Answer:
t≈8.0927
Step-by-step explanation:
h(t) = -16t^2 + 128t +12
We want to find when h(t) is zero ( or when it hits the ground)
0 = -16t^2 + 128t +12
Completing the square
Subtract 12 from each side
-12 = -16t^2 + 128t
Divide each side by -16
-12/-16 = -16/-16t^2 + 128/-16t
3/4 = t^2 -8t
Take the coefficient of t and divide it by 8
-8/2 = -4
Then square it
(-4) ^2 = 16
Add 16 to each side
16+3/4 = t^2 -8t+16
64/4 + 3/4= (t-4)^2
67/4 = (t-4)^2
Take the square root of each side
±sqrt(67/4) =sqrt( (t-4)^2)
±1/2sqrt(67) = (t-4)
Add 4 to each side
4 ±1/2sqrt(67) = t
The approximate values for t are
t≈-0.092676
t≈8.0927
The first is before the rocket is launched so the only valid answer is the second one
B is the correct answer.
8 x 35 = 280
Step-by-step explanation:
theta = 180 - 123 = 57 degree
arc length = 2 * pi * R * ( theta / 360)
= 2 * pi * R * ( 57 / 360)
= 0.994R cm
Answer:4
Step-by-step explanation:2x+y+z=2(-2)+3+5=(-4)+3+5=(-1)+5=4
Answer:
Step-by-step explanation: . y - 7 = -3⁄4 (x +5)
1. y = -3/4(x+5) + 7
2. y = -3/4x + -15/4 + 7
3. y = -3/4x + 13/4
