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find indicated sum, difference, product or quotient: f(x)=16-x^2; g(x)=4-x find (f+g)(x)

Mathematics
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g(f(x)) = 4-(16-x^2) right so far?
yes.
so then it's 4 -16 +x^2

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Other answers:

and -12 +x^2
yup! :)
cool, now do I plug in that g(x) for f(x)= 16-x^2
quite horrible, but just sub 4-x into the x.
like 16 -(4-x)
\(16 -(4-x)^2\)
oh shoot, I forgot the exponent
16-(4-x)^2 is 16 (-4-x)(-4-x) which is 32 +8x+x^2
and then add that to the -12+x^2 from g(x)?
hmm, very close, but something's off isn't it?
16-(4-x)^2 is not 16 (-4-x)(-4-x) it's 16 - (4-x)(4-x) though.
ohhh, that may fix it
yeah. something will cancel out.
yea, it left +8x -x^2
but I still don't have it exactly right. I should be getting -x^2 +x +12 I think, but I've got a -12
note that (f+g)(x)=16-x^2 +4-x
(f-g)(x)=16-x^2 -(4-x)
hmm, so what do I need to do to solve it now?
well, first, lol which are you doing? addition or subtraction?
well, it says f+g, so I would assume addition
so, simplify this one and you're done. (f+g)(x)=16-x^2 +4-x
oh ok
16-(12+x^2) +4-(12+x^2)? I think I screwed it up
or is it 16-(+8x -x^2) +4-(+8x -x^2) I'm not too sure of anything anymore
you don't need sub anything, just simplify as it's (f+g)(x) =f(x)+g(x)
oh, cuz I've already done all that. duh.
but when I add them together, I only get 20, the x^2's cancel out
I have an option for the answer to be -x^2-x+20
add 16-x^2 and 4-x only.
ok, well, thats it, obviously. but what happened to the 8x and +/-2x^2's? were those just wrong?
they weren't needed? :)
jesus christ. haha. well ok. sorry for being such a pain in the retrice thanks for the help though
hmm, well thats an odd substitution for @$$.... whatever lol
lol you're welcome

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