Holt Algebra 1

• Chapter 1 Exercise 1.1
• Chapter 1 Exercise 1.2
• Chapter 1 Exercise 1.3
• Chapter 1 Exercise 1.4
• Chapter 1 Exercise 1.5
• Chapter 1 Exercise 1.6
• Chapter 1 Exercise 1.7

• Chapter 1 Exercise 1.8
• Chapter 2 Exercise 2.1
• Chapter 2 Exercise 2.2
• Chapter 2 Exercise 2.3
• Chapter 2 Exercise 2.4
• Chapter 2 Exercise 2.5
• Chapter 2 Exercise 2.6
• Chapter 2 Exercise 2.7

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

Question. Find the square root of √36.

Given:−√36

To find the square root of the given number. We will first resolve the number as the product of primes, then take one factor out of each pair.

Writing the number as a product of primes

−√36=−√2×18

−√36=−√2×2×9

−√36=−√2×2×3×3

So,−√36=−(2×3)

∴−√36=−6

Hence, The square root of−√36 is −6.

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Holt Algebra 1 Homework and Practice Workbook Chapter 1 Exercise 1.5 solutions

Question. Find the square root of √1/49

Given:√1/49

To find the square root of the given number.

We will first resolve the number as the product of primes, then take one factor out of each pair.

First we will find the square root of numerator √1= 1

Now, finding the square root of denominator

Writing the number as product of primes

49=7×7/√49

49 =√7×7

∴√49=7

Thus,√1/49

=√1/7

Hence, the square root of √1/49 is √1/7.

Chapter 1 Exercise 1.5 Answers Holt Algebra 1 Practice Workbook 1st Edition Question. Find the square root of √196

Question. Find the square root of √4/25

Given:−√4/25

To find the square root of the given number.

We will first resolve the number as the product of primes, then take one factor out of each pair.

First we will find the square root of numerator

√4=√2×2

∴√4/=2​

Now, finding the square root of denominator

√25=√5×5

∴√25

√25 =5​

So,−√4/25=−2/5

Hence, the square root of−√4/25

=−2/5.

Question. A contractor needs to cut a piece of glass to fit a square window. Area of the window is 12 ft2. Find the length of the side of the window to the nearest tenth of a foot and find the square root.

Given: A contractor needs to cut a piece of glass to fit a square window.

The area of the window is 12 ft².

To find the length of the side of the window to the nearest tenth of a foot.

We will first resolve the number as the product of primes, then take one factor out of each pair.

Since, the area of square window is 12 ft²

To find the length of the side of the window we have to find the square root.

Writing the number as product of primes

√12=√2×2×3

√12=2√3

√12=2×1.732

√12=3.464

​To round to the tenths, we need to look at the hundredths

∴√12=3.5

Hence, the length of the side of the window to the nearest tenth of a foot is 3.5 feet.

Step-By-Step Solutions For Exercise 1.5 Chapter 1 Holt Algebra 1 Homework Workbook

Question. A piece of cloth must be cut to exactly cover a square table. The area of the table is 27ft². Find the length of the side of the table to the nearest tenth of a foot and find the square root.

Given: A piece of cloth must be cut to exactly cover a square table.

The area of the table is 27ft²

To find the length of the side of the table to the nearest tenth of a foot.

We will first resolve the number as the product of primes, then take one factor out of each pair.

Since, the area of table is 27ft²

To find the length of the side of the table we have to find the square root.

Writing the number as product of primes

√27=√3×9

√27=√3×3×3

√27=3√3

√27=3×1.732

∴√27=5.196

​To round to the tenths, we need to look at the hundredths

∴√27=5.2

Hence, the length of the side of the table to the nearest tenth of a foot is 5.2 feet.

Examples Of Problems From Exercise 1.5 Chapter 1 In Holt Algebra 1 Workbook

Exercise 1.5 Solutions for Chapter 1 Holt Algebra 1 Homework and Practice Workbook Question. Write all the classifications that apply to √2 is a real number.

Given: √2

To write all the classifications that apply to given real number.

The given number √2 is irrational

Hence, the given number √2 is irrational.

Question. Write all the classifications that apply to 2/3 is a real number.

Given: 2/3

To write all the classifications that apply to given real number.

The given number 2/3 is rational.

Hence, the given number 2/3 is rational.

Common Core Exercise 1.5 Chapter 1 solutions detailed Holt Algebra 1 Practice Workbook 1st Edition

Question. Write all the classifications that apply to 10 is a real number.

Given: 10

To write all the classifications that apply to given real number.

The given number−10 is rational and integer.

Hence, the given number−10 is rational and integer.

Student Edition Chapter 1 Exercise 1.5 Holt Algebra 1 Homework Workbook solutions guide” Question. Write all the classifications that apply to √81 is a real number.

Given: √81

To write all the classifications that apply to given real number.

The given number √81 is rational as the square root of √81 is 9 and it is integer, natural and a whole number also.

Hence, the given number √81 is rational, integer, natural and a whole number.

Question. Write all the classifications that apply to 0 is a real number.

Given: 0

To write all the classifications that apply to given real number.

The given number 0 is rational, integer and whole number.

Hence, the given number 0 is rational, integer and whole number.

Step-by-step answers for Exercise 1.5 Chapter 1 in Holt Algebra 1 HomeworkAnd Practice Workbook

Question. Write all the classifications that apply to 1 is a real number.

Given: 1

To write all the classifications that apply to given real number.

The given number 1 is rational, integer , whole and natural number.

Hence, the given number 1 is rational, whole, integer and natural number.

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

Holt Algebra 1 Homework And Practice Workbook Chapter 1 Exercise 1.6 Solutions Page 6 Problem 1 Answer

Question. Simplify the 5 . 3 + 2(4) expression. Find the value of expression.

Given: Algebraic expression is 5⋅3+2(4)

To Find: Simplify given expression.

For finding the value of expression we apply BODMAS rule.

It is given that 5⋅3+2(4)

Now, using BODMAS we get,

⇒5⋅3+2(4)

5⋅3+2(4) =5⋅3+8

5⋅3+2(4) =15+8

5⋅3+2(4) =23

​Thus, the value of 5⋅3+2(4) is 23

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Question. Simplify the -2[7 + 6(3 – 5)] expression. Find the  value of expression.

Given: Algebraic expression is −2[7+6(3−5)]

To Find: Simplify given expression.

For finding the value of expression we apply BODMAS rule.

It is given that,−2[7+6(3−5)]

Now, using BODMAS we get,

⇒−2[7+6(3−5)]

−2[7+6(3−5)] =−2[7+6(−2)]

−2[7+6(3−5)] =−2[7−12]

−2[7+6(3−5)] =−2[−5]

−2[7+6(3−5)] =10

​Thus, the value of −2[7+6(3−5)] is 10

Question. Simplify the -7-(24 ÷ 8) expression. Find the value of the expression.

Question. Simplify the -6.3+1 expression. Find the value of ethe xpression.

Given: Algebraic expression is∣ −6⋅3+∣

∣−3(−4+23)∣

To Find: Simplify given expression.

For finding the value of expression we apply BODMAS rule.

It is given that,∣−6⋅3+∣

∣−3(−4+23)∣

Examples of absolute value are ​

∣−1∣=1

∣−14∣=14

∣1∣=1

∣0∣=0

​Now, using definition of absolute value & BODMAS we get,

​⇒∣−6⋅3+∣

∣−3(−4+23)∣

=−6⋅3+∣−3(−4+8)∣

=−6⋅3+∣−3(4)∣

=−6⋅3+∣−12∣

=−6⋅3+12

=−18+12

=−6

​Thus, the value of −6⋅3+∣−3(−4+23) is −6

Chapter 1 Exercise 1.6 Answers Holt Algebra 1 Practice Workbook 1st Edition

Question. Simplify the −16+4/2(√13−4) expression. Find the  value of expression.

