Question

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

Answer 1

do you have the general formula?

Answer 2

No I do not

Answer 3

use:( y" -y')?(x"-x1)

Answer 4

The temperature dropped 18 degrees (42-24) 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

Answer 5

Is the y2= 42 or 24?
Is the y1 = 42 or 24?

Answer 6

y2=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

Answer 7

sorry -3/2 not -3/4