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YES!!! hehehe...

Oh yeah woops...the answers well two so far are wrong..hehe.. Well i dunno if theyre are multiple solutions.. but i dont see how their answer is obtainable...

Could somebody check my answer to this question (i again got a differnet answer). I=∫cos2xcos4xdx =∫cos2x(1-2sin22x)dx Let u = sin2x du/dx = 2cos2x du/2 = cos2x .:. I = (1/2)∫(1-2u2)du = (1/2)(u - 2u3/3) + c = (1/2)sin2x - (2/6)sin32x + c

yeah see thats what i got...thanks...hehe

Can someone actually do the question.. I used subsitution of u = cos2x, and got the answer i put (yeah i meant c) and then i used subs. of cos x as trev suggested...and i get the same answer. Can somebody else try it..thanks..

Can somebody please help me with this question. ∫sinxcos2xdx. Its probably very easy but i just cant get the answer in the back of the book.. i get (-2/3)cos3x + cosx + x, where the answer in the back is (-1/6)cos3x + (1/2)cosx + c. Thanks.

man i hate it when that happens!! makes me feel so stupid..

I dont really think these questions were that hard, but the conditions would probably be different under exam conditions especially with the strict time limit. Here are my solutions if anyone wants/needs them..

You could more simply (and probably cheaper) buy a microphone for you computer. I knew a shop that solds headsets with alright quality for like $7, so im sure you might be able to find one around...

These questions are pretty hard actually... im considering to not actually do it as i might just make a fool of myself. Heres some of the questions from past years: Problem 1. A disc of radius 1 unit is cut into quadrants (identical quaters), and the quadrants are place in a square of...

fuuuuuuuuckk... man were still on conics!!!!! We have yet to do integration, volumes and mechanics. People say integration is easy..which i hope cuz then ill be able to do it myself. How hard is volumes? Cuz i wanna go ahead and learn it myself... but i dont really see a point, cuz if my...

2xy - 3x - y - 2 = 0 2x(y - (3/2)) - y = 2 2x(y - (3/2)) - y + (3/2) = (7/2) 2x(y - (3/2)) -(y - (3/2)) = (7/2) (2x - 1)(y - (3/2)) = (7/2) (x - (1/2))(y - (3/2)) = (7/4) :) From what i learnt from someone else on the forum, to show its a rectangular hyperbola, you can show that the...

but the parametric form of (x-3)2/sup]/64 - (y + 1)[sup]2/ 36 = 1 is x = 8sec@ + 3, y = 6tan@ - 1, so by selecting these parametric equations arent you already using the definition of a hyperbola to show its a hyperbola? I was thinking to show it's eccentricity was greater than one.. but that...

How bout the second part.. i thought to show that an equation was a hyperbola, it could be written in the form x2/a2 - y2/b2 = 1.. but if you see the first question is already in this form :| so im confused...

Well for the first question (rect. hyperbolae), x2 - y2 = 12 2xy - 3x - y - 2 = 0 etc... and for the other question (just hyperbolae) (x -3)2/64 - (y + 1)2 = 1 etc..

These questions are weird! Sketch the following and show that they represent rectangular hyperbolas. and the next one, Show that the following equationjs represent hyperbolas and find their centre, foci and the equation of their directrices and asymptotes. How do you show an equation...

I remember during year 10, our school gave us a presentation on all the HSC english subjects. They had a chart which showed the highest UAI gained from students with ESL english, and on the chart it was 100. So you can definately get 95+.

woah..haha i didnt even think of subbing in the value of the point (1,3) into the equation y = mx + k :| i dunno whats wrong with me today... thanks!

A question im stuck on: THe first part is: Show y = mx + k is a tangent to the hyperbola (x2/a2) - (y2/b2) = 1, then m2a2 - b2 = k2. I can do that one but the second part follows on from this, Hence find the equations of the tangents from the point (1,3) to the hyperbola (x2/4) -...

Not really... the only information i know i got whilst walking out of the building for recess, so i didnt get to talk for much. I did however find a website (as i was looking for past questions), but its not very good... Here it is anyway...