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Source: — Data Sufficiency |
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by Gurpinder » Mon Sep 20, 2010 5:56 pm
Veronica wrote:If vmt ≠ 0, is v^2*m^3*t^-4 > 0?
1. m > v^2
2. m > t^-4


Please explain this question, thanks![/list][/spoiler][/list]
v^2*m^3*1/t^4 > 0

(1) since m is being cubed, it will keep the sign of its original #. so since its bigger than v^2 which will be positive, v^2*m^3*1/t^4 is greater than 0.

(2) 1/t^4 is going to be a fraction. a positive fraction. since m is bigger than it, its a positive. hence v^2*m^3*1/t^4 is greater than 0

hence IMO (D).
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by debmalya_dutta » Mon Sep 20, 2010 6:01 pm
Veronica wrote:If vmt ≠ 0, is v^2*m^3*t^-4 > 0?
1. m > v^2
2. m > t^-4


Please explain this question, thanks![/list][/spoiler][/list]
think the answer is d here

In the question ... you really need to prove that m is positive because v^2 and t^-4 are always positive

statement 1 : m > v^2 ..this means that m is positive since v^2 is positive
sufficient

statement 2 :m > t^-4 ..this means that m is postive since t^-4 is positive
sufficient

And BTW, m,v or t cannot be 0 because it has been stated that vmt ≠ 0
@Deb
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