<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>
2p+6a=$14
3p+9a=$21
3p+9a=21
Subtract 9a
3p=21-9a
divide all by 3
p=7-3a
plug it into start equations
2p+6a=14
2(7-3a)+6a=14
14-6a....+6a=14
this zeroes out...
Answer:
A is 2 out of 6
Step-by-step explanation:
Probability of him pulling the one with parallel sides is 2/6 because there are only two out of 6 with parallel lines
Answer:
=0.000342
Standard form is 3.4237×10
Step-by-step explanation:
------------------
Answer:
1,260
Step-by-step explanation:
6x3=18 7000÷18=1,260
