<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>
(6(x^2-1))*((6x-1)/(6(x+1))
(6((x+1)(x-1)))((6x-1)/(6(x+1))
(6(x-1))*(6x-1)/(6)
(x-1)(6x-1)
6x^2-x-6x+1
6x^2-7x+1
Answer:
?=14
Step-by-step explanation:
12+5=17
8+8=16
8+4х8=40
4+6х4=28
9-5=4
9-12+5+8+4=14