Search results

120

I cant read the handwriting and I dont know the question, but I assume you would if you get the right answer and its not one of those "prove volume = something" questions you'd get full marks. Does anyone know if you need the preamble to do with \delta x etc? Like the limiting sum thing...

yeah I initially forgot to end with [ /tex]

I go to sydney grammar Suppose it had a repeated root w. Then taking the derivative, the following two are true: w^n - w^{n-1} = 1 nw^{n-1} - (n-1)w^{n-2} = 0 Note eqn (1) implies w is nonzero. From that we can divide by w^{n-2} in (2) to obtain: nw - (n-1) = 0 w = \frac{n-1}{n} < 1 Now...

w*w^2*w^3*...*w^{5k-1} = w^{1+2+...+5k-1} = w^{\frac{5k(5k-1)}{2}} = (w^5)^{\frac{k(5k-1)}{2}} = 1 and w+w^2+w^3+...+w^{5k-1} = \frac{w(w^{5k-1}-1)}{w-1} = \frac{w(\frac{w^{5k}}{w}-1)}{w-1} = \frac{w(\frac{1}{w}-1)}{w-1} = -1 So combining the answer is 2. Now try this question, it's from my...

This thing has a massive problem with rigour towards the end of the sum. (The reasoning, take n->infinity and then just let the 1+k/n term be zero doesnt work when the k youre dealing with is really big). It's probably enough for HSC though. Also theres a minus sign wrong , should be 1 - (k-1)/n...

You can look at this geometrically as well. If O is the origin, A represents z, and B represents z+1, and C represents 1, then because z has unit modulus (implied by z = cistheta), and because then AB has length 1 (try tail-to-tip addition or whatever) then OAB is isosceles. Another way of...

You're missing a solution that you'd obtain from the 3(y+5) = -4(x-2) case (i.e. 4x + 3y + 7 = 0). Also you got the arithmetic wrong, -8-15= -23 :P

You dont need all that stuff listed earlier, you can do this question almost straight out using the AM-GM for two variables. \noindent (1-a)(1-b)(1-c) = (b+c)(a+c)(a+b) \ge (2\sqrt{bc})(2\sqrt{ac})(2\sqrt{ab}) = 8abc The first step is just using the a+b+c=1 condition, whereas the second step...

Yep that's right.

rawr try this Find the integral (dx) from 0 to 1 of [ln(1+x)]/(1+x^2) (sorry I dunno how to use latex)

Is this question from a past SGS trial? It looks very familiar.

It's not a paradox, it's probability zero. The problem is that "probability zero" and "never occurs" are not always equivalent statements when dealing with infinite sets. Similarly "probability 1" actually means "almost always occurs" rather than "always occurs", see...

If the points are A, B, C; find the equations of the perpendicular bisectors of AB and AC (not that hard with either the distance formula or just find midpoint AB and -1/gradient AB, same for AC) and then find the point where they intersect. Call this point O, its the centre of your circle...

The vast majority of HSC geometry problems are either special cases or direct proofs of theorems not in the syllabus; compare last year's q7 from the 4u paper with http://en.wikipedia.org/wiki/Ptolemy's_theorem. Clearly the question is no more or less stupid as a result.

In what sense?

The first question is a direct consequence of http://en.wikipedia.org/wiki/Desargues'_theorem The second question is a well known result, http://en.wikipedia.org/wiki/Radical_axis. To do part (iv): Let AB and CD intersect at X; it remains to prove X lies on EF. From part (iii), the following...