finding inverse equation, please.?
920396
Now that you have the variables switched, you're supposed to solve for 'y' (where you have an equation that reads y=...), and that should give you the inverse of the original equation
768593
To find the inverse of this relation, reverse the coordinates of the ordered pairs.\n\nThe answer to this would be\n {(18, -14), (-18, 3), (-1, -2)}
874355
Using matices to do this\n\n[ 1 -2 1] * [x] = [-5]\n[ 2 -1 1]   [y]    [ 1]\n[2  1  2]   [z]    [10]\n\nAX = B\nA*A^(-1)X = A^(-1)B\nX = A^(-1)B\n\nSO we need the inverse of A this can be worked out using a long and tedious method or just use a calculator\n\nA^-1 = [ -.6 1 -.2]\n           [ -.4 0   2]   \n           [.8   -1 .6]\n\nA^-1B = [2}\n              [4}\n              [1]\nso x =2  y=4 z=1\n\n now you do the other one\nCheck\n2 - (4*2) +1 = 2 - 8 +1 = -5
822532
This is an interesting problem, because we are solving for x instead of y (given the equation).  First divide by dy, thus you get:\n\ndx/dy = me^(-ky) - x \n\nWe cannot use separation of variables for this O.D.E. because it is not of the form dx/dy = g(y)h(x).  Thus, the easiest way is to utilize an integrating factor.  Rewrite the equation as:\n\ndx/dy + x = me^(-ky)\n\nNow multiply through by e^y:\n\ne^y(dx/dy) + xe^y = me^(-ky)e^y = me^(-2ky)\n\nWhen utilizing an integrating factor, you try to make one side look like the product rule of the derivative.  By inspection, you can see that the LHS of this equation is:\n\nd[xe^y]/dy\n\nWe can double-check this:\n\nd[xe^y]/dy = e^y(dx/dy) + x(d[e^y]/dy) = e^y(dx/dy) + xe^y\n\nTherefore, our equation simplifies to:\n\nd[xe^y]/dy = me^(-2ky).  Now taking the integral of both sides:\n\nxe^y = [-1/(2k)]me^(-2ky) + C\n\nDividin by e^y and get your solution:\n\nx = (-m/[2k])e^(-3ky) + C\n\nIf you are looking for y, just take the inverse of this equation (interchange x and y and solve for x).\n\n-----------\n\nHope this helps
458222
0 1 0 0\n0 0 1 0\n1 0 0 0\n0 0 0 1\nPut the identity matrix behind your matrix.\nTurn you matrix into the identity matrix using row operations.  The matrix in the back will be the inverse.\n0 1 0 0  1 0 0 0 \n0 0 1 0  0 1 0 0 \n1 0 0 0  0 0 1 0 \n0 0 0 1  0 0 0 1\nSwap row 1 with row 3\n1 0 0 0  0 0 1 0 \n0 0 1 0  0 1 0 0 \n0 1 0 0  1 0 0 0 \n0 0 0 1  0 0 0 1\nSwap row 2 with row 3\n1 0 0 0  0 0 1 0 \n0 1 0 0  1 0 0 0\n0 0 1 0  0 1 0 0 \n0 0 0 1  0 0 0 1\nThus, the inverse is\n0 0 1 0\n1 0 0 0\n0 1 0 0 \n0 0 0 1
646401
it's the inverse x^-1=1/x
273264
Lets call this matrix A.\n[ 2 0 4 ]\n[ 3 1 5 ]\n[ -1 1 -2 ] = A\n\nNow we need to solve the equation AV = I\nthen V is the inverse matrix.  Lets write this out\n\n[ 2 0 4 ] ` ` ` ` [ 1 0 0 ]\n[ 3 1 5 ] ` ` ` ` [ 0 1 0 ]\n[ -1 1 -2 ] V = [ 0 0 1 ]\n\nNow just multiply elementary matrices (row operations) to the left hand side of both sides of the equation until the left-hand entity is the identity.  Multiply the following matrix on the left-hand-side both sides [[1/2 0 0][0 1 0][0 0 -1]], that is replace the first row with 1/2 times the first row, and replace the third row with -1 times the third row:\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 3 1 5 ] ` ` ` ` [ 0 ` 1 0 ]\n[ 1 -1 2 ] V = [ 0 ` 0 -1 ]\n\napply [[1 0 0][[-3 1 0][0 0 1]] both sides, that is replace second row with second row plus negative-three times first row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 -1 ] ` ` ` ` [ -3/2 1 0 ]\n[ 1 -1 2 ] V = [ 0 ` 0 -1 ]\n\nmultiply both sides on the left by the matrix [[1 0 0][0 1 0][ -1 0 1]], that replace third row with third row plus negative one times first row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 -1 ] ` ` ` ` [ -3/2 ` 1 0 ]\n[ 0 -1 0 ] V = [ -1/2 ` 0 -1 ]\n\nmultiply by [[1 0 0][0 0 -1][0 1 0]], that is place opposite of row 3 for row 2 and row 2 for row 