{1}{3}\)x + \(\frac{5}{6}\)(-2) = 2
\(\frac{1}{3}\)x – \(\frac{10}{6}\) = 2
2x – 10 = 12
2x = 12 + 10
2x = 22
x = 11

Question 22.
\(\frac{0.5(x-3)}{y}\) = 10
Answer:
Given,
\(\frac{0.5(x-3)}{y}\) = 10
y = -2
\(\frac{0.5(x-3)}{-2}\) = 10
0.5(x – 3) = 10 × -2
0.5x – 1.5 = -20
0.5x = -20 + 1.5
0.5x = -18.5
x = -12.33

Question 23.
0.25(x + y) = 15 – x
Answer:
Given,
0.25(x + y) = 15 – x
0.25x + 0.25y = 15 – x
0.25x + 0.25(-2) = 15 – x
0.25x + x = 15 – 0.25(-2)
1.25x = 15 + 0.50
1.25x = 15.50
x = 15.50/1.25
x = 12.4

Question 24.
\(\frac{y}{2}\) – \(\frac{2 x+y}{5}\) = 7
Answer:
Given,
\(\frac{y}{2}\) – \(\frac{2 x+y}{5}\) = 7
y = -2
\(\frac{-2}{2}\) – \(\frac{2 x – 2}{5}\) = 7
-1 – \(\frac{2 x – 2}{5}\) = 7
– \(\frac{2 x – 2}{5}\) = 7 + 1
– \(\frac{2 x – 2}{5}\) = 8
-(2x – 2) = 8 × 5
-2x + 2 = 40
-2x = 40 – 2
-2x = 38
x = -19

Problem Solving

Solve. Show your work.

Question 25.
A circular pizza with a radius of 7 inches is cut into four quadrants. The perimeter Q of each quadrant can be found using the formula Q = d(1 + \(\frac{\pi}{4}\)), where d is the diameter of the pizza. Find the perimeter of each quadrant of the pizza. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
Given,
A circular pizza with a radius of 7 inches is cut into four quadrants.
The perimeter Q of each quadrant can be found using the formula Q = d(1 + \(\frac{\pi}{4}\)), where d is the diameter of the pizza.
pi = \(\frac{22}{7}\)
Q = d(1 + \(\frac{\pi}{4}\))
Q = d(1 + \(\frac{\pi}{4}\))
Q = d(1 + 0.785)
Q = 2(7)(1.785)
Q = 24.99 or 25 approx
Thus the perimeter of each quadrant of the pizza is 25 inches.

Question 26.
Some students painted a design on the wall of the cafeteria using the school colors. The middle section of the design is 4.2 feet tall and is painted white. The top section is red, and the bottom section is blue. The ratio of the height of the blue section to the height of the red section is 1: 2. The total height of the design is 10.5 feet. Find the height of the red section of the design.
Answer:
Given,
The middle section of the design is 4.2 feet tall and is painted white.
The top section is red, and the bottom section is blue.
The ratio of the height of the blue section to the height of the red section is 1: 2.
The total height of the design is 10.5 feet.
x + 4.2 + 2x = 10.5
3x + 4.2 = 10.5
3x = 10.5 – 4.2
3x = 6.3
x = 2.1
2x = 4.2

Question 27.
In a grocery store, each pound of green beans costs one and a quarter times the price of each pound of potatoes. Mrs. Gomez bought 4 pounds of green beans and 5 pounds of potatoes. Miss Jacobs bought 10 pounds of potatoes. They paid the same amount.
a) Write a linear equation to find the cost of each pound of potatoes, p dollars.
Answer: 4 . \(\frac{5}{4}\)p + 5p = 10p

b) Tell whether the equation has one solution, is inconsistent, or is an identity. Explain your reasoning.
Answer:
Since 0 = 0 is always true, the linear equation is true for any value of p.
So, this equation has infinitely many solutions and it is an identity.

Question 28.
A department store is offering a percent discount on the original price of a watch. The original price of the watch is $80.
a) Let r represent the percent discount, written as a decimal. Write an equation to find the discounted price, y dollars, of the watch.
Answer: y = 80 – 80r

b) The store gave discounts of 5%, 10%, and 15% during this sale in the three previous years. Create a table of r- and y-values for the equation.
Answer:

Question 29.
The company Jake uses for Internet service charges $25 each month plus $0.04 for each minute of usage time.
a) Write a linear equation for the monthly charge, M dollars, in terms of usage time, t minutes.
Answer: M = 25 + 0.04t

b) Express t in terms of M.
Answer: t = 25(M – 25)

c) Calculate his usage time in hours if he paid $49 for his Internet bill in November.
Answer: 10h

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 3 Lesson 3.4 Solving for a Variable in a Two-Variable Linear Equation to finish your assignments.

Math in Focus Grade 7 Course 3 A Chapter 3 Lesson 3.4 Answer Key Solving for a Variable in a Two-Variable Linear Equation

Math in Focus Grade 8 Chapter 3 Lesson 3.4 Guided Practice Answer Key

Solve. Show your work.

Question 1.
Express x in terms of y for the equation 2(x – 3) =3y -1. Find the value of x when y = 3.

Answer:
x = 7,

Explanation:

Question 2.
Express x in terms of y for the equation y = \(\frac{2 x+3}{2}\). Find the value of x when y = -13.
Answer:
x = -29/2,

Explanation:

Solve. Show your work.

Question 3.
The mean (or average) of three numbers x, x\(\sqrt{3}\) and 2 is M.

a) Express x in terms of M.

Answer:
x = 3M-2/2.732
Explanation:

b) Create a table of values for M and x when M = 0, 1, 2, and 3.
Round each x-value to the nearest hundredth.
Substitute 0, 1, 2, and 3 for M into the equation x = .

Answer :
When M = 0, X = -0.732,
M= 1, x =0.366, M=2 , x = 1.464 and M = 3 , x= 2.562,
Explanation:

Math in Focus Course 3A Practice 3.4 Answer Key

Express y in terms of x. Find the value of y when x = -1.

Question 1.
5 – y = 3x
Answer:
y = 5 – 3x,y = 8

Explanation:
Given 5 -y = 3x, x=-1,
Expressing y in terms of x as y = 5-3x and as x = -1
y = 5 – 3(-1)= 5 +3 = 8.

Question 2.
-3(x + 2) = 5y
Answer:
y =-3/5(x+2), y = -3/5,

Explanation:
Given -3(x+2) = 5y and x = -1
Expressing y in terms of x as y = -3/5(x +2),
y = -3/5(-1+2),
y= -3/5(1),
y = -3/5.

Question 3.
6(x – y) = 19
Answer:
y= x -19/6, y = -25/6,

Explanation:
Given 6(x-y)=19,
6x – 19 =  6y,
y = x – 19/6,when x = -1,
y = -1 -19/6 = -25/6.

Question 4.
4x – 3 = 0.4x – 2y
Answer:
y= 1.5 – 1.8 x, y = 3.3,

Explanation:
Given 4x – 3 =0.4x -2y,
4x – 3-0.4x = -2y
3.6x -3 =-2y
y = -3.6x/2 + 3/2
y = 1.5 – 1.8x, when x = -1
y = 1.5 -1.8(-1)= 1.5+1.8 = 3.3.

Question 5.
\(\frac{1}{6}\)x + \(\frac{3}{4}\)y = 4
Answer:
y = (96-4x)/18,y = 5.5

Explanation:
Given\(\frac{1}{6}\)x + \(\frac{3}{4}\)y = 4 multiplying both sides with 24,
4x +18 y =96,
18 y = 96-4x
y = (96- 4x)/18,when x = -1,
y = (96-4(-1))/18 = 96+4/18 = 100/18 = 5.5.

Question 6.
0.5y – 2 = 0.25x
Answer:
y = 4+0.5x, y = 3.5,

Explanation:
Given 0.5y – 2 = 0.25x,
0.5 y = 2 +0.25x
y = 4 +0.5 x, when x = -1,
y = 4 +0.5(-1) = 4 -0.5 = 3.5.

Express x in terms of y. Find the value of x when y = 5.

Question 7.
5x – y = 3(x + y)
Answer:
x = 2y, x = 10,

Explanation:
Given 5x – y = 3(x+y),
5x – y = 3x +3y,
5x -3x = 3y+y,
2x = 4y,
x = 2y,when y = 5,
x = 2 X 5 = 10.

Question 8.
3(x + 2y) = 2x + 5y
Answer:
x = -y, x = -5,

Explanation:
Given 3(x+2y)=2x +5y,
3x +6y = 2x +5y,
3x – 2x = 5y-6y,
x = -y,when y =5, x = -5.

Question 9.
1.5(x – y) = 1
Answer:
x= 0.66 +y, x = 5.66,

Explanation:
Given 1.5(x-y)=1,
1.5x = 1+ 1.5 y,
x = 0.66 + y,when y = 5, x = 0.66 + 5,
x = 5.66.

Question 10.

2y + 8 = \(\frac{1}{4} x\)
Answer:
x =(8y+32), x = 72,

Explanation:
Given 2y+ 8 = \(\frac{1}{4} x\),
4(2y+8)= x,
x = 8y + 32, when y = 5,
x = 8 X 5+ 32= 40 +32 = 72.

Question 11.
\(\frac{2(x-3)}{y}\) = 5
Answer:
x = (5y+6)/2, x = 31/2,

Explanation:
Given \(\frac{2(x-3)}{y}\) = 5,
2(x -3)=5y,
2x – 6 = 5y,
2x = 5y +6,
x =(5y +6)/2,when y = 5,
x = (5 X 5 +6)/2,
x = 31/2.

Question 12.
\(\frac{1}{3}(6 x-1)\) = \(\frac{6 y}{5}\)
Answer:
x= (18y+5)/30, x =95/30,

Explanation:
Given\(\frac{1}{3}(6 x-1)\) = \(\frac{6 y}{5}\),
5(6x-1) = 18y,
30 x- 5 = 18y,
30x = 18y +5,
x=(18y+ 5 )/30, when y =5,
x = (18X 5+5)/30= 95/30.

Solve. Show your work.

Question 13.
The perimeter, P inches, of a semicircle of diameter, d inches, is represented by P = 0.5πd+ d.

a) Express d in terms of P.
Answer:
p = d(0.5π +1) inches,

Explanation:
Given the perimeter, P inches, of a semicircle of diameter,
d inches, is represented by P = 0.5πd+ d so p= d(0.5π +1) inches.

b) Find the diameter if the perimeter is 36 inches.
Use \(\frac{22}{7}\) as an approximation for π.
Answer:
The diameter is 14.007 inches,

Explanation:
Given to find the diameter if the perimeter is 36 inches.
Using \(\frac{22}{7}\) as an approximation for π= 3.14,
we have p = d(0.5π +1) inches,
36 = d(0.5 X 3.14 +1),
36 = d(1.57 +1),
36 = d (2.57),
d = 36/2.57= 14.007 inches.

Question 14.
The horizontal distance, X inches, and vertical distance,
Y inches, of each step of a staircase are related by the linear equation
X = \(\frac{1}{2}\)(20 + Y).

a) Express Y in terms of X.
Answer:

Y = 2X – 20,

Explanation:
Given X = \(\frac{1}{2}\)(20 + Y),
2 X = 20 + Y,
Y = 2X – 20.

b) Complete the table below.

