Search results

Re: Re: Re: Q. circle geometry ahh my mistake, i've edited my post now... with the second, it's a max-min problem, so you differentiate and find when dy/dc = 0 to find turning pt. it's clearly the max there.

Q1: if you draw the nth fraction of the circle out: length of chord = 2r * sin(pi/n) length of arc = r * 2pi/n thus p = {nsin(pi/n)} / pi area of triangle = (1/2)r^2 * sin(2pi/n) area of sector = r^2 * (pi/n) so a = {nsin(2pi/n)} / 2pi Q2: Using sine rule PK/ sinc = KL / sinb PK...

*** exclude all soluble ions. i.e. the ions that are soluble in both before and after reaction (i) potassium oxide and sulfuric acid exclude K and SO4_2- because they are soluble in both cases. 2H+ + O_2- -> H2O(l) here I highly suspect K2O is in solid form, because oxides exist in...

:rolleyes: dun be so hard on urself :D

just a note - you CAN'T work out the charge on polyatomic anions by using the periodic table. This is because atoms like N and S have multiple oxidation states. e.g. you can't explain just using the periodic table why NO2- and NO3- both exist, as with SO3_2- and SO4_2-, using abdoooo's method...

then how're u learnin differentiation 'a couple of yrs ago'?

well that's wot i heard from the guy who used to write the 4u papers (unless my memory's goin bonkers again)

just a note - the top 3u guy was practically the best result in the 3u paper out of all the people who didn't do 4u as well. so like bout a dozen 4u ppl beat that guy in 3u but weren't eligible for it.

no there isn't remembering them off by heart is the best option. this is all i remember you need: strong acids: HNO3, HCl, H2SO4 strong bases: NaOH, KOH all others are weak. diprotic: H2SO4 triprotic: H3PO4, citric acid all others are monoprotic.

Not quite the same the other question is for roots (1-a)/a, (1-b)/b, (1-c)/c = 1/a - 1, 1/b - 1, 1/c - 1 this basically boils down to doing it two steps 1) finding the polynomial roots 1/a, 1/b, 1/c 2) from that, shift the polynomial to the right by one unit. i.e. find the polynomial...

yes, on the ordinary school level asians usually do well because they can work for their mark. just an interesting thing: i found that once you look at the people at the absolute top in the state, nation, etc. you'd find that there's actually a normal balance between asians and skips. What I...

The long explanation: 1. Low electronegativity: the metal-oxygen bond is ionic. The oxide acts as a O^2- ion. This reacts with water (hydrates) to form hydroxide, which then dissociates: e.g. Na2O(s) + H2O(l) -> 2Na+ + 2OH- 2. High electronegativity: the non-metal-oxygen bond is...

shaynO, which year's yr12 were u from?

if i were to get this question in an exam, i'd seriously skip it

Use implicit differentiation (because you really really want to log the thing) y = x^x lny = xlnx diff. r.t.x: (1/y)dy/dx = (x*1/x + 1*lnx) dy/dx = y(1+lnx) = x^x(1+lnx)

The method of working backwards: Assume the cubic with roots a+1/a, b+1/b, c+1/c has the equation Ax^3 + Bx^2 + Cx + D = 0 Now let x = y + 1/y (note that there are two values of y: if x=a+1/a, y=a or y=1/a both work) Hence the expression: A(y + 1/y)^3 + B(y + 1/y)^2 + C(y + 1/y) + D =...

3rd Q: rewrite RHS to = sin({2n+1}@/2) - sin(@/2) / 2sin(@/2) from sums-to-products this = [2sin(n@/2)cos({n+1}@/2)] / 2sin(@/2) = sin(n@/2)cos({n+1}@/2) / sin(@/2) Case n=1 LHS = cos@ RHS = sin(n@/2)cos({n+1}@/2) / sin(@/2) = sin(@/2)cos(2@/2)/sin(@/2) = cos@ = LHS Suppose...

Just a little thought for the 2nd question. The number 8888....88 (n digits) = 8 * 1111....11 (n digits) and the number 1111....11 (n digits) is a sum of a geometric progression 1 + 10 + 100 + 1000 + ... + 10^(n-1)

I'm only gonna do the first one today Well we know 1 + 2 + 3 + ... + n = n(n+1)/2 So RHS = 4 * {n(n+1)/2}^3 = (1/2) * n^3 * (n+1)^3 Now RTP: LHS = RHS for all n n=1 LHS = 1 + 3 = 4 RHS = (1/2) * 1^3 * (1+1)^3 = 4 LHS = RHS Assume result true n=k i.e. LHS(k) = (1/2) * k^3 *...

Technique1: Find the polynomial with roots a^2, b^2 and c^2 Then you can find (a^2)^2 + (b^2)^2 + (c^2)^2 using the normal way of finding the sum-of-squares method. Technique 2: Be an absolute freak and remember by heart that: a^4 + b^4 + c^4 = (a^3 + b^3 + c^3)(a+b+c) - (a^2 + b^2 +...