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: 101

Answer

$16x^{2}$

Work Step by Step

According to the definition of negative exponents, $a^{-n}=\frac{1}{a^{n}}$ (where $a\ne0$). Therefore, $(\frac{1}{4x})^{-2}=\frac{1}{(\frac{1}{4x})^{2}}=(\frac{4x}{1})^{2}=\frac{4^{2}x^{2}}{1^{2}}=\frac{16x^{2}}{1}=16x^{2}$.
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