Question

PLEASE HELP ME

Answer 1

I think A or B

Answer 2

B or C*

Answer 3

ignore the answer choices for a moment

Answer 4

what does x^2 - 4 factor to?

Answer 5

ok (x+2) (x-2)

Answer 6

correct

Answer 7

so what we do is factor the denominator and you'll see a pair of terms cancel

Answer 8

\[\Large \frac{-5}{5x+15} \cdot \frac{2(x+2)}{x^2-4}\]
\[\Large \frac{-5}{5x+15} \cdot \frac{2(x+2)}{(x+2)(x-2)}\]
\[\Large \frac{-5}{5x+15} \cdot \frac{2\cancel{(x+2)}}{\cancel{(x+2)}(x-2)}\]
\[\Large \frac{-5}{5x+15} \cdot \frac{2}{x-2}\]

Answer 9

making sense so far?

Answer 10

yes

Answer 11

now, what does 5x+15 factor to?

Answer 12

can I tell you what I think it is I tried doing the math

Answer 13

what do you think 5x+15 factors to

Answer 14

5x + 15

Answer 15

try factoring out the GCF

Answer 16

one sec

Answer 17

5(x+3)

Answer 18

yep

Answer 19

:)

Answer 20

\[\Large \frac{-5}{5x+15} \cdot \frac{2}{x-2}\]
\[\Large \frac{-5}{5(x+3)} \cdot \frac{2}{x-2}\]
\[\Large \frac{-\cancel{5}}{\cancel{5}(x+3)} \cdot \frac{2}{x-2}\]
\[\Large \frac{-1}{x+3} \cdot \frac{2}{x-2}\]

Answer 21

then you multiply the fractions (straight across: numerator with numerator, denominator with denominator)

\[\Large \frac{-1}{x+3} \cdot \frac{2}{x-2}\]
\[\Large \frac{-1*2}{(x+3)(x-2)}\]
\[\Large \frac{-2}{(x+3)(x-2)}\]

Answer 22

so

\[\Large \frac{-5}{5x+15} \cdot \frac{2(x+2)}{x^2-4}\]

simplifies to

\[\Large \frac{-2}{(x+3)(x-2)}\]

Answer 23

the question is now: what is the domain?

Answer 24

to answer that sub-question, you need to set every denominator in the original expression equal to zero and solve

Answer 25

5x+15 = 0 ----> x = ???

x^2 - 4 = 0 ----> x = ???

Answer 26

its C thank you !

Answer 27

yep, correct