Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Prerequisites - Review Exercises - Page 62: 37


(a) $\frac{(5a)^{-2}}{(5a)^2}=\frac{1}{(5a)^4}=\frac{1}{625a^4}$ (b) $\frac{4(x^{-1})^{-3}}{4^{-2}(x^{-1})^{-1}}=4^3x^2=64x^2$

Work Step by Step

(a) $\frac{(5a)^{-2}}{(5a)^2}=\frac{1}{(5a)^2(5a)^2}=\frac{1}{(5a)^4}=\frac{1}{625a^4}$ (b) $\frac{4(x^{-1})^{-3}}{4^{-2}(x^{-1})^{-1}}=\frac{4(4)^2x^3}{x}=4^3x^2=64x^2$
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