Intermediate Algebra (6th Edition)

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

Chapter 5 - Review: 24

Answer

$\frac{1}{16x^{2}}$

Work Step by Step

$(-4x)^{-2}=\frac{1}{(-4x)^{2}}=\frac{1}{(-4\times-4)\times(x^{1+1})}=\frac{1}{16x^{2}}$
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