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: 31

Answer

$(a)$ $$z=k\frac{x^3}{y^2}$$ $(b)$ $$z=k\frac{27x^3}{4y^2}$$

Work Step by Step

$(a)$ Applying direct variation of $x^3$ and inverse variation of $y^2$ gives us the following: $z=k\times x^3 \times \frac{1}{y^2}=k\frac{x^3}{y^2}$ $(b)$ If we triple $x$ and double $y$, we will get: $z=k\frac{(3x)^3}{(2y)^2}=k\frac{27x^3}{4y^2}$
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