Answer:

Explanation:
We have Y = 2X – 20,
when Y = 10, 2X = Y + 20 = 2X = 30, X =15,
When X = 16, Y = (2x-20) = 2X 16 – 20 = 32 – 20 =12,
When Y = 14, 2X = Y +20 ,2 X = 14 + 20 = 34 , X =17,
When Y =16, 2X = 16+20 = 36, X = 36/2 = 18,
When X = 19,Y =(2x -20)= 2X 19 – 20 = 38 – 20 = 18,
Completed the table above.

c) Find the value of X and Y if X = Y.
Answer:
If X = Y then Y = 20,

Explanation:
We have Y =2X – 20,
Y = 2 Y -20,
2Y -Y =20,
Y = 20.

Question 15.
Use the isosceles triangle.

a) Write an equation for y in terms of x.

Answer:
x + y = 90,

Explanation:
Given the triangle is isosceles is 2(x +y)+ 90 = 180,
2(x+y) = 180-90= 90,
x + y = 90/2 = 45.

b) Find the value of y when x = 24.5.
Answer:
y = 65.5,

Explanation:
We have x +y = 90, when x = 24.5 ,
y = 90-24.5 = 65.5.

Question 16.
At a travel agency, the cost of a trip to Mexico is $350 for an adult and $200 for a child.
One month, the agency sold 50 trips. Of these trips, y trips were for children.
The travel agency collected C dollars from selling all 50 trips.

a) Write a linear equation for C in terms of y.
Answer:
C =350(50 -y)+ 200 y,

Explanation:
Given at a travel agency, the cost of a trip to Mexico is $350 for an adult and $200 for a child.
One month, the agency sold 50 trips. Of these trips, y trips were for children.
The travel agency collected C dollars from selling all 50 trips. is
C =350(50 -y)+ 200 y.

b) Express y in terms of C.
Answer:
y = (17,500 – C)/150,

Explanation:
We have C =350(50 -y)+ 200 y,
C = 17,500 – 350y + 200y,
C = 17,500 – 150y,
150 y = 17,500 -C
y = (17,500 – C)/150.

c) If $15,250 was collected, find the number of children going on the trip.
Answer:
The number of children going on the trip is 15,

Explanation:
If $15,250 is collected the number of children going on the trip,
we have y= (17,500-c)/150,
y= (17,500-15,250)/150,
y= 2,250/150= 15.

Question 17.
A trash disposal company charges accordinq to the weight of trash it disposes.
The charge, y dollars, for x pounds of trash is represented by the equation
y = 2(\(\frac{1}{4}\)x – 10).

a) Write an equation for x in terms of y.
Answer:
x = 2x +40,

Explanation:
y = 2(\(\frac{1}{4}\)x – 10),
y = 1/2 x – 20,
1/2 x = y +20,
x = 2y +40.

b) Create a table of x- and y-values for y = 60, 80, 100, 120, and 140.
Answer:

Explanation:
We have x = 2y + 40,
when y =60, x = 2X 60+ 40 =120 + 40 = 160,
when y = 80, x = 2 X 80 + 40 = 160 + 40 = 200,
when y = 100, x = 2 X 100+ 40 = 200 + 40 = 240,
when y = 120 , x = 2 X 120 + 40 = 240 + 40 = 280,
when y = 140, x = 2 X 140 + 40 = 280 + 40 = 320,
Created a table of x and y values above.

Question 18.
The mean, or average, of three numbers 17.4, 23.8, and x is M.

a) Write an equation for x in terms of M.
Answer:
x = 3M – 41.2,

Explanation:
Given the mean or average of three numbers 17.4, 23.8, and x is M so
M = (17.4 + 23.8 +x)/3,
3 M = x + 41.2,
x = 3M – 41.2.

b) Create a table of values for M = 15, 17, 19, 21, and 23.
Answer:

Explanation:
Created the tables of values M = 15,17,19,21 and 23,
so x = 3.8,9.8,15.8,21.8,27.8 respectively.

Question 19.
A rectangle has a width of w units and a length of 5 units.
Its perimeter is given by P = 2(w + 5). Solve for w in terms of P.
Create a table of values for P = 12, 14, 16, and 18.
Answer:

Explanation:
Given P = 2(w+5),
2w +10 = P,
2w = P + 10,
w = (p+10)/2,
Craeted a table of values for P=12,14,16 and 18 as shown above.

Question 20.
In the number pattern 3, 7, 11, 15, 19, …, each new number is
4 greater than the previous number. To find the number L in the nth position,
use the formula L = 3 + 4(n – 1).

a) What number is in the tenth position?
Answer:
39 is in the tenth position,

Explanation:
Given the number pattern 3, 7, 11, 15, 19, …, each new number is
4 greater than the previous number. To find the number L in the nth position,
we use the formula L = 3 + 4(n – 1) when n = tenth position,
L = 3 +4(10-1),
L = 3 + 4(9),
L = 3 + 36= 39.

b) In which position is the number 63?
Answer:
In 16th position is the number 63.

Explanation:
We have L= 3+4(n -1)  where L is the number and n will be position
if the number is 63 the position is 63 = 3+4(n-1),
63 -3 = 4(n-1),
60 = 4(n-1),
n -1 = 60/4,
n -1 = 15,
n = 15+1 = 16.

Brain @ Work

Question 1.
Lynnette runs a private tutoring business. She rents a room for $500 a month, which is her only expense. She charges $50 an hour per student, and gives each student two lessons per month. Each lesson lasts 1.5 hours.

a) Write an equation for her monthly profit, P, in terms of s, the number of students she has.
Answer:
P = $50s – $500,

Explanation:
Given Lynnette runs a private tutoring business.
She rents a room for $500 a month, which is her only expense.
She charges $50 an hour per student, equation for her monthly profit P
in terms of s, the number of students she has is P = $50s – $500.

b) Find her monthly profit if she has 40 students.
Answer:
If she has 40 students her montly profit is $1,500,

Explanation:
We have P = $50s – $500, if Lynnette has 40 students
P = $50 X 40 – $500,
P = $2,000 – $500 = $1500.

c) Find the minimum number of students she needs if she wants to make a monthly profit of at least $4,600.

Answer:
102 minimum number of students she needs if she wants to make a monthly profit of at least $4,600,

Explanation:
We have P = $50s – $500 if Lynnette wants to make a monthly profit of at least $4,600,
the number of students s is $4,600 = $50s – $500,
$50s = $4,600+ $500,
$50s = $5,100,
s = $5,100/50 = 102.

Question 2.
Stefanie’s train will leave her train station in 24 minutes and
she is y miles from the station. To catch the train, she walks at a speed of
4 miles per hour and later runs at a speed of 8 miles per hour.
a) Write an equation in terms of y for the distance she has to walk, w,
to reach the station in 24 minutes.
Answer:
w = (16-5y)/5 or  (3.2-y)

Explanation:
Given Stefanie’s train will leave her train station in 24 minutes and
she is y miles from the station. To catch the train, she walks at a speed of
4 miles per hour and later runs at a speed of 8 miles per hour.
Walk is w/4 and run (y-w)/8,
so w/4+(y -w)/ 8 = 24/60,
2w+y -w = 24 X 8/60,
w + y = 16/5,
w = (16-5y)/5 or 3.2 -y,

b) Solve for y in terms of w. How far is she from the station if she has to
walk 1 mile to reach the station on time?
Answer:
y =2.2,

Explanation:
As we have w = (16-5y)/5 or 3.2 -y if she is from the station if she has to
walk 1 mile to reach the station on time then y = 3.2 – 1= 2.2.

c) Why do the values of y have to be between 1.6 and 3.2?
Answer:
As walk cannot not zero or less than 4 miles,

Explanation:
If y values are not to be between 1.6 and 3.2 As walk cannot not zero or less than
4 miles if y = 3.2 then w = 3.2 – 3.2 = 0.

Question 3.
A polygon has n sides. The sum of the measures of a polygon’s
interior angles is equal to the sum of the measures of r right angles.
A table of r- and n-values is shown below.
Explain how you would find a linear equation involving r and n.

Answer:
The sum of the measures of interior angles of a polygon of n sides is equal to
$\Rightarrow 2(n-2)$ right angles,

Explanation:
If a polygon has $n$ sides, then it is divided into $(n-2)$ triangles.
As we know that the sum of the angles of triangle = 180∘.
Therefore, the sum of the angles of $(n-2)$ triangles = 180 $\times$ (n−2),
$\Rightarrow$ 2$\times$right angles$\times$(n−2),
The sum of the measures of interior angles of a polygon of n sides is equal to
$\Rightarrow 2(n-2)$ right angles.

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 3 Lesson 3.1 Solving Linear Equations with One Variable to finish your assignments.

Math in Focus Grade 7 Course 3 A Chapter 3 Lesson 3.2 Answer Key Identifying the Number of Solutions to a Linear Equation

Math in Focus Grade 8 Chapter 3 Lesson 3.2 Guided Practice Answer Key

Tell whether each equation is consistent or inconsistent. Justify your answer by simplifying each equation. Write +, ×, or ÷ for each ?

Question 1.
7(x – 3) – 7x – 21 = 0

Answer:
The equation has a solution,
the equation is consistent.

Explanation:
Given 7(x-3)-7x – 21 ≠ 0,
7x – 21 – 7x -21 ≠ 0 (using the distributive property),
-42 ≠ 0 because -42 ≠ 0 the equation has a solution,
the equation is consistent.

Question 2.
\(5\left(x+\frac{1}{5}\right)\) = 5x + 3
Answer:
Inconsistent,

Explanation:

Question 3.
x + \(\frac{1}{4}\) = –\(\frac{1}{4}\)(4x – 1)
Answer:
The equation has a solution,
the equation is consistent.

Equation:

Tell whether each equation is an identity. Justify your answer by simplifying each equation.
Write +, -, ×, or ÷ for each Write = or ≠ for each

Question 4.

Answer:
Left side is equal to right side,
Equation has an identity,

Explanation:

Question 5.

Answer:
Left side is not equal to right side,
Equation has no identity,

Explanation:

Try substituting some values of x into the equation. If you find that the left side is always equal to the right side, the equation is an identity.

Math in Focus Course 3A Practice 3.2 Answer Key

Tell whether each equation has one solution, no solution, or an infinite number of solutions. Justify your answer.

Question 1.
2x – 3 = -2\(\left(\frac{3}{2}-x\right)\)
Answer:
Equation has one solution,

Explanation:

Question 2.
2x + 5 = -4\(\left(\frac{3}{2}-x\right)\)
Answer:
Equation has one solution,

Explanation:

Question 3.
3x + 5 = 2x – 7
Answer:
x = -12, Equation has one solution,

Explanation:
Given 3x + 5 = 2x -7,
3x – 2x = -7 -5,
x = -12,
Equation has one solution.

