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: 14

Answer

$\color{blue}{z=\frac{1}{17}(x^3+y^2)}$

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. $z$ varies directly with the sum of the cube of $x$ and the square of $y$. Thus, the equation of the variation is: $z=k(x^3+y^2)$ Since $z=1$ when $x=2$ and $y=3$, substituting these into the tentative equation above gives: $\require{cancel} z=k(2^3+3^2) \\1=k(8+9) \\1=k(17) \\\frac{1}{17}=\frac{k(17)}{17} \\\frac{1}{17}=k$ Thus, the equation of the inverse variation is: $\color{blue}{z=\frac{1}{17}(x^3+y^2)}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.