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
Answer:
C.
Step-by-step explanation:
pretty positive! :) hope it helps you
Given:
A space shuttle can travel at 28968 km/h.
To find:
The distance it can cover in 8 hrs.
Solution:
We have,
Speed of space shuttle = 28968 km/h.
So,
Distance covered by the space shuttle in 1 hour = 28968 km/h.
Distance covered by the space shuttle in 8 hour = (28968 × 8) km
= 231744 km
Therefore, the distance it can cover in 8 hrs is 231744 km.
Answer:
A) w=3
Step-by-step explanation:
4 x 3 = 12
This is the work when you plug in the numbers in the varible w