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Mathematics: Post your doubts here!

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I'm stuck on this question, from O580/43/O/N/13 (October November 2013, 43, IGCSE)

It's Question 9ci and 9cii

c) The total number of lines in the first n diagram is 1/2n^3 + pn^2 + qn

i) When n=1, show that p+q=8.5


ii) By choosing another value of n and using the equation in part (c)(i), find the values of p and q.

Need help ASAP.

Link to the paper: http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf
 
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can
1 part d) 540(1+6%)^2 = 606.744
They say it was put 2 years ago means compound interest for 2 years has been given thus the formula
Q3 part aiii) Use englargement matrix and multiply it with the coordinates or make x and y lines at (4,5) and mark coordinates of the triangle along this line then simply mutiply coordinates by 1.5
i use tracing paper for enlaregement!
 
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Q 6 part c) TSA = curved surface area of the frustum i.e area of the larger cone - area of the smaller cone + pie x 1.5^2 ( area of the base of smaller cone) +pie x 3^2 + 2 x pie x 3 x 12
i dont get it can u do the workings plz leave sum space after one part of working!?
 
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For Q 9) a)
6, 9, 12, 15, 18 (Because of adding 3) nth term would be 6+3(n-1) = 6+ 3n -3 so 3+3n
For second one you have the formula so substitute 4 and 5 with n to find their answers
For Number of triangles its the squares, like 2^2, 3^2, so for 4th and 5th term it would be 4^2 and 5^2 and formula would be (1+n)^2 (Simple way is to find the nth term of underoot of these terms i.e 2, 3, 4, 5 and then square the whole term)
 
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In detail
To find number of triangles: it is 2^2, 3^2, 4^2
so the underoots are 2, 3, 4, 5 etc and their nth term is 2+1(n-1) = 1+n
so the nth term would be squared ie (n+1)^2
 
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Part b Solution
360 x 2/3 = n^2 +3n +2
240= n^2 + 3n +2
so
n^2 + 3n + 2 = 240
n^2 + 3n + 2 -240= 0
n^2 +3n -238 = 0
n^2 +17n -14n - 238 =0
n(n +17)-14(n+17) = 0
(n+17)(n-14)=0
Since n cant be negative so n-14= 0 so n is 14
 
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