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Mathematics: Post your doubts here!

Discussion in 'Cambridge IGCSE' started by XPFMember, Sep 24, 2010.

  1. delta.charlie321

    delta.charlie321

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    Thank you so much. I just have one more question. In the examiner report they say another method is to find the average of 25 and 35 as the width of the water. Can you explain what is the mathematical basis for using this method? I cant find it in any textbooks or on the internet?

    [​IMG]
     
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  3. delta.charlie321

    delta.charlie321

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    I just found what the Examiners meant when they said the width of the water can be found out by the average of 25 and 35. They are talking about the median of the trapezoid, which is a line halfway between two bases. The median is always average of two bases.
     
  4. NamChanachon

    NamChanachon

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    Ok guys, I'm taking the International Mathematics (code 0607) and I'm a little stuck on rotating images. For example, if the question asks us to "Rotate point A by 90 degrees" does that mean 90 degrees clockwise or counter clockwise? (normally the question specifies this but I would like to know JUST in case). Thanks in advance!
     
  5. delta.charlie321

    delta.charlie321

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    For rotating an image by 90 degrees the direction must be specified to avoid ambiguity otherwise the marking scheme must allow for both directions. I dont know about 0607 Mathematics but in IGCSE 0580 a direction is always specified for 90 degree rotations so I would assume this to be true for other syllabuses as well.
     
  6. Nabzz_96

    Nabzz_96

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    Heyya guys!! Could u help me with questiom no. 22, 27 and 28 in the image.
     

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  7. delta.charlie321

    delta.charlie321

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    It is very difficult to answer the question without knowing which aspect of the question u find difficult? It probably looks like homework and you should try it on your own. But I will give you some tips.
    Q.22 find the gradient and then use y=mx+c and find the value of c
    Q.27 and 28 You need to setup simultaneous equations in terms of a and b using the coordinates given

    The methods may differ depending on your syllabus but this is how I would approach it.
     
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  8. Nabzz_96

    Nabzz_96

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    Thanks for getting back to me........nope its not homework, i am a private candidate having my exams in 2 weeks. :D

    Anyhow i figured out Q22 and i used the same method as you suggested. but Q27 is a bit confusing which is why i left it aside for now......i will solve it once i finish learning about the curves and graphs.....
    thnx once again :) I have a bad habit of panicking :p
     
  9. delta.charlie321

    delta.charlie321

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    Which syllabus are you following. I am mostly familiar with 0580 IGCSE mathematics.

    PS
    In questions 27 and 28, notice that they have given a set of coordinates where x is 0 like (0,5). Use this set of coordinates to find the value of c (which is equal to 5 in both q.27 and 1.28). From there you can easily setup simultaneous equations with remaining set of coordinates. Let me know if you need more help.
     
    Last edited: Oct 17, 2016
  10. Nabzz_96

    Nabzz_96

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    OMG!!! I hate working with X now... :(
     
  11. Nabzz_96

    Nabzz_96

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    thnx for the help :D i will definitely come back to trouble you :D
     
  12. Hamna naseer

    Hamna naseer

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    upload_2017-3-18_0-5-34.png
    pzz help
    0580/s10/22
     
  13. Ayesha Asif333

    Ayesha Asif333

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    a) sin(45) - cos(45)= 0

    because we want an two angles when subtracted will give us 0, we take the point where both sin x curve and cos y curve intersect which is exactly 45 degrees
    b) sin(66) - cos(66)=0.5

    check where on the graph is the difference between two curves o.5...so at 66 the position of two points on the curve is such that the distance between the two is 10 small blocks...for sinx curve the value at y axis is 0.9 and for cosx curve the value y axis(from that point) is 0.4, so when we subtract the two we get the difference which is 0.5

    i can't explain any better then this so hope u get it
     
  14. bassamkhan35

    bassamkhan35

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    Hi
    Can anyone help in these questions.
    0580/42/M/J/14
    Q. 6 part (b) and (c)
    Q. 7 part (a) ii
    Q. 11 full
     
  15. Ayesha Asif333

    Ayesha Asif333

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    180 - 115= 65

    FEG= 90- 65=25

    FGE= 90 so GFE= 180- 25-90= 65...all the angles in a triangle add up to 180

    GHE= 180- 65= 115 coz opp angles in a cyclic quadrilateral are supplementary

    angle EGH = GEF....25=25

    so angle GEH is 180- 25-115=41

    41's the answer


    (c)...make a line joining the point c with the center, u'll get a triangle

    then find AOC which is 180-14-14= 152

    360-152=208 is the angle at the center

    so 208/2 will give u 104....this is angle ADC

    base angles in an isosceles are equal so get ACD we subtract 104 with 180 then divide it by 2

    the ans u'll get should be 38
     
  16. Ayesha Asif333

    Ayesha Asif333

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    multiply the values of the frequency densities with its class width(from the graph) to get all the frequencies …the formula is f=f.d * c.w

    since the class intervals are given and we need x so take the midvalue(x) for each which is calculated by adding the highest and least value of masses in a class eg for the first one we have 0 and 30 so 30/2

    once you’ve got all the f and x values, multiply the two to get fx

    once you’ve got all the fx values, add all of them to get ∑fx

    the last step will be the mean formula which is, in case you don’t know ∑fx / ∑f, and you’ll get the right answer
     
  17. jamesosama

    jamesosama

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    can some one help me with o580/21/mj/10
    QUESTION 7,15
    0580/21/O/n/2010 (A)QUESTION 18(A)PART 2
    O580/21/M/j/11 QUESTION13,question 16 part (a),18PART(B)
    0580/21/0/n/11 QUESTION 18
    0580/21/m/J/12 question 19 full and question 20 and21
    o580/21/mj/2013 QUESTION 20(B) AND 26
    0580/21/O/n/2013 QUESTION 18 AND 21
     
  18. Ayesha Asif333

    Ayesha Asif333

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    q7) SA= ( 0.8* 1.4) + ( 90/360* π 0.8square)

    q15) a) method 1: search for the pattern in the points they’ve given, since if of one mark you’re not required to do much working

    method 2: to find the gradient m, then substitute any value of x and y in the equation y= mx +c to obtain the value for c, then the last step should be the substitution of (1,k) in the final equation

    b) same equation you made earlier
     
  19. Ayesha Asif333

    Ayesha Asif333

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    method 1 : draw a tangent then fill in any two values of x and y in the gradient equation

    method 2: draw a tangent, make a big triangle, count the points on vertical axis and the horizontal axis then divide the two….this is the rise over run rule
     
  20. Ayesha Asif333

    Ayesha Asif333

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    q13) a) 1 : 20000

    0.000027: x


    b) 1 :20000square

    x : 644000000



    q16) a) the shaded area is A

    and A =’s the area of the square minus area of the circle

    area of the square is k square

    the area of the circle is π(k/2)whole square ….where k/2 is the radius of the circle

    now you can make the equation


    q18) b) MX = MR + RX

    r/2 + 3/4 q – r

    Simplify it further and you'll get the ans
     
  21. Ayesha Asif333

    Ayesha Asif333

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    i'm tired now
    maybe i'll do them some other day
     

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