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2x+x is 3x
Amelie Runte
Wiki User
∙ 11y agox - 2x = -x
Wiki User
∙ 12y agoAssuming you mean
(2x) (-x)
the answer is -2x^2
Wiki User
∙ 11y agox times 2x is 2x squared or 2x2
Wiki User
∙ 15y ago2x+x is 3x
Wiki User
∙ 11y ago4x2
Anonymous
x-2x=x
Anonymous
3x
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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
Why is 2x cubed different from 2x times 2x times 2x?
2X * 2X * 2X = 8X^3 2X^3 is saying 2*( X * X * X) = 2X^3
What is 2x minus x?
2x - x = x
What is the combined term of 2x plus x?
2x+x2x = x+xso 2x+x isx+x+x = 3x
What is X if 2x 4 36?
What is X if 2x + 4 = 36 What is X ? 2x + 4 = 36
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
Why is 2x cubed different from 2x times 2x times 2x?
2X * 2X * 2X = 8X^3 2X^3 is saying 2*( X * X * X) = 2X^3
How do you solve x plus 2x equals 3?
2x + 3 = x 2x + 3 - 3 = x - 3 2x = x - 3 2x - x = x - 3 - x x = -3
Simplify 2x x 2x x 2x?
8x4
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
Factor 2x squared plus 2x plus xy plus y?
4
What is 2x minus x?
2x - x = x
How do you factor out 2x2 -2x?
2x2- 2x = 2x*(x - 1)
2x -6 and x - 43?
2x - 6 = x - 43 2x = x - 37 x = -37
What is the answer to 2x equals X plus 3?
2x = x+3 2x-x = 3 x = 3
What is 7 plus x equals 2x?
x+7 = 2x 2x = x+7 2x-x = 7 x = 7
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
