/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 $$\left(-2 w^{-5}\right)^{3}$$... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 22

$$\left(-2 w^{-5}\right)^{3}$$

Short Answer

Expert verified
-8 w^{-15}

Step by step solution

01

Understand the base and exponent

Identify the base \((-2 w^{-5})\) and the exponent \(3\). The expression \((-2 w^{-5})^3\) means that \(-2 w^{-5}\) is raised to the power of \(3\).
02

Apply the power to the coefficient

Raise the coefficient \(-2\) to the power of \(3\). This means calculating \((-2)^3\). \((-2)^3 = -8\).
03

Apply the power to the variable

Raise the variable part \(w^{-5}\) to the power of \(3\). Use the exponent rule \((a^m)^n = a^{m \times n}\). So, \( (w^{-5})^3 = w^{-5 \times 3} = w^{-15} \).
04

Combine the results

Combine the results from Steps 2 and 3. The coefficient is \(-8\) and the variable part is \w^{-15}\. Hence, \((-2 w^{-5})^3 = -8 w^{-15}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebraic Expressions
Algebraic expressions are a combination of numbers, variables, and operations. In the expression \((-2w^{-5})^{3}\), \(-2\) is the coefficient and \(w^{-5}\) involves the variable \(w\) with an exponent of \(-5\). Knowing how to manipulate such expressions is key in algebra.
Here are the components you need to identify:
  • Coefficients: Numerical part, here it is \(-2\).
  • Variables: Symbols representing numbers, here it is \(w\).
  • Exponents: They tell you how many times to multiply a number by itself, here it is \(-5\) and afterwards applied as \(3\).
By understanding these parts, solving the expression becomes straightforward.
Exponent Rules
Exponent rules help simplify expressions involving powers. One important rule used in this problem is the power of a power rule.
Following are the main exponent rules:
  • Product of Powers Rule: \(a^{m} \times a^{n} = a^{m+n}\)
  • Power of a Power Rule: \((a^{m})^{n} = a^{m \times n}\) 鈥� This is used to solve the variable part of \((w^{-5})^{3} = w^{-5 \times 3} = w^{-15}\).
  • Power of a Product Rule: \((ab)^{n} = a^{n}b^{n}\)
Mastering these rules allows you to confidently manipulate and simplify expressions with exponents.
Negative Exponents
A negative exponent shows that the base is on the opposite side of the fraction. To change a negative exponent into a positive one, you can take the reciprocal of the base.
For example: \(\frac{1}{a^{n}} = a^{-n}\).
In our given expression, \(w^{-5}\) means \(w\) raised to the power of \(-5\). When raised to another power, apply the exponent multiplication rule: \(w^{-5 \times 3} = w^{-15}\). Remember:
  • A negative exponent indicates division, not multiplication.
  • To convert, use reciprocal: \(x^{-n} = 1/x^{n}\).
Dealing with negative exponents correctly simplifies complex algebraic expressions.
Mathematical Operations
Mathematical operations, like addition, subtraction, multiplication, and division, are the backbone of algebra.
In the expression \((-2 w^{-5})^{3}\), two main operations are used:
  • Exponentiation: Raise one number to the power of another. First to coefficient \((-2)^{3} = -8\).
  • Multiplication: Combine results from exponentiation. Join \(-8\) and \w^{-15}\ to get final answer \(-8 w^{-15}\).
Understanding these operations, and when to apply each, is necessary to solve algebraic problems effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve each equation. $$(3 x-5)\left(25 x^{2}-4\right)=0$$

Factor each polynomial by grouping. $$ 2 m w-5 w-2 n+5 $$

Factor each polynomial by grouping. $$ 2 q h-h+8 q-4 $$

Solve each equation for \(y .\) Assume a and b are positive numbers. $$4 y^{2}+4 b y+b^{2}=0$$

Factor each polynomial using the trial-and-error method. $$ 7 a^{2}-8 a+1 $$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.