LoyalWater wrote:
If vmt ≠ 0, is \((v^2)*(m^3)*(t^(-4)) > 0\)
(1)\(m > v^2\)
(2) \(m > t^(-4)\)
If vmt ≠ 0, is \(v^2*m^3*t^{-4} > 0\)
Lets analyse the question quickly
V*M*T is not zero ==> none of the digit is zero. But they can be -ve or +ve
IS \(v^2*m^3*t^{-4} > 0\)? ---------------> The expression \(>0\) means
positive; so the question is asking is the expression
positive ?
ok...\(v^2\) cannot be ever negative; squaring kills the polarity and always yield a positive number.
\(t^{-4}\) cannot be ever negative . \(t^{-4}\) is actually \(\frac{1}{t^4}\) ; Diving 1 by a positive number will yield a positive number (fraction or otherwise but definitely positive)
\(m^3\) can be positive or negative depending upon whether M is +ve or -ve
SO THE REAL QUESTION IS :- IS M POSITIVE?
(1)\(m > v^2\)
m = positive since squaring will give a positive number
SUFFICIENT
(2) \(m > t^{-4}\)
m will again be positive\(t^-4\) means \(\frac{1}{t^4}\); again positive fraction or integer
SUFFICIENT
ANSWER IS D
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