document.write( "Question 250576: Suppose that $6000 is invested in an account where interest is compounded continuously at 3.7% per year. What is the balance after 1 year? after 2 years? \n" ); document.write( "
Algebra.Com's Answer #182466 by nerdybill(7384)\"\" \"About 
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Suppose that $6000 is invested in an account where interest is compounded continuously at 3.7% per year. What is the balance after 1 year? after 2 years?
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\n" ); document.write( "The equation for \"continuously compounded interest\" is:
\n" ); document.write( "\"A+=+Pe%5E%28rt%29\"
\n" ); document.write( "where
\n" ); document.write( "A is the amount after t years
\n" ); document.write( "P is the initial principal
\n" ); document.write( "r is the interest rate
\n" ); document.write( "t is the time
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\n" ); document.write( "Your equation, with the parameters from your problem:
\n" ); document.write( "\"A+=+Pe%5E%28rt%29\"
\n" ); document.write( "\"A+=+6000e%5E%28.037t%29\"
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\n" ); document.write( "For 1 year:
\n" ); document.write( "\"A+=+6000e%5E%28.037%281%29%29\"
\n" ); document.write( "\"A+=+6000e%5E%28.037%29\"
\n" ); document.write( "\"A+=+6000%281.03769%29\"
\n" ); document.write( "A = $6226.16
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\n" ); document.write( "For 2 years:
\n" ); document.write( "\"A+=+6000e%5E%28.037%282%29%29\"
\n" ); document.write( "\"A+=+6000e%5E%28.074%29\"
\n" ); document.write( "\"A+=+6000%281.07681%29\"
\n" ); document.write( "A = $6460.84
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