Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 9 - Cumulative Review: 3

Answer

a. $\frac{16}{625}$ b. $\frac{1}{16}$

Work Step by Step

a. $(\frac{2}{5})^{4}$ $=\frac{2^{4}}{5^{4}}$ $=\frac{2\times2\times2\times2}{5\times5\times5\times5}$ $=\frac{4\times2\times2}{25\times5\times5}$ $=\frac{8\times2}{125\times5}$ $=\frac{16}{625}$ b. $(-\frac{1}{4})^{2}$ $=\frac{1^{2}}{4^{2}}$ $=\frac{1\times1}{4\times4}$ $=\frac{1}{16}$
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