Really need some understanding this the temperature of a cup of coffee varies according to newton's law of cooling: dt/dt = -k (t-a), where t is the water temperature, a is the room temperature, and k is a positive constant. if the coffee cools from 180°f to 100°f in 10 minutes at a room temperature of 75°f, how long (to the nearest minute) will it take the water to cool to 80°f? ,

If you're studying differential equations, you can find the solution to the given equation using the boundary conditions given.

Some of us would rather cut to the chase. We recognize that this is an exponential decay problem in which the initial temperature difference from the 75 °F room temperature is decaying to zero.

In 10 minutes, the temperature difference has decayed from 180-75 = 105 to 100-75 = 25, so we can write the temperature function of time as
.. T(t) = 75 +105*(25/105)^(t/10)

We can solve this for
.. T(t) = 80
using logarithms, but it may be easier to do it graphically.

To the nearest minute, it will take 21 minutes for the coffee to cool to 80 °F.

Answer d; a supplementary angle = 180°, so if that angle = 150°, y would = 30°.