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Recent content by Drongoski

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    If you can come to Epping, I can do Ext 1 for $95/lesson (1.5 hrs). I can explain difficult...

    If you can come to Epping, I can do Ext 1 for $95/lesson (1.5 hrs). I can explain difficult concepts better than most teachers. I've been posting efficient solutions on Bored since 2009. If interested, message me on 0490 - 780 347. Dr Drongoski
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    Non-English speaking background

    But please read (well-written material, good books, old & new) widely, and, critically.
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    Non-English speaking background

    I'm sure there are many many from non-English speaking background who do extremely well in their English. Just because you speak Mandarin or Hindi, Korean, Vietnamese , Urdu or Hokkien at home doesn't mean your English is handicapped. In fact I find so many spelling, grammar and punctuation...
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    3U Maths HSC 2019 Solutions

    \int \frac{sin 2x}{4+cos^2 x} dx =\int \frac {sin 2x}{4 + \frac{1}{2} (1 + cos 2x)}dx = \int \frac {2sin 2x}{9 + cos 2x}dx \\ \\ = \int \frac {-d(9 + cos 2x)}{9 + cos 2x} = -ln|9+cos 2x| + C\\ \\ \therefore \int _0 ^{\frac {\pi}{4}} = -[ln(9 + 0) - ln(9 + 1)] = ln \frac {10}{9} Is this right??
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    Square root of complex numbers... I need... help...

    I didn't mean to criticise your method. It is, after all, the standard method. But for simple cases, the "Terry Lee" method is really short. But I bet many will find it hard, unless they are handy with algebraic fiddling.
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    Square root of complex numbers... I need... help...

    Well - you can compare the "normal" method with the "abnormal" method. If you can manage the shorter method, would you want to go through the rigmarole of the "normal" method, unless the question mandates this approach.
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    Square root of complex numbers... I need... help...

    -5 + 12i = 4 + 2 \times 2 \times 3i -9 = 2^2 + 2\times 2 \times 3i + (3i)^2 = (2 + 3i)^2 \\ \\ \therefore \sqrt {-5 + 12i} = \pm(2 + 3i) Take your pick: which is shorter??
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    Square root of complex numbers... I need... help...

    I like the method in Terry Lee's text: z^2 = 2 - \sqrt 5 i = \frac {1}{2}(4 - 2\sqrt 5 i) = \frac {1}{2}(5 - 2\sqrt 5 i -1) \\ \\ = \frac{1}{2}((\sqrt 5)^2 +2\times\sqrt 5 \times (-i) +(-i)^2)) = \frac{1}{2} (\sqrt 5 - i)^2\\ \\ \therefore z = \pm \frac {1}{\sqrt 2} (\sqrt 5 - i) @Zoee...
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    Anyone else in a small class for MX1 / MX2??

    Aren't you lucky. You have the ideal Socratic learning situation: 1 student & 1 teacher!
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    IB Maths Tutor - Sydney

    Musings of an IB Maths tutor . . . .
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    Solving equations with exponents

    Fact she cannot frame her question correctly shows she has gaps in her basic algebra. Could have typed: 2^(x-3)/4^(1-x) = 1
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    box on inclined plane with friction Q, thanks

    Unfortunately I don't know how to provide diagrams ( which would have helped greatly). So I need to explain as best I can. The weight 40g, can be resolved into 2 components: 40g sin(30) down slope and 40g cos(30) perp to (and against) the slope. The force P can be resolved into 2 comps: P...
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    box on inclined plane with friction Q, thanks

    Is the answer about 349.5 Newton (using g = 9.8) or 351.53 N (g = 10)??
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    SHOULD I DO MATHS EXT2???? (ONE LAST TIME)

    Ask your teacher, for example, to check you out and advise you. Maybe you can handle MX2, given the right guidance. I presume you are beginning Yr 12 now - so you don't have much time to seek help/advice. People on this forum cannot advise you meaningfully because they do not know your true...
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    SHOULD I DO MATHS EXT2???? (ONE LAST TIME)

    Jimmy - whether or not you should do MX2 is hard to say. We don't really know your maths. But I have the feeling you are not approaching your maths the "right" way. You yourself may not be able to tell. You need someone else to see if this is the case.
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