Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Prerequisites - Review Exercises - Page 62: 36

Answer

(a) $\frac{6^2u^3v^{-3}}{12u^{-2}v}=\frac{3u^5}{v^4}$ (b) $\frac{3^{-4}m^{-1}n^{-3}}{9^{-2}mn^{-3}}=\frac{1}{m^2}$

Work Step by Step

(a) $\frac{6^2u^3v^{-3}}{12u^{-2}v^1}=\frac{36}{12}u^{3-(-2)}v^{-3-1}=3u^5v^{-4}=\frac{3u^5}{v^4}$ (b) $\frac{3^{-4}m^{-1}n^{-3}}{9^{-2}m^1n^{-3}}=\frac{3^{-4}}{(3^{2})^{-2}}m^{-1-1}n^{-3-(-3)}=\frac{3^{-4}}{3^{-4}}m^{-2}=\frac{1}{m^2}$
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.