Statement 1 tells us x^3 < x^2. Since x is an odd integer, we know x is not 0, and thus x^2 is always positive. Thus, we can divide both sides by x^2 without flipping the inequality. Dividing both sides by x^2 gives us x<1. Since x is an odd integer, it must be a negative odd integer.

Let's say x=-1. Then we have to find out the inequality in -1^-1 ? -1^1. In this case, both sides equal -1, so it is not true that x^x > x^-x.

Now lets say x=-3. Then we have -3^-3 > -3^3. In this case, it is true that x^x > x^-x.

Thus, statement 1 is not sufficient to determine whether x^x > x^-x.

Statement 2 tells us xy is not equal to – y. In other words, x does not equal -1, but it can equal any other odd integer.

Let's say x=1. Then we have to find out the inequality in 1^1 ? 1^-1. In this case, both sides equal 1, so it is not true that x^x > x^-x.

Now lets say x=-3. Then we have -3^-3 > -3^3. In this case, it is true that x^x > x^-x.

Thus, statement 2 is not sufficient to determine whether x^x > x^-x.

However, putting the statements together, we know that x is a negative, odd integer that is not equal to -1. We saw what happened when we plugged -3 into the formula. x^x > x^-x because x^x is between -1 and 0, and x^-x is something much less than -1. We can see this will be the case for any negative integer smaller than -3 as well.

Thus, both statements together tell us that x^x > x^-x, and we have sufficient info. The answer is C.

Thanks for the explanation VP_Tatiana..That helped a lot.

Hi, sorry for the newbie question, what does the notation ^ mean?

Thank you!