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: 33

Answer

$(a)$ $$z=kx^3y^5$$ $(b)$ $$864$$

Work Step by Step

$(a)$ According to the problem, $z$ is jointly proportional to $x^3$ and $y^5$, so we can write: $z=kx^3y^5$ $(b)$ If we triple $x$ and double $y$, we will get: $z=k(3x)^3(2y)^5=k\times27x^3\times 32y^5=864\times kx^3y^5$ It changes by a factor of $864$
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