anonymous
  • anonymous
Suppose that f(x)=(2x)/(x^2-9) and that g(x)=(5x^2)/(2x-6) If h(x)=f(x)/g(x), what do you think will be the simplest rule that gives the correct values of H(x)? I know how to do other problems like this but everytime I have tried doing this one, the original and simplified versions never come out the same.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Just to check...is H(x) the same as h(x) for you
anonymous
  • anonymous
yea sorry typo
anonymous
  • anonymous
Oh okay haha only because depending on the context like in calculus, \(H(x)=\int{h(x)}\) So I just wasn't sure if this was the same thing lol okay so lets see...

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anonymous
  • anonymous
\[h(x) = \frac{ f(x) }{ g(x) }\] \[h(x) = \frac{ \frac{ (2x) }{ (x^2-9) } }{ \frac{ (5x^2) }{ (2x-6) } }\] \[h(x) = \frac{ \frac{ (2x) }{ (x+3)(x-3) } }{ \frac{ (5x^2) }{ 2(x-3) } }\] \[h(x) = \frac{ \frac{ 2(2x) }{ 2(x+3)(x-3) } }{ \frac{ (x+3)(5x^2) }{ 2(x+3)(x-3) } }\] \[h(x) = \frac{2(2x)}{(x+3)(5x^2)}\] \[h(x) = \frac{4x}{5x^2(x+3)}\] Not sure if this was the easiest way to go about this, but I think that's the right answer.
anonymous
  • anonymous
Basically multiplying each separate fraction by fancy sorts of 1, like 2/2 or x+3/x+3 to get equivalent denominators is the way I usually go about these things. (see step 4 of my reply)
anonymous
  • anonymous
I get that same answer too but when i plug it into the calculator they don't turn up the same.
anonymous
  • anonymous
Fixed it... I was putting it into the calculator wrong.

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