Given: Algebraic expression is −16+4/2(√13−4)

To Find: Simplify given expression.

For finding the value of expression we apply BODMAS rule.

It is given that −16+4/2(√13−4)

Now, using BODMAS we get,

⇒−16+4/2(√13−4)

−16+4/2(√13−4) =−12/2(√13−4)

−16+4/2(√13−4) =−12/2(√9)

−16+4/2(√13−4) =−12/2(3)

−16+4/2(√13−4) =−12/6

−16+4/2(√13−4) =−2

​Thus, the value of −16+4/2(√13−4) is −2

Question. Simplify the 3−y²+7 expression. Find the  value of expression.

Given: Algebraic expression is 3−y²+7

To Determine: Evaluate given expression for the given value of the variable.

For finding the value of expression we put the y=5 in given expression.

The given expression is 3−y²+7

Now put y=5 & apply BODMAS rule then we get,

⇒3−y²+7

3−y²+7 =3−5²+7

3−y²+7 =3−25+7

3−y²+7 =10−25

3−y²+7 =−15

​Thus, the value of3−y²+7 for y=5 is −15

Question. Simplify the −3(x+12⋅2) expression. Find the  value of expression.

Given: Algebraic expression is −3(x+12⋅2)

To Determine: Evaluate given expression for the given value of the variable.

For finding the value of expression we put the x=−8 in given expression.

The given expression is −3(x+12⋅2)

Now put x=−8 & apply BODMAS rule then we get,

⇒−3(x+12⋅2)

−3(x+12⋅2) =−3(−8+12⋅2)

−3(x+12⋅2) =−3(−8+24)

−3(x+12⋅2) =−3(16)

−3(x+12⋅2) =−48

​Thus, the value of −3(x+12⋅2) for x=−8 is −48

Step-By-Step Solutions For Exercise 1.6 Chapter 1 Holt Algebra 1 Homework Workbook

Question. Evaluate the (m+6)÷(2-5) expression for the value of the variable.

Given: Algebraic expression is (m+6)÷(2−5)

To Determine: Evaluate given expression for the given value of the variable.

For finding the value of expression we put the m=9 in given expression.

The given expression is(m+6)÷(2−5)

Now put m=9 & apply BODMAS rule then we get,

⇒(m+6)÷(2−5)

(m+6)÷(2−5) =(9+6)÷(2−5)

(m+6)÷(2−5) =(15)÷(2−5)

(m+6)÷(2−5) =(15)÷(−3)

(m+6)÷(2−5) =15/−3

(m+6)÷(2−5) =−5​

Thus, the value of (m+6)÷(2−5) for m=9 is −5

Question. Evaluate the -5t + 12 – 1/2t expression for the value of the variable.

Given: Algebraic expression is −5t+12−1/2t

To Determine: Evaluate given expression for the given value of the variable.

For finding the value of expression we put the t=−10 in given expression.

The given expression is−5t+12−1/2t

Now put t=−10 & apply BODMAS rule then we get,

⇒−5t+12−1/2t

−5t+12−1/2t =−5(−10)+12−1/2(−10)

−5t+12−1/2t =50+12−1/2(−10)

−5t+12−1/2t =50+12−(−5)

−5t+12−1/2t =50+12+5

−5t+12−1/2t =50+17

−5t+12−1/2t =67

​Thus, the value of −5t+12−1/2t for t=−10 is 67

Question. The product of 6 and the sum of 3 and 20 and translate each word phrase into a numerical or algebraic expression.

Given: The product of 6 and the sum of 3 and 20

To Determine: Translate each word phrase into a numerical or algebraic expression.

For translating the word phrase into a numerical or algebraic expression we use different operation such as Addition, Subtraction, Multiplication & Division.

It is given that The product of 6 and the sum of 3 and 20 Now, the word phrase into a numerical or algebraic expression using different operation then we get,sum of 3 and 20 is denoted as 3+20

Therefore,The product of 6 and the sum of 3 and 20 is denoted as 6×(3+20)

Now, apply BODMAS rule then we get,

⇒6×(3+20)

6×(3+20) =6×(23)

6×(3+20) =138

​Thus, the word phrase the product of 6 and the sum of 3 and 20 into a numerical or algebraic expression is denoted as 6×(3+20) and its value is 138

Question. The absolute value of the difference of m and -15 and translate each word phase into a numerical or algebraic expression.

Given: The absolute value of the difference of m and −15

To Determine: Translate each word phrase into a numerical or algebraic expression.

For translating the word phrase into a numerical or algebraic expression we use different operation such as Addition, Subtraction, Multiplication & Division.

It is given that The absolute value of the difference of m and −15

Now, the word phrase into a numerical or algebraic expression using different operation then we get, difference of is denoted as m−(−15)

Therefore, The absolute value of the difference of m and −15  is denoted as ∣m−(−15)∣

Thus, the word phrase the absolute value of the difference of m and −15 into a numerical or algebraic expression is denoted as ∣m−(−15)∣

Exercise 1.6 Solutions For Chapter 1 Holt Algebra 1 Homework And Practice Workbook

Question. The hottest recorded day in Florida history was 109°F, which occurred on June 29, 1931 in Monticello. Convert this temperature to degrees Celsius. Round your answer to the nearest tenth of a degree.

Given: The hottest recorded day in Florida history was 109∘F, which occurred on June 29, 1931 in Monticello.

To Determine: Convert this temperature to degrees Celsius. Round your answer to the nearest tenth of a degree.

For converting this temperature to degrees Celsius we use 5/9(F−32)

It is given that the hottest recorded day in Florida history was 109∘F, which occurred on June 29, 1931 in Monticello.

Hence, Put F=109∘ in 5/9(F−32)

then we get,

⇒5/9(F−32)

=5/9(109−32)

​Now, apply PEDMAS rule then we get,

=5/9(109−32)

=5/9(77)

=0.5555(77)

=42.7777

≈42.8∘

C (Rounding off to the nearest tenth)

​Thus, the temperature in degrees Celsius is 42.8∘C

Question. The coldest recorded day in Florida history was about -18.°C, which occurred on February 13m 1899 in the city of Tallahassee. Convert this temperature to degrees Fahrenheit. Round  your answer to the nearest tenth of a degree.

Given: The coldest recorded day in Florida history was about −18.9∘C, which occurred on February 13, 1899 in the city of Tallahassee.

To Determine: Convert this temperature to degrees Fahrenheit. Round your answer to the nearest tenth of a degree.

For converting this temperature to degrees Fahrenheit we use 9/5c+32

It is given that the coldest recorded day in Florida history was about −18.9∘C, which occurred on February 13,1899 in the city of Tallahassee.

Hence,put c=−18.9∘C in 9/5c+32 then we get,

⇒9/5c+32

=9/5(−18.9)+32

​Now, apply BODMAS rule then we get,

=9/5(−18.9)+32

9/5(−18.9)+32 =1.8(−18.9)+32

9/5(−18.9)+32 =−34.02+32

9/5(−18.9)+32 =−2.02

9/5(−18.9)+32 ≈−2∘F

​Thus, the temperature in degrees Fahrenheit is −2∘F

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

Holt Algebra 1 Homework And Practice Workbook Chapter 1 Exercise 1.7 Solutions Page 7 Problem 1 Answer

Given: 18+9+1+12

To Find: Sum of the given equation

In order to solve the equation, we will divide the equation in two parts and then solve it 18+9+1+12

Lets divide the equation in two parts in order to solve the equation.

=(18+9)+(1+12)

Now, adding the 2 equations separately, we get: =(27)+(13)

To get the final answer, remove the brackets and add the final two digits.

=27+13

=40

​The solution of the given equation is 40.

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Page 7 Problem 2 Answer

7×15×12

Lets divide the equation in two parts in order to solve the equation.

=(7×15)×2

Now, multiplying the bracket equation, we get:

=(105)×2

To get the final answer, remove the brackets and multiply the final two digits.

