12.04 meters is 474 inches so 12 would be closest
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
Okay, so a general rule for finding perpendicular lines in the form of y = mx + b is y = (-1/m) + b.
First, let's ignore b (-7) because we're going to find that later.
A perpendicular line to y = 4x + b is y = -1/4x + b.
Alright, so now let's plug in the values. They are in the form of (x,y), so let's plug them in accordingly.
3 = -1/4(4) + b
3 = -1 + b
b = 4
y = -1/4x + 4
So a line perpendicular to y = 4x - 7 is y = -1/4x + 4.
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