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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
no abcde u made a mistake q=3 not 0 therefore p =2
no abcde u made a mistake q=3 not 0 therefore p =2
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Mathematics: Post your doubts here!
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
heyy can anyone help me answer this question, its been irritating me since i've never seen a question like this before!
http://www.xtremepapers.me/CIE/Cambridg ... _qp_41.pdf
question 9 (a) (iv) (v)
heyy can anyone help me answer this question, its been irritating me since i've never seen a question like this before!
http://www.xtremepapers.me/CIE/Cambridg ... _qp_41.pdf
question 9 (a) (iv) (v)
XPFMember
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Assalamoalaikum!!
hey...i saw ur question...and smtng cliked my mind
welll u r lucky...although i didnt have time to solve tis.bt check this viewtopic.php?f=24&t=2842&start=480#p67049 go from this post onwards...this question's discussed
Assalamoalaikum!!
hey...i saw ur question...and smtng cliked my mind
welll u r lucky...although i didnt have time to solve tis.bt check this viewtopic.php?f=24&t=2842&start=480#p67049 go from this post onwards...this question's discussed
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Asalam-O-Alaikum,
Hey guys tomorrow is my paper 4 of maths and it seems that I have got many doubts over here. I would be very much pleased and Thankful if someone could help me out!
The questions are Maths p4 2010 M/J 41
Q 2 (b)
Q 3 (d)
Q 7 (c) (i)
(iii)
(iv)
Q 11 (d)
Maths p4 2010 M/J 42
Q 1 (b) i
Q 2 (b) iii
Q 3 (d)
Thanks alot guys!
Asalam-O-Alaikum,
Hey guys tomorrow is my paper 4 of maths and it seems that I have got many doubts over here. I would be very much pleased and Thankful if someone could help me out!
The questions are Maths p4 2010 M/J 41
Q 2 (b)
Q 3 (d)
Q 7 (c) (i)
(iii)
(iv)
Q 11 (d)
Maths p4 2010 M/J 42
Q 1 (b) i
Q 2 (b) iii
Q 3 (d)
Thanks alot guys!
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
guys u have got to help me here!!! Question 10 (a)!!!
http://www.xtremepapers.me/CIE/Cambridg ... _qp_43.pdf
guys u have got to help me here!!! Question 10 (a)!!!
http://www.xtremepapers.me/CIE/Cambridg ... _qp_43.pdf
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Check it out on the link Math_Angel gave you please and then let us know if you still have doubts. Do read throughly! :wink:
Hehe, I've answered this question before!soccerprinc3 said:
Check it out on the link Math_Angel gave you please and then let us know if you still have doubts. Do read throughly! :wink:
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
yes here we go
if the mean is 10 and there are six numbers then the total is 60
then the smallest number is more than 6 and the lagest is 16
the median is 9 and you have only six numbers so 9=the average of the third and fourth integer
so now arrange x x 8 10 x 16 then the qsn says that the mode is 8 so you have more than one 8 and it should be replaced by the first two x's
x 8 8 10 x 16 then it says that the first number is greater than 6 so it would be 7 or 8 so try the 7 and you'll have 7 8 8 10 11 16 or 8 8 8 10 10 16
hope this helped
yes here we go
if the mean is 10 and there are six numbers then the total is 60
then the smallest number is more than 6 and the lagest is 16
the median is 9 and you have only six numbers so 9=the average of the third and fourth integer
so now arrange x x 8 10 x 16 then the qsn says that the mode is 8 so you have more than one 8 and it should be replaced by the first two x's
x 8 8 10 x 16 then it says that the first number is greater than 6 so it would be 7 or 8 so try the 7 and you'll have 7 8 8 10 11 16 or 8 8 8 10 10 16
hope this helped
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Q2)(b) In the previous part you must have got this expression for the mean : 190/40
For the other n set of students, the total number of people will be 3n (mean = sum of all people in all families/no of families)
So the expression will be (190 + 3n)/(40 + n).
Q3) (d) For a 90* clockwise rotation about the origin, the transformation matrix is always (0 1)
(-1 0)
Q7) (c)(iii)The perimeter consists of three straight lines and three arcs. Each arc is the same as arc DE (found in previous part to be 0.6492625 m).
Each straight line actually is made of 6 radii of the circles. So perimeter = 3(0.6492625) + 6(3)(0.31) = 7.53 m (correct to 3 sig. fig.)
(iv) simply multiply the perimeter with the length = 7. 5277875 x 15 = 113 m^2 (correct to 3 sig. fig.)
Q11) (d) (i) Just consider the first diagram. n = 1 and the number of lines is 6.
=> k (1^2 + 1 + 1) = 6
=> 3k = 6
so k = 2
(ii) Insert 100 into the formula.
2 (100^2 + 100 + 1 ) = 2 (10 000 + 100 + 1 ) = 2 x 10, 101 = 20, 202.
Hope this cleared your confusion!
Wa- alikum-a-salam!Martee100 said:
Q2)(b) In the previous part you must have got this expression for the mean : 190/40
For the other n set of students, the total number of people will be 3n (mean = sum of all people in all families/no of families)
So the expression will be (190 + 3n)/(40 + n).
