Answer:
Do you want to be extremely boring?
Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?
is a valid solution.
Want something more fun? Why not a parabola? 
.
At this point you have three parameters to play with, and from the fact that 
we can already fix one of them, in particular 
. At this point I would recommend picking an easy value for one of the two, let's say 
(or even 
, it will just flip everything upside down) and find out b accordingly:
Our function becomes
Notice that it works even by switching sign in the first two terms: 
Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: 
Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need 
, and at that point the first condition is guaranteed; using the second to find k we get 
Or how about a sine wave that oscillates around 2? with a similar reasoning you get
Sky is the limit.
Answer:
y = 20
Step-by-step explanation:
y = kx
24 = 6k
24/6 = 6k/6
k = 4
y = kx
y = 4(5)
y = 20
15+22(4+11)=x
4+11=15
22*15=330
15+330=345
x=345
Answer:
x=20
Step-by-step explanation:
inverse operation
add 7 to both sides
x=20
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.
