Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 423: 14

Answer

The required equation is $x=k\frac{\sqrt[3]{z}}{y}$ and value of $y$ is $k\frac{\sqrt[3]{z}}{x}$.

Work Step by Step

Since, the value of $x$ varies directly as $\ \sqrt[3]{z}$ and inversely as $\ y$. $x=k\frac{\sqrt[3]{z}}{y}$ Where $k$ is a constant. Now, solve the equation for $y$. $y=k\frac{\sqrt[3]{z}}{x}$
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