Question 4.
5y + (86 – y) = 86 + 4y
Answer:
The equation is identity and
an identity equation has infinetly many solutions,

Explanation:
Given 5y + (86 – y) = 86 + 4y upon solving
5y +86 -y = 86 + 4y,
4y + 86 = 86 + 4y ,the equation is identity and
an identity equation has infinetly many solutions.

Question 5.
0.5(6x – 3) = 3(1 + x)
Answer:
No solution,

Explanation:
Given 0.5(6x -3) = 3(1 + x),
(0.5 X  6x) – (0.5 X 3) = 3 +3x,
3x -1.5 = 3 + 3x
-1.5 ≠ 3, So the equation has no solution,

Question 6.
4(18a — 7) + 40 = 3(4 + 24a)
Answer:
The equation is identity and
an identity equation has infinetly many solutions,

Explanation:
4 X 18a – 4 X 7 + 40 = 12 + 72a
72 a – 28 +40 = 72a+ 12
72a + 12 = 72a+12,
both are equal so infinite solutions.

Question 7.
\(\frac{1}{7}\)(7x – 21) = 8x + 7x – 24
Answer:
Equation has one solution,

Explanation:
Given \(\frac{1}{7}\)(7x – 21) = 8x + 7x – 24,

Question 8.
\(\frac{1}{6}\)(12x – 18) = 2\(\left(x-\frac{3}{2}\right)\)
Answer:
The equation is identity and
an identity equation has infinetly many solutions,

Question 9.
7 – 0.75x = -7\(\left(\frac{3}{28} x+1\right)\)
Answer:
No solution,

Explanation:

Question 10.
6 + 0.5y = -2\(\left(3-\frac{1}{4} y\right)\)
Answer:
No solution,

Explanation:

Question 11.
\(\frac{x-3}{4}\) = 0.25x – 0.75
Answer:
The equation is identity and
an identity equation has infinetly many solutions,

Explanation:

Question 12.
\(\frac{1}{3}\)x + 5 = \(\frac{1}{6}\)(2x – 5)
Answer:
No solution,

Explanation:

Solve. Show your work.

Question 13.
Cabinet A is 5 inches taller than Cabinet B. Cabinet C is 3 inches taller than Cabinet B whose height is x inches.

a) Write algebraic expressions for the heights of cabinets A and C.
Answer:
Cabinet A = 5 + x,
Cabinet C = 3 + x,

Explanation:
Given Cabinet B height is x inches as Cabinet A is 5 inches taller than Cabinet B
Cabinet A = 5 + x and Cabinet C is 3 inches taller than Cabinet B so Cabinet C = 3 + x.

b) If the total height of the three cabinets is (3x + 8) inches, can you solve for the height of Cabinet B? Explain.
Answer:
We cannot solve for the height of Cabinet B
we need numbers.

Explanation:
Given the total height of the three cabinets is (3x + 8) inches,
we have Cabinet A = 5 + x, Cabinet B = x and Cabinet C = 3 + x,
5 + x + x +3 + x = 3x +8, We cannot solve for the height of Cabinet B
we need numbers.

Question 14.
A room’s floor is y meters long. Its width is 5 meters shorter.
If the perimeter of the floor is (4y + 1) meters, can you solve for its length? Explain.
Answer:
There is not answer. Or, the problem needs more definition,

Explanation:
Room Floor Length(L) = y m.Width(W) y – 5 m.
Perimeter = 4y + 1,Find: Length(L),
Using perimeter formula P = 2L + 2W to get an equation to solve.
P = 4y + 1 = 2(y) + 2(y – 5),
4y + 1 = 2y + 2y – 10 => 1 = -10,
There is not answer. Or, the problem needs more definition.

Question 15.
Math Journal
Grace gave her sister a riddle: I have a number x. I add 15 to twice x to get result A.
I subtract 4 from x to get result B. I multiply result B by 3 to get result C.
Result A is equal to result C. Grace’s sister said the riddle cannot be solved,
but Grace thought otherwise. Who is right? Explain.
Answer:
Grace is right she solved the riddle,

Explanation:
Given Grace gave her sister a riddle: I have a number x. I add 15 to twice x to get result A.
A= 15 + 2x, I subtract 4 from x to get result B. B = x -4,
I multiply result B by 3 to get result C. 3B = C ,
Result A is equal to result C, 15 + 2x = C = 3(x-4)
15 + 2x = 3x – 12,
15 + 12 = 3x -2x,
27 = x.
Grace is right she solved the riddle.

Question 16.

Math Journal Look at this “proof” that 2 = 0.

What is wrong with this proof? How can a true statement lead to an inconsistent equation?
Answer:
Substituting values of a and b and a true statement lead to an inconsistent equation,
Inconsistent equations is defined as two or more equations that are
impossible to solve based on using one set of values for the variables.

Explanation:
As when a =1 and b= 1 then (a-b)(a+b) = 0 means
(1-1)(1+1) = 0,0=0 true statement,
0= 0 but if (a-b)(a+b) =0  dividing both sides by (a-b) we get
a + b = 0 substitute for a and b we get 1 + 1 = 0 and 2 = 0 which is inconsistent.

Math in Focus Grade 7 Course 3 A Chapter 3 Lesson 3.1 Answer Key Solving Linear Equations with One Variable

Math in Focus Grade 8 Chapter 3 Lesson 3.1 Guided Practice Answer Key

Solve each linear equation.

Question 1.

Answer:
x = –\(\frac{2}{15}\)

Explanation:
(2x/3) -(2+x)/3 = -4,
2x/3 -2/3 + x /3 = -4,
(2x/3+ x/3) = -4 +(2/3),
(2x+x)/3 = (-12+ 2)/3,
x = -10/3

Question 2.
00.6(1 – x) + 0.2(x – 5) = 10
Answer:
x = -26

Explanation:
Given 00.6(1 – x) + 0.2(x – 5) = 10 upon solving
0.6 – 0.6x + 0.2 x – (0.2 X 5) = 10,
0.6 – 0.4x -1.0 = 10,
0.6 – o.4x =  11,
-0.4 x = 11 – 0.6,
-0.4x = 10.4
x = -10.4 ÷ 0.4
x = -26

Question 3.
\(\frac{3 x}{5}\) + \(\frac{x-1}{3}\) = \(\frac{2}{15}\)
Answer:
\(\frac{1}{2}\),

Explanation:

Write each repeating decimal as a fraction. Show your work.

Question 4.
\(0 . \overline{09}\)
Answer:
9/100,

Explanation:

Question 5.
\(0 . \overline{8}\)
Answer:
8/10,

Explanation:

Question 6.
\(0.0 \overline{6}\)
Answer:
6/100,

Explanation:

Write an equation and solve. Show your work.

Question 7.
Mr. Johnson wants to add a circular pond to his backyard. The backyard is 20\(\frac{1}{2}\) yards long, and the pond wilt be
6\(\frac{1}{4}\) yards across. He wants the pond set back from the house, so that the distance from the pond to the back
fence is half the distance from the pond to the back of the house. How far should the pond be from the back of the house?

Let the distance from the pond to the house be x yards.
So, the distance from the pond to the fence is \(\frac{1}{2}\)x yards.

The length of the backyard is about yards and the width of the pond is about yards. So the total distance of the pond to the house and of the pond to the gate is around yards. of yards is about yards. So, the answer is reasonable.

Answer:
The distance from the pond to the house is 13.666 yards,

Explanation:
Given let the distance from the pond to the house be x yards.
So, the distance from the pond to the fence is \(\frac{1}{2}\)x yards,

The length of the backyard is about 20 1/2 yards and
the width of the pond is about 6 1/4 yards.
So the total distance of the pond to the house and of the pond to the gate is around 201/2yards.
13.66 yards and 6 1/4  yards is about  20 1/2 yards. So, the answer is reasonable.

Question 8.
A packager of tea leaves blends 3.5 pounds of Tea Leaf A with 1.5 pounds of
Tea Leaf B to make a special blend. One pound of Tea Leaf B costs $2 less than
one pound of Tea Leaf A. The packager finds that the
cost of making the blend is $3 per pound, Find the cost of one pound of Tea Leaf B.
Answer:
$3 is the cost of one pound of tea Leaf B,

Explanation:
Given A packager of tea leaves blends 3.5 pounds of Tea ,
Leaf A with 1.5 pounds of Tea,
Leaf B to make a special blend.
One pound of Tea Leaf B costs $2 less than one pound of Tea Leaf A.
The packager finds that the cost of making the blend is $3 per pound,
So the packager has 3.5 pounds means
Leaf A + Leaf B = 3.5 pound,
1.5 pound + Leaf B = 3.5 Pound,
Leaf B = 2 Pounds to make blend,
Given the packager finds that the cost of making the blend is $3 per pound,
therefore $3 is the cost of one pound of tea Leaf B.

Math in Focus Course 3A Practice 3.1 Answer Key

Solve each linear equation. Show your work.

Question 1.
4x – (10 – x) = \(\frac{15}{2}\)
Answer:
x= 7/2,

Explanation:

Question 2.
0.5(x + 1) – 1 = 0.2
Answer:
x= 1.4,

Explanation:

Question 3.
2(x – 1) – 6 = 101(1 – x) + 6
Answer:
x = \(\frac{1}{2}\)

Explanation:
Given 2(x-1)-6 = 101(1-x) + 6,
2x -2 -6 = 101 -101x +6,
2x -8 = 107-101x,
2x +101 x = 107+8,
103 x = 115,
x = 115/103.

Question 4.
8(x – 3) – (x – 3) = 0.7
Answer:
x = 3.1,

Explanation:
Given 8(x – 3) – (x – 3) = 0.7,
8x – 24 – x +3 = 0.7,
7x – 21 = 0.7,
7x = 0.7 + 21,
7x = 21.7,
x = 21.7 ÷ 7,
x = 3.1.

Question 5.
2(x – 4) + 0.5(2 + 8x) = 0
Answer:
x = 7/6,

Explanation:
Given 2(x – 4) + 0.5(2 + 8x) = 0,
2x – 8 + 1.0 + 4x = 0,
6x -7 = 0,
x = 7/6.

Question 6.
5 – 3(x -7 ) = 2(2 – x) – 8
Answer:
x = 30,

Explanation:
Given 5-3(x-7)= 2(2-x)-8,
5 – 3x +21 = 4 -2x -8,
-3x +26 = -2x -4,
26+4 = -2x +3x
30 = x.

Question 7.
3x – 0.4(5 – 2x) = 5.6
Answer:
x = 2,

Explanation:
Given 3x – 0.4(5 – 2x) = 5.6,
3x – 2 + 0.8 x = 5.6,
3x + 0.8 x = 5.6 + 2,
3.8 x = 7.6,
x = 7.6 ÷ 3.8,
x = 2.

Question 8.
6 + \(\frac{1}{3}\)(x – 9) = \(\frac{1}{2}\)(2 – x)
Answer:
x = -12/5,

Explanation:

Question 9.
\(\frac{3 x-2}{8}\) + \(\frac{2-x}{4}\) = \(-\frac{1}{2}\)
Answer:
x = -6,

Explanation:

Question 10.
\(-\frac{x+1}{6}\) -\(\) = \(\frac{1}{3}\)
Answer:
x = 3,

Explanation:

Question 11.
\(\frac{5(x+2)}{3}\) – \(\frac{x-1}{3}\) = 1
Answer:
x = -2,

Explanation:

Question 12.
\(\frac{4(2 x+3)}{5}\) – \(\frac{x+1}{4}\) = \(\frac{31}{5}\)
Answer:
x = 3,

Explanation:

Express each repeating decimal as a fraction. Show your work.

