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uncool

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Everything posted by uncool

  1. Another answer: The horse is at one end of the circle, and the hay at another end... -Uncool-
  2. The answer should be (98 49) - 98 - 99. If I'm wrong, please pots it. -Uncool-
  3. 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-
  4. Umm...isn't that triangle a 3-4-5 triangle? -Uncool-
  5. 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-
  6. 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-
  7. uncool

    Area of triangle

    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|
  8. 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-
  9. uncool

    Mathematics

    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-
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