Question

The function 't' which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by: $t(C)=\frac {9C}{5}+32$
Find:

1. t(0)
2. t(28)
3. t(-10)
4. The value of C, when t(C) = 212

Answer 1

The given function is $t(C)=\frac {9C}{5}+32$
Therefore,
(i) $t(0)$ = \frac {9\times 0}{5}$$+32 = $0+32$

---

(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(C) = 212

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

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

$\frac {9C}{5}=180$

$9C=180\times 5$

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

$C = 100$

Thus the value of t when t(C) = 212 is 100.