=105×2

=210

​Given:7×15×2

To Find: Multiplication of the given equation

In order to solve the equation, we will divide the equation in two parts and then solve it.

The solution of the given equation is 210.

Chapter 1 Exercise 1.7 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 7 Problem 3 Answer

​The Solution to the given equation is 24.

Page 7 Problem 4 Answer

Given:−5×7×20

To Find: Multiply the given equation

In order to solve the equation, we will divide the given equation in two parts and multiply them separately. Since one digit is in negative, then the final answer will also be in negative

−5×7×20

Let’s divide the equation in two parts.

=−5×(7×20)

Now, by multiplying the bracket equation, we get,

−5×(7×20) =−5×(140)

Since, one digit is in negative, we will apply the rule of Negative number×

Positive number=Negative number.

Therefore, the final solution will also be in negative. We will remove the bracket and simply multiply the equation.

=−5×140

−5×140 =−700

​The Solution of the given equation is −700.

Step-By-Step Solutions For Exercise 1.7 Chapter 1 Holt Algebra 1 Homework Workbook Page 7 Problem 5 Answer

Given:−12+3+12+19

To Find: Sum of the given equation

In order to solve the equation,We will divide the equation in two partsWe will keep the negative number on the left side and positive on the right sideAdd the positive numbers first and then subtract with the negative number.

The final solution sign should be the same sign as the number with the greater absolute value.

let’s divide the equation in two parts and keep the negative number on the left and positive number on the right.

=(−12)+(3+12+19)

Now, add all the positive numbers:

(−12)+(3+12+19) =(−12)+(34)

Remove the bracket, and then solve the equation. Since12<34, then the final answer will be in positive number. We just need to subtract the final equation.

=−12+34

−12+34 =22

​The Solution of the given equation is 22

Exercise 1.7 Solutions For Chapter 1 Holt Algebra 1 Homework And Practice Workbook Page 7 Problem 6 Answer

Given:−1×5×9×2

To Find: Multiply the given equation

In order to solve the equation, we will divide the given equation in two parts and multiply them separately.

Since one digit is in negative, then the final answer will also be in negative.

−1×5×9×2

Let’s divide the equation into two parts. Keep the negative number on the left and the positive number on the right.

=(−1)×(5×9×2)

Now, we will multiply the positive number first.

(−1)×(5×9×2) =(−1)×(45×2)

=(−1)×(90)

Since, one digit is in negative, we will apply the rule of Negative number×

Positive number= Negative number.

Therefore, the final solution will also be in negative. We will remove the bracket and simply multiply the equation.

=−1×90

=−90

​The Solution of the given equation is −90

Examples Of Problems From Exercise 1.7 Chapter 1 In Holt Algebra 1 Workbook Page 7 Problem 7 Answer

Given: 14(12)

To Find: Solution of the equation by distributive property.

In order to solve the equation, we have to remove the bracket and multiply the given digits.

14(12)

In order to solve this equation, we will use distributive property which means it is an algebraic property that is used to multiply a single value and two or more values within a set of parenthesis.

We will multiply the number immediately outside parentheses with those inside:

=14×12

Now we will multiply the equation, and the solution will be:

=168

The Solution of the given equation is 168

Page 7 Problem 8 Answer

Given: 5(47)

To Find: Solution of the equation by distributive property.

In order to solve the equation, we have to remove the bracket and multiply the given digits.

5(47)

In order to solve this equation, we will use distributive property which means it is an algebraic property that is used to multiply a single value and two or more values within a set of parenthesis.

We will multiply the number immediately outside parentheses with those inside:

=5×47

Now we will multiply the equation, and the solution will be:

=235

The solution of the given equation is 235.

Common Core Exercise 1.7 Chapter 1 Solutions Detailed Holt Algebra 1 Practice Workbook 1st Edition Page 7 Problem 9 Answer

Given: 4(106)

To Find: Solution of the equation by distributive property.

In order to solve the equation, we have to remove the bracket and multiply the given digits.

4(106)

In order to solve this equation, we will use distributive property which means it is an algebraic property that is used to multiply a single value and two or more values within a set of parenthesis.

We will multiply the number immediately outside parentheses with those inside:

=4×106

Now we will multiply the equation, and the solution will be:=424

The Solution of the given equation is 424

Page 7 Problem 10 Answer

Given 16x+27x

To Find: Sum of the given equation.

In order to solve the equation, we can combine them into a single term by adding their coefficients.

16x+27x

Let’s combine the equation into a single term by adding their coefficients and taking the x outside the bracket.

=(16+27)x

Add the equations which are in the brackets.=(43)x

Now, remove the bracket to get the final answer.=43x

The Solution of the given equation is 43x.

Student Edition Chapter 1 Exercise 1.7 Holt Algebra 1 Homework Workbook Solutions Guide Page 7 Problem 11 Answer

Given: 6t²−2t²

To Find the difference in the given equation,

In order to solve the equation, we can combine them into a single term by subtracting their coefficients.

6t²−2t²

Let’s combine the equation into a single term by adding their coefficients and taking the t² outside the bracket.

=(6−2)t²

Subtract the equation which are in the brackets.=(4)t²

Now, remove the bracket to get the final answer.=4t²

The Solution of the given equation is 4t².

Step-By-Step Answers For Exercise 1.7 Chapter 1 In Holt Algebra 1 Homework And Practice WorkbookPage 7 Problem 12 Answer

Given: −5w³+18w³

To Find: Solution of the given equation.

In order to solve the equation, we can combine them into a single term by adding their coefficients.

We have to find the difference of the absolute values by giving the sum the same sign as the number with the greater absolute value.

−5w³+18w³

Let’s combine the equation into a single term by adding their coefficients and taking w3 outside the bracket.

=(−5+18)w³

Since 5<18 then the answer will be in positive number, we just have to subtract the given digits.

=(13)w³

Now, remove the bracket to get the final answer.

=13w³

The Solution of the given equation is 13w3

Page 7 Problem 13 Answer

Given:−2.6d−3.4d

To find: Solution of the equation

In order to solve the equation, we can combine them into a single term by adding their coefficients.

We will also follow the rule of −Negative numbers−Negative numbers =Negative Solution, but we have to add it

−2.6d−3.4d

Let’s combine the equation into a single term by adding their coefficients and taking the outside the bracket.

=(−2.6−3.4)d

When you add two negative integers together, the sum is a more negative number.

=(-6.0)d

Now, remove the bracket to get the final answer.

=−6.0d

The Solution of the given equation is −6.0d.

Page 7 Problem 14 Answer

Given: −12d+3+14d+18

To Find: Solution of the given equation

In order to solve the equation, we can combine them into a single term by adding their coefficients.

We have to find the difference of the absolute values by giving the sum the same sign as the number with the greater absolute value.

−12d+3+14d+18

First, Combine the like terms and divide it with the bracket.

=(−12d+14d)+(3+18)

Add both the equation separately.

Since 12<14 then the answer will be in positive number, we just have to subtract the first equation.

=(2d)+21

Now, remove the bracket to get the final answer.

=2d+21

The solution of the given equation is2d+21.

Page 7 Problem 15 Answer

Given 42x+36x+42x+36x

To Find: Perimeter of the given Parallelogram

In order to solve the equation, we have to add all the sides of the parallelogram.

The given diagram is parallelogram. To know the perimeter we have to add the sides of the parallelogram.

42x+36x+42x+36x

Let’s divide the equation in two parts in order to solve the equation.

=(42x+36x)+(42x+36x)

Now, adding the 2 equations separately, we get:

=(78x)+(78x)

Let’s combine the equation into a single term by adding their coefficients and taking the x outside the bracket.

=(78+78)x

=(156)x

​Now, remove the bracket to get the final answer.

=156x

The perimeter of the given parallelogram is 156x.

Page 7 Problem 16 Answer

Given : 3x+4x+8+3(x−2)

To Find: Perimeter of the given diagram.

In order to solve the equation, we have to add all the sides of the triangle.

