4764 to the nearest ten

r24.

Step-by-step explanation:

it woud be

Here we will use the relation:

Ln(e^x) = x

The Newton's law of cooling can be written as:

T(t) = Te + (T(0min) - Te)*e^(-t/k)

Where t is time in minutes, Te is the environment temperature, T(0min) is the initial temperature and k is a constant.

In this case, we have that:

Te = 60°F

T(0min) = 150°F

now we can replace those in our equation and get:

T(t) = 60°F + (150°F - 60°F)*e^(-t/k)

T(t) = 60°F + 90°F*e^(-t/k)

Now, we know that after 15 minutes the temperature of the tea is 100°F.

T(15min) = 60°F + 90°F*e^(-15min/k) = 100°F

Whit this we can find the value of k

60°F + 90°F*e^(-15min/k) = 100°F

90°F*e^(-15min/k) = 100°F - 60°F = 40°F

e^(-15min/k) = 40°F/90°F = (4/9)

Now we can apply the Ln(x) in both sides, to get:

Ln(e^(-15min/k) ) = Ln(4/9)

-15min/k = Ln(4/9)

k = -15min/Ln(4/9) = 18.497 min.

Now we can replace that in our equation and we get:

T(t) = 60°F + (90°F)*e^(-t/18.497 min)

Now we want to find the time x such that the temperature of the tea is 80°F.

T(x) = 80°F =  60°F + (90°F)*e^(-x/18.497 min)

Now let's solve this for x.

80°F =  60°F + (90°F)*e^(-x/18.497 min)

80°F - 60°F =  (90°F)*e^(-x/18.497 min)

20°F/90°F = e^(-x/18.497 min)

Now we apply Ln(x) in both sides:

Ln(2/9) = Ln(e^(-x/18.497 min)) = -x/18.497 min

ln(2/9)*18.497min = -x = -27.82 min.

x = 27.82 mins.

So 27.82 minutes after you pour the cup of tea, will be at 80°F.

But the question is:

"How long before the tea is at 80 degrees * F"

And this is asked 15 minutes after the tea was poured, so at this point the time left will be:

27.82min -15min = 12.82 min