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[2014-10-16. : 12:50 am]  Azrael -- All possibilities aren't infinite anyways. Just arbitrarily large.[2014-10-16. : 12:49 am] jjf28 -- Mini Moose 2707Mini Moose 2707 shouted: jjf28 If there are an infinite number of parallel universes spanning all possibilities, is there a parallel universe where parallel universes don't exist? no, as that would require there being a parallel universe for that universe to be parallel to; in addition the format of your question implies that parallel universes have a containment relationship, when that defies their very nature[2014-10-16. : 12:46 am]  Moose -- jjf28jjf28 shouted: you forgot to say something clever moose If there are an infinite number of parallel universes spanning all possibilities, is there a parallel universe where parallel universes don't exist?[2014-10-16. : 12:44 am] Azrael -- " What's the second derivative test? I've done all this before, but it was two years ago and I had no interest in it and have forgotten a lot."[2014-10-16. : 12:44 am] jjf28 -- like "you could always call in a bomb threat, but I think you get sent to GBay nowadays... how do you feel about cockmeat sandwiches?"[2014-10-16. : 12:44 am] Azrael -- JackJack shouted: demon man didn't you go to school http://webcache.googleusercontent.com/search?q=cache:aolLEw3Y7C0J:www.staredit.net/%3Fp%3Dshoutbox%26start_shout%3D1039234+&cd=1&hl=en&ct=clnk[2014-10-16. : 12:33 am] Moose -- Do you know how much more I could be doing with six? Okay, not much I'm not doing already actually, lel.[2014-10-16. : 12:21 am] Roy -- It's interesting that the 5820K is a 6-core processor, though; the previous generation (4820K) was only 4 cores.[2014-10-16. : 12:03 am] Roy -- It's kinda like how the GTX 980 is cheaper than a GTX 680, even though, well... http://gpuboss.com/gpus/GeForce-GTX-980-vs-GeForce-GTX-680[2014-10-16. : 12:02 am] Roy -- Vrael Old tech vs new tech: Sandy Bridge chips are really expensive. The 4790K is a Haswell Refresh chip, and is cheaper than the 5820K, as it should be.[2014-10-15. : 11:42 pm] Moose -- Oh, I think I see it. The absolute value bounds the equation from below by 0, that's where x^2 - 4 >= 0 comes from.[2014-10-15. : 11:39 pm] Moose -- Because it changes signs from left to right, from + to - (reversed by the absolute value) to + again[2014-10-15. : 11:39 pm] trgk -- else, |a| computes to -a, so |a| < b means -a < b, which is equivilant to -b < a[2014-10-15. : 11:38 pm] Moose -- Uh, you would have to know how the parabola works, that x^2 - 4 is a parabola with zeroes at -2 and 2[2014-10-15. : 11:38 pm] trgk -- http://upload.wikimedia.org/math/8/c/b/8cb53c58631c94cc12e446129810669d.png[2014-10-15. : 11:37 pm] trgk -- If b>0, then (a >= 0 and a < b) or (a < 0 and a > -b) can be simplified to (-b < a < b)[2014-10-15. : 11:36 pm] Moose -- x^2 - 4 < 5 is describing the parts from -3 to -2 and 2 to 3, where the parabola isn't reversed and opens upwards[2014-10-15. : 11:35 pm] Moose -- x^2 - 4 >= 0 is describing that upwards parabolic curve from -2 to 2[2014-10-15. : 11:32 pm] Moose -- Since you only have x^2 in the function, any solution x > 0 has a mirror solution -x[2014-10-15. : 11:32 pm] Moose -- Oh, that's using symmetry of the graph to make life easier. Did you graph the inequality?[2014-10-15. : 11:31 pm]  Dem0n -- And then we solve those four inequalities and find their intersection[2014-10-15. : 11:31 pm] Dem0n -- But the way we did it in class is we had two cases: one was where x^2 - 4 >= 0 and x^2 - 4 < 5, and the other was the opposite of that[2014-10-15. : 11:27 pm] Dem0n -- Yeah I know how to do it that way, but I have to use stupid cases to solve them, and I don't get that[2014-10-15. : 11:26 pm] Moose -- You would just need to solve (x+2)(x-2) < 5 and -5 < (x+2)(x-2) and intersect the solutions of those.[2014-10-15. : 11:24 pm] Dem0n -- Like |x2-4| < 5 has two cases, one being x2-4 >= 0 and x[sup]2-4 < 5, and the other being the opposite of the first one[2014-10-15. : 11:23 pm] Moose -- Tell your TA that he is a scrub and some moose on the internet rekt him.[2014-10-15. : 11:21 pm] Dem0n -- My TA was showing us how to solve them with cases and I had no clue what he was doing[2014-10-15. : 11:21 pm] Moose -- ex |x-3| < 5 becomes -5 < x-3 and x-3 < 5 and combine the results: intersect if the original equality was < or <=, union if the original inequality was > or >= |
