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I still don't really understand what a parametric coordinate is? I've never learnt that.

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2012exams/pdf_doc/2012-hsc-exam-maths.pdf This is probably really obvious, but: How do you find the coordinates of T?

Are you in prelim or HSC? You don't need to know about this for the HSC...

About 5-6 minutes.

Thanks man, I notice you always reply to my topics which is nice, so thanks for helping me out!

I do know all of them, but I always take myself in the wrong direction when applying them. The easier - medium ones I can do, but HSC level ones I can't. Aaaand yeah, kinda wish I didn't do Industrial Technology or English Standard since they scale poorly, and also since I want to do...

Oh so it is 2x, thanks :) There is one thing which I absolutely cannot do no matter how much I practice: proving trig identities. That will be the reason I fail HSC maths.

Btw when you pull the power out, is it 2x = .... or x^2 = ....?

Oh fuck woops You still get x = 0 and imaginary though

Oh wait, you mean just let p = e^2x or something? so p^2 + p -2 = 0 P: -2 S: 1 F: 2,-1 p^2 + 2p -p - 2 = 0 p(p+2) -(p+2) = 0 (p-1)(p+2) = 0 p = 1, -2 e^x = 1, e^x = -2 x = ln(1), x= ln(-2) x = 0, x = imaginary?

Where do you put the 2 though?

x = 0 from just looking at it but doing that the long way -- how do you factorise properly?

Oh you're right, it's in Ex 4.6, I think I skipped over questions 3+ and forgot to do them. Guess I'll do them now lol.

lol Maths in Focus doesn't have any of these though... that's why I couldn't do it since we never learnt how to. Thanks for helping though

Ohhhhh so basically: ln(e^x) is xln(e), which is x *1 = x So x = ln(1/2) = ln1 - ln2 = 0 - ln(2) = -ln(2) Does anyone have heaps of these so I can practice?

Yeah so that gets me e^x = 1/2, but i don't know what to do to get x

So I came across a question in the 2007 HSC that required you to solve an exponential equation. I haven't learnt this at all...more the reason to love Maths in Focus. Seriously thinking about picking up a better textbook, it's trash. Anyway, so if anyone could teach me how to solve this...

Ok thanks! Also, sorry I couldn't get the CSSA trial for you

2011 or 2012 CSSA exam not sure which though cbf looking. "Sample answer: Weight force on CD due to mass, W = mg = 0.005 * 9.8 = 0.049 N Since length is 0.30 m, distance of P to axis is 0.15 m, hence torque due to P = length * w = 0.15 * 0.049 = 0.00735 Nm (up the page) Since length = 0.30 m...

The answer's 1.01 A anticlockwise through the circuit... edit: but seeing's they were wrong in another question, they're probably wrong here lol