Q3) (d) For a 90* clockwise rotation about the origin, the transformation matrix is always (0 1)
(-1 0)
Q7) (c)(iii)The perimeter consists of three straight lines and three arcs. Each arc is the same as arc DE (found in previous part to be 0.6492625 m).
Each straight line actually is made of 6 radii of the circles. So perimeter = 3(0.6492625) + 6(3)(0.31) = 7.53 m (correct to 3 sig. fig.)
(iv) simply multiply the perimeter with the length = 7. 5277875 x 15 = 113 m^2 (correct to 3 sig. fig.)
Q11) (d) (i) Just consider the first diagram. n = 1 and the number of lines is 6.
=> k (1^2 + 1 + 1) = 6
=> 3k = 6
so k = 2
(ii) Insert 100 into the formula.
2 (100^2 + 100 + 1 ) = 2 (10 000 + 100 + 1 ) = 2 x 10, 101 = 20, 202.
Hope this cleared your confusion!
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Q1(b) (i) Interest = Amount - principal = 99 - 90 = $9.
Using the formula, I = PRT /100
=> 9 = (90 x r x 2)/100
=> r = 5.
Q2) (b) (ii) OM = OC + CM
Q3) (d) Probabilty that the spinner stops on 2 = 2/6 = 1/3
probability that it does not stop on 2 is = 4/6 = 2/3
(2/3)^(n-1) x 1/3 = 16/243 = 2^4/3^5.
Now you do the simple arithmetic! Hope this helped a lot!
Martee100 said:
Q1(b) (i) Interest = Amount - principal = 99 - 90 = $9.
Using the formula, I = PRT /100
=> 9 = (90 x r x 2)/100
=> r = 5.
Q2) (b) (ii) OM = OC + CM
Q3) (d) Probabilty that the spinner stops on 2 = 2/6 = 1/3
probability that it does not stop on 2 is = 4/6 = 2/3
(2/3)^(n-1) x 1/3 = 16/243 = 2^4/3^5.
Now you do the simple arithmetic!
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Shearing is parallel to the invariant line. The image was to the right of the original. So inv. line was x-axis.henry407 said:
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Well its quite simple !!
As the triangle inside the circle is equilateral, all 3 sides are 60 degrees
so the angles of the half triangle is 30 degrees
Therefore : 90-30=60 degrees
so both sides are 60*
umm well dats how it is I guess !!
Correct me if I am wrong
abcde said:
Well its quite simple !!
As the triangle inside the circle is equilateral, all 3 sides are 60 degrees
so the angles of the half triangle is 30 degrees
Therefore : 90-30=60 degrees
so both sides are 60*
umm well dats how it is I guess !!
Correct me if I am wrong
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Yes, it is this way.
Yes, it is this way.
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Make a new topic, if you want to.
Make a new topic, if you want to.
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
AoA!
Can you kindly look at Q22(c) of this paper: http://www.xtremepapers.me/CIE/Cambridg ... 4_qp_2.pdf
It's a one mark question but please tell me how you figure it out.
JazakAllah!
and also
Can you kindly help me with question 9 part (f) of this paper: http://www.xtremepapers.me/CIE/Cambridg ... 4_qp_4.pdf
Here's the mark scheme: http://www.xtremepapers.me/CIE/Cambridg ... s04_ms.pdf
Your help will be very much appreciated. Thank you!
AoA!
Can you kindly look at Q22(c) of this paper: http://www.xtremepapers.me/CIE/Cambridg ... 4_qp_2.pdf
It's a one mark question but please tell me how you figure it out.
JazakAllah!
and also
Can you kindly help me with question 9 part (f) of this paper: http://www.xtremepapers.me/CIE/Cambridg ... 4_qp_4.pdf
Here's the mark scheme: http://www.xtremepapers.me/CIE/Cambridg ... s04_ms.pdf
Your help will be very much appreciated. Thank you!
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Re: Maths made easy =) stuck sumwhere?Ask ur doubts here!!
Hey math angel cn u gv me some advice plzzz...i am jst normal at maths i have my exam on monday and wednesday...and i have tried to get da hang ov mathz since 3 yearz but still i gt confuzed at questionz or make silly mistakez...i practised too but somehow wen da question is chnged i jst make a blunder in it or am thinking nt accordingly to the solution ov da question or fall into the examinerz trap in a particular question and cm up wd a wrong answer...wat cn i do in deze dayz before da exam...lyk itz tomorrow???practice iznt making me perfect at it and nw i am actually very disheartened and scared and i have even abandoned my mathz bookz in hopelessness...cn u tell me wat to do thnx a lot....
Hey math angel cn u gv me some advice plzzz...i am jst normal at maths i have my exam on monday and wednesday...and i have tried to get da hang ov mathz since 3 yearz but still i gt confuzed at questionz or make silly mistakez...i practised too but somehow wen da question is chnged i jst make a blunder in it or am thinking nt accordingly to the solution ov da question or fall into the examinerz trap in a particular question and cm up wd a wrong answer...wat cn i do in deze dayz before da exam...lyk itz tomorrow???practice iznt making me perfect at it and nw i am actually very disheartened and scared and i have even abandoned my mathz bookz in hopelessness...cn u tell me wat to do
thnx a lot....