anonymous
  • anonymous
6 POINTS (A,B,C,D,E,F) LIE ON A CIRCLE
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
That's cool. Also, nice caps.
anonymous
  • anonymous
|dw:1342406130806:dw|
anonymous
  • anonymous
Find the maximum no. of Points of Intersection of chords inside the circle? LOL @SmoothMath !

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
fe || ad || bc ?
anonymous
  • anonymous
huh?
anonymous
  • anonymous
Virtus, why not draw in all of the possible chords?
anonymous
  • anonymous
|dw:1342406517682:dw|
anonymous
  • anonymous
oh then just count it
anonymous
  • anonymous
Truedat, sab.
anonymous
  • anonymous
no but you are supposed to use combinations to solve it, Like in the exam you can't just say refer to the diagram. You actually have to show them how you can get the answer through combinations!
UnkleRhaukus
  • UnkleRhaukus
well for example (a) can join with 5 other point to make a cord,
anonymous
  • anonymous
Is there any statement about where the points are on the circle?
anonymous
  • anonymous
no there isn't but the answer is 6C4 in case you were wondering
UnkleRhaukus
  • UnkleRhaukus
i got \[^6C_2\]
UnkleRhaukus
  • UnkleRhaukus
\[^6C_2=15=^6C_4\]
anonymous
  • anonymous
@UnkleRhaukus but then there was a question prior to that that asked How many chords can be drawn by joining these different pairs of points and the answer was 6C2 Why couldn't they have written 6C2 for the second question?
UnkleRhaukus
  • UnkleRhaukus
hmm im not sure
anonymous
  • anonymous
do you know where the 4 may have come from then in 6C4?
anonymous
  • anonymous
because i think i understand if it is 2, cause its kinda like that hand shake question then
anonymous
  • anonymous
ncr = nc(n-r)
UnkleRhaukus
  • UnkleRhaukus
is suppose the question is asking the number of intersections the cords make
anonymous
  • anonymous
huh?
anonymous
  • anonymous
i guess we should calculate # intersections per one point, and multiply it with 6 then we are using counting... not combintions lol
anonymous
  • anonymous
I honestly don't see how it's a combination question, but hey.
anonymous
  • anonymous
im seeing a pattern maybe we can work backward # intersecting points for point "a" inside the circle : 1.3 + 2.2 + 3.1 = 10 total # of intersecting points possible = 10*6 = 60
anonymous
  • anonymous
thats the answer. lets see if the pattern means something in combination terms 1.3 + 2.2 + 3.1 r(n) + (r+1)(n-1) + (r+2)(n-2) + ....
anonymous
  • anonymous
thats same as SUM (r)(n-r)
anonymous
  • anonymous
|dw:1342408544273:dw|
anonymous
  • anonymous
this looks more compact than any combination formula... but it would be good to see if this can be converted to NCR s...

Looking for something else?

Not the answer you are looking for? Search for more explanations.