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Did you get anything in the realms of 2547a³π/20 units cubed? I banged that up pretty quickly so chances are it's wrong but if not let me know and I'll type up any part of the solution that you're after.

That was the course that I really wanted for while but recently I've been leaning more towards UNSW arts/medicine which means I only need to be in the 99's to remain competitive (heh, "only"...) as aposed to the 99.9's. Setting your sights on med is a good way to make your life more difficult...

Très sneaky.

For this one you express (cosθ + isinθ )⁵ in two ways and then you equate real parts. I'll show you what I mean: Using de moivre's (cosθ + isinθ )⁵ = cos5θ + isin5θ Using binomial expansion (cosθ + isinθ )⁵...

The equation of the curve is y = -3 + 4x - x². Rearranging you get: x² - 4x + (y+3) = 0 ... then treat it like a quadratic with the y acting as a constant as per the example you gave and use the quadratic equation to get: x = 2 ± √(16 -4[y+3])/2 =...

What they're saying is that x1 and x2 are the roots of the equation x^2 - 4x + y = 0. This makes sense since for (almost) every y value you have two solutions for x. Using sum of roots: x1 + x2 = 4 Using product of roots: x1x2 = y

Well, so far I havn't dotted my axes for any complex questions on my school assessment papers and I havn't been marked down for it. As you mentioned most textbooks don't dot them either. Basically I don't think they're expecting you to do that. You could if you wanted to go the extra mile but I...

You should try messaging 'ngai' on this forum. I don't believe the board of studies ever released his raw mark to him but you could get him to estimate what he thinks he got.

Your school must look pretty :)

que? ...........

Here's something to twist your noodle: technically ⁸C₄ includes the mirror cases since out of the 8 people you can pick: {p1, p2, p3, p4} or {p5, p6, p7, p8} If you removed either grouping from the inital 8 you will create the same two groups. Hence it would make...

Do you dot your axes when you graph y=x²?

Tip of the day: Learn to use a calculator :p ⁸C₄/(⁸C₁ + ⁸C₂ + ⁸C₃ + ⁸C₄) = 0.43209 ... = 35/81

Oh and sorry, my reasoning for this is: ⁸C₁ = the number of ways to pick 1 person from 8. In pickign 1 from 8 you establish two groups (of 1 and of 7 respectively). This accounts for the number of 1 and 7 group combinations. ⁸C₂ = The number of...

⁸C₄/(2⁸ - 2) is what I was thinking at first but I got caught up on something. This situation is different to, say, distributing 8 people amongst two distinct rooms since being in one room is different to being in the other. However, in this situation using...

I reckon Gman's solution simplified this: http://community.boredofstudies.org/showthread.php?t=85487 (remembering of course that e^iθ = cosθ + isinθ )

High risk --> high returns. If you want cash go black market and start selling drugs.

Get out while you still can. Drop out of highschool and become a rock guitarist.

Personally I find that questions 1 and 2 aren't the ones which make an exam hard or easy.

- miss - miss ^ spot the difference