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anonymous
 5 years ago
The temperature outside dropped from 42 degrees at 5:00 am to 24 degrees at 5:00 pm. Find the rate of change in the temperature
anonymous
 5 years ago
The temperature outside dropped from 42 degrees at 5:00 am to 24 degrees at 5:00 pm. Find the rate of change in the temperature

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you have the general formula?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use:( y" y')?(x"x1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The temperature dropped 18 degrees (4224) in 12 hours, so the rate of change was a decrease of 1.5 degrees per hour. (18/12) or Newton's Law of Cooling? T(t) = Ts + (To  Ts)*e^(k*t) ; where: t is the time in the preferred units (seconds, minutes, hours, etc.) T(t) is the temperature of the object at time t Ts is the sorrounding constant temperature To is the initial temperature of the object k is a constant to be found

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is the y2= 42 or 24? Is the y1 = 42 or 24?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y2=5pm,y1=5am subtrac you get 12 x2=24, x1= 42 subtrac you det negative =18 slope= 12/18=3/4 the rate 3/4
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