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Hey guys, I'm wondering which of those subjects to do for my second semester. I'm a first year doing Civil Engineering/Commerce. I was considering doing CHEM1011 since I've heard it's easier than MATS1101, but MATS1101 is the recommended elective so I'm not sure if I should. Also, instead of...

Re: UNSW chit chat thread 2016 Which lecturer do you have?

i) \begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \left (\frac{3x}{9+x^2}+\tan^{-1}{\frac{x}{3}} \right ) &= \frac{\left (\dfrac{\mathrm{d}}{\mathrm{d}x}3x \right )(9+x^2)- \left (\dfrac{\mathrm{d}}{\mathrm{d}x}9+x^2 \right )\times 3x}{(9+x^2)^2} + \frac{3}{9+x^2}\\ &=\frac{3(9+x^2)-2x \times...

The sports uniform has been changed recently. A survey has been going around regarding changing the normal uniform, so there is a chance that it'll be revised before the school moves.

http://www.dailytelegraph.com.au/newslocal/macarthur/hurlstone-agriculture-high-school-to-relocated-to-hawkesbury-in-2020/story-fngr8h70-1227613681204?nk=397c885e7bd4ed622c08b1f4bb127d82-1447815725...

Re: HSC 2016 3U Marathon \begin{align*}f'(x)&=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \textup{ and }g'(x)=\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}. \\\frac{\mathrm{d}}{\mathrm{d}x}\left (f(x)g(x) \right )&=\lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} \\&=\lim_{h \to...

Re: HSC Physics - Exam Paper (download) Wouldn't the force due to gravity be different as they have different masses?

lol woops threw my copy away with all the rest of my HSC stuff. I think Fizzy_Cyst might upload it if he manages to get a paper.

Re: Carrotsticks' Solutions 2015 Extension 2 HSC1 Shouldn't it be 2.57 for 12(d)(iii)?

Re: HSC 2015 3U Marathon In the second triangle tan(x/2)=t. Using this triangle, you can find sinx, cosx and tanx in terms of t. The first triangle is the sides of a triangle with sinx, cosx and tanx in terms of t.

So there's 13 ways of choosing 2 cards from a group of 4 cards with the same value. The 12C3 is then choosing 3 cards of different value from the 12 remaining card values that haven't been chosen. But why the 4³?

Re: HSC 2015 3U Marathon \begin{align*}\textup{If we let }x&=e^{-2t} \textup{ we have:} \\S&=x+2x^2+3x^3+4x^4+... \\Sx&=x^2+2x^3+3x^4+4x^5+... \\S-Sx&=x+2x^2+3x^3+4x^4+...-x^2-2x^3-3x^4-4x^5-... \\S(1-x)&=x+x^2+x^3+x^4+... \\&=\frac{x}{1-x} \textup{ using }S_\infty=\frac{a}{1-r}...

Re: HSC 2015 3U Marathon $I'm assuming you've got the first part done. For the projectile released at $O$: $x=Vt\cos{\alpha}$, $y=\frac{-gt^2}{2}+Vt\sin{\alpha}$. For the projectile released at $A$: $x=Ut\cos{\beta}$, $y=\frac{-gt^2}{2}+Ut\sin{\beta}+20$. When the particles collide at time $T$...

Re: HSC 2015 3U Marathon Someone correct me if I'm wrong, but I think polynomials can not have negative indices. So since you know P(x)=0, you should times through by x3 at the end to leave only positive indices.

Not confident about this, but I think it's because each is considered a unique way of getting the required configuration (1 pair, 3 singles).

The probability of getting a pair is: 1\times\frac{3}{51}=\frac{1}{17}. Once we take a card, there are only 3 of the same value in the rest of the deck, and our sample space reduces by one. For the next 3 cards, we need 3 cards of different values (e.g. 4,5,6). The probability of getting this...

$(a) $f(x)=ax^3+bx^2+cx+d \to f'(x)=3ax^2+2bx+c$. There is a stationary point at at $x=0$, which means $f'(0)=0$. Solving this gives you $c=0$. \\ \indent (b) $f''(x)=6ax+2b$. There is a point of inflexion at $x=1$, which means $f''(1)=0$. Solving this gives you $b=-3a$. \\ \indent (c) We know...

Re: HSC 2015 3U Marathon $1. The roots of the polynomial $P(x)=x^3 - 6x^2 + 11x - 6$ are $\alpha$, $\beta$ and $\gamma$. Without finding the values of these roots, find the polynomial with roots: \\ \indent (i) $\alpha + \beta$, $\alpha + \gamma$, $\beta + \gamma$ \\ \indent (ii) $\alpha...

Re: HSC 2015 3U Marathon $1. The real number $x$ is a solution of the equation $x^2-x-1=0$. Use the Binomial Theorem to show that the sum $S$ of the series $1+x+x^2+...+x^{2n-1}$ ($n=1,2,3...$) is given by $S=\sum^n_{r=1}\binom{n}{r}x^{r+1}$. $2. (i) Show that $\binom{n}{r}=\binom{n}{n-r}$...

Re: HSC 2015 3U Marathon I think your discriminant should be \Delta=b^2-4ac not \Delta=\sqrt{b^2-4ac}. I'll look for binomial questions now.