Prove that

Solution:

Given: $\mathrm{F}(\mathrm{x})=\frac{9}{5} \mathrm{x}+32$ ....(i)

To find: (i) F(0)

Substituting the value of x = 0 in eq. (i), we get

$\mathrm{F}(\mathrm{x})=\frac{9}{5} \mathrm{x}+32$

$\Rightarrow \mathrm{F}(0)=\frac{9}{5} \times 0+32$

$\Rightarrow \mathrm{F}(0)=32$

It means $0^{\circ} \mathrm{C}=32^{\circ} \mathrm{F}$

To find: (ii) $F(-10)$

Substituting the value of $x=-10$ in eq. (i), we get

$\mathrm{F}(\mathrm{x})=\frac{9}{5} \mathrm{x}+32$

$\Rightarrow \mathrm{F}(-10)=\frac{9}{5} \times(-10)+32$

$\Rightarrow \mathrm{F}(-10)=9 \times(-2)+32$

$\Rightarrow \mathrm{F}(-10)=-18+32$

$\Rightarrow \mathrm{f}(-10)=14$

It means $-10^{\circ} \mathrm{C}=14^{\circ} \mathrm{F}$

To find: (iii) the value of $x$ when $F(x)=212$

It is given that $\mathrm{F}(\mathrm{x})=\frac{9}{5} \mathrm{x}+32$

Substituting the value of F(x) = 212 in the above equation, we get

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

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

$\Rightarrow 180=\frac{9}{5} \mathrm{x}$

$\Rightarrow \mathrm{x}=180 \times \frac{5}{9}$

$\Rightarrow \mathrm{x}=20 \times 5$

$\Rightarrow \mathrm{x}=100$

It means $212^{\circ} \mathrm{F}=100^{\circ} \mathrm{C}$