Question 13.
\(0.8 \overline{3}\)
Answer:
5/6,

Explanation:

Question 14.
\(0.0 \overline{8}\)
Answer:
8/99,

Explanation:

Question 15.
\(0 . \overline{1}\)
Answer:
1/9,

Explanation:

Question 16.
\(0.08 \overline{3}\)
Answer:
1/12

Explanation:

Question 17.
\(0.0 \overline{5}\)
Answer:
1/20,

Explanation:

Question 18.
\(0.0 \overline{45}\)
Answer:
9/200,

Explanation:

Solve each problem algebraically. Show your work.

Question 19.
Logan saves $5.50 in dimes and quarters over a week.
He has 20 more dimes than quarters.
Find the number of dimes and quarters he saves.
Answer:
Logan saves 10 quarters and  30 dimes each week,

Explanation:

Question 20.
Maggie makes some fruit punch. She mixes 2\(\frac{1}{2}\) quarts of grape juice with 1\(\frac{1}{2}\) quarts of orange juice. One quart of grape juice costs $1 less than one quart of orange juice. She finds that the total cost of making the fruit punch is $12.50. Calculate the cost of each quart of grape juice and each quart of orange juice.
Answer:
The cost of each quart of orange juice=$3.75,
The cost of each quart of grape juice=$2.75,

Explanation:
The expression for the total cost of making fruit punch is as follows;
Total cost of making fruit punch=
total cost of making grape juice + total cost of making orange juice where;
Total cost of making fruit punch=$12.50,
Total cost of making grape juice=Cost per unit of grape juice×total quantity of grape juice,
Total cost making grape juice=(2 1/2)×g=2.5 g,
Total cost of making orange juice=Cost per unit of orange juice×total quantity of orange juice,
Total cost of making orange juice=(1 1/2)×o=1.5 o,replacing;
12.50=2.5 g+1.5 o…equation 1,
But price per unit of grape juice is one dollars less that the price per unit  of orange juice;
Price per unit of orange juice=o,
Price per unit of grape juice=(o-1), g=o-1
replace the value of g=o-1 in equation 1;
12.50=2.5 g+1.5 o…equation 1
12.50=2.5(o-1)+1.5 o,
12.50=2.5 o-2.5+1.5 o,
2.5 o+1.5 o=12.50+2.50,
4 o=15,
o=15/4=3.75,
g=o-1=3.75-1=2.75,
The cost of each quart of orange juice=o=$3.75,
The cost of each quart of grape juice=g=$2.75.

Question 21.
Ms. Handler walks to work at an average speed of 5 kilometers per hour.
If she increases her speed to 6 kilometers per hour, she will save 10 minutes.

a) Complete the table.

Answer:

Explanation:
Completed the table as shown above,

b) Find the distance she walks.
Answer:
Ms.Hadler walks 5 km,

Explanation:
I assume that the increase in speed is for the whole distance and
that we want to know far she has to walk.
Let X be the distance,
60min/5Km * X km = 12X minutes,
60min/6km * X km = 10X minutes,
10X +10 = 12X
10 = 2X
X = 5km,

Question 22.
Jane is x years old today. Her brother Kenny is 4 years older.
After seven years, their total combined age will be 24 years.
a) Write a linear equation for their total combined age after 7 years.
Answer:
Linear equation for their total combined age after 7 years is 2x+ 18 =24,

Explanation:
Given Jane is x years old today. Her brother Kenny is 4 years older.,
x and x+4,After seven years, their total combined age will be 24 years,
x +7 +x+4+7 =24,
2x + 18 = 24,
so linear equation for their total combined age after 7 years is 2x +18 = 24.

b) Find Jane’s age today.
Answer:
Janes age today is 3,

Explanation:
Equation is 2x + 18 = 24,
2x = 24-18 = 6,
x = 6/2 = 3 years.

Question 23.
Casper bought some pencils at 50¢ each. He had $3 left after the purchase.
If he wanted to buy the same number of note pads at 80.
Answer:
.50p+3 = .80p-1.50,

Explanation:
.50p+3 = .80p-1.50,
.3p = 4.50,
p = 15,
so, 15 pencils cost $7.50, meaning he started out with $10.50
15 notepads cost $12.00, so he’d be $1.50 short.

Question 24.
Alexis earns 2\(\frac{1}{2}\) times as much as Gary in a day.
James earns $18 more than Gary in a day. If the total daily salary of all three people is $306,
find Alexis’s salary.
Answer:
Alexis salary is $160,

Answer:
Number of shirts sold are 40,

Explanation:
Given a store has y shirts. It sold most of them for $16 each, and the last dozen
were sold on sale for $14 each. If it sold all the shirts for $616,
y=(616-(14 X 12))/16 + 12,
y = (616-168)/16 + 12,
y = 448/16 +12,
y = 28 +12 = 40,
therefore number of shirts sold are 40.

Question 26.
There are 40 questions on a class test. Six points are given for each
correct answer and three points are deducted for each wrong or missing answer.
Find the number of correct answers for a test score of 105.
Answer:
The number of correct answers are 25,

Explanation:
Given there are 40 questions on a class test. Six points are given for each
correct answer and three points are deducted for each wrong or missing answer.
Let x be correct answers incorrect are 40 – x,
so 6x -3(40-x)=105,
6x -120 +3x = 105,
9x= 105+120,
9x = 225,
x = 225 ÷ 9,
x = 25, therefore correct answers are 25 and wrong answers are (40-25)= 15.

Question 27.
Math Journal Georgina was given that the length of a rectangle was
2.5 inches longer than its width, and that the perimeter of the rectangle was 75.4 inches.
She found the length and width algebraically.
How could she use estimation to check if her answers were reasonable?
Answer:
length is 20.1 inches and width is 17.6 inches,

Explanation:
Given Georgina was given that the length of a rectangle was
2.5 inches longer than its width, and that the perimeter of the rectangle was 75.4 inches.
Let length is and width is w so
perimeter = 2(l+w),
75.4 = 2(2.5 +w + w),
75.4 =2(2.5+2w),
37.7 = 2.5 +2w,
2w = 37.7-2.5,
2w = 35.2,
w = 35.2 ÷ 2,
w = 17.6, therfore width is 1
so length = 17.6 + 2.5 = 20.1 inches.

Question 28.
Math Journal Consider the decimal \(0 . \overline{9}\).
a) Find the fraction equivalent of \(0 . \overline{9}\).
Answer:
The fraction eqivalent is 9/10,

Explanation:
0.9 is a decimal fraction, 1 place means tenths,
0.9 is 9/10.

b) The decimal \(0 . \overline{9}\) can be thought of as being equal to the following sum, in which the pattern shown continues forever.
0.9 + 0.09 + 0.009 + 0.0009 + …
How can thinking about this sum help you explain the result you saw in a)?
Answer:
Thinking about this sum helped in explaining the result in tenths place,

Explanation:
Every place value after the decimal point can be expressed as a fraction
with $10$ to some power in the denominator.
The first place after the decimal point is called the “tenths place” If I had $0.9$,
I would have $9$ tenths, or as a fraction $\frac{\mathrm{9/}}{10}$.
The second place after the decimal is called the “hundredths place”.
The question above asks for $0.09$. This would be $9$ hundredths, or as a fraction $\frac{\mathrm{9/}}{\mathrm{100\; and\; so\; on\; thinking\; about\; this\; sum\; helped\; in\; explaining\; the\; result\; in\; tenths\; place.}}$

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 3 Algebraic Linear Equations to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 3 Answer Key Algebraic Linear Equations

Math in Focus Grade 8 Chapter 3 Quick Check Answer Key

Explain why each pair of equations is equivalent or not equivalent.

Question 1.
x + 4 = 10 and x – 1 = 3
Answer:
Not equivalent,

Explanation:
Given x + 4 = 10 and x-1 =3 means x = 3+1 =4,
if we substitute x = 4 in x + 4 = 4 + 4 = 8 ≠ 10,
so equations are not equivalent.

Question 2.
\(\frac{1}{5}\)x = 4 and x = 20
Answer:
Equivalent,

Explanation:
Given \(\frac{1}{5}\)x = 4 and x = 20 upon solving
\(\frac{1}{5}\)x = 4 we get x = 4 X 5 = 20 therefore
pair is equivalent as x = 20.

Question 3.
0.5x + 1 = 1.5x and 2x = 2
Answer:
Equivalent,

Explanation:
Given 0.5x +1 = 1.5x and 2x =2 means x= 2/2 equal to 1,
so 0.5 (1) + 1 = 1.5 X 1 is 0.5 +1 = 1.5 , So 1.5 = 1.5,therefore
both equations are equivalent.

Question 4.
2(x + 9) = 14 and 2(x – 7) = -18
Answer:
Equivalent,

Explanation:
Given 2(x -7) = -18 upon solving
2x – 14 = -18,
2x = -18+14,
2x = -4 therfore x = -4/2 = -2 now substituting
2(-2 -7) = – 4 -14 = -18 which is equal to – 18
So both equations are equivalent.

Write a linear equation for each situation. State the independent and dependent variables for each equation.

Question 5.
A manufacturer produces beverages in small and large bottles.
Each small bottle contains s liters of beverages.
Each large bottle contains t liters, which is 1 more liter than the
quantity in the small bottle. Express t in terms of s.
Answer:
t = (s + (sX1)) liters,

Explanation:
Given A manufacturer produces beverages in small and large bottles.
Each small bottle contains s liters of beverages.
Each large bottle contains t liters, which is 1 more liter than the
quantity in the small bottle. Expressing t in terms of s as t = (s + (sX1))  liters.

Question 6.
Hazel is 4 years younger than Alphonso. Express Alphonso’s age, a, in terms of Hazel’s age, h.
Answer:
h=(a-4),

Explanation:
Given Hazel is 4 years younger than Alphonso.
Expressing Alphonso’s age, a, in terms of Hazel’s age, h.
h=(a-4).

Question 7.
A bouquet of lavender costs $12. Find the cost, C, of n bouquets of lavender.
Answer:
C = $12n

Explanation:
Given a bouquet of lavender costs $12, the cost, C, of n bouquets of lavender is
c = $12 X n = $12n.

Question 8.
The distance, d miles, traveled by a bus is 40 times the time,t hours,
used for the journey. Find d in terms of t.
Answer:
d = 40t,

Explanation:
Given the distance, d miles, traveled by a bus is 40 times the time,t hours,
used for the journey is d = 40t.

Solve each equation.

Question 9.
4x = 14 + 2x
Answer:
x =7,

Explanation:
Given 4x = 14 + 2x, upon solving
4x – 2x = 14,
2x = 14,
x = 14/2 = 7.

