Intermediate Algebra (12th Edition)

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

Chapter 4 - Review Exercises: 14

Answer

$\frac{25}{m^{18}}$

Work Step by Step

$(5m^{-3})^{2}(m^{4})^{-3}=5^{2}(m^{-3})^{2}(m^{4})^{-3}=25m^{-3\times2}m^{4\times-3}=25m^{-6}m^{-12}=25m^{-6+(-12)}=25m^{-18}=\frac{25}{m^{18}}$
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