#### Answer

(a) $F=\dfrac {9}{5}C+32$
(b) $95 ^\circ F$

#### Work Step by Step

(a) As we are given that $C=\dfrac {5}{9}(F-32)$
$9C=5(F-32)$ [ Cross multiplication ]
$\dfrac {9}{5}C=(F-32)$
$F-32=\dfrac {9}{5}C$
$F=32+\dfrac {9}{5}C$
or, $F=\dfrac {9}{5}C+32$
(b) From part (a), we have : $F=\dfrac {9}{5}C+32$
Thus, at $C=35$, the equation becomes
$F=\dfrac {9}{5}(35)+32$
$F=63+32=95 ^\circ F$