If you pour a cup of coffee that is 200f , and set it in a room that is 68f, and 10 minutes later it is 145f, what temperature will it be 15 minutes after you originally poured it?

This is not as simple as it looks, and the obvious answer is wrong.

You might think that since the coffee cooled (200 - 145) = 55 degrees
in 10 minutes, it would cool another (55/2) = 27.5 degrees in the next
5 minutes, and arrive at 117.5F .  Not only would you be wrong, but
your coffee would not do that.  It would be warmer than 117.5F . 

This is a classic example of a process whose speed depends on
how far it still has to go.  Just like a little kid walking to school,
the closer it gets to its final destination, the slower it goes. 
As the temperature of the coffee gets closer and closer to 68F,
it cools slower and slower.

The equation that describes the temperature of your coffee is

           Temp = 68 + ( 132 e^-minutes/18.55) .

After 10 minutes,  Temp = 68 + (132 e^-0.539) =
                                      68 + (132 x 0.583) =
                                      68 + (76.99)          =    145F .

After 15 minutes,  Temp = 68 + (132 e^-0.809) =
                                      68 + (132 x 0.445)  =
                                      68 + (  58.8 )          =   126.8 F .  

than do 23.50 times 18 and then divide your answer by 100 than you will get x.