Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises: 59

Answer

16

Work Step by Step

According to the negative exponent rule, $a^{-n}=\frac{1}{a^{n}}$. Therefore, $\frac{1}{4^{-2}}=\frac{1}{\frac{1}{4^{2}}}=1\div\frac{1}{4^{2}}=1\times4^{2}=1\times16=16$.
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