Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.6 Rational Exponents - R.6 Exercises - Page 63: 10

Answer

(a) to D (b) to A (c) to B (d) to C

Work Step by Step

The matches are: (a) to D $(\frac{8}{27})^{2/3}=(\frac{2^3}{3^3})^{2/3}=(\frac{2}{3})^{2}=\frac{4}{9}$ (b) to A $(\frac{8}{27})^{-2/3}=(\frac{2^3}{3^3})^{-2/3}=(\frac{2}{3})^{-2}=\frac{9}{4}$ (c) to B $-(\frac{27}{8})^{2/3}=-(\frac{3^3}{2^3})^{2/3}=-(\frac{3}{2})^{2}=-\frac{9}{4}$ (d) to C $-(\frac{27}{8})^{-2/3}=-(\frac{3^3}{2^3})^{-2/3}=-(\frac{3}{2})^{-2}=-\frac{4}{9}$
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