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Chikis

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Posts posted by Chikis

  1. [MATH]\frac{1}{m-1}+\frac{9}{2m+3}-\frac{8}{m+4}[/MATH]

    =

    [MATH]\frac{(2m+3)(m+4)+9(m-1)(m+4)-8(2m+3)(m-1)}{(m-1)(2m+3)(m+4)}[/MATH]

    =

    [MATH]\dfrac{2m^2+8m+3m+12+9m^2+36m-9m-36-16m^2+16m-24m+24}{(m-1)(2m+3)(m+4)}[/MATH]

     

    =

    [MATH]\frac{30m-5m^2}{(m-1)(2m+3)(m+4)}[/MATH]

    =

    [MATH]\frac{5m(6-m)}{(m-1)(2m+3)(m+4)}[/MATH]

    Am confident that this one is very correct.

    Just go through it slowly. You will have to show your working out here for us to spot a mistake. Typically mistakes are due to minus signs and silly mistakes with multiplication. You are on the right lines with this, so don't give up.

    Thanks for the caoching.

    [math]\dfrac{(2m+3)(m+4)+(9m-9)(m+4)-(16m+24)(m-1)}{(m-1)(2m+3)(m+4)}[/math]Be sure to watch the signs as ajb pointed out.In the above, (2m+3)(m+4) is correct.For (9)(m-1)(m+4) I suggest doing the (m-1)(m+4) first and then multiply through by 9.Same for the (-8)(2m+3)(m-1). Do the (2m+3)(m-1) first and then multiply through by the -8.You laid it out right in the first post with this:[math]\frac{(2m+3)(m+4)+9(m-1)(m+4)-8(2m+3)(m-1)}{(m-1)(2m+3)(m+4)}[/math]

    Thanks for the tip.
  2. Check your numerator again to be sure, I think it could be wrong. I also don't think (mod the possible mistake) that you can simplify this any further.

    I go again:

    [MATH]\dfrac{(2m+3)(m+4)+(9m-9)(m+4)-(16m+24)(m-1)}{(m-1)(2m+3)(m+4)}[/MATH]

    =

    [MATH]\dfrac{46m+11m^2-32}{(m-1)(2m+3)(m+4)}[/MATH]

    Is the numerator okay now?

  3. [MATH]\frac{1}{m-1}+\frac{9}{2m+3}-\frac{8}{m+4}[/MATH]

    =

    [MATH]\frac{(2m+3)(m+4)+9(m-1)(m+4)-8(2m+3)(m-1)}{(m-1)(2m+3)(m+4)}[/MATH]=

    [MATH]\dfrac{2m^2+8m+3m+12+9m^2-9m+36m-36-16m^2+m-24m+24}{(m-1)(2m+3)(m+4)}[/MATH]=

    [MATH]\frac{5m(3-m)}{(m-1)(2m+3)(m+4)}[/MATH]

    What next should be done?

  4. I have managed to complete the diagram. The diagram has the description:

    triangle QPR. <QPR = [latex]140^0[/latex]. QP = 20km and PR = 8km.

     

    I first used cosine rule to find PR, which I got as 14.79km. I again used cosine rule to find <PQR - bearing of R from P, which I got as [latex]20^0[/latex] to the nearest degree. Now, how do you see my work?

     

    As for the second part of the question (b), am yet to understand it fully. How do they mean by "how far north of P, R is". From the way I look at it, in the diagram Q is 20km north of P. Or do they mean the distance PR, which I got as 14.79km?

  5. Example 1

    [latex]5/3 /4|5[/latex]

    Example 2

    [latex]4/8/2|17[/latex]

    Example 3

    [latex]13/7/4|5[/latex]

    What are the relationship in the example?

    Use the relationship to solve for the missing number?

    [latex]?/5/3|17[/latex]

    just a repost:-)

  6. That's right. You got it! That is to say that 7 is a kind of identity in the relationship.

    The rule given seems not to work with?

    [latex]1.2+5 => y ?[/latex]

     

    [latex]3+2 => z ?[/latex]

    You cannot use the rule to find y and z respectively.

  7. Town Q is 20km due north of P. The bearing of town R from Q is [latex]140^0[/latex]. If R is 8 km from Q. calculate:

     

    (a) the bearing of R from P, to the nearest degree:

     

    (b) how far north of P, R is, correct to 2 significant figures;

    Drawing the diagram is my main problem. If you can help me with the drawing, I can attempt the working.

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