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Math Help - Several maths questions

  1. #1
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    Several maths questions

    Alright, at uni they make us sit a skills test of stuff we did back at school. Problem is i can't remember half of it! Struggling big time, goign to see a tutor on Monday and got till august to sit but i want to learn as much as i can now. I can pratice on the web this test as many times as i want but the test is supervised the questions that follow are from the practice so am not asking for you guys to do my homework just need help. The test is always the same qurestions just diffrent numbers so i need to find out how to do things. So I have 10 of the 20 questions am not sure of can you please show the steps how to do it or for the 3 graph ones can you tell me how to go around doing them. Since it's alot of questions might be best if you pick one or two to show me and then someone else can pick another few to do or what not, if you want. Al label the ones i can do.
    A. Explained!


    B. Explained!

    C. Explained!


    D. Explained!


    E. Explained!


    F.

    G.


    H.Explained!


    I.


    I know it's alot so any help will be much appricated. Any questions on the above ask and i will explain what the questions asking. The really frustrating thing is i have already passed the degree exam and i still need to sit this . Oh well just need to do it.


    Thanks guys
    Last edited by Johnt447; June 19th 2009 at 08:53 AM.
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  2. #2
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    A & E) (they are the same problem)
    Find the common denominator.
    \frac{x}{x + 1} + \frac{2}{x - 1}
    \begin{aligned}<br />
&= \frac{x(x - 1)}{(x + 1)(x - 1)} + \frac{2(x + 1)}{(x + 1)(x - 1)} \\<br />
&= \frac{x^2 - x}{(x + 1)(x - 1)} + \frac{2x + 2}{(x + 1)(x - 1)} \\<br />
&= \frac{x^2 - x + 2x + 2}{(x + 1)(x - 1)} \\<br />
&= \frac{x^2 + x + 2}{(x + 1)(x - 1)}<br />
 \end{aligned}


    B)
    \begin{aligned}<br />
2x - z &= -x(yz + 1) \\<br />
2x + x(yz + 1) &= z \\<br />
x(2 + yz + 1) &= z \\<br />
x(yz + 3) &= z \\<br />
x &= \frac{z}{yz + 3}<br />
\end{aligned}


    D)
    \frac{x^3 - 3x^2 + 3x - 1}{x^2 - 2x + 1}
    \begin{aligned}<br />
&= \frac{(x - 1)^3}{(x - 1)^2} \\<br />
&= x - 1<br />
\end{aligned}


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  3. #3
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    Thanks
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  4. #4
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    Problem C

    I suppose I could help out with problem C, since that one hasn't been done yet. Start by factoring (2x+1)^2 and (6x-1)^3 from the numerator, since they are common to both terms in the numerator.

    \frac{(2x+1)^2(6x-1)^3[(2x+1)-(6x-1)]}{(2x+1)^6}

    Now there are two things you can do. You can combine like terms inside the brackets, and you can remove the common factor of (2x+1)^2 from the numerator and denominator. It doesn't matter which one you do first. When finished you should come up with

    \frac{(6x-1)^3(-4x + 2)}{(2x+1)^4}

    At this pioint you can also factor out a negative two from the numerator to finish the problem off. I come up with

    \frac{-2(6x-1)^3(2x-1)}{(2x+1)^4}

    Hope this helps.
    Last edited by mr fantastic; June 18th 2009 at 07:36 PM. Reason: Fixed latex, removed color (it was causing problems so easier to get rid of it)
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  6. #6
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    H.
    y = (x - 2)(x^2 - 4x + 3)

    This is the same as
    y = (x - 2)(x - 1)(x - 3)

    We have a cubic, with three real roots, x = 1, 2, 3. That means that the graph crosses the x-axis at (1, 0), (2, 0), and (3, 0). If you were to multiply the factors out, you would see that the leading coefficient is positive (+1), and since the degree of the polynomial (3, cubic) is odd, you're graph is such that as x-> -∞, y -> -∞, and as x -> ∞, y -> ∞.

    x = 1 is the first root, and the curve to the left of this root would be under the x-axis. You can see that this is correct by plugging in x = 0 (part of the interval [0, 4]) -> y = -6. The next place that the graph crosses the x-axis is at x = 2, which means that between x = 1 and x = 2, there is a small "hump" above the axis. And between x = 2 and x = 3 (the final place where the graphs crosses the x-axis), there must be a small "hump" below the axis. Finally, after x = 3, the curve is above the x-axis, and when x = 4, y = 6.



    Just so you know, it's possible to draw a rough sketch of a polynomial without graphing it on a calculator first, if you know things like degree, end behavior, roots, etc.


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  7. #7
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    And again thank you
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