A) 14
b) 16
c)-8
d) -37
e) 15
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.
__
<em>Additional comment</em>
The quadratic formula tells us the solution to ...
ax² +bx +c = 0
is given by ...

Here, we have a=16, b=-25, c=2. Of course, our variable is t, not x, but the relation is the same.
Sry this is completely off topic, but where do you go to ask questions? I can't seem to find it. The answer is C I think
Answer:
25, 21, 13, 9
Step-by-step explanation:
All you do here is plug in the X from the table to the X in the equation so,
y=29 - 4 x 1 = 25
y=29 - 4 x 2 = 21
y=29 - 4 x 4 = 13
y=29 - 4 x 5 = 9