Step-by-step explanation:
Domain of a rational function is everywhere except where we set vertical asymptotes. or removable discontinues
Here, we have

First, notice we have x in both the numerator and denomiator so we have a removable discounties at x.
Since, we don't want x to be 0,
We have a removable discontinuity at x=0
Now, we have

We don't want the denomiator be zero because we can't divide by zero.
so


So our domain is
All Real Numbers except-2 and 0.
The vertical asymptors is x=-2.
To find the horinzontal asymptote, notice how the numerator and denomator have the same degree. So this mean we will have a horinzontal asymptoe of
The leading coeffixent of the numerator/ the leading coefficent of the denomiator.
So that becomes

So we have a horinzontal asymptofe of 2
I believe that would be b
Answer:
6x-108
Step-by-step explanation:
-3 · 2(7 · 2 - x + 4)
= ((−3)(2))((7)(2) +− x + 4)
= −84 + 6x − 24
= 6x − 108
The answer is 6x - 108
A for both of them add all the numbers up and divide by the amount of numbers to get your iQR but A is right for both
Answer:
f(6) = 53
Step-by-step explanation:
f(x) = 10x - 7
We want to find f(6)
To do so simply substitute 6 for x in the equation
Equation: 10x - 7
x = 6
f(6) = 10(6) - 7
multiply 10 and 6
f(6) = 60 - 7
subtract
f(6) = 53