how do you do this complex equation?
914544
4x+3/2-2x-2=3-x-4/3+2x-5        then you add everthing up\n\n2x-1/2=x-3and2/3           Sorry for the and....i didn't want it to look like 32 divided by 3....it's 3 and 2/3\nmultiply each side by 6...\n12x-3=6x-22\nThen you have to simplify \n6x=19  Divide each side by 6 to get the answer\nx=19/6\nI'm pretty sure but im not absolutly sure....
821031
I=I1+I2; I1=int{for x=pi/6 until pi/2} of x*dx = x^2/2{for x=pi/6 until pi/2}= pi^2/9;\nI2=int{for x=pi/6 until pi/2}of  25*dx/(sinx)^2; \nLet t=tg(x), then 1/(sinx)^2=1/t^2+1 and dx=(1+t^2)*dt;\nI2=int{for t=1/sqrt(3) until +inf}of 25*t^(-2)*dt = -25/t = 25*sqrt(3)
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
1124422
[(y^2 + y )/(y+ 2)] / [(y^2 + 3y)/(y^2-4)]\n= [(y^2 + y )/(y+ 2)] times (y^2-4)/(y^2+3y)\nFactor everything\ny^2 + y = y(y+1)\ny^2 + 3y = y(y+3)\ny^2 - 4 = (y+2)(y-2)\nCancel anything on the top with anything on the bottom\nAnswer: (y+1)(y-2)/(y+3)
429835
ok, u have to use distance formula\n\nsqrt[(-4-(-7))^2+(c-7)^2]=5   \nsquare both sides\n[(-4-(-7))^2+(c-7)^2]=25      \nadd the stuff in the parentesies\n[(-3)^2+(c-7)^2]=25           \nexponents\n9+(c-7)^2=25                  \nsubtract from both sides\n(c-7)^2=16                    \nsquare root both sides\nc-7= +/-4                     \nadd to both sides\nc=11 or c=3
508029
The equation is:\n(x+a+b+c)^2 = 2x^2+(x+a)^2-(x+b)^2-(x+c)^2\nA simple rearrangement yields\n(x+a+b+c)^2 - x^2 = (x+a)^2-(x+b)^2 + x^2-(x+c)^2\nFactorize using the identity a^2-b^2=(a+b)(a-b)\n(x+a+b+c+x)(x+a+b+c-x) = (x+a+x+b)(x+a-x-b) + (x+x+c)(x-x-c)\nwhich can be written as\n(2x+a+b+c)(a+b+c) = (2x+a+b)(a-b)-c(2x+c)\n==> 2x(a+b+c)+(a+b+c)^2 = 2x(a-b)+(a-b)(a+b)-2xc-c^2\n==> 2x(a+b+c-a+b+c) = (a+b+c)^2 + a^2-b^2-c^2\n==> 4x(b+c) = a^2+b^2+c^2+2ab+2bc+2ca+a^2-b^2-c^2\n            = 2(a^2+ab+bc+ca) = 2(a+b)(c+a)\n==> x = (a+b)(c+a)/[2*(b+c)]
708943
You have to complete the square.\n\ny = a(x^2 + b/a x) + c\ny = a(x^2 +b/a x + b^2/4a^2) + c - b^2/4a\ny = a(x + b/2a)^2 + (4ac - b^2)/4a\n\n[4ay - (4ac - b^2)]/4a = a(x + b/2a)^2\n[4ay - 4ac + b^2]/(4a^2) = (x + b/2a)^2\n\n+/- sqrt(4ay - 4ac + b^2)/2a = x + b/2a\n\nx = (-b +/- sqrt(4ay - 4ac + b^2))/2a
1027604
[(1/y)+4)]/[(1/y^2) -16]\nFirst simplfy the numerator.\n(1/y) +4\n= (1/y) +4y/y\n= (1+4y)/y\nthen simplify the denominator.\n(1/y^2) -16\n= (1/y^2) - 16y^2/y^2\n= (1-16y^2)/y^2 \n= [(1-4y)(1+4y)]/y^2\nNow multiply the numerator by the reciprocal of the denominator.\n(1/y+4)/(1/y*2)-16\n= [(1+4y)/y] * y^2/[(1-4y)(1+4y)]\n(1+4y) and y cancel\nAnswer: y/(1-4y)
193529
if your equation is of the form\nax^2 +bx+c=0 then the general formula for the solution is\nthe two roots are:\n(-b+(b^2-4ac)^(1/2))/(2a)\n(-b-(b^2-4ac)^(1/2))/(2a)\nif b^2-4ac is less than zero then you will have imaginary roots, in such a case ur roots would be\n(-b+i(4ac-b^2)^(1/2))/(2a)\n(-b-i(4ac-b^2)^(1/2))/(2a)
247800
it is basically:\na2 + b2 = c2   -->  (A squared)+(B squared)=(C squared)\n  \n\nif you have a right triangle.you have three sides.the two legs.and the hypotenus(sp?).a and b are the two legs and c is the hypotenus.\n\nin an equation.if you are given a and b.you can square each of them and then add the two values together.the will give you c squared.to find c you then need to take the square root of the value(a squared + b squared).\n\nif you have c and a leg(example:a).you square c.and then subtract a squared from that value(c squared).that will give you b squared.to get b you just get the square root of c squared - a squared.\n\n\nhope that helps.
605867
(x^2)/2 - 3(x^(3/2))/(3/2) + c = (x^2)/2 - 2x^(3/2) + c
227193
∫[(2x^4) - (4/x^5)]dx=\n\n∫(2x^4)dx - ∫(4/x^5)dx=\n\n2∫x^4dx - 4∫[x^(-5)]dx=\n\n2*[x^5/5] - 4*[x^-4/-4]=\n\n(2/5)x^5 + x^-4 + C=\n\n(2/5)x^5 + 1/x^4 + C
730173
x = (-b +/- sqrt(b^2 - 4ac))/2a\n\nwhere a= 1, b= -6 , c=8\n\nx1 = (6 + sqrt(36 - 32))/2 = 4\nx2 = (6 - sqrt4)/2 = 2
965376
z=w+whx\nz=w(1+hx)\nz/(1+hx) = w &#92;
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
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
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4