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The line of intersection will be parallel to both planes. We can write the equations of the two planes in 'normal form' as r.(2,1,-1)=4 and r.(3,5,2)=13 respectively. We can then read off the normal vecto...

Consider the case n=1. Then 1(1+1)(2*1+1)/6 = 1 = 1^2 and so the claim is true for n=1. Suppose the claim is true for some positive integer n, so that 1+4+9+...+n^2 = n(n+1)(2n+1)/6. Then by the inductive...

The general method is: 1)write down what needs to be shown (the claim) 2)check it holds for the lowest value of n required (normally n=1 but check question) 3)write down sentence: 'Suppose when n=m the cl...

The first step in tackling this problem is to find the intersection points of the graph with the x and y axes. The y-intercept can be found by substituting x=0 into the function, so y=(0-1)/(0+1)=-1 and t...

Let z be a cube root of 1 such that: z^3 = 1 z^3 - 1 = 0 Factorise: (z-1)(z^2 + z + 1) = 0 Then, z=1, the real root, or: z^2 + z + 1 = 0 with z not equal to 1 Use quadratic equation: z = (-1 +- sqrt(1-4))...