The given diagram is triangle. To know the perimeter we have to add the sides of the triangle.

3x+4x+8+3(x−2)

Multiply out the brackets and remember that everything inside the brackets will get multiplied by 3

=3x+4x+8+3x−6

Let’s divide the equation in two parts in order to solve the equation.

=(3x+4x+3x)+(8−6)

Now, adding the 2 equations separately, we get:

=10x+2

The perimeter of the given triangle is 10x+2

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

Holt Algebra 1 Homework And Practice Workbook Chapter 1 Exercise 1.8 Solutions Page 8 Problem 1 Answer

Given: the point G(2,2)

To draw the graph of the point.

For this locate the coordinate on the set of axes.

Draw the graph of the point.

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Start from the origin (0,0)

Move 2 units right and 2 units up to locate the point G.

The graph of the point is shown below:

Chapter 1 Exercise 1.8 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 8 Problem 2 Answer

Move 3 units right and 8 units up to locate the point.

The graph of the point is shown below:

Step-By-Step Solutions For Exercise 1.8 Chapter 1 Holt Algebra 1 Homework Workbook Page 8 Problem 3 Answer

Given: the point X(4,−7)

To draw the graph of the point.

For this locate the coordinate on the set of axes.

Draw the graph of the point.

Start from the origin (0,0)

Move 4 units right and 7 units down to locate the point.

The graph of the point is shown below:

Exercise 1.8 Solutions For Chapter 1 Holt Algebra 1 Homework And Practice Workbook Page 8 Problem 4 Answer

Given: the point L(−6,−1)

To draw the graph of the point.

For this locate the coordinate on the set of axes.

Draw the graph of the point.

Start from the origin(0,0)

Move 6 units left and 1 unit down to locate the point.

The graph of the point is shown below:

Exercise 1.8 Solutions For Chapter 1 Holt Algebra 1 Homework And Practice Workbook Page 8 Problem 5 Answer

Given: the point K(8,0)

To draw the graph of the point.

For this locate the coordinate on the set of axes.

Draw the graph of the point.

Start from the origin (0,0)

Move 8 units right on the x−axis to locate the point.

The graph of the point is shown below:

Examples Of Problems From Exercise 1.8 Chapter 1 In Holt Algebra 1 Workbook Page 8 Problem 6 Answer

Given: the point T(−2,5)

To draw the graph of the point.

For this locate the coordinate on the set of axes.

Draw the graph of the point.

Start from the origin (0,0)

Move 2 units left and 5 units up to locate the point.

The graph of the point is shown below:

Common Core Exercise 1.8 Chapter 1 Solutions Detailed Holt Algebra 1 Practice Workbook 1st Edition Page 8 Problem 7 Answer

Given the graph of the point.

To find the name of the quadrant where the point lies.

For this, determine the location of a point from the origin.

Determine the quadrant of point A.

Here, in the graph starting from the origin (0,0)

the point A is located 2 units right and 9 units up

Therefore, the coordinate is A(2,9)

Both values are positive which means the point lies in Quadrant I.

The point A lies in Quadrant I.

Page 8 Problem 8 Answer

Given the graph of the point.

To find the name of the quadrant where the point lies.

For this, determine the location of a point from the origin.

Determine the quadrant of point B.

Here, in the graph starting from the origin (0,0) point B is located 5 units left and 4 units up

Therefore, the coordinate is (−5,4)

Therefore, the point lies in Quadrant II.

The point B lies in Quadrant II.

Student Edition Chapter 1 Exercise 1.8 Holt Algebra 1 Homework Workbook Solutions Guide Page 8 Problem 9 Answer

Given the graph of the point.

To find the name of the quadrant where the point lies.

For this, determine the location of a point from the origin.

Determine the quadrant of point C.

Here, in the graph starting from the origin (0,0)

point C is located 3 units left and 2 units down

Therefore, the coordinate is C(−3,−2)

Both values are negative which means the point lies in Quadrant III.

The point C lies in Quadrant III.

Page 8 Problem 10 Answer

Given the graph of the point.

To find the name of the quadrant where the point lies.

For this, determine the location of a point from the origin.

Determine the quadrant of point D.

Here, in the graph starting from the origin (0,0).

Point D is located 4 units right and 7 units up

Therefore, the coordinate is D(4,7)

Both values are positive which means the point lies in Quadrant I.

The point D lies in Quadrant I.

Page 8 Problem 11 Answer

Given the graph of the point.

To find the name of the quadrant where the point lies.

For this, determine the location of a point from the origin.

Determine the quadrant of point E.

Here, in the graph starting from the origin (0,0)

Point E is located 8 units right and 6 units down

Therefore, the coordinate is (8,−6)

Therefore, the point lies in Quadrant IV.

The point E lies in quadrant IV.

Step-By-Step Answers For Exercise 1.8 Chapter 1 In Holt Algebra 1 Homework And Practice Workbook Page 8 Problem 12 Answer

Given the graph of the point.

To find the name of the quadrant where the point lies.

For this, determine the location of a point from the origin.

Determine the quadrant of point F.

Here, in the graph starting from the origin (0,0)

Point F is on y−axis.

Therefore, the point does not lie in any Quadrant.

The point F lies on y−axis which means it does not lie in any quadrant.

Page 8 Problem 13 Answer

Given the function y=∣x−4∣

To find the ordered pairs for the function using x=2,3,4,5,6. Then, draw the graph of the ordered pairs.

For this, substitute the value of x into the function and evaluate the ordered pairs.

Create a table for the ordered pairs.

Plot the ordered pairs on the set of axes.

The graph shows that points form a V-shape.

The ordered pairs are:

(2,2),(3,1),(4,0)(5,1)(6,2)

The graph of the ordered pairs is shown below:

The graph shows that points form a V-shape.

Page 8 Problem 14 Answer

Given that the number of chaperones at a school field trip are 1/5 the number of students attending, plus the 2 teacher sponsors.

To find the rule for the number of chaperones that must be on the trip. Also, determine the ordered pairs to represent the number of chaperones that must attend the trip when there are 120,150,200, and 210 students.

For this, determine the rule for the problem.

Then substitute the values of x to determine the ordered pairs.

Let y be the number of chaperones and x represents the number of students attending.

Determine the rule for y that is 1/5 times of x plus the 2 teacher sponsors.

y=1/5x+2

Create a table for the ordered pairs.

The rule for the number of chaperones that must be on the trip is y=1/5x+2.

The ordered pairs to represent the number of chaperones who attend the trip are:

(120,26), (150,32), (200,42) and (210,44)

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

Holt Algebra 1 Homework And Practice Workbook Chapter 1 Exercise 1.3 Solutions Page 3 Problem 1 Answer

Given: Algebraic expression is −24÷−8

To Find: the value of expression.

For finding the value of expression we use the division.

The given expression is −24÷−8

Now,−24÷−8=−24/−8=24/8=3​

Thus, the value of −24÷−8  is 3

Holt Algebra 1 Chapter 1 Page 3 Problem 2 Answer

Give: Algebraic expression is 24(−5)

To Find: the value of expression.

For finding the value of expression we use the multiplication.

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

The given expression is 24(−5)

Now, 24(−5)

=−24×5

=−120

​Thus, the value of 24(−5) is −120

Chapter 1 Exercise 1.3 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 3 Problem 3 Answer

Given: Algebraic expression is −6(20)

To Find: the value of expression.

For finding the value of expression we use the multiplication.

The given expression is −6(20)

Now,−6(20)

=−6×20

=−120

​Thus, the value of−6(20) is −120

Holt Algebra 1 Chapter 1 Page 3 Problem 4 Answer

Page 3 Problem 5 Answer

Given: Algebraic expression is t÷(−1.5) for t=6

To Find: the value of expression.

For finding the value of expression we put the value of t in given expression.

