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Re: English, Modern History, Visual Arts Tutoring - 99.55ATAR, Extended Eastern Subur This guy would pick up the soap from the shower floor in a prison for you. He is dedicated, committed and knows his stuff I would definitely pm him if you're looking to lift up your marks!

Hey BOS, this may sounds like a stupid question but in what circumstances would i not use phenolphthalein as my indicator in a titration? According to wiki: An acid-base titration is the determination of the concentration of an acid or base by exactly neutralizing the acid/base with an acid...

yes..yes you should

This may be an easy questions but I'm just not seeing it. Any help is greatly appreciated, cheers. http://i.imgur.com/cpA9K.jpg

but doesnt give the opposite result wouldnt 2arg(z-3) = argz edit: oops double post

wow, i didnt know you oculd do that, should my answer be 1 + 5i?

http://i.imgur.com/t5YSB.jpg Half Yearly's are tomorrow so any help would be great I have no idea how to do d (ii), something to do with an angle subtended at the centre of a circle is double that at the circ??? for e) how do i decide whether the vector between 2 points is z1 - z2 or z2 - z1?

since there is a multiple root then nx^n-1 + m = 0 therefore x^n-1 = -m/n - [1] we also know that x(x^n-1 + m) = b subbing in [1]: x = b/(-m/n + m) = bn/m(n-1) =) x^n-1 also = [bn/m(n-1)]^n-1 solving simultaneously with [1]: [bn/m(n-1)]^n-1 = -m/n therefore [b/(n-1)] ^ n-1 . [n/m] ^ n-1...

http://i.imgur.com/BBY7H.jpg Sorry for all these questions, I definitely need more practice with complex no.'s. Any help is greatly appreciated.

Thanks for the help guys but i dont see how the angle at z1 made by z4 and z2 = arg(z1-z2) - arg (z1 - z4) isnt the argument take from the origin? I guess what im try to say is how can you find an argument of a vector?

sorry man, i'm still not seeing it... arg(a-b)/arg(c-b) = arg(a-b) -arg(c-b) do you have to use the result that opposite angles of a cyclic quad are supplementary? I still cant put the pieces together.

http://i.imgur.com/rXlgT.jpg Comes from 2002 SBHS trial have no clue how to solve part ii

nah, man it's its just plain old Jeff here. I made this account ages ago with Zach and I'm not sure why we chose scardizzle, i guess we thought it was kinda funny but anyway sorry for any misconceptions caused.

your question isn't making much sense to me. Where did you get the b term from? The only advice i can give atm is this q probably involves differentiating and making x the subject

P is a variable point on the ellipse with equation x^2/a + y^2/b = 1 and S and S' are the foci. Show that PS and PS' are equally inclined to the tangent at P.

I would say the year 11 course and the year 12 course a fairly similar. Both are content based with a little bit of maths. If you're struggling with understanding the basic principles e.g. conservation laws, Newtons laws etc. I would consider dropping as these are built upon in the year 12 course.

I think he's been learning ahead, i remember seeing him last year in the 3U forums

well w1,w2...wn are all roots of the equation z^n = 1 using the sum of roots, we know that w1 + w2 +...wn = -b/a in this case b = 0 hence w1 + w2 +...+wn = 0 (sorry for the lack of latex i cbf to learn at this point in time)

Prove that w, a complex cube root of unity, is a repeated root of P(x) = 3x^5 + 2x^4 + x^3 - 6x^2 - 5x - 4. EDIT(left out the other half of the q):Hence find the zeroes to the equation over the complex number field

so (x,y) lie on the parabola y = x^2 therefore (x,y) can be written as (x,x^2) assuming the q is talking about its perpendicular distance to that line distance = | x - x^2 -1| x (1 + 1)^-1/2 = |x^2 -x +1| x 2^-1/2 (dividing by -1 this wont affect the answer since there are absolute value...