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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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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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do you have the general formula?
No I do not
use:( y" -y')?(x"-x1)

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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
Is the y2= 42 or 24? Is the y1 = 42 or 24?
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
sorry -3/2 not -3/4

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