Recent content by Kingom

Yo guys what's the method of finding kinetic energy for last q using binding energies

I felt like for many of the section 2 questions, they were designed for you to argue against the question or at least to a non-significant extent. Personally I did Dobson, and i argued that it was true to a medium extent (did for the question for Young girl, cock crow and then went against for...

so just marking q11 of the 4u papers I seem to notice a couple of common trends: -for the first integral many methods and manipulations were used with most of them being fruitless (and someone even told us to research an identity smh) -for the volumes question, many people got the correct bounds...

Hi everyone, please sign my petition to unblock cool maths games at https://www.change.org/p/students-unblock-cool-math-games-on-det-internet?recruiter=891913833&utm_source=share_petition&utm_medium=copylink&utm_campaign=share_petition Thank you!!

Show that \int_0^{\tfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \dfrac{\ln x}{(x-1)\sqrt{x^2-2(15+8\sqrt{3})x+1}}dx=\dfrac{2}{3}(2-\sqrt{3})G where G is Catalan's constant.

Re: HSC 2018 MX2 Marathon Continuing on, prove that the area of the ellipse defined by \frac{(a x + b y)^2}{p^2} + \frac{(cx + dy)^2}{q^2} = 1 has area \frac{\pi p q}{|ad - bc|}

Re: Extracurricular Integration Marathon Can you please show working out? I can't really learn from this and what is this "slight manipulation" that you have stated.

Re: HSC 2018 MX2 Integration Marathon Yup. FYI the mistranscribed integral is equivalent to 1/root(2) integral of root(tanx) which has been calculated to exhaustion.

Re: HSC 2018 MX2 Integration Marathon Further hint: take something out of the root and simplify, then substitute

Well done! This is basically the solution I have as well.

and peru

Can a cube be inscribed in a cone so that 7 vertices of the cube lie on the surface of the cone?

These questions are very easy compared to standard HSC-type examination questions. Most of these are doable by inspection and are thus left as exercises to the reader

Re: HSC 2017 MX2 Marathon ADVANCED ewwww formula pleb jkjk

Re: HSC 2017 MX2 Marathon ADVANCED It's just \frac { { a }^{ 3 } }{ 6 } . Consider a cube with side length a. If you cut along its diagonals u get 6 identical pyramids with height half the base.