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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.

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http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...