1.
If no changes are made, the school has a revenue of :
625*400$/student=250,000$
2.
Assume that the school decides to reduce n*20$.
This means that there will be an increase of 50n students.
Thus there are 625 + 50n students, each paying 400-20n dollars.
The revenue is:
(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)
3.
check the options that we have,
a fee of $380 means that n=1, thus
250(n+50)(20-n)=250(1+50)(20-1)=242,250 ($)
a fee of $320 means that n=4, thus
250(n+50)(20-n)=250(4+50)(20-4)=216,000 ($)
the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.
Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.
Vertex form for a parabola with vertex (p,q) is
![y = a(x-p)^2 + q](https://tex.z-dn.net/?f=y%20%3D%20a%28x-p%29%5E2%20%2B%20q)
We have (p,q)=(-4,5) and another point (-3,2) so we can solve for <em>a:</em>
![2 = a(-3 - -4)^2 + 5](https://tex.z-dn.net/?f=2%20%3D%20a%28-3%20-%20-4%29%5E2%20%2B%205)
![2 = a + 5](https://tex.z-dn.net/?f=2%20%3D%20a%20%2B%205)
![a = -3](https://tex.z-dn.net/?f=a%20%3D%20-3)
Putting it together,
Answer: ![y = -3(x + 4)^2 + 5](https://tex.z-dn.net/?f=y%20%3D%20-3%28x%20%2B%204%29%5E2%20%2B%205)
We could multiply that out if we like but I won't bother.
9514 1404 393
Answer:
16 seconds
Step-by-step explanation:
You want to find t such that h(t) = 0
h(t) = -16t^2 +4096 . . . equation for height
0 = -16t^2 +4096 . . . . .equation for height = 0
16t^2 = 4096 . . . . add 16t^2
t^2 = 256 . . . . . . . divide by 16
t = 16 . . . . . . . . . . . take the positive square root
It will take 16 seconds for the camera to hit the ground.