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HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Exercise 1.2 Adding and Subtracting Integers

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HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Adding and Subtracting Integers

Page 3 Problem 1 Answer

In the question, the expression is given as 2+(−3). The following number line is given:

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 1

It is required to show the addition on the number line and find the sum.

To solve this, take the first integer given in the expression and show it on the number line.

Use the sign of the second integer to show the addition on the number line and find the sum.

The first integer in the expression 2+(−3) is 2. The second integer in the expression is −3.

Place 2 on the number line and go −3 units to the left to show the addition on the number line.

From the number line, the result of the sum of 2+(−3) is −1, which can be expressed as:

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 1 1

The addition of 2+(−3) in the number line is shown below:

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 1 2

The sum of the expression 2+(−3) is −1.

Page 3 Problem 2 Answer

The given statement is −3+4.

It is required to show addition on number line and find sum of given expression.

To do so, simulate the addition on a number line, begin at a point −3 and advance 4 to the right and perform the addition.

Begin at −3 and advance 4 units to the right, as 4 is a positive number.

This gives

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 2

Therefore the sum is,−3+4=1.

−3+4=1 is the sum and the representation on number line is given below,

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 2 1

Page 3 Problem 3 Answer

The given statement is,−4+9.

It is required to show addition on number line and find sum of given expression.

To do so, simulate the addition on a number line.  Begin at −4 and advance 9 units to the right, as 9  is a positive number and obtain the required sum.

The expression is −4+9.

Consider the number line and then, begin at −4 and advance 9 units to the right, as 9 is a positive number and obtain the required sum.

This gives,

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 3

Therefore,−4+9=5.

Thus, the sum obtained is 5.

The sum of given expression −4+9 is found to be 5.

Page 3 Problem 4 Answer

The given statement is,6+(−9).

It is required to show addition on number line and find sum of given expression.

To do so, simulate the addition on a number line. Begin at 6 and advance 9 units to the left, as 9  is a negative number and obtain the required sum.

The given expression is 6+(−9).Consider the number line and then, begin at 6 and advance 9 units to the left, as 9 is a negative number and obtain the required sum.

This gives,

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 4

Therefore,

6+(−9)=−3.

The sum of given expression 6+(−9) is found to be −3.

Page 3 Problem 5 Answer

The given statement is,5+(−7).

It is required to show addition on number line and find sum of given expression.

To do so, simulate the addition on a number line.  Begin at 5 and advance −7 units to the left, as −7 is a negative number and obtain the required sum.

The expression is 5+(−7).

Consider the number line and then, begin at 5 and advance −7 units to the left, as −7 is a negative number and obtain the required sum.

This gives,

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 5

Therefore, 5+(−7)=−2.

The sum of given expression 5+(−7) is −2.

Page 3 Problem 6 Answer

The given statement is,9+(−5).

It is required to show addition on number line and find sum of given expression.

To do so, simulate the addition on a number line. Then, begin at 9 and advance −5 units to the left, as −5 is a negative number.

The expression is 9+(−5).

Consider the number line and then, begin at 9 and advance −5 units to the left, as −5 is a negative number.

This gives,

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 6

Therefore,

9+(−5)=4.

The sum of given expression 9+(−5) is found to be 4.

Page 3 Problem 7 Answer

The given statement is,(−1)+9.

It is required to show addition on a number line and find the sum of the given expression.

To do so, simulate the addition on a number line.

Then, begin at −1 and advance 9 units to the right, as 9 is a positive number, and obtain the required sum.

This gives,

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 7

Therefore,

(−1)+9=8.

The sum of given expression (−1)+9 is found to be 8.

Page 3 Problem 8 Answer

The given statement is 9+(−7).

It is required to show addition on number line and find sum of given expression.

To do so, simulate the addition on a number line.

Then, begin at 9 and advance −7 units to the left, as −7  is a negative number and obtain the required sum.

The expression is 9+(−7).

Consider the number line and then, begin at 9 and advance −7 units to the left, as −7 is a negative number and obtain the required sum.

This gives,

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 8

Therefore,9+(−7)=2.

The sum of given expression 9+(−7) is found to be 2.

Page 3 Problem 9 Answer

The given statement is, 50+(−7).

It is required to show addition on number line and find sum of given expression.

To do so, simulate the addition on a number line.

Then, begin at 50 and advance −7 units to the left, as −7 is a negative number and obtain the required sum.

The expression is 50+(−7).

Consider the number line and then, begin at 50 and advance −7 units to the left, as −7 is a negative number and obtain the required sum.

This gives,

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 9

Therefore,50+(−7)=43.

The sum of given expression 50+(−7) is found to be 43.

Page 3 Problem 10 Answer

It is given that there is an expression 1+(−30).

It is required to evaluate the sum of the expression.

To evaluate the expression, consider the number line then, begin to advance from 0 to −30 and then shift 1 places right, then evaluate and obtain the required answer.

The expression is 1+(−30).

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 10

Consider the number line and start from 0 to −30 in the negative direction and then shift +1 to the right to represent the required sum on the number line.

Further, evaluate and check the answer.

The expression 1+(−30) is evaluated as,​

1+(−30)=1−30

1+(−30) =−29​

The sum of the expression 1+(−30) is calculated as −29.

Page 3 Problem 11 Answer

It is given that there is an expression 15+(−25).

It is required to evaluate the sum of the expression.

To do this, consider the number line then, begin to advance from 0 to +15 and then shift 25 places left, then evaluate and obtain the required answer.

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 11

Consider a number line and then, begin to advance from 0 to 27 and then shift 6 places left, to obtain the required sum.

The expression 15+(−25) is evaluated as​

15+(−25)=15−25

15+(−25) =−10

​The sum of the expression 15+(−25) is calculated as −10.

Page 3 Problem 12 Answer

It is given that the outside temperature dropped by 13∘F in seven hours.

The final temperature is recorded as −2∘F.

It is required to find the starting temperature.

To find the starting temperature, assume the starting temperature as x.

Then make an equation showing the difference between the starting and final temperature. Finally, solve for x

in the linear equation and find the value of starting temperature.

Solve the equation obtained in step 1

x−(−2)=13

x+2=13

x=13−2

x=11

​The starting temperature outside when temperature dropped by 13∘F in seven hours to −2∘F was 11∘F.

Page 3 Problem 13 Answer

It is given that a football team gains eight yards in one play and then loses five yards in the next play

It is required to find the total yardage from the two plays.

To find the total yardage, add the 8 yards that the team went forward and the 5 yards that they were forced back to gain the total yardage of the team.

It is given that the team went forward 8 yards and came back 5 yards.

Hence the total yardage of the team is given as 8+(−5).

Solve this expression to obtain the total yardage as

​8+(−5)=8−5

​8+(−5) =3​

The total yardage of the football team is calculated as 3 yards.

Page 4 Problem 14 Answer

It is given that there is an expression 3+(−9).

It is required to decide whether this expression is to be added or subtracted and also state the reason.

The expression is 3+(−9).

The expression has the addition operator.

Hence, the two integers will be added.

However, since the nature of the bigger number 9 is negative, the addition operation changes into subtraction and becomes 3−9=−6.

Thus, 3+(−9) results in −6.

The numbers 3,9 in the expression 3+(−9) is to be subtracted as 3−9=−6 because the nature of the bigger number 9 is negative, the addition operation changes into subtraction.

Page 4 Problem 15 Answer

It is given that there is an expression 3+(−9)

It is required to decide whether this expression is to be added or subtracted and also state the reason.

