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yeh, just getting the graph right solves half the problem! ;)

I sure do! ;)

I just read over it, there are so many! I fixed them up by the way. Sorry for that, it must have made it really hard to understand :S I had to write it really really fast because i had to go to dinner :)

Oh, yeh, sorry. The last part is just AZ = BZ\,\,$(corresponding sides of congruent triangles)$

I will just do this one quick, if you dont understand im happy to elaborate. $Construct the radii PO and QO$ AO = BO $(equal radii)$ PO = QO $(equal radii)$ \therefore \frac{AO}{PO} = \frac{BO}{QO} \angle AOB = \angle POQ\,\,$(common angle)$ \therefore \triangle AOB \parallel \triangle...

I will assume that it is: $We have:$ \triangle OAC \cong \triangle OBC \therefore \angle AOZ = \angle BOZ\,\, $(corresponding angles in congruent triangles)$ AO = BO\,\,$(equal radii)$ $OZ is common to both $By SAS,$ \triangle AOZ \cong \triangle BOZ \therefore \angle AZP = \angle...

Also, maybe, look at the values of the trigonometric ratios and how they occur. That is, trigonometric tables (for example, the values of sine as the angle increases). There is also alot of information regarding the History of Trigonometry.

Is Z where AB intersects with OC?

When you draw the graph, it becomes immediately obvious. But the line y = x+1 is only greater than y= (x^2/4)-2 between -2 and 6, for the rest, the quadratic is greater than y= x+1.

$You have to find where f(x) and g(x) intersect, drawing the graphs would also help$ $You would find that the only area that is bounded by the three curves is between 1 and 3$ $Between 1 and 3, f(x) is greater than g(x)$ \therefore A = \int_{1}^{3} \left({x^2 - \frac{1}{x^2} \right) dx =...

$Solve these two graphs simultaneously to determine where they intersect$ x+1 = \frac{x^2}{4} - 2 \frac{x^2}{4} -x -3 = 0 x^2 -4x -12 = 0 (x-6)(x+2) = 0 \therefore x = 6,-2 Between -2 and 6, the line is greater than the quadratic <x><x> \therefore A = \int_{-2}^{6} \left( (x+1) -...

Re: 回复: Re: What mark are you expecting in 4 unit? I know sen couldn't have accelerated (because he came from Ruse and they don't accelerate students in maths), not to say that he was not at that level. And out of the 6 people who when to IMO 2008, 4 didnt come from Ruse, and of those four...

T_{n} = a + (n-1)d \therefore T_{n} = 2.9 + (n-1)\times 0.33 $n is the year, so$\,\,n=1\,\,$is the first year, etc$ $So in the$\,\, n^{th}\,\,$year, the diameter is$\,\, 19.4 \therefore 19.4 = 2.9 + (n-1)\times 0.33 $Solving yeilds: n = 51 $The age of the tree when it was 19.4cm was 51...

I also got 1331 th edit: OMG! yet again! My brain must be turned off :{

I got 63.7 also. edit: I think i made the same mistake as you (???). I put 10/5 instead of 10/2. Our brains must have done the 5 calculation with out thinking ;P

\frac{du}{dx} = 1 + \frac{1}{x^2} \therefore dx = \frac{du}{1+\frac{1}{x^2}} \int \frac{1+x^2}{1+x^4}\,\, dx = \int \frac{ \frac{1+x^2}{1+x^4}}{ 1+\frac{1}{x^2} }\,\, du = \int \frac{1+x^2}{1+x^4} \times \frac{x^2}{1+x^2}\,\, du = \int \frac{x^2}{1+x^4}\,\, du u^2 = x^2 + \frac{1}{x^2}...

LOL, i completely forgot ;P

$The profit will be$\,\,$(x - 2) $Therefore, the total amount of profit per month will be:$ (\frac{800}{x^2})\times(x-2) = 800x^{-1} -1600x^{-2} \frac{dP}{dx} = -800x^{-2} + 3200x^{-3} $To find the maximum profit,$\,\, \frac{dP}{dx} = 0 $ie$\,\, -800x^{-2} + 3200x^{-3} = 0 $By...

All you can do now is try to do better in the next examination. Instead on dwelling on this mark, look to the future, and improve. This is all that you can do. Goodluck =)

$for internal division$ x= \frac{mx_{2} +nx_{1} }{m+n}\;\;\;\;y=\frac{my_{2} +ny_{1} }{m+n} $for external division$ x= \frac{mx_{2} -nx_{1} }{m-n}\;\;\;\;y=\frac{my_{2} -ny_{1} }{m-n}