<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>
V=lengthxwidthxheight
v=2 1/2x2 1/2x 5
=31.25
35-31.25=3.75
The answer is 3.75cm³
Answer:
k = 5
Step-by-step explanation:
Since the equation is y = kx and x equals 3 while y equals 15, in order to solve this equation we are going to have to substitute the variables with their answers:
x = 3
y = 15
15 = 3k
Now we divide 3 from both 3x and 15 to segregate the variable x:
15/3 = 3k/3
5 = k
Now to check our work we are going to substitute 5 for the variable k, 3 for the variable x and 15 for the variable y.
y = kx
15= (5)(3)
15 = 15
Tada, I hope this helps you!
Answer:
a = 14
Step-by-step explanation:
28[-14]=14;
-14
