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Well i got 97.8 last year, but the new Mx1 score and the Mx2 will add on, so i dk.

Ahh ofcourse, thanks for the clarification Treb Yeah, that was my Q, the original Q couldn't have been done using 3U (since no substitution given is 4U, except for this Q: int. sin^2@cos@d@). But as you can see the Q was written incorrect, solutions provided by Unthouchablecuz and Nomad would...

Say you get: MX2: 98 MX1: 99 ESL: 92 Physics: 95 Chemistry: 97 Gee that's gotta be close to 99.95 bro + I didn't really love the Mx2 paper either, i don't like when they spread the difficulty throughout the paper, like q8, apart from probability Q was simple. I prefer when papers get...

Well, i suppose i'm only judging off your maths. Since i read you think 83/84 in MX1 and about 110 in MX2, i mean thats 100 or 99 in Mx1 and 98 (high end)-99 in MX2. How do you think you went at your other subkjects? I think i got 78/84 in MX1 (just one mark stuff dropped here and there) and...

Ahh ok man :) lol Goodluck with marks btw, as you saw on my 99.95 list, i expect big things lol

Wolfram Mathematica Online Integrator How did they do that? Is it something simple that i'm missing?

The kid thought that rt(-1) was a "false concept" lol, i wasn't going to start on about how x,y E R but R is a set of C etc, wasn't required.

yeah, i know, i was just trying to give the kid a quick idea, didn't want to go into details about systems etc

i think it's (3x-2)[(2x-3)^-1/2] that would make sense, but don't think it would be 3U Either of your above is pretty ugly to solve, especially i) lol

a) Let x=asinu [same thing, jsut rearranged] dx/du=acosu ---> dx=acosu du Change limits: when x=b, u=sin^-1(b/a) x=a, a=asinu therefore u=pi/2 int. (sin^-1(b/a)-->pi/2) rt (a^2-a^2sin^2@) . acosu du int. (sin^-1(b/a)-->pi/2) rt [a^2(cos^2@] . acosu du int. (sin^-1(b/a)-->pi/2)...

i don't know if this is serious or not, it's a true concept. We let i=rt(-1) [such that i^2=-1], which is called an 'imaginary number'. Sometimes numbers have both real and imaginary components, these are called 'Complex numbers'. They are in the form z=x+iy. It's a concept dealt that's dealt...

Are you sure that's right? I can't get it using only 3U knowledge

z1 = 2 cis (5pi/6) and z2 = 3cis (2pi/3). z1z2=6cis(5pi/6+2pi/3)=6cis(3pi/2) b) it just wants it in a,b form (ie. not mod-arg) so you just gotta convert the above 6[cos(3pi/2)+isin(3pi/2)] since in quad 3, cos 3pi/2=-cos(pi+@)=-cos(pi/2)=0 sin3pi/2=-sin(pi/2)=-1 therefore...

lol 10mins earlier ay, that must be killing you, such a change

During school: went to bed at about 11, work up at 7. Now: Bed at 12-1am, up at 8

Just let a, b, and c be the roots of the equation therefore a+b+c=-b/a ab+ac+bc=c/a abc=-d/a then resubstitute for x, doesn't require much complex knowledge at all if you attack it like that. Complex number way: x=cos^2x, Let 4@= 32x^3-48^2+18x-1 which is real (cos@)...

lol ofcourse, he's undoubtedly the best maths 09er on this site. Heres the link: - http://community.boredofstudies.org/members/redfield/ - http://community.boredofstudies.org/members/jacob1991/

we heard nothing, ssssssssshhhhhhhhhhh

I'm like 100% Redfield and Jacob1991 will get 99.95. Look at Redfield's sig, hes got what i consider similar marks to Laurie Fields, + in the 3U exam thoughts he said "Got it all out. My marks are the same as Terry Lee, so here's hope, full marks", so 99 or 100.

Here are few names off the top of my head: 99.95: Redfield, Studyfreak, Jacob 1991, Azeaurus88 (dnt think spelling is right), 99+(no order): lolokay, Namu, Tim Sui, Aerath, Untouchablecuz, Jetblack2007, Gurmies, Clintmyster, Annabackwards, GUSSSSSSS, I'll add more as i think of them