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i suspect i may have answered a slightly different problem since my pidgeons are perhaps indistinguishable and the people are not. i'm going back to sleep :P

Maybe I'm missing something but you are simply solving an equation of the form: x1 + x2 + x3 + ... + xr = n ie. placing n pidgeons into r holes (d/w im not gonna use pidgeon hole principle, its just an analogy :P) now since we have the condition that xi > 0 we solve: y1 + y2 +...

That actually leads to a 'cute' result. You can easily see that the integral must be positive. Thus, 22/7 - pi > 0 22/7 > pi pi != 22/7 It's cute in the sense that 22/7 is a 'well-known' approximation for pi.

I've never been myself so I can't comment on how many students are actually in each lecture but it is not too hard to see how such a low price could be viable. The bigger lecture theatres hold 200 students. Multiply that by 20 kids and you've got $4000. I imagine hiring a room / lecturer...

I guess this is open to a certain amount of interpretation. Personally, I would suggest that you are both wrong. ^_^ I will explain what he means though. A handy definition: The x-coordinate of a point in a two dimensional coordinate system. In other words, the distance along...

Could you please confirm that you typed the question correctly. It would be a little odd for an interval, PN, to be the abscissa of a point, P. Well... more impossible than odd.

In the fourth year of a Commerce/Science degree at UNSW which should be the last year but due to work commitments (a tutoring college of all places) I'm doing part-time this year and will be completing my degree in 2007 instead.

I didn't realise I would spark such discussion! I was only looking for reference! Anyone who would actually consider using this method is out of their mind. :P It's cumbersome and unknown. The giant, very loud klaxons are already sounding in my head at the thought of anyone using...

If I recall correctly, the question is equally as interesting (and perhaps just as difficult) if you instead create the constraint that the circle must pass through the origin. From memory, in this case, the locus of 1/z will actually be a straight line. Disclaimer: I may not recall...

Found the document I was looking for. http://www.users.bigpond.net.au/gotmaths/ Link #3

You're all pretty much correct. But I'm still fairly sure there is a "name" for this method (I tried researching under the name Riviet suggested) and I know there was a document here at some stage about it. If you ask me, the method is uncommon and should be avoided. Most teachers would...

I recall someone posting a link to a document about a different method of integration but I have forgotten the 'name' of this method. I was hoping someone would be able to provide a link or some information. An example of how it works: Integral of tan x . sec^2 x dx = Integral of tan x...

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I'll assume you are talking about for an ellipse. |PS| + |PS'| = e|PD| + e|PD'| = e(a/e - x) + e(a/e + x) = 2a And you're done. We can say that |PS| + |PS'| = e|PD| + e|PD'| due to the focus-directrix definition of an ellipse, where D is the point on the directrix. It can also be...

in fact your given information: -------------------------------------------------------------------------------- let z=a+bi where a^2+b^2 is not equal to 0 -------------------------------------------------------------------------------- all that information specifies is that you have...

i think your question is broken partner i could set a = 2 and b = 0, and a² + b² would be greater than 0

Some of these assessments are quite interesting. Some of the questions are quite common. Some have been ripped straight from HSC papers. In response to Riviet's assessment: You could do Question 3 using sum of a geometric series btw and it is perhaps more intuitive to think of it...

p(z) = z³ + (1 - i)z² + (1 - i)z - i = (z - i)(z² + z + 1) by inspection You can now factorise that last quadratic yourself. Alternatively, if you aren't comfortable with 'by inspection' you could: 1. Use long division 2. Rearrange the polynomial...

Remember how there is no formula for the sum of two squares?! Make it into a difference of two squares! :) z⁴ + 64 = 0 (z²)² - (8i)² = 0 (z² - 8i)(z² + 8i) = 0 From here, there are a few ways you could go, but I'll...

I suspect people get confused because there are several colleges with variations of the name 'Intuition' that are completely unrelated. I know of at least three such colleges in NSW. The one that is being advertised in this thread is the one that is in Chatswood. There is also one in...