Okay, here we have this:
Considering the provided information, we are going to calculate the value of f, so we obtain the following:
We'll replace in the function with one of the provided points, so we have:
[tex]\begin{gathered} d(v)=\frac{2.15v^2}{64.4f} \\ 38=\frac{2.15\left(20\right)^2}{64.4f} \\ 38=\frac{860}{64.4f} \\ 38\cdot\: 64.4f=\frac{860}{64.4f}\cdot\: 64.4f \\ 2447.2f=860 \\ \frac{24472f}{24472}=\frac{8600}{24472} \\ f=\frac{1075}{3059} \\ f\approx0.35 \end{gathered}[/tex]
Finally we obtain that f is approximately 0.35.
