Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 1 Exercise 1.4

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.

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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.