[tex]\begin{gathered} f(x)=0.0775x^2-0.7x+65.8 \\ g(x)=0.0775x^2+2.9x+12.9 \end{gathered}[/tex]
a) find f and g at x=65
Stopping distance on dry pavement (when x=65):
[tex]\begin{gathered} f(65)=0.0775(65)^2-0.7(65)+65.8 \\ f(65)=0.0775(4225)-45.5+65.8 \\ f(65)=327.4375-45.5+65.8 \\ f(65)=347.7375 \end{gathered}[/tex]
The stopping distance on dry pavement is 347.7375 feet
Stopping on wet pavement (when x=65):
[tex]\begin{gathered} g(65)=0.0775(65)^2+2.9(65)+12.9 \\ g(65)=0.0775(4225)+188.5+12.9 \\ g(65)=327.4375+188.5+12.9 \\ g(65)=528.8375 \end{gathered}[/tex]
The stopping distance on wet pavement is 528.8375 feet
