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Are you sure that's supposed to be 20m? Cause I think if you use 200m it gives (a), using your method.

Simplest way to do it is to notice that ∠TRQ = 150 degrees and since triangle TRQ is isoceles (TR = RQ) ∠RTQ = ∠STP (RTQ ||| STP)(obv. they're also congruent) = 15 degrees. The result follows. You could also set the length of the square to 1 unit on an x-y cartesian co-ordinate...

Have only finished english thus far. We still have a final module to complete for both physics and chemistry and a couple of topics left for MX1 and MX2. Trials for all subjects are mid-way next term afaik.

http://integrals.wolfram.com/index.jsp?expr=x%2F%288%2B%28sinx%29%5E2%29&random=false ...not pleasant.

ma.tan@ = b.sec@ -> ma.tan@ - bsec@ = 0 ...(1) b.tan@ - ma.sec@ = k ...(2) (2)2 - (1)2 -> b2.tan2@ - 2mba.tan@sec@ + m2a2sec2@ - m2a2.tan2@ + 2mba.sec@tan@ - b2.sec2@ = k2 Result follows.

Who here uses dot-point responses when answering short-response questions? Wondering, 'cause I probably should try this.

OR alternatively: 2sin(105)cos(105) = sin(210) = sin(180+30) = -1/2

It says that he deposited $500 until Tom's 18th birthday. This means that he did not deposit any money on Tom's 18th birthday. If he had of done so, the question would explicitly say so.

y=ex x=lny => x2=(lny)2 V = pi[3-->5]∫(lny)2dy Using simpson's rule, we first evaluate [3-->5]∫(lny)2dy: Hence: I=2/12[(ln3)2+(ln5)2+4((ln3.5)2+(ln4.5)2)+2(ln4)2]=3.827919825 Now to find the volume, we need to multiply by pi which gives: 12.0257648

:wave: I got 4/7 as well.

Nah, I exercise way more now than I ever did. I find it an excellent way to relieve stress.:D

Series: {21 + (2(1)-1), 22 + (2(2)-1), 23 + (2(3)-1) ...} S10 = S10[AP] + S10[GP] = 10/2[2 + (9)2] + 2(210 - 1)/(2 - 1) = 2146, as required.

Yeah, that's correct. Post up a question ngogiathuan.

Exams!

1/(1+sinx) = (1-sinx)/cos2x = sec2x - secxtanx I = [-pi/4 --> pi/4]∫sec2xdx - [-pi/4 --> pi/4]∫secxtanxdx Second integral is odd, therefore = 0. And the first integral is even: I = [0 --> pi/4]∫2sec2xdx Next Question: Find: [0 --> pi/2] ∫sinnxdx/(sinnx + cosnx) Not very...

I went up around 420 ranks ...lol.

QFT.

This question is literally impossible. Don't bother trying to prove it. There is no restriction on L, and so it's like trying to prove that ∠KLM is a right angle wherever we place L, since L can be moved anywhere. That obviously is impossible.

Win. :ninja:

Using t=tan(x/2): int. dx/(1+sinx) = int. [2dt/(1+t2)]/[1+(2t/(1+t2))] = int. 2dt/(t2+2t+1) = 2[int. dt/((t+1)2)] = -2/(t+1) + c = -2/(tan(x/2) +1) + c Alternatively: int. dx/(1+sinx) Multiplying the numerator and the denominator by [1-sinx]: I = int. dx(1-sinx)/cos2x = int. sec2 dx -...