Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.2 - Rational Exponents - Concept and Vocabulary Check - Page 521: 7

Answer

$\dfrac{1}{64}$

Work Step by Step

$16^{-\frac{3}{2}}=\dfrac{1}{16^{\frac{3}{2}}}=\dfrac{1}{\sqrt{16}^3}=\dfrac{1}{4^3}=\dfrac{1}{64}$
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