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Re: 2012 HSC MX2 Marathon Prove that, for any triangle with sides a, b, c and area A, a^2 + b^2 + c^2 >= 4sqrt(3)A

Then the answer probably has that typo. I got the left hand side as my derivative.

To expand on Math Man's answer, the proof of this theorem depends on an important corollary: A polynomial with degree of at most n with more then n roots must be the zero polynomial. With this the proof is simple and was outlined above: We form a new polynomial, G(z) say, such that G(z) = p1(z)...

Hint: rearrange the equation into the form x+y = a(xy)^1/2 Edit: I think he made a typo on the right side and means dy/dx instead of y/x

Yes, In-fact all scaling of the triple {3,4,5} by a real number form an AP. Since r x {3,4,5} = {3r,4r,5r}={3r, 3r + r, 3r + 2r}. i.e, when r = 1.5, then the triple {4.5,6,7.5} satisfy both the Pythagorean theorem and form an AP.

Who are you, nameless poster?

Solve the equation as a quadratic in terms of x and it becomes obvious the graph is those two lines.

\int tan^3xdx =\int(sec^2x-1)tanxdx \\ let \; u=cosx, \;du=-sinxdx \\then \;= \int(1-\frac{1}{u^2})\frac{du}{u} =ln\left | u \right | + \frac{1}{4u^4} + c \\ =ln\left | cosx \right | + \frac{1}{4cos^4x} + c

Paranoid: High Schizoid: Very High Schizotypal: High Antisocial: High Borderline: Moderate Histrionic: Low Narcissistic: Moderate Avoidant: Low Dependent: Low Obsessive-Compulsive: High

Paranoid: High Schizoid: Very High Schizotypal: High Antisocial: High Borderline: Moderate Histrionic: Low Narcissistic: Moderate Avoidant: Low Dependent: Low Obsessive-Compulsive: High

isn't it 0?

I just looked at some of the earlier papers. Damn the old syllabus looks like alot of fun. ):

For question 5, so far i can only prove \frac{1}{\prod_{i=1}^{n}a_i}\geq n^\frac{n}{k}. But im having difficulty in showing that, \sum_{i=1}^{n}a_i \geq n^{1-1/k} Infact, i think it could be the opposite. ): Buchanan, any hints on how to show it? Also, where do these questions come from?

Interesting question construction: let the midpoint of XY by P. let the radius be x units long. Proof: To prove XY>AB and CD is the same as proving OP < OM and OL. as lines from origin to midpoints of chords are perpendicular to the chord and the closer the chord is to the origin the...

sum \;of \;roots: a+b=-p \\product\;of\;roots:ab=q\\also, \; (a+b)^2=a^2+b^2+2ab \\ \therefore a^2+b^2 =(a+b)^2-2ab=p^2-2q \\ \\ subbing \; (a-b) \;into \;x^2-p^2+4q=0 \\LHS=(a-b)^2-p^2+4q \\ =a^2+b^2-2ab -p^2+4q \\=p^2-2q-2q-p^2+4q\\=0 \\similarly for x=(b-a), \;noting\; (b-a)=-(a-b) \;and \...

Yea, your calculations for team 1 and team 2 are correct but they also overcount. *tries to explain it better* your method allows for the same two teams to be picked in different order. ie, your pick mrA and msB for the first team and then you pick mrC and ms D for the second. but you also...

badquinton, wouldn't your solution count each pair of teams twice? as in, your also ordering the pairs of team on the court, ie team A vs Team B is different to Team B vs Team A. to fix this devide 260 by two for the overcounting. ...And this leads to 130, the answer?

I got, \frac{4sin(\frac{n\pi}{2})}{n^2\pi^2}-\frac{2cos(\frac{n\pi}{2})}{n\pi}

x^2+px+q=0 \\x=\frac{-p\pm \sqrt {p^2-4q}}{2} \\\Rightarrow a-b=\sqrt {p^2-4q}, \;b-a=-\sqrt {p^2-4q} \\\therefore equation \; is \; (x-\sqrt {p^2-4q})(x+\sqrt {p^2-4q})=0 \\ x^2-p^2+4q=0

1. Depends on the school. 2. Depends on the person. Personally i find hsc physics easy, it requires basic maths and some memorisation of names, experiments and effects on society, etc... But as long as you can understand the basic idea you should be fine. For chemistry i noticed some people...