Alan

I guess this is meant to be csc(3*pi/14).  Use the calculator:

$${csc}{\left({\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{14}}}}\right)} = {\mathtt{85.111\: \!752\: \!545\: \!185\: \!742}}$$

Dude when in life would you have to answer this.....

If you were in the nuclear industry, or in astrophysical research, to name a couple, you would have to know this.  You should get out more, your view of life is too narrow!

That's nonsense!  c is the velocity of light.

I guess this isn't a serious question!

1. Does the 5 million refer to mass or energy?

2. What are its units?

Thank you for the tip rosala - I will do that in future.

Google translation from Indonesian gives: length of the diagonal of a square sqrt3 (2), the circumference of the square is

If this is meant to say the diagonal is (√3)/2 then, if the side of the square is x, from Pythagoras we have 

√(2x²) = (√3)/2 

(√2)x = (√3)/2

x = (√3)/(2*√2) = √2*√3/4 = √6/4

So circumference (better to use perimeter for a square I suppose!) = 4x = √6

I'm afraid your analogy is misleading you Melody!  There is only one way of "choosing" a boy.  There are two ways of choosing a red side from a card with two red sides.

I just lost sight of the fact that the last term was negative while concentrating on all the LaTeX stuff!  Thanks for correcting this rosala.

Express each term with a common denominator.  Easiest way to find a common denominator is to multiply all three denominators together.

\frac{8}{25}*\frac{25*7*2}{25*7*2}+\frac{5}{7}*\frac{25*7*2}{25*7*2}+\frac{1}{2}*\frac{25*7*2}{25*7*2} $$\\\\ $$ \frac{8*7*2}{25*7*2}+\frac{5*25*2}{25*7*2}+\frac{1*25*7}{25*7*2} $$\\\\ $$ \frac{8*7*2+5*25*2+1*25*7}{25*7*2} $$\\\\ $$ \frac{112+250+175}{350} $$\\\\ $$ \frac{537}{350}$$ 

$${\frac{{\mathtt{537}}}{{\mathtt{350}}}} = {\mathtt{1.534\: \!285\: \!714\: \!285\: \!714\: \!3}}$$

or  

  $$1\frac{187}{350}$$

Edit:  I put a "+" instead of a "-" for the last term, so the final answer is incorrect!  See rosala's post for the correct answer. 

I'm afraid you're wrong guys!  This is a well known problem in recreational maths.  

P(A and B) = Probability of two sides red = 1/3 (there are 3 cards, one of which has two red sides).  

P(A) = Probability of one side red = 1/2 (there are 6 sides, 3 of which are red).

P(A and B)/P(A) = (1/3)/(1/2) = 2/3.