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:
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Step-by-step explanation:
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Answer:
D. zero
Step-by-step explanation:
Since the graphs do not intersect, there are zero solutions.
Answer:
$ 5674.076
Step-by-step explanation:
The question is on compound interest
The formulae = A= P(1+ r/n) ^nt .......where P is the principal amount, r is the rate of interest in decimal, n is number of compoundings per year and t is the total number of years.
Given; P= $4,000.00 , r=12/100=0.12, n=2 and t=3
Substituting values in the equation A= P(1+ r/n) ^nt
A= 4000 ( 1+0.12/2)^2×3
A=4000(1.06)^6
A=$ 5674.08