document.write( "Question 759570: A square with an area of 49 in. To the second power had a perimeter of________.
\n" ); document.write( " I tried solving it i don't know if its correct.
\n" ); document.write( "49 in. To the second power =2401/4= 600.25·4= 2401 so the perimeter is 2401. Is that correct?
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Algebra.Com's Answer #462220 by Cromlix(4381)\"\" \"About 
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A square with an area of 49 ins^2
\n" ); document.write( " has a perimeter of 28 ins.
\n" ); document.write( "Area of square = length^2
\n" ); document.write( "Length^2 = 49
\n" ); document.write( "Therefore length = square root of 49
\n" ); document.write( " length = 7 ins
\n" ); document.write( "Perimeter = 2*length + 2* width
\n" ); document.write( "(With a square, length = width)
\n" ); document.write( " = 2*7 + 2*7
\n" ); document.write( " = 28 ins.
\n" ); document.write( "I think you are confusing the second power.
\n" ); document.write( "The second power is associated with inches
\n" ); document.write( "i.e. 49 ins^2 was the area.
\n" ); document.write( "Hope this helps,.
\n" ); document.write( ":-)
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