Solution:
Given;
The area, A, of the patch is
[tex]16.1\text{ cm}^2[/tex]
a) The area of the patch in km² is
[tex]\begin{gathered} Where \\ 1\text{ cm}^2=10^{-10}\text{ km}^2 \\ 16.1\text{ cm}^2=16.1\times10^{-10}=1.61\times10^{-9}\text{ km}^2 \end{gathered}[/tex]
Hence, the answer is
[tex]1.61\times10^{-9}\text{ km}^2[/tex]
b) If the patching material cost $2.94/in^2, i.e
[tex]1\text{ in}^2=\text{ \$2.94}[/tex]
Where
[tex]\begin{gathered} 1\text{ cm}^2=0.155\text{ in}^2 \\ Converting\text{ into in}^2 \\ 16.1\text{ cm}^2=0.155\times16.1=2.4955\text{ in}^2 \end{gathered}[/tex]
The cost of the patching material will be
[tex]\begin{gathered} 1\text{ in}^2=\text{ \$2.94} \\ 2.4955\text{ in}^2=2.4955\times2.94=\text{ \$}7.33677=\text{ \$7.34 \lparen nearest cent\rparen} \end{gathered}[/tex]
Hence, the cost of the patching material is $7.34 (nearest cent)
