Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Exercises - Page 134: 18

Answer

$a)$ $16x^{3}$ $b)$ $r^{10}s^{2}$

Work Step by Step

$a)$ $\dfrac{x^{2}(2x)^{4}}{x^{3}}$ Evaluate the power in the numerator: $\dfrac{x^{2}(2x)^{4}}{x^{3}}=\dfrac{x^{2}(16x^{4})}{x^{3}}=...$ Evaluate the product in the numerator: $...=\dfrac{16x^{6}}{x^{3}}=...$ Evaluate the division and simplify: $...=16x^{6-3}=16x^{3}$ $b)$ $\Big(\dfrac{r^{2}s^{4/3}}{r^{1/3}s}\Big)^{6}$ Evaluate the power: $\Big(\dfrac{r^{2}s^{4/3}}{r^{1/3}s}\Big)^{6}=\dfrac{r^{12}s^{8}}{r^{2}s^{6}}=...$ Evaluate the division and simplify: $...=r^{12-2}s^{8-6}=r^{10}s^{2}$
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