(i) We haveF=(59​)C+32 When C=0, F=(59​)×0+32=32 When C=15, F=(59​)(−15)+32=−27+32=5 When C=−10, F=59​(−10)+32=−18+32=14 We have the following table:Plotting the ordered pairs (0,32),(–15,5) and (–10,14) on a graph paper. Joining these points, we get a straight line A B.(ii) From the graph, we have 86∘F corresponds to 30∘C.(iii) From the graph, we have 95∘F corresponds 35∘C.(iv) We have, C=0 From (1), we getF=\\left(\\frac{9}{5}\\right) 0+32=32\\mathrm{Also}, \\mathrm{F}=0From (1), we get0=(59​)C+32⇒9−32×5​=C⇒C=−17.8 (V) When From (1), we getF=59​F+32⇒F−59​F=32 ⇒5−4​ F = 32 ⇒ F=-40 ∴ Temperature is –40∘ both in F and C