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Yeah it does seem kinda random. However, it's essentially a simplification of the term a^2(e^2 - 1), nothing else. Even without b you would still have the same graph, except you'd notate b as a√(e^2 - 1) Regarding subbing in B(0,b): at this point in your proof you give no indication of...

Wait... those percentiles got 99???? What. Free bump. But what.

Wait... those percentiles got 99???? What. Free bump. But what.

HI5 Same here. >.<

There are only 4 spectral lines within the spectrum of visible light - there are an infinite number of possible energy states but they all produce EM radiation outside the spectrum of visible light.

Hey, at least you guys are getting something done <_<

Well, there is no extreme point denoted by 'b' on the hyperbola, it only applies to circles and ellipses :/ b is basically a function of the eccentricity of the hyperbola - as e > 1, it can reach infinity so your assertion that b is the extreme point of the hyperbola is kinda irrelevant...

Construct lines AE, CD, BC (extend BC to X) Let the intersection of AE and BC be F, < ABC = a, < CBE = b So < ABC = < ACB = a (equal <'s opp equal sides) But < ACB = < AEB = a (<'s subtended by same arc) Considering ∆ BEF, < BFE = 180 - (a + b) (< sum of ∆) So < AFC = 180 - (a + b) (vert opp...

Why did North Korea get disqualified?

Tafe

You don't derive b^2 = a^2 (e^2 - 1), you postulate it :/

Shame China beat them :( Nice work to Aaron Chong

Solids and liquids cannot be used for equilibrium calculations - in an aqueous solution the concentration of water is assumed constant regardless ([H2O] = 1)

For the locus bit: Let C be midpoint of PS, ie. C (ap, a/2(p^2 + 1)) ie. x = ap, y = a/2(p^2 + 1) ie. p = x/a ie. y = a/2((x/a)^2 + 1) 2y = x^2/a + a x^2 = 2ay - a^2

Whoops haha it simplifies to that. The first thing you have to do is get the parametric function: x = cosø dx/dt = -(dø/dt)sinø dx/dt = -(dø/dt)√(1-x^2) Then you have to work out dø/dt, which is angular velocity. Given tangential velocity = 2 m/s Period T = (2π)/2 = π But you know angular...

For the second part, the centre C of the circle is the midpoint M of P and S (circle geometry) - you can derive the locus from there.

Is the answer to part one -2sin(arccosx)?

1 is just a constant substituted in place of the focal length (a)?

The slightly more challenging SGS papers usually crop up in 3U/4U, not 2U. Sorry bud

:L means he's laughing bru