Answer:
options b ... on the y axis only the x coordinate is always 0
Answer:
A
Step-by-step explanation:
-5(2) + (-5)(1/16)
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.
Answer:
18500 is the initial value
Every year they increase by a factor of 1.03
Step-by-step explanation:
This is written in the form
y = ab^x where a is the initial value and b is the growth rate/decrease
y = (18500) (1.03)^t
The initial value is 18500
The students will increase by 1.03 - 1 = .03
The factor of increase is what they multiply by which is 1.03
