gauss-jordan elimination?
468863
Here's an article about it:\nhttp://en.wikipedia.org/wiki/Gaussian_elimination
1166133
Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
920734
2xy(-15x^2y^4)+8x^2)-5/-4xy^3\n(-1/2)[-15x^2y^4+8x^2-5)/y^2\n(-1/2y^2)[-15x^2y^4+8x^2-5]
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
627220
I don't grasp this sphere scenario but \n\n-----\n\nLet n be this normal vector. Choose any vector v with\n\nv ≠ 0 and v, n not collinear\n\ncompute\n\na = v x n and b = a x n\n\nthen {n,a,b} is a orthogonal basis. If necessary normalize a, b to get an orthonormal basis.\n\n-----\n\nAnother way: Take the standard basis {e1,e2,e3}, replace one base vector with n (avoid coplanarity) and apply the Gram–Schmidt process:\n\nhttp://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process\n\n-----\n\nremarks: Both methods leave one degree of freedom (rotation around n). In each case you have to choose. That's quite annoying. If you find an algorithm without such a decision please tell me.\n\n-----\n\ncross product: a x b
492690
S(I)+3f (g)--> SF (g)\n..........2...........6
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
906026
Well dear \n\n▪gof(x) = g(f(x)) = g ( √x ) = (√x) - 1 = x - 1\n▪Dgof = { x I x Є  Df , f(x)  Є  Dg }\n▪Df = [ 0 , + ∞ )\n▪Dg = IR\n▪{ x I x Є  [ 0 , + ∞ ) , x Є  IR = [ 0 , + ∞ ) \n\nGood Luck\n{am sure its not your homework , but try to find why the answer would be that }
1075376
wow..that site http://www.algebra.com/services/renderin... is really amazing. You can also check out mupad. It will do simplifications and give answers to integration and much more. There's a demo available at http://www.sciface.com/
818852
First, we perform a substitution:\nu=5e^x+2, du=5e^x dx, dx=du/(u-2)\nSo our integral becomes:\n∫1/(u(u-2)) du\nNow this is the perfect form for partial fraction decomposition. We solve:\n\nA/u + B/(u-2) = 1/(u(u-2))\nA(u-2) + Bu = 1\n(A+B)u - 2A = 1\n-2A = 1, A+B=0\nA=-1/2, B=1/2\n\nSo our integral becomes:\n\n∫-1/2/u + 1/2/(u-2) du\n1/2∫1/(u-2) - 1/u du\n1/2(ln (u-2) - ln u) + C\n\nSubstituting our original value for u:\n\n1/2(ln (5e^x) - ln (5e^x + 2)) + C\n\nAnd simplifying:\n\nln (5e^x/(5e^x+2))/2 + C
102340
I assume you mean invertible, not convertible.\n\nIt depends on how detailed you want to get. If you do not want to dig in extremely deeply, you can just do it by construction. \n\n1. A symmetric matrix M has an orthogonal decomposition\n\nM=O'DO\n\nwhere O is an orthogonal matrix, O'O=I, the identity, and D is a diagonal matrix with eigenvalues of M down the diagonal.\n\n2. If M is positive definite, then the diagonal elements of D are >0\n\n3. Produce the matrix \n\nA=O'(D^(-1))O\n\n\n4. Verify that A is the inverse of M\n\ni.e., \n\nA*M=M*A=I
194771
1. Write down a couple of Jordan blocks.\n\n2. Write down their squares, cubes, maybe even a fourth power.\n\n3. Gaze at each set of identity matrix, J, J^2, J^3, ... until you see a simple linear combination of the matrices that sums to the all zero matrix.  That is your minimal polynomial.\n\n4. Generalize.\n\n5. Now it is obvious. :-)\n\nAlternative:\n\n1.  Google for some combination of:\n\nminimal polynomial Jordan block\n"minimal polynomial" "Jordan block"\nJordan block minimal polynomial\n"Jordan block" "minimal polynomial"\n\n2. Read answer.\n\n3. Too bad, it still isn't obvious. :-(
725218
Modified Gram-Schmidt algorithm\n------------------- -----------------------\nA  is m x n,  with m ≥ n  s.t. rank(A) = n\n\n(a) For i = 1, ... , n \n`````````w_i = a_i\nEnd (a) For\n(b) For i = 1, ... , n\n`````````r_ii = ||w_i||\n`````````q_i = (w_i)/(r_ii)\n`````````(c) For j = i + 1, ... , n\n``````````````````r_ij = <q_i, w_j>\n``````````````````w_j = w_j - (r_ij)(q_i)\n`````````End (c) For\nEnd (b) For
77120
It doesn't appear to simply significantly from the original expression.  I was able to get it to the following:\n\n(tan^3-tan)*sec\n\nDo the multipling by csc-1 top-n-bottom and then use the fact that cot^2=csc^2-1.  Bring the cot^2=cos^2/sin^2 to the top to get the above.\nGood luck.
72362
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
584514
Use 'Gaussian elimination' to obtain a basis for the span of matrix columns or rows.\n\nhttp://en.wikipedia.org/wiki/Gaussian_elimination
1108682
U•V = |U||V|cosα  Where |U| is the norm of U and α is the angle between the vectors.\n\nIt also boils down to:\nU•V = x1y1+x2y2\nU•V= (-4)(1) + (6)(5)\nU•V = 26