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  1. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Yeah but it doesn't explicitly state that D HAS to be in the first quadrant it just says find D such that ABCD forms a parallelogram. I remember questions like this in the past where students have got marks for 3 different solutions.
  2. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon So is (-4, 1) still a solution?
  3. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Can't you just find difference of C from A, and apply that from B to D such that D(-4, 1)
  4. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon \int_0^{\pi/4} 1 + \sin^2 x + \sin^4 x + \sin^6x + \dots \ dx \\ \\ \indent = \int_0^{\pi/4} \frac {1}{1-sin^{2}x} \ dx \\ \\ \indent = \int_0^{\pi/4} \frac {1}{cos^{2}x} \ dx \\ \\ \indent = \int_0^{\pi/4} sec^{2}x \ dx \\ \\ \indent = tan\frac {\pi}{4} - tan0 \\ \\...
  5. Shazer2

    Q9) 2012 Maths past paper

    Yes true! I've just learnt that way for some reason.
  6. Shazer2

    Q9) 2012 Maths past paper

    \int _{ 1 }^{ 4 }{ \frac { 1 }{ 3x } dx } = \frac { 1 }{ 3 } \int _{ 1 }^{ 4 }{ \frac { 3 }{ 3x } } dx \\ \\ \indent = \frac { 1 }{ 3 } (\log _{ e }{ 12 } -\log _{ e }{ 3 } ) \\ \\ \indent = \frac { 1 }{ 3 } \log _{ e }{ 4 }
  7. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon \frac {1}{1-x} = 5x \\ \\ \indent 5x-5x^{2} = 1 \\ \\ \indent 5x^{2}-5x+1=0 \\ \\ \indent x = \frac {5 \pm \sqrt {(-5)^{2}-4(5)(1)}}{10} \\ \\ \indent x = \frac {5 \pm \sqrt {5}}{10}
  8. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon x(49-x^2)^{1/2} \\ \\ \indent \frac {d}{dx} \quad x(49-x^{2})^{1/2} \\ \\ \indent = x \times -x(49-x^{2})^{-\frac {1}{2}} + (49-x^{2})^{1/2} \\ \\ \indent = \frac {-x^{2}}{\sqrt {49-x^{2}}} + \sqrt {49-x^{2}}
  9. Shazer2

    HSC 2013 Maths Marathon (archive)

    Yeah I didn't mean that :) was my solution correct though? Sent from my iPhone using Tapatalk 2
  10. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon x^{2} - 6x -8y = 23 \\ \\ \indent (x-3)^{2} = 8y + 32 $ (Completing the square)$ \\ \\ \indent (x-3)^2 = 8(y+4) \\ \\ \indent \therefore $ Vertex at $(3, -4) \\ \\ \\ \indent 4a = 8 \\ \\ \indent a = 2 \\ \\ \indent \therefore $ Focus at $(3, -2). \\ \\ \indent...
  11. Shazer2

    How many past papers have you done ?

    Re: How many past papers have yous done ? Maths - like 6 maybe (not all completed, up to last question usually) Eco - 2 short answers, 1 MC Business - 1 short answers
  12. Shazer2

    BOS Trials 2013 2U Maths Solutions and Results.

    Very hard paper, and 96%, that's impressive.
  13. Shazer2

    BOS Trials 2013 2U Maths Solutions and Results.

    It'd be interesting to see who got first. :)
  14. Shazer2

    SDD Marathon 2013

    I'm not sure what sort of thing you're looking for for 6 marks, but I'd start with something like: 1. A computer input device has a number of operations. Most notably, it allows the user to interact and interface with the system. 2. Primary and secondary storage are two separate methods of...
  15. Shazer2

    SDD Marathon 2013

    I can't really remember covering data stream error checking either.
  16. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon I'd write questions if I had the knowledge!
  17. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Yeah it was a pretty stupid mistake. :(
  18. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon Woah that's heaps neat!
  19. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon I'm not sure, all I know is 1/x integrated is ln(x), not sure about 1/ln(x).
  20. Shazer2

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon I think this is right? \int \frac {1}{xln(x)} dx \\ \\ \indent = x \int \frac {\frac {1}{x}}{xln(x)} dx \\ \\ \indent = xln(xln(x)) + c
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