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Skim Milk>Milk>yogurt>>>>>>Cheese (cheese is mostly fat with some water almost opposite of milk) If you want to eat cheese only eat "cottage cheese"

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.