Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Review Exercises: 18

Answer

$\dfrac{2,025}{8r^{4}}$

Work Step by Step

Using the laws of exponents, the given expression, $ \left( \dfrac{3r^5}{5r^{-3}} \right)^{-2}\left(\dfrac{9r^{-1}}{2r^{-5}}\right)^3 ,$ simplifies to \begin{array}{l}\require{cancel} \left( \dfrac{5r^{-3}}{3r^5} \right)^{2}\left(\dfrac{9r^{-1}}{2r^{-5}}\right)^3 \\\\= \left( \dfrac{5^2r^{-3(2)}}{3^2r^{5(2)}} \right)\left(\dfrac{9^3r^{-1(3)}}{2^3r^{-5(3)}}\right) \\\\= \left( \dfrac{5^2r^{-6}}{3^2r^{10}} \right)\left(\dfrac{9^3r^{-3}}{2^3r^{-15}}\right) \\\\= \dfrac{5^29^3r^{-6}r^{-3}}{3^22^3r^{10}r^{-15}} \\\\= \dfrac{5^2(3^2)^3r^{-6+(-3)-10-(-15)}}{3^22^3} \\\\= \dfrac{5^23^6r^{-6-3-10+15}}{3^22^3} \\\\= \dfrac{5^23^{6-2}r^{-4}}{2^3} \\\\= \dfrac{5^23^{4}}{2^3r^{4}} \\\\= \dfrac{(25)(81)}{8r^{4}} \\\\= \dfrac{2,025}{8r^{4}} .\end{array}
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