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Maths, Addmaths and Statistics: Post your doubts here!

ICT

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Where can i get the mark schemes of maths (4024) for the papers from 1990-2000?
 
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sorry but i havent checked through marking scheme , if you found some thing wrong in it pls correct me. !!!

I think u have to take half the are of cylinder which u did then take quarter the area of a sphere and also add half the area of the circle add these up and u get the surface area :p
 
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I think u have to take half the are of cylinder which u did then take quarter the area of a sphere and also add half the area of the circle add these up and u get the surface area :p

I could not understand i have took the area of 1/4 sphere and 1/2cylinder , but havent took area of circle bcuz that circle is already included in cylinder!!! o_O:p
 
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Its only released when syllabus or paper format is changed this year it wasnt :D

,

Have u checked ur answer with the ms because i think theres something fishy in it :p

do you know the answer ??
actually the answer not the solution! :D
 
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I could not understand i have took the area of 1/4 sphere and 1/2cylinder , but havent took area of circle bcuz that circle is already included in cylinder!!! o_O:p

Actually no its not unless u take the full formula that is 2(pi)rh + 2(pi)r^2
now here 2(pi)rh means the area of the curved surface of the cylinder and not the circle while the circle formula u use gives the area of circle in this way u get the area of whole figure !
Now the formula doesnt mean that u cant play with it :D u can remove 2 from the area of circle when u only require one circle and divide multiply according to ur need !
 
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Actually no its not unless u take the full formula that is 2(pi)rh + 2(pi)r^2
now here 2(pi)rh means the area of the curved surface of the cylinder and not the circle while the circle formula u use gives the area of circle in this way u get the area of whole figure !
Now the formula doesnt mean that u cant play with it :D u can remove 2 from the area of circle when u only require one circle and divide multiply according to ur need !


okay.. thnx for that..I would get better understand my mistake by your solution to query . . and thanx in advance for correcting me !!!!:)
 
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ok then how do you get the surface area of the liquid in the sphere...
i got all the rest i got stuck on that part...:/

when does it ask abt that part,now can u tell me are u talking abt the same question then tell me the part ?? :p
 
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b) i)
There are different ways of doing this... You can use the formula if you wish...

a + (n-1)d
a is the first term...
d is difference...

So 10+(n-1)4
10+4n-4
6+4n

OR

find the difference between the values...

4 in this case...

Then write it as 4n+-(x) = 10

10 is the first value... S0 n =1

4(1)+-(x)=10

You have to find such x value which satisfies the equation... So it it is 4 in this case... Put it in the first equation and just take the left side of it... 4n+6

ii) The first difference is not the same for this part... The second difference is...

So its a quadratic sequence..

The formula for quadratic sequence is a+(n-1)d1+(1/2)(n-1)(n-2)d2
a is the first term of sequence
d1 1st difference
d2 2nd difference...

a is 2
d1 is 4
d2 is 2

2+(n-1)4+(1/2)(n-1)(n-2)2
Solve and you get n^2+n

c) Add the equations of b) i and b) ii

n^2+n +4n + 6
n^2+5n+6
n^2+2n+3n+6
n(n+2)+3(n+2)
(n+3)(n+2)
(n+2)(n+3)
i want to ask a little question in part b in second part how is d2 2 ... d1=(6-2)=4....wont d2 be (12-6)??...
 
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In O level matrix transformation what is the invariant line for shear parallel to both
x axis.....
y axis....
 
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i want to ask a little question in part b in second part how is d2 2 ... d1=(6-2)=4....wont d2 be (12-6)??...
2, 6, 12
First difference between the consecutive numbers : 4, 6
Second difference between the new numbers is now 6-4= 2.
 
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2, 6, 12
First difference between the consecutive numbers : 4, 6
Second difference between the new numbers is now 6-4= 2.
ohh thnku so much for ur help bro....will this formula be applied to any sequence where the difference is different?? (a+(n-1)d1.+1/2(n-1)(n-2)d2)
 
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Cricle has five sectors with the same angle...

360/5= 72

Try finding the shaded area :

Area of one small circle = (72/360) π r^2 - π R^2

As each small circle is equal in area to the shaded region that makes the total area 10 times the area above. 5 times the small circle... And 5 times the shaded area...
Total area = 10( (72/360) π r^2 - π R^2)

Total area is also equal to π R^2

So

π R^2 = 10( (72/360) π r^2 - π R^2)

Solve it you will get 10 : 1
 
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