Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.6 Exponents - R.6 Exercises - Page R-25: 49

Answer

$\displaystyle \frac{h^{1/3}t^{1/5}}{k^{2/5}}$

Work Step by Step

$ \displaystyle \frac{k^{-3/5}\cdot h^{-1/3}\cdot t^{2/5}}{k^{-1/5}\cdot h^{-2/3}\cdot t^{1/5}}\qquad$ ....... use $\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}$ $=k^{-3/5-(-1/5)}\cdot h^{-1/3-(-2/3)}\cdot t^{2/5-1/5}$ ... simplify exponents $=k^{-2/5}h^{1/3}t^{1/5}\qquad$ ....... use $a^{-n}=\displaystyle \frac{1}{a^{n}}=(\frac{1}{a})^{n}$ $= \displaystyle \frac{h^{1/3}t^{1/5}}{k^{2/5}}$
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.