forum.virtualchallengemeets.com

56. Usually i graph then look at table to count them, is there a better way? answers are pretty large..
57. sorry i'm jsut bad at these things. there is at lease one question like this on every math test... skip

Read for 56.



57: Treat each couple as one, figure out the number of permutations, then multiply by 8 (as you have 2 permutations per couple), so you would have 3!(8) = 48.

thanks!!
1 last question: #45 seems impossible

Ugh! #45 reminds me of the stupid question I skipped on that practice test for no reason
Cause: Overlooking questions! I just saw it nd was like - this is hard
but I sit here and think for 2 seconds and get the answer... Sad stuff :(

And in high school geometry class. Just saying.

I mean it is kinda obvious right - the formula?

I mean i was not aware of it until sophomore yr but I am sure I could have derived it if I had a chance...