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Old December 2nd, 2008, 07:42 PM
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Default 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!
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  #2  
Old December 2nd, 2008, 08:21 PM
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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 \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \, .....

You're expected to recognise that this is an infinite geometric series ....
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Old December 2nd, 2008, 08:32 PM
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Originally Posted by mr fantastic View Post
Area of a triangle is 1/2 (base)(height).

So the total area (starting from right most triangle) is \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \, .....

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?
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Old December 2nd, 2008, 08:45 PM
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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?
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Originally Posted by mr fantastic View Post
Area of a triangle is 1/2 (base)(height).

So the total area (starting from right most triangle) is \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \, .....

You're expected to recognise that this is an infinite geometric series ....
*Ahem* ....
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Old December 2nd, 2008, 09:14 PM
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*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.
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Old December 2nd, 2008, 09:26 PM
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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.
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Old December 2nd, 2008, 09:38 PM
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a = 1/2, r = 1/2.
so.... S=1?
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Old December 2nd, 2008, 09:49 PM
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so.... S=1?
Yes.
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Old December 2nd, 2008, 09:49 PM
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Yes.
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Old December 2nd, 2008, 11:16 PM
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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
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Old December 2nd, 2008, 11:35 PM
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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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