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Hi all, I'm Jathu and I will be offering online tutoring for upcoming Year 12 (2025 HSC) students in Chemistry, Mathematics Advanced and Mathematics Extension 1. Classes will be in small groups of 2-6 students. I have both the knowledge and experience to help guide you in your HSC studies, as...

I concur :D

FINS1612: 55-60 ECON1101: 83 MATH2018: 98-100 Hopefully math can pull my wam lol

https://www.smh.com.au/national/nsw/hsc-2019-how-your-school-ranked-20191212-p53j72.html

Does anyone have the pdf of the exam? Can't seem to find it on any thread (or im just blind haha)

I graduated from St Marys 2 yrs ago so here's my 2c from what I can remember Now to get in, you just hand them in your application form and reports/NAPLAN/other docs etc and they'll assess it and will let you know, so no need for exams or anything. Now me being so bad at English, couldn't get...

\noindent \text{Begin by writing the long division statement:} \\ P(x)=(x^2-5)Q(x)+x+4 \\ \text{We also require P(-x), and we can find this out from the expression above:} \\P(-x)=((-x)^2-5)Q(-x)-x+4 \\ \\ \text{Adding both these expressions, we get:} \\...

The hazard perception test is exactly like the practice Q's (unless it's changed from last yr) so all I did was just spam the practice qs until I could get then all correct every try, and then did the test and was all g You could go over the handbook if you have time/bothered, but yeh most of...

1^2 +2^2 +3^2+ ... +k^2 =\frac{1}{6}k(k+1)(2k+1) \qquad (*)\\ \\$ Prove true for $ n = k+1, ie: \\ 1^2 +2^2 +3^2+ ... +k^2 + (k+1)^2 =\frac{1}{6}(k+1)(k+2)(2k+3) \\ LHS = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2 \qquad $ from $(*) \\$ Now factorse $ \frac{1}{6}(k+1): \\ =\frac{1}{6}(k+1)\left...

$ \noindent Since $ |z|=1, $ we can use the identity: $z^n+z^{-n}=\cos n\theta $ (you can prove this by subbing in $z= $ cis$\theta $ and using DMT$. \\ \\$ Dividing the LHS by $z^3, $ we obtain $: \\ z^3+z^{-3} = 2\cos 3\theta \qquad $ Using the above identity$ \\ \\ $ On the RHS, we divide...

$\noindent we can write the sum of two positive integers as $ x+y=k,$ where $x,y,k $ are positive integers$ \\ \Rightarrow y= k -x \qquad (1) \\ \\ $ now the product, P, is equal to $xy \\ P=xy = x(k-x) =x^2 -kx \qquad $ from eqn 1$ \\ \frac{dP}{dx}=2x - k \\ \therefore $ max at $x =...

yeh it can be pretty confusing and dodgy in the beginning, but basically once you draw up all the curves, try to identify which closed region they 'form.' So once we draw all the curves/lines (in red), they'll only form a triangle on top (try not to think of the axes if it helps)...

For part a) the region bounded by all those three 3 curves is actually the upper triangle (ie limits of integration is from 1 to 2 ). You can figure this out if you draw all those 3 curves, you'll see that only the upper triangle is fully bounded/enclosed So subbing the new limits in we...

What an absolute legend pikachu! Great job on the excellent mark mate!

ooft legend! Thanks for your effort mr fluffchuck! These solutions are super helpful :D

$ Let the roots be $ \alpha, \beta, \gamma $ and $ \alpha\beta=\gamma,$ then: $ \\ d=\alpha\beta\gamma =\gamma^2 \\ c=\alpha\beta + \alpha\gamma + \beta\gamma=\gamma +\alpha\gamma+\beta\gamma \\ \therefore c+d=\gamma^2 + \gamma + \alpha\gamma+ \beta\gamma = \gamma(\alpha+\beta+\gamma+1) \\c+d =...

no worries! lol idk if im blind or my laptop display is being dodgy

pretty sure it should be 1/2 cos(2x) for (c) z^4 + 1 = (z^2+1)^2-2z^2=(z^2-\sqrt{2}z+1)(z^2+\sqrt{2}z+1) \\ $Dividing by $ z^2, $ we get :$ \\ z^2 + z^{-2}=(z+z^{-1}-\sqrt{2})(z+z^{-1}+\sqrt{2}) \\ $ Applying DMT and cancelling outs the sines: $ \\ 2 \cos{2\theta}=(2 \cos \theta - \sqrt2)(2...

I agree :D

join societies, go to random interesting events and meet new people! or just break the ice with someone during lectures if you dont know anyone, and build up a group of outgoing people. A healthy soocial life makes uni much less stressful and a fun, worthwhile experience!