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Let A,B,C,D be the vertices of the rhombus, named clockwise so that ∠ABC = ∠CDA = 120°. Construct diagonal AC: First we want to find an expression for the area of triangle ABC; A = (1/2)*BC*AB*sin ∠ABC As we are dealing with a rhombus, AB=BC=CD=DA. Let the side length be x...

y = 1/x2 dy/dx = -1/x Therefore this graph is monotonically decreasing for positive values of x, and likewise monotonically increasing for negative values of x. Note that the graph must always be above the x-axis, and that the lim (x-->0-) = +∞ and that the lim (x-->0+) = +∞. We...

Complex numbers and graphs.

A parking orbit is simply a temporary orbit.

d/dx (xe^x) = xe^x + e^x ∫(xe^x + e^x) dx = ∫(xe^x) dx + ∫e^x dx => xe^x = ∫xe^x dx + e^x => ∫xe^x dx = [xe^x - e^x] Remembering this is all considered with the upper and lower bounds omitted for simplification. Finally just evaluate the expression [xe^x - e^x] from x = 2...

Holy shit, well done man ;)

Lol WTF? :lol:

The Cambridge Year 11 Extension Maths textbook has excellent questions on factorising in the extension section.

a^4 + 4b^4 = (a^2 + 2b^2)^2 - 4a^2b^2 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab) <---using the difference of two squares = [(a+b)^2 + b^2][(a-b)^2 + b^2] for a more neat and concise factorisation. You just need to expose yourself to a variety of questions in order to develop your methods and...

http://library.thinkquest.org/20991/geo/circles.html This may help. It is very simple and it provides you with the basics.

Why do you have to make the decision NOW? Why not wait it out, see how you cope with extension 1 maths and then decide what will be best for your studies.

Umm, your maths test was out of 14? :confused: We normally have tests worth at least 60 marks. How long did you have to do this test?

http://en.wikipedia.org/wiki/Polylactic_acid :D

y=x^1/2 y^2=x ...Which is the inverse graph of y=x^2, so in other words just draw y=x^2, but on its side, concave in the direction of the x-axis, with turning pt. at (0,0). As no value of x will produce a negative value for y, simply omit anything below the x-axis and that is your graph...

Bah, of course! I see now, thanks :)

The only way I could think of simplifying i^i: i^i = (cos(pi/2) + isin(pi/2))^i = (e^(ipi/2))^i = e^((i^2)(pi/2)) = e^(-pi/2) Is that the correct answer undalay? Just curious :D

If you can't understand it then maybe you should just wait until you do it in class.

Don't bother with Maths in Focus ...WAY too simple IMO. Doesn't prepare you sufficiently. The Challenge exercises aren't bad, but you'd rather go for something more thorough like Cambridge, which is what I use often at home.

That match was intense, I was sitting on the edge of my seat (literally) for the entire last two sets. Holy shit. :eek: jannny, it's Kohlschreiber, just so you know. :wave:

QFT. That is precisely how I interpreted his post. :wave: