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Re: HSC 2013 2U Marathon { \int _{ 0 }^{ \frac { \pi }{ 4 } }{ ({ sec }^{ 2 }x-1)\quad dx } }\\ \indent = { [tanx-x] }_{ 0 }^{ \frac { \pi }{ 4 } }\\ \indent = [1-\frac { \pi }{ 4 } ]\\ \indent = \frac { 4-\pi }{ 4 }

I'd say data streams, considering the previous 2 questions are data stream related.

I'd say data streams, considering the previous 2 questions are data stream related.

Re: HSC 2013 2U Marathon Thanks! Hope it's useful :)

Yeah interpretation is reading in code line by line and executing but compilation is translating the whole code to object code then executing. (I think). Imagine the translation of an English book to Spanish, interpretation you translate each line, compilation you translate the whole book.

A company may decide to use an interpreted language for a number of reasons. Firstly, the use of an interpreted language allows for great flexibility. It also makes syntax checking a much easier process and makes the location of syntax errors far easier than a compiled language. Another...

I posted a question :)

26.125 = 11010.001 Change to scientific notation to get 1.1010001 x 2^4 so the bold part is our mantissa. Now we need to get the exponent by adding 127 to the power, so we get 131 which is 10000011. We can write our final number as: Sign Exponent Mantissa 0 10000011...

Re: HSC 2013 2U Marathon I made this so you can check your LaTeX output: JSLaTeX :)

Re: HSC 2013 2U Marathon LHS = 2cos^{2}\theta + 1 \\ \\ \indent = 2(1-sin^{2}\theta) + 1 \\ \\ \indent = 2 - 2sin^{2}\theta + 1 \\ \\ \indent = 3 - 2sin^{2}\theta = RHS

Exchange rates would be good!

I hope it's not policies :(

Re: HSC 2013 2U Marathon Answer is D. Since concavity at a is upwards, so f''(a) > 0, and the gradient at a is negative, so f'(a) < 0.

Eco is hard though.

I also had the 1 week gap.

Thanks heaps :)

Thanks! Any more estimates? :)

Re: HSC 2013 2U Marathon I think he was thinking ln(-1).

Re: HSC 2013 2U Marathon In that case, x = \frac {1}{e} {S}_{\infty} = \frac {a}{1-r} \\ \\ \\ \indent \frac {1}{1-\frac {1}{2} \log _{e}{x}} = \frac {2}{3} \\ \\ \\ \indent 3 = 2-\log_{e}{x} \\ \\ \\ \indent e^{-1} = x \\ \\ \\ \indent \therefore x = \frac {1}{e}

Re: HSC 2013 2U Marathon Well I thought you'd let the sum of geometric series = 2/3, but I don't know number of terms or x. I got this far: \frac {1(\frac {1}{2}lnx^{n}-1)}{\frac {1}{2}lnx-1} Yeah. If you give n, it's solvable I think.