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help needed!

by xcise_science » Thu Dec 06, 2007 6:57 am
Can you someone help me with these 2 questions?

(I keep getting a '500 internal server' error each time I try to add my image, whats up with that?)


Thanks
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by chuongbill » Thu Dec 06, 2007 7:53 am
Question 1.
We have the length of wire used to form a circle is circumference of the circle which means 2*pi*r, so the remains of wire used to form a square is (40 - 2*pi*r).
The area of the circle is pi*r^2 and the square is (40 - 2*pi*r)^2. It means that represent the total area of the circular and the square regions in terms of r is D: pi*r^2 + (40 - 2*pi*r)^2.

Question 2.
2^x - 2^(x-2) = 3(2^13)
-> 2^x - (2^x)/(2^2) = 3(2^13)
-> (2^2)*(2^x) - 2^x = (2^2)*3(2^13)
-> 4(2^x) - (2^x) = 3(2^15)
-> 3(2^x) = 3(2^15)
-> 2^x = 2^15
-> x = 15
The answer for this question is D:15.

Hope it's useful!
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by xcise_science » Thu Dec 06, 2007 8:59 am
Hi, thanks for the quick response.
For the first question (cut wire), the answer is not D, which is what I chose. The answer is E.


Thanks
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by xcise_science » Thu Dec 06, 2007 9:55 am
For the wire question, maybe it has something to do with 'the total area in square meters' part of the question? Because the original length was given as 40meters.
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by sirikesav » Thu Dec 06, 2007 12:51 pm
2pr + 4s = 40

s= 40-2pr/4

Area = pir^2 + (40-2pir)^2/16
=(pi)r^2 + (10 -1/2 (pi)r)^2

please note : here pi = 22/7

hence ans E
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by xcise_science » Thu Dec 06, 2007 4:21 pm
great thanks.
I started off that path as well, but somehow dropped it....
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by khurram » Sun May 25, 2008 11:04 am
2pr + 4s = 40

s= 40-2pr/4

Area = pir^2 + (40-2pir)^2/16
=(pi)r^2 + (10 -1/2 (pi)r)^2

please note : here pi = 22/7

hence ans E

Best way to solve this question

Thanks
khurram
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