A small hole in the wing of a space shuttle requires a 17.7cm² patch
a.) What is the patch's area in space kilometers?
b.) If the patching material costs NASA USD 3.25/in², what is the cost of the patch?
The patch's area in space kilometers
1cm=0.00001km1cm = 0.00001km1cm=0.00001km
1cm2=(0.00001km)2=10−10km21cm^2 = (0.00001km)^2=10^{-10}km^21cm2=(0.00001km)2=10−10km2
17.7cm2=17.7×10−10km2=2.07×10−9km−217.7cm^2=17.7×10^{-10}km^2=2.07×10^{-9}km^{-2}17.7cm2=17.7×10−10km2=2.07×10−9km−2
b)
1cm=0.393701inch1cm=0.393701inch1cm=0.393701inch
1cm2=0.1550inch21cm^2=0.1550inch^21cm2=0.1550inch2
17.7cm2=17.7×0.1550inch2=2.7435inch217.7cm^2=17.7×0.1550inch^2=2.7435inch^217.7cm2=17.7×0.1550inch2=2.7435inch2
Cost of patch =2.7435×3.25=8.916= 2.7435×3.25=8.916=2.7435×3.25=8.916
Hence cost = NASA USD 8.916
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