## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 91

#### Answer

a. B. 3 b. D. -3 c. D. -3 d. B. 3

#### Work Step by Step

According to the definition of negative exponents $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$). Therefore, a. $(\frac{1}{3})^{-1}=\frac{1}{(\frac{1}{3})}=3$ b. $(-\frac{1}{3})^{-1}=\frac{1}{(-\frac{1}{3})}=-3$ c. $-(\frac{1}{3})^{-1}=-\frac{1}{(\frac{1}{3})}=-3$ d. $-(-\frac{1}{3})^{-1}=-(\frac{1}{-\frac{1}{3}})=-(-3)=3$

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