The given expression is t÷(−1.5)

Now put t=6 then we get,t÷(−1.5)

=6/−1.5

=−6/1.5

=−60

15 (multiply the numerator and denominator by 10)=−4

​Thus, the value of t÷(−1.5) for t=6 is −4

Holt Algebra 1 Chapter 1 Page 3 Problem 6 Answer

Given: Algebraic expression is −8/9÷2/3

To Find: the value of expression.

For finding the value of expression we use the division.

The given expression is −8/9÷2/3

Now, −8/9÷2/3

=−8/9×3/2 (∵a/b÷c/d=a/b×d/c)

​therefore,=−8/9×3/2=−4/3

Thus, the value of −8/9÷2/3 is −4/3

Step-By-Step Solutions For Exercise 1.3 Chapter 1 Holt Algebra 1 Homework Workbook Page 3 Problem 7 Answer

Given: Algebraic expression is −12÷(−6/25)

To Find: the value of expression.

For finding the value of expression we use the division.

The given expression is −12÷(−6/25)

Now, −12÷(−6/25)=−12×(−25/6)(∵a/b÷c/d=a/b×d/c)

​therefore,=−12×(−25/6)

=2×25

=50

​Thus, the value of −12÷(−6/25) is 50

Holt Algebra 1 Chapter 1 Page 3 Problem 8 Answer

Given: Algebraic expression is 21/4÷(−51/3)

To Find: the value of expression.

For finding the value of expression we use the division.

=2 ÷ 5

=1/4 (−1/3)

= 8 + 1/4÷  (−15 + 1/3 )

= 9/4÷  (−16/3 )

=  9/4× (−3/16)

therefore, =  9/4 ×(−3/16)

Thus, the value of 21/4÷(−51/3) is −27/64.

Exercise 1.3 Solutions For Chapter 1 Holt Algebra 1 Homework And Practice Workbook Page 3 Problem 9 Answer

Given: Expression is 0⋅4.75

To Find: The value of expression (using multiplication).

For identifying the value, we would multiply the two values given in the expression.

It is given 0⋅4.75

Here,  we need to multiply 0 and 4.75, as ⋅ this signifies multiplication.

Hence, the operation is: 0⋅4.75=0.

Therefore, the value of 0⋅4.75 is 0.

Holt Algebra 1 Chapter 1 Page 3 Problem 10 Answer

Given: Expression is 0÷10

To Find: Multiply or divide.

For identifying multiply or divide we see the mathematical operator.

It is given 0÷10

Here, we can see that between 0 and 10 the mathematical operator is division.

Hence, It is divide and its value is 0÷10=0/10=0

​Thus, 0÷10 will divide.

OR The mathematical operator between 0 and 10 is divide.

Examples Of Problems From Exercise 1.3 Chapter 1 In Holt Algebra 1 Workbook Page 3 Problem 11 Answer

Given: Expression is −1/3÷0

To Find: The value of expression (using division).

For identifying the value, we would divide the two values given in the expression.

It is given−1/3÷0

Here, we we need to divide −1/3 and 0, as ÷ this signifies division.

Hence, The value −1/3÷0 is undefined, since the division by 0 is not defined.

Therefore, the value of −1/3÷0 is undefined.

Holt Algebra 1 Chapter 1 Page 3 Problem 12 Answer

Given: When Brianna’s first CD sold a million copies, her record label gave her a $5000 bonus.

She split the money evenly between herself, her agent, her producer, and her stylist.

To Find: Multiply or divide. How much money did each person receive?

For identifying multiply or divide we see the mathematical operator.

It is given that when Brianna’s first CD sold a million copies, her record label gave her a $5000 bonus.

She split the money evenly between herself, her agent, her producer, and her stylist.

Therefore, For distributing the money between herself, her agent, her producer, and her stylist the divide symbol will come.

There are total of 4 person.

So, Money each person receive is =5000/4=$1250

​Thus, divide symbol will come and Money each person receive is $1250

Common Core Exercise 1.3 Chapter 1 Solutions Detailed Holt Algebra 1 Practice Workbook 1st Edition Page 3 Problem 13 Answer

Given: Expression is (0.3)(−1.8)

To Find: The value of expression (using multiplication).

For identifying the value, we would multiply the two values given in the expression.

It is given, (0.3)(−1.8).

Here, we need to multiply 0.3 and −1.8.

Hence, the operation is: (0.3)(−1.8)=−0.54.

Therefore, the value of (0.3)(−1.8) is −0.54.

Holt Algebra 1 Chapter 1 Page 3 Problem 14 Answer

Given: Expression is 2/5(−5/2)

To Find: Multiply or divide.

For identifying multiply or divide we see the mathematical operator.

It is given 2/5(−5/2)

Here, we can see that between 2/5 and (−5/2) the mathematical operator is multiplication.

Hence, It is multiply and its value is ​=2/5(−5/2)=−2×5/5×2=−1​

Thus, the mathematical operator between 2/5 and (−5/2) is multiply.

Common Core Exercise 1.3 Chapter 1 solutions detailed Holt Algebra 1 Practice Workbook 1st Edition Page 3 Problem 15 Answer

Given: Expression is −15÷(−6)

To Find: Multiply or divide.

For identifying multiply or divide we see the mathematical operator.

It is given −15÷(−6)

Here,we can see that between −15 and −6 the mathematical operator is division.

Hence, It is division and its value is −15÷(−6)

−15÷(−6) =−15/−6

−15÷(−6) =15/6

−15÷(−6) = 5/2Thus, the mathematical operator between −15 and (−6) is divide.

Holt Algebra 1 Chapter 1 Page 3 Problem 16 Answer

Given: Algebraic expression is x⋅y

To Find: Evaluate expression for x=16,y=−4, and z=−2

For finding the value of expression we put the value of x=16,y=−4 in given expression.

The given expression is x⋅y

Now put x=16,y=−4 then we get,

⇒x⋅y

=x⋅y

=16⋅(−4)

=16×(−4)

=−64

​Thus, the value of x⋅y for x=16,y=−4 is −64

Student Edition Chapter 1 Exercise 1.3 Holt Algebra 1 Homework Workbook Solutions Guide Page 3 Problem 17 Answer

Given: Algebraic expression is xz

To Find: Evaluate expression for x=16,y=−4, and z=−2.

For finding the value of expression we put the value of x=16, and z=−2.  in given expression.

The given expression is xz

Now put x=16, and z=−2 then we get,

⇒xz

=16×(−2)

=−32

​Thus, the value of xz for x=16, and z=−2 is −32

Holt Algebra 1 Chapter 1 Page 3 Problem 18 Answer

Given: Algebraic expression is z÷y

To Find: Evaluate expression for x=16,y=−4, and z=−2

For finding the value of expression we put the value of y=−4, and z=−2 in given expression.

The given expression is z÷y

Now put y=−4, and z=−2

then we get,

⇒z÷y

z÷y =−2÷−4

z÷y =−2/−4

z÷y =2/4

z÷y =1/2

Thus, the value of z÷y for y=−4, and z=−2 is 1/2

Page 3 Problem 19 Answer

Given: Algebraic expression is (y)(z)

To Find: Evaluate expression for x=16,y=−4, and z=−2

For finding the value of expression we put the value of y=−4, and z=−2 in given expression.

The given expression is (y)(z)

Now put y=−4, and z=−2

then we get,

⇒(y)(z)

(y)(z) =(−4)(−2)

(y)(z) =8​

Thus, the value of(y)(z) for y=−4, and z=−2 is 8

Holt Algebra 1 Chapter 1 Page 3 Problem 20 Answer

Given: Algebraic expression is x÷z

To Find: Evaluate expression for x=16,y=−4, and z=−2

For finding the value of expression we put the value of x=16 and z=−2 in given expression.

