Question

question 5 (mandatory) (1 point) saved 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? 1. temperature after 10 minutes : 24 c 2. time to reach 30c :16 minutes.

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 to reach 30°C

Set $T(t)=30$, so $30 = 59(0.5)^{\frac{t}{10}}+18$.
First, subtract 18 from both sides: $30 - 18=59(0.5)^{\frac{t}{10}}$, which gives $12 = 59(0.5)^{\frac{t}{10}}$.
Then, $\frac{12}{59}=(0.5)^{\frac{t}{10}}$.
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 have $\ln(\frac{12}{59})=\frac{t}{10}\ln(0.5)$.
Solve for $t$: $t = 10\times\frac{\ln(\frac{12}{59})}{\ln(0.5)}$.
$\ln(\frac{12}{59})\approx\ln(12)-\ln(59)\approx2.4849 - 4.0775=-1.5926$, $\ln(0.5)\approx - 0.6931$.
$t = 10\times\frac{-1.5926}{-0.6931}\approx10\times2.3\approx23$ minutes.

Answer:

1. Temperature after 10 minutes: $48^{\circ}C$
2. Time to reach $30^{\circ}C$: $23$ minutes