Question

help!?!?!?!?

Answer 1

\[\frac{ 8x+16 }{ x^2-13+22 }\]

Factor both numerator and denominator
\[\frac{ 8(x-2) }{ (x-2)(x-11) }\]

The (x-2) cancel each other
\[\frac{ 8 }{ x-11 }\]

To find the second part of the answer, we need to find what makes the statement undefined. (0 in the denominator) Simply solve for x

\[x - 11 \neq 0\]
Add 11 to both sides
\[x \neq 11\]

So C

Answer 2

thank you

Answer 3

x cannot be 2 either. So answer is D.

Answer 4

x can be any number. We are looking for a restriction that would make the statement undefined.

2-11 = 9 not 0

Answer 5

The original function, f(x) is not defined at x = 2 or at x = 11.

When you factored out (x-2) and cancelled the term from the numerator and the denominator you are actually cancelling out 0/0 if x = 2. You cannot do that. That is why x cannot be 2 or 11.

Answer 6

Plug both numbers into the original equation

Answer 7

The denominator is x^2 - 13x + 22
At x =2 the denominator becomes: 2^2 - (13)(2) + 22 = 0
Same with x = 11. That is how you factored it.

BTW, there is a typo in your first reply the denominator is: x^2 - 13x + 22 (the x is missing).

Answer 8

Also the numerator should be 8x - 16 and not 8x + 16

Answer 9

@lowcard2 @ranga >>> x cannot be 2 either. So answer is D.

I agree with you.

The (x-2) in the denominator divided out with the (x-2) in the numerator provided that x is not equal to 2. Otherwise, we would be dividing by zero.

Then, in the simplified form of the answer, x cannot be 11 for the same reason of dividing by zero.

Answer 10

Yeah, I overlooked that. I messaged her the correction. Been doing math all day, so im kind of out of it lol