Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.1 - Exponents and Scientific Notation - Exercise Set - Page 262: 35



Work Step by Step

We are given the expression $4^{-2}$. In general, $a^{-n}=\frac{1}{a^{n}}$, where a is a nonzero real number and n is a positive integer. Therefore, $4^{-2}=\frac{1}{4^{2}}=\frac{1}{4\times4}=\frac{1}{16}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.