find the 6th term in the sequence...\n-3 + sqroot of 2, 0, 3- sqroot of 2
123410
-3 + sqrt(2) or -(3 - sqrt(2))\nthats like saying\n\n-1x, 0x, 1x\n\nthe answer is problem found like this\n\n2x, 3x, 4x\n\nso i believe the answer is \n\n6 - 2sqrt(2), 9 - 3sqrt(2), 12 - 4sqrt(2)
352784
All you need is the quadratic formula:\n\n(-b +/- sqrt(b^2 - 4ac))/(2a)\n\nso the answer is:\n\n(-6 +/- sqrt(36 - 4(-2)(-3)))/(-4)\n= (6 +/- sqrt 12) / 4\n= (3 +/- sqrt 3) / 2\n\nwhich is approximately equal to:\n2.36603 and 0.63397
54139
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.
483408
6x^2(x-3)+x(x-3)-2(x-3)=0\n=>(x-3)(6x^2+x-2)=0\n=>(x-3)(6x^2+4x-3x-2)=0\n=>(x-3){2x(3x+2)-1(3x+2)}=0\n=>(x-3)(3x+2)(2x-1)=0\nTherefore x-3,3x+2 or 2x-1 is equal to zero\nTherefore x=3,-2/3 or 1/2 ans.
339128
determine the square of 6 - sqrt[(x-2)^2 + y^2]\n\n-->\n\n36 + [(x-2)^2 + y^2] - 12*sqrt[(x-2)^2 + y^2]\n= 40 + x^2 + 2x + y^2 - 12*sqrt[(x-2)^2 + y^2]
474457
well sqrt(2/3) = sqrt(2)/sqrt(3)\nsame thing for the 3/2\n\nso we have\nsqrt(2)/sqrt(3) + sqrt(3)/sqrt(2)\n\nmultiply the 2/3 term by sqrt(3)/sqrt(3) to rationalize the denominator and the 3/2 term by sqrt(2)/sqrt(2)\n\nsqrt(6)/3 + sqrt(6)/2\n2sqrt(6)/6 + 3sqrt(6)/6\n5sqrt(6)/6 ------- your answer
524880
1.adding-2f-7\n-6f-2f+7-7<2f-2f-9-7\n-8f<-16\ndividing by -8\nx>2\n\n2.5/8g-4>2/3g+6\nadding 4\n5/8g>2/3g+10\nmultiplying by 24\n15g>16g+240\nadding -16g\n-g>240\ndividing by -1\ng<-240\n\n3.sum incomplete\n\n4.-2 1/9d+3/5<-3 2/9d-4 2/5\n-19/9d+3/5<-29/9d-22/5\nmultiplying by45\n-95d+27<-145d-198\nadding 145d\n50d+27<-198\nadding -27\n50d<-225\ndividng by 50\nd<-5 1/2
374422
it's A\n\n4st+6t -10s-15\n2t(2s+3) -5(2s+3)    \n(I factored out the negative for both numbers in the -10s-15)\n\n(2t-5)(2s+3) or (2s+3)(2t-5)
380864
Sqrt(3x/2) = Sqrt(3x/2) * SQRT(2/2) = SQRT(6x/4) = 1/2 * SQRT(6x)\n\n6*SQRT(7) - SQR(28) = 6*SQRT(7) - SQR(4*7) = 6*SQRT(7) - 2*SQR(7) = 4 * SQRT(7)
348733
you almost got it !!\n(x^2 - 3x)= x(x-3)\nand (x^2 - 5x + 6 ) = (x -3 )(x - 2)\ntherefore you have 4/[x(x-3)]   +  3 / [(x -3 )(x - 2)]\nnow you take the LCM\nthat is, you multiply (x-2)to the nr & dr of the first term .\nyou multiply x to the nr and dr of the 2nd term\nyou will thus get ..\n[4(x-2)]/[x(x-3)(x-2)]  +  3x/[x(x-3)(x-2)]\n=[4x+3x-2]/[x(x-3)(x-2)]\n=[7x-2]/[x(x-3)(x-2)]
595739
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
489690
abc(-24a^2b+43c)/6abc\n=(1/6)(-24a^2b+43c)
5945
its 1, 2, sqrt (3). i dont know what u mean by 6,6 over 3,12
342519
Just plug in the numbers and solve for s...\ns=6^2-4^2\n    -----------\n       2(2)\ns=36-16\n     -------\n         4\n\ns=20/4\ns=5
102518
the answer is \n\n(sq.rt 3)(sq.rt 3) + 3(sq.rt 3) - (sq.rt 3) - 3\n\n= 3 + 2(sq.rt 3) - 3\n\n= 2(sq.rt 3)\n\nand that is worth 10 marks not 2
520859
What??\n\n3x2 - x +6 = 0\n6 -x +6 = 0\n12 - x = 0\nwhen x is 12 the equation is zero.\nwhen x is less than 12, the equation is less than zero.\n\n\n\n(-1)2 - 4(3)(6) = x\n-2 - 72 = x\n-74 = x
435368
x^3 - 6x^2 - 3x + 18 \n= x^2 (x -6) - 3(x-6) \n= (x^2 - 3)(x - 6)