math help with inverse variation?
344744
I went to Nasa's website, which has an explanation about this. I calculated it out and it turned out to be 145 pounds, rounded to the nearest pound. I myself do not exactly understand the equation fully, but I do understand the concept. Hope this helps!
381290
I have an iterative method that is good for at least 80% of these types of problems.\n\nFirst, write down the series.\n\nNext, above the line you just wrote, write the difference between each pair of elements. In this case, it will be 1, 2, 4, 8, .... In this case, the answer is obvious - they are powers of two. The next one in the differences line should be 16, so the next on the original line should be 15+16=31. This will allow you to get any member in the sequence.\n\nIf the first order of differences doesn't give rise to a discernable pattern, try the differences of differences and so on.
951729
See this site for details of lagrange polynomials, then the program shouldn't be hard to write:\n\nhttp://en.wikipedia.org/wiki/Lagrange_interpolation
400015
By newron's ii law\nMa=F-f\nf=10-10*0.6\nf=4 N\nf= -4 i (vector notation)
789574
from the ⌠udv = uv - ⌠vdu, let u=(lnx)^2 and dv=x^4dx\nhence du=2dx/x and v=x^5/5\nsubstituing values we have:\nx^4 (lnx)^2 dx = (lnx)^2(x^5/5) - ⌠(x^5/5)(2dx/x)\n                      = (lnx)^2(x^5/5) - ⌠(2/5)x^4dx\n                      = (lnx)^2(x^5/5) - x^5/10\nthen using the limits; this yields to 1.796442507
436335
To find the opposite, or the additive inverse, just change the sign to the opposite sign.\nIf it's positive, then make it negative.\nIf it's negative, then make it positive.
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
745743
For the first problem, subtract 2/3x on both sides:\n6y=-2/3x+1\n\nDivide 6 on both sides:\ny=-2/18x+1/6\ny=-1/9x+1/6\nSlope:-1/9\n\nFor the other problem, the inverse of -1/3 is 3:\n-1/(-1/3)\n-1*-3=3
797737
Ok, take your equation \nf(x) = 2x^2 - 10\n\nreplace f(x) with y\n\ny = 2x^2 - 10\n\nswap y and x\n\nx = 2y^2 - 10\n\nsolve for y\nto do this first add 10 to both sides\n\nx + 10 = 2y^2\n\ndivide both sides by 2\n\n(1/2)x + 5 = y^2\n\ntake the sqaure root of both sides\n\nsqrt((1/2)x + 5) = y\n\nchange y to f(x)^-1 [the inverse function notation]\n\nf(x)^-1 = sqrt((1/2)x + 5)\n\nhope this helps
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
944835
Yes, you're right.  The only way to get a parralel line w/ x=6 is another vertical line.  So if x=4, the only possible perp line is y=6. Always hated those "trick" questions!  They seem so hard, yet are ridiculously easy.
929836
v = b - u/k\nv - b = u/k\nk(v-b) = u
567979
inverse of f(x) is (x-1)/2\n\nworking:\nlet inverse of f(x) be y\n       2x+1  = y\n       2x    = y-1\n        x    = (y-1)/2\nthus, inverse of f(x)= (x-1)/2
69686
I think the answer is x=1/2.\n\nIf x and y vary inversely, then xy = k, which is some fixed number. Taking the first two values, (3)(14)=k, so k=42.\n\nIf y=84, and k=42, then (84*x)=42, or x = 0.5
747397
Checking by method of differences\n\n. x ............. f(x) ...... Δ1 ...... Δ2\n. 0 .......... 1104\n. 1 .......... 1104 ........ 0\n. 2 .......... 1100 ........ 4 ....... -4\n. 3 .......... 1092 ........ 8 ....... -4\n. 4 .......... 1080 ...... 12 ....... -4\n. 5 .......... 1064 ...... 16 ....... -4\n. 6 .......... 1044 ...... 20 ....... -4\n. 7 .......... 1020 ...... 24 ....... -4\n. 8 ............ 992 ...... 28 ....... -4\n. 9 ............ 960 ...... 32 ....... -4\n10 ............ 924 ...... 36 ....... -4\n11 ............ 884 ...... 40 ....... -4\n12 ............ 840 ...... 44 ....... -4\n13 ............ 792 ...... 48 ....... -4\n14 ............ 740 ...... 52 ....... -4\n15 ............ 684 ...... 56 ....... -4\n16 ............ 624 ...... 60 ....... -4\n17 ............ 560 ...... 64 ....... -4\n18 ............ 492 ...... 68 ....... -4\n19 ............ 420 ...... 72 ....... -4\n20 ............ 344 ...... 76 ....... -4\n21 ............ 264 ...... 80 ....... -4\n22 ............ 180 ...... 84 ....... -4\n23 .............. 92 ...... 88 ....... -4\n24 ................ 0 ...... 92 ....... -4\n\nSo the relation is quadratic (as Δ2 is constant which imples the 2nd derivative is constant)\n\nSo the relation is of the form\n\nf(x) = ax^2 + bx + c\n\nNow f(0) = c = 1104\nf(1) = a + b + c = 1104\nSo a + b + 1104 = 1104\nie a + b = 0\n\nf(2) = 4a + 2b + c = 1100\n4a + 2b + 1104 = 1100\nSo 4a + 2b = -4\n\nie 2a + 2a + 2b = 2a + 2(a + b) = -4\nSo 2a + 0 = -4\nie a = -2 and b = 2\n\nThus f(x) = -2x^2 + 2x + 1104\n= -2(x^2 - x - 552)\n= -2(x - 24)(x + 23)\n\nTest f(10) = -2*(10 - 24)(10 + 23)\n= -2 * (-14) * (33)\n= 28 * 33\n= 924 YESSSSSSSSSSS!!
1033116
A X = D\nX = A^(-1) D. (A inverse * D)\n Do it ur self.
307198
It is linear on intervals, on (- infinite, 0) and on (0, +infinite) but not on the whole R. \n\nTry plotting it, you get a reversed V, not a straight line.