Can you check if this is correct?

Question

ohmybookness · Answer

attachment

ohmybookness · Answer

attachment

ohmybookness · Answer

@phi

phi · Answer

I think the part where you find the temperature at t=0  (you found 200 degrees) , should go into part a

phi · Answer

also, they probably want a number for 72 + 128*e^0.6
type 
72 + 128*e^0.6=
that into google or use a calcuator

ohmybookness · Answer

what about the second attachment?

ohmybookness · Answer

Thank you!!

phi · Answer

the 22 minutes is correct

For the bonus, you want to change the 72 to 40

but you also want the equation to give 200 at t=0
so you have to change the 128 to a number so that at t=0 you get 200

phi · Answer

the range in the graph looks ok, but it looks like you show the graph at -3 not at y=-2

ohmybookness · Answer

Oh i see. Yeah the bonus question is confusing for me :/

phi · Answer

at infinite time  e^-0.04t  becomes tiny (call it zero)
and you want the temperature to be the outside temperature (it has "cooled down")
that means you want the 72 to be 40

you also want the temp to be 200 at t=0
in other words, solve for A in
200 = 40 + A*e^0

ohmybookness · Answer

oh okay!! thank you :)