document.write( "Question 150874This question is from textbook number power
\n" ); document.write( ": A patio in the shape of a half circle and is 4 inches thick is being built. The radius is 12 feet.
\n" ); document.write( "To the nearest cubic yard, how much concrete will be needed?
\n" ); document.write( "How did they come up with 3 cubic yds of concrete?
\n" ); document.write( "Could you please explain this for me, I am so lost. I am so grateful for any help.
\n" ); document.write( "Thank-You
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Algebra.Com's Answer #110781 by stanbon(75887)\"\" \"About 
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A patio in the shape of a half circle and is 4 inches thick is being built. The radius is 12 feet.
\n" ); document.write( "To the nearest cubic yard, how much concrete will be needed?
\n" ); document.write( "How did they come up with 3 cubic yds of concrete?
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\n" ); document.write( "The half circle has a surface area of (1/2)(pi)r^2
\n" ); document.write( "Your radius is 12 feet so the surface is (1/2)pi^12^2 = 226.19 sq. ft.
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\n" ); document.write( "The depth of the patio is 4 incles or (1/3) ft
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\n" ); document.write( "So the volume of the patio is (1/3)(226.19) = 75.4 cu. ft.
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\n" ); document.write( "A cubic yard of cuncrete is 3 ft x 3 ft x 3ft = 27 cu. ft.
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\n" ); document.write( "Dividing 75.4 cu. ft by 27 cu. ft you get 2.79 cu. yd of concrete.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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