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December 2nd, 2008, 08:42 PM
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| | Need Massive Help!!! With Scanned Image I scanned this because I didn't even know how to explain this problem or where to start. Let me know what you guys think!
Thanks For your help! | 
December 2nd, 2008, 09:21 PM
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Originally Posted by jermio I scanned this because I didn't even know how to explain this problem or where to start. Let me know what you guys think!
Thanks For your help! | Area of a triangle is 1/2 (base)(height).
So the total area (starting from right most triangle) is
You're expected to recognise that this is an infinite geometric series ....
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December 2nd, 2008, 09:32 PM
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Originally Posted by mr fantastic Area of a triangle is 1/2 (base)(height).
So the total area (starting from right most triangle) is
You're expected to recognise that this is an infinite geometric series .... | Thanks. I have been working on it as a geometric sequence, but cannot figure out what n is. I have to find n before I can find the sum of the geometric sequence.
so a(base n) = a(base 1) (r^n-1)
2=1/8(2^n-1)
divide both sides by 1/8
1/4=2^n-1
and now I am stuck. I feel stupid, but I can't figure out how to get n from here. Any thoughts? | 
December 2nd, 2008, 09:45 PM
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Originally Posted by jermio Thanks. I have been working on it as a geometric sequence, but cannot figure out what n is. I have to find n before I can find the sum of the geometric sequence.
so a(base n) = a(base 1) (r^n-1)
2=1/8(2^n-1)
divide both sides by 1/8
1/4=2^n-1
and now I am stuck. I feel stupid, but I can't figure out how to get n from here. Any thoughts? | Quote:
Originally Posted by mr fantastic Area of a triangle is 1/2 (base)(height).
So the total area (starting from right most triangle) is
You're expected to recognise that this is an infinite geometric series .... | *Ahem* ....
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti) To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
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December 2nd, 2008, 10:14 PM
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Originally Posted by mr fantastic *Ahem* .... | So now I am really confused...lol.
So... n --> infinity ?
or
s=a/1-r
s=(1/8)/1-2
s= -1/8?
I am really starting to lose my mind. Thanks for your patience. | 
December 2nd, 2008, 10:26 PM
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Originally Posted by jermio So now I am really confused...lol.
So... n --> infinity ?
or
s=a/1-r
s=(1/8)/1-2
s= -1/8?
I am really starting to lose my mind. Thanks for your patience. | a = 1/2, r = 1/2.
__________________ There are two things you should never try to prove: the impossible and the obvious. The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti) To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
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December 2nd, 2008, 10:38 PM
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Originally Posted by mr fantastic a = 1/2, r = 1/2. | so.... S=1? | 
December 2nd, 2008, 10:49 PM
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Originally Posted by jermio so.... S=1? | Yes.
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| | The following users thank mr fantastic for this useful post: | |  | 
December 2nd, 2008, 10:49 PM
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Originally Posted by mr fantastic Yes. | SAWEET! | 
December 3rd, 2008, 12:16 AM
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Originally Posted by jermio I scanned this because I didn't even know how to explain this problem or where to start. Let me know what you guys think!
Thanks For your help! | The area enclosed by the triangles is obviously half the area of the rectangle that encloses them which is 1 by 2, so the required area is 1.
(each triangle encloses half the area of the rectangle on the same base and of the same height that encloses it and the sum of the areas of these rectangles is the area of the big rectangle)
CB
__________________ Truth does not change because it is, or is not, believed by a majority of the people.
Giordano Bruno | 
December 3rd, 2008, 12:35 AM
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Originally Posted by CaptainBlack The area enclosed by the triangles is obviously half the area of the rectangle that encloses them which is 1 by 2, so the required area is 1.
(each triangle encloses half the area of the rectangle on the same base and of the same height that encloses it and the sum of the areas of these rectangles is the area of the big rectangle)
CB | Hmmmm .... And I suppose when you go to France you get the boat or the plane rather than swim.
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