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anonymous
 one year ago
If f(x)=(x+1)^−1 and g(x)=x−2, what is the domain of f(x)÷g(x)?
anonymous
 one year ago
If f(x)=(x+1)^−1 and g(x)=x−2, what is the domain of f(x)÷g(x)?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0all values of x (−∞,−1),(−1,2), and (2,∞) (−∞,2) and (2,∞) (−∞,−1] and [2,∞)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first off, can you write the function f(x) ÷ g(x)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x) \div g(x)=(x+1)^{1}\div (x2)=\frac{ 1 }{ x+1 }\div (x2)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do the division to right that as a single fraction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have no clue how to do any this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know how to divide fractions? because that's what this is...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you need to flip the (x2) part to it's reciprocal and change the division sign to multiplication, then just multiply across like these examples. dw:1436549101824:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok what? I'm waiting on a response from you.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont know how to do it. or what to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not really sure what more I can do here if you're not going to try. I put an example up there that's pretty much your problem with different numbers. You have to at least make an attempt for me to evaluate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0omg, then dont comment. if i knew how to do it i would

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what don't you know how to do? You're not giving me anything to work with at all
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