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If \(v*m*t\) not \(= 0\), is \(v^2*m^3*t^{-4} > 0\)?

1) \(m>v^2\) 2) \(m>t^{-4}\)

Shalva's solution is correct.

First of all: knowing that \(v*m*t\neq{0}\) implies that none of the variables equals to zero.

Second of all: \(v^2*m^3*t^{-4}>0\)? --> \(\frac{v^2*m^3}{t^4}> 0\)? Now when this inequality holds true? Obviously as \(v^2\) and \(t^4\) are positive 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.

in both conditions the only variable that can change the sign of the expression is M. Since in both conditions we are told that M most be greater than some always positive number than in both condition we can answer the question if M is positive and thus the expression is positive.

Re: If v*m*t not = 0 , is v^2*m^3*t^{-4} > 0 ? [#permalink]
18 Dec 2013, 04:30

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