<p>A set has property X if t^2-t is in the set whenever t is in the set. Which of the following sets has property X?</p>
<p>(A) {-2, -1, 0}
(B) {-1, 0}
(C) {-1, 0, 1}
(D) {0,1}
(E) {1,2}</p>
<p>Is that actually t^(2-t)?</p>
<p>The question is asking, for which of the following sets does each value come out as its own answer when you put it into the expression?</p>
<p>Assuming that the expression is what I’ve written above, you can approach it like this.</p>
<p>It can’t be A. If you evaluate (-2) in the expression, (-2)^[2-(-2)] = (-2)^4 = 16. Thus, even though (-2) is in Set A, t^(2-t) (i.e., 16), is not. Set A then is lacking Property X: t is in the set, but t^(2-t) is not.</p>
<p>It’s Set D that works. 0 is in the set; t^(2-t) = (0)^(2-0) = 0^2 = 0. Similarly, 1 is in the set; t^(2-t)= (1)^(2-1) = 1^1 = 1. When t = 0 or 1, t^(2-t) is also in the set. That’s Property X; Set D has it.</p>
<p>“Is that actually t^(2-t)?”</p>
<p>No, it’s t^2-t</p>
<p>But the answer is D though…</p>
<p>Same principle: it can’t be A, because if t = (-2), then (-2)^2 - (-2) = 4 + 2 = 6, and 6 is not in set A.</p>
<p>It’s D because (0)^2 - 0 = 0 (0 is in Set D), and (1)^2 - (1) = 0 (and 0 is still in the Set). When I said in my previous post, “The question is asking, for which of the following sets does each value come out as its own answer when you put it into the expression?” that was just plain sloppy of me, and I apologize. It is not necessary that f(x) = x. It is only necessary for every x in the Set, that f(x) also be an element of the set.</p>
<p>Thank you, I now see what was this question about.</p>
<p>Oh, BTW, maybe you can tell me what mode means? (I’m an international and I don’t quite sure if my assumption is right)</p>
<p>In a set of data, the mode is the one that occurs most often.</p>
<p>Sent from my DROIDX using CC App</p>
<p>Yeah, I thought so 
Thanks again!</p>