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The topic's look good enough but I am interested as to how hard they will be and the depth which they will be examined in.

I never implied it did, I said it makes the repressive nature of the institution more disgusting. When reading an argument it helps to finish reading the said argument before drawing a conclusion on its meaning.

If using the distance from the foci definition of the ellipse or hyperbola and the eccentricity is 1, the locus is the interval of a straight line and not a circle!

Go use your sarcasm on some pre-teens who will appreciate it, this is a discussion.

Schizophrenic's, people with Bi-polar disorder and depressed people should not be allowed to bear arms for obvious reasons, but where does that stop? Really, weapons have only ever increased the volatility of a culture and if the only reason to bear arms is to defend oneself against someone else...

Well, as far as I am concerned the evidence for the existence of a christian god is as strong as the evidence for the existence of Odin. With that said I plan to live my life as an Athiest because the "godless" universe is the one that I have experienced and until I experience anything else that...

So what I am curious about is whether Priests ever molest adults, or attempt to seduce them. This could shed some light on whether the desire stems from a repressed sexuality or something else. If adults aren't molested or abused then I really think that the religious institutions are...

Re: 2012 HSC MX2 Marathon Okay. Question 1: The ellipse (x^2)/2 +y^2 = 1 is intercepted at two distinct points P(x1,y1) and Q(x2,y2) by the line y=mx+3. Show that points x1 and x2 are the solutions to the equation (1+2m^2)x^2 +12mx +16 =0 Question 2: Find the values of m for which x1 and x2...

Re: 2012 HSC MX2 Marathon Question 1: ß and ∂ are complex roots of the equation x^3 + 5x + 1=0. Show that the other root is -1/|ß|^2 Question 2: Show that ß∂ is a root of the equation x^3 -5x^2 -1=0.

Im not sure, the typical way is seemingly drawn out. I think there must be a easier way but i cant quite think of it, other than finding the equations of RS and QS' and subbing in the given x value of k to see you if get the given y value.

Okay, I hope I dont sound stupid but I dont understand how cabc takes the form of root/product of two other roots. Am i being incredibly dumb? edit: yes i was, misread the question. Im such a dumbass.

The moderators do? and I don't know how to approach this but it reminds me of this comic

for f(x)<0 there are no values of ln(f(x)). for f(x)=0 there is a vertical asymptote. as f(x)->0 ln(f(x))-> -inf

By the way you phrased the question, i think it sounds like R and M are the same point. Is the Question right?

Prime numbers have to be greater than 1. So i think the answer is n=-1 and -3. The sign on n is not important to determine whether q(n) is prime. Just as P(x)=3+x is prime for x=-1 even though -1 is negative.

But what if we define a series of terms such that a term is equal to 1 raised to the power of the previous term? The value of the first term is one

Re: HSC 2012 Marathon :) smallest, lowest length.

Look if your good you will get a good mark. Now go study!

he means (x+iy)^5=(x-iy) find x+iy.