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
You first need to break each shape down. for example in the top right corner you have a pyramid and a triangular prism. find the area for those two shapes and add them together and you will have the surface area for the entire shape
First, distribute the 12
156-168-15=16+4+x
-27=20+x
-47=x
Answer:
Step-by-step explanation:
I think it is she sped up and passed susana since it shows it in the graph.
<span>associative property of addition
(a+b) + c = a + (b +c)
hope that helps.....</span>
