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: 11

Answer

$\color{blue}{z=\dfrac{1}{5}(x^2+y^2)}$

Work Step by Step

$z$ varies directly with the sum of the squares of $x$ and $y$. The sum of the squares of $x$ and $y$ in symbols is $x^2+y^2$. Thus, the equation that represents the variation is: $z=k(x^2+y^2)$ Since $z=5$ when $x=3$ and $y=4$, substituting these into the tentative equation above gives: $z=k(x^2+y^2) \\5=k(3^2+4^2) \\5=k(9+16) \\5=k(25) \\\dfrac{5}{25}=\dfrac{k(25)}{25} \\\dfrac{1}{5}=k$ Thus, the equation of the inverse variation is: $\color{blue}{z=\dfrac{1}{5}(x^2+y^2)}$
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