document.write( "Question 73441This question is from textbook algebra and trigonometry structure and method
\n" ); document.write( ": one thousand dollars is invested at 12% interest compounded annually. determine how much the investment is after 2 years \n" ); document.write( "
Algebra.Com's Answer #52563 by bucky(2189)\"\" \"About 
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The equation for compound interest is:
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\n" ); document.write( "\"P+=+C%2A%281+%2B+%28r%2Fn%29%29%5E%28n%2At%29\"
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\n" ); document.write( "where the variables are defined as:
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\n" ); document.write( "P is the future value
\n" ); document.write( "C is the initial value
\n" ); document.write( "r is the annual interest expressed as a decimal (6% = 0.06)
\n" ); document.write( "n is the number of times per year the interest is compounded
\n" ); document.write( "t is the number of years invested.
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\n" ); document.write( "For this problem:
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\n" ); document.write( "P is unknown
\n" ); document.write( "C is 1000
\n" ); document.write( "r is 0.12
\n" ); document.write( "n is 1 (annually is once per year)
\n" ); document.write( "t is 2 years
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\n" ); document.write( "Substitute these numbers into the equation
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\n" ); document.write( "\"P+=+1000%2A%281+%2B+%280.12%2F1%29%29%5E%281%2A2%29\"
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\n" ); document.write( "Simplify:
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\n" ); document.write( "\"P+=+1000%2A%281.12%29%5E2=+1000%2A1.2544+=+1254.40\"
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\n" ); document.write( "The answer to this problem is that at 12% compounded annually for 2 years $1,000 will grow
\n" ); document.write( "to be $1254.40.
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\n" ); document.write( "Hope this helps you to understand compound interest and how your money can work for you.
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