Question

question 5 (mandatory) (1 point) a student records the temperature of an art project that has been left to cool on a counter. the room temperature is 18°c. an equation that models this situation is (t(t)=59(0.5)^{\frac{t}{10}} + 18) where (t) is the temperature in degrees celsius and (t) is the time in minutes. determine the temperature, to the nearest degree, of the art project after 10 minutes. how much time did it take for the project to reach a temperature of 30°c?

Explanation:

Step1: Find temperature after 10 minutes

Substitute $t = 10$ into $T(t)=59(0.5)^{\frac{t}{10}}+18$.
$T(10)=59(0.5)^{\frac{10}{10}}+18$
$=59\times0.5 + 18$
$=29.5+18$
$=47.5\approx48^{\circ}C$

Step2: Find time when $T = 30^{\circ}C$

Set $T(t)=30$, so $30 = 59(0.5)^{\frac{t}{10}}+18$.
First, subtract 18 from both sides:
$30 - 18=59(0.5)^{\frac{t}{10}}$
$12 = 59(0.5)^{\frac{t}{10}}$
Then, divide both sides by 59:
$\frac{12}{59}=(0.5)^{\frac{t}{10}}$
Take the natural - logarithm of both sides:
$\ln(\frac{12}{59})=\ln((0.5)^{\frac{t}{10}})$
Using the property $\ln(a^b)=b\ln(a)$, we get:
$\ln(\frac{12}{59})=\frac{t}{10}\ln(0.5)$
Solve for $t$:
$t = 10\times\frac{\ln(\frac{12}{59})}{\ln(0.5)}$
$t = 10\times\frac{\ln(12)-\ln(59)}{\ln(0.5)}$
$t\approx10\times\frac{2.4849 - 4.0775}{- 0.6931}$
$t\approx10\times\frac{- 1.5926}{- 0.6931}\approx23$ minutes

Answer:

The temperature of the art - project after 10 minutes is $48^{\circ}C$. It took approximately 23 minutes for the project to reach a temperature of $30^{\circ}C$.