## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 7 - Functions and Graphs - 7.5 Formulas, Applications, and Variation - 7.5 Exercise Set - Page 488: 69

#### Answer

$\color{blue}{y=\frac{1}{2}x^2}$

#### Work Step by Step

RECALL: (1) When $y$ varies directly as $x$, the variation is direct and is represented by the equation $y=kx$ where $k$ is the constant of variation. (2) When $y$ varies inversely as $x$, the variation is inverse and is represented by the equation $y=\frac{k}{x}$ where $k$ is the constant of variation. $y$ varies directly as he square of $x$ so the equation is $y=kx^2$ with $k$=constant of variation. To find the value of $k$, substitute the given values into $y=kx^2$ to obtain: $$y=kx^2 \\50=k(10^2) \\50 =k(100) \\\frac{50}{100}=\frac{k(100)}{100} \\\frac{1}{2}=k$$ Thus, the equation of the variation is: $\color{blue}{y=\frac{1}{2}x^2}$.

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