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i think electrical engineering has the hardest maths... that is my first preference at UNSW with a master in biomedical

yea that is what i got :) and with the bending force question, the equation stress=my/I needs I in m^4 units, not mm^4 units, so you just had to convert the units of I and you got the right answer, there wasn't anything wrong with the question

it wasn't too bad, it's like every other year with questions you can do and ones that are unexpected. Just alot of drawing this year

the L : D ratio; tan(theta)=drag/lift, you had drag and the angle..

draw a parabola positioned on the origin, the parabola passes through the point (2,0.4) because at 0.4m high the radius is 2. Then i just let the parabolas equation be x^2=4ay, subbed in the point (2,0.4) to find a (a=2.5) then we want to find what y is when x=1.75 (half the diameter) so just...

is the answer 0.3m?

can they ask for two related texts in module A? for standard english

we had all of these questions on that page in our trial haha, what paper is it?

ohh i get it now, i wasn't sure why w^2+w+1=0 and why it couldn't be w-1=0 but i forgot that w is complex thanks :)

Could someone explain how to work these sort of questions out please? even when i look at the answers i don't understand haha <a href="http://www.codecogs.com/eqnedit.php?latex=if~ \omega ~ is~a~complex~cube~root~of~unity,show~that~ (1@plus;\omega )^3(1@plus;2\omega@plus;2\omega^2)=1"...

Re: MX2 Integration Marathon <a href="http://www.codecogs.com/eqnedit.php?latex=\noindent let \space\ u=lnx \\\therefore du=\frac{1}{x}dx\\ \therefore \int \frac{ln(lnx)}{xlnx}=\int \frac{lnu}{u}du\\ let\space\ I= \int \frac{lnu}{u}du\\ integrating \space\ by \space\ parts:\\ I=(lnu)^2-\int...

ohh okay thanks!

how did you attach the image like that without creating a link?

for ii) i got k:l, not sure if its right though and have you done i)?

Is this what it looks like?

ahh yeap, my bad

pretty sure its 10 multiple choice and 6 questions worth 15 marks each, 100 marks total

couldn't be bothered to write full working out but let me know if you don't get any of it aha

i think this is it for question 4 <a href="http://www.codecogs.com/eqnedit.php?latex=Since \, the\, two\,lines\,are\,parallel\,\\\therefore m(L_2)=m(L_1)\\\therefore y-0=m(x-Q)\\y=m(x-Q)\\ at\,y\,intercept\,x=0\\y=m(0-Q)\\\therefore y=-mQ \\ Hence\,y\,intercept\,of\,L_2\,is (0,-mQ)"...

thankyou ! :)