You didn’t post the question. If you post the questions, I can help.
Also, I think you can search the question and find the answers… I remember doing this one.
For this case, we can model the problem as a triangle with a right angle.
Where,
x, y: lengths of the sides of the triangle
d: hypotenuse of the triangle
We then have the following relationship:

Clearing d we have:
Answer:
A functions that would best model the situation above is:
D square root
Split up the interval [2, 5] into

equally spaced subintervals, then consider the value of

at the right endpoint of each subinterval.
The length of the interval is

, so the length of each subinterval would be

. This means the first rectangle's height would be taken to be

when

, so that the height is

, and its base would have length

. So the area under

over the first subinterval is

.
Continuing in this fashion, the area under

over the

th subinterval is approximated by

, and so the Riemann approximation to the definite integral is

and its value is given exactly by taking

. So the answer is D (and the value of the integral is exactly 39).
Let a = Adult tickets, s = Student tickets; we are going to set up a system of equations.
8a + 4s = 880
a - s = 20
We are going to multiply the second equation by so that we can cancel out one of the variables and solve for the other. This gives you:
8a + 4s = 880
4a - 4s = 20
Notice the s-variables will cancel:
12a = 900
a = 75
Now we plug in the amount of adult tickets to solve for the student ones.
75 - s = 20
- s = - 55
Divide both sides by - 1
s = 55.
There were 55 student tickets sold and 75 adult tickets.