If a room temperature roast cools from 68°f to 25°f in 5 hours at freezer temperature of 20°f,
how long (to the nearest hour) will it take the roast to cool to 21°f?

9 hours

Step-by-step explanation:

According to Newton's laws of cooling

dT/dt = -k(T - A)

Let U = T - A

dU/dT = 1

dU = dT

dt/dt = -kU

dT = -kU(dt)

dU/U = -kdt

On integration

ln(U) = -kt + C

U = Ce^-kt

T - A = Ce^-kt

T(0) = 68

T(5) = 25

68 - 20 = Ce^-k(0)

C = 48

and

25 -20 = 48e^-k(5)

5 = 48e^-5k

e^-5k = 5/48

-5k = ln (5/48)

k = - ln(5/48) / 5

k = - 0.4524

T - 20 = 48e^-0.4524t

When T = 21

21 -20 = 48e^-0.4524t

1 = 48e^-0..4524t

e^-0.4524t = 1/48

-0.4524t = ln (1/48)

t = - ln(1/48) / 0.4524

t = 8.5570

t= 9 hours ( to the nearest hour)

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.