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that happened to me too :( have u tried any other program? with me I just reduced the printing resolution but it's still larger than the original file. didn't bother trying another tool. I only need PDF conversion once or twice.

well I havent seen the paper... but I can download it, but I prefer to try my deductive skills first :p reading from ur description and how they used cosine rule, it seems that they gave u the chord length, not the arc length. also, it's l=r@ ==> @=l/r, not @=lr but if u used @=lr, your...

do a google search on laurence field. in case you didn't know who he is. if you did, I think it's quite normal for someone like him to expect higher marks.

That doesn't make sense. Given the A formula, A won't be constant as the particle moves. Also this is not exponential growth and decay.

M is the midpoint of PQ x =1/2(2ap+2aq)=...=2 y =1/2(ap^2+aq^2) =1/2a[(p+q)^2-2pq] =2a-apq =2a-ap(2-p) then find dy/dp to find minimum point, which is (1,a) in (p,y) format so y>=a Do you mean geometrical inspection? I needed to do the x=1/2(2ap+2aq)=...=2 thing (algebraic). If...

(1): 3=asin@ (2): -6root3=2acos@ (1)^2: 9=a^2 sin^2@ (2)^2: 108=4a^2 cos^2@, which becomes 27=a^2 cos^2@ (1)^2 + (2)^2: 9 + 27 = a^2 sin^2@ + a^2 cos^2@ 9 + 27 = a^2 (sin^2@ + cos^2@) 9 + 27 = a^2 (1) a=6 since we define a to be positive (well you can use the negative if you want...

why am I getting y>=a ??? :confused: and how does M and the parabola intersect at y=2??

looking at dictionary, it means motion in a straight line motion in 2 dimention... and in the 3U context it assumes there is no air resistance and friction :p in 4U it can have resistance

what?? year 12 and only 14 years old??

well firstly its (m+3)Cr not (m+n)Cr now: (1+x)^m+3=(1+x)^m(1+x)³ general term of LHS = (m+3)Cr x^r general term of RHS = mCk x^k (1+3x+3x²+x³) = mCk x^k + 3x mCk x^k + 3x² mCk...

oh ok. thank you.

we did complex (the whole term 4 :P), graphs, conics, polynomial, integration, volumes, mechanics, harder 3U so yea.. usually its complex then random 4 topics then volumes, mechanics and 3U

ok, I just did part i and iii which I didnt get to do during the exam and u dont need that thing in the pic and in the wolfram site theres less number theory compared with previous years I think.

well probably the trick is not everyone has done partial fraction where it has a linear factor raised to a power of greater than 1. In this Q it's the (x-1)^2 factor. So if the identity is not given not everyone may be able to do it (and btw the syllabus says this case is not to be considered...

yea.. I said x^3 as "example" in the edited post. can you think of a better approach? mine still takes significant work.

A circle just touches the positive x and y axes, and also passes through the point (1,2). (i) Draw a diagram to represent this information (ii) Determine the centre and radius of all such circles For (i), there are 2 possible locations of (1,2) on the circle right?? How do we do (ii)?

The approach I can think of atm: Find the general Tn formula for (x-2)^6 and for (x-1)^6. Then find the term with x^0, with x^1, x^2, x^3. Example with x^3: (x-2)⁶(x-1)⁶ power: 0 3 (so u want x^0 from the (x-2)^6, and x^3 from the other) 1...

if anyone still needs confirmation yes I meant in the state

ooh.. sorryy... its x=0 and x=2a also dont forget the radius is (x+2a) because its rotated about x=-2a am I correct? (hoping so...)

no thats not true there should at least 10 people with >110 in any year and this year there will be more because this year's exam is easier relative to past ones and it's likely that >1 person will get full marks this year