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: 32

Answer

$(a)$ $$z=k\frac{x^2}{y^4}$$ $(b)$ $$\frac{9}{16}$$

Work Step by Step

$(a)$ Applying direct variation of $x^2$ and inverse variation of $y^4$ gives us the following: $z=k\times x^2 \times \frac{1}{y^4}=k\frac{x^2}{y^4}$ $(b)$ If we triple $x$ and double $y$, we will get: $z=k\frac{(3x)^2}{(2y)^4}=k\frac{9x^2}{16y^4}=\frac{9}{16}k\frac{x^2}{y^4}$ It changes by a factor of $\frac{9}{16}$
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