$$
T_C=frac{5}{9}(T_F-32)
$$

Here $T_F=110text{textdegree}$F, therefore we have

$$
T_C=frac{5}{9}(110-32)=43.3text{textdegree}mathrm{C}
$$

So the corresponding temperature in Celsius is 43.3$text{textdegree}$ C.

$$
T_F = dfrac{9}{5}T_C + 32
$$

Where $T_C = – 89.2text{ }^circ text{C}$, so

$$
begin{align*}
T_F &= dfrac{9}{5}left( { – 89.2} right) + 32\
\
T_F &= dfrac{ – 802.8}{5} + 32\
\
T_F &= -160.56 + 32\
\
T_F &= boxed{-128.56text{ }^circ text{F}}\
end{align*}
$$

$$
T_C=frac{5}{9}(T_F-32)
$$

So the change in the temperature scale is given by

$$
Delta T_C=frac{5}{9}Delta T_F=frac{5}{9}27=15text{textdegree}mathrm{C}
$$

So the corresponding change in temperature in Celsius scale is 15$text{textdegree}$ C.

$$
begin{align*}
& T_F=frac{9}{5}T_C+32\
Rightarrow & T=frac{9}{5}T+32\
Rightarrow & 5T=9T+160\
Rightarrow & -4T=160\
Rightarrow & T=-40
end{align*}
$$

So the $-40text{textdegree}$C temperature is the same in both scale.