Question

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Divide
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Answer 1

those are my options @MARC_D

Answer 2

@ParthKohli

Answer 3

@karim728

Answer 4

@UnkleRhaukus

Answer 5

@.Sam.

Answer 6

@triciaal

Answer 7

where is the question? @18jonea

Answer 8

the link after divide

Answer 9

oh ok..

Answer 10

\[\frac{ x^2+7x+12 }{ x^2-9 }\div \frac{ x+4 }{ x-4 }\]\[=\frac{ (x+3)(x+4) }{ x^2-3^2 }\div \frac{ x+4 }{ x-4 }\]

Answer 11

that doesnt match any of my answers though

Answer 12

\[=\frac{ (x+3)(x+4) }{ (x-3)(x+3) }\div \frac{ x+4 }{ x-4 }\]

Answer 13

\[=\frac{ (x+3)(x-4) }{ (x-3)(x+3) }\times \frac{ x-4 }{ x+4 }\]

Answer 14

to divide by a fraction invert and multiply
x^2 - 9 can be factored as the difference of 2 squares

Answer 15

simplify it

Answer 16

restrictions you do not want to have 0 in the denominator

Answer 17

always keep in mind that\[a^2-b^2=(a+b)(a-b)\]

Answer 18

x-4/x-3 but i dont know the resprictons

Answer 19

yup, what @triciaal said was correct.. restrictions means you do not want to have 0 in the denominator

Answer 20

i think the answer would be b then

Answer 21

for example\[\frac{ 1(numerator) }{ 0(denominator) }\]

Answer 22

would that be right or would it be a

Answer 23

@MARC_D

Answer 24

@triciaal

Answer 25

\[=\frac{ (x+3)(x+4) }{ (x-3)(x-3) }\times \frac{ (x-4) }{ (x+4) }\]simplify it\[=\frac{ \cancel{(x+3)}\cancel{(x+4)} }{ (x-3)\cancel{(x+3)} }\times \frac{ (x-4) }{ \cancel{(x+4)} }\]

Answer 26

what is your question?

Answer 27

right i got the x-4/x-3 now i need to do the restrictions

Answer 28

do you understand what a restriction means? read above

Answer 29

restrictions means you do not want to have 0 in the denominator..for example\[\frac{ x+9 }{ x-10 },x \neq10\]

Answer 30

ok so it is just 3

Answer 31

you want to exclude x-values that will make the denominator zero you do not divide by 0

Answer 32

answer is a

Answer 33

correct! @18jonea