Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.2 - Exponents and Radicals - Exercises - Page 16: 28

Answer

$(\frac{x^2y^{-1}z^0}{xyz})^2 = \frac{x^2}{y^4z^2}$

Work Step by Step

By the formulas given on page eight and nine, $(\frac{x^2y^{-1}z^0}{xyz})^2 = \frac{(x^2y^{-1})^2}{(xyz)^2} = \frac{x^4y^{-2}}{x^2y^2z^2} = x^{4-2}y^{(-2)-2}z^{0-2} = x^2y^{-4}z^{-2} =x^2\frac{1}{y^4} \frac{1}{z^2} = \frac{x^2}{y^4z^2}$.
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