Abody was found at 10 a. m. in a warehouse where the temperature was 40of. the medical examiner found the temperature of the body to be 80of. what was the approximate time of death? use newton's law of cooling, with k = 0.1947.

The formula we use for this is , where T(t) is the temp after the cooling has taken place, is temp of the room, is the initial temp of the body, and t is the time in hours. "e" is Euler's number.  Out T(t) = 80°, our is 40°, the initial temp of the body is 98.6°, and t is what we are solving for.  So our equation then looks like this: .  We will simplify inside the parenthesis first, of course, to get this: .  Subtract 40 from both sides to get .  Divide both sides by 58.6 to get .  You're probably in logs right now in math, so you should be aware of the fact that a natural log "undoes" Euler's number, so we will take the natural log of both sides to move that t down into a place where we can work with it a bit easier.  ln(.68259385)=-.1947t.  Take the natural log of that decimal and get -.3818552424= -.1947t.  Divide both sides by -.1947 to get a t value of 1.96 hours.  If we subtract 1.96 hours from 10am we would get that the time of death was just about 8:03am

62926281916 inches of rain: )