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my firefox can display this but IE can't http://www.howtocreate.co.uk/sidehtmlentity.html Did you use IE Xayma?

rectangles? :P

98.60 and no invite haha maybe they dont like me. I think u need to have Usyd in ur first few preferences

eng adv 83 chem 90 physics 95 business studies 92 religion 1U 46 maths X1 97 maths X2 92 SAM 2003: 99.20 2004 UAI: 98.60 SAM counts 2U of chem whilst UAC counts religion and 1U of business.

LOL another bad example from mojako :p y=(x^3+1)/x y=x^2 + 1/x so y=x^2 is what the curve would behave far on the right and left of the coordinate axis, but I'm not really sure if it's called asymptote. when I went to dictionary.com it mentions a "line". Also it defines asymptote as "A...

oblique asymptote is a slanted line to find out if its oblique or horiz: u find the limiting value of y as x approaches infinity if the limiting value is a constant (number) then the asymptote is horizontal if the limiting value is infinity or negative infinity then it's not horizontal...

probability that it's faulty = p = 0.013 probability that it works = q = 0.987 (a) q8 = 90.1% (b) 8C6 p2 q6 = 0.4% (c) P(7 will work) + P(8 will work) = 8 p q7 + q8 = 99.6%

Yes I did realise that. My point is that you can't be completely sure that tywebb is buchanan and can't make it as a conclusion. Do you?

Just remember that simple tasks such as calculating the value of "1 + 1" actually require some maths knowledge and skills.

so you like physics or not?? uni physics has a LOT of maths... maths also has a lot of physics so you still hate physics? then you also hate maths :p

ok I'm getting out I won't help you ever again :P

haha.. who wants to keep deriving that formula everytime maths methods are lame :p

be aware that this "w, w^2, w^3, ..." thing only applies to the roots of 1 (doesn't apply to -1, i and any other number in the complex field)... I'm quite sure of it. Hmm... if I'm not mistaken (and probably I am mistaken) someone once said in this forum that they can't penalise you for using...

I think it's in BIOS (press Del when you start your comp) Just click the options one by one and you'll find it eventually :D ...or you can read the manual

since nobody has answered this and this (http://www.boredofstudies.org/community/showpost.php?p=1054668&postcount=18) I'll attempt it :p Let A_k be the amount still owing after k years. A_0 = 2000 A_1 = A_0*1.06 - 300 (* is for times) A_2 = A_1*1.06 - 300 continuing and rearranging, A_k...

actually that one works too, sorry. its just that I've never done it that way before so I thought it was wrong. z = r cis (@ + k 2pi/n) where @ is any argument for which z = r cis @ is one of the roots. in z = cis (pi + k 2pi/3), @=pi and z = cis pi = -1 is one of the roots in z = cis (pi...

yeap, n is the power so for cube roots n is 3 but the cube roots of -1 is z=cis (pi + k 2pi/3) where k is an integer (the 3 is inside the brackets)

apparently complex numbers are useful in electrical engineering and some other real-world applications. and btw the complex number field doesnt cover all numbers that "exist" (or have been created haha...) I think this file explains it pretty well...

if it is k 2pi then they all correspond to the same angle and the exact same solution, hehe... maybe we're thinking of different things

root3 + i -- converting to mod-arg form: mod = root ([root3]^2 + 1^2) = 2 arg = inverse tan of 1/root3 = pi/6 (btw if it was -root3 + i, the arg is 5pi/6.., just want to make sure you know how to find arg) -- so we have 2 cis pi/6 De Moivre's theorem says: (rcis@)^k = r^k cis(k@) (2...