DS exponents

[tonebeeze]

Can someone please walk me through this. Thanks


If vmt ≠ 0 , is v^2 m^3 t^(-4) > 0?

1. m > v^2

2. m >t^-4

I selected B

OA = D

[Tani]

This is all about positives and negatives. You need to recognize that a negative number raised to an even power will be positive and raised to an odd power will be negative.

the first statement tells us that m^3 is greater than v^2. Since v^2 must be positive, m must be positive. T^(-4) is equivalent to (1/T)^4. Therefore it must be positive as well. The product of three positive terms is positive. Sufficient

The second statement does the same thing, except that it starts by telling us that t^(-4) is positive. Since is greater than t^(-4) again it must be positive. Again we have the product of three positive terms. Sufficient