/*! 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 65 Find the indicated products and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 65

Find the indicated products and quotients. Express final results using positive integral exponents only. \(\left(-7 a^{2} b^{-5}\right)\left(-a^{-2} b^{7}\right)\)

Short Answer

Expert verified
The simplified expression is \(7b^2\).

Step by step solution

01

Recognize the operation

The problem requires us to multiply two expressions: \[(-7a^2b^{-5})(-a^{-2}b^7)\] We will apply the associative and commutative properties of multiplication to rearrange terms.
02

Multiply coefficients

Identify and multiply the coefficients (numerical parts) of the expressions. Here, the coefficients are -7 and -1 (remember the invisible -1 in front of \(-a^{-2}\)).\[-7 \times (-1) = 7\].
03

Multiply the variable terms with exponents

Next, apply the law of exponents that states \(x^m \times x^n = x^{m+n}\) to the variable parts of the expressions.- For the \(a\) terms: \[a^2 \times a^{-2} = a^{2 + (-2)} = a^0 = 1\].- For the \(b\) terms:\[b^{-5} \times b^7 = b^{-5 + 7} = b^2\].
04

Combine the results

Combine the results of steps 2 and 3:\[7 \times 1 \times b^2 = 7b^2\] Thus, the expression simplifies to \(7b^2\).
05

Express result with positive exponents

The result \(7b^2\) already has positive integral exponents for all terms. Therefore, the expression is simplified to its final form.

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.

Multiplication of Expressions
Multiplication of expressions involves combining expressions systematically to achieve a single simplified form. When tackling a problem like \((-7a^2b^{-5})(-a^{-2}b^7)\), follow these simple guidelines:
  • Understand the expressions: Break down and identify each part of the expressions to see both the coefficients (numbers) and variables (like \(a\) and \(b\) with their exponents).
  • Perform operations methodically: Use the commutative property, which allows you to switch the order of multiplication, enabling a focused approach on like terms (numbers with numbers, and similar bases together).
  • Handle coefficients separately: Multiply the numerical coefficients directly. In our case, multiply \(-7\) and \(-1\) to get a positive \(7\).
This structured approach ensures every part of the expression receives proper attention, leading to a correct and streamlined outcome.
Integral Exponents
Integral exponents are simply the powers to which a number or variable is raised. They can be positive, negative, or zero. It's essential to understand how they function in operations like \((-7a^2b^{-5})(-a^{-2}b^7)\):
  • Positive exponents: Indicate how many times a base number is used as a factor. For example, \(b^2\) means \(b\) is multiplied by itself.
  • Negative exponents: Indicate a reciprocal. For instance, \(b^{-5}\) is seen as \(\frac{1}{b^5}\). When multiplying, they can switch to positive based on the combined total power.
  • Zero exponents: Lead to the result being \(1\), as shown in \(a^0 = 1\).
Recognizing how integral exponents work helps simplify expressions by ensuring all components are properly managed, ultimately simplifying into an expression with positive exponents.
Laws of Exponents
The laws of exponents provide a toolkit for handling expressions involving powers effectively. When working through the multiplication \((-7a^2b^{-5})(-a^{-2}b^7)\), these laws are invaluable:
  • Product of powers: When bases are alike, add the exponents: \(x^m \times x^n = x^{m+n}\). For example, the \(b\) terms combine as \(b^{-5+7}=b^2\).
  • Power of zero: Any base to the power of zero equals one: \(x^0 = 1\). The \(a\) terms illustrate this with \(a^{2-2} = a^0 = 1\).
  • Negative exponents: Convert to positive by finding the reciprocal. Though not needed in our final result \(7b^2\), understanding this ensures clarity when we begin with expressions like \(b^{-5}\).
By using these laws, expressions become more manageable, ensuring that we simplify them while maintaining clarity and accuracy. This results in clear and concise expressions with only positive integral exponents.

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

(a) Because \(\frac{4}{5}=0.8\), we can evaluate \(10^{\frac{4}{5}}\) by evaluating \(10^{0.8}\), which involves a shorter sequence of "calculator steps." Evaluate parts (b), (c), (d), (e), and (f) of Problem 96 and take advantage of decimal exponents. (b) What problem is created when we try to evaluate \(7^{\frac{4}{3}}\) by changing the exponent to decimal form?

Convert each number to scientific notation and perform the indicated operations. Express the result in ordinary decimal notation. $$ (0.00003)(0.00025) $$

Convert each number to scientific notation and perform the indicated operations. Express the result in ordinary decimal notation. $$ \sqrt{0.00000009} $$

For Problems \(1-18\), write each of the following in scientific notation. \(0.00005\)

Explain how you would evaluate \(27^{\frac{2}{3}}\) without a calculator.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

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.