The expression is 3+(−9).

The expression has the addition operator. Hence, the two integers will be added.

However, since the nature of the bigger number 9 is negative, the addition operation changes into subtraction. This can be also defined as adding 3 in −9.

Hence the expression becomes 3−9=−6.

Here, the sum −6 is negative.

The numbers 3,9 in the expression 3+(−9) is to be subtracted as 3−9=−6 and the sum of this expression is negative.

Page 4 Problem 16 Answer

It is given that −2+(−3).

It is required to find the sum of the given integers.

In order to do so, first consider the given integers.

After that, find the difference of their absolute values.

Further, assign the required sign, positive or negative, based on the higher absolute value.

First consider the given integers,

−2+(−3)

Then, determine the absolute values that is,

∣−2∣=2 and∣−3∣=3

So, 2+3=5

The sign of the total has the sign of the terms since the terms are of the same sign. So, the sign of answer will be negative.

Thus,−2+(−3)=−5

The sum of −2+(−3)=−5.

Page 4 Problem 17 Answer

It is given that −5+4.

It is required to find the sum of the given integers.

In order to do so, first consider the given integers.

After that, find the difference of their absolute values.

Further, assign the required sign, positive or negative, based on the higher absolute value.

First consider the given integers,−5+4

Then, determine the absolute values that is,

∣−5∣=5 and∣4∣=4

So, 5−4=1

Since, ∣−5∣>∣4∣, because the sign of the total is the sign of the term whose absolute value is bigger that is 5>4, the sign of answer will be negative.

Thus,−5+4=−1

The sum of −5+4=−1.

Page 4 Problem 18 Answer

It is given that −3+(−1).

It is required to find the sum of the given integers.

In order to do so, first consider the given integers.

After that, find the difference of their absolute values.

Further, assign the required sign, positive or negative, based on the higher absolute value.

First consider the given integers,−3+(−1)

Then, determine the absolute values that is,

∣−3∣=3 and∣−1∣=1

So, 3+1=4

The sign of the total has the sign of the terms since the terms are of the same sign.

So, the sign of answer will be negative.

Thus,−3+(−1)=−4

The sum of −3+(−1)=−4.

Page 4 Problem 19 Answer

It is given that −7+9.

It is required to find the sum of the given integers.

In order to do so, first consider the given integers.

After that, find the difference of their absolute values.

Further, assign the required sign, positive or negative, based on the higher absolute value.

First consider the given integers,−7+9

Then, determine the absolute values that is,

∣−7∣=7 and∣9∣=9

So, 9−7=2

Since, ∣9∣>∣−7∣, because the sign of the total is the sign of the term whose absolute value is bigger that is 9>7, the sign of answer will be positive.

Thus,−7+9=2

The sum of −7+9=2.

Page 4 Problem 20 Answer

It is given that 4+(−9).

It is required to find the sum of the given integers.

In order to do so, first consider the given integers.

After that, find the difference of their absolute values.

Further, assign the required sign, positive or negative, based on the higher absolute value.

First consider the given integers,

4+(−9)

Then, determine the absolute values that is,

∣4∣=4 and∣−9∣=9

So, 9−4=5

Since, ∣−9∣>∣4∣, because the sign of the total is the sign of the term whose absolute value is bigger that is 9>4, the sign of answer will be negative.

Thus,4+(−9)=−5

The sum of 4+(−9)=−5.

Page 4 Problem 21 Answer

It is given that 16+(−7).

It is required to find the sum of the given integers.

In order to do so, first consider the given integers.

After that, find the difference of their absolute values.

Further, assign the required sign, positive or negative, based on the higher absolute value.

First consider the given integers,

16+(−7)

Then, determine the absolute values that is,

∣16∣=16 and∣−7∣=7

So, 16−7=9

Since, ∣16∣>∣−7∣, because the sign of the total is the sign of the term whose absolute value is bigger that is 16>7, the sign of answer will be positive.

Thus,16+(−7)=9

The sum of 16+(−7)=9.

Page 4 Problem 22 Answer

It is given that −21+11.

It is required to find the sum of the given integers.

In order to do so, first consider the given integers.

After that, find the difference of their absolute values.

Further, assign the required sign, positive or negative, based on the higher absolute value.

First consider the given integers,−21+11

Then, determine the absolute values that is,

∣−21∣=21 and∣11∣=11

So, 21−11=10

Since, ∣−21∣>∣11∣, because the sign of the total is the sign of the term whose absolute value is bigger that is 21>11, the sign of answer will be negative.

Thus,−21+11=−10

The sum of −21+11=−10

Page 4 Problem 23 Answer

It is given that 3+(−9).

It is required to find the sum of the given integers and how to know if the sum is negative.

In order to do so, first consider the given integers.

After that, find the difference in their absolute values.

Further, assign the required sign, positive or negative, based on the higher absolute value.

First consider the given integers,3+(−9)

Then, determine the absolute values that is,

∣3∣=3 and∣−9∣=9

So, 9−3=6

Since, ∣−9∣>∣3∣, because the sign of the total is the sign of the term whose absolute value is bigger that is 9>3, the sign of answer will be negative.

Thus, 3+(−9)=−6

Hence, the sum is negative.

The sum of 3+(−9)=−6 and the sum is negative.

HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Exercise 1.1 Adding and Subtracting Integers

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HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Adding and Subtracting Integers

Page 1 Problem 1 Answer

In the given question the following expression is given,−4+(−2).

And the given number line is as follows

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 1

It is required to model their addition on the number line and find its sum.

To solve, draw a number line and model the addition of the expression on the number line.

First, draw the given number line.

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 1 2

Then, locate the first integer on the number line.

If the second integer is positive, move that many units on the number line to the right from the location of the first integer and if the second integer is negative, move that many units on the number line to the right from the location of the first integer.

For modelling the addition on the number line, use the given number line.

Start from −4 on the number line and move 2 units to the left because the number −2 in the given expression −4+(−2) is negative.

Therefore, the sum of the expression −4+(−2) from the number line is −6.

Go Math!, Middle School Grade 7, Practice Fluency Workbook, 1st Edition, Chapter 1 Adding and Subtracting Integers 1 3

Therefore, the addition of the given integers −4 and −2 in the given expression −4+(−2) after modelling them on the number line, is −6.

Page 1 Problem 2 Answer

In the given question following expression is given,−7+(−1)

It is required to find the sum of the given integers.

To calculate the sum first check the signs of the integers. If both are positive then add the integers.

And if both the integers are negative, then first ignore the negative signs and add the integers.

Then at the end write the sum as a negative number.

Check the signs in the given expression −7+(−1) and observe if they are negative or not.

Here −7 and −1 are both negative integers.

Ignore the negative signs from the expressions −7+(−1) and add the integers using simple addition properties.

7+1=8

Write the obtained sum as a negative number.

−7+(−1)=−8

Therefore, the sum of the given expression −7+(−1) is obtained as −8.

Page 1 Problem 3 Answer

In the given question following expression is given,−5+(−4)

It is required to find the sum of the given integers.

To calculate the sum first check the signs of the integers.

If both are positive then add the integers.

And if both the integers are negative, then first ignore the negative signs and add the integers.

Then at the end write the sum as a negative number.

Check the signs in the given expression −5+(−4) and observe if they are negative or not.

Here −5 and −4 are both negative integers.

Ignore the negative signs from the expressions −5+(−4) and add the integers using simple addition properties.

