## Algebra 2 (1st Edition)

$\approx0.286$ dollars
From part a): $f(x)=\frac{-0.27x^3+1.4x^2+1.05x+39.4}{-8.25x^3+53.1x^2-7.82x+138}$ Thus if we plug in $x=0$ we get: $f(0)=\frac{-0.27(0)^3+1.4(0)^2+1.05(0)+39.4}{-8.25(0)^3+53.1(0)^2-7.82(0)+138}\approx0.286$