Again, the quarterback throws the football, but this time the model [tex]y=-16 x^2+35 x+6.5[/tex] describes the height ([tex]y[/tex]), in feet, of the football [tex]x[/tex] seconds after it is thrown. If the ball is not caught, how long will it be in the air?
[tex]
\begin{array}{l}
0=-16 x^2+35 x+6.5 \\
x=\frac{-35 \pm \sqrt{35^2-4(-16)(6.5)}}{2(-16)}
\end{array}
[/tex]
Rounded to the nearest tenth, the solutions of the equation are $\square$
Answer (1)
It's 70(x)2.5 = 175 I think I might be wrong
