where
s = arc length
n = measure of central angle
r = radius
6t+12-4t
Use the distribution property.
g = -3,5
8 - |2g - 5| = 8 - |2 · (-3,5) - 5| =
= 8 - |-7 - 5| = 8 - |-12| = 8 - 12 = <u>-</u><u>4</u>
<span>Formula: H(t) = 56t – 16t^2
</span>
H(t) = - 16t^2 + 56t
<span><span>A.
</span>What is the height of the ball after 1 second? H
(1) = 56(1) – 16(1) ^2 = 40 pt.</span>
<span><span>B.
</span>What is the maximum height? X = - (56)/2(- 16) =
1.75 sec h (1.75) = 56(1.75) – 16(1.75) ^2 h (1.75) = 49ft.</span>
<span><span>C.
</span><span>After how many seconds will it return to the
ground? – 16t^2 + 56t = 0 - 8t =0 t = 0</span></span>
<span><span>-
</span><span>8t (2 + - 7) = 0 2t – 7 = 0 t = 7/2
Ans: 3.5 seconds</span></span>
H/4=7/14
we cross multiply , so
14*h = 7*4
14h=28
divide by 14 on both sides
h=28/2=2