The given expression is x÷z

Now put x=16 and z=−2 then we get,

⇒x÷z

x÷z =x/z

x÷z =16/−2

x÷z =−8​

Thus, the value of x÷z for x=16 and z=−2 is −8

Step-By-Step Answers For Exercise 1.3 Chapter 1 In Holt Algebra 1 Homework And Practice Workbook Page 3 Problem 21 Answer

Given: Algebraic expression is x÷y

To Find: Evaluate expression for x=16,y=−4, and z=−2

For finding the value of expression we put the value of x=16,y=−4 in given expression.

The given expression is x÷y

Now put x=16,y=−4 then we get,

​⇒x÷y

x÷y =x/y

x÷y =16/−4

x÷y =−4

​Thus, the value of x÷y for x=16,y=−4 is −4

Holt Algebra 1 Chapter 1 Page 3 Problem 22 Answer

Given: Algebraic expression is z÷x

To Find: Evaluate expression for x=16,y=−4, and z=−2

For finding the value of expression we put the value of x=16 and z=−2 in given expression.

The given expression is z÷x

Now put x=16, and z=−2  then we get,

⇒z÷x

z÷x =z/x

z÷x =−2/16

z÷x =−1/8​

Thus, the value of z÷x for x=16, and z=−2 is−1/8

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

Holt Algebra 1 Homework And Practice Workbook Chapter 1 Exercise 1.2 Solutions Page 2 Problem 1 Answer

Given: Expression is −6+(−8).

To add or subtract the given expression using a number line.

Start at 0. Move left to -6.To add -8, move left 8 units.

The given expression can be added using a number line as below. 

Mathematically,−6+(−8)=−14

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Chapter 1 Exercise 1.2 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 2 Problem 2 Answer

Given: Expression is 10+(−4).

To add or subtract the given expression using a number line.

Start at 0. Move right to 10.To add -4, move left 4 units.

The given expression can be added using a number line as below. 

Mathematically,10+(−4)=6

The value of the given expression is 10+(−4)=6.

Page 2 Problem 3 Answer

Given: Expression is−2−(−6).

To add or subtract the given expression using a number line.

Start at 0. Move left to -2. To subtract -6, move right 6 units.

The given expression can be added using a number line as below. 

Mathematically,−2−(−6)=4

The value of the given expression is−2−(−6)=4.

Step-By-Step Solutions For Exercise 1.2 Chapter 1 Holt Algebra 1 Homework Workbook Page 2 Problem 4 Answer

Given: Expression is−7+7.

To add or subtract the given expression using a number line.

Start at 0. Move left to -7.To add 7, move right 7 units.

The given expression can be added using a number line as below. 

Mathematically,−7+7=0

The value of the given expression is−7+7=0.

Page 2 Problem 5 Answer

Given: Expression is 24.6+(−45.5).

To add the given expression.

When the signs of the numbers are the different, find the difference of their absolute values.

Use the sign of the number with the greater absolute value.

Absolute difference of 45.5 and 24.6 is − 45.5−24.6=20.9

Using the sign of a number with greater value we get -24.6+(−45.5)=−20.9

The addition of 24.6+(−45.5)=−20.9 .

Exercise 1.2 Solutions For Chapter 1 Holt Algebra 1 Homework And Practice Workbook Page 2 Problem 6 Answer

Given: Expression is −3/8+5.

To add the given expression.

When the signs of the numbers are the different, find the difference of their absolute values.

Use the sign of the number with the greater absolute value.

Absolute difference of 3/8 and 5 is -3/8−5=−37/8

Using the sign of a number with greater value we get-

−3/8+5=37/8

The addition of−3/8+5=37/8.

Page 2 Problem 7 Answer

Given: Expression is a+(−14) and a=16.

To add the given expression for the given value of a.

First substitute 16 for a.

When the signs of the numbers are the different, find the difference of their absolute values.

Use the sign of the number with the greater absolute value.

On substituting the value of a in the given expression we get -16+(−14)

Absolute difference of 16 and 14 is − 16−14=2

Using the sign of a number with greater value we get -16+(−14)=2

The addition of a+(−14) for a=16 is 2.

Examples Of Problems From Exercise 1.2 Chapter 1 In Holt Algebra 1 Workbook Page 2 Problem 8 Answer

Given: Expression is −3.3+x and x=−9.1.

To add the given expression for the given value of x.

First substitute -9.1 for x.When the signs of the numbers are the same, find the sum of their absolute values.

Both numbers are negative, so the sum is negative.

On substituting the value of x in the given expression we get -−3.3+(−9.1)

Absolute sum of 3.3 and 9.1 is -3.3+9.1=12.4

Both numbers are negative, so the sum is negative.

Therefore, −3.3+(−9.1)=−12.4

The addition of −3.3+x and x=−9.1 is −12.4

Page 2 Problem 9 Answer

Given: Expression is 12−(−16).

To subtract the given expression.

First write the expression in the form of addition.

When the signs of the numbers are the same, find the sum of their absolute values.

Both numbers are positive, so the sum is positive.

On writing the given expression as addition we get -12+16

Absolute sum of 12 and 16 is -12+16=28

Both numbers are positive, so the sum is positive.

Therefore,12−(−16)=28

The subtraction of 12−(−16)=28.

Page 2 Problem 10 Answer

Given: Expression is 8.3−10.7.

To subtract the given expression.

First write the expression in the form of addition.

When the signs of the numbers are the different, find the difference of their absolute values.

Use the sign of the number with the greater absolute value.

On writing the given expression in the form of addition we get -8.3+(−10.7)

Absolute difference of 8.3 and 10.7 is -10.7−8.3=2.4

Using the sign of a number with greater value we get-8.3−10.7=−2.4

The subtraction of 8.3−10.7=−2.4.

Page 2 Problem 11 Answer

Given: Expression is −2/3−51/3.

To subtract the given expression.

First convert the given expression into proper fraction and then write it in the form of addition.

When the signs of the numbers are the same, find the sum of their absolute values.

Both numbers are negative, so the sum is negative.

On writing the given expression as proper fraction and in the form of addition we get -−2/3+(−16/3)

Absolute sum of 2/3 and 16/3 is -2/3+16/3=6

Both numbers are negative, so the sum is negative.

Therefore,−2/3−51/3=−6

The subtraction of −2/3−51/3=−6.

Page 2 Problem 12 Answer

Given: Expression is z−3.5 and z=1.

To subtract the given expression.

First substitute the value of z in the given expression then write it in the form of addition.

When the signs of the numbers are the different, find the difference of their absolute values.

Use the sign of the number with the greater absolute value.

On substituting the value of z and writing it in the form of addition we get -1+(−3.5)

Absolute difference of 1 and 3.5 is -3.5−1=2.5

Using the sign of a number with greater value we get-z−3.5 for z=1 is −2.5

The subtraction of z−35 for z=1 is −2.5.

Page 2 Problem 13 Answer

Given: The record high temperature for Asheville, North Carolina was 99∘F and the record low was −17∘F.

To find The difference between these two temperatures.

Subtract the two temperatures.

Then, write it in the form of addition.

When the signs of the numbers are the same, find the sum of their absolute values.

Both numbers are positive, so the sum is positive.

On subtracting the two temperature and writing it in the form of addition we get -99−(−17)=99+17

Absolute sum of 99 and 17 is -99+17=116

Both numbers are positive, so the sum is positive.

Therefore, 99−(−17)=116

The difference between the two temperatures is 116∘F.

Page 2 Problem 14 Answer

Given: The balance in Mr. Sanchez′s bank account was $293.74 and he accidentally wrote a check for $300.

To find His balance now.

Subtract the check  Mr. Sanchez  accidentaly wrote from his balance bank account.

When the amount that is subtracted is greater than the amount from which it is subtracted, then you can subtract the smaller amount from the larger amount and put the negative sign in the result.

Mr. Sanchez’s bank account balance is $293.74.

He wrote a check of $300.

The new balance will be the difference of the starting balance minus the amount of the chek:

So, his new balance is: 293.74−300=−6.96.

Therefore, the balance amount in Mr. Sanchez’s bank account after the accicentally wrote a check for $300

when the balance of his bank account was $293.74 is −$6.26.

