Q: Is 2 X 2 4

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(x^2-4) (x^2+6x+8) __________ x _________ which is (x^2+2x-8) (x^2+4x+4) (x+2)(x-2) (x^2+4x+2x+8) ___________ x _____________ which is (x^2+4x-2x-8) (x^2+2x+2x+4) (x+2)(x-2) x(x+4)+2(x+4) ___________ x ____________ which is x(x+4)-2(x+4) x(x+2)+2(x+2) (x+2)(x-2) (x+2)(x+4) ________ x _________ which is 1 as all the terms in the num & den cancel (x-2)(x+4) (x+2)(x+2)

x^(2-x-6)/x^(2-4) x^(-4-x)/x^-2(x^-2)*(x^(-2-6))= x^(-2-x)

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x/(x + 4) = 4/(x + 4) + 2 (x - 4)/(x + 4) = 2 x - 4 = 2x + 8 -x = 12 x = -12

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x-2(x)+4/x^2 -4=x-2x+4/x^2 -4=-x-4+4/x^2

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(-9x^2/√x) + 4= [-9x^2/x^(1/2)] + 4= (-9x^2)[x^(-1/2)] + 4= -9x^[2 + (-1/2)] + 4= -9x^(2 - 1/2) + 4= -9x^(3/2) + 4= -9√x^3 + 4= -9√[(x^2)(x)] + 4= -9x√x + 4Or,(-9x^2/√x) + 4= [(-9x^2)(√x)/(√x)(√x)] + 4= [(-9x^2)(√x)/√x^2] + 4= [-9(x)(x)(√x)/x] + 4 simplify x= -9x√x + 4

x^2/x +2=4/x +2 (x^2/x)-(4/x)=0 (x^2-4)/x=0 x^2-4=0 (x+2)(x-2)=0 x=±2

x+4=2 x=2-4 (taking 4 to other side ) x=-2 ------------------------------------------------------------------------------------------------------------------ substituting x=-2 x+4=2 -2+4=2 2=2

(x^2-4) (x^2+6x+8) __________ x _________ which is (x^2+2x-8) (x^2+4x+4) (x+2)(x-2) (x^2+4x+2x+8) ___________ x _____________ which is (x^2+4x-2x-8) (x^2+2x+2x+4) (x+2)(x-2) x(x+4)+2(x+4) ___________ x ____________ which is x(x+4)-2(x+4) x(x+2)+2(x+2) (x+2)(x-2) (x+2)(x+4) ________ x _________ which is 1 as all the terms in the num & den cancel (x-2)(x+4) (x+2)(x+2)

2 x 8 4 x 4 2 x 2 x 4 2 x 2 x 2 x 2

x/4 - 4 = -2 Therefore, x/4 = -2 + 4 = 2 Therefore, x = 2 x 4 x = 8

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2 x 2 x 2 x 2 2 x 2 x 4 2 x 8 4 x 4

x^(2-x-6)/x^(2-4) x^(-4-x)/x^-2(x^-2)*(x^(-2-6))= x^(-2-x)

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