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^ that

The reasoning is not so striaght forward but bear with me. Now arg(z-5i)=-theta and arg(z+5i) = phi. subbing this into locus gives phi + theta = pi/4. however this statement is only true in the sense of subtracting the arguments of the vectors. Whereas in plane geometry it is obvious that both...

Math has various limits, Obviously we are infinite limits That work at all hours

Nah I love paint

I would be good if I knew who this was

Re: 2012 HSC MX1 Marathon evaluating the area gives: \int_{-1}{1} \frac{1}{x}dx =\int_{0}{1} \frac{1}{x}dx + \int_{-1}_{0} \frac{1}{x}dx \\ \\ = ln|x| |^{1}_{0} + ln|x| |^{0}_{-1} = ln(1)-ln(1)+2ln(0)=\inf this is true as \int_{0}{1}\frac{1}{x}dx diverges to infinity, therefore it does...

Re: 2012 HSC MX2 Marathon cause this is a proof question i started with LHS: \!\!\!\!\!\!\!\!\!\!\!\!\!\! LHS= 3|z-1|^{2} \\ = 3(x-1)^{2} + 3y^{2} \\ = 3(x^{2}-2x+1) + 3y^{2} \\ = 3((x-2)^{2} +2x -3) +3y^{2} \\ = (x^{2}-4x+4 +6x -9)+y^{2} +2((x-2)^{2}+y^{2}) \\ = x^{2}+2x -5 +y^{2} +6 \\...

yes the angle you marked is theta, but theta represents a pos value and clearly the arg is neg

just do 3(eqn1)+(eqn2) and (eqn3)-2(eqn1) to eliminate y and then solve the two new eqns simulatenously to find x and z

I have been researching which topics are apart of the discrete course of math 2969 and found from 2006 these and was wondering if it was taught like this in 2010/11 in this order: Recursion, induction and generating functions Asymptotics and analysis of algorithms Number theory Combinatorics...

on the LHS of the plane arg(z+5i)-arg(z-5i) = pi/4, try that if you havent already

i would have to say if it is bounded by axis then yes

The proof for 3pi/4 is:

EDIT: in this case angle between two vectors isnt pi/4 i worked it out to be 3pi/4 which changes the answer

This

To make proof more concrete you shouldn't just conclude r is a factor a_{0} , you should add: Since we have found that a_{0} is the product of r and another number, say B, where B=-(r^{n-1} +a_{n-1}r^{n-2}+...+a_{1}) r must be a factor of a_{0} then add QED or little square or just even...

Whenever it says to prove a line touches a curve it means prove it is a tangent and in general the easiest and most straight forward method is to sub the straight line into the curve and form a quadratic and then show the discriminant equals 0 as this proves the line and curve only intersect at...

You don't have to expand it, jus factorise immediately As difference and sum of cubes instead

i cant be bothered typing it so long, but for 2 just apply difference of cubes on first two terms and sum of two cubes on second two terms and after that complete the square and keep factorising