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IGCSE starting on Friday... pls.post doubts, important notes etc.

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well.i think it'll be easy for you..anyway..i somehow got stucked in here..
x:y = 6:5 and y:z = 4:5 find x:y:z in integer form.
 
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wrong 3rd second
substitute 3rd second and 2nd second find the difference
the ans is 35m.
 
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sorry I forgot 'bout that
x:y...........y:z
6:5...........4:5
make y values same..by multiplying x:y by 4
& multiplying y:z by 5
x:y:z=24:20:25
it's in one of the 2006 papers rite??
 
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1)Simplify the ratio 1 km : 800 m : 50 cm, giving your answer in the form
a : b : 1.

2)A length of wire encloses a square of area 144cm^2
a)Calculate the perimeter of the square.
The length of wire is now bent it to the shape of a regular hexagon.
b)Calculate the area, in cm^2,to 3 sgf, of the hexagon.

3)A bag contains 3 blue marbles and 2 red marbles. A marble is to be taken at random from
the bag and then replaced. A second marble is then to be taken at random from the bag.
Calculate the probability that the colours of the two marbles will be different.

4)A right pyramid with a square base and a right circular cylinder have the same height of 12cm and the same volume.The length of a side of the base of the pyramid is 9cm.Calculate the radius, in cm,to 3sgf, of the cylinder.
 
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sorry bro..i got to go..i'll talk with you later..
thanks for the questions and answers
good luck :)
 
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gud but 4 is wrong...

4)Volume of Sphere=(1/3 * base area * height)............(sq,base=9*9)
.................................=1/3 * 9 * 9 * 12
.................................=324cm^3
Volume of Sphere=Volume of cylinder
324=pi * r^2 * h
r= root {324/(pi * 12)}
r=2.93cm
 
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An arithmetic series has first term "a" and common difference "d".
a. Prove that the sum of the first "n" terms is = n/2[2a+(n-1)d]
 
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Recall that an arithmetic series looks like:
(a) + (a + d) + (a + 2d) + ...

Where the nth term is given by:
A[n] = a + (n - 1)d.

The sum of the first n terms is then, counting from term 1 to term n:
S[n] = (a) + (a + d) + (a + 2d) + ... + (a + (n - 3)d) + (a + (n - 2)d) + (a + (n - 1)d).

Also, counting from term n to term 1:
S[n] = (a + (n - 1)d) + (a + (n - 2)d) + (a + (n - 3)d) + ... + (a + 2d) + (a + d) + (a).

Adding the two together term by term:
S[n] + S[n] = 2 S[n]
= ((a) + (a + (n - 1)d)) + ((a + d) + (a + (n - 2)d) + ... + ((a + (n - 2)d) + (a + d)) + ((a + (n - 1)d) + (a)).

There are two as in each term, add them together:
2 S[n] = (2a + (n - 1)d) + (2a + d + (n - 2)d) + ... + (2a + (n - 2)d + d) + (2a + (n - 1)d).

Now factor the ds together:
2 S[n] = (2a + (n - 1)d) + (2a + (1 + n - 2)d) + ... + (2a + (n - 2 + 1)d) + (2a + (n - 1)d).

Simplify each of the d factors:
2 S[n] = (2a + (n - 1)d) + (2a + (n -1)d)) + ... + (2a + (n - 1)d) + (2a + (n - 1)d).

Since there are n terms all the same, we can write this as a multiplication by n:
2 S[n] = n (2a + (n - 1)d).

Halve to get the sum of n terms:
S[n] = (n / 2) (2a + (n - 1)d).


P.S. i copied it from answers.yahoo.com :D
 
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