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The first step is to find a common denominator. Since x^2-9 can be expanded to (x-3)(x+3) and 1/(3-x) can be written as -1/(x-3), we can see that all the terms contain either (x-3), (x+3), or both. Theref...

x^2+4x-5=0(x+2)^2-(2)^2-5=0subtract 2^2 as (x+2)^2 = x^2+4x+4 when we want x^2+4x(x+2)^2-4-5=0(x+2)^2-9=0(x+2)^2=9square root both sidesx+2=+/-3x=-2+/-3x=-2+3 or x=-2-3x=1 or x=-5

To calculate the answer to this question we need to know the formula for the area of the circle. The area of a circle is:

A = Pi x r^2

The value of r is given in the question; r = 12. We the...

This was the final question in AQA's 2011 non-calculator paper, and most top candidates will be looking to score all 5 marks that are on offer. Our usual technique for simultaneous equations isn't going t...

x^2 + y^2 =1 y = 3x + 1 x^2 + (3x + 1)^2 = 1 x^2 + 9x^2 + 6x+ 1 = 1 10x^2 + 6x = 0 x(10x+6) = 0 x= 0 or x = -6/10 = -3/5 When x = 0, y = 1 When x = -3/5, y = -4/5