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I was doing some questions on euler's formula when i noticed the following eiπ = - 1 => e2iπ = (- 1)2 .... (squaring both sides) => e2iπ = 1 => ln [ e2iπ ] = ln(1) .... (taking natural logs of both sides) .'. 2iπ = 0 ... (since ln(1) = 0 ) Now, how can 2iπ = 0? Since both i and Pi are a...

omfg, I've figured it out, without using the alcalder's LHS = RHS arguement. Proof: dv/dt = a(1 - t/b) v = a(t - t2/2b) x = a(t2/2 - t2/6b) Vmax when acceleration = 0, t = b Vmax = ab/2 Now we find the displacement when the car reaches maximum velocity, which occurs when t = b Therefore: x =...

Re: when did you lose your virginity? on BoredOfStudies yesterday

Why does the US think they're the cops of the world?

wtf is interpretative dance?!

man, that question is ridiculous, i get, 2b = 3(T - T1) EDIT: ==== Damn, screwed up the working out, but i finally got the correct answer: look at my next post below

if you implicity differentiate your hyperbola equation, you should see that a2 and b2 are constants.

x2/a2 - y2/b2 = 1 Tangent Eqn => xx'/a2 - yy'/b2 = 1 ......... (sub P(asec@,btan@) => ab2xsec@ - ba2ytan@ = a2b2 ... (1) Asymptotes are +- bx/a .... (sub y = bx/a) Hence (1) = > ab2xsec@ - ab2xtan@ = a2b2 => ab2x[sec@ - tan@] = a2b2 => x = a/(sec@ - tan@) Similarly...... if y = -bx/a, then...

The easier way would be to prove that P(asec@, btan@) is the midpoint of MN.

This sounds sus. are you sure it wasn't YOU who came after 10 minutes with ur gf?

this is an Inverse Tan Integration I = ∫ 1/(x2 + 4) {x = 0 ; 2} I => arctan(x/2) / 2 {x = 0 ; 2} Therefore I = 1/2 [arctan(1) - arctan(0)] = 1/2 [π/4] Thus I = π/8

Here are some hints: a) y = arcsin(x2) => x2 = sin(y) b) y = xarccos(x) => x = cos(y/x) c) y = arcsin(tan(x)) => tan(x) = sin(y) go from there

Ok, off the top of my head: The Beatles scrupulously moulded an entire new music generation appropriate to the historical and social context at the time. Discuss.

i think you mean: a^(b+c)=a^b * a^c not a^(b+c)=a^c.b^c yea, i'm always used to using the same constant letter throughout, saves me alot of confusion and time ~ as if you introduce a new constant, you'll have to state, again; "where [ ] is a constant" would i get marked down for this or...

2ex+(dy/dx)*(1-ex)tan(y)=0 Rearraging like Iruka said: (dy/dx)*(1-ex)tan(y) = - 2ex Hence (dy/dx)*tan(y) = [ 2ex ] / [ ex - 1 ] ∫tan(y)dy = ∫[ 2ex ] / [ ex - 1 ] dx => - ln[cos(y)] = 2ln[ ex - 1 ] + C ............. Take exp of both sides.. => 1 / cos(y) = C(ex - 1)2 => sec(y) = C(ex - 1)2

Let sin(x) = cos(x) Therefore tan(x) = 1 Hence x = pi/4 A = ∫[cos(x) - sin(x)]dx between x = 0 and x = pi/4 A = [sin(x) + cos(x)] {pi/4 ~ 0} A = (1/Rt(2) + 1/Rt(2)) - (1) Therefore A = Rt(2) - 1 For volume, use whatever method you've learnt, like slicing, or cylindrical shells. etc...

1. There were two classes, both examined on the same question. 2. My class is slightly larger, in a smaller room, hence more cramped. 3. As I stated before, only instructions were "be on time" 4. How can you justify a comparison between an Uni exam, with specific designated testing facilities...

It is highly unlike that we need to utilise the General Solution of Cubic equations anytime soon, as its completely out of the syllabus, and creates more working compared to other methods. however, there is a interesting method here which solves cubic equations if coefficient before x2 is zero...

This week, I had to write an English essay on Hamlet and Ros&Guil Are Dead. and how values are the result of contextual interpretations.... It was like an open book test, and we were allowed 3 pages of notes with us to aid our work~ however, it is best to partially remember some of tbe essay...

pianos dont hurt people, people hurt people ;)