Question

Be sure to answer all parts.A small hole in the wing of a space shuttle requires a 16.1 cm² patch.(a) What is the patch's area in square kilometers (km²)? Enter your answer in scientific notation.✔,²1.61ol(b) If the patching material costs NASA $2.94/in², what is the cost of the patch to the nearest cent?

Answer 1

Given;

The area, A, of the patch is

`16.1\text{ cm}^2`

a) The area of the patch in km² is

`\begin{gathered} Where \\ 1\text{ cm}^2=10^(-10)\text{ km}^2 \\ 16.1\text{ cm}^2=16.1*10^(-10)=1.61*10^(-9)\text{ km}^2 \end{gathered}`

Hence, the answer is

`1.61*10^(-9)\text{ km}^2`

b) If the patching material cost $2.94/in^2, i.e

`1\text{ in}^2=\text{ \$2.94}`

Where

`\begin{gathered} 1\text{ cm}^2=0.155\text{ in}^2 \\ Converting\text{ into in}^2 \\ 16.1\text{ cm}^2=0.155*16.1=2.4955\text{ in}^2 \end{gathered}`

The cost of the patching material will be

`\begin{gathered} 1\text{ in}^2=\text{ \$2.94} \\ 2.4955\text{ in}^2=2.4955*2.94=\text{ \$}7.33677=\text{ \$7.34 \lparen nearest cent\rparen} \end{gathered}`

Hence, the cost of the patching material is $7.34 (nearest cent)