Given relationship is
[tex]T(h)=38-1.25h[/tex]
Suppose,
[tex]x=T(h)[/tex]
(b). Now,
[tex]\begin{gathered} x=38-1.25h \\ 1.25h=38-x \\ h=\frac{38-x}{1.25} \end{gathered}[/tex]
Thus,
[tex]T^{-1}(x)=\frac{38-x}{1.25}[/tex]
(a). The function
[tex]T^{-1}(x)[/tex]
Represents the height above the surface when the temperature is x degree celsius.
(c). Putting x=25
[tex]\begin{gathered} T^{-1}(25)=\frac{38-25}{1.25} \\ T^{-1}(25)=10.4 \end{gathered}[/tex]
Hence, the desired value is 10.4.
