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So to answer your first question, if you took the negative, a is unreal. Remember if "a" is real then a^2 is positive and real. Type 5-\sqrt{29} into you calculator and you will find that the answer is negative so you HAVE to accept the positive answer for a^2. I can't seem to see what you...

BUMP bump BUMP!

My advice is to analyse your texts (analyse techniques and how they show certain ideas of belonging or whatever module you are doing now) and type this down into your notes instead of just flipping back at your text to find the appropriate quote and wasting your time. But most importantly keep...

I think this is best way to do this question (Y)

$Ok so I tried doing it this way and got $z=2i$ and $w=-1+i\sqrt3\\\\$Now, $z-w=2i-(-1+i\sqrt3)\\\\\indent\indent\indent=1+i(2-\sqrt3)\\\\$To find arg (z-w), we use $tan \theta=\frac{2-\sqrt3}{1}$ where \theta=arg (z-w)\\\\tan\theta=2-\sqrt3\\\\\theta\approx 0.26\\\\\therefore arg (z-w)=0.26

It's ok we can all learn from our mistakes :)

this is because you forgot to change the limits

\int ^4_0x\sqrt{16-x^2} dx\\\\$Using the substitution $u=16-x^2$ we can figure out that $\frac{du}{dx} =-2x$ which you can make into $xdx=-\frac{du}{2}$ since we can use this in the integral we are given.$\\\\$For the new limits: when $x=0, u=16$ and when $ x=4, u=0.\\\\$So we can now write it...

For now, I would pick advanced if I were you. Yr 10 is nothing like Yr 11 and 12, you'll see its a big jump but you can work hard on your writing (constantly write practice essays!) and analysis of texts so by the end of Yr 11 you can make the decision to stay in advanced or move to english...

bumpity bump bump

$So we know that the axis of parabola is parallel to the y-axis, which means that it is a vertical parabola so it is in the form:$\\\\(x – h)^2 = 4a (y – k)$ where (\textit{h,k}) is the vertex and we know that the vertex is (3,-1) so it is:\\\\ $(x – 3)^2 = 4a (y + 1)$\\\\$$Now, we want to...

do you mean it touches the line 4x+y-7=0?

i can't seem to find the video, it's usually live streaming on smh but it's not there:/

$Ok so we know the formula is $T_{k+1}=\binom{n}{k}a^{n-k}b^k \\\\$so looking at $\left(\frac{x^{1/2}}{2}-\frac{4}{x^{1/4}}\right)^9$ we can deduce that$\\\\T_{k+1}=\binom{9}{k}\left (\frac{x^{1/2}}{2} \right )^{9-k} \left (\frac{-4}{x^{1/4}} \right )^{k}\\\\$Now if we simplify (by collecting...

what interests do you have? do you have any favourite TV shows? because you can relate you jersey name to that :D

ahhh thanks for the advice, i'll remember next time:)

oh yeh and most of the diagrams were taken from cambridge solutions just so you know:)

since i don't know how to draw diagrams here, i'll just attach these images of my solution I hope it makes sense:)

\therefore z^5-1\\\\=(z-{z_{1}})(z-{z_{2}})(z-{z_{3}})(z-{z_{4}})(z-{z_{5}})\\\\=(z-1})(z-{z_{2}})(z-{z_{3}})(z-\overline{z_{3}})(z-\overline{z_{2}})$ since $ {z_{1}}=1, {z_{4}}=\overline{z_{3}}$ and ${z_{5}}=\overline{z_{2}}\\\\$and we know that: $ (z-{z_{2}})(z-\overline{z_{2}})=z^2-2\...