The Celsius-to-Fahrenheit function is \(t(C)=9\frac{C}{5}+32\). Find \(t(0),t(28),t(-10)\), and the Celsius value giving \(212^\circ F\).
Solution to Question 1061
Complete working\(t(0)=32\). \[\begin{aligned}t(28) &= 9(28)/5+32\\&= 82.4\end{aligned}\]. \[\begin{aligned}t(-10) &= -18+32\\&= 14\end{aligned}\].
For \(t(C)=212\):
\(32^\circ F,82.4^\circ F,14^\circ F\), and \(100^\circ C\).
