College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.3 - Page 135: 74

Answer

$R_{1}= \frac{RR_{2}}{R_{2}-R}$

Work Step by Step

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ Solve for $R_{1}$ $\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ $\frac{1}{R}-\frac{1}{R_{2}}=\frac{1}{R_{1}}$ Take LCD, $\frac{R_{2}-R}{RR_{2}}=\frac{1}{R_{1}}$ $R_{1}= \frac{RR_{2}}{R_{2}-R}$
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