anonymous
  • anonymous
Can someone PLEASE explain how to solve this? The fraction at the end completely throws me off!! Pic related: http://prntscr.com/12xgei
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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terenzreignz
  • terenzreignz
Okay... So, essentially \[\huge \frac{f}g(x) = \frac{f(x)}{g(x)}\] is the key to understanding this...
anonymous
  • anonymous
I deducted it and f(x) = 3 and g(x) = 7 But, my answer isn't 7... My answers are: 3 -3 -1 1 I tried dividing 3 by 7 and 7 by 3, but that didn't work.
terenzreignz
  • terenzreignz
f(x) = 3? I take it you meant f(-2) indeed, f(-2) = 3 I suggest you retry g(-2)

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anonymous
  • anonymous
I thought: \[\frac{ f }{ g }(-2) = \frac{ f(-2) }{ g }\] Then I'd find f(-2) Then I'd plug in -2 to g(x) and get 7.
terenzreignz
  • terenzreignz
Ahh. No. \[\huge \frac{f}g(x) = \frac{f(x)}{g(x)}\\\huge \frac{f}g(-2)=\frac{f(-2)}{g(-2)}\]
terenzreignz
  • terenzreignz
It does bug me when they use shorthand notation resulting into unclear stuff... but yeah, that's Maths for you :P
anonymous
  • anonymous
Here is what I did: f(x) = -3x - 3 f(x) = -3(-2) - 3 f(x) = 6 - 3 f(x) = 3 Plug it into g g(x) = 2(3) + 1 g(x) = 7 But the answer isn't seven. Am I doing it wrong?
terenzreignz
  • terenzreignz
because you're not supposed to get g(f(-2)) Just g(-2)
anonymous
  • anonymous
\[\frac{ -3(-2) - 3 }{ 2(-2) + 1 }\] Is that it?
terenzreignz
  • terenzreignz
Yup.
anonymous
  • anonymous
\[\frac{ 3 }{ -3 } = -1\]
terenzreignz
  • terenzreignz
There you go.
anonymous
  • anonymous
;D
anonymous
  • anonymous
Okay, and can you explain the g(f[4]) thingy?
terenzreignz
  • terenzreignz
That's when you first get the result of f(4) and then plug it into g(x)
anonymous
  • anonymous
And if you had f(g[4]) You'd first get g and plug it into f?
terenzreignz
  • terenzreignz
You'd first get g(4) and then plug it into f(x) , yes :)
anonymous
  • anonymous
I see. It's all about order. But sometimes it's so unclear ;-;
terenzreignz
  • terenzreignz
Hopefully clearer now, if only a little bit :)
anonymous
  • anonymous
Yup! Thanks

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