## Intermediate Algebra (6th Edition)

$\text{approximately } 2.9 \text{ times}$
Substituting $x=5$ in the given function, $F(x)=0.6\sqrt{49-x^2} ,$ results to \begin{array}{l}\require{cancel} F(x)=0.6\sqrt{49-5^2} \\ F(x)=0.6\sqrt{49-25} \\ F(x)=0.6\sqrt{24} \\ F(x)=0.6\sqrt{4\cdot6} \\ F(x)=0.6\sqrt{2^2\cdot6} \\ F(x)=0.6(2)\sqrt{6} \\ F(x)=1.2\sqrt{6} \\ F(x)\approx2.9 .\end{array} Hence, the demand per week, $F(x),$ of an older released DVD is $\text{approximately } 2.9 \text{ times} .$
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