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BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
\[(1/(2t)) + (1/(2+t)) = 2\]
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
i got to the point where 2 = 4/((2t)(2+t))
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
but i dont know how to go from here, and i dont even know if im right about this step
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
i tried to simplify the denominator and got an answer 4/(4t) am i correct?
 3 years ago

Alive Group TitleBest ResponseYou've already chosen the best response.0
I'm not so good at these either, re arranging things to get the right look. The best thing to do is to practice a lot of prealgebra questions. Download some ebooks about that and it will show you how to do these easily. But don't spend months trying to find the best ebooks or videos on the internet. Just find a few and do them then suddenly you're on the right track. It's easy after that. And whatever you don't understand, you should immediately ask here like what you're doing right now. Good Luck :) And keep on persisting.
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
ok, thx for the study tip
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
SO, DOES anyone know if this is correct, what is the correct answer?
 3 years ago

BecomeMyFan=D Group TitleBest ResponseYou've already chosen the best response.0
i found t = 2 is this right?
 3 years ago

mstud Group TitleBest ResponseYou've already chosen the best response.0
No, this is not correct, when you simplify the denominator, you get4/(4t^2)=2 , because t*t=t^2. Then multiply both sides by (4t^2), this leaves you with 4=2(4t^2) > 82t^2=4 > 2t^2=4 > 2t^2=4 t^2=2 > \[t=\pm\ \sqrt{2}\]
 3 years ago

radar Group TitleBest ResponseYou've already chosen the best response.0
mstud, nice explanation, you've been fanned
 3 years ago

mstud Group TitleBest ResponseYou've already chosen the best response.0
Thank you ...
 3 years ago
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