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: 94

Answer

25

Work Step by Step

According to the definition of negative exponents, $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$). Therefore, $(\frac{1}{5})^{-2}=\frac{1}{(\frac{1}{5})^{2}}=\frac{5^{2}}{1^{2}}=\frac{25}{1}=25$.
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