what is the laplace inverse of 2?
353367
if f(t) = [delta](0) (unit impulse at t = 0), its Laplace transform is F(s) = 1, therefore, the inverse transform of F(s) = 1 is f(t) = [delta](0).\n\n And given that {L(a*f(t)) = a*L(f(t))}\n\nL(2*[delta](0)) = 2 * (1).\n\nThe answer is:\n  2*[delta](0)
705951
b) 2/X, since x/2 * 2/X = 1, which is the definition of the multiplicative inverse.
609321
One is the inverse of the other
446101
switch the y and x\n\nx = 2y - 3\n\nThen solve for y\n\nx + 3 = 2y\n\n(x+3)/2 = y\n\nnow you have the inverse.
126194
d=(-infinity, +infinity)\nr= [-1,1]
739570
d2 - =int(c2/10)\n\neasy
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
786717
its 1\n\nsince i = (-1)^1/2\n\nso i^2 = (-1)^(2*(1/2))\n\nwhich is 1
169494
The answer is:\n2y^3 - y^2\nOR\n(2y-1)*y^2
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
1072090
There seems to be a typo in your function.  I'll assume you want the Laplace transform of:\n\n f(t)=t^2*exp(ln*5t)\n\n(If this is wrong, try reposting the correct function.)\n\nFirst, simplify the function\n\n f(t)= t^2*5t\n f(t) = 5*t^3\n\nThe Laplace transform of x^n, for integral n>0 is given by (see source):\n\nn!/(s^(n+1))\nso L[f(t)](s) = 5*3!/s^(4) = 30/s^4
966844
f(0) = -i\nf(1) = 1² - i = 1-i\nf(2) = 2² - i = 4-i\n\nI take it you are dealing with complex numbers?  i is the square root of -1.\n\nNRT
291300
Try this:\nhttp://flash.lakeheadu.ca/~yzhang/myText/dspGuide/CH32.PDF#search='laplace%20transform'
1072243
Domain: (-infinity, +infinity)\nRange: (-infinity, +infinity)
650764
Remember that i^2 is -1 so\n-3i+2(-1)-2+i=-4-2i
732315
This one's a little tricky.\n\nAllow me to use "t" instead of "delta"\n\n\nint ((cos(2t))^3)dt\n\nSince (cos(2t))^2 = 1 - (sin(2t))^2\n\nSubstitute.  this becomes:\n\nint ((1 - (sin(2t))^2) * cos(2t))dt\n\nSubstitute u = sin(2t).  Then du = cos(2t)dt.\nSo dt = du / cos(2t)\nThe key here is when you plug this in, the cos(2t) values cancel.\nSo we get:\n\nint (1 - u^2)du\n= u - (1/3)u^3 + C\n= sin(2t) - ((sin(2t))^3)/3 + C
347039
F = G*m1*m2/d^2\ninverse square law