5+4=9

Write the obtained sum as a negative number.

−5+(−4)=−9

Therefore, the sum of the given expression −5+(−4) is obtained as −9.

Page 1 Problem 4 Answer

In the given question following expression is given−51+(−42).

It is required to find the sum of the given integers.

To calculate the sum first check the signs of the integers.

If both are positive then add the integers.

And if both the integers are negative, then first ignore the negative signs and add the integers.

Then at the end write the sum as a negative number.

Check the signs in the given expression −51+(−42) and observe if they are negative or not.

Here −51 and −42 are both negative integers.

Ignore the negative signs from the expressions −51+(−42) and add the integers using simple addition properties.

51+42=93

Write the obtained sum as a negative number.

−51+(−42)=−93

Therefore, the sum of the given expression −51+(−42) is obtained as −93.

Page 1 Problem 5 Answer

In the given question following expression is given 98+126.

It is required to find the sum of the given integers.

To calculate the sum first check the signs of the integers.

If both are positive then add the integers.

And if both the integers are negative, then first ignore the negative signs and add the integers.

Then at the end write the sum as a negative number.

Check the signs in the given expression 98+126 and observe if they are negative or not.

Here 98 and 126 are both positive integers.

Add the given integers 98 and 126 by using simple addition properties.

98+126=224

Therefore, the sum of the given expression 98+126 is obtained as 224.

Page 1 Problem 6 Answer

In the given question following expression is given,−20+(−75)

It is required to find the sum of the given integers.

To calculate the sum first check the signs of the integers.

If both are positive then add the integers.

And if both the integers are negative, then first ignore the negative signs and add the integers.

Then at the end write the sum as a negative number.

Check the signs in the given expression −20+(−75) and observe if they are negative or not.

Here −20 and −75 are both negative integers.

Ignore the negative signs from the expressions −20+(−75) and add the integers using simple addition properties.

20+75=95

Write the obtained sum as a negative number.

−20+(−75)=−95

Therefore, the sum of the given expression −20+(−75) is obtained as −95.

Page 1 Problem 7 Answer

In the question the expression −350+(−250) is given.

It is required to find the sum of the integers.

To calculate the sum first observe that the integers have the same negative sign or not.

At first add the integers ignoring the negative signs.

Then at the end write the sum as a negative number.

Check the signs in the given expression  −350+(−250)  and observe if they are negative or not.

Here−350 and−250 are both negative.

Ignore the negative signs from the expression  −350+(−250) and  add the integers using simple addition properties.

350+250=600

Write the sum as a negative number.

−350+(−250)=−600

The sum of the given expression  −350+(−250) is −600.

Page 1 Problem 8 Answer

In the question the given expression is−110+(−1200).

It is required to find the sum of the integers

To calculate the sum first observe that the integers have the same negative sign or not.

First, add the integers  ignoring the negative signs.

Then at the end write the sum as a negative number.

Check the signs in the given expression −110+(−1200) and observe if they are negative or not.

Here −110 and −1200 are both negative numbers.

Ignore the negative signs from the expression −110+(−1200) and add the integers using simple addition properties.

110+1200=1310.

Write the sum as a negative number.

−110+(−1200)=−1310

The sum of the given expression −110+(−1200) is −1310.

Page 1 Problem 9 Answer

It is given that a construction crew is digging a hole.

Now on the first day, they dug a hole 3 feet.

Then on the second day, they dug 2 more feet and on the third day, they dug 4  more feet.

It is required to find a sum of negative numbers to represent this situation and also it is required to find the total sum and its relation with the problem.

To calculate the total sum observe all the integers have the same negative sign since all the integers are representing the depth of hole.

Add the integers ignoring the negative signs. Then at the end write the sum as a negative number.

Check the signs of the integers in the given problem (−3)+(−2)+(−4) and observe if they are negative or not.

Here −3 ,−2 and−4 are all negative integers since the integers are expressing the depth of the hole.

Ignore the negative signs from the expression and  add the integers using simple addition properties.

3+2+4=9

Write the sum as a negative number.

(−3)+(−2)+(−4)=−9

The total sum is −9 and here 9 represents the depth of hole which has been dug in three days and the negative sign − before 9 represents the fact that the value is representing depth.

The sum of the negative numbers is −9.

The total sum is −9 and here 9 represents the depth of hole which has been dug in three days and the negative sign − before 9 represents the fact that the value is representing depth.

Page 2 Problem 10 Answer

In the question the given expression is 3+6 .

It is required to find if the integers are both positive or both are negative.

To find if the integers are both positive or negative, check if the integers are with positive sign or the integers are with negative sign.

Check the signs of the integers  in the given expression 3+6 and observe if they are negative or positive.

Both 3 and 6 have positive signs.

Therefore the integers 3 and 6 are both positive integers.

The integers 3 and 6 in the given expression3+6 are both positive integers.

Page 2 Problem 11 Answer

In the question the given expression is 3+6 .

It is required to add the integers.

To find the addition of integers  3 and 6 in the given expression  3+6 use simple addition properties.

Add the integers 3 and 6 in the given expression 3+6 using simple addition properties.

3+6=9

Therefore the value of the addition is 9.

The required value of addition of 3 and 6 in the expression 3+6 is 9 .

Page 2 Problem 12 Answer

In the question the given expression is 3+6.

It is required to find the sum of the integers.

To calculate the sum observe if the integers have the same positive sign. Then add the integers and at the end write the sum as positive number.

Check the signs of the integers 3 and 6 in the given expression is 3+6 if  the integers are both positive or the integers are both negative.

The integers 3 and 6 are both positive.

Add the integers 3 and 6 in the given expression is 3+6 using simple addition properties.

3+6=9

The required value of the integers 3 and 6 in the given expression is 3+6  is 9.

Page 2 Problem 13 Answer

In the question the given expression is −7+(−1) .It is required to find if the integers are both positive or both negative.

To find if the integers are both positive or negative, check if the integers are with positive sign or the integers are with negative sign.

Check the signs of the integers  in the given expression −7+(−1) and observe if they are negative or positive.

Both −7 and −1 have negative signs.

Therefore the integers −7 and −1 are both negative integers.

The integers −7 and −1 in the given expression −7+(−1) are both negative integers.

Page 2 Problem 14 Answer

In the question the given expression is −7+(−1).

It is required to add the integers.

To find the addition, ignore the negative sign of the integers −7 and−1 and add them using simple addition properties.

Ignore the negative sign of the integers −7 and −1 in the given expression is −7+(−1)  and add them using simple addition properties.

7+1=8

Therefore the addition of the integers is 8.

The value of the addition of the integers −7 and−1 in the given expression −7+(−1) is – 8.

Page 2 Problem 15 Answer

In the question, the expression is given as −7+(−1).

It is required to write the sum of the given expression.

To solve this, add the integers given in the expression and write it as a negative number to write the sum of the expression.

The sum of the two integers 7 and 1 is 8.

The sum of −7+(−1) will be the negative of 8 and can be written as: −7+(−1)=−8

The sum of the expression −7+(−1) is −7+(−1)=−8.

Page 2 Problem 16 Answer

In the question, the expression is given as −5+(−2).

It is required to state if both the integers are positive or negative.

To solve this, write if the integers given in the expression are positive or negative by noting the sign before the integers.

The two integers given in the question are −5 and −2. Both the integers have a negative sign before them.

Hence, both the integers are negative.

The integers in the expression −5+(−2) are both negative.

