A cake is removed from an oven. The temperature, in degrees Fahrenheit, of the cake after t minutes can be modeled by the equation, f(x)=70+140e^-0.024t. What is the temperature of the cake at the moment it is removed from the oven and about how long will it take for the cake to reach a temperature of 80F ? Round your answer to the nearest whole minute.

Thank you (:

9514 1404 393

Step-by-step explanation:

A graphing calculator answers the questions easily.

At t=0, f(0) = 210 . . . degrees F

f(t) = 80 at about t=109.96, so it takes about 110 minutes to reach 80 °F.

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Working "by hand"

The exponential term is 1 at t=0, so the temperature then is ...

  f(0) = 70 +140·1 = 210 . . . degrees F

The temperature of the cake from the oven is 210 °F.

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The time when the temperature is 80 °F will be ...

  80 = 70 +140e^(-.024t)

  10 = 140e^(-.024t)

  10/140 = e^(-.024t)

  ln(1/14) = -0.024t

  t = ln(1/14)/-0.024 ≈ 109.9607

  t ≈ 110 when f(t) = 80

It takes about 110 minutes for the cake to cool to 80 °F.

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.