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Q47 is a standard property of determinants and the proof is a bit long, if not tedious. For Q48, if $ $Q$ $ is an $ $n\times{n}$ $ orthogonal matrix, then $ $Q^TQ=I_{n}$ $ . Take the determinant of both sides and use the result of Q47 to show that $ $\det{Q}=\pm{1}$ $ . Hope it helps :)

From my experience, it's all on the final exam, pre-exam mark doesn't matter too much as long as it is not too low (you can still get 90+ with pre-exam mark of 24+ (out of 36), noting that final is weighted 64%) so there is a decent amount of scaling in the course. This is not the easiest thing...

http://community.boredofstudies.org/136/australian-school-business/280993/how-much-scaling-math1151.html

No.

LOL

Lol, then you would probably just leave it at 18 then, can't be too fussy about it.

Round up to 19 years.

^ Never mind, r = 0.05 is correct for each 6 month period, since I realised the 10% rate was a nominal and not an effective rate.

Yep, I should have divided my answer by 2 to get 7.2725.., so it would be at the end of 8 years.

Is it after 14.545.. (round up to 15) years?

Re: MX2 2015 Integration Marathon Use the property that $ $\int_{0}^{\frac{\pi}{2}} f(\theta)\text{ d}\theta=\int_{0}^{\frac{\pi}{2}} f(\frac{\pi}{2}-\theta)\text{ d}\theta$. $ then use auxiliary angle method to solve the rest.

Write out the series as you normally would, and see if you can identify that the series is in fact geometric with r = 1/4.

The 'r' refers to the ratio of the geometric series, in this case each subsequent term is being multiplied by a factor of 1/4, hence the ratio r = 1/4.

Sorry, I edited my original post, didn't realise it said '1/4' not '1/2'. Do you understand it now? You just need to use the limiting sum formula.

For the first one, use the transformation method (auxillary angle method) and you should be able to sketch a simple sine (or cosine) graph. For the second part you could sketch each graph and then subtract the corresponding co-ordinates of one graph from the other. Also sub. in points to see...

Use limiting sum formula, answer should be \\ $ $5\times(1+\frac{1}{4}+\frac{1}{16}+...)=5\cdot \frac{4}{3}=\frac{20}{3}.$ $

It should be the greatest integer value less than or equal to k , so in this case, your greatest coefficient would be the coefficient $ $T_{3}$ $ .

I get 37o (36o52' to be accurate) using my calculator, so I don't know if it's your calculator causing problems.

I am getting dimensions of $ $10\sqrt{10}-5$ $ by $ $6\sqrt{10}-3$ $ . Method: Let x and y be the sides of the rectangle so that xy=600. Then, since you only want to maximise an area where each side is shrunken by a margin, you would have (x-5) and (y-3) as your required dimensions for the...

Re: MX2 2015 Integration Marathon $ Substitute $x=2\tan{\theta}$ and use the property that for some $a\in\mathbb{R}, \\ \int_{0}^{a} f(\theta)\text{ d}\theta=\int_{0}^{a} f(a-\theta)\text{ d}\theta.$ \\ Answer is $\frac{3\pi}{16}\cdot\ln2$ $