Alarge object with an initial temperature of 150 degrees fahrenheit is dropped into the ocean. the current temperature of the ocean water in the area is 77 degrees. the function f(t)=ce(−kt)+77 represents the situation, where t is time in minutes, c is a constant, and k is a constant. after 10 minutes the object has a temperature of 120 degrees. after how many minutes will the temperature of the object be equal to 80 degrees?

t = 60.3 minutes.

Step-by-step explanation:

The function ........ (1)  

Now, at t = 0, f(0) = 150 = C + 77

⇒ C = 73

So, the function (1) becomes ......... (2)

Now, it is given that at t = 10 minutes, f(10) = 120 degree.

Therefore, from equation (2),  

⇒

⇒

Now, taking ln both sides we get -10k (ln e) =ln (0.589)

⇒ k = 0.0529

Therefore, the equation (2) becomes   ......(3)

Now, putting f(t) = 80 degree, we have fro equation (3),

⇒

⇒

Taking ln both sides we get, -0.0529t = - 3.19, ⇒ t = 60.3 minutes. (Answer)

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!