Answer:
Step-by-step explanation:
B and c
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
Solve for the value of Y using the first equation, 3x-7, and then plug in 3x-7 wherever you see Y, because Y=3x-7. You can find X and then use that value of X and plug it into the equation to solve for Y
We won't have a function if for same value of x in (x,y) we get different values y.
So first step: figure out k so that the first coordinate (x) is the same:
3k-4=4k | solve for k
k = -4
no check the values y for the elements of the relation
x = 3k-4 = -12-4=-16
so at -16 we get (-16,16) and (-16, 32), which mean for k=-4 the relation is not a function.
Let me know if you have any questions.