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is it the temperature inside the oven that is increasing or its film surface outside temperature?

yeah right

its been ages since ive done this um..

have you learnt logistic functions?

Yes

It is just saying the rate is dropping , not the max temp.

perhaps solve \[\frac{ dT }{ dt }=0.1(500-T)\]

i'm only throwing ideas here

yes, solving the differential equation sounds good.

i think thats a logarithm function you learn in final year in high school...

Sorry! Had to have dinne. This is the answer
\[T=500-480\times0.9^m\]

correct, although we tend to use the exponential function in the equation.

\[T(t)=500-480e ^{-0.1t}\]

so by simply solving the differential equation, we can get to this solution

can you do this?

thats how you solve the differential function of \[\frac{ dT }{ dt }=0.1(500-T)\]

have you learnt how to solve differential equations?

we express rates as a differential function, i.e\[\frac{ dT }{ dt}\]

this simply means, the change in temperature with respect to the change in time.

i.e \[\frac{ dT }{ dt}=0.1(500-T)\]

0.1 is simply a decimal form which represents 10% right?

Yup

have you sort of grasped this concept yet?

I get that. But I am confused where the 480 comes into play :/

its comes out when you solve it...

\[\frac{ dT }{ dt }=0.1(500-T)\]
\[\frac{ dT }{ 500-T }=0.1dt\]

\[\int\limits_{}^{}\frac{ dT }{ 500-T }=\int\limits_{}^{}0.1dt\]

remember separable differential equations?

We haven't done that. This is meant to be WAY simpler than th

yea interesting :/

im not sure if i can help you then :/