document.write( "Question 1210359: The dimensions of a square is given as 6.25 cm. A student measured one side of the square as 6.12 cm to calculate the perimeter and the area of the square.
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Algebra.Com's Answer #852046 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Although this problem can be solved exactly by algebra, as the other tutor has\r\n" );
document.write( "solved it, I think this might be a calculus problem demonstrating using\r\n" );
document.write( "differentials to approximate error.\r\n" );
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document.write( ">>The dimensions of a square is given as 6.25 cm. A student measured one side of\r\n" );
document.write( "the square as 6.12 cm to calculate the perimeter and the area of the square.<<  \r\n" );
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document.write( "ds = 6.12 - 6.25 = -0.13\r\n" );
document.write( "P = 4s\r\n" );
document.write( "dP = 4ds = 4(-0.13) = -0.52\r\n" );
document.write( "A = s2\r\n" );
document.write( "dA = 2s(ds) = 2(6.25)(-0.13) = -1.625\r\n" );
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document.write( ">>Find the percentage error in:  \r\n" );
document.write( "i. Measured length<< \r\n" );
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document.write( "\"ds%2Fs+=-0.13%2F6.25+=+0.208+=+-%222.08%25%22\"\r\n" );
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document.write( ">>ii. Calculated perimeter<<\r\n" );
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document.write( "\"dP%2FP=%28-0.52%29%2F%284s%29=%28-0.52%29%2F%284%2A6.25%29=-0.0208=%22-2.08%25%22\"\r\n" );
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document.write( ">>iii. Calculated area<<  \r\n" );
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document.write( "Edwin
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