Page 2 Problem 17 Answer

In the question, the expression is given as −5+(−2).

It is required to add the integers.

To solve this, add the integers given in the expression by ignoring the negative numbers.

Ignoring the negative signs, the two integers given in the expression −5+(−2) are 5 and 2.

Simplify the expression 5+2 using simple addition operations.

5+2=7

The addition of the integers in the expression −5+(−2) is 7.

Page 2 Problem 18 Answer

In the question, the expression is given as −5+(−2).

It is required to write the sum of the given expression.

To solve this, add the integers given in the expression and write it as a negative number to write the sum of the expression.

The sum of the two integers 5 and 2 is 7.

The sum of −5+(−2) will be the negative of 7 and can be written as:

−5+(−2)=−7

The sum of the expression −5+(−2) is −5+(−2)=−7.

Page 2 Problem 19 Answer

In the question, the expression is given as 6+4.

It is required to state if both the integers are positive or negative.

To solve this, write if the integers given in the expression are positive or negative by noting the sign before the integers.

The two integers given in the question are 6 and 4. Both the integers have no negative sign before them.

Hence, both the integers are positive.

The integers in the expression  6+4 are both positive.

Page 2 Problem 20 Answer

In the question, the expression is given as 6+4.

It is required to add the integers.

To solve this, add the integers given in the expression using addition operations.

Simplify the expression 6+4 using simple addition operations.

6+4=10

The addition of the integers in the expression 6+4 is 10.

Page 2 Problem 21 Answer

In the question, the expression is given as 6+4.

It is required to write the sum of the given expression.

To solve this, add the integers given in the expression to write the sum of the expression.

The two integers 6 and 4 have a sum of 10, which is a positive number.

The sum of the expression 6+4 can be written as:

6+4=10

The sum of the expression 6+4 is 6+4=10.

Page 2 Problem 22 Answer

In the question, the expression is given as −10+(−3).

It is required to find the sum of the expression.

To solve this, add the integers given in the expression ignoring the negative numbers.

Write the sum as a negative number to write the sum of the expression.

Both the integers −10 and −3 in the expression −10+(−3) have a negative sign before them.

Hence, the integers are both negative.

Ignoring the negative signs, the two integers given in the expression −10+(−3) are 10 and 3.

Simplify the expression 10+3 using simple addition operations.

10+3=13

The sum of −10+(−3) will be the negative of 13 and can be written as:−10+(−3)=−13

The sum of the expression −10+(−3) is −13.

Page 2 Problem 23 Answer

In the question, the expression is given as −4+(−12).

It is required to find the sum of the expression.

To solve this, add the integers given in the expression ignoring the negative numbers.

Write the sum as a negative number to write the sum of the expression.

Both the integers −4 and −12 in the expression −4+(−12) have a negative sign before them.

Hence, the integers are both negative.

Ignoring the negative signs, the two integers given in the expression −4+(−12) are 4 and 12.

Simplify the expression 4+12 using simple addition operations.

4+12=16

The sum of −4+(−12) will be the negative of 16 and can be written as: −4+(−12)=−16

The sum of the expression −4+(−12) is −16.

Page 2 Problem 24 Answer

In the question, the expression is given as 22+15.

It is required to find the sum of the expression.

To solve this, add the integers given in the expression ignoring the negative numbers.

Write the sum as a negative number to write the sum of the expression.

Both the integers 22 and 15 in the expression 22+15 have no negative sign before them.

Hence, the integers are both positive.

The two integers 22 and 15 have a sum of 37, which is a positive number.

The sum of the expression 22+15 can be written as:

22+15=37

The sum of the expression 22+15 is 37.

Page 2 Problem 25 Answer

In the question, the expression is given as −10+(−31).

It is required to find the sum of the expression.

To solve this, add the integers given in the expression ignoring the negative numbers.

Write the sum as a negative number to write the sum of the expression.

Both the integers −10 and −31 in the expression −10+(−31) have a negative sign before them.

Hence, the integers are both negative.

Ignoring the negative signs, the two integers given in the expression −10+(−31) are 10 and 31.

Simplify the expression 10+31 using simple addition operations.

10+31=41

The sum of −10+(−31) will be the negative of 41 and can be written as:

−10+(−31)=−41

The sum of the expression −10+(−31) is −41.

Page 2 Problem 26 Answer

In the question, the expression is given as −18+(−6).

It is required to find the sum of the expression.

To solve this, add the integers given in the expression ignoring the negative numbers.

Write the sum as a negative number to write the sum of the expression.

Both the integers −18 and −6 in the expression −18+(−6) have a negative sign before them.

Hence, the integers are both negative.

Ignoring the negative signs, the two integers given in the expression −18+(−6) are 18 and 6.

Simplify the expression 18+6 using simple addition operations.

18+6=24

The sum of −18+(−6) will be the negative of 24 and can be written as:−18+(−6)=−24

The sum of the expression −18+(−6) is −24.

Page 2 Problem 27 Answer

In the question, the expression is given as 35+17.

It is required to find the sum of the expression.

To solve this, add the integers given in the expression ignoring the negative numbers.

Write the sum as a negative number to write the sum of the expression.

Both the integers 35 and 17 in the expression 35+17 have no negative sign before them. Hence, the integers are both positive.

The two integers 35 and 17 have a sum of 52, which is a positive number.

The sum of the expression 35+17 can be written as 35+17=52

The sum of the expression 35+17 is 52.

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2 Exercise 2.7

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

Page 15 Problem 1 Answer

Given ΔABC∼ΔDEF

To Find : x in the given diagram

In order to find the solution, we need to find EF by applying similar to rule.

ΔABC∼ΔDEF∼ is known as similar sign, where we now know that,

AB/DE=BC/EF=AC/DF But since, the value of AB and DE

has not been provided, then the equation will change into,

BC/EF=AC/DF

Now replace the equation with the values given in the diagram:

5/x=9/27

Use cross multiplication, where N1 is multiplied by D2 and D1 is multiplied by N2. The new equation will be

5×27=9×x

5×27=9×x

Now, to find the x solve the equation

=135=9x

Divide 135 with 9

=135/9

=x

=x=15 cm

​The value of x=15 cm

Page 15 Problem 2 Answer

FGHJK∼MNPQR

To Find: x

In order to find the solution, we need to find PQ by applying similar to rule.

Since FGHJK∼MNPQR∼ is known as similar sign, where we now know that

FG/MN=GH/NP

=HJ/PQ

=JK/QR

But since, HJ, JK, PQ and QR values has been provided, then the equation will change into,

HJ/PQ=JK/QR

We need to find the value of PQ.

Replace the equation with the values given in the diagram:

2/x=5/8

Use cross multiplication, where is N1 multiplied by D2 by and D1 is multiplied by N2 . The new equation will be

=2×8=5×x

=2×8=5×x

Now, to find the x solve the equation.

=16=5x

Divide 5 with 16

=16/5

=x

=x=3.2 cm

​The value of x=3.2cm

Page 15 Problem 3 Answer

Given the height of the first pole and both the shadow.

To find: height of the second pole.

In order to find the height of the second pole:

we will substitute the height with the xconvert the equation into fraction and cross multiply it to find the solution of x

Since we know the both shadow lengths, let’s convert it into fraction, as,4ft/20ft

Let’s make the height of the second utility pole be x, and convert it into fraction,5.5ft/x

To find x, combine the equation together

4ft/20ft=5.5ft/x

In order to find x, Cross multiply the given equation

=4×x=20×5.5

=4x=110​

Divide 110 with 4

x=110/4ft

x=27.5ft

​The height of the utility pole is 27.5 ft.

