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Why is x plus x equal to 2x?
It is equal to 2x because if you were to replace the "x" with a number say, 2 for example so 2 + 2 would be 4 and 2 times 2 would also be 4 which is what "2x" stands for (2 times x). So then if you had x + x you would get 2x.
Factor 4x -2x - 12?
As written, that's 2(x - 6) You may have meant 4x^2 - 2x - 12 which would be 2(2x + 3)(x - 2)
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
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
What is the answer for this equation 2x-3 equals x plus 2?
2x-3=x+2 2x-x=2+3 x=5
What is x times -2?
X * - 2= - 2X========that simple
What is x 2x 3?
x^2 x^3? That would be x^5 If you mean x X 2x X 3, then what would be 2x^2+3
Why is x plus x equal to 2x?
It is equal to 2x because if you were to replace the "x" with a number say, 2 for example so 2 + 2 would be 4 and 2 times 2 would also be 4 which is what "2x" stands for (2 times x). So then if you had x + x you would get 2x.
What is the answer to 2x squared minus 11x plus 14?
so you are talking about 2x^2 - 11x + 14 ... So then you found the Common factor from the two pairs. Then you would do 2x^2 - 4x - 7x + 14. 2(x^2-2x) - (7x-2). If you rewrite it in factor form, then it would be (2x−7)(x−2). If you try to find what x is, then you would have two different sets now. 2x - 7 and x - 2. Move the number to the other side... 2x = 7 and x = 2. (2x = 7) = (x = 7/2). Simply the equation: (2x−7)(x−2) Solving for x: x = 7/2 or 2
Factor 4x -2x - 12?
As written, that's 2(x - 6) You may have meant 4x^2 - 2x - 12 which would be 2(2x + 3)(x - 2)
How would you factor 4x2-2x-12?
2(2x2 - x - 6) = 2(2x + 3)(x - 2)
What is x x 2x?
2x+x is 3x
What is the sum of 2x plus 1 and x times x minus 2x plus one?
(2x + 1) + (x*x - 2x + 1) = x^2 + 2x - 2x + 1 + 1 = x^2 + 2
Find the derivative of y x2 sin x 2xcos x - 2sin x?
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
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
How do you factor 2x plus 5x plus 2?
If that was 2x^2 + 5x + 2, it would factor to (2x + 1)(x + 2)
What is the answer for this equation 2x-3 equals x plus 2?
2x-3=x+2 2x-x=2+3 x=5
What is x times -2?
X * - 2= - 2X========that simple
