How do you find the corrdinates of an equation like this?
91777
y=-6\n4x+y=2\n\nsubstitute -6 for Y\n\n4x-6=2\nadd 6 to both sides\n\n4x=8\nDivide both sides by 4\n\n\nx=2\ny=-6 (taken from above)
843247
You have enough information.\n\nIf I understand you correctly, you have the (N,E) coordinates for points A and F.  So given what you know so far, you should be able to compute coordinates for points B, C, and E.  All you need for a complete picture is coordinates for D.  Lets call them (N(D), E(D)).\n\nNow you know the distance DE. Using the distance formula,\n\n(DE)^2 = [N(E) - N(D)]^2 + [E(E) - E(D)]^2 \nthis is one equation with two unknowns.\n\nAlso, you know the azimuth for CD.  Call it θ.\nThen tan(θ) =   [E(D) - E(C)] / [N(D) - N(C)]\nthis is a second equation with the same two unknowns.\n\nTwo equations, two unknowns.  Solve simultaneously to get\nN(D) and E(D).\n\nOnce you have the coordinates you can compute missing azimuth and missing distance.\nIf you would like to email me the numbers I'll take a look.
332600
to find areas of parallelograms use the formula: \n\nA = base×height.
749030
y = m (x-x1) + y1\nm=slope (so m=2)\nx1 and y1 are the coordinates. Just plug your numbers in and expand the m(x-x1) part.
1122517
sin^2+cos^2=1\nsin=1/cosec\nsin=opp side/hypotheses side\ncos=adj side/hypotheses side\nhyp^2=opp^2+adj^2\n\nSin*cosec*cos=cos
191332
Find the dementions and multiply.
697247
C = (W^2*L^2) / (24T^2)\nC(24T^2) = W^2*L^2\n[C(24T^2)]/L^2 = W^2\nAnswer:\nW = sq.rt {[C(24T^2)]/L^2}
1010902
This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
627342
ok, this is found in the 4th quadrant, the order is as follows\n\n(x,y) 1st\n(-x,y) 2nd\n(-x,-y) 3rd\n(x,-y) 4th\n\nhope this helps\n\nmatttlocke
935340
= cosx[cosxcosy+sinxsiny]+sinx[sinxcosy-cosxsiny]\n\n= (Cosx)^2cosy+cosxsinxsiny+(sinx)^2cosy-cosxsinxsiny\n\n= (Cosx)^2cosy + (Sinx)^2cosy\n\n= Cosy [ (cos)x^2 + (sinx)^2]\n\n= Cos(y)
324456
give parameters for x, y, and z.  Otherwise you cannot prove it, as no parameters are given for the equation as you described it.
736379
for this just substitute for x and y in eqn and see if LHS=RHS\nLHS=4\nRHS=(-1/2)(-3)-5/2\n=3/2-5/2\n=-2/2\n=-1\nnot = RHS\nso pt. not on line
880047
I can find this on a TI-84+, is that how you want to get it or do you need an algebraic solution?
750949
I'm not sure of the exact equations, but I know the side that's 4 times longer should be 4 + x, and the one side should be x.  Then using the Pathagorem thm., x^2+ ?^2= (4+ x)^2\n\nI'm just not sure how to find the third side.\n\nHope that helps :/
895922
That equation is just the following when written as a matrix (// = new line)\n[ 2.7 -3.5 // 6.3 4][x // y] = [13.35 // 12.9]\nNow, you need to find the inverse of the left hand matrix. The inverse of [a b // c d] is 1/(ad-bc) * [d -b // -c a].\nSo, the inverse is 1/32.85 * [4 3.5 // -6.3 2.7].\nSo, multiplying both sides by that:\n[x // y ] = 1/32.85 * [4 3.5 // -6.3 2.7][13.35 // 12.9].\nMultiplying that out gives [x // y] = [3 // -1.5]
772048
cot²(x)\n------------ =  ............ (NOTE cot²x + 1 = csc²x)\ncsc(x) - 1\n\ncsc²(x) -1\n------------ =\ncsc(x) - 1\n\n(csc(x) - 1)(csc(x) + 1)\n----------------------------- =\ncsc(x) - 1\n\ncsc(x) + 1
830875
You have an equation in the form of y = ax^2 + bx + c, and you want the form (y - k) = a(x - h)^2\nManipulating all those decimals can be very confusing, so let's try manipulating letters, instead.\na = -4.97\nb + 10.91\nc = -5.16\ny = a(x^2 + (b/a)x) + c\ny = a(x^2 + (b/a)x + (b/2a)^2x^2 - (b/2a)^2) + c\ny = a(x^2 + (b/a)x + (b/2a)^2x^2) - a(b/2a)^2 + c\ny + a(b/2a)^2 - c = a(2 + (b/2a))^2\ny + (b^2/4a) - c = a(x + (b/2a))^2\n(y + (b^2 - 4ac)/4a) = a(x + (b/2a))^2\nNow, substituting the numbers back in:\n(y +((10.91)^2  -4(-4.97)(-5.16))/(4(-4.97) = -4.97(x + (10.91/(2(-4.97))^2\n(y + (119.0281 - 102.5808)/-19.88) = -4.97(x + (10.91/-9.94))^2\n(y -16.4473/19.88) = -4.97(x - (10.91/9.94))^2\n(y - 0.82733) ≈ -4.97(x - 1.0976)^2