Questions: A small hole in the wing of a space shuttle requires a 16.1 cm^2 patch. What is the patch's area in square kilometers (km^2)? Be sure your answer has the correct number of significant figures. If the patching material costs NASA 3.64/in^2, what is the cost of the patch to the nearest cent? Be sure your answer has the correct number of significant figures.

Solution Steps

1. Understand that \(1 \, \mathrm{m}^2 = 10,000 \, \mathrm{cm}^2\).
2. Convert \(16.1 \, \mathrm{cm}^2\) to \(\mathrm{m}^2\) using the conversion factor: \[ 16.1 \, \mathrm{cm}^2 \times \frac{1 \, \mathrm{m}^2}{10,000 \, \mathrm{cm}^2} = 0.00161 \, \mathrm{m}^2 \]

1. Understand that \(1 \, \mathrm{km}^2 = 1,000,000 \, \mathrm{m}^2\).
2. Convert \(0.00161 \, \mathrm{m}^2\) to \(\mathrm{km}^2\) using the conversion factor: \[ 0.00161 \, \mathrm{m}^2 \times \frac{1 \, \mathrm{km}^2}{1,000,000 \, \mathrm{m}^2} = 1.61 \times 10^{-9} \, \mathrm{km}^2 \]

1. Understand that \(1 \, \mathrm{in}^2 = 6.4516 \, \mathrm{cm}^2\).
2. Convert \(16.1 \, \mathrm{cm}^2\) to \(\mathrm{in}^2\) using the conversion factor: \[ 16.1 \, \mathrm{cm}^2 \times \frac{1 \, \mathrm{in}^2}{6.4516 \, \mathrm{cm}^2} \approx 2.496 \, \mathrm{in}^2 \]

1. Use the cost per square inch to find the total cost: \[ 2.496 \, \mathrm{in}^2 \times \frac{\$3.64}{\mathrm{in}^2} \approx \$9.08 \]
2. Round the cost to the nearest cent.

Final Answer