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Are you really getting married, I fell in love so I tried livestreaming, Mute girl and her new friend, Alya sometimes hides her feeling in russian, The fragrant flower blooms with dignity, Nisekoi, Kimi wa Yakamashi Tojite yo Kuchi wo, A Bouquet for an ugly girl, The omniscient point of view of...

yeah but then you have to find out \tan\left(\frac{\pi}{2}\right)=????? :speechless:

\\LHS=\frac{1}{\sin(4x)}+\frac{\cos(4x)}{\sin(4x)}\\\\=\frac{\cos(4x)+1}{sin(4x)}\\\\=\frac{2\cos^2(2x)-1+1}{2\sin(2x)\cos(2x)}\\\\=\frac{(\cos^2(x)-\sin^2(x))^2}{2\sin(x)\cos(x)(\cos^2(x)-\sin^2(x))}\\\\=\frac{\cos^2(x)-\sin^2(x)}{2\sin(x)\cos(x)}\\\\=\frac{1}{2}(\cot(x)-\tan(x))\\\\=RHS

Question 1 part i does not make sense x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+y^{n-1}) not what the question says. Also for question 2, since we are choosing 3 points from a total of n (assuming that N=n) points in the space to make a plane (order does not matter), we can do...

dw I am fucked anyway

man that place is always crowded, only good food in UNSW are the free sausage sizzles

The question did not specify whether the ratio was was np:ns or ns:np, you'd have to deduce that yourself. I don't think it is a good idea to just assume the ratio will always be np:ns, make sure to check if they haven't specified it.

Since there are two possible values for a and c, we might need to consider a quadratic equation. Lets work backwards a little to see what quadratic we need. \\a=c\implies{\frac{a}{c}}=1\\\\a=9c\implies{\frac{a}{c}=9} So this quadratic has roots of 1 and 9...

Since the transformer is step down, there will be more coils in the primary coil than the second. Thus by using the formula \frac{V_p}{V_s}=\frac{N_p}{N_s} \begin{align*}\frac{V_p}{V_s}&=13\\ V_s&=\frac{V_p}{13}\\ V_s&=\frac{240}{13}\end{align} so now by conservation of energy and also...

Flowers for Angernon is an interesting short novel which might suit the question. Here is a link to the pdf version https://www.sdfo.org/gj/stories/flowersforalgernon.pdf

I think you misinterpreted the question, it was asking for how many ways you could keep the two women separated from the man, while they are also arranged in alternating order. I think the best way to start is to find the arrangements just for the men in a circle which is 4!, given that one man...

interesting graph, have a look at this graph which probably shows the question better and also helps demonstrate what @Drongoski was saying about the anser being independent of the argument of z1.

what else could z_1 equal to?

using the diagram from part i or just by solving for the roots of unity z_3=cis\left(\frac{2\pi}{3}\right) and z_2=cis\left(-\frac{2\pi}{3}\right) and z_1=1 now the expression |z_1-1|^2+|z_2-1|^2+|z_3-1|^2 can be simplified to |z_2-1|^2+|z_3-1|^2 cause z_1=1 then we just sub in z_2 and z_3 in...

Yeah cause for triangle AOC is congruent to triangle ABC, so angle DAB is 60 degrees too.

if you subbed n = 1 into the LHS of the original, you would get 1+t, so yeah the original is wrong.

wait na, the original is wrong \\\frac{t^{n+1}-1}{t-1}\\\\=\frac{(t-1)(t^n+t^{n-1}+...+t+1)}{t-1}\\\\=t^n+t^{n-1}+...+t+1