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anonymous
 one year ago
HELP PLEASE?! VERY DESPERATE!
anonymous
 one year ago
HELP PLEASE?! VERY DESPERATE!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Id asked this question last night.... but i dont completely understand how they answered it. I'm a bit deperate right now, can soeone please help!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so for every point i start with, i can make 7 triangles that go over the center

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if i remove the over countin i get 21 different triangles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the probability should be 21/ 85?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0You see, from AB , you can form 7 triangles but only 1 of them can conclude the center, it it ABF , you have to prove it!! (Fortunately, I did it for you on the previous post)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Now, calculate sample space.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0How many sample space do you have?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i can create 84 different triangles in total

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0among them, how many of the triangles include the center?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0at first I found that for any point, i can make 7 triangles from that point that satisfies the restrictions. Because we can make the the same triangle from three different points, i reduced the overcounting by a factor of three to get 21/84 = 1/4.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but now now im getting12 per point, and 12*9/3= 36 different triangles in total. so 36/84

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because from each point i counted how many cases for each different possible second point., does that make sense? am i thinking right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0*sorry i counted how many different triangles i can make that satisfy the restriction for each different second point (which are my cases)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0and ganeshie gave you the general form also.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, but i, not sure how he got it and i need to include an explanation...

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0So, you have to wait for him, hehehe..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks so much once again! you're a great teacher!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just need part b. Any help is appreciated :D

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I'll tag @ganeshie8 for you, that way I get to see the post as well hehe ;).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0part \(a\) : number of acute triangle in a regular nonagon is \(30\). Lets start by fixing one vertex, call it \(A\). Then observe that the other two vertices of required triangle must necessarily be on opposite sides of diameter through \(A\) : dw:1439637980907:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0If you choose the second vertex on right side of semicircle at \(1\) vertex away from \(A\), dw:1439638852437:dw then there is exactly \(1\) choice for the third vertex : dw:1439638864355:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Similarly, if you choose the second vertex on right side of semicircle at \(2\) vertices away from \(A\), dw:1439639089591:dw then there are exactly \(2\) choices for the third vertex : dw:1439639098201:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Similarly see if you can count the number of acute triangles with one vertex fixed at \(A\) and the second vertex at \(3\) vertices away from \(A\) : dw:1439639671473:dw
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