Questions: A small hole in the wing of a space shuttle requires a 16.3 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. 1.63 x 10^-9 km^2 If the patching material costs NASA 2.85 / in^2, what is the cost of the patch to the nearest cent? Be sure your answer has the correct number of significant figures.

A small hole in the wing of a space shuttle requires a 16.3 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. 1.63 x 10^-9 km^2 If the patching material costs NASA 2.85 / in^2, what is the cost of the patch to the nearest cent? Be sure your answer has the correct number of significant figures.
Transcript text: A small hole in the wing of a space shuttle requires a $16.3 \mathrm{~cm}^{2}$ patch. What is the patch's area in square kilometers $\left(\mathrm{km}^{2}\right)$ ? Be sure your answer has the correct number of significant figures. \[ 1.63 \times 10^{-9} \mathrm{~km}^{2} \] If the patching material costs NASA $\frac{\$ 2.85}{\text { in }^{2}}$, what is the cost of the patch to the nearest cent? Be sure your answer has the correct number of significant figures.
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Solution

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Solution Steps

Step 1: Convert \(\mathrm{cm}^2\) to \(\mathrm{km}^2\)
  1. The area of the patch is \(16.3 \, \mathrm{cm}^2\).
  2. Convert \(\mathrm{cm}^2\) to \(\mathrm{m}^2\): \(1 \, \mathrm{cm}^2 = 10^{-4} \, \mathrm{m}^2\).
  3. Therefore, \(16.3 \, \mathrm{cm}^2 = 16.3 \times 10^{-4} \, \mathrm{m}^2 = 0.00163 \, \mathrm{m}^2\).
  4. Convert \(\mathrm{m}^2\) to \(\mathrm{km}^2\): \(1 \, \mathrm{m}^2 = 10^{-6} \, \mathrm{km}^2\).
  5. Therefore, \(0.00163 \, \mathrm{m}^2 = 0.00163 \times 10^{-6} \, \mathrm{km}^2 = 1.63 \times 10^{-9} \, \mathrm{km}^2\).
Step 2: Convert \(\mathrm{cm}^2\) to \(\mathrm{in}^2\)
  1. The area of the patch is \(16.3 \, \mathrm{cm}^2\).
  2. Convert \(\mathrm{cm}^2\) to \(\mathrm{in}^2\): \(1 \, \mathrm{cm}^2 = 0.155 \, \mathrm{in}^2\).
  3. Therefore, \(16.3 \, \mathrm{cm}^2 = 16.3 \times 0.155 \, \mathrm{in}^2 = 2.5265 \, \mathrm{in}^2\).
Step 3: Calculate the cost of the patch
  1. The cost of the patching material is \(\$2.85\) per \(\mathrm{in}^2\).
  2. The area of the patch is \(2.5265 \, \mathrm{in}^2\).
  3. Therefore, the cost of the patch is \(2.5265 \, \mathrm{in}^2 \times \$2.85/\mathrm{in}^2 = \$7.198525\).
  4. Round to the nearest cent: \(\$7.20\).

Final Answer

Part 1: \( \boxed{1.63 \times 10^{-9} \, \mathrm{km}^2} \)

Part 2: \$ \( \boxed{7.20} \)

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