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 488: 28

Answer

$0.88^{\circ} \; Celsius$.

Work Step by Step

Step 1:- Translate the statement to form an equation. Let the water temperature be $T$ and the water's depth be $D$. Because $T$ varies inversely as $D$ we have: $\Rightarrow T=\frac{k}{D}$ ...... (1) Step 2:- Substitute the first set of values into the equation to find the value of $k$. The given values are $D=1000\; meters$ and $T=4.4^{\circ}\;Celsius$. Substitute into the equation (1). $\Rightarrow 4.4=\frac{k}{1000}$ Multiply both sides by $1000$. $\Rightarrow 1000\cdot 4.4=1000\cdot \frac{k}{1000}$ Simplify. $\Rightarrow 4400=k$ Step 3:- Substitute the value of $k$ into the original equation. Substitute $k=4400$ into the equation (1). $\Rightarrow T=\frac{4400}{D}$ ...... (2) Step 4:- Solve the equation to find the required value. Substitute $D=5000\; meter$ into the equation (2). $\Rightarrow T=\frac{4400}{5000}$ Simplify. $\Rightarrow T=0.88$ Hence, the temperature is $0.88^{\circ} \; Celsius$.
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