Question 10.
\(\frac{1}{3}\)v = 2 – \(\frac{2}{9}\)v
Answer:
v = \(\frac{18}{5}\),

Explanation:
Given \(\frac{1}{3}\)v = 2 – \(\frac{2}{9}\)v,
\(\frac{1}{3}\)v = \(\frac{18 -2v}{9}\),
\(\frac{9}{3}\)v = 18 -2v,
3 v = 18 -2v,
5v = 18, therefore v = \(\frac{18}{5}\).

Question 11.
c + 2(1 – c) = 10 – 3c
Answer:
c = 4,

Explanation:
Given c + 2(1 – c) = 10 – 3c,
c + 2 – 2c = 10 -3c,
2 – c = 10 – 3c,
3c-c = 10-2,
2c = 8,
c = 4.

Question 12.
3(2 + 3x) = 13(x + 2)
Answer:
x = -5,

Explanation:
Given 3(2+3x) = 13(x +2) upon solving
6 + 9x = 13x + 26,
6-26 = 13x – 9x,
-20 = 4 x,
x = -5.

Write the decimal for each fraction. Use bar notation.

Question 13.
\(\frac{3}{18}\)
Answer:

Explanation:
A bar notation, which is a way to show repeating numbers after a decimal point,
here when 3 divides by 18 we get 0.166666 where 6 is repeating so we keep a bar on 6
as shown above.

Question 14.
\(\frac{14}{99}\)
Answer:

Explanation:
When 14 divides by 99 we get 0.14141414 where 14 is repeating so we keep a bar on 14
as shown above.

Question 15.
\(\frac{13}{12}\)
Answer:

Explanation:
When 13 divides by 12 we get 1.083333 where 3 is repeating so we keep a bar on 3
as shown above.

Question 16.
\(\frac{5}{27}\)
Answer:

Explanation:
When 5 divides by 27 we get 0.185185 where 185 is repeating so we keep a bar on 185
as shown above.

Go through the Math in Focus Grade 8 Workbook Answer Key Cumulative Review Chapters 1-2 to finish your assignments.

Math in Focus Grade 8 Course 3 A Cumulative Review Chapters 1-2 Answer Key

Concepts and Skills

Write the prime factorization of each number in exponential notation. (Lesson 1.1)

Question 1.
16,807
Answer:
Prime factorization of 16,807 is 7×7×7×7×7×7
The number in the exponential notation is 7⁶.

Question 2.
25,920
Answer:
Prime factorization of 25,920 is 2×2×2×2×2×2×3×3×3×3×5.
The number in the exponential form is 2⁶ × 3⁴ × 5.

Simplify each expression. Write your answer in exponential notation. (Lessons 1.2, 1.3)

Question 3.
\(\frac{\left[\left(\frac{3}{5}\right) \cdot\left(\frac{3}{5}\right)^{3}\right]^{4}}{\left[\left(\frac{3}{5}\right)^{2}\right]^{2}}\)
Answer:
((3/5) • (3/5)³)⁴/((3/5)²)²
⅗ • ( ⅗)³)⁴ • ((⅗)²)²
⅗ • (⅗)¹² • (⅗)⁴
Bases are equal powers should be added.
⅗ • (⅗)¹⁶
(⅗)¹⁷

Question 4.
(a⁶ • a⁷)³ ÷ (4a³)²
Answer:
Given (a6 • a7)3 ÷ (4a3)2
(a⁶)³ • (a⁷)³ ÷ (4a)⁶
a¹⁸ • a²¹ ÷ 4a⁶
Bases are equal powers should be added.
a³⁹/4a⁶
¼ × a³⁹/a⁶
¼ × a³³

Simplify each expression. Write your answer using a positive exponent. (Lessons 1.2, 1.3, 1.4, 1.5)

Question 5.
\(\frac{6^{3} \cdot 15^{3}}{\left(7^{0}\right)^{3}}\)
Answer:
6³ • 15³/(7⁰)³
= 90³/7⁰
= 90³/1
= 90³

Question 6.
\(\frac{2^{8} \cdot(-3)^{8} \cdot 3^{0}}{5^{-8}}\)
Answer:
Given 2⁸ • (-3)⁸ • 3⁰/5^-8
= -6⁸ • 1/5^-8
= -6⁸/5^-8

Question 7.
[12² • 3²]³ ÷ 3⁶
Answer:
Given [122 • 32]3 ÷ 36
12⁶ • 3⁶ ÷ 3⁶
36⁶/3⁶
12⁶/1⁶

Question 8.
(16⁷ ÷ 16⁴) • \(\frac{\left(5^{0}\right)^{3}}{2^{3} \cdot 4^{3}}\)
Answer:
Given (167 ÷ 164) • (5⁰)³/2³ • 4³
16⁷/16⁴ • 5⁰/2³ • 4³
16³ • 1/2³ • 4³
64³ • (½)³

Question 9.
8^-2 • \(\frac{3^{0} \cdot 8^{-3}}{4^{-5}}\) .
Answer:
Given 8-2 • 3⁰ • 8-3/4^-5
8-2 • 1 • 8-3/4^-5
8-2 • 8-3/4^-5
8-5/4^-5
2-5/1^-5

Question 10.
6^-4 • (5⁰)^-4 • (\(\frac{1}{3}\))^-4 ÷ 3^-4
Answer:
6^-4 (5⁰)^-4 ∙ (\(\frac{1}{3}\))^-4 ÷ 3^-4 We are given the expression:
6^-4 ∙ (5⁰)^-4 ∙ (\(\frac{1}{3}\))^-4 ÷ 3^-4 Simplify:
= 6^-4 ∙ \(\frac{1}{3^{-4}} \cdot \frac{1}{3^{-4}}\)
= \(\frac{6^{-4}}{3^{-4}} \cdot \frac{1}{\frac{1}{3^{4}}}\)
= \(\left(\frac{6}{3}\right)^{-4} \cdot 3^{4}\)
= 2^-4 ∙ 3⁴
= \(\frac{3^{4}}{2^{4}}\)
= \(\left(\frac{3}{2}\right)^{4}\)

Evaluate the square roots of each number. Round your answer to the nearest tenth when you can. (Lesson 1.6)

Question 11.
576
Answer:
576 = (24)²
Square of 576 is 24
24 rounded to the nearest tenth is 20

Question 12.
1,003.4
Answer:
1,003.4 = (31.6764897046)²
Square of 1,003.4 is 31.6764897046
31.6764897046 rounded to the nearest tenth is 30

Evaluate the cube root of each number. Round your answer to the nearest tenth when you can. (Lesson 1.6)

Question 13.
\(\frac{27}{216}\)
Answer:
27/216
0.125
0.125 = (0.5)³
Cubic root of 27/216 is 0.5

Question 14.
-629.5
Answer:
-629.5 = (-8.57035039349)³
Cubic root of -629.5 is -8.57035039349

Evaluate each expression and write your answer in scientific notation. Identify the greater number. (Lessons 2.1, 2.2, 2.3)

Question 15.
3.27 • 10¹¹ + 3.13 • 10¹¹ and 9.28 • 10¹¹ – 4.15 • 10¹¹
Answer:
3.27 • 1011 + 3.13 • 1011
3.27 + 3.13 • 10¹¹
6.4 • 10¹¹
And 9.28 • 1011 – 4.15 • 1011
9.28 – 4.15 • 10¹¹
5.13 • 10¹¹
6.4 • 10¹¹ is the greater number

Question 16.
9.1 • 10^-5 – 8.2 • 10^-6 and 1.2 • 10^-6 – 5.5 • 10^-7
Answer:
9.1 • 10^-5 – 8.2 • 10^-6
9.1 • 10^-5 – 0.82 • 10^-5
9.1 – 0.82 • 10^-5
8.28 • 10^-5
And 1.2 • 10-6 – 5.5 • 10^-7
1.2 • 10-6 – 0.55 • 10^-6
1.2 – 0.55 • 10^-6
0.65 • 10^-6
8.28 • 10^-5 is the greater number.

Question 17.
8.4 • 10⁵ • 2 • 10⁵ and 3.2 • 10^-7 • 2 • 10^-5
Answer:
8.4 • 10⁵ • 2 • 10⁵
8.4 • 10⁵ • 2 • 10⁵
16.8 • 10¹⁰
And 3.2 • 10^-7 • 2 • 10^-5
6.4 • 10^-12
16.8 • 10¹⁰ is the greater number.

Question 18.
9.1 • 10³ ÷ (7 • 10⁵) and 7.2 • 10^-4 ÷ (1.2 • 10^-4)
Answer:
9.1 • 10³ ÷ (7 • 10⁵)
9.1/7 • 10³/10⁵
1.3 • 1/10²
1.3 • 10-2
And 7.2 • 10^-4 ÷ (1.2 • 10^-4)
7.2/1.2 • 10^-4/10^-4
6 • 1
6 = 0.6 • 10¹
0.6 • 10¹ is the greater number.

Write each measurement in the appropriate unit in prefix form. (Lesson 2.2)

Question 19.
0.000020 meter
Answer:
0.02 • 10³ meter
0.02 millimeter

Question 20.
0.070 gram
Answer:
Given 0.070 gram
0.07 •10^-3 gram

Question 21.
35,000,000 bytes
Answer:
Given 35000000 bytes
35 • 10⁶ bytes
0.035 kilobytes

Question 22.
42,000 volts
Answer:
Given 42,000 volts
42 • 10³ volts
42 millivolts.

Problem Solving

Solve. Show your work.

Question 23.
The total surface area of a cube is 4,704 square inches. What is the length of each side? (Chapter 1)

Answer:
From the given question
Total surface area = 4,704 in².
We know that
The total surface area of a cube is 6a²
Where a = side of the cube
6a² = 4704
a² = 4704/6
a² = 784
a = √784
a = 28 in

Question 24.
The volume of a spherical balloon is 12.348π cubic feet. (Chapter 1)
a) Find its radius. Round to the nearest tenth.
Answer:
Given that the volume of a spherical balloon is 12.348π cubic feet.
We know that
The volume of the spherical balloon is 4/3 × πr³
4/3 × πr³ = 12.348π cubic feet.
r³ = 12.348π/4/3× π
r³ = 12.348/ 1.33
r³ = 9.261
r = 2.1
2.1 rounded to the nearest tenth is 2.1

b) Air is pumped into the balloon, so that its radius doubles every 10 seconds. Using 3.14 as an approximation for n, find its surface area after 30 seconds. Round to the nearest tenth.
Answer:
Let r be the radius of the balloon
The radius of the balloon doubles for every 10 seconds.
For 10 seconds radius = r²
For 20 seconds radius = r³
For 30 seconds radius = r⁴
Therefore r⁴ = 30
r = ± 2.340
2.340 rounded to the nearest tenth is 2.3

Question 25.
An oxygen atom has a total of 8 protons. If the mass of one proton is 1.67 • 10^-24 gram, find the total mass of the protons in the oxygen atom. Write your answer in scientific notation. Round the coefficient to 3 significant digits. (Chapters 1, 2)
Answer:
Given that the oxygen atom has a total of 8 protons
Mass of one proton = 1.67 • 10-24
Mass of 8 protons = 8
Total mass of a protons = 8 × 1.67 • 10-24
= 13.36• 10-24
13.36• 10-24 Round the coefficient to 3 significant digits is 13.4

