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tiph
Group Title
Part 1: Simplify the complex fractions. (4 points)
the quantity x¨Csquared minus x minus 20 over 4 all over the quantity x minus 5 over 10 and the quantity x¨Csquared minus x minus 20 over x minus 5 all over the quantity 4 over 10
Part 2: Are the complex fractions equivalent? Explain why or why not. (4 points)
 2 years ago
 2 years ago
tiph Group Title
Part 1: Simplify the complex fractions. (4 points) the quantity x¨Csquared minus x minus 20 over 4 all over the quantity x minus 5 over 10 and the quantity x¨Csquared minus x minus 20 over x minus 5 all over the quantity 4 over 10 Part 2: Are the complex fractions equivalent? Explain why or why not. (4 points)
 2 years ago
 2 years ago

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tiph Group TitleBest ResponseYou've already chosen the best response.0
I get the first part I just don't get if its equivalent or not??
 2 years ago

cathyangs Group TitleBest ResponseYou've already chosen the best response.1
Is what you're trying to say: (C^2(x(20/4)))/x  5/10 ? Could you please write it out in numbers?
 2 years ago

tiph Group TitleBest ResponseYou've already chosen the best response.0
dw:1346611358335:dw
 2 years ago

cathyangs Group TitleBest ResponseYou've already chosen the best response.1
Even though there are quadratic formulae there, it's still just dividing fractions. Think of it as: \[\frac{ x ^{2}x2 }{ 4 } \times \frac{ 10 }{ x5}\] and \[\frac{ 10 }{ x5 } \times \frac{ x ^{2}x2 }{ 4 }\] are they equal to each other :)?
 2 years ago

tiph Group TitleBest ResponseYou've already chosen the best response.0
i think are equal but for some reason im just second guessing myself?
 2 years ago

cathyangs Group TitleBest ResponseYou've already chosen the best response.1
:) Well, if you're uncertain, you could try solving both sides. I'd advise doing the fraction division using the equations I wrote in LaTeX up there ^. Fyi, I would say they are equal.
 2 years ago

tiph Group TitleBest ResponseYou've already chosen the best response.0
ok thanks. I will do that (:
 2 years ago

cathyangs Group TitleBest ResponseYou've already chosen the best response.1
You're welcome! I'm glad I could help.
 2 years ago
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