Q: Would x 2 2x

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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.

As written, that's 2(x - 6) You may have meant 4x^2 - 2x - 12 which would be 2(2x + 3)(x - 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-3=x+2 2x-x=2+3 x=5

X * - 2= - 2X========that simple

Related questions

x^2 x^3? That would be x^5 If you mean x X 2x X 3, then what would be 2x^2+3

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.

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

2(2x2 - x - 6) = 2(2x + 3)(x - 2)

As written, that's 2(x - 6) You may have meant 4x^2 - 2x - 12 which would be 2(2x + 3)(x - 2)

(2x + 1) + (x*x - 2x + 1) = x^2 + 2x - 2x + 1 + 1 = x^2 + 2

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

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+x is 3x

If that was 2x^2 + 5x + 2, it would factor to (2x + 1)(x + 2)

2x-3=x+2 2x-x=2+3 x=5

X * - 2= - 2X========that simple