Question 26.
The table lists the energy in Calories contained in 100 grams of fruits. (Chapter 2)

a) Calculate the total energy of the three fruits. Write your answer in scientific notation.
Answer:
Given that the energy in calories contained in 100 grams of apples = 4.9 • 10⁴
The energy in calories contained in 100 grams of oranges = 6.2 • 10⁴
The energy in calories contained in 100 grams of pear = 3.5 • 10⁴
Total energy in all the three fruits = 4.9 • 10⁴ + 6.2 • 10⁴ + 3.5 • 10⁴
= 4.9 + 6.2 + 3.5 • 10⁴
= 14.6 • 10⁴

b) Find the difference in energy contained between 100 grams of apple and 100 grams of pear.
Answer:
Given that the energy in calories contained in 100 grams of apples = 4.9 • 10⁴
The energy in calories contained in 100 grams of pear = 3.5 • 10⁴
Difference = 4.9 • 10⁴ – 3.5 • 10⁴
= 1.4 • 10⁴
The difference in the energy contained between 100 grams of apple and 100 grams of pear is 1.4 • 10⁴

c) How many times more energy does 100 grams .of orange have compared to 100 grams of apple? Round to the nearest tenth.
Answer:
Given that the energy in calories contained in 100 grams of oranges = 6.2 • 10⁴
The energy in calories contained in 100 grams of apples = 4.9 • 10⁴
6.2 • 10⁴ – 4.9 • 10⁴
6.2 – 4.9 • 10⁴
1.3 • 10⁴
The energy in calories contained in 100 grams of oranges is 1.3 • 10⁴ times more than the 100 grams of apple

Question 27.
Jim deposits $2,000 in a bank, which gives 6% interest, compounded yearly. Use the formula A = P (1 + r )ⁿ to find the amount of money in his account after 15 years. A represents the final amount of investment, P is the original principal, r is the interest rate, and n is the number of years it was invested. (Chapter 1)
Answer:
Given that
Jim deposits $2,000 in a bank
It gives 6% interest
Using the formula A = P (1 + r )n
P is the original principal
r is the interest rate
n is the number of years
A = 2000(1+6)¹⁵
A = 2000(7)¹⁵
A = 2000 • 7¹⁵

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 2 Review Test to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 2 Review Test Answer Key

Concepts and Skills

Tell whether each number is written correctly in scientific notation. If incorrectly written, state the reason.

Question 1.
10 • 10²
Answer:
10 • 10² is correct in scientific notation, because the given notation is equal to 10.

Question 2.
0.99 • 10¹²
Answer:
The 0.99 • 10¹² is incorrect in scientific notation.
The coefficient is less than 1

Question 3.
1.4 • 10²
Answer:
1.4 • 10² is Correct in scientific notation, because the given notation is in between 1 to 10.

Question 4.
0.4 • 10²⁵
Answer:
0.4 • 10²⁵ is incorrect in scientific notation.
Because the coefficient is less than 1.

Write each number in scientific notation.

Question 5.
714,000
Answer:
714,000 in the scientific notation is
7.14 •10⁵

Question 6.
0.00087
Answer:
0.00087 in the scientific notation is 0.87 • 10^-3

Write each number in standard form.

Question 7.
3.46 • 10²
Answer:
3.46 • 10²
10² = 100
3.46 × 100 = 346.

Question 8.
5.4 • 10⁴
Answer:
5.4 • 10⁴
5.4 × 10,000
= 54,000

Identify the greater number in each pair of numbers.

Question 9.
7.8 • 10^-5 and 5.4 • 10^-7
Answer:
7.8 • 10^-5 is greater than the 5.4 • 10^-7
Because 7.8 • 10^-5 has a big exponent.

Question 10.
1.4 • 10^-5 and 6 • 10^-4
Answer:
6 • 10^-4 is greater than the 1.4 • 10^-5
Because 6 • 10^-4 has a big exponent.

Question 11.
6.5 • 10^-15 and 9.3 • 10^-12
Answer:
9.3 • 10^-12 is greater than the 6.5 • 10^-15
Because 9.3 • 10^-12 has a big exponent.

Question 12.
3.5 • 10^-2 and 4 • 10^-3
Answer:
3.5 • 10^-2 is greater than the 4 • 10^-3
Because 3.5 • 10^-2 has a big exponent.

Evaluate. Write your answer in scientific notation. Round the coefficient to the nearest tenth.

Question 13.
2.44 • 10³ + 1.9 • 10⁵
Answer:
2.44 • 10³ + 1.9 • 105
Rewrite the numbers to have the same power of 10
0.0244 • 10⁵ + 1.9 • 10⁵
Now add the numbers
0.0244 + 1.9 = 1.9244
Rewrite the number in scientific notation is
1.9244 • 10⁵
1.9244 rounded to the nearest tenth is 1.9

Question 14.
3.12 • 10^-3 – 3 • 10^-3
Answer:
3.12 • 10^-3 – 3 • 10^-3
Now subtract the numbers
(3.12 – 3) • 10^-3
Rewrite the number in scientific notation is
0.12 • 10^-3

Question 15.
2.4 • 10^-2 • 5 • 10^-1
Answer:
2.4 • 10^-2 • 5 • 10^-1
Here bases are equal powers should be added.
2.4 • 5 • 10^-3
12 • 10^-3

Question 16.
3.2 • 10⁸ ÷ (1.6 • 10⁴)
Answer:
3.2 • 10⁸ ÷ (1.6 • 104)
3.2/1.6 × 10⁸/10⁴
3.2/1.6 × 10⁴
2 • 10⁴

Express each of the following in prefix form. Choose the most appropriate unit.

Question 17.
2.8 • 10³ meters
Answer:
2.8 • 10³ meters
We have to using the prefix form
2.8 • 10¹ • 10²
2.8 • 10¹ centimeters
28 centimeters

Question 18.
1.5 • 10^-6 meter
Answer:
1.5 • 10^-6 meter
We have to use the prefix form
1.5 • 10^-6 meter
1.5 micrometers

Question 19.
6.4 • 10⁹ bytes
Answer:
6.4 • 10⁹ bytes
We have to use the prefix form
64 • 10⁸ bytes

Question 20.
4.8 • 10^-9 gram
Answer:
4.8 • 10^-9 gram
We have to use the prefix form
4.8 • 10^-6 •10^-3 gram
4.8 •10^-3 micrograms
0.0048 micrograms

Problem Solving

The table shows the length of two organisms. Use the table to answer questions 21 to 23.

Question 21.
Which organism is longer?
Answer:
Given that the length of eriophyid mite = 250 micrometers
The length of patiriella parvivipara = 5 millimeter.
1 micrometer = 0.001 millimeters
250 micrometers = 0.001 • 25 = 0.025 millimetres
Therefore, The length of patiriella parvivipara is longer.

Question 22.
Express the length of the eriophyid mite in millimeters.
Answer:
Given that the length of the eriophyid mite 250 micrometers
1 micrometer = 0.001 millimeters
250 micrometers = 0.001 • 25 = 0.025 millimeters
Therefore the length of the eriophyid mite in millimeters is 0.025

Question 23.
Write each length in scientific notation using the basic unit.
Answer:
Given that the length of eriophyid mite = 250 micrometers
0.025 • 10⁴
The length of patiriella parvivipara = 5 millimeter.
5 millimetres = 0.5 • 10¹

The top five materials used in the automotive industry in the United States in 2000 are as shown in the table. Use the table to answer questions 24 to 26. Write your answers in scientific notation. Round coefficients to the nearest tenth.

Question 24.
How much more plastic was used than aluminum?
Answer:
Given that the plastic material used in the automotive industry is 46240 = 46.240 • 10³
The aluminum Material used in the automotive industry is 11,320 = 11.320 • 10³
Plastic material – aluminium material = 46.240 • 10³ – 11.320 • 10³ = 34.92 = 3.492 • 10¹
Plastic material is 3.492 • 10¹ consumption is used more than aluminum.
34920 rounded to the nearest tenth is 34920

Question 25.
How much more steel was used than glass?
Answer:
Given that the steel material used in the automotive industry is 9.894 • 10⁷
9.894 • 10⁷
The glass material used in the automotive industry is 54,17,000 = 0.5417000 • 10⁷
Steel material – glass material = 9.894 • 10⁷ – 0.5417000 • 10⁷ = 9.3523 = 93.523 • 10^-1
93.523 • 10^-1 rounded to the nearest tenth is 93.5 • 10^-1

Question 26.
Find the total consumption of these materials used by the automotive industry in 2000.
Answer:
Given that the plastic material used in the automotive industry is 46,240 = 46.240 • 10³
The aluminum material used in the automotive industry is 11,320 = 11.320 • 10³
The steel material used in the automotive industry is 9.894 • 10⁷
The glass material used in the automotive industry is 54,17,000 = 0.5417000 • 10⁷
The rubber material used in the automotive industry is 2.86 • 10⁶
Total consumption = 46.240 • 10³ + 11.320 • 10³ + 9.894 • 10⁷ + 0.5417000 • 10⁷ + 2.86 • 10⁶ = 70.8547 • 10²⁶
70.8547 • 10²⁶ rounded to the nearest tenth is 70.85 • 10²⁶

Question 27.
The table shows the weights of some animals. Round your answers to the nearest tenth.

a) About how many times greater than the weight of a walrus is the weight of a hippopotamus?
Answer:
Given that the weight of a walrus = 4.4 • 10³
And the weight of a hippopotamus = 9.9 • 10³
weight of hippopotamus – weight of walrus = 9.9 • 10³ – 4.4 • 10³ = 5.5
Weight of a hippopotamus is 5.5 times greater than that of a walrus.
5.5 rounded to the nearest tenth is 5.5

b) About how many times greater than the weight of a hippopotamus is the weight of an African bush elephant?
Answer:
Given that the weight of a hippopotamus = 9.9 • 10³
The weight of a African bush elephant = 2.706 • 10⁴ = 27.06 • 10³
Weight of African bush elephant – weight of a hippopotamus = 27.06;• 10³ – 9.9 • 10³ = 17.16
The weight of an African bush elephant is 17.16 times greater than the hippopotamus.
17.16 rounded to the nearest tenth is 17.2

Question 28.
The floor of the Palace of the Parliament in Romania measures 8.9 • 10² feet by 7.9• 10² feet. The building reaches 2.82 • 10² feet above ground, and 3.02 • 10² feet below ground. Use the formula for the volume of a rectangular prism to estimate the volume inside the palace.
Answer:
Length: 8.9 • 10² feet We are given the dimensions of the rectangular prism:
Width: 7.9 • 10² feet
Above height: 2.82 • 10² feet
Below height: 3.02 • 10² feet
Length • Width • (Above height + Below height) We determine the volume of the prism:
= 8.9 • 10² • 7.9 • 10² • (2.82 • 10² + 3.02 • 10²)
= 8.9 • 10² 7.9 • 10² • [(2.82 + 3.02) • 10²]
= 8.9 • 10² • 7.9 • 10² • 5.84 • 10²
= 8.9 • 7.9 • 5.84 • 10² • 10² • 10²
= 410.6104 • 10²⁺²⁺²
= 410.6104 • 10⁶
= 4.106104 • 10⁸
≈ 4.1 • 10⁸ cube feet

Question 29.
The Stockholm Globe Arena in Sweden resembles a hemisphere. It has an inner diameter of about 1.1 • 10² meters. Find the approximate volume of the building. Use 3.14 as an approximation for π.
Answer:
Given that the inner diameter of a hemisphere is 1.1 • 10².
Assume that the building is in the shape of a hemisphere.
D = 2r
1.1 • 10² = 2r
r = 1.1 • 10²/2 = 110/2 = 55
Volume of a hemisphere = ⅔ πr³ cubic units
= ⅔ • 3.14 • (55)³
= 348,278.33

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Lesson 8.4 Dilations to score better marks in the exam.

