uncool

Each of them mean that there are two cards that are different. Let us renumber and rearrange the cards so that the first n-2 cards are in order, and the last two cards are switched in one deck. We then must figure out how many ways the cards can be renumbered and rearranged. There are n!/2 ways of renumbering the cards (n! because there are n cards, /2 because switching the order of the two switched cards is not allowed). There are n(n-1) ways of placing the cards (n-1 for the first switched card to be placed, n for the second). So the total number of ways is n*(n-1)*n!/2. For the first, it is 4*3*24/2 = 144 For the second, it is 9*8*362880/2 possibilities. Divide by n!^2. 9*8/2*9! = 1/2*7! = 1/10080 Someone check if I'm right? -Uncool-

According to the general theory of relativity, light has constand speed. According to quantum physics, velocity cannot be known exactly. According to recent research, the speed of light has been decreasing (but only slightly). -Uncool-

If you notice, the whole thing is not a triangle. You can check the slopes of the two triangle hypoteni: One of them is 3/8 = 0.375 The other is 2/5 = 0.4 The missing area in the second one is compensated for on top, but you can't see it because the slopes are so close. -Uncool-

Redshift being equal in all directions only points to it being a sphere, not the surface of a hypersphere. It expanding equally in all directions for all points is just like a circle expanding and all the points on it expanding equally. Just look at it. You'll see what I mean -Uncool-

This whole thread is extremely funny to me because I learned it PEMDAS Parentheses Exponents Multiplication/Division Addition/Subtraction -Uncool-

7,24,25 8,15,17 Etc.

Just multiply both sides by C on the left side. Only one x can be the solution to Ix = Cb. -Uncool-

Hey guys...I posted the answer. anyone wanna check it? -Uncool-

I'm taking this as what happens over time, and I'm assuming an ideal water. In those cases, the water will occupy the same level in the end. However, should the water be less than ideal (which we can assume it is), then, because of capillary action, it will rise more in the 15mm pipe than the 40mm pipe. -Uncool-

What I mean is that one stays right here, while the other one goes off in a spaceship. At some time when the two are together (for example, let's say the first goes around once), they synchronize their watches. The one in the spaceship zooms off into the distance, and eventually comes right back around. Therefore, there will have been no acceleration, and the standard objection cannot be used. As the spaceship twin's time is slower (to me), his time should read less when he comes back around. However, it would appear the same way to him, and therefore it would seem to him as if mine should read less. As no acceleration occurred between the synchronization and the re-reading, each view seems to be correct. How is this possible? -Uncool-

Martinez: A perfect square is a way in number theory to denote any number that is equal to some integer times itself. For example, 4 is a perfect square because it is equal to 2*2. The impossibility of a perfect circle and a perfect square in reality is not because of mathematics, but because of physics and things like atoms and quantum physics. Algebracus, notice that the ending of 51^2 = 2601, 52^2 = 2704, etc. My proof: Let there b c 0's. The square root must end in 0000...1, 2, or 3 (otherwise, the ending of the square could not be 000...0b). The square would therefore have 3 parts: the square of the part before the 0, the geometric mean of the two parts, and the part after the 0's. Therefore, you can't have only 0's in the middle. For example: 1000001^2 = 1000002000001 - the 3 parts are the 1 at the beginning, the 2 in the middle, and the 1 at the end. -Uncool-

That is almost what I had been thinking. However, I want to do away with the acceleration, just leave the velocity. For example: he accelerates from earth, goes around once, and then once he sees the earth again, he synchronizes the watches. He then passes the earth, and goes in another orbit. However, I just thought of something else. Since they would be the surface of a hypersphere, the two would automatically be accelerating when compared to the other - in another dimension. -Uncool-

Well, first, decide what is meant by the "measures used at Guantanamo Bay" Second, decide what is easiest to defend, and what you can throw away. Third, find ways to separate the two groups. Fourth, find ways to defend the first group while condemning the second. -Uncool-

No force is required for you to move through time at a constant rate. Remember, moving through space and moving through time are directly related - so accelerating in space will deccelerate your movement in time. -Uncool-

A small question I thought of recently: If the universe is the surface of a hypersphere, then what happens if an item goes 'around' the hypersphere and meets back up with the earth? Who will have aged more and why? (Assuming constant velocity and all that) Both will think that the other was moving and therefore slowed down, and each one would be equally correct. -Uncool-

Just write it out from what hte problem says. You know the foci. You know the difference between the distance to those foci. So try to figure out: what equation will I need to do this? -Uncool-

Not quite, demosthenes. It is possible that there was an infinite regress of life (alien scientists) or that there was an immaterial creator. Not saying that I believe this, but it is possible. -Uncool-

The two things must come together again for the two to be comparable, andthat is only possible (assuming a flat universe)with acceleration. it is the acceleration, NOT THE SPEED, which causes the difference. Should the earth come back to the other thing, then the earth would be the younger one. -uncool-

RNA is able to synthesize itself, so it does not need the ribosomes. Also, there is RNA in prokaryotic organisms, which don't have ribosomes, so the RNA must have come first. -Uncool-

Callipygous, it is possible. [hide]1 quarter = 25 cents 2 dimes = 20 cents 2 nickels = 10 cents 45 pennies = 45 cents 2 quarters, however, are not possible to do. 0 quarters is possible. 2 dimes = 20 cents 8 nickels = 40 cents 40 pennies = 40 cents [/hide] Therefore, this question has no definite answer. -Uncool-

f''(x) jumps from -2 to 2 at 0. d is therefore anwered incorrectly. The answer is that the second derivative does necessarily vanish. -Uncool-

However, what I showed can be used to find the sums - as each of the Pascal's triangle parts is a polynomial. So, for example, to find the sum of squares: Pascal(x,2) = x^2/2-x/2 Pascal(x+1,3) = x(x+1)(x-1)/6 = x^2/2-x/2+(x-1)^2/2-(x-1)/2+... x(x-1)(x+1)/3 = x^2+(x-1)^2+... - x - (x-1) - ... x(x-1)(x+1)/3+x(x+1)/2 = x^2+(x-1)^2+... x(x+1)(2x+1)/6 = x^2+(x-1)^2+... -Uncool-

g(x) = f(x+2) + 5 = -5(x+2)/2 + 9/2 = -5x/2 -5 + 9/2 + 5 = -5x/2 + 9/2 It is because you are moving the line in the same direction as the line is going - the slope is -5/2, or 5 units up over -2 units to the left - which is what you said. However, (1,2) goes to (3,7). -Uncool-

Personally, I find that the easiest way to remember is that the sum along a diagonal of Pascal's triangle is equal to the number in the next diagonal down from the last number in the sum. Pascal's triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 + 2 + 3 + 4 + 5 = 15 (so go down from the 5, and you have your sum) 1 + 4 + 10 = 15 ... Now, remember the formula for each part in Pascal's triangle: Let r = the row number, and c = the number of columns over from either 1 (where 0 means you're using the 1) Pascal(r,c) = r!/(c!(r-c)!) So for the third column over, Pascal(r,2) = r*(r-1)/2 For the fourth, Pascal(r,3) = r*(r-1)*(r-2)/3*2 And so on. -Uncool-

Err, chatha, "someone had to create that first bacteria anyway?" Can you back that up? There have been many experiments pointing to the possibility of naturalistic abiogenesis... -Uncool-