Answer: 1
Explanation:
1 to any power will always be 1
We have been given the steps to solve the equation as
Step 1: 2x = 18
Step 2: x = 18 − 2
Step 3: x = 16
There is an error in step 2.
We have 
x is in multiplication with 2. So in order to get rid of 2 from the left hand side, we perform opposite operation of multiplication i.e. division.
So we must divide both sides by 2.
Therefore, A is the correct option.
A) She did not divide 18 by 2.
F(X) = 6/X
F(X) = 6 • 1/X
F(X) = 6 • x^-1
F(X) = 6x^-1
F'(X) = 6 • d(x^-1)/dx
F'(X) = 6 • -1x^-1-1
F'(X) = 6 • -1x^-2
F'(X) = -6x^-2
F'(X) = -6/x^2
F'(-2) = -6/(-2)^2
F'(-2) = -6/4
F'(-2) = -3/2
The solution would be C. -3/2.
Answer:
So starting off with 1a f(4)
We simply substitute x with 4 for the function
f(4)=2(4)-5
f(4)=8-5
f(4)=3
Next 1b, here it's a little different since you have to multiply the functions I think ( please correct me if I'm wrong).
gf(4), basically means functions g and f are being multiplied and their variables are being substituted with 4.
g(4)=4^2+3
g(4)=16+3
g(4)=19
and we also know from before that f(4)=3, so we'll do 19 times 3 which is 57
For the next two, I may be wrong, but I'm still going to try:
anything to the power of a negative is a fraction so for f^-1(x)=2x-5 we'll have 1/f(x)=x/2+5/2
And then for g^-1, we'll have 1/g(x)=
Hope I helped :)