Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.12 - Modeling Variation - 1.12 Exercises - Page 127: 34

Answer

$(a)$ $$z=\frac{k}{x^2y^3}$$ $(b)$ $$\frac{1}{72}$$

Work Step by Step

$(a)$ Applying inverse variation of $x^2$ and $y^3$ gives us the following: $z=k\times\frac{1}{x^2}\times\frac{1}{y^3}=\frac{k}{x^2y^3}$ $(b)$ If we triple $x$ and double $y$, we will get: $z=\frac{k}{(3x)^2(2y)^3}=\frac{k}{9x^2\times8y^3}=\frac{1}{72}\times\frac{k}{x^2y^3}$ It changes by a factor of $\frac{1}{72}$
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.