9514 1404 393
Answer:
a. 1.48 seconds
Step-by-step explanation:
You want to find the larger value of t such that h(t) = 10.
-16t^2 +25t +8 = 10
16t^2 -25t +2 = 0 . . . . subtract the left side to get standard form
Using the quadratic formula, we find the values of t to be ...
t = (-(-25) ± √((-25)^2 -4(16)(2)))/(2(16)) = (25±√497)/32
t ≈ 0.08 or 1.48
The ball goes in the hoop about 1.48 seconds after it is thrown.
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Additional comment
The quadratic formula tells us the solution to ...
ax² +bx +c = 0
is given by ...
[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Here, we have a=16, b=-25, c=2. Of course, our variable is t, not x, but the relation is the same.