Page 2 Problem 15 Answer

Given: Expression is 18−n and n=−13.

To evaluate the given expression for each value of n.

Substitute the value of n in the given expression and write it in the form of addition.

When the signs of the numbers are the same, find the sum of their absolute values.

Both numbers are positive, so the sum is positive.

On substitute the value of n in the given expression and writing it in the form of addition we get −18−(−13)=18+13

Absolute sum of 18 and 13 is -18+13=31

Both numbers are positive, so the sum is positive.

Therefore, the value of the expression 18−n for n=−13 is 31.

The value of the expression 18−n for n=−13 is 31.

Page 2 Problem 16 Answer

Given: Expression is 18−n and n=8.55.

To evaluate the given expression for each value of n.

Substitute the value of n in the given expression and write it in the form of addition.

When the signs of the numbers are the different, find the difference of their absolute values.

Use the sign of the number with the greater absolute value.

On substitute the value of n in the given expression and writing it in the form of addition we get -18−8.55=18+(−8.55)

Absolute difference of 18 and 8.55 is -18−8.55=9.45

Using the sign of a number with greater value we get-The value of the expression 18−n for n=8.55 is 9.45.

The value of the expression 18−n for n=8.55 is 9.45.

Page 2 Problem 17 Answer

Given: Expression is 18−n and n=201/5.

To evaluate the given expression for each value of n.

Substitute the value of n in the given expression and write it in the form of addition.

When the signs of the numbers are the different, find the difference of their absolute values.

Use the sign of the number with the greater absolute value.

On substitute the value of n in the given expression and writing it in the form of addition we get −18−201/5=18+(−101/5)

Absolute difference of 18 and 101/5 is -101/5−18=11/5

Using the sign of a number with greater value we get-The value of the expression 18−n for n=201/5 is −11/5.

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

Holt Algebra 1 Homework And Practice Workbook Chapter 1 Exercise 1.1 Solutions Page 1 Problem 1 Answer

Given: Algebraic expression is 15−b

To Find Two ways to write the given algebraic expression in words.

Word expressions for 15 – b :

(1) b is subtracted from 15

(2) 15 is added to negative b

Thus, the two ways to write 15−b algebraic expression in words are

(1) b is subtracted from 15

(2) 15 is added to negative b

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Chapter 1 Exercise 1.1 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 1 Problem 2 Answer

Given: Algebraic expression is x/16

To Find: Two ways to write the given algebraic expression in words.

Word expressions for x/16:

(1) Divide x by 16

(2) Divide x by the square of four.

Thus, the two ways to write x/16 algebraic expression in words are

(1) Divide x by 16

(2) Divide x by the square of four.

Page 1 Problem 3 Answer

Step-by-step solutions for Exercise 1.1 Chapter 1 Holt Algebra 1 Homework Workbook Page 1 Problem 4 Answer

Given: Algebraic expression is 4y

To Find: Two ways to write the given algebraic expression in words.

Word expressions for 4y :

(1) “4 is multiplied by y”

(2) “the product of 4 and y”

Thus, the two ways to write 4y algebraic expression in words are

(1) “4 is multiplied by y”

(2) “the product of 4 and y”

Page 1 Problem 5 Answer

Given: Sophie’s math class has 6 fewer boys than girls, and there are ‘g’ girls.

To Find: Write an expression for the number of boys.

For expressing in mathematical form we have to take some variable for Sophie’s math class.

Let b & g represents the number of boys and girls in the math class respectively.

Then, according to the given information, we get b=g−8

Thus, the required expression for the number of boys in the math class is b=g−8

Exercise 1.1 Solutions For Chapter 1 Holt Algebra 1 Homework And Practice Workbook Page 1 Problem 6 Answer

Given: A computer printer can print 10 pages per minute.

To Find: Write an expression for the number of pages the printer can print in m minutes.

For expressing in mathematical form we have to take some variable and simplify according to given condition.

Given that, 10 pages /1 minute

Suppose, x number of pages a printer can print in m minutes.

⇒ Thus,10 pages /1 minute =x pages m minutes

⇒10=x/m

x=10m

​Hence, the number of pages printer can print in ‘m’ minutes is : 10m

Page 1 Problem 7 Answer

Given: Expression st

To Find: Evaluate expression for r=8,s=2, and t=5

For evaluating the expression we have to put the given data in st and then simplify.

Given expression st put s=2, and t=5 then we get,

⇒st

=2(5)

=10

​Thus, the value of st is 10

Examples Of Problems From Exercise 1.1 Chapter 1 In Holt Algebra 1 Workbook Page 1 Problem 8 Answer

Given: Expression s+t

To Find: Evaluate expression for r=8,s=2, and t=5

For evaluating the expression we have to put the given data in s+t and then simplify.

Given expression s+t

put s=2, and t=5 then we get,

⇒s+t

=2+5

=7

​Thus, for s=2 and t=5 the value of s+t  is 7

Page 1 Problem 9 Answer

Given: Expression r−t

To Find: Evaluate expression for r=8,s=2, and t=5

For evaluating the expression we have to put the given data in r−t and then simplify.

Given expression r−t

put r=8 and t=5 then we get,

⇒r−t

=8−5

=3

​Thus, the value of r−t is 3

Common Core Exercise 1.1 Chapter 1 Solutions Detailed Holt Algebra 1 Practice Workbook 1st Edition Page 1 Problem 10 Answer

Given: Expression r⋅s

To Find: Evaluate expression for r=8,s=2, and t=5

For evaluating the expression we have to put the given data in r⋅s and then simplify.

Given expression r⋅s

put r=8,and s=2 then we get,

⇒r⋅s

=8⋅2

=16

​Thus, the value of r⋅s is 16

Student Edition Chapter 1 Exercise 1.1 Holt Algebra 1 Homework Workbook Solutions Guide Page 1 Problem 11 Answer

Given: Expression t−s

To Find: Evaluate expression for r=8,s=2, and t=5

For evaluating the expression we have to put the given data in t−s and then simplify.

Given expression t−s

put s=2, and t=5 then we get,

⇒t−s

=5−2

=3

​Thus, the value of t−s is 3

Page 1 Problem 12 Answer

Given: Paula always withdraws 20 dollars more than she needs from the bank.

To Find: Write an expression for the amount of money Paula withdraws if she needs d dollars.

For expressing in mathematical form we have to take some variable for withdrawal amount.

Let the withdrawal be denoted by w and dollars by d then according to given condition we get,w=d+20

Thus, the expression for the amount of money Paula withdraws if she needs d dollars is w=d+20

Step-By-Step Answers For Exercise 1.1 Chapter 1 In Holt Algebra 1 Homework And Practice Workbook Page 1 Problem 13 Answer

Given: Paula always withdraws 20 dollars more than she needs from the bank.

To Find: The amount of money Paula withdraws if she needs 20,60, and 75 dollars.

For finding the amount of money Paula withdraws if she needs 20,60, and 75 dollars we have to put 20,60, and 75 dollars in w=d+20

From first part we have w=d+20

Now,

for d=20

we get,

w=d+20

∴w=20+20

∴w=40

for d=60

we get,

w=d+20

∴w=60+20

∴w=80

​for d=75

we get,

w=d+20

∴w=75+20

∴w=95

​Thus, the amount of money Paula withdraws if she needs 20,60, and 75 dollars are 40,80 and 95 respectively.

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

Olt Algebra 1 Homework And Practice Workbook Chapter 1 Exercise 1.4 Solutions Page 4 Problem 1 Answer

Given:

To Find: The power represented by the geometric model.

Method Used: The power represented by the geometric model is equal to volume of the given cube.

We know, the length of cube is 5 units.

Volume of cube =a³ , where a is the length of the cube.

So, volume of cube =5³.

Hence, the power represented by the geometric model is 5³.

Therefore, the power represented by the geometric model is 5³.

