The function ' $t$ ' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $t(\mathrm{C})=\frac{9 \mathrm{C}}{5}+32$.
Find
(i) $t$ ( 0 )
(ii) $t$ (28)
(iii) $t(-10)$
(iv) The value of $\mathrm{C}$, when $t(\mathrm{C})=212$
The given function is $t(\mathrm{C})=\frac{9 \mathrm{C}}{5}+32$.
Therefore,
(i) $t(0)=\frac{9 \times 0}{5}+32=0+32=32$
(ii) $t(28)=\frac{9 \times 28}{5}+32=\frac{252+160}{5}=\frac{412}{5}$
(iii) $t(-10)=\frac{9 \times(-10)}{5}+32=9 \times(-2)+32=-18+32=14$
(iv) It is given that $t(\mathrm{C})=212$
$\therefore 212=\frac{9 \mathrm{C}}{5}+32$
$\therefore 212=\frac{9 C}{5}+32$
$\Rightarrow \frac{9 C}{5}=180$
$\Rightarrow 9 C=180 \times 5$
$\Rightarrow C=\frac{180 \times 5}{9}=100$
Thus, the value of $t$, when $t(C)=212$, is 100 .
$\Rightarrow \frac{9 C}{5}=212-32$
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