Page 15 Problem 4 Answer

Given: Radius of cylinder=3cm

Length of cylinder=10cm

and every dimension of cylinder is multiplied by 3 to form a new cylinder.

To find the ratio of the volumes related to the ratio of corresponding dimension

We find the new dimension and using that we find new volume and proceed.

Original radius=3cm;Length=10cm

New Radius =3× 3 cm =9cm

new length = 10× 3 = 30cm

we know Volume of cylinder = 2πrh2

Therefore, ratio of volume=  2π(3)2(10)2π(9)2(30)=1/27=1/33

Hence the ratio of volume is 3 times the ratio of dimension

Hence the ratio of volume is 32 times the ratio of dimension

Page 15 Problem 5 Answer

Given: Area of both the rectangle.

To find: Scale factor of the rectangle.

In order to find the solution,

Compare both the rectangle.Making the scale factor xFinding the x, we will find the scale factor

Since, the two rectangles will be similar figures, we can determine the scale factor that was used by comparing the squares of the ratio of the sides with the ratio of their areas.

Let b be the original breadth

Let the scale factor be x

b2(b×x)2=48/12

=(b/bx)2

=4/1

=(1/x)2

=4/1 [Find the square root of both sides]

=1/x

=2/1 [Invert to find x]

=x=1/2 The scale factor of the rectangle is 1/2.

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2 Exercise 2.6

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

Page 14 Problem 1 Answer

Given: The ratio of freshman to sophomores in a drama club is 5: 6. There are 18 sophomores in the drama club.

To Find: How many freshmen are there?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

The ratio of freshman to sophomores in a drama club is Freshman/Sophomores=5/6.

There are 18 sophomores in the drama club.

Freshman/Sophomores=5/6 Freshman

18=5/6

⇒Freshman=5/6×18

⇒Freshman=15

​Number of freshman=15

Page 14 Problem 2 Answer

Given:  Four pounds of apples cost1.96

To Find: unit rate

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

As we have four pounds of apples cost 196

Cost of 4 apples=1.96

Cost of 1 apples=1.96/4

Cost of 1 apple=0.49

​Therefore, cost of 1 apple=0.49 dollars

Page 14 Problem 3 Answer

Given: Sal washed 5 cars in 50 minutes

To Find: unit rate

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

According to the question we have

Minutes for washing 5 cars=50

Minutes for washing 1 car=50/5

Minutes for washing 1 car=10​

Therefore, unit is 10 minutes per car

Page 14 Problem 4 Answer

Given: A giraffe can run 32 miles per hour.

To Find: What is this speed in feet per second?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

According to question a giraffe can run 32 miles per hour.

Using 1 mile =5280 feet and 1 hour =60×60 second

we get Then a giraffe can run

32×miles hour

=32×5280

60×60×feet second

=47 feet per seond

​32×miles per hour =47 feet per seond

Page 14 Problem 5 Answer

Given: proportion is y/4=10/8

To Find: value of variable

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Cross multiplying the given proportion

Y/4=10/8

⇒8y=10×4

⇒y=10×4/8

⇒y=5

​Required value is y=5

Page 14 Problem 6 Answer

Given: proportion is 2/x=30−6

To Find: value of variable

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Cross multiplying the given proportion

2/x=30−6

⇒−12=30×x

⇒x=−12/30

⇒x=−2/5

⇒x=−1/2.5

⇒x=−0.4

​Required value is x=−0.4

Page 14 Problem 7 Answer

Given: proportion is 3/12=−24/m

To Find: value of variable

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Cross multiplying the given proportion

3/12=−24m

⇒1/4 =−24m

⇒m=−24×4

⇒m=−96

​Required value ism=−96

Page 14 Problem 8 Answer

Given: proportion is 3t

10=1/2

To Find: value of variable

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Cross multiplying the given proportion

3t/10=1/2

⇒3t×2=10

⇒t=10/6

⇒t=1.67

​Required value is t=1.67

Page 14 Problem 9 Answer

Given: proportion is 7x=1/0.5

To Find: value of variable

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Cross multiplying the given proportion

7/x=1/0.5

⇒7×0.5=x

⇒x=3.5

​Required value is x=3.5

Page 14 Problem 10 Answer

Given: The scale of his model to the actual car is 1:10

and his model is 181/2 inches long

To Find: How long is the actual car?

Method Used: use of formula of scale

Here, 1 inch model/10 inches actual=18.5 inch model/x inches actual

x=18.5×10/1

=185 inches

​Actual length is=185 inches

Page 14 Problem 11 Answer

Given: The scale on a map of Virginia shows that 1 centimeter represents 30 miles.

The actual distance from Richmond, VA to Washington, DC is 110 miles.

To Find: how many centimeters are between the two cities?

Method Used: use of formula of scale.

Scale is given as dimension on map actual dimension

Here, 1 centimeters represents 30 miles then

30 miles =1 cm

1 mile =1/30cm

110 miles =1/30×110

110 miles =3.67 cm

​Required distance in map is110 miles =3.67 cm

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2 Exercise 2.5

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

Page 13 Problem 1 Answer

Given: C=2πr

To Find: Solve the equation for r.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

C=2πr

⇒C/2π

=r

⇒r=C/2π

Required equation is r=C/2π

Page 13 Problem 2 Answer

Given: y=mx+b

To Find: Solve the equation for m.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

y=mx+b

⇒y−b=mx

⇒m=y−b/x

Required equation ism=y−b/x

Page 13 Problem 3 Answer

Given: 4c=d

To Find: Solve the equation for c.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

4c=d

⇒c=d/4

Required equation is c=d/4

Page 13 Problem 4 Answer

Given: n−6m=8.

To find: equation for n

Method Used: Simplification using addition, subtraction, division or multiplication of integers

Given equation can be written as

n−6m=8

⇒n=8+6m

​Required equation is n=8+6m

Page 13 Problem 5 Answer

Given: 2p+5r=q

To Find: Solve the equation for p.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

2p+5r=q

⇒2p=q−5r

⇒p=q−5r/2

Required equation is p=q−5r/2

Page 13 Problem 6 Answer

Given: −10=xy+z

To Find: Solve the equation for x.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

−10=xy+z

⇒−10−z=xy

⇒x=−(10+z)/y

Required equation is x=−(10+z)/y

Page 13 Problem 7 Answer

Given: a/b=c

To Find: Solve the equation for b.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

A/b=c

⇒a=bc

⇒b=a/c

Required equation is b=a/c

Page 13 Problem 8 Answer

Given: h−4/j=k

To Find: Solve the equation for j.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

h−4/j=k

⇒h−4=kj

⇒j=h−4/k

Required equation is j=h−4/k

Page 13 Problem 9 Answer

Given: c=5p+215

To Find: Solve the equation for p.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

c=5p+215

⇒c−215=5p

⇒p=c−215/5

⇒p=c/5−43

​Required equation is p=c/5−43

Page 13 Problem 10 Answer

Given: c=5p+215 where c=300

To Find: Solve the equation for p

Method Used: Put all the values in p=c−215/5

Given equation can be written asp=c−215/5

p=c−215/5

⇒p=300−215/5

⇒p=85/5

⇒p=17

​Required value is p=17

Page 13 Problem 11 Answer

Given: A=1/2bh

To Find: Solve the equation for b.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Given equation can be written as

