Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 5 - Section 5.7 - Negative Exponents and Scientific Notation - Exercise Set: 19

Answer

$16$

Work Step by Step

Using $x^{-a}=\dfrac{1}{x^a}$ the given expression, $ \left( \dfrac{1}{4} \right)^{-2} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{1}{\left( \dfrac{1}{4} \right)^{2}} \\\\= \dfrac{1}{\left( \dfrac{1}{16} \right)} \\\\= 1\div\dfrac{1}{16} \\\\= 1\cdot\dfrac{16}{1} \\\\= 16 .\end{array}
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