3\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 1 -1 ] V = [ -3/2 1 0 ]\n\nnow apply [[1 0 0][0 1 0][ 0 -1 1]] both sides, that is replace third row with third row plus -1 times second row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 0 -1 ] V = [ -2 ` 1 -1 ]\n\napply [[ 1 0 2][0 1 0][0 0 -1]] both sides, replace row 1 with row 1 plus 2 times row 3, and replace row 3 with negative 1 times row 3\n\n[ 1 0 0 ] ` ` ` ` [ -3/2 2 -2 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 0 1 ] V = [ 2 -1 1 ]\n\nNow, since the left hand item is the identity, the inverse matrix is\n\n[ -3/2 2 -2 ]\n[ 1/2 0 1 ]\n[ 2 ` -1 1 ]\n\nThe product of all of the elementary operation matrices specified is also this matrix, and the reader can verity that AV = VA = I
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
699429
c'mon dude, these are easy.\n\nAdd second to first.\n\n2x = 2, x = 1, therefore y = -2.
886265
gotta put the equation into y=mx+b first\ny= 1/3x - 1\n\nperpendicular would be y = -3x (+ or - any #)
195858
Here is an easy way. Say you want to find the inverse of the 2x2 matrix\n\n4 1\n1 2\n\nTo type this matrix into a TI-89, you can type in a list of lists using the square brackets, like this:\n\n[[4,1],[1,2]]\n\nYou can also use semi colons to separate the rows instead of using sublists, which can be a bit easier to read, like this:\n\n[4,1;1,2]\n\nNow to find the inverse, just type ^(-1) onto the end of the matrix you typed in (and make sure you use the unary minus key, which is the gray (-) key on the number pad, not the black subtraction key which is on the right side of the calculator)\n\nAll put together, to find the inverse of the matrix above, you can type:\n\n[4,1;1,2]^(-1)\n\nand then push enter of course.\n\nAs another example, to find the inverse of the 3x3 matrix\n\n1 2 3\n5 4 8\n9 6 7\n\ntype in:\n\n[1,2,3;5,4,8;9,6,7]^(-1)
830737
I do not understand what it is you are trying to solve. Are yoiu trying to find the inverse fuction? If you are that will be extremely difficult. If you seek the derivative then it has been doone above for you.\n\nIf you seek the integral then it would probably requiire a power series But as it is your problem is not a problem as you have not stated what you want solved.\n\ny = x^x\n= e^xlnx\n\ndy/dx = (lnx + 1)x*x\n= (lnx +1)*y\nSo lnx = 1/y *dy/dx - 1\nie x = 1/e * e^(1/lny * dy/dx)\nWhence x = 1/e * e^ [1/(lny* dx/dy)]\n\nNow I find that unconvincing as to get dx/dy I need to know x in terms of y Its looped onto itself!!!! In fact it is a differential equation not a function. ( x = 1/e * e^ [1/(x' . lny)]
744474
x=2y\n\n-2 -4\n0 0\n1 2\n3 6
21539
If you have a system of equations with two variables of the form ax + by = c, dx + ey = f, where a,b,c,d,e,f are the coefficients, and x and y are variables, you can set up the equation as \n\n[a b][x] = [c]\n[d e][y] = [f]\n\n(ignore the fact there are two equals signs there, they're only there for formatting reasons)\n\nwhere the square brackets are a crude attempt on by behalf to 'draw' matrices and column vectors.\n\nTo solve the system you need to find the inverse of the matrix of the right hand side (assuming it's invertible to begin with), and "right multiply" (that is multiply from the right of each side of the equation) both sides of the equation.\n\nThat should give you:\n\n[1 0][x] = [ inverse of matrix ][c]\n[0 1][y] = [ ........................ ][f] \n\nThen it's just a matter of performing the calculation.\n\n(sorry for the crappy formatting of the post)
916864
Ord(e^(4x)-1 -4x) ~ Ord(e^(4x) > Ord(x^2)\nThen:\n\nlim(e^(4x)-1 -4x)/x^2 {x--> oo} = oo\n\nRemember\n\nln x < x^a < e^x < x^x
978722
You can't, algebraically.  You have two equations and three variables.  Sorry.  You need one more equation, such as a = b.
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.