College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.5 - Variation - 2.5 Assess Your Understanding: 16

Answer

$\color{blue}{z^3=\frac{8}{97}(x^2+y^2)}$

Work Step by Step

RECALL: (1) If $y$ varies directly as $x$, then $y=kx$ where $k$ is the constant of proportionality. (2) If $y$ varies inversely as $x$, then $y=\dfrac{k}{x}$ where $k$ is the constant of proportionality. Notice that when the variation is direct, the variable is on the numerator while if the variation is inverse, the variable is in the denominator. $z^3$ varies directly with the sum of the squares of $x$ and $y$. Thus, the equation of the variation is: $z^3=k(x^2+y^2)$ Since $z=2$ when $x=9$ and $y=4$, substituting these into the tentative equation above gives: $\require{cancel} z^3=k(x^2+y^2) \\2^3=k(9^2+4^2) \\8=k(81+16) \\8=k(97) \\\frac{8}{97}=\frac{k(97)}{97} \\\frac{8}{97} = k$ Thus, the equation of the inverse variation is: $\color{blue}{z^3=\frac{8}{97}(x^2+y^2)}$
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