3 years ago

PLEASE HURRY!!!!

Mathematics

1 answer:

svp [43]3 years ago

5 0

1. D

2. D

3. c

4. A

5. C

I'm not 100% sure about these. Just letting you know.

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3 years ago

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ANSWER

See below

EXPLANATION

Given

$f(x) = \frac{ {x}- 9 }{x + 5}$

and

$g(x) = \frac{ - 5x - 9}{x - 1}$

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$(f \circ \: g)(x)= \frac{ \frac{ - 5x - 9 - 9(x - 1)}{x - 1}}{\frac{ - 5x - 9 + 5(x - 1)}{x - 1} }$

Expand:

$(f \circ \: g)(x)= \frac{ \frac{ - 5x - 9 - 9x + 9}{x - 1}}{\frac{ - 5x - 9 + 5x - 5}{x - 1} }$

$(f \circ \: g)(x)= \frac{ \frac{ - 5x - 9x + 9 - 9}{x - 1}}{\frac{ - 5x + 5x - 5 - 9}{x - 1} }$

$(f \circ \: g)(x)= \frac{ \frac{ - 14x }{x - 1}}{\frac{ -14}{x - 1} }$

Since the denominators are the same, they will cancel out,

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3 years ago

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