CPhill

Call  the common difference between the angles,  D

And call the measure of the  smallest angle, A

And we  know  that the sum of the interior angles =  (5-2) (180)  = 3 *180  = 540°

So  we  have  that

A + (A + D )  + (A + 2D)  + (A + 3D) + (A + 4D)  = 540

5A + 10D  = 540    (1)

And we know that    (A + 4D)  - A  = 48

So, simplifying.......4D  = 48

D = 12    (2)

Sub  (2)  into (1)  and we have that

5A  + 10(12)  = 540

5A  + 120  = 540

5A  =  540 -120

5A  = 420

A = 420/5  =  84°

I just took a lengthy break.....I'm still hanging in here  ...LOL!!!!

Here's my  best  attempt :

Call  the vertices  (A,B,C,D,E,F,G,H)

The  number of possible  combos of choosing any two vertices  (without regard to order)  =  8C2  = 28

Note  that  the  vertices chosen - again disregarding order - where there  is only one  vertex bt\etween them  are :

ABC , BCD, CDE, DEF, EFG, FGH, GHA, HAB   =  

AC, BD, CE, DF, EG, FH , GA, HB

So....the probability =  8/28 =   2/7

Hey, GM......good to see you  !!!

Here's  the first one

Mean  age of the first  3 teachers   =  (30 + 40 + 50)/3 =  40

Mean age  of 4 teachers ( 30 + 40 + 50 + 39) / 4  = 39.75 

Variance of  three teachers

[(50  - 40)^2  + (40 -40)^2  + (30 -40)^2 ]  =  [10^2  + 0  + (-10)^2]    = 200

Variance  with 4th teacher

[ (50 - 39.75)^2  + (40 -39.75)^2  + (39 - 39.75)^2  + (30 - 39.75)^2 ]   =

[ 10.25^2  + .25^2  + (-.75)^2  + ( -9.75)^2  ]  = 200.75

So.....the mean  decreases and  variance  increases

Note that this could be a "trick" question.....

Look at the Canadian Time  zones, here : https://www.timeanddate.com/time/zone/canada

Note that the time  difference between Edmonton  and St. Johns  =  3 and 1/2   hrs

Thus when it is   8:45 PM  in St Johns, it is actually  just  5:15 PM in Edmonton

So....he  has  actually been traveling from   1:20PM to  5:15 PM  as viewed in  terms of  Edmonton time 

So....he   actually  only travels  5  minutes less than 4 hours =   3 hrs 55 min

Note that   √(xy)  =  x^(1/2) * y^(1/2)

So  we have

x^(1/2) * y^(1/2)  =  1  + x^2y

Using implicit differentiation

(1/2)x^(-1/2)* y^(1/2)  + x^(1/2) * (1/2) y^(-1/2) y'  =  0  + 2xy  + x^2 y'  =

y^(1/2) / [ 2x^(1/2) ] +  (x^(1/2) / [2 y ^(1/2) ]) y'  = 2xy + x^2 y'

Gather the y' terms on the left and everything else on the right we have that

y'  [ x^(1/2)/ [ 2y^(1/2)]  - x^2 ]   =  2xy  - y^(1/2) / [ 2x^(1/2) ]

y' =  dy/dx =   (  2xy  - y^(1/2) / [ 2x^(1/2) ] )  /  [ x^(1/2)/ [ 2y^(1/2)]  - x^2 ]

sec u    =  1 /cos u     

Sorry....I should have made that a  little  clearer   !!!

Look at the graph, here  : https://www.desmos.com/calculator/40z8d53zzr

The triangle  formed by  L   has a base of  6  and a height of 6...so....its area = (1/2) (6)(6)  = 36/2  = 18

The two lines intersect at (2,4)

So.....the  area  between  L and M   and below L  is a triangle with a base of 2 and a height of 4

And this area  =  (1/2)(2) (4)  =  8/2  = 4

So....the probability of a point  being below L and between L and M  =   4/18  = 2/9   ≈ 0.22

First  blank  we have   1 /tan u   

And 1/ tan u  = 1 / (sin u / cos u)  =  cos u / sin u

So now we have 

cos u  * sec u  *  cos u  / sin u  =       cos u  * (1 /cos u)  *  (cos u /sin u) 

(The red part goes into the  second blank)

And this   evaluates to     cot u   .....which is what we are trying to show  !!!!

Nice, Eternity  !!!!