Short answer: I don't know, but that doesn't mean I can't give you something that you can decide for yourself.
y = 4*2^(2n - 2) is the pattern.
Go for broke. Try n = 4. You should get 256. Let's try it.
y = 4 * 2^(2*4 - 2)
y = 4 * 2^(8 - 2)
y = 4 * 2^6
y = 4 * 64
y = 256 yup it works.
The other end is just as important. Suppose n = 1
Then y = 4 * 2^(2*1 - 2) = 4 * 2^0 = 4*1 = 4 Both work.
If this formula is correct, we can abbreviate it to make your task easier.
y = 4 * 2^(2n - 2)
y = 2^2 * 2^(2n - 2)
y = 2^(2n - 2 + 2)
y = 2^(2n) Now try the two end points again.
n = 4
y = 2^(2*4)
y = 2^8
y = 256
n = 1
y = 2^(2*1)
y = 2^2
y = 4 which again checks.
so y = 2^(2n) I think is an exponential function.
Sorry my explanation is so long.
The way to solve is using order of operation PEMDAS. Its very easy. You should be able to solve it yourself. I'll give you the first 3 answers:
1. g/6=x
2. x=1/2u+1
3. z-m=x
I would give you all the answers but it would be best if you did it yourself. You teacher and parents would be much happier if you learnt the skill. Hope this answer helps! Mark as brainliest! :)
