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$Let$\ x=3\sin\phi \\dx=3\cos\phi\, d\phi \int \sqrt{9-x^2}dx\\=\int (3\cos\phi )(3\cos\phi )d\phi \\=9\int \cos^2\phi d\phi \\=\frac{9}{2}\int (1+\cos2\phi) d\phi \\=\frac{9}{2}(\phi +\frac{1}{2}\sin2\phi...

What are some dependent and independent variables for the heat of combustion experiment and the titrations experiment? Thanks

How do you know that from the question?

A railway line around a circular arc of radius 800m is banked by raising the outer rail h metres above the inner rail, where the distance between the rails is 1.5 metres. When the train travels around the curve at 10m/s, the lateral thrust on the inner rail is equal to the lateral thrust on the...

You could even start some of it now, since some of the 4u stuff will help you in 3u, like integration and polynomials. As for the other topics, you could learn them during the period between hsc and uni. I'm pretty sure you'll be set for uni after that.

ok, consider this: We have 3 numbers, A, B and C, then A+C=B+C if and only if A=B so proving A+C=B+C is the same as proving A=B, no?

Use the fact that exterior angle of cyclic quad is equal to opposite interior angle. You will get: arg(z_2-z_3)-arg(z_3)=arg(z_1-z_2)-arg(z_1)\\arg(z_2-z_3)-[arg(z_3)+arg(z_2)]=arg(z_1-z_2)-[arg(z_1)+arg(z_2)]\\arg(\frac{z_2-z_3}{z_2z_3})=arg(\frac{z_1-z_2}{z_1z_2})\\\therefore...

However, if the equality doesnt hold, then adding ln0.5 to both sides WONT preserve the equality. I agree with this. But notice how the guy actually goes on to prove that it does actually preserve the quality (and hence that the original equality is true). He goes this in the step...

nah, im pretty sure you can do that, he's just adding ln0.5 to both sites. If the original equality holds, then adding ln0.5 to both sides should preserve the equality.

This question is hard...something got to do with the inclusion-exclusion principle thing, which i dont really get.

nah, i didnt read it wrong, that cosx part was a typo. It's been fixed.

\int_{0}^{\pi/2}\sin^nx\cos^2xdx let u=sinx du=cosxdx x=0, u=0 x=pi/2, u=1 I_n=\int_{0}^{1}u^n\sqrt{1-u^2}du\\=\int_{0}^{1}u^{n-1}(u\sqrt{1-u^2})du\\=\int_{0}^{1}u^{n-1}\frac{\mathrm{d} }{\mathrm{d}...

Lol, i was doing the same question this morning. For 1a), continuing on from tim, we have x=(\frac{2b}{n})^\frac{1}{n-2} as the root...

Is there a way to prove this without induction? 1^2 + 2^2 + 3^2 +...+ n^2 = n(n+1)(2n+1)/6

not really, its just: asec@ - acos@ =acos@(sec^2@ - 1) =acos@tan^2@

2/3 is not the absolute minimum. The answer in your book is probably wrong as you'll notice 2/(3sqrt5) is smaller than 2/3.

To find perpendicular distance, we need the gradient of y=2x-1 to be the same as the gradient of the tangent to y=3x^2. dy/dx = 6x = 2 so x=1/3 subbing it back into y=3x^2, we get coordinates (1/3, 1/3) using perpendicular distance with 2x-y-1=0, we get 2/(3sqrt5) as suggested by Drongoski.

Tim, the "minimum" distance you found was only the minimum difference in y values between y=3x^2 and y=2x-1. But the real minimum distance is the perpendicular distance. I think Drongoski is correct.

(i) ln(6!)>6 so true for S(6) Assume S(n) true. ie. ln(n!)>n Consider S(n+1): ln(n+1)! =ln[(n+1)n!] =ln(n+1)+ln(n!) >ln(n+1)+n >n+1 since ln(n+1)>1 for n>6 S(n+1) is true if S(n) is true but S(6) is true so it is true for n>=6 (ii) ln(n!)>n n!>e^n (1/(n!))<(1/e^n)...

I dont like working with U_n and @ so i changed it to I_n and x. Anyway, here it is: I_n=\int_{0}^{\pi/2}x\sin^nxdx\\=\int_{0}^{\pi/2}\sin^{n-1}x(x\sin x)dx\\=\int_{0}^{\pi/2}\sin^{n-1}x\frac{\mathrm{d} }{\mathrm{d} x}(-x\cos x+\sin x)dx\\=[-x\cos...