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:
(1/5)x-5.4=y
Step-by-step explanation:
mx+b=y
m=(1/5)
(-8,-7)=(x,y)
(1/5)(-8)+b=(-7)
-1.6+b=-7
b=-5.4
(1/5)x-5.4=y
Pour 4 times of 124-gallon into a big jug.
Amount of water = 124 x 4 = 496 gallons
Then use the 45-gallon jug to scoop out the water 11 times.
Amount of water scooped out = 11 x 45 = 455 gallons
Water remaining = 496 - 450 = 1 gallon.