Answer:
37 - 5 = 32
32/4 = 8
8 students are in each of the 4 rows
The height at t seconds after launch is
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.
Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s
The projectile reaches a height of 192 ft at 3 s on the way up, and at 4 s on the way down.
Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s
When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.
Answer: 7 s
Answer:
The answer is y=1/3x-1
Step-by-step explanation:
m=1/3
y-y=m(x-x1)
y-(-1)=1/3(x-0)
y+1=1/3x-0
-1 -1
y=1/3x-1
For 12 divided by 587, I got 0.02044293.
For 587 divided by 12, I got 48.9166666666666.
I hope this helps.
Answer: D
I just took the test and the answer is D.