A=1/2bh

⇒2A=bh

⇒b=2A/h

Required equation is b=2A/h

Page 13 Problem 12 Answer

Given: A=1/2bh, area is 192 mm2 and height is 12 mm

To Find: Solve the value of base

Method Used: put all the values in b=2Ah

Given equation can be written as b=2A/h

Putting all the values as

b=2A/h

⇒b=2×192/12

⇒b=2×16

⇒b=32 mm

​Required equation is b=32 mm

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2 Exercise 2.4

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

Page 12 problem 1 Answer

Given: 3d+8=2d−17

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

3d+8=2d−17

⇒3d−2d=−17−8

⇒d=−25

​Required solution is d=−25

Page 12 problem 2 Answer

Given: 2n−7=5n−10

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

2n−7=5n−10

⇒5n−2n=−7+10

⇒3n=3

⇒n=1

​Required solution is n=1

Page 12 problem 3 Answer

Given: p−15=13−6p

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

p−15=13−6p

⇒p+6p=13+15

⇒7p=28

⇒p=28/7

⇒p=4

​Required solution is p=4

Page 12 problem 4 Answer

Given: −t+5=t−19

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

−t+5=t−19

⇒t+t=5+19

⇒2t=24

⇒t=12

​Required solution is t=12

Page 12 problem 5 Answer

Given: 15x−10=−9x+2

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

15x−10=−9x+2

⇒15x+9x=10+2

⇒24x=12

⇒x=1/2

Required solution is x=1/2

Page 12 problem 6 Answer

Given: 4n+6−2n=2(n+3)

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

4n+6−2n=2(n+3)

⇒4n+6−2n=2n+6

⇒4n−2n−2n=6−6

⇒0=0

​The equation will always be true but the solution is not present.

Page 12 problem 7 Answer

Given: 6m−8=2+9m−1

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

6m−8=2+9m−1

⇒9m−6m=−8−2+1

⇒3m=−9

⇒m=−3​

Required solution is m=−3

Page 12 problem 8 Answer

Given: −v+5+6v=1+5v+3

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

−v+5+6v=1+5v+3

⇒5v+5=5v+4

​This equation is wrong.

the solution is not present as the equation is wrong.

Page 12 problem 9 Answer

Given: 2(3b−4)=8b−11

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

2(3b−4)=8b−11

⇒6b−8=8b−11

⇒2b=3

⇒b=3/2

Required solution is b=3/2

Page 12 problem 10 Answer

Given: 2(3b−4)=8b−11

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Simplifying and solving the equation as

2(3b−4)=8b−11

⇒6b−8=8b−11

⇒2b=3

⇒b=3/2

Required solution is b=3/2

Page 12 problem 11 Answer

Given:  One company offers a starting salary of 28000 with a raise of 3000 and other company offers 36000 with a raise of 2000.

To Find:  After how many years would Janine’s salary be the same with both companies?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Let  us consider x represent the number of years then

At the first company, after x years Jennie’s salary will be: 28,000+3,000x

At the second company, after x years Jennie’s salary will be: 36,000+2,000x

Equating both values as

28,000+3,000x=36,000+2,000x

⇒1,000x=36,000−28,000

⇒x=8

​Required number of years is x=8

Page 12 problem 12 Answer

Given:  One company offers a starting salary of 28000 with a raise of 3000 and other company offers 36,000 with a raise of 2000 .

To Find:  After how many years would Janine’s salary be the same with both companies?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

As we have calculated that after 8 years, salaries are same.

Putting this in any one of the salaries as

28,000+3,000∗x=28,000+3,000∗8

⇒28,000+3,000∗8=52,000.

​After 8 years the salary will be: 28,000+3,000∗8=52,000.

Page 12 problem 13 Answer

Given:  Xian and his cousin both collect stamps. Xian has 56 stamps, and his cousin has 80 stamps.

Both have recently joined different stamp-collecting clubs.

Xian’s club will send him 12 new stamps per month, and his cousin’s club will send him 8 new stamps per month.

To Find:   After how many months will Xian and his cousin have the same number of stamps?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Let us consider the number of months be x

According to the question56+12x=80+8x

56+12x=80+8x

⇒12x−8x=80−56

⇒4x=24

⇒x=6

​After 6 many months will Xian and his cousin have the same number of stamps.

Page 12 problem 14 Answer

Given:  Xian and his cousin both collect stamps. Xian has 56 stamps, and his cousin has 80 stamps.

Both have recently joined different stamp-collecting clubs.

Xian’s club will send him 12 new stamps per month, and his cousin’s club will send him 8 new stamps per month.

To Find:  How many stamps will that be?

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

As we have calculated, After 6 many months will Xian and his cousin have the same number of stamps.

Then number of stamps is

56+12∗x=56+12∗6

⇒56+12∗6=128

80+8x=80+8∗4

⇒80+8∗4=128

​After six months number of stamps is 128

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2 Exercise 2.3

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

Page 11 Problem 1 Answer

Given: −4x+7=11.

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know, −4x+7=11

Subtract both sides by 7,

−4x+7−7=11−7

−4x=4

−4x=4

Divide both sides by −4,−4x

−4=4/−4

x=−1

Therefore, x=−1.

Page 11 Problem 2 Answer

Given: 17=5y−3.

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know,

17=5y−3

Adding both sides by 3,

17+3=5y−3+3

20=5y

20=5y

Divide both sides by 5,

20/5=5y/5

y=4

Therefore, y=4.

Page 11 Problem 3 Answer

Given: −4=2p+10.

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know, −4=2p+10

Subtract both sides by 10,

−4−10=2p+10−10

−14=2p

−14=2p

Divide both sides by 2,−14/2

=2p/2

p=−7

Therefore, p=−7.

Page 11 Problem 4 Answer

Given: 3m+4=1.

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know, 3m+4=1

Subtract both sides by 4,

3m+4−4=1−4

3m=−3

3m=−3

Divide both sides by 3,3m/3

=−3/3

m=−1

Therefore, m=−1.

Page 11 Problem 5 Answer

Given: 12.5=2g−3.5.

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know, 12.5=2g−3.5

Add both sides by 3.5,

12.5+3.5=2g−3.5+3.5

16=2g

16=2g

Divide both sides by 2,16/2

=2g/2

g=8

Therefore, g=8.

Page 11 Problem 6 Answer

Given: 7/9=2n+1/9.

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know, 7/9=2n+1/9

Subtract both sides by 1/9,7/9−1/9

=2n+1/9−1/9

7−1/9=2n/6

9=2n/6

9=2n

Divide both sides by 2,6/9×2

=2n/2

n=1/3

Therefore, n=1/3.

Page 11 Problem 7 Answer

Given: −4/5t+2/5=2/3.

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know, −4/5t+2/5=2/3

Subtract both sides by 2/5,−4/5t+2/5−2

5=2/3−2/5−4/5

t=10−6/15−4/5

t=4/15−4/5

t=4/15

Multiple both sides by −5/4,−4/5t×−5/4

=4/15×−5/4

t=−1/3

Therefore, t=−1/3.

Page 11 Problem 8 Answer

Given: −2(b+5)=−6.

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know, −2(b+5)=−6

−2b−10=−6

Add both sides by 10,

−2b−10+10=−6+10

−2b=4

−2b=4

Divide both sides by −2,

−2b/−2=4/−2

b=−2

Therefore, b=−2.

