what equation would i use to model:z varies jointly as x and y and inversely as w?
81607
z = kxy/w where k is the constant of proportionality
1178662
a (x+y) + b (x*y) = z
152046
y=|x+2| \n this shifts the "v" shaped graph 2 units to the left from the origin.
776321
the question is unclear.\nDo you want an explicit formula for Y[n or do you just want to write out the convolution]
92874
FOR THE MOST PART EQUATIONS WITH VARIABLES LOOK LIKE THIS F(IV)=DV FOR EXAMPLE Y=2X. X IS INDEPENDENT (OR IV IN MY EQ TO THE RIGHT) AND Y IS THE DEPENDENT OR DV THE INDEPENDENT VARIABLE IS THE ONE THAT YOU CONTROL
714493
This *is* the way to write it so that x is the input and w is the output. If you want to solve for x in terms of w (i.e. have w the input and x the output), you are better off doing numerical approximations than using the cubic formula (it is really, really nasty).
952689
5x+3y+z+5w=-2\n3x+2y+2z+4w=2\n2x+4y+3z-3w=-11\n4x-3y-2z+2w=-3\n\n=>\n5     3     1     5   I  -2\n3     2     2     4   I   2\n2     4     3    -3   I  -11 \n4    -3    -2     2   I  -3\n\nR4 - (4/5)*R1 , R3 - (2/5)*R1 and R2 - (3/5)*R1 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    2.8000    2.6000   -5.0000  -10.2000\n         0   -5.4000   -2.8000   -2.0000   -1.4000\n\nR4 - (-5.4/0.2)*R2 , R3 - (-5.2/0.2)*R2 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000  -17.0000  -19.0000  -55.0000\n         0    0.0000   35.0000   25.0000   85.0000\n\nR4 - (-37/17)R3 will give\n\n5.0000   3.0000    1.0000    5.0000   -2.0000\n         0   0.2000    1.4000    1.0000    3.2000\n         0   0.0000  -17.0000  -19.0000  -55.0000\n         0   0.0000    0.0000  -14.1176  -28.2353\n\nR4 / -14.1176 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000  -17.0000  -19.0000  -55.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR3 - (-19)R4 will give\n\n    5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0   -0.0000  -17.0000         0  -17.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR3 /(-17) will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000    1.0000    3.2000\n         0    0.0000    1.0000    0.0000    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR2- R4 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    1.4000             0    1.2000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR2- (1.4)R3 will give\n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    0.2000    0.0000             0   -0.2000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR3 /(0.2) will give \n\n5.0000    3.0000    1.0000    5.0000   -2.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000    0.0000    1.0000    2.0000\n\nR1- 5R4 will give\n\n5.0000    3.0000    1.0000             0  -12.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1-R3 will give \n\n5.0000    3.0000             0             0  -13.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1-3R2 will give \n\n5.0000             0   -0.0000             0  -10.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nR1/5 will give \n1.0000             0   -0.0000             0   -2.0000\n         0    1.0000    0.0000             0   -1.0000\n         0    0.0000    1.0000             0    1.0000\n         0    0.0000   -0.0000    1.0000    2.0000\n\nAnd thats the final answer.\n\nSo, your answer is x=-2, y=-1,z=1, w=2\n\nVerification\na=\n     5     3     1     5\n     3     2     2     4\n     2     4     3    -3\n     4    -3    -2     2\nb= \n    -2\n     2\n   -11\n    -3\ninverse of a is\n\n    0.0250   -0.0250    0.1250    0.1750\n    0.4000   -0.4000    0.0000   -0.2000\n   -0.5125    0.6792    0.1042    0.0792\n    0.0375    0.1292   -0.1458   -0.0708\n\nThus,\ninv(a)*b = [-2 -1 1 2]'\n\nI hope i answered your question.\n\nAll the best
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
749264
Usually there won't be an answer. \n\nFor planes you have:\nz=ax+by+c\nz=dx+ey+f --->\nax+by+c=dx+ey+f\n(a-d)x+(b-e)y=c-f ---> gives a line for an answer.\n\nHowever if you look at two lines, you will have that the x and y coordinates are dependent on each other:\nI.e.\nz=f(x),y=g(x)\nz=h(x),y=j(x)\nSo --- f(x)=h(x) and g(x)=j(x) ---- which is over determined.
840346
I think that you need to use the chain rule on that second term of your f(x,y,z)
22881
Its a linear equation with d as the variable.
787521
Example:\nIf z=f(u,v),\nu=g(x,y) and\nv=h(x,y), then\n(all derivatives are supposed to be partial derivatives)\ndz/dx=dz/du*du/dx+dz/dv*dv/dx\ndz/dy=dz/du*du/dy+dz/dv*dv/dy\n\nAlternatively, you can set of the matrices\n[[dz/du dz/dv]]\nand\n[[ du/dx du/dy]\n[dv/dx dv/dy]]\nand get that\n[[dz/dx dz/dy]]\nis the product of the two matrices.
108506
y=az+(-6/49)\n(6/49)a+ya=z\n\nam i right or close i hope
1084156
The answer to that is a line.  If you want a unique solution to a two-variable problem, you need two equations.
830984
You have to first eliminate one variable from two of the equations which would leave two equations with two variables.\nThen eliminate one of those two variables to get one value.  Plug this in to get a second value; plug these in to get the third.\n\nFor example in your problem b, if you multiply the second row by -1 and add it to the first row, that would eliminate z\n\n(2x - 1y = 1)\n\nThen divide the third row by 2 and add it to the (original) first row, which also eliminates z\n\n(20x - 4y = 16)\n\nNow take those 2 rows, eliminate either x or y, and start plugging in
190850
also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
966181
This equation is plainly the result of a least squares regression in the form y = ax + b.  The formula is exactly correct for the mean values of X and Y, so if xbar = sum(x)/n and ybar = sum(y)/n, then\n\nybar = a xbar + b\n\nxbar = (ybar - b) / a\n\nsum(x) / n = [sum(y) / n - b] / a\n\nsum(x) = [sum(y) - nb] / a\n\nSo to get the sum of the x's, you need to know n and the sum of the y's.  That's as close as you can get.