document.write( "Question 1176972: A mainframe computer whose cost is k220,000 will depreciate to a scrap value of k12,000in 5 years.
\n" ); document.write( "(1)what is the book value of the computer at the end of the third year?
\n" ); document.write( "(2)how much more would the book value be at the end of the third year of the straight line method of depreciation had been used? \n" ); document.write( "
Algebra.Com's Answer #805261 by Boreal(15235)\"\" \"About 
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exponential
\n" ); document.write( "P=Ae^kt
\n" ); document.write( "12000=220000e^5k
\n" ); document.write( "12/220=e^5k
\n" ); document.write( "ln both sides
\n" ); document.write( "-2.909=5k
\n" ); document.write( "k=-0.5817
\n" ); document.write( "-
\n" ); document.write( "after 3 years
\n" ); document.write( "P=220000*e^(-1.745)
\n" ); document.write( "P=k38,412.96 rounding at end
\n" ); document.write( "-
\n" ); document.write( "straight line
\n" ); document.write( "(0, 220)
\n" ); document.write( "(5, 12)
\n" ); document.write( "slope is 208/-5=-41.6 thousands
\n" ); document.write( "It decreases k41,600 a year for 3 years or 124,800 less.
\n" ); document.write( "220,000-124,800=K95,200
\n" ); document.write( "The book value would be k56,787 more\r
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