## Related questions with answers

A city's main well was recently found to be contaminated with trichloroethylene, a cancer-causing chemical, as a result of an abandoned chemical dump leaching chemicals into the water. A proposal submitted to city council members indicates that the cost, measured in millions of dollars, of removing x% of the toxic pollutant is given by $C(x)=\frac{0.5 x}{100-x}$ $C(x)=\frac{0.5 x}{100-x}$

Find the cost of removing 50%, 60%, 70%,80%, 90%, and 95% of the pollutant.

Solution

VerifiedGiven:

$C(x)=\frac{0.5x}{100-x}\text{ with }0<x<100$

Thus

$\begin{align*} C(50)&=\frac{0.5\cdot 50}{100-50}=\color{#4257b2}{0.5}\\ C(60)&=\frac{0.5\cdot 60}{100-60}=\color{#4257b2}{0.75}\\ C(70)&=\frac{0.5\cdot 70}{100-70}=\color{#4257b2}{1.17}\\ C(80)&=\frac{0.5\cdot 80}{100-80}=\color{#4257b2}{2}\\ C(90)&=\frac{0.5\cdot 90}{100-90}=\color{#4257b2}{4.5}\\ C(95)&=\frac{0.5\cdot 95}{100-95}=\color{#4257b2}{9.5}\\ \end{align*}$

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