Page 11 Problem 9 Answer

Given: 8=4(q−2)+4.

To Find: Solve the equation.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know,

8=4(q−2)+4

8=4q−8+4

8=4q−4

Add both sides by 4,

8+4=4q−4+4

12=4q

12=4q

Divide both sides, by 4,12/4=4q/4

q=3

Therefore, q=3.

Page 11 Problem 10 Answer

Given: 3x−8=−2.

To Find: The value of x−6.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know,

3x−8=−2

Add both sides by 8,

3x−8+8=−2+8

3x=6

Divide both sides by 3,3x/3=6/3

x=2

We know, x=2.

Substitute x=2 in x−6,

x−6=2−6

x−6=−4.

Therefore, the value of x−6=−4.

Page 11 Problem 11 Answer

Given: −2(3y+5)=−4.

To Find: The value of 5y.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

We know,

−2(3y+5)=−4

−6y−10=−4

Add both sides by 10,

−6y−10+10=−4+10

−6y=6

Divide both sides by −6,−6y/−6=6/−6

y=−1

We know, y=−1.

Substitute y=−1 in 5y,

5y=5×−1=−5

Therefore, the value of 5y=−5.

Page 11 Problem 12 Answer

Given: The two angles shown form a right angle.

To Find: Write and solve an equation to find the value of x.

Method Used: Simplification using addition, subtraction, division or multiplication of integers.

Holt Algebra 1 Homework and Practice Workbook, 1st Edition, Chapter 12

We know, the right angle is equal to 90∘.

So, as per the given information in the question,

3x−5+2x=90

5x−5=90

Add both sides by 5,

5x−5+5=90+5

5x=95

Divide both sides by 5,5x/5=95/5

x=19

Therefore, the value of x=19∘.

Page 11 Problem 13 Answer

Given: For her cellular phone service, Vera pays $32 a month, plus $0.75 for each minute over the allowed minutes in her plan.

Vera received a bill for $47 last month.

To Find: For how many minutes did she use her phone beyond the allowed minutes?

Method Used: Using the information in the question form a linear equation and simplify using addition, subtraction, division or multiplication of integers.

We know, Vera pays $32 a month, plus $0.75 for each minute over the allowed minutes in her plan. Vera received a bill for $47 last month.

Let us assume that Vera used her phone for x minutes beyond allowed minutes.

So, as per the question equation,

0.75x+32=47

Subtract both sides by 32,

0.75x+32−32=47−32

0.75x=15

0.75x=15

Divide both sides by 0.75

0.75x/0.75=15/0.75

x=20

Therefore, Vera used her phone for 20 minutes beyond allowed minutes.

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

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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.

Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2 Exercise 2.1

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

Page 9 Problem 1 Answer

Given: g−7=14

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

g−7=15

Add 7 on both the sides of the equation,

g−7+7=15+7

g=22

​After solving the given expression, the value of g is 22.

Page 9 Problem 2 Answer

Given: t+4=6

To Solve each equation.

To solve the given expression, we will subtract the same thing on both the sides.

t+4=6

Subtract 4 from both the sides,

t+4−4=6−4

t=2

​After solving the given expression, the value of t is 2.

Page 9 Problem 3 Answer

Given: 13=m−7

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

13=m−7

Add 7 on both the sides,

13+7=m−7+7

20=m

​After solving the given expression, the value of m is 20.

Page 9 Problem 4 Answer

Given: x+3.4=9.1

To Solve each equation.

To solve the given expression, we will subtract the same thing on both the sides.

x+3.4=9.1

Subtract 3.4 from both the sides,

x+3.4−3.4=9.1−3.4

x=5.7

After solving the given expression, the value of x is 5.7.

Page 9 Problem 5 Answer

Given: n−3/8=1/8

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

Given: n−3/8=1/8

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

Simplify,

n−3/8+3/8=1/8+3/8

n=4/8

On simplifying 4/8 we get, 1/2.

The value of n is 1/2.

Page 9 Problem 6 Answer

Given: p−1/3=2/3

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

p−1/3=2/3

Add 1/3 on both the sides,

p−1/3+1/3=2/3+1/3

Simplify,

p−1/3+1/3=2/3+1/3

p=3/3

On simplifying 3/3 we get 1.

The value of p is 1.

Page 9 Problem 7 Answer

Given: −6+k=32

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

−6+k=32

Add 6 on both the sides.

−6+k=32

−6+6+k=32+6

k=38

​The value of k is 38.

Page 9 Problem 8 Answer

Given: 7=w+9.3

To Solve each equation.

To solve the given expression, we will subtract the same thing on both the sides.

7=w+9.3

Subtract 9.3 from both the sides,

7−9.3=w+9.3−9.3

−2.3=w

​The value of w is −2.3.

Page 9 Problem 9 Answer

Given: y−57=−40

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

y−57=−40

Add 57 on both the sides,

y−57+57=−40+57

y=17

​The value of y is 17.

Page 9 Problem 10 Answer

Given: −5.1+b=−7.1

To Solve each equation.

To solve the given expression, we will add the same thing on both the sides.

−5.1+b=−7.1

Add 5.1 on both the sides,

−5.1+5.1+b=−7.1+5.1

b=−2

​The value of b is −2.

Page 9 Problem 11 Answer

Given: a+15=15

To Solve each equation.

To solve the given expression, we will subtract the same thing on both the sides.

a+15=15

Subtract 15 from both the sides,

a+15−15=15−15

a=0

​The value of a is 0.

Page 9 Problem 12 Answer

To write and solve an equation to determine Marietta’s hourly wage before her raise.

Show that your answer is reasonable.

Assume hourly wage and solve the resulting equation.

Let x be her hourly wage before her raise. She was given a raise of $0.75 an hour, which bought her hourly wage to $12.25.

So, the equation, which represents this situation, will be x+0.75=12.25.

Solve for x to get

x=12.25−0.75

x =11.5

​So, her hourly wage before the raise was $11.5.

So, her hourly wage before the raise was  $11.5.Check if this is true or if you answer is reasonable, by plugging x=11.5 back in the said equation 11.5+0.75=12.25.

Page 9 Problem 13 Answer

Given: Brad grew 41/4 inches this year and is now 567/8 inches tall.

To find Brad’s height at the start of the year.

Show that your answer is reasonable.

We will convert both the given mixed fractions into improper fractions and then we will subtract the value of height grown from the total height to get Brad’s height at the start of the year.

Brad grew 41/4 inches this year. We will convert the mixed fraction to an improper fraction.

We will multiply the denominator of the fraction with the whole number and add whatever there is in the numerator, it can now be written as,

41/4=(4×4)+1/4

On simplifying, we get: 41/4=17/4

Therefore, Brad grew 17/4 inches this year.

Currently Brad is 567/8 inches tall. We will convert the mixed fraction to an improper fraction.

We will multiply the denominator of the fraction with the whole number and add whatever there is in the numerator, it can now be written as,

567/8=(56×8)+7/8

On simplifying, we get:567/8=455/8

Therefore, Brad is 455/8 tall.

Now brads height at the start of the year will be the subtraction of the current height and the height grown.

It can be written as,

⇒455/8−17/4

On taking the lowest common multiple in the denominator, we get,

⇒455/8−34/8

On taking the denominator common, we get,

⇒455−34/8

On simplifying the terms in the numerator, we get,

⇒421/8

Brad’s height at the start of the year is 421/8.

