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If we assume that it equals zero and you wish to solve for x, then the answer is:-x2 + 2x - 3 = 0x2 - 2x + 3 = 0x2 -2x + 1 = -2(x - 1)2 = -2x - 1 = ± √(-2)x = 1 ± i√2
2x + 1 = 9 2x = 9 - 1 2x = 8 2x/2 = 8/2 x=4
x2 - 2x + 2 = 0(x-1)2 - 1 + 2 = 0(x-1)2 = -1x-1 = +- ix = 1+- i
2x + 2/3= 5/3 2x= 3/3 2x=1 x=1/2
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
The solution to the equation 2x plus 2 is 2(x + 1).
If we assume that it equals zero and you wish to solve for x, then the answer is:-x2 + 2x - 3 = 0x2 - 2x + 3 = 0x2 -2x + 1 = -2(x - 1)2 = -2x - 1 = ± √(-2)x = 1 ± i√2
2x + 1 = 9 2x = 9 - 1 2x = 8 2x/2 = 8/2 x=4
If: x+1+2x = 1x+5 Then: x = 2
When using substitution the answer to y0.5x plus 2 -y-2x plus 4 is y = -2 (x-1.
2x + 7 = 1 Therefore, 2x = -6 x = -6/2 x = -3
2x+8 = 6 2x = 6-8 2x = -2 x = -1
x2 - 2x + 2 = 0(x-1)2 - 1 + 2 = 0(x-1)2 = -1x-1 = +- ix = 1+- i
2x + 2/3= 5/3 2x= 3/3 2x=1 x=1/2
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
2x + 5x - 1 = 9x 7x - 1 = 9x -1 = 9x - 7x -1 = 2x -1/2=x
2x+8=10... Subtract 8 on both sides, then devide by 2... 2x=2... x=1