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Math Help - please help me with this exam practise

  1. #1
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    please help me with this exam practise

    Please help me with this as I would like to check my answers...

    Any help is really appreciated...
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Paulo1913 View Post
    Please help me with this as I would like to check my answers...

    Any help is really appreciated...
    It would be easier if you posted your work and then we could comment on it
    directly.

    RonL
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  3. #3
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    Hello, Paulo1913!

    Please help me with this as I would like to check my answers...

    It's hard to see your work from here,
    . . but it looks like you played the 7\heartsuit instead of K\spadesuit,
    . . you transposed to the key of G major instead of E\flat minor,
    . . and your units are in fathoms/fortnight instead of meters/second.

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  4. #4
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    I can't post my answers here as they would take forever to write out... even if you could do one or two questions it would be appreciated...
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  5. #5
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    please help

    I really need help with this...
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  6. #6
    Senior Member DivideBy0's Avatar
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    Oh, you can't be serious. Just post your final answer. Whatever it might be. 204? 39.5? 1958357? It'll only take a minute.
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  7. #7
    Grand Panjandrum
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    1.a 1.961505..

    1.b 0.718242..

    RonL
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  8. #8
    Super Member angel.white's Avatar
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    For question2.a: use the Pythagorean theorem, a^{2}+b^{2}=c^{2}
    where a= length of triangle
    b = height of triangle
    c = hypotenuse of triangle

    For example, if you had a triangle with a length of 3, and a height of 4 find the hypotenuse:

    1. 3^{2}+4^{2}=c^{2}

    2. 9+16=c^{2}

    2. 25=c^{2}

    2. \sqrt{25}=c

    2. 5=c

    So the hypotenuse would be 5.

    So figure out what you know (hint: you know the length and the height) and solve for what you want to know (the paper will tell you what you want to know)


    -----
    For 2.b
    The rectangle will have the same length on it's top side as it has on it's bottom side, so the top of the box is 2.8m long. The triangle shares this side, so the triangle has a length of 2.8 and a height of 1.2

    You should know that tangent of an angle tells you the opposite side divided by the adjacent side (in this case, height divided by width).

    So tan(EHD)=\frac{ED}{FG}

    Plug in the values you know:
    tan(EHD)=\frac{1.2}{2.8}

    Solve for EHD by taking the arctangent of both sides:
    EHD=arctan(\frac{1.2}{2.8})

    Now just plug it into your calculator:
    EHD\approx 23.2

    -----

    2.c
    Look at the problem and see that if you draw a horizontal line straight left from point K, it will intersect line EF, label that point "P" Now angle PKH is a 90 angle. This means you only have to solve for PKF, then add 90.

    Now PKF will be the same as KFG, (I can explain that if you need, or you can think about it for a while, remembering that there are 180 degrees in a triangle, and 90 degrees at the corner of a rectangle)

    So we need to solve for KFG, and add 90. Why don't you try, using the same technique from 2.b (but this time it won't be tangent, think about what values you know, then decide which trigonometric function you need to use)

    -----

    2.d
    Look at the problem and see that CB=JI

    Since you have AIand JI, you can use the Pythagorean theorem to find AJ (as we did in question 2.1)

    Now, with AJ and DA, you can use the Pythagorean theorem to find DJ, I already told you how to find AJ, can you figure out how to find DA? (hint, GC=HJ)
    Last edited by angel.white; November 10th 2007 at 01:27 AM.
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