Intermediate Algebra (12th Edition)

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

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises - Page 278: 133



Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{(2k)^2m^{-5}}{(km)^{-3}} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{2^2k^2m^{-5}}{k^{-3}m^{-3}} \\\\= 4k^{2-(-3)}m^{-5-(-3)} \\\\= 4k^{2+3}m^{-5+3} \\\\= 4k^{5}m^{-2} \\\\= \dfrac{4k^{5}}{m^2} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.