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The function ' $t$ ' maps temperature in degree Celsius into temperature in Fahrenheit. It is defined by $t(C)=\frac{9 C}{5}+32$.
Find (i) $t(0)$, (ii) $t(28)$, (iii) $t(-10)$, (iv). The value of $C$ when $t(C)=212$.
Find (i) $t(0)$, (ii) $t(28)$, (iii) $t(-10)$, (iv). The value of $C$ when $t(C)=212$.
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We have $t(C)=\frac{9 C}{5}+32$
$\therefore$ (i) $t(0)=\frac{9(0)}{5}+32=32$
(ii) $t(28)=\frac{9(28)}{5}+32=\frac{252}{5}+32$ $=\frac{412}{5}=82.4$
(iii) $t(-10)=\frac{9(-10)}{5}+32$
$=\frac{-90}{5}+32=-18+32=14$
(iv) When $t(C)=212$, i.e., $212=\frac{9 C}{5}+32$
$\begin{aligned}
&\Rightarrow 212=\frac{9 \mathrm{C}+160}{5} \Rightarrow 1060=9 \mathrm{C}+160 \\
&\Rightarrow \quad 9 C=900 \Rightarrow C=100
\end{aligned}$
$\therefore \quad$ The value of $C$ is 100 .
$\therefore$ (i) $t(0)=\frac{9(0)}{5}+32=32$
(ii) $t(28)=\frac{9(28)}{5}+32=\frac{252}{5}+32$ $=\frac{412}{5}=82.4$
(iii) $t(-10)=\frac{9(-10)}{5}+32$
$=\frac{-90}{5}+32=-18+32=14$
(iv) When $t(C)=212$, i.e., $212=\frac{9 C}{5}+32$
$\begin{aligned}
&\Rightarrow 212=\frac{9 \mathrm{C}+160}{5} \Rightarrow 1060=9 \mathrm{C}+160 \\
&\Rightarrow \quad 9 C=900 \Rightarrow C=100
\end{aligned}$
$\therefore \quad$ The value of $C$ is 100 .
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