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It shld be D Is \(v^2 *m^3*t^-4>0?\) Since \(v^2\) and \(1/t^4\)are positive, so it depends upon whether m is positive or not (1) m>\(v^2\) >0 hence sufficient (2) m> \(1/t^4\) >0 hence sufficient
Can anyone please explain in Stmt2, how m> (1/(t^4))?
Thanks, Weirdo
If \(v*m*t\) not \(= 0\), is \(v^2*m^3*t^{-4} > 0\)?
\(v*m*t\neq{0}\) means that none of the unknowns equals to zero.
Is \(v^2*m^3*t^{-4}>0\)? --> is \(\frac{v^2*m^3}{t^4}> 0\)? As \(v^2\) and \(t^4\) are positive (remember none of the unknowns equals to zero) this inequality will hold true if and only \(m^3>0\), or, which is the same, when \(m>0\).
(1) \(m>v^2\) --> \(m\) is more than some positive number (\(v^2\)), hence \(m\) is positive. Sufficient.
(2) \(m>t^{-4}\) --> \(m>\frac{1}{t^4}\) --> Again \(m\) is more than some positive number (\(\frac{1}{t^4}\) ), hence \(m\) is positive. Sufficient.
Re: If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink]
30 Dec 2013, 21:27
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Re: If vmt ≠ 0, is v^2m^3t^-4 > 0? [#permalink]
29 Apr 2015, 08:38
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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