## College Algebra (10th Edition)

Published by Pearson

# Chapter 2 - Section 2.5 - Variation - 2.5 Assess Your Understanding: 12

#### Answer

$\color{blue}{T=d^2\sqrt[3]{x}}$

#### 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. (3) If $y$ varies jointly as $x$ and $z$, then $y = kxz$ where $k$ is the constant of proportionality. Notice that when the variation is direct or joint, the variable/ are on the numerator while if the variation is inverse, the variable is in the denominator. $T$ varies jointly with the cube root of $x$ and the square of $d$. Using the formula in (3) above, the equation of the joint variation is: $T=k\cdot \sqrt[3]{x} \cdot d^2 \\T=kd^2\sqrt[3]{x}$ Since $T=18$ when $x=8$ and $d=3$, substituting these into the tentative equation above gives: $\require{cancel} T=kd^2\sqrt[3]{x} \\18=k(3^2)\sqrt[3]{8} \\18=k(9)(2) \\18=k(18) \\\frac{18}{18}=k \\1=k$ Thus, the equation of the joint variation is: $T=1\cdot d^2\sqrt[3]{x} \\\color{blue}{T=d^2\sqrt[3]{x}}$

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