math 163
I messed up. I confused the f(t) = 17.7782 with the f(17.07933). Reading too many things too late at night, and those numbers are very similar.
I've changed the file at your access site, and included the correction here:
The second problem states evaluate f(17.07933) and f(2a+b) for the function y=f(t)=.03t^2+-2.1t+47 What is the value of the function for clock time t=25.619? What equation would you solve to determine the value of t for which f(t)=17.7782?(You need not actually evaluate the equation).
ok so heres what I did, y=.3(17.07933)^2+ -2.1(17.07933)+47
This is correct.
By contrast note that the solution to the equation f(t) = 17.8 is obtained by solving the equation
17.8 = .03t^2+-2.1t+47.
You solve this by subtracting 17.8 from both sides then using the quadratic formula.