rewrite in function form: 2x+-y=0?
549653
Subtract 2x from each side\n-y = 0 - 2x\nMultiply each side by -1\ny = 2x
156464
u can't devide by (a-b) as (a-b)=0
869140
Just graph it like an equality, and then your answer is everything either above (>) or below (<) the line.  For example, in the last inequality, the answer is everything above the line y = -2, not including the line itself.
930650
cos(0) = 1. so you can plug in x = 0 to get the limit as 0*1 = 0
909269
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
367158
(0! + 0! + 0! + 0! + 0!)!\n\n\n0^0 is undefined.
200150
for all of those, all u do is set once variable at a time equal to zero. for instance..\n\n-x+0=2\n-x=2  <--- multiply x and 2 by -1\nx=2\n\nthen..\n\n0+y=2\ny=2
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
733627
Rewrite the equations as:\ny = -x + 2 and\ny = (x+10)/2 = 1/2*x +5\n\nNow, if you insert x=0 in each equation, you will get a y-value.\nIf you insert y=0 in each equation, you will get an x-value.\nFor each line, you can use those two intercepts to graph the line.
771254
First, you use the point slope formula:\ny-y1=m(x-x1)\n\ny-2=3(x-0)\ny-2=3x\ny=3x+2\n\nStandard form:\nAx+By=C\n\nSubtract 3x on both sides:\n-3x+y=2\nThat's your standard form!
934633
By substitution, the first equation when substituting in y=2+x becomes:\n4x - 4(2+x) = -8 ==> -8 = -8 which is always true.  This means that what you have are two equations with an infinite number of solutions, as the two lines are the same line.  So any pair of the form (x,x+2) such as (0,2) or (1000,1002) will work.
870259
You can do it by hand or with a calculator. Either way, it would look like this:\n\n2x + y = 3 \n(move the 2x to the right side of the equation, so only y is on the left side)\ny = 3 - 2x\n\nNow put 1, 2, 3, etc. in for x, and see what y equals:\n\nwhen x=1, y=1\nwhen x=2, y=(-1)\nwhen x=3, y=(-3)\nand so on.....\n\nNow put these values into a table:\n\nx --- y\n1 --- 1\n2 --- (-1)\n3 --- (-3)
1082168
y=x^2+10x+8\ny=(x+5)^2 - 17\n\nVertex form: (5,-17)
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
897443
2x + y = - 9\n\n2x + y - y = - 9 - y\n\n2x = - 9 - y\n\n2x/2 = - 9/2 - y/2\n\nx = - 9/2 - y/2\n\n- - - - - - - -\n\ny values\n\n2x + y = - 9\n\n2x + y - 2x = - 9 - 2x\n\ny = - 9 - 2x\n\n- - - - - - - - - - - - -\n\n2).\n\nx values\n\n-x - y = - 8\n\n- x - y + y = - 8 + y\n\n- x = - 8 - y\n\n- 1(- x) = - 1(- 8) + (- 1)(y)\n\nx = 8 - y\n\n- - - - - - - - - - - -\n\ny values\n\n- x - y = - 8\n\n- x - y + x = - 8 + x\n\n- y = - 8 + y\n\n- 1(- y) = -1(- 8) + (- 1)(x)\n\ny = 8 - x\n\n- - - - - - - - - - - -\n\n3).\n\nx values\n\n- 3x + 4y = 2\n\n- 3x + 4y - 4y = 2 - 4y\n\n- 3x = 2 - 4y\n\n- 3x/ - 3 = 2/- 3 - 4/- 3\n\nx = - 2/3 + 4/3\n\n- - - - - - - -\n\ny values\n\n- 3x + 4y = 2\n\n- 3x + 4y + 3x = 2 + 3x\n\n4y = 2 + 3x\n\n4y/4 = 2/4 + 3x/4\n\ny = 1/2 + 3x/4\n\n- - - - - - - - - -\n\n4).\n\nx values\n\n- 2x - y = - 3\n\n- 2x - y + y = - 3 - y\n\n- 2x = - 3 - y\n\n- 2x/ -2 = - 3/- 2 - y/-2\n\nx = 3/2 + y/2\n\n- - - - - - - - - \n\ny values\n\n- 2x - y = - 3\n\n- 2x - y + 2x = - 3 + 2x\n\n- y = - 3 + 2x\n\n- 1(- y) = -1 (- 3) + (- 1) (2x)\n\ny = 3 - 2x\n\n- - - - - - - -s-
840688
Not sure what you were doing. I would rewrite f(x) in this form:\n\nf(x) = (2x-1)^(-1/5)\n\nThen you're dealing with a u^n situation, so\n\nf'(x) = (-1/5)(2x-1)^(-6/5)*(2)\n\nf'(x) = (-2/5)(2x-1)^(-6/5) \n\nThat's equivalent to what you had, but a little simpler.\n\nLet the underboss know that this is a "site" and I did answer your question.