Question

The function $t$ which maps temperature in degree Celcius into temperature in degree Fahrenheit is defined by $t\left(C\right)=\frac{9C}{5}+32.$
Find $\left(i\right)t\left(0\right)\left(ii\right)t\left(28\right)\left(iii\right)t(-10)$
$\left(iv\right)$The value of $C,$ when $t\left(C\right)=212$

Answer 1

$\left(i\right)$ Given:$\left(t\right(C)=\frac{9C}{5}+32$
Evaluate the value of function
Put $C=0$ in the given function.
$t\left(0\right)=\frac{9\left(0\right)}{5}+32=32$
Therefore, the value of $t\left(0\right)$ is $32$.

$\left(ii\right)$ Given:$\left(t\right(C)=\frac{9C}{5}+32$
Evaluate the value of function
Put $C=28$ in the given function.
$t\left(28\right)=\frac{9\left(28\right)}{5}+32$
$=\frac{252}{5}+32=\frac{252+160}{5}=\frac{412}{5}$
Therefore, the value of $t\left(28\right)$ is $\frac{412}{5}$.

$\left(iii\right)$ Given:$\left(t\right(C)=\frac{9C}{5}+32$
Evaluate the value of function
Put $C=-10$ in the given function.
$t\left(0\right)=\frac{9(-10)}{5}+32$
$=9(-2)+32=-18+32=14$
Therefore, the value of $t(-10)$ is $14$.

$\left(iv\right)$ Given: $t\left(C\right)=\frac{9C}{5}+32$
The value of $C$, when $t\left(C\right)=212,$ simplify for $C$
$\Rightarrow \frac{9C}{5}+32=212$
$\Rightarrow \frac{9C}{5}=212-32$
$\Rightarrow \frac{9C}{5}=180$
$\Rightarrow C=\frac{5}{9}\times 180$
$\Rightarrow C=100$
Therefore, the value of $C$ when $t\left(C\right)=212$ is $100$