when x vries inversely as the square root of Y. What is X, when Y=3,  and Y, when X=6 ?
339722
when X vries inversely as the square root of Y,\nX = k/sqrtY for some constant k\nX, when Y=3, is\nX = k/sqrt3\nX = ksqrt3/3\nand Y, when X=6, is\nk/sqrtY = 6\n6 sqrt Y = k\n36Y = k^2\nY = k^2/36\n\n^_^
1116670
Nope, not quite.  sqrt(x^4*y^3) = x^2 * sqrt(y^3) = x^2 * y^(3/2)\n\nWhen you're not sure about problems like this one, try:\nhttp://www.quickmath.com/\nIt will solve most problems like that.
878479
To find the inverse function switch the x's and y's\n\ny=x^2-3\nbecomes\nx=y^2-3\nand after simplifying\ny=+/-sqrt(x+3) -- don't forget the plus or minus, you need both for this one!
89471
idk wat the square root is.... the square root of a negative is not a negative though... a negative times a negative is a positive.... also, this is the 4th ? i have answered of urs
827399
V.A. : none\nH.A. : y=3 (  because  if the degree of the numerator x^2 is equal to the degree of the denominator ,you should divide the coefficient of the numerator by the coefficient of the denominator,so 3/1 = 3)
939650
x+y=-3\nx=-y-3\nx -3  -2  -1  0  1  2  3 \ny  0  -1  -2 -3-4-5 -6\n\nx^2+2y^2-12y-18=0\nwhen y=-1\nx^2+2+12-18=0\nx^2-4=0\nx^2=4\nx=+/-2\nso (-2,-1),(2,-1)\nwhen y=3\nx^2+18-36-18=0\nx^2-36=0\nx=+/-6\nso(-6,3),(6,3)\n\npoints of intersection\n(-2,-1) and (-6,3)
835948
function Cos^ (-1)x is an inverse of function Cos x\n(like square(x) and squareroot(x) act oppostite\nsquare(sqrootx) = x or sqroot(square(x) = x\nTherefore Cos[Cos^(-1)x = x\nNow to your question\nLet [Cos^(-squareroot3/2) = x\nTaking Cos on both side\nCos[Cos^(-1)(-squareroot3/2)] = Cos x\nor -squareroot3/2 = Cos x\nThis equation has many solutions as you can see (for every cycle of 360 deg you will have two solutions - one each in second quadrant and  third quadrant where Cosine (or cos) has negative values\nPrimarily we take the solution which between zero and 360 degrees\nTwo solutions are\nCos (180 -30) = Cos (180+30) = (- squareroot 3/2)\ntherefore 150 deg and 210 deg\nconverted to 150xpi/180 = 5/6 pi radians\nand 210xpi/180 = 7/6 pi radians\nSimilarly we can find a set of two solutions between 360 &720 deg, between 720 &1080 deg etc and so on\neg [360 + (180+30)] deg and [360 +180-30] deg [720 + (180+30)] deg [720+ (180-30)] deg and so on\nCheers\nSubhash
1093922
if y = 6-x\nthen you just put that in for y\n2x-3(6-x) = -2\n2x- 18 + 3x = -2\n5x - 18 = -2\n5x = -20\nx = -4
801012
inverse square root of a number say"x" is\nx^[-(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
871743
Since we're given y = x+8, let's solve the first equation for x by plugging in y.\n\n3x*3y=y*x^2+96 <==> 3x*3(x+8)=(x+8)*x^2+96\nMove everything to one side and expand.\n(x+8)*x^2+96-3x*3(x+8) = 0\nx^3+8x^2 -9x^2 -24x + 96 = 0\nx^3 - x^2 -24x + 96 = 0\n\nThis is easiest solved by plugging in to the solver of a graphic calculator. Once you get values for x, plug in to y = x + 8 and you have values for y.
1160946
If you mean sqrt(x+4) = 3, then the answer is x = 5 since\n\nsqrt(x+4) = 3\nSo x+4 = sqrt(x+4)^2 = 3^2 = 9.\nSo x+4 = 9\nSo x = x+4-4 = 9-4 = 5.\n\nIf you mean sqrt(x) + 4 = 3, then there is no solution since sqrt must be positive.
747583
y= √(1/(x² + 1) – 2) \n\nWell 1/(x² + 1) – 2 ≥ 0\nie 1/(x² + 1) ≥ 2\nSince x² + 1 ≥ 0 for all x\nOn taking reciprocals\nx² + 1 ≤ ½\nx² ≤ -½ This is not possible as x² ≥ 0 for all x \nSo there is no domain\n----------------------- -----------------------\nWell that makes sense as  x² + 1 ≥ 1\nSo 1/(x² + 1) ≤ 1\nSo 1/(x² + 1) - 2 ≤ -1\n\nSo taking square roots gives no real solutions.
593847
The answer is zero isn't it?  The square root of (7 squared) is 7.  If you minus 14 you get -7. Add another 7, brings you up to zero.
362883
16^(3/4) means take a fourth root of 16, and then cube it.  Anytime you take an even root (square root, fourth root, etc.), you always have +/- answers.\n\nConsider that the square root of 4 is +/- 2.  That makes sense because if you square 2, it is 4, and if you square -2, it's 4 as well.\n\nNow consider the fourth root of 16.  That gives you +/- 2.  If you reverse it, you'll see it makes sense: 2^4 = 2*2*2*2 = 16, and (-2)^4 = (-2)(-2)(-2)(-2) = 16, since all the negatives end up cancelling out.\n\nSo for your original problem,\n\n(3x-1)^(4/3) = 16\n3x - 1 = 16^(3/4)\n3x - 1 = +/- 8\n3x = 1 +/- 8\nx = (1 +/- 8)/3\n= 3 or -7/3\n\nAnd if you plug in either value into your original equation, it works out.
241669
Y=x^3(x+2)(3x-1)\n we have three possibility for x\nx^3=0 than X=0\nx+2=0 than X=-2\n3x-1=0 than x=1/3\n\nso the answer is a
319669
did you mean the square root of x-1 = 3?\nin that case it will be 2:\nadd positive 1 to both sides:\n   x-1+1= 3+1\nthen solve the remaining equation:\n   x=4\nget the square root of 4:\n   x=2