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

Holt Algebra 1 Chapter 1 Page 4 Problem 2 Answer

Given:

To Find: The power represented by the geometric model.

Method Used: The power represented by the geometric model is equal to the area of the given square.

We know,

The given figure is of 7 rows and 7 columns.

Thus, 7×7 which means that the factor 7 is used 2 times.

Therefore, therefore, the power of the geometric model is 72.

Holt Algebra 1 Chapter 1 Page 4 Problem 3 Answer

Given:

Chapter 1 Exercise 1.4 Answers Holt Algebra 1 Practice Workbook 1st Edition Page 4 Problem 4 Answer

Given: (−3)³.

To Find: Evaluate the expression.

Method Used: From left to right, evaluate all powers.

To raise a number to a power, simply multiply it by itself that many times.

Therefore, (−3)³=−27.

Holt Algebra 1 Chapter 1 Page 4 Problem 5 Answer

Given: (2/5)².

To Find: Evaluate the expression.

Method Used: From left to right, evaluate all powers.

To raise a number to a power, simply multiply it by itself that many times.

We know, (2/5)².

Simplify,(2/5)²

(2/5)²=2/5×2/5

=(2/5)²=4/25

Therefore, (2/5)²=4/25.

Step-By-Step Solutions For Exercise 1.4 Chapter 1 Holt Algebra 1 Homework Workbook Page 4 Problem 6 Answer

Given: 3⁵.

To Find: Evaluate the expression.

Method Used: From left to right, evaluate all powers.

To raise a number to a power, simply multiply it by itself that many times.

Simplify,3⁵=3×3×3×3×3

3⁵=243

Therefore, 3⁵=243.

Holt Algebra 1 Chapter 1 Page 4 Problem 7 Answer

Given: (−10)⁴.

To Find: Evaluate the expression.

Method Used: From left to right, evaluate all powers.

To raise a number to a power, simply multiply it by itself that many times.

We know, (−10)⁴.

Simplify,(−10)⁴=−10×−10×−10×−10

(−10)⁴=10000

Therefore, (−10)⁴=10000.

Exercise 1.4 Solutions For Chapter 1 Holt Algebra 1 Homework And Practice Workbook Page 4 Problem 8 Answer

Given: (3/4)².

To Find: Evaluate the expression.

Method Used: From left to right, evaluate all powers.

To raise a number to a power, simply multiply it by itself that many times.

We know, (3/4)².

Simplify,(3/4)²=3/4×3/4

=(3/4)²=9/16

Therefore, (3/4)²=9/16.

Holt Algebra 1 Chapter 1 Page 4 Problem 9 Answer

Given: 16; base 2.

To Find: Write each number as a power of the given base.

Method Used: Take the given base and put given power on top of it.

We know, given number is 16 and base is 2.

An expression written with an exponent and a base, or the value of such an expression, is referred to as a power.

A power is represented by the number 6³, where 6 is the base and the number that is factored; on the other hand, 3 is the exponent, which indicates how many times the base, 6, is factored.

So, 16=2×2×2×2

16=24

Therefore, the number 16 can be written as 24.

Examples of problems from Exercise 1.4 Chapter 1 in Holt Algebra 1 Workbook Page 4 Problem 10 Answer

Given: 1,000,000; base 10.

To Find: Write each number as a power of the given base.

Method Used: Take the given base and put given power on top of it.

We know, given number is 1,000,000 and base is 10.

An expression written with an exponent and a base, or the value of such an expression, is referred to as a power.

Power is represented by the number 63, where 6 is the base and the number that is factored; on the other hand, 3 is the exponent, which indicates how many times the base, 6, is factored.

So,1000000=10×10×10×10×10×10

1000000=10⁶

Therefore, the number 1,000,000 can be written as 10⁶.

Holt Algebra 1 Chapter 1 Page 4 Problem 11 Answer

Given: −216; base −6.

To Find: Write each number as a power of the given base.

Method Used: Take the given base and put given power on top of it.

We know, given number is −216 and base is −6.

An expression written with an exponent and a base, or the value of such an expression, is referred to as a power.

A power is represented by the number 6³, where 6 is the base and the number that is factored; on the other hand, 3 is the exponent, which indicates how many times the base, 6, is factored.

So,−216=−6×−6×−6

−216=(−6)³

Therefore, the number −216 can be written as (−6)³.

Common Core Exercise 1.4 Chapter 1 solutions detailed Holt Algebra 1 Practice Workbook 1st Edition Page 4 Problem 12 Answer

Given: 2401; base 7.

To Find: Write each number as a power of the given base.

Method Used: Take the given base and put given power on top of it.

We know, given power is 2401 and base is 7.

An expression written with an exponent and a base, or the value of such an expression, is referred to as a power.

A power is represented by the number 6³, where 6 is the base and the number that is factored; on the other hand, 3 is the exponent, which indicates how many times the base, 6, is factored.

Thus, number is 7²⁴⁰¹.

Therefore, the number for the given base and power is 7²⁴⁰¹.

Holt Algebra 1 Chapter 1 Page 4 Problem 13 Answer

Given: 256; base −4.

To Find: Write each number as a power of the given base.

Method Used: Take the given base and put given power on top of it.

We know, given power is 256 and base is −4.

An expression written with an exponent and a base, or the value of such an expression, is referred to as a power.

A power is represented by the number 6³, where 6 is the base and the number that is factored; on the other hand, 3 is the exponent, which indicates how many times the base, 6, is factored.

Thus, number is (−4)²⁵⁶.

Therefore, the number for the given base and power is (−4)²⁵⁶.

Student Edition Chapter 1 Exercise 1.4 Holt Algebra 1 Homework Workbook solutions guide Page 4 Problem 14 Answer

Given: 8/27; base 2/3.

To Find: Write each number as a power of the given base.

Method Used: Take the given base and put given power on top of it.

We know, given power is 8/27 and base is 2/3.

An expression written with an exponent and a base, or the value of such an expression, is referred to as a power.

A power is represented by the number 6³, where 6 is the base and the number that is factored; on the other hand, 3 is the exponent, which indicates how many times the base, 6, is factored.

Thus, number is (2/3) 8/27.

Therefore, the number for the given base and power is (2/3) 8/27.

Holt Algebra 1 Chapter 1 Page 4 Problem 15 Answer

Given: Anna called two people and asked each of them to call two other people, and so on.

It takes one minute to call two people.

To Find: How many phone calls were made during the fifth minute?

Method Used: From left to right, evaluate all powers.

To raise a number to a power, simply multiply it by itself that many times.

We know, Anna called two people and asked each of them to call two other people, and so on.

It take one minute to call two people.

In the first minute, anna calls two people, that means 21.

In the second minute, each of two people calls 2 people, that means 2².

Similarly, In the third minute, number of calls =2³.

Similarly, In the fourth minute, number of calls =2⁴.

So, in the fifth minute, the number of calls =2⁵

Simplify,

2⁵=2×2×2×2×2

2⁵=32.

Therefore, the number of calls made during fifth minute is 32.

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

Holt Algebra 1 Homework And Practice Workbook Chapter 2 Exercise 2.7 Solutions 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

Read and Learn More Holt Algebra 1 Homework and Practice Workbook Solutions

=135=9x

Divide 135 with 9

=135/9

=x

=x=15 cm

​The value of x=15 cm

Chapter 2 Exercise 2.7 Answers Holt Algebra 1 Practice Workbook 1st Edition 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

Step-By-Step Solutions For Exercise 2.7 Chapter 2 Holt Algebra 1 Homework Workbook Page 15 Problem 3 Answer

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

To find: height of the second pole.

Divide 110 with 4

x=110/4ft

x=27.5ft

​The height of the utility pole is 27.5 ft.

Exercise 2.7 Solutions For Chapter 2 Holt Algebra 1 Homework And Practice Workbook 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

Examples Of Problems From Exercise 2.7 Chapter 2 In Holt Algebra 1 Workbook 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.