CPhill

OK....but.....I suspect hectictar might have gotten the same answer  !!!

I just beat her to the punch   ....LOL!!!!

Strictly by a little trial and error :

2^5 * 9^2  =  2592

So

x * y  =   5 * 2  =  10

Draw altitude CM

Then CM  must  = 8    because  sin A  = CM / AC  =   8/10   = 4/5

Using the Pythagorean Theorem   BM  =  sqrt  (BC^2 - CM^2)  = sqrt ( 17^2 - 8^2)  = sqrt ( 289 - 64) =

sqrt ( 225)  =  15

And cos B  =   BM / BC    =   15/ 17

If we have the point  (x, y)

Then reflecting this across the x axis results in the point  ( x, -y)

So    

(3, 4) ⇒   (3, - 4)

No big deal, HL.....we all make mistakes on here

Here's my best attempt....I'm not good at these, so...I can't guarantee that I'm correct....

He will not need change if

(1) He selects the 1.75 toy first   ....or....

(2) He selects the .25 toy first and the 1.75 toy second

The probability of the first  =  1/8

And the probability of the second is  (1/8) (1/7)  =  1/56

So...the probability that he will need change  is  1  - P(that he won't need change )  =

1 -  ( 1/8 + 1/56)  =

1 - ( 8/56)  =

56/56 - 8/56  =

48/56  =

6/7

m angle 1  =  (1/2) ( m arc AB  + m arc CD)

We have  5 possibilites for the every space in the top row  except the middle space which only has 4 possibilities

So

5^4 * 4  =   2500 possibilities

Yep....that's better   !!!

Let the teams be   (A, B)  (C, D)  (E, F)

So....the number of handshakes are

(A,C) (A,D) (A,E) (A, F)

(B, C) (B,D) (B,E) (B, F)

(C, E) ( C, F)

(D, E) (D, F)   =        12