Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.2 - Exponents and Radicals - 1.2 Exercises: 38


a) $\frac{1}{yz^{3}}$ b) $\frac{y^8}{x^{12}}$

Work Step by Step

a) $\frac{y^{-2}z^{-3}}{y^{-1}}$ = $y^{-1}$$z^{-3}$= $\frac{1}{yz^{3}}$ b) ($\frac{x^3y^{-2}}{x^{-3}y^{2}}$)$^{-2}$ = (${x^{3+3}y^{-2-2}}$)$^{-2}$ = $x^{(6)(-2)}$$y^{(-4)(-2)}$ = $x^{-12}$$y^{8}$ = $\frac{y^8}{x^{12}}$
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.