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This can be seen as the same as (x)(x + 5) + (-2)(x + 5). So you are just multiplying the first value in the first brackets with both the values in the second brackets. This is xx + x5 which equa...

Roots = where line touches x-axis. Set equation equal to 0 as y=0 at the x-axis. Now we need to solve for the x values at which y=0. Quadratic is now 0=x^2+5x+6. Factorise quadratic: 0=(x+3)(x+2). Now you...

To solve simulatenous equations there are two main methods, substitution and elimintion. The first method requires the principle where if x is equal to a number, say x=2 then we can substitute this in, fo...

f'(x)= 3x^2+1/3x^(-2/3)

To differentiate you need to multiply the coefficant of the x dependent terms by the powers and then the power of x goes down by one. 

For example: differentiate f(x)...

To maintain the equality everything that is done to one side needs to be done also to the other. First, we square both sides to get rid of the power on the right-hand side. What we get is p^2 = (x+y)/5. N...