Question

The function t which maps temperature in degree Celsius to temperature in degree Fahrenheit is defined by $t\left(C\right)=\frac{9C}{5}+32.$ Find

(i) t(0) (ii) t(28) (iii) t(-10)

(iv) The value of C when t(C) = 212.

Answer 1

Here $t\left(C\right)=\frac{9C}{5}+32$

(i) Putting C = 0

$\therefore t\left(0\right)=\frac{9\times 0}{5}+32=32$

(ii) Putting C = 28

$\therefore t\left(28\right)=\frac{9\times 28}{5}+32$

$=\frac{252\times 160}{5}=\frac{412}{5}$

(iii) Putting C = -10

$\therefore (-10)=\frac{9\times (-10)}{5}+32$

$=-18+32=14$

(iv) Here t(C) = 212

$\therefore 212=\frac{9C}{5}+32$

$\Rightarrow \frac{9C}{5}=212-32$

$\Rightarrow 9C=180\times 5$

$\Rightarrow C=\frac{180\times 5}{9}=100$