uncool
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Everything posted by uncool
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Azure, yes. -Uncool-
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Yes, coquina. -Uncool-
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Coquina, no Benson, irrelevant. -Uncool-
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Azure, no Calli, yes -Uncool-
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Irrelevant, dak -Uncool-
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I do trust you, Phi, I was just saying that from the question, it sounded like you knew the riddle in another form, and that the question was your way of recognizing it. Dak: no. -Uncool-
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Dak: yes, but it was simply the normal waiter conversation - that is, pretty much irrelevant to this story. Phi: I think you know the riddle... -Uncool-
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Azure: no Dak: define significant/relevant (sorry, but I will probably ask this question a lot...) Phi: no -Uncool-
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No. I can see this getting really long, really fast... -Uncool-
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Yes...you guessed that quickly... -Uncool-
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Yes, boat, not ship. -Uncool-
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Two emaciated men jump out of a ship onto a beach. One simply walks away. The other goes into a nearby restaurant and orders seagull. He takes one bite, and then commits suicide. Why? You may ask yes or no questions, I may answer yes, no, irrelevant, or pretty much whatever I please . If you have heard this before, please, don't answer for others. -Uncool-
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Who was the greatest scientist or inventor that ever lived ?
uncool replied to vrus's topic in Other Sciences
I would say whatever guy discovered fire, and discovered how to get it repeatedly. I mean, that's what really sparked the advance of the human race...either that or the wheel. -Uncool- -
Mathematical induction is a way of saying "If it works for some integer, and we know that if it works for one integer, it works for the integer after it, then it works for all integers." Here, mathematical induction can be used in the following manner: Let f(n) = 1 + 2 + 3 + ... + n I say that f(n) = n*(n+1)/2 It works for 1, because f(1) = 1 = 1*2/2. f(n+1) = 1 + 2 + 3 + ... + n + n+1 = f(n) + n So if f(n) = n*(n+1)/2, then f(n+1) = n*(n+1)/2+n+1 = (n+1)*(n/2 + 1) = (n+1)*(n+2)/2 = (n+1)*(n+1+1)/2. Therefore, the formula is correct for all numbers past 1. -Uncool-
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sin(FBE)/sin(FEB) = EF/BF sin(FCD)/sin(FDC) = DF/CF sin(FBC)/sin(FCB) = CF/BF sin(FED)/sin(FDE) = DF/EF So: sin(FED)/sin(FDE) = DF/EF = (CF*sin(FCD)/sin(FDC))/(BF*sin(FBE)/sin(FEB)) = (sin(FEB)*sin(FCD)*sin(FBC))/(sin(FBE)*sin(FDC)*sin(FCB)) = (sin(40)*sin(20)*sin(50))/(sin(30)*sin(50)*sin(60)) FED = 110-x sin(110-x)/sin(x) = (sin(40)*sin(20)*sin(50))/(sin(30)*sin(50)*sin(60)) sin(110)*tan(x) - cos(110) = (sin(40)*sin(20)*sin(50))/(sin(30)*sin(50)*sin(60)) tan(x) = ((sin(40)*sin(20)*sin(50))/(sin(30)*sin(50)*sin(60)) + cos(110))/sin(110) Find x. -Uncool-
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difference between brains and computers
uncool replied to gib65's topic in Anatomy, Physiology and Neuroscience
invert nexus, that is exactly what I meant with the Chinese room. I was just saying an extremely inefficient way in which human brain behavior could be modeled - but much less efficient than most ways. Now, to read the paper against determinism...thanks for the link. -Uncool- -
Jordan, remember, sine and cosine go from -1 to 1. And a sine can be made into a cosine, with a shift of 90 degrees. Therefore, if you can use one, you can use the other. As for the formula: y = amplitude (170) * sine (or cosine) (2*pi*frequency*t+someshift) + average -Uncool-
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difference between brains and computers
uncool replied to gib65's topic in Anatomy, Physiology and Neuroscience
DNA, I believe that, should we be able to find all the reactions in the brain/body, it would be a simple (though extremely tedious) task to replicate a human. All chemical reactions can be modeled by switches of a type or computers, I believe, and therefore, I believe that eventually, computers will be able to create "emotions" of a type. Our brain has its own source code that tells it to create new code - and that has been done with computers, just not quite to the same extent. Sorry for rambling here! -Uncool- -
erm...what's the BT paradox? -Uncool-
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Just that it makes the 360 neither too high, nor too low, Edward. -Uncool-
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Hmm, can't see the LATEX stuff - it's giving me a "couldn't open URL" error. -Uncool- So to reiterate: To find the first solution: Take out the x^2 term by substituting y - A/3 for x. Let this be y^3 + p*y+q = 0 Let y = z - p/3z (I think) This should form an equation of the form: z^3 + q + r/z^3 = 0 z^6 + qz^3 + r = 0 Find z^3, take the cube root, and substitute back into y, then into x. That should be the first solution, and, as you showed, finding the other two is easy. -Uncool-
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Edward, 360 is the first number with 24 factors - that's what I'm saying. -Uncool-
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360 is a "highly divisible number" - that is, it is the first number with at least as many factors as it has. 360 has 24 factors - 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360. -Uncool-
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Correct, I was showing how to find the first root of the equation. -Uncool-