Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.8 - Modeling Using Variation - Exercise Set - Page 489: 34

Answer

$2.4\; footcandles $.

Work Step by Step

Step 1:- Translate the statement to form an equation. Let the illumination be $I$ and the distance from the headlight be $D$. Because $I$ varies inversely as $D^2$ we have: $\Rightarrow I=\frac{k}{D^2}$ ...... (1) Step 2:- Substitute the first set of values into the equation to find the value of $k$. The given values are $I=3.75\; footcandles$ and $D=40\; feet$. Substitute into the equation (1). $\Rightarrow 3.75=\frac{k}{40^2}$ $\Rightarrow 3.75=\frac{k}{1600}$ Multiply both sides by $1600$. $\Rightarrow 1600\cdot 3.75=1600\cdot \frac{k}{1600}$ Simplify. $\Rightarrow 6000=k$ Step 3:- Substitute the value of $k$ into the original equation. Substitute $k=6000$ into the equation (1). $\Rightarrow I=\frac{6000}{D^2}$ ...... (2) Step 4:- Solve the equation to find the required value. Substitute $D=50\;feet$ into the equation (2). $\Rightarrow I=\frac{6000}{(50)^2}$ Simplify. $\Rightarrow I=2.4$ Hence, the illumination is $2.4\; footcandles $.
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