Find all of the solutions of the equation.?
954106
arccos .91=24.5 degrees\nsince this is positive, it must be in the 1st & 4th quadrants.\n24.5 degrees & 335.5 degrees.\n56^0=1 so\n1-3x=24.5 degrees\n-3x=23.5\nx=-7.8 degrees\n\n1-3x=335.5 degrees\n-3x=334.5\nx=-111.5 degrees\n\nWhat-s with 0^0? That is undefined.
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
289077
add all the equations, you get\n2(r+s+t) = 56\nso,\nr+s+t = 56/2 = 28\ndone!!
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
905030
1. y-axis\n\n2. 2nd quadrant\n\n3. 2nd or 3rd quadrants\n\n4. 2x + 3y = 6 => y = 2 - (2/3)x\n\n5. x - 2y = 8 => y = (1/2)x - 4\n\n6. x - 3y = 4 => x = 4 + 3y\n   Only (a) is a solution\n\n7. y - 2x = -1 => y = -1 + 2x\n   (b), (c) are solutions
761571
1.(4x+1)(16x^2-4x+1)\n\n2.(1-5x)(1+5x+25x^2)\n\n3.(2xy^2+3)(4x^2y^4-6xy^2+9)\n\n4.b^3(a^3-b^3)\nb^3(a-b)(a^2+ab+b^2)\n\n5.(3x-2y)(9x^2+6xy+4y^2)
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
123571
add the two equations on left hand side and right hand side\n\n3x - y + 5x + y = 4 + 4\n\n8x = 8\n\nx = 1\n\nput this into the first equation\n3(1) - y = 4\n-y = 4 - 3 = 1\ny = -1\n\nHence x= 1 y = -1\nconfirm it by checking the 2nd equation \n5(1) + (-1) = 4 \nCORRECT!
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.
990365
No solutions my ascii, haven't these people ever heard of complex numbers?\n\nAs has been previously mentioned, for the first part of the equation you can set U=e^x and solve for U. Thus:\nU^2+6U+8=0\n(U+2)(U+4)=0\nU=-2 or -4\ne^x=-2 or -4\n\nNow, here's where we demonstrate that this equation has solutions. Remember Euler's identity: e^(iπ)+1=0. Thus, e^(iπ)=-1. Thus:\ne^x=-2\n(-1)e^x=2\ne^(iπ)e^x=2\ne^(x+iπ)=2\nx+iπ=ln 2\nx=ln 2+πi\n\nAnd from e^x=-4, we get x=ln 4+πi. So x=ln 2+πi or  x=ln 4+πi. And that's your solution.\n\nActually, for the sake of completeness, since e^(2πi)=1, x could also be any of these solutions plus an integral multiple of 2πi. Thus, x=ln 2 + (2K+1)πi or x=ln 4 + (2K+1)πi, where K is an arbitrary integer.
565039
For this you are asked to take the the equation y = 2n + 3 and substitute the rest of the values in for n and solve the equation. The implied operation between 2 and n is multiplication.\n\nSo you take the n out of the equation and put in the values that you have listed.\n\nFirst Value:\ny = 2(2) + 3  Do the multiplication first then add the 3\ny = 4 + 3\ny = 7\n\nSecond Value:\ny = 2(1/2) + 3\ny = 1 + 3\ny = 4\n\nThird Value:\ny = 2(0) + 3\ny = 3
610726
there is a solution, all x,y that satisfy this equation:\ny = 4 - 2x\n\nwhich is a straight line with intercepts (0,4) & (2,0)\n\nIt has an infinite amount of solutions. To be able to solve for actual numbers, you need as many equations as you have variables.
289962
every one that passes by (-1,6). there are infinite equations. You must give some more clues.\n\n Let me explan myself: \nif your equation were x=-1 (completely horizontal)\nif your equation were y=6 (completely vertical)\nand so on f you have a crcle, a hperbolic, a sinoidal, a straight line with some inclination.... there are just a lot!!!!\n\nhope i help\n\nthanks for the details\n\nJust substitute x=-1 on each equaton, and the one that results on y=6 is the one you are looking for.
683580
y^2 + 7y + 3y + 21\ny^2 + (7 + 3)y + 21\ny^2 + 10y + 21\n(y + 7)*(y + 3)
1142710
Use this formula, you can find millions of them\n\nL x W = 2(L + W)\n\nuse any number for L and solve for W
859187
I'll do one for u.\nIn the first,\nMultiplying both sides by x*,we get\nx+3x+2x*=7x*\n4x+2x*=7x*\nMoving 4x and 2x* to the R.H.S.\n5x*-4x+0=0\nThus 5x*-4x=0\nAttempt the others in the same manner.
140559
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O