Math in Focus Grade 7 Course 3 B Chapter 8 Lesson 8.4 Answer Key Dilations

Math in Focus Grade 8 Chapter 8 Lesson 8.4 Guided Practice Answer Key

Solve.

Question 1.
Which triangles are dilations of one another? Explain.

a)

Answer:
Δ STU and Δ PTR. They have a center of dilation T and the sides of Δ PTR are twice as long as the sides of Δ STU.

b)

Answer:
Δ PQS and Δ PUT. They have a center of dilation, P, and the sides of Δ PUT are 1 1/2 times as long as the sides of Δ PQS.

Copy and complete.

Question 2.
A rectangle has coordinates A (5, 1), B (3, 1), C (3, 4), and D (5, 4).

a) Find the length and width of ABCD.
The length of ABCD is units. Its width is units.
Answer:

The length of ABCD is 3 units. Its width is 2 units.

b) Find the length and width of the image of ABCD when dilated with scale factor 2.
Length of image: • = units
Width of image: • = units
Answer:
scale factor = 2
Length of image: 3 • 2 = 6 units
Width of image: 2 • 2 = 4 units

c) Find the length and width of the image of ABCD when dilated with scale factor \(\frac{1}{2}\).
Length of image: • = units
Width of image: • = units
Answer:
Scale factor \(\frac{1}{2}\).
Length of image: 3 • \(\frac{1}{2}\) = 3\(\frac{1}{2}\). units
Width of image: 2 • \(\frac{1}{2}\). = 1 units

d)
What are the coordinates of the image rectangle under each dilation if the center of dilation is at the origin?

Answer:
Scale factor 2:
A (5, 1) = 5 × 2 , 1 × 2 = (10, 2)
B (3, 1) = 3 × 2, 1 × 2 = (6, 1)
C (3, 4) = 3 × 2, 4 × 2 = (6, 8)
D (5, 4) = 5 × 2, 4 × 2 = (10, 8)
Scale factor 1/2:
A (5, 1) = 5 × 1/2 , 1 × 1/2 = (2.5, 0.5)
B (3, 1) = 3 × 1/2, 1 × 1/2 = (1.5, 0.5)
C (3, 4) = 3 × 1/2, 4 × 1/2 = (1.5, 2)
D (5, 4) = 5 × 1/2, 4 × 1/2 = (2.5, 2)

You may want to draw the rectangle and its images on the coordinate plane to solve c).

Technology Activity

Materials:

• geometry software

Explore The Properties Of Dilations With Geometry Software

Step 1.
Draw a line segment AB using a geometry software program.

Step 2.
Select the Dilate function, within the Transform menu. Enter the scale factor 2 to dilate the line segment about the origin. Record your results in a table of coordinates.

Step 3.
Describe how the image of \(\overline{A B}\) is related to \(\overline{A B}\).

Step 4.
Repeat Step 1 to Step 3 using a rectangle as the original figure. For Step 2, enter the scale factor \(\frac{1}{2}\).

Step 5.
Repeat Step 1 to Step 3 using a rectangle as the original figure. For Step 2, enter the scale factor -2.

Math Journal Observe any changes in the size or shape of the figure after the dilation. Which of these properties does a dilation preserve: lengths, shape, parallel lines, or perpendicular lines? Explain.

Copy and complete on graph paper.

Question 3.
The management of a swimming pool built a springboard above the pool. The height of the springboard is a dilation of the depth of the pool with center at the origin, O and scale factor –\(\frac{1}{3}\). The depth of the pool is 4.5 meters, represented by \(\overline{S T}\) on the coordinate plane. The floor is represented by the positive x-axis and the surface of the water is represented by the negative x-axis. Draw the location and height of the stand for the springboard, \(\overline{U V}\), on a copy of this vertical cross section of the pool.

Answer:

Use graph paper. Use 1 grid square on both axes to represent 1 unit for the interval from -7 to 4.

Question 4.
The triangles are each mapped onto their images by a dilation. Draw each triangle and its image on a coordinate plane. Then mark and label C as the center of dilation. Find the scale factor for each triangle.
a)

Answer:

C(-1, 1), scale factor = -2

b)

Answer:

C(3, 4) and scale factor = 3

Math in Focus Course 3B Practice 8.4 Answer Key

Tell whether each transformation is a dilation. Explain.

Question 1.

Answer:
Yes, Δ ABC and Δ ADE have center of dilation at A and the sides of Δ ADE are twice as long as the sides of Δ ABC.

Question 2.

Answer: No it is not a dilation

Solve. Show your work.

Question 3.
Nikita wants to make a mosaic for a T-shirt’s design. She makes some dilated copies of a drawing with a photocopier. The drawing is 6 inches by 4 inches. Find the length and width of each copy with the scale factor given in a) to d). State whether each copy is an enlargement or reduction of the drawing.
a) 1.5
Answer:
Scale factor = 1.5
(6, 4) = 6 × 3/2, 4 × 3/2 = 9/2 in, 6 in, Enlargement

b) 2
Answer:
Scale factor = 2
(6, 4) = 6 × 2, 4 × 2 = (12, 8), Enlargement

c) \(\frac{1}{4}\)
Answer:
Scale factor = \(\frac{1}{4}\)
(6, 4) = 6 × \(\frac{1}{4}\), 4 × \(\frac{1}{4}\) = 1.5 in, 1 in, Reduction

d) 140%
Answer:
Scale factor = 140% = 1.4
(6, 4) = 6 × 1.4, 4 × 1.4 = 8.4 in, 5.6 in. Enlargment

Copy and complete on graph paper.

Question 4.
Timothy uses a lens to view a 2-inch pencil that is represented by \(\overline{\mathrm{AB}}\) on the coordinate plane. \(\overline{\mathrm{AB}}\) is mapped onto \(\overline{A^{\prime} B^{\prime}}\) by a dilation with center at the origin, O. Draw each image for the given scale factor.

a) Scale factor -0.5

Answer:
Observe that the center of dilation is at the origin then the method of multiplying the coordinates of \(\overline{A B}\) to the scale factor can be used. Let j be the coordinates after the dilation of A whose coordinate is (3, 2).
j = A ·  (-0.5)    Formula.
= (3, 2) · (-0.5)    Substitute the Coordinates of A.
= (-1.5, -1)    Multiply.
Thus, point A’ is on (-1.5, -1). Next, Let k be the coordinates after the dilation of B whose coordinate is (3,0).
k = B · (-0.5)    Formula.
= (3, 0) · (-0.5)   Substitute the Coordinates of B.
= (-1.5. 0)    Multtiply.
Thus, point B’ is on (-1.5, 0). Therefore, the points of \(\overline{A^{\prime} B^{\prime}}\) are on (-1.5, -1) and (-1.5, 0). The illustration for the image of AB would be as depicted below.

b) Scale factor 0.5

Answer:
Since the center of dilation is at the origin then the method of multiplying the coordinates of \(\overline{A B}\) to the scale factor can be used. Let l be the coordinates after the dilation of A whose coordinate is (2, 3).
l =  A · 0.5    Formula.
= (2, 3) · 0.5    Substitute the Coordinates of A.
= (1, 1.5) Multiply.
Thus, point A’ is on (1, 1.5). Next, Let m be the coordinates after the dilation of B whose coordinate is (2, 1).
m = B · 0.5    Formula.
= (2. 1) · 0.5    Substitute the Coordinates of B.
= (1, 0.5)    Multiply.
Thus, point B’ is on (1, 0.5). Therefore, the points of \(\overline{A^{\prime} B^{\prime}}\) are on (1, 1.5) and (1, 0.5). The illustration for the image of \(\overline{A B}\) wouLd be as depicted below.

Question 5.
Each figure is each mapped onto its image by a dilation with its center at the origin, O. On a copy of the coordinate plane, draw each image.

a) Triangle LMN is mapped onto triangle L’M’N’ with scale factor –\(\frac{1}{2}\).

Answer:

b) Rectangle PQRS is mapped onto rectangle P’Q’R’S’ with scale factor 3

Answer:

Solve on graph paper. Show your work.

Question 6.
In a room, a flashlight is used to illuminate objects and cast their shadows on a wall. Each shadow is a dilation of the object’s profile.
a) A pen at point (4, 2) is mapped onto its shadow at (6, 3) with the origin as the center of dilation. Find the scale factor.
Answer:
To find the factor of the dilation, solve for the ratio of the dilated points to the originaL points. Take the x-coordinates of the two points. The x- coordinate of the shadow is 6 and x- coordinate of the original object is 4, then
Scale Factor = \(\frac{x-\text { coordinate of Dilated Image }}{x-\text { coordinate of the Original Point }}\) Formula.
= \(\frac{6}{4}\)   Substitution.
= \(\frac{3}{2}\)    Simplify.
Therefore, the scale factor is \(\frac{3}{2}\). If the y-coordinates of the two points is the chosen one to find the scale factor, then the y- coordinate of the shadow is 3 and y- coordinate of the original object is 2, so
Scale Factor = \(\frac{y-\text { coordinate of Dilated Image }}{y-\text { coordinate of the Original Point }}\) Formula.
= \(\frac{3}{2}\)   Substitution.
Therefore, the scale factor is \(\frac{3}{2}\).

b) The shadow of a circular disc has its center at (6, 2) and radius 3 units. The circular disc has center at (2, 2) and radius 1 unit. Find the center of dilation.
Answer:
To identify the center of dilation, let x₁ be the x-coordinate of the center of dilation and y₂ for the y-coordinate. Then, solving the x-coordinate of the center of dilation is given by the scale factor, f multiplied by the x-coordinate of the original point denoted by x₂ minus x-coordinate of the dilated image denoted by x₃ divided by scale factor minus 1.
x₁ = \(\frac{f x_{2}-x_{3}}{f-1}\)    Formula.
= \(\frac{3(2)-6}{3-1}\)    Substitution.
= \(\frac{6-6}{3-1}\)    Multiply.
= \(\frac{0}{2}\)   Subtract.
= 0    Simplify.
Thus, the center of dilation is on 0 of x Next, solving the y-coordinate of the center of dilation is given by the scale factor, f multiplied by the y-coordinate of the original point denoted by y₂ minus y-coordinate of the dilated image denoted by y₃ divided by scale factor minus 1.
y₁   = \(\frac{f y_{2}-y_{3}}{f-1}\)    Formula.
= \(\frac{3(2)-2}{3-1}\)    Substitution.
= \(\frac{6-2}{3-1}\)    Multiply.
= \(\frac{4}{2}\)   Subtract.
= 2    Simplify.
Thus, the center of dilation is on 2 of y. Therefore, the center of dilation lies on point (0, 2).

