Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2 Exercise 2.2

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2

Page 10 Problem 1 Answer

Given,d/8=6

To find : the value of d.

Solve the above equation and find the value of d.

Consider the given equation.

d/8=6

Cross multiplying, we get

d=8×6

d=48

Now, check our answer

48/8=6

6=6​

Hence, the value of d is 48.

Page 10 Problem 2 Answer

Given,−5=n/2

To find : the value of n.

Solve the above equation and find the value of n.

Consider the given equation.

−5=n/2

Cross multiplying, we get

−5×2=n

n=−10

Now, check our answer

−5=−10/2

−5=−5

​Hence, the value of n is −10.

Page 10 Problem 3 Answer

Given,−t/2=12

To find : the value of t.

Solve the above equation and find the value of t.

Consider the given equation.

−t/2=12

On solving the above equation, we get

−t=2×12

−t=24

t=−24

​Now, check my answer

−(−24)/2=12/24

2=12

12=12

​Hence, the value of t is −24.

Page 10 Problem 4 Answer

Given,−40=−4x

To find : the value of x

Solve the above equation and find the value of x.

Consider the given equation.

−40=−4x

On dividing the above equation by −4, we get

−40−4=−4x

−4/10=x

x=10

Now, check my answer

−40=−4(10)

−40=−40

​Hence, the value of x is 10.

Page 10 Problem 5 Answer

Given,2r/3=16

To find : the value of r

Solve the above equation and find the value of r.

Consider the given equation.

2r/3=16

On solving the above equation, we get

2r=3×16

2r=48

r=48/2

r=24

​Now, check my answer

2(24)/3=16/48

3=16

16=16

​Hence, the value of  r is 24.

Page 10 Problem 6 Answer

Given,−49=7y

To find : the value of y

Solve the above equation and find the value of y.

Consider the given equation.

−49=7y

On dividing the above equation by 7, we get

−49/7=7y

7−7=y

y=−7

​Now, check our answer

−49=7(−7)

−49=−49

​Hence, the value of y is −7.

Page 10 Problem 7 Answer

Given,−15=−3n/5

To find : the value of n.

Solve the above equation and find the value of n.

Consider the given equation.

−15=−3n/5

On solving the above equation, we get

−15×5=−3n

−3n=−75

n=−75/−3

n=25

Now, check our answer

−15=−3(25)/5

−15=−15

​Hence, the value of n is 25.

Page 10 Problem 8 Answer

Given,v−3=−6

To find : the value of v.

Solve the above equation and find the value of v.

Consider the given equation.

v−3=−6

On solving the above equation, we get

v=−6×−3

v=18

Now, check my answer

18−3=−6

−6=−6

​Hence, the value of v is 18.

Page 10 Problem 9 Answer

Given,

2.8=b/4

To find: the value of b

Solve the above equation and find the value of b.

Consider the given equation.

2.8=b/4

Cross multiplying, we get

b=2.8×4

b=11.2

Now, check our answer

2.8=11.2/4

2.8=2.8

​Hence, the value of b is 11.2.

Page 10 Problem 10 Answer

Given, 3r /4=1/8

To find: the value of r

Solve the above equation and find the value of r.

Consider the given equation.

3r/4=1/8

On solving the above equation, we get

3r=4/8

3r=1/2

r=1/6

Now, check my answer

3(1/6)/4=1/8

1/2/4=1/8

1/8=1/8

Hence, the value of r is 1/6.

Page 10 Problem 11 Answer

Given, The price of the package = $2.07

The first-class rate at the time = $0.23 per ounce

To find the weight of the package.

Apply the concept of linear equation. And solve the equation.

Given, The price of the package =$2.07

The first-class rate at the time =$0.23  per ounce

Let ′x′ be the weight of the package.

According to the question,

0.23x=2.07

x=2.07/0.23

x=9

​Hence, the weight of the package is 9 ounce.

Page 10 Problem 12 Answer

Given, Lola spends $8 per at the movies.

To find : Lola’s weekly allowance.

Apply the concept of linear equation.

And solve the equation.

Given, Lola spends $8 per at the movies.

Let ′x′ be Lola’s weekly allowance.

According to the given question,

1/3

x=8

x=8×3

x=24

​Hence, Lola’s weekly allowance is $24.

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