Asked by Lanise
A small hole in the wing of a space shuttle requires a 12.8cm^2 patch. What is the patch's area in square kilometers (km^2)? if the patching material costs NASA 3.21/in^2, what is the cost of the patch?
Answers
Answered by
DrBob222
There are 100 cm in a m and 1000 m in a km; therefore,
12.8 cm x (1 m/100 cm)^2 x (1 km/1000 m)^2 = ?? or go to google and type in
12.8 square cm to square km and hit enter.
For the cost, I would convert cm^2 to in^2 first. There are 2.54 cm in 1 inch.
12.8 cm^2 x (1 inch/2.54 cm)^2 = ??, then find the cost. By the way, you don't have units for the cost per unit area. You may mean $3.21/in^2.
12.8 cm x (1 m/100 cm)^2 x (1 km/1000 m)^2 = ?? or go to google and type in
12.8 square cm to square km and hit enter.
For the cost, I would convert cm^2 to in^2 first. There are 2.54 cm in 1 inch.
12.8 cm^2 x (1 inch/2.54 cm)^2 = ??, then find the cost. By the way, you don't have units for the cost per unit area. You may mean $3.21/in^2.
Answered by
Anonymous
12.8cmx10^-9km^2
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