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.
Each next term is being multiplied by 5. Assuming there are only 4 terms to the sum, you have
4x + 5832x = 5836x
If that's not correct put the exponents back which then this would be your answer
4x^2 + 5832x^3
Hope this helps, good luck my friend.
Answer:
x+y = 50
Step-by-step explanation:
We can find the slope of the line
m = (y2-y1)/(x2-x1)
= (50-41)/(0-9)
= 9/-9
=-1
We can use the point slope form of the equation to make an equation for a line
y-y1 = m(x-x1)
y-50 = -1(x-0)
y-50 = -x
To get it in standard from, subtract y from each side
y-y-50 = -x-y
-50 = -x-y
Multiply each side by -1
50 = x+y
x+y = 50