document.write( "Question 332248: Suppose that $2000 is invested at a rate of 6% per year compounded continuously. What is the balance after 1 yr? After 2 yrs? \n" ); document.write( "
Algebra.Com's Answer #238116 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Recall that the formula for continuous compounding interest is\r
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\n" ); document.write( "\n" ); document.write( "\"A=Pe%5E%28rt%29\"\r
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\n" ); document.write( "\n" ); document.write( "where A is the return, P is the principal (amount invested), r is the interest rate (in decimal form) and t is the time in years.\r
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\n" ); document.write( "\n" ); document.write( "Since \"$2000 is invested at a rate of 6% per year compounded continuously\", we know that \"P=2000\" and \"r=0.06\" (the decimal equivalent of 6%).\r
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\n" ); document.write( "\n" ); document.write( "Now let's compute the return when \"t=1\" (ie find the balance after one year)\r
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\n" ); document.write( "\n" ); document.write( "\"A=Pe%5E%28rt%29\" Start with the continuous compounding formula.\r
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\n" ); document.write( "\n" ); document.write( "\"A=2000%2Ae%5E%280.06%2A1%29\" Plug in \"P=2000\", \"r=0.06\", and \"t=1\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=2000%2Ae%5E%280.06%29\" Multiply \"0.06\" and \"1\" to get \"0.06\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=2000%2A1.06183654654536\" Raise 'e' (which is approximately 2.71828) to the power \"0.06\" to get \"1.06183654654536\" (this value is approximate).\r
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\n" ); document.write( "\n" ); document.write( "\"A=2123.67309309072\" Multiply \"2000\" and \"1.06183654654536\" to get \"2123.67309309072\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=2123.67\" Round to the nearest hundredth (ie to the nearest penny).\r
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\n" ); document.write( "\n" ); document.write( "So after 1 year, you'll have about $2,123.67\r
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\n" ); document.write( "\n" ); document.write( "Now let's compute the return when \"t=2\" (ie find the balance after two years)\r
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\n" ); document.write( "\n" ); document.write( "\"A=Pe%5E%28rt%29\" Start with the continuous compounding formula.\r
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\n" ); document.write( "\n" ); document.write( "\"A=2000%2Ae%5E%280.06%2A2%29\" Plug in \"P=2000\", \"r=0.06\" (the decimal equivalent of 6%), and \"t=2\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=2000%2Ae%5E%280.12%29\" Multiply \"0.06\" and \"2\" to get \"0.12\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=2000%2A1.12749685157938\" Raise 'e' (which is approximately 2.71828) to the power \"0.12\" to get \"1.12749685157938\" (this value is approximate).\r
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\n" ); document.write( "\n" ); document.write( "\"A=2254.99370315875\" Multiply \"2000\" and \"1.12749685157938\" to get \"2254.99370315875\".\r
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\n" ); document.write( "\n" ); document.write( "\"A=2254.99\" Round to the nearest hundredth (ie to the nearest penny).\r
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\n" ); document.write( "\n" ); document.write( "So after two years, you'll have $2,254.99
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