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: 9

Answer

(a) to E (b) to G (c) to F (d) to F

Work Step by Step

The matches are: (a) to E $(\frac{4}{9})^{3/2}=(\frac{2^2}{3^2})^{3/2}=(\frac{2}{3})^{3}=\frac{8}{27}$ (b) to G $(\frac{4}{9})^{-3/2}=(\frac{2^2}{3^2})^{-3/2}=(\frac{2}{3})^{-3}=\frac{27}{8}$ (c) to F $-(\frac{9}{4})^{3/2}=-(\frac{3^2}{2^2})^{3/2}=-(\frac{3}{2})^{3}=-\frac{27}{8}$ (d) to F $-(\frac{4}{9})^{-3/2}=-(\frac{2^2}{3^2})^{-3/2}=-(\frac{2}{3})^{-3}=-\frac{27}{8}$
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