Intermediate Algebra (6th Edition)

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

Chapter 5 - Review: 27

Answer

$\frac{9}{16}$

Work Step by Step

$(\frac{4}{3})^{-2}=\frac{1}{(\frac{4}{3})^{2}}=\frac{1}{\frac{4\times4}{3\times3}}=\frac{1}{\frac{16}{9}}=\frac{9}{16}$
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