To show the answer reasonable add Brad’s height at the start of the year in his grown height i.e.,

421/8+17/4=421+34/8

421+34/8=455/8

From Step 2 we can write 455/8 as 567/8 which is the current height of Brad’s.

Page 9 Problem 14 Answer

To solve an equation to find Heather’s practice time.

Heather’s finishing time is 2.6 seconds less than her practice time, so ,

Finishing time=Practice time−2.6 seconds.

Let P= practice time.

58.4=P−2.6

Add 2.6 to both sides,61=P.

Heather’s practice time is 61 seconds.

Page 9 Problem 15 Answer

Given: Radius of Earth 6378.1 km

To determine the radius of Mars.

Let the radius of Mars be R

As it is given that the radius of Earth is 2981.1 Km bigger than the radius of Mars,

So,6378.1km=R+(2981.1km)

Subtract (2981.1 km) from both sides of the equation.

6378.1−2981.1=R+2981.1−2981.1

3397=R

​The Radius of Mars is 3397Km.

To show the answer reasonable add the radius of Mars in 2981.1 we will get the Radius of earth.

Geometry Homework Practice Workbook 1st Edition Chapter 2 Exercise 2.8 Inductive Reasoning and Conjecture

by

Geometry Homework Practice Workbook 1st Edition Chapter 2 Inductive Reasoning and Conjecture

Page 29 Problem 1 Answer

We are given a figure, in which m∠5=22 and one angle measures 90.

We are required to find the measure of m∠6.

Here, we will use the fact that total sum of angles that lie on a line is 180.

As all the angles lie on the same line, by angles on a straight line property we have,

90+m∠5+m∠6=180

90+22+m∠6=180

m∠6=180−112

m∠6=68

​In the given figure, by angles on a straight line property, we have m∠6=68 when m∠5=22.

Page 29 Problem 2 Answer

Here in the question, we have been given a figure in which m∠1=38∘.

We have to find the measure of each angle given in the figure.

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 2

In the figure, we can see that we only have two angles ∠1,∠2.

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 2 1

And by the vertical opposite angle m∠1=m∠2.Implies that m∠2=38∘.

Hence from the given figure and m∠1=38∘, by the vertically opposite angles, we get m∠2 as 38∘.

Page 29 Problem 3 Answer

In the question, we have been given a figure and also we have:

m∠13=4x+11

m∠14=3x+1

​We have been asked to find the measure of each numbered angle and name the theorems that justify our work.

Using the linear pair theorem, we will find the result.

Here in the given figure, by the linear pair, we get

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 3 1

∠13+∠14=180∘

4x+11+3x+1=180∘

7x=180−12

x=168/7

x=24∘

Now substituting the value of x=24∘ in the angles, we get

m∠13=4×24+11

=107∘m∠14=3×24+1

=73∘

So, by linear pair theorem, we get the measure of each angle.

Hence we get the measure of each angle by the linear pair in the given figuer as:

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 3

m∠13=107∘

m∠14=73∘​

Page 29 Problem 4 Answer

We have been given that in the given figure, ∠9,∠10 are complementary angles, ∠7 congruent to ∠9, m∠8=41.

We have to find the measure of each angle.

Using the linear pair theorem, we will get the measure.

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 4

By the linear pair, we get

∠7+∠8+∠9+∠10=180

∠7+41+90=180

∠7=49

Now given that ∠7≅∠9.

Thus we get 49∘=∠9.

Also, we have given that ∠9,∠10 are complementary angles.

Thus we get

∠9+∠10=90∘

49+∠10=90∘

∠10=41∘

So, we get the measure of all the angles.

Hence we get the measure of all the angles in the figure as: ∠7=49∘

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 4 1

by linear pair ∠10=41∘

by complement ∠9=49∘by congruence

Page 29 Problem 5 Answer

In the question, we have been given a figure and also given that ∠QPS≅∠TPR and also an incomplete table.

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 5

We have to complete the table with the proof of ∠QPR≅∠TPS.

Using the given information and figure we will complete it.

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 5 1

Here the complete table is:

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 5 2

Hence with the help of the given figure, we had completed the table and proved that ∠QPR≅∠TPS

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 5 3

Page 30 Problem 6 Answer

Here, we have given a figure in which a straight line is divided by a line into two angles, that are ∠1,∠2.Further, we have given

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 6

m∠1=x+10

m∠2=3x+18.

Thus we just have to find the value of m∠1,m∠2.

As we have given

m∠1=x+10……(1)

m∠2=3x+18……(2)

So we also know according to the property of straight line,m∠1+m∠2=180∘

i.e.x+10+3x+18=180∘

4x+28=180∘

4x=180−28

x=152/4

We get

x=38∘

Now putting x in eq. 1 and 2, we will have

m∠1=x+10

=38+10

=48∘and

m∠2=3x+18

=3⋅38+18

=114+18

=132∘

Hence by solving

m∠1=x+10

m∠2=3x+18​

by using the given Fig.1 we get m∠1=48∘

m∠2=132∘.

Page 30 Problem 7 Answer

Here, we have given a figure in which we have given three angles, that are ∠3,∠4,∠5 .Further, we can say we have given

m∠3=90∘

m∠4+m∠5=90∘.

Thus we just have to find the value of m∠4,m∠5.

As we have given ​m∠4=2x−5……(1)

m∠5=4x−13……(2)​

And from the given figure we also have m∠4+m∠5=90∘

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 7

Thus we will have

2x−5+4x−13=90∘

6x−18=90

6x=108

x=18∘

Now we will just put the value of x in eq. 1 and 2 respectively,

we will have

m∠4=2x−5

=2⋅18−5

=36−5

=31∘and

m∠5=4x−13

=4⋅18−13

=59∘

Hence by solving

m∠4=2x−5

m∠5=4x−13

​by using the given Fig.1.

we get m∠4=31∘

m∠5=59∘

Page 30 Problem 8 Answer

Here, we have given

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 8

And we have to find the unknown value of the angles m∠6 and m∠7 by using the given figure.

As we know that a theorem states that the two opposite angles are equal when two straight lines intersect, forming four angles.

So now we can write m∠6=m∠7

That implies

7x−24=5x+14

7x−5x=14+24

2x=38

x=19∘

Thus putting x in eq. 1 and 2, we will have

m∠6=7x−24

=7⋅19−24

=133−24

=109∘and

m∠7=5x+14

=5⋅19+14

=95+14

=109∘

Hence by solving m∠6=7x−24

m∠7=5x+14​ by using the given Fig.1.we get m∠6=m∠7=109∘.

Page 30 Problem 9 Answer

Here we have given straight lines referred to as the road names.

So as the figure says, Barton rode and Olive tree lane is making 90∘angle where Tryon street and Olive tree lane is making 57∘angle.

So as we have been told, we just need to find the acute angle made by Tryon Street with Barton Road.

As we know that

Geometry, Homework Practice Workbook, 1st Edition, Chapter 2 Inductive Reasoning and Conjecture 9

The angle between Barton Road and Olive tree lane+The angle between Barton Road and Tyron St.+

Tyron St. and Olive tree lane=180∘

⇒90∘+The angle between Barton Road and Tyron St.+57∘

⇒The angle between Barton Road and Tyron St.=180∘−90∘−57∘

=33∘

So the measure of the acute angle Tryon Street forms with Barton Road will be

= 90∘+The angle between Barton Road and Tyron St.

=90∘+33∘

=123∘

Hence by using the given Fig.1, we have measured the acute angle Tryon Street forms with Barton Road is 123∘.