Question 7.
Each figure is mapped into its image by a dilation. Draw each figure and its image on the coordinate plane. Then mark and label C as the center of dilation. Find the scale factor for each figure. Use 1 grid square on both axes to represent 1 unit for the interval from —8 to 6.

a) Quadrilateral WXYZ

Answer: C(1, 2) Scale factor = -0.5

b) Triangle PQR

Answer:

C(1, 2), Scale factor is 2.5

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Lesson 8.3 Rotations to score better marks in the exam.

Math in Focus Grade 7 Course 3 B Chapter 8 Lesson 8.3 Answer Key Rotations

Math in Focus Grade 8 Chapter 8 Lesson 8.3 Guided Practice Answer Key

Solve. Show your work.

Question 1.
P is rotated counterclockwise to P’ about the origin. State the angle of rotation.

a)

Answer:

b)

Answer:

Question 2.
The tip of a fan blade for a ceiling fan rotates from position W to W. The angle of rotation is 180°.

Answer:

A 180° rotation is also called a half turn. You do not have to include the direction of rotation for half-turn, because 180° clockwise is the same as 180° counterclockwise.

a) On a copy of the graph, mark and label the center of rotation as T.
Answer:

T = (x1+x2 / 2, y1+y2 / 2)
T = (-2 + 2/2, 1+(-3) / 2)
T = (0/2, -0/2)
T = (0, -1)
Therefore, the centre of rotation T has a coordinate (0, -1)

b) W’ is rotated 90° clockwise to W” about the center T. Label W” on the graph in a).
Answer:

Complete.

The hour hand of a clock turns through an angle from 12 noon to 4 P.M. State the following.

a) The center of rotation
Answer: The hands of the clock are placed on the center while moving from 12 to 4. While undergoing a rotation, this point where the other end of the hands does not move. Thus the center of the rotation is the center of the clock.

b) The angle and direction of rotation
Answer: The complete rotation of the hands of the whole clock is 360°. The clock is divided into 12 representing the 12 hours. By dividing 360° to 12, then for every hour the hand of the clock rotates at an angle of 30°. Thus, turning from 12 to 4, then the angle of rotation of the hand of the clock is 120°. Also, the hand moves from left to right direction then the rotation is in a clockwise direction.

Technology Activity

Explore The Properties Of Rotations With Geometry Software

Materials:

• geometry software

Step 1.
Draw a line segment AB using a geometry software program.

Step 2.
Select the Rotate function, within the Transform menu. Enter 90° to rotate \(\overline{A B}\) counterclockwise about the origin. Call the image \(\overline{A^{\prime} B^{\prime}}\).

Step 3.
Make a table of coordinates for the segment endpoints and their images. How are the lengths of the two segments related?

Step 4.
Rotate \(\overline{\mathrm{AB}}\) 90° clockwise by changing the angle of rotation to -90°. Call this image \(\overline{A^{\prime \prime} B^{\prime \prime}}\). Repeat the activity in Step 3.

Step 5.
Choose a different angle of rotation and rotate \(\overline{\mathrm{AB}}\) once again. Repeat . How do the lengths of \(\overline{\mathrm{AB}}\) and \(\overline{A^{\prime} B^{\prime}}\) compare?

Step 6.
Repeat Step 2 to Step 5 using a rectangle as the original figure. Observe how the image is related to the original rectangle.

Math Journal Which of these properties does a rotation seem to preserve: length, shape, parallel lines, or perpendicular lines? Explain.

Draw images after rotations about the origin.

In the activity, you may have observed the following:

Rotations preserve shape and size. They also preserve parallelism and perpendicularity.

You have seen how a line segment can be rotated. A rotation can be described as a transformation that rotates all points on a line or figure clockwise or counterclockwise angle about the center of rotation. You will see in the next example how to find the image of a figure after a rotation.

Copy and complete on graph paper.

Question 4.
A rotation of ∆ABC 90° clockwise about the origin, O, produces the image ∆A’B’C’. Draw and label the image ∆A’B’C’.

Answer:

Copy and complete on graph paper.

Question 5.
An animation artist draws a fish on the coordinate plane and marks the points A, B, C, and D. Then the artist rotates the fish 180° about the origin, O. Complete the table of coordinates to show the coordinates of the image points A’, B’, C’, and D’.

Answer:

Question 6.
DEFG is rotated 90° counterclockwise about O.
a) Draw and label the image D’E’FG’.
Answer:

b) Complete the table of coordinates for DEFG and its image D’E’F’G’.

Answer:

Math in Focus Course 3B Practice 8.3 Answer Key

Solve. Show your work.

Question 1.
A rotation of point P clockwise about O maps onto P’. State the angle of rotation.

a)

Answer:

b)

Answer:

Question 2.
\(\overline{O N}\) is rotated about the origin, O to form the image \(\overline{O N^{\prime}}\). State the angle and direction of each rotation.

a)

Answer:
The point N is on (1,3), N’ is on (-1,-3) and the point of rotation is point O. The line created from N to O and O to N’ is a straight line and has an angle of 180°. To ensure the angle of rotation is 180°, the coordinates should have changed from (x,y) to (-x,-y). Since N is on (1,3) and its image N’ is on (-1,-3) then the angle of rotation is indeed 180°.

b)

Answer:
The point N is on (3, -2), N’ is on (2, 3) and the point of rotation is point O. The lines created from P to O and O to P’ makes a right angle. To ensure that the angle of rotation is 90° counterclockwise, the coordinate should have changed from (x, y) to (-y, x). Since N is on (3, -2) and its image N’ is on (2, 3) then the angle of rotation is indeed 90° in a counterclockwise direction.

Solve. Show your work.

Question 3.
At an amusement park, Olivia is riding the carousel at point P. She is rotated from P by each of the following rotations. Mark and label her position after each rotation from P on a copy of the graph.

a) A: 90° counterclockwise about the origin
Answer: P is at (3, 5). The rotation at 90° counterclockwise changes the coordinates from (x, y) to (-y, x). Therefore, Olivia at (3, 5) will be on point A(-5, 3) when the carousel rotates at this degree and at this direction.

b) B : 90° clockwise about the origin
Answer: The rotation at 90° clockwise shifts the coordinates from (x, y) to (y, -x). So Olivia at (3, 5) will be at B (5, -3) after the rotation at this angle and in this direction.

c) C: 270° counterclockwise about the origin
Answer: A 270° rotation counterclockwise is basically just a 180° plus 90° counterclockwise direction. A 180° angle of rotation moves the coordinates from (x, y) to (-x, -y). Thus, P(3, 5) will be on (-3,-5). Then rotating it again 90°counterclockwise, the new point should be at the form (-y, x). Therefore, Olivia will be on C(5, -3) after this rotation at this direction.

d) D: Half turn about the origin
Answer:

Answer:

Question 4.
A cam of an automobile rotates about a shaft at the origin, O. Point P on the cam rotates to point Q.
a) Describe the rotation.
Answer: The point is on (2, 2) and Q is on (-2, 2). The coordinate (x, y) shifted to (-y, x). Therefore, the angle of rotation is 90° counterclockwise direction because the coordinates changed from (x, y) to (-y, x).

b) A point (-5, 4) undergoes the same rotation. Find the coordinates of the image.

Answer: (-5,4) is rotated counterclockwise at 90° then the coordinates (x, y) should shift to (-y, x). Therefore, the image of (-5, 4) will have the coordinates (-4, -5).

Question 5.
The hinges on a door are at (0, 0), looking down from above. Its keyhole is at position (2, 4) when the door is closed. The door swings open. Find the position of the keyhole under each rotation below.

a) 90° clockwise
Answer: The point is on (2,4) should changed from (x, y) to (-y, x) for 90° clockwise rotation. Therefore, the keyhole at (2, 4) will be on point (4,-2) when the door is open 90° clockwise.

b) 90°counterclockwise
Answer: (2, 4) is rotated counterclockwise at 90° then the coordinates (x, y) should shift to (-y, x). Therefore, the keyhole at (2, 4) will have the coordinates (4, -2) when the door is open 90° clockwise.

c) 180°
Answer: Point (2, 4) has a 180° angle of rotation then the coordinates should have changed from (x, y) to (-y, x). Therefore, the keyhole at (2, 4) will have the coordinates from (-2, -4) when the door is open rotates 180°.

Question 6.
Pentagon ABCDE is drawn on the coordinate plane.

a) ABCDE is rotated 90° clockwise about the origin, O. Draw and label the image A’B’C’D’E’.
Answer:

b) State the coordinates of A’, B’, C’, D’, and E’.
Answer: The rotation changes the coordinates from (x,y) to (y,-x). Thus, the coordinates of points A’, B’, C’, D’ and E’ are (-1, -2), (-2, -3), (-3, -3), (-3, -1) and (-2, -1) respectively.

c) ABCDE is rotated 90° counterclockwise about the origin, O. Draw the image A”B”C”D”E”. State the coordinates for A”B”C”D”E”.
Answer: A 90° rotation counterclockwise direction changes the coordinates (x,y) to (-y,x). Thus, A at (2,-1) will have an image A” at (1, 2), B(3, -2) will be mapped onto B”(2, 3), C at (3, -3) will have an image C” at (3, -3) D(1, -3) will be mapped onto D”(3, 1) and E at (1, -2) will have an image E” at (2, 1). So, plotting the points of the images, A”, B”, C”, D”, E”.

d) How are A’B’C’D’E’ and A”B”C”D”E” related?
Answer: The coordinates of points A’B’C’D’E’ are (-1, -2), (-2, -3), (-3, -3), (-3,-1) and (-2, -1) respectively while the coordinates of points A”B”C”D”E” (1, 2), (2, 3), (3, 3), (3, 1) and (2, 1) respectively. On the other hand, the 180° rotation changes the coordinates from (x, y) to (-x, -y) which is the case if pentagon A’B’C’D’E’ will be the original figure and the image will be A”B”C”D”E”. Therefore A”B”C”D”E” is the image of A’B’C”D’E’ if it is rotated at an angle of 180°.

Question 7.
A regular hexagon ABCDEF is rotated about its center, O, so that its appearance stays the same, but the vertices are rotated to different positions. For example in one rotation, A moves to B, B to C, and so on. Which clockwise rotations will cause this effect? Which points are invariant under a rotation? Explain.

Answer: A whole hexagon has a complete angle of 720°. The figure has 6 equal angles, dividing 720° to 6 than a angle to angle rotation has 120° which is the case in this problem. Therefore, the angle of rotation is 120°.

Question 8.
Math Journal Which points are invariant under a rotation? Explain.
Answer: The points that are invariant are those that did not change even after the rotation. If a point or line or figure is rotated at any angle at any direction about (0, 0) then the invariant is (0, 0) since this point will not change at any rotation the point or line or figure will undergo. If it is rotated at about (1, 2) then the invariant is (1, 2). Therefore, the centers of rotation are the invariant points.