CPhill

cos (theta)  =   [ u dot v ] /  [ length of u * length of v ]

u  = (-13, -1)   v   = (19, -3.1)

u dot v  =  ( -13 * 19  +   -1 * -3.1)  =  (-247 + 3.1)  = -243.9

Length of u   =   sqrt[ 13^2 + 1^1) =  sqrt (170)

Length of v  = sqrt ( 19^2 + 3.1^2)   = sqrt (370.61)

cos (theta  =   -243.9 / [ sqrt (170) * sqrt ( 370.61) ]

arccos  [ -243.9 / [ sqrt (170) * sqrt (370.61) ] = theta  ≈  166.33°

We can  use the cosine....

cos (12 ° )  =  H / 150       where H is the horizontal distance.....multiply both sides by 150

150 * cos (12°)   = H  ≈  146.7 m

Use the cosine

cos ( 22°)  = x  / 11    multiply both sides by 11

11 * cos ( 22°)  = x  ≈   10.2

Circumference   =  pi * diameter   .....so....

16  =  pi  *   diameter           divide both sides by  pi

16 / pi   = diameter  ≈  5.093 

sin x  =  opp/ hyp  =  sqrt (119)  / 12

cos x   =  adj / hyp  =  5 /12

(6 /25) * 4000  =

24000 / 25   =

960   males

5 5/6   =   35/6

So

28  /  ( 35/6)  =

(28/1) / (35/6)  =  

(28/1) (6/35)   =    swap denominators

(28 / 35)  *  (6 /1)   =

(4 / 5)   *  ( 6 /1)   =

24 / 5    =

4  4/5

If 6^2 + 8^2  = 10^2....we have a right triangle.....so

36  +  64   = 100    is true.......thus....a right triangle.....

61^2=29^2+50^2-2(29)(50)cos x   

Solving for cosx  we have

[ 61^2 - 29^2 - 50^2 ]  /  [ -2 * 29 * 50 ]    = cos x

Using the cosine inverse, we have

arccos ( [ 61^2 - 29^2 - 50^2 ]  /  [ -2 * 29 * 50 ]  ) = x   ≈  97.53°

Pi  is irrational......thus.....it has no fixed digit length.....

And the square root of an irrational is also irrational.....so   sqrt (pi)   also has no fixed digit length....