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