Form5 add maths project 2009

[cschuah92]

can tel me why is part 3(b) πx^2/4? I mean why is the power 2/4?

Nope.
If the power is 2/4, then it should be -πx^(2/4)
-πx^2/4 means (-πx^2)/4.
It is just same like (-1/4)*π*x^2.
Note that "*" represents multiply.

[azfarnds]

PLease someone give me the full conclusion please........

[cris]

urm....can u just give the full conclusion please(gomen nasai)PLEASE FULL CONCLUSION

ok here you are http://www.4shared.com/file/103736687/df88d863/conclusion.html

well im sorry if theres a bad grammar.hope you can fix it if im wrong..

[cschuah92]

urm....can u just give the full conclusion please(gomen nasai)PLEASE FULL CONCLUSION

ok here you are http://www.4shared.com/file/103736687/df88d863/conclusion.html

well im sorry if theres a bad grammar.hope you can fix it if im wrong..

Why don't you just post it here?
Or you don't know how to post it here?
I found many people having problems to download files from file hosting website including me.
xD

[singeryoon92]

I just rewrote the conclusion that was done by cris.
I don't know whether the content is correct or it have some grammar error or typing error.

Part 1
Not all objects surrounding us are related to circles. If all the objects are circle, there would be no balance and stability. In our daily life, we could related circles in objects. For example: a fan, a ball or a wheel. In Pi(π), we accept 3.142 or 22/7 as the best value of pi. The circumference of the circle is proportional as pi(π) x diameter. If the circle has twice the diameter, d of another circle, thus the circunference, C will also have twice of its value, where preserving the ratio =Cid

Part 2
The relation between the length of arcs PQR, PAB and BCR where the semicircles PQR is the outer semicircle while inner semicircle PAB and BCR is Length of arc=PQR = Length of PAB + Length of arc BCR.The length of arc for each semicircles can be obtained as in length of arc = 1/2(2πr). As in conclusion, outer semicircle is also equal to the inner semicircles where Sin=Sout .
Part 3
In semicircle ABC(the shaded region), and the two semicircles which is AEB and BFC, the area of the shaded region semicircle ADC is written as in Area of shaded region ADC =Area of ADC – (Area of AEB + Area of BFC). When we plot a straight link graph based on linear law, we may still obtained a linear graph because Sin=Sout where the diameter has a constant value for a semicircle.

[ErrorlnDisplay]

umm.. is there conjecture for part 3 ?

[izarimaurice]

3 (e) Arithmetic Progression
The keywords are:
The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society. (n = 12)
The sum of the diameters of the semicircular flower beds is 10 m. (S12 = 10 m)
The diameter of the smallest flower bed is 30 cm. (a = 30 cm = 0.3 m)
The diameter of the flower beds are increased by a constant value successively. (d = ?)
S12 = (n/2)[2a + (n - 1)d]
10 = (12/2)[2(0.3) + (12-1)d]
= 6(0.6 + 11d)
= 3.6 + 66d
66d = 6.4
d = 16/165
Since the first flower bed is 0.3 m,
Hence the diameters of remaining 11 flower beds expressed in arithmetic progression are:
131/330 m, 163/330 m, 13/22 m, 227/330 m, 259/330 m, 97/110 m, 323/330 m, 71/66 m, 129/110 m, 419/330 m, 41/30 m

If there is any mistake please kindly PM or e-mail me.
I haven't done my project work yet but this is my brief working on it.
Let me know my mistake so that I won't get wrong when I started doing it.

Yo, they want the answers in metre or centimetre?? Urgent reply asap!!

[cschuah92]

3 (e) Arithmetic Progression
The keywords are:
The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society. (n = 12)
The sum of the diameters of the semicircular flower beds is 10 m. (S12 = 10 m)
The diameter of the smallest flower bed is 30 cm. (a = 30 cm = 0.3 m)
The diameter of the flower beds are increased by a constant value successively. (d = ?)
S12 = (n/2)[2a + (n - 1)d]
10 = (12/2)[2(0.3) + (12-1)d]
= 6(0.6 + 11d)
= 3.6 + 66d
66d = 6.4
d = 16/165
Since the first flower bed is 0.3 m,
Hence the diameters of remaining 11 flower beds expressed in arithmetic progression are:
131/330 m, 163/330 m, 13/22 m, 227/330 m, 259/330 m, 97/110 m, 323/330 m, 71/66 m, 129/110 m, 419/330 m, 41/30 m

If there is any mistake please kindly PM or e-mail me.
I haven't done my project work yet but this is my brief working on it.
Let me know my mistake so that I won't get wrong when I started doing it.

Yo, they want the answers in metre or centimetre?? Urgent reply asap!!

The question don't mention anything about that.
We can present the answer in any way we prefer.