Answer
$\frac{16}{9}$
Work Step by Step
$(\frac{27}{64})^{-\frac{2}{3}}=(\frac{3^{3}}{4^{3}})^{-\frac{2}{3}}=\frac{(3^{3})^{-\frac{2}{3}}}{(4^{3})^{-\frac{2}{3}}}=\frac{3^{-2}}{4^{-2}}=\frac{4^{2}}{3^{2}}=\frac{16}{9}$
Algebra: A Combined Approach (4th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321726391
ISBN 13:
978-0-32172-639-1
Chapter 10 - Review - Page 749: 148
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