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SC English literacy - 95 Mathematics - 98 Science - 93 Geography - 90 History - 83 HSC English (Advanced) 89 Math 3U 97 Math 4U 92 Physics 91 Business Studies 95 So a large drop in English - which I think is due to the multiple choice present in the SC and not the HSC Computing - 94

Einstein: 99.35 Umax: 99.30 Actual: 99.00 Quite pleased.

Thank you Affinity, that really helped. Yeah, stupid of me not seeing that I had to use (x1, y1) and (x2, y2)...

This is question 20, Exercise 6C. "Find the equations of the four tangents common to the hyperbola x^2 - 2y^2 = 4 and the circle x^2 + y^2 = 1. Find the points of contact of these tangents with the circle." [Hint: Let xx1 + yy1 = 1 be tangent to x^2 + y^2 = 1 at P(x1, y1)] Here's what I tried...

My mistake, thanks for clarifying Trebla. Didn't occur to me for a sec that we consider z to be real when graphing.

@ shaon0, why does 5[(z^2+3a/10)^2+b/5-9a^2/100] >= 5[9a^2/100+b/5-9a^2/100] ? You're suggesting that (z^2 + 3a/10)^2 >= (3a/10)^2 but that is not necessarily true as z^2 may well be negative (since a complex number squared may be smaller than zero).

1/(3x+1) > 1/(2x-3) 1/(3x+1) - 1/(2x-3) > 0 Using common denominator: (-x-4)/[(3x+1)(2x-3)] > 0 Multiply inequality on both sides by denominator squared: -(x+4)(3x+1)(2x-3) > 0 Sketch curve noting the branches above x-axis. This will give the answer of x<-4, -1/3 < x < 3/2

I'm quite annoyed that I didn't recheck my physics exam before it finished... I realised at the last 30 seconds that i had left two short answer questions worth a mark each. Started scribbling but was stopped by "pens down." But otherwise I think I did pretty good.

The number of arrangements = (6!/3!) x (4!/2!) = 1440 Consider this: Group the vowels (I, O, E, E) together as a single entity. Then we can arrange this entity within the group of letters 6! ways. But note that the there are 3 letter 'S' within this larger group, therefore we need to divide by...

Thanks guys I got it. LHS = 1 + cosec^2 (A) tan^2 (C) ______________________ 1 + cosec^2 (B) tan^2(C) = 1 + [1 + cot^2 (A)] tan^2 (C) _________________________ 1 + [1 + cot^2 (B)] tan^2 (C) = 1 + tan^2 (C) + cot^2 (A)...

Yeah I tried but it wouldn't work unfortunately...

Hey, I just can't seem to prove this identity: [1 + cosec^2 (A) tan^2 (C)] [1 + cot^2 (A) sin^2 (C)] ______________________ = ___________________ [1 + cosec^2 (B) tan^2 (C)] [1 + cot^2 (B) sin^2 (C)] It'll be of great help if someone could solve this. Thanks.

For part (iv), you square the equation on both sides and arrive at the same equation as in part (iii). We proved before that the equation only holds if [z - 2]^2 = 3. (using [ ] as modulus symbol) Hence [z - 2]= sqrt 3 The locus of z is circle with radius sqrt.3, centre (2 + 0i). For part (v)...

For the last question, arg[(z-z1)/(z-z2)] = B This is the same as writing arg(z - z1) - arg(z - z2) = B The locus z is the arc of a circle with z1 and z2 as the endpoints, the two subtending an angle of B at the circumference. Note that z =/= z1 or z2.

For the first question, this method is a bit tedious and I'm sure there's an easier way, but here it goes: Let z = x + iy (I'll be using [ ] as the modulus symbol) 3[(x - 1) + iy]^2 = [(x + 1) + iy]^2 Remember that the modulus squared is equal to the complex no. times its conjugate. (I'll...

Thank you scardizzle. It hadn't occured to me to do it your way... nice work (Y).

I'm stuck on question 28 Exercise 5E on the Patel 4 Unit book. "Show that the polynomial x^n +mx - b = 0 has a multiple root provided (m/n)^n + (b/(n-1))^(n-1) = 0. Find this root." I've tried solving it simultaneously but didn't arrive at the relation stated by the book. Any help...

Hi Jess! I think you know who this is :P

Noted. Thanks for clarifying.

Thanks vafa, very clear working. But may I ask, how does (cos pi/7)(cos 2pi/7)(cos 3pi/7) equate to (cos 2pi/7)(cos 4pi/7)(cos 6pi/7) as ninetypercent stated? Now I know that both yield 1/8, but exactly can they be proved to be equivalent (other than just stating that their values are the same)?