For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.

We can model that in this way

$\frac{dT}{dt}=k*(T_r-T)$

If we rearrange and integrate

$\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}$

We know that at time 0, the temperature of the drink was 52°F. Then we have:

$T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20$

We also know that at t=2, T=55°F

$T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08$

The model for the temperature of the drink can be written as

$T=72-20e^{-0.08t}$

7 0

3 years ago

Brandon has a box 9.5 inches tall, 3.8 inches wide , and 4.2 inches long how many square inches of wrapping paper does he need t

Flauer [41]

The answer to your question is 151.6200 hope I helped

7 0

3 years ago

Please helpppppppppp

zysi [14]

Answer:

77 cubed in.

Step-by-step explanation:

I think it is 77 cubed inches because the formula is V-bh

4 0

2 years ago