Solve the equation.  Explain?
444311
The area of a triangle is\n\n1/2 * Base * Height\n\nSo it the max base is 2 ft, and the max area is 20 ft^2, then the max height would have to be 20ft
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
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
1144396
you would take the 2x-4y=3 and solve for a variable, such as x=(3-4y)/3   you now have what x equals and you sub that into the other equation so you now have 2[(3-4y)/3]-3y=0 and you should be able to solve that equation
413333
let suppose that \nr(@) = 4/ sqrt( 4cos^2(@) +sin^2(@) )\nthen x(@(t) ) = r(@) . cos (@) emplies\n\nx^2 (t) = r^2 (@) . cos^2(@)\n\nx^2 = 16 / (4 cos^2 + sin^2)  . cos^2\n\nso x^2(t) = 16/ (4+ tan^2(t) )\n\nBut tan( @(t) ) = y(t) / x(t)\n\nThus: x^2 = 16 / (4+ y^2 / x^2 )\nThus: \nEQ : x^2/4 + y^2/16 =  1\n\nCheck @=0  --> r(@=0) = 2\nAnd (x=2 , y=0 ) is solution of EQ
505936
What exactly are you supposed to do with this equation?  Are you suppose to graph it, find solutions, factor or what?  You need to be more specific.
598425
Solving for x in the first equation gives us...\n\nx + 5 - 5 < 4 - 5\nx < -1\n\nSolving for x in the second gives us...\n\n2 - 7x - 2 < 16 - 2\n-7x < 14\n---   ---   Here, don't forget to swap the sign\n -7    -7   since we are dividing by a negative.\nx > -2\n\nSo use that to find your solution set.  We know x must be less than -1 but greater than -2.  \n\nHope this is helpful.
607838
where is the rest of the equation?\n\n2x+5-x+3x+2-2x+5x-3x+8-4x+2+1x\n\nCollect like terms:\n\n2x-x+3x-2x+5x-3x-4x+1x+5+2+8+2\n\nRelocate the terms:\n\n1x+2x+3x+5x-x-2x-3x-4x+5+2+8+2 = 37\n\nAdd like terms:\n\n11x - 10x + 17 = 37\n\nNow, subtract all terms and solve from there.\n\nx + 17 = 37\nx = 20
647266
(Ok, by the way, this is not a linear equation. A linear equation is when the equation includes "x" and "y")\n\nBefore you can get the answer, you have to obviously solve it. \nSo for this equation, you want to get the variable alone; to one side. Now looking at the equation, everything isn't simplified. The left side is, you can not do anything with "x" and "-15"...Though, on the right side, there are like terms (expressions/terms that can be simplified)..."x" and "2x" become 3x. See 1x (1x is same as x) plus 2x equals 3x. So now your equation is x-15=3x+1 Now that everything is simplifed, you can solve it. As I said, get the variable to one side. Usually, the variable is on the left side of the equation, but it doesn't really matter which side its on...So to get x alone from the left side, you have to do what to get rid of NEGATIVE 15?...Do the opposite, POSTIVE 15, which then cancel each other out and x will be left alone. However when you add 15 to the left side, you have to add 15 to the right side. So adding 15 to the right side, 3x+1+15= 3x+16...Now the equation should be x=3x+16. So now, you have to get the 3x from the right side to the left side. How do you get sixteen alone? You minus 3x from both sides. The right side will result with having just 16 as the other side will have -2x. The equation is ow -2x=16. Again, get x alone. And how to do that is divide -2 to both sides. So -2x divided by -2 is 1x or just x and 16 divided by -2 is -8. At the end you should have x=-8\n\nHere's the work:\nx-15=x+2x+1\nx-15=3x+1\nx-15+15=3x+1+15\nx=3x+16\nx=3x+16-3x\nx-3x=16\n-2x=16\n-2x/-2=16/-2\nx=-8\n\nAnd the "check" works too(when you plug in -8 as x)\n\nx-15=x+2x+1\n-8-15=-8+2(-8)+1\n-23=-8-16+1\n-23=-23
961595
umm..are you missing an equals sign somewhere or am i really dumb? how do you figure out what x is unless you have both sides of the equation?
464039
Your equation would be:\n\n= ((1/p) - (1/q))/(1/pq)\n\n= ((q - p)/pq) * pq\n\n= q - p\n\nHope I am clear.
685042
A=2LW+2LH+2WH\n\nA-2LW=2LW-2LW+2LH+2WH\n\nA-2LW=2LH/2L+2WH\n\nA-2LW\n------------ = 2WH\n    2L\n\nA-2LW ~~~(1)  ~~~~~~(1)\n------------  -------- = 2WH   --------\n    2L~~~~~ (2W) ~~~~~(2W)\n\nA - 2LW\n------------- = H\n    2L(2W)
561262
u jammed the equation together.\ndo u mean\n6x^3 + ( 6/x - 2 * x^2 - x/3x2 - 3x + 3)\nits really hard to see ur question. can u write it out with spaces...\nhurry up i am waiting... lol
895454
There is no ordered pair {a,b} which solves both equations\n\nif a+b = a-b , then 2b = 0 or b = 0 and a is anything\n\nBut the second equation is not defined for b=0, no matter what a is (specifically, a/b is not defined for b=0)
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)]