Recent content by Drongoski

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    HSC Maths Ext2 Questions + Answers

    "practice" unless you are American.
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    How to solve this?

    Cherry Finally found my LaTeX error for my solution to e) above.
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    How to solve this?

    For e): \therefore $ sum of roots: $ \frac {1}{k}\alpha + \alpha + k\alpha = - b \implies (\frac {1}{k} + 1 + k)\alpha = -b \\ \\ $ sum of root pairs: $ \frac {1}{k} \alpha \times \alpha + \frac {1}{k} \alpha \times k \alpha + \alpha \times k \alpha = (\frac {1}{k} + 1 + k)\alpha ^2 = c...
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    How to solve this?

    For c): P(x) = x^3 + 4x^2 + 2x + k You are told: remainder upon division by (x-1) = 2 times remainder upon division by x or (x-0). That means P(1) = 2 P(0) i.e. 1 + 4 + 2 + k = 2 x k .: k = 7
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    How to solve this?

    Cherry: I know you want a very gentle step-by-step help. Not helpful hints here and there. For b) P(x) = (x^2-1)Q(x) + kx + 2 If x-1 is a factor of P(x), then the Remainder Thm tells you that P(1) = 0 i.e. P(1) = (1^2 - 1)Q(1) + k + 2 = 0 ==> 0 + k + 2 = 0 so that k = -2. You now know that...
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    How to solve this?

    Or doing the usual way: \frac {2x(x-4)}{(x-1)} > 7\\ \\ \therefore (x-1)^2\times \left [ \frac {2x(x-4)}{(x-1)}\right ] > (x-1)^2 \times 7 \\ \\ \therefore 2x(x-4)(x-1) > 7(x-1)^2\\ \ \therefore (x-1)[2x(x-4) - 7(x-1) > 0] \\ \\ \therefore (x-1)(2x^2-15x+8) > 0 \\ \\ \therefore (x-1)(2x-1)(x-7)...
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    How to solve this?

    If question is to solve, say: \frac {2x(x-4)}{x-1} > 7 can do this in a number of ways. One is using sign diagrams. Note because of the denominator, note x must not equal 1. The above inequality is equivalent to: \frac{(2x - 1)(x-7)}{(x-1)} > 0 3 relevant factors are: 2x-1, x-7 and x-1...
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    How to solve this?

    Don't understand your question. What is that small circle??
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    Should I apply to Ruse?

    Just pulling your leg.
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    Should I apply to Ruse?

    When you say you are pretty all rounded, does it mean you are rotund?
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    induction question

    Love your very nice handwriting too, harrowed2.
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    It looks like you are unable to "get it" by yourself. So you will need someone to help you, if you can be helped at all. Going to a Chemistry coaching centre isn't the place for you. You will need someone, like jazz519, to give you personalised help. Getting a "right person" to help you is very...
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    To all the people who have done the hsc or are in year 12 right now...

    Lots of people seek advice. Say just a few things about their problem. Do we really have enough background info about you to give you meaningful advice?
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    One suggestion: In the end-of-school-term Lifeline book sales, you can often pick up a good uni 1st year American Calculus text book - usually for less than $10. (I once picked up an unused hardcover university text with about 1000 pages: Physics - by Young & Freedman for only $2!). This...
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    IB Maths Tutor - Sydney