Prealgebra (7th Edition)

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

Appendix B - Exercise Set: 30

Answer

$\frac{3}{64}$

Work Step by Step

$4^{-2}-4^{-3}=\frac{1}{4^2}-\frac{1}{4^3}=\frac{1\times4}{4^2\times4}-\frac{1}{4^3}=\frac{4}{4^3}-\frac{1}{4^3}=\frac{3}{4^3}=\frac{3}{64}$
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