uncool

Right, yeah, was thinking A*x^3+B*x^2+C*x+D. Well, you can let a calculator do it, or you can finish out the calculation. It takes less time than you might expect. -Uncool-

One way to do this: ratio test. Take the absolute value of the ratio of each term to the one before it. Take the limit of this as n goes to infinity. If the limit is less that 1, then the series converges. If it is greater, it diverges. If it hits exactly, then you have to figure out what happens. -Uncool-

Actually, I meant to assume each separately. E.g. assume y = x^r first. This will work because it is a linear equation. In this case: 3*r*(r-1)*(r-2)+9*r-1 = 0 3*r^3 - 9*r^2+15*r-1 = 0 Solve that, and then plug the three r's as a1, a2, and a3 into the equation. -Uncool-

First thing to assume is that y is a simple power function - because each term is in the form x^a*(ath derivative of y). Find all working a's. There should be 3 of them. Let the function be b1*x^a1 + b2*x^a2 + b3*x^a3, where b1, b2, and b3 are any constants. This should solve the equation. If there are two equal solutions (i.e. a1 = a2), then instead use b1*x^a1+b2*x^a1*ln(x) + b3*x^a3, etc. -Uncool-

Doesn't look right, should be 3*x |x| dy/du = 3u^2, not 2u^2 -Uncool-

Algebracus, 0^0 cannot be defined except for the limit which is being used. In this case, I believe that the limit successfully gets 0^0 = 0. -Uncool-

Now, to find the radius of the circumcircle, you can use this: a/sin A = b/sin B = c/sin C = 2R iirc where a, b, and c are the sides, and A, B, and C are the opposite angles of the corresponding sides, and R is the radius of the circumcircle -Uncool-

Would you accept proof through law of cosines? -Uncool-

Say that by the definition of a line, since the increase is proportional to the x move, then it must be a constant plus a number times x. I.e. give them the basics of mathematical induction If it works for some x-value a, then it must work for any x-value a + c. Proof: y(a) = ma + b y(a+c)-y(a) must be proportional to c. y(a+c)-y(a) = m*c for all c. y(a+c) = m*c + y(a) = m*c + m*a + b = m*(a+c) + b. Since only one formula can work for a function, if there is an a for which this works, this must be the formula. It works for x = 0, y = b, so it must work for any value -Uncool-

usamo - American competition, part of a 4-part competition. Starts with the AMC (American math competition) 10/12, goes to AIME (American invitational mathematics exam), goes to USAMO (USA mathematical olympiad) goes to IMO (International math olympiad). IMO is a math competition with 6 people from each country competing on teams. USAMO has 250 kids coming from various parts of the US. Site is: http://www.unl.edu/amc -Uncool-

Hey guys, I'm about to take the USAMO. what would you suggest for me to study? -Uncool-

That's not usually how they do 3 digit multiplication in their heads. I have done two-digit multiplication by using difference of square things, and it seems to work out well, usually. I expect that that could be extended to 3 digit multiplication. Divisible by 7: Take the last digit, remove it from the number. Multiply the digit by 2, and subtract it from the rest of the number. Repeat until it's one digit long. This should leave you with a number divisible by 7 if the original is divisible by 7. My own generalization: For any number a not divisible by 2 or 5, to determine if a number b is divisible by it: Take the lowest multiple c of a that ends in one. Then replace the 2 in the example for 7 with (c-1)/10, and 7 with a. For example: To determine if a number is divisible by 17, take the last digit and remove it from the number. Multiply the digit by 5 (from 51 = 3 * 17) and subtract that product from the rest of the original number. Repeat until it is down to a small number. if the original is divisible by 17, then this last number, and all intermediate numbers will be divisible by 17. Ask me for proof of this later if you want. Example: 65569 = 17*3857 6556 9 6556 45 6511 = 17*383 651 1 651 5 646 = 17*38 64 6 64 30 34 = 17*2 Oh, and 99...9 (a 9's)^2 = 999... (a-1 9's)8000...(a-1 0's)1 -Uncool-

I would have to disagree with this one...we would then evolve from that. A non-aging species would still evolve to get past other threats to life. -Uncool-

Another answer: The horse is at one end of the circle, and the hay at another end... -Uncool-

The answer should be (98 49) - 98 - 99. If I'm wrong, please pots it. -Uncool-

First thing: get rid of A. divide the equation by A. Second thing: get rid of B. Do this by a constant shift - y = x - B/3. Then, let y = z - p/3z where p is the y-coefficient. This will leave you with an equation with z^3, constand, and 1/z^3 terms. Multiply by z^3. Let u = z^3. Solve for u by the quadratic formula. either solution works - they reveal the same final answer. Solve for z. Substitute z into the equation for y. Find x. Done. -Uncool-

Umm...isn't that triangle a 3-4-5 triangle? -Uncool-

The formula is supposedly the McClauren (SP?) series for sine - that is x - x^3/3! + x^5/5! -... + (-1)^k*x^(2k+1)/(2k+1)! -Uncool-

Just look for basic parts, like divisibility by 2, 3, 5, and 11. Another thing to know: if a number is not divisible by any of the primes up to its square root, then it itself is prime. -Uncool-

A = (a, d) B = (b, e) C = (c, f) A(triangle) = length of cevian * width of triangle with respect to cevian - that is, the length of the projection of the opposite side onto the perpendicular to the cevian BC: y = ((f-e)/(c-b))(x-b)+e = ((f-e)/(c-b))x - b((f-e)/(c-b)) + e = ((f-e)/(c-b))x - ((bf - be)/(c-b)) + e = ((f-e)/(c-b))x - ((bf - be + be - ec)/(c-b)) Length of vertical cevian: d - ((f-e)/(c-b))a + ((bf - ec)/(c-b)) Width of triangle = c - b Area = d(c-b) - a(f-e) + bf - ec = d(c-b) + e(a-c) + f(b-a) = |a d 1| |b e 1| |c f 1|

I'm not sure about the sum part, but the others come from taking differences. Here iswhy it works: ((x+1)^n-x^n) = nx^(n-1)+n*(n-1)*x^(n-2)/2 + ... So the coefficient of the first term will be n. You then subtract again, because you eventually want to get a constant number times the first term without the x value. This leaves you with n*(n-1). Then you subtract again, leaving you with n*(n-1)*(n-2) . . . Then you subtract again, leaving you with n*(n-1)*(n-2)*....*3*2*1 = n! All the other coefficients cancel out. Hope this helps. -Uncool-

I assume you mean that lim (x->a) [xf(x) = aL] If lim(x->a) [f(x) = L] There is a neighborhood around a where f(x) is continuous Therefore, there is a neighborhood around a where x f(x) is continuous In the neighborhood, the function returns x f(x) Since both approach finite values, the limit of the product is the product of the limits. Therefore, the limit is aL. -Uncool-