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 - Page 192: 13



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. $M$ varies directly with the square of $d$ and inversely with the square root of $x$. This means that $d^2$ will be on the numerator while $\sqrt{x}$ will be in the denominator. Thus, the equation of the variation is: $M=k\cdot \frac{d^2}{\sqrt{x}}$ Since $M=24$ when $x=9$ and $d=4$, substituting these into the tentative equation above gives: $\require{cancel} M=k\cdot \frac{4^2}{\sqrt{9}} \\24=k \cdot \frac{16}{3} \\24=\frac{16}{3}k \\\frac{3}{16} \cdot 24=\frac{16}{3}k \cdot \frac{3}{16} \\\frac{3}{\cancel{16}2} \cdot \cancel{24}3=\frac{\cancel{16}}{\cancel{3}}k \cdot \frac{\cancel{3}}{\cancel{16}} \\\frac{9}{2}=k$ Thus, the equation of the inverse variation is: $M=\frac{9}{2}\cdot \frac{d^2}{\sqrt{x}} \\\color{blue}{M=\dfrac{9d^2}{2\sqrt{x}}}$
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