4. Take your values from the previous chart (In question 2) and convert them fromCelsius to Fahrenheit. Follow the example below, and use the conversion rule to fillout the chart for degrees Fahrenheit when t = 12 and t = 24. (2 polnts: 1 point foreach row)FO-c«) +32Example(0,-32)C(O)- -3.20FCO) - (-32)+32F(0) - -5.76 +32FO) - 26.24 F12

[tex]\begin{gathered} F(t)=\frac{9}{5}C(t)\text{ + 32 }\ldots eqn(1) \\ \text{Making C(t) the subject of the formula, we get} \\ \text{Subtract 32 from both sides, we get} \\ F(t)\text{ -32=}\frac{9}{5}C(t)+32-32 \\ F(t)\text{ -32 =}\frac{9}{5}C(t) \\ \text{Multiplying both sides by }\frac{5}{9}\text{ , we get} \\ \frac{5}{9}\lbrack F(t)-32\rbrack=\frac{5}{9}\times\frac{9}{5}C(t) \\ \\ \frac{5}{9}\lbrack F(t)\text{ - 32\rbrack = C(t)} \\ C(t)=\frac{5}{9}\lbrack F(t)\text{ -32\rbrack}\ldots eqn(2) \end{gathered}[/tex]

Substitute 12 for t in eqn(1), we get

[tex]\begin{gathered} t=12 \\ C(12)=40 \\ F(12)=\frac{9}{5}C(12)\text{ + 32} \\ F(12)=(\frac{9}{5}\times40)\text{ }+32 \\ F(12)=(9\times8)+32 \\ F(12)=72+32=104^oF \end{gathered}[/tex]

When t =24,

C(24) = 0 , ( does not exist since the graph did not touch the graph)

[tex]\begin{gathered} t=24 \\ C(24)=0 \\ F(24)=\frac{9}{5}C(24)\text{ + 32} \\ F(24)=(\frac{9}{5}\times0)\text{ +32} \\ F(24)=32^oF \end{gathered}[/tex]

Hence, the correct answer are:

t =12, C(12) = 104 degrees Fehrenheit, and

t =24, C(24) = 32 degrees Fehrenheit