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(x + 1)(x + 2) = x^2 + 2x + x + 2 = x^2 + 3x+2therefore(x + 1)(x + 2)(x+3)= (x^2+3x+2)(x+3) = x^3 + 3x^2 + 2x + 3x^2 + 9x + 6 = x^3 + 6x^2+ 11x + 6where a=1 b=6 c=11 and d=9expansion of brackets can be do...

7xy+3y2=10x 3y2=10-7xy 3y2=x(10-7y) 3y2/10-7y=x

solve 64^2/3 ... note X^2/3 = X^{2 x 1/3} therefore 64^2/3= (64²)^1/3= (4096) ^1/3= 16

Find the value of 9^-1/2. For this question we need to know indices rules. We apply two rules here as the power is negative and because it's a fraction. First, let's tackle the negative power ru...

We know the presence of the quadratic means there must be two solutions for x. We can solve this problem by treating the formulas for the graphs as simultaneous equations. We can substitute in the formula...