document.write( "Question 1120055: Use graphical approximation techniques or an equation solver to approximate the desired interest rate. A person makes annual payments of $ 1000 into an ordinary annuity. At the end of 5 ​years, the amount in the annuity is $ 5703.88. What annual nominal compounding rate has this annuity​ earned? \n" ); document.write( "
Algebra.Com's Answer #735729 by ankor@dixie-net.com(22740)\"\" \"About 
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A person makes annual payments of $ 1000 into an ordinary annuity.
\n" ); document.write( " At the end of 5 years, the amount in the annuity is $ 5703.88.
\n" ); document.write( " What annual nominal compounding rate has this annuity earned?
\n" ); document.write( ":
\n" ); document.write( "We can use the annuity formula
\n" ); document.write( "P(\"%28%281%2Bi%29%5Et-1%29%2Fi\") = Fv; where
\n" ); document.write( "P =payment
\n" ); document.write( "i = interest rate
\n" ); document.write( "t = periods
\n" ); document.write( "Fv = Final value
\n" ); document.write( ":
\n" ); document.write( "1000(\"%28%281%2Bi%29%5E5-1%29%2Fi\") = 5703.88
\n" ); document.write( "Put in the form that we can graph; i = x
\n" ); document.write( "1000(\"%28%281%2Bx%29%5E5-1%29%2Fx\") = 5703.88
\n" ); document.write( "divide both sides by 1000
\n" ); document.write( "\"%28%281%2Bx%29%5E5-1%29%2Fx\" = \"5703.88%2F1000\"
\n" ); document.write( "Approx solution; x = .066 or 6.6% interest
\n" ); document.write( "\"%28%281%2Bx%29%5E5-1%29%2Fx\" = 5.704
\n" ); document.write( "multiply both sides by x
\n" ); document.write( "(1+x)^5 - 1 = 5.704x
\n" ); document.write( "A quadratic equation we can graph
\n" ); document.write( "(1+x)^5 - 5.704x - 1 = 0
\n" ); document.write( ":
\n" ); document.write( "\"+graph%28+300%2C+200%2C+-.1%2C+.1%2C+-.05%2C+.05%2C+%281%2Bx%29%5E5-5.704x-1%29+\"\r
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