if m not equal to zero is m^3 > m^2 ?
2) m^2 > m
The question asks if m^3 > m^2. We can safely divide by m^2 on both sides without worrying about whether to reverse the inequality, because m^2 can never be negative. So the question is just asking "Is m > 1?"
Statement 1 is now clearly not sufficient. Statement 2 is true for every negative number, but is also true when m > 1, so is not sufficient. When we combine the statements, we know m is positive from Statement 1, so we can safely divide both sides by m in Statement 2, and we find m > 1, which is what we wanted to prove. So the answer is C.
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