<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>
I can give you the answer right of the top of my head here it is 1 and 3/4 x 2 and 1/2 is 4 and 3/8
Answer:
m<2
Step-by-step explanation:
To get to this, you first need to move the variable to one side, so you could subtract m from both sides to make the equation 16>7m+2. Next, you subtract 2 from both sides to make the equation into 14>7m. Finally, you would divide 7 from both sides to get 2>m, or m<2.
Hope this helps!
Answer:
1/6×12/13
2/13
Step-by-step explanation:
thats the answer
Answer:
59
Step-by-step explanation:
(360-92-150)/2
= 118/2
= 59
Answered by GAUTHMATH
