Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Review: 14

Answer

$\frac{1}{2a^{17}}$

Work Step by Step

$\frac{9a(a^{-3})}{18a^{15}}=\frac{9}{18}\times a^{1+(-3)-15}=\frac{1}{2}\times a^{-17}=\frac{1}{2a^{17}}$
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