document.write( "Question 765749: At one boat dealer, she found a boat she likes that sells for $15,000 and depreciates at a rate of
\n" ); document.write( "30% per year. What will be the value of the boat after 3 years?
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Algebra.Com's Answer #466375 by MathLover1(20849)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "you can do it this way:\r
\n" ); document.write( "\n" ); document.write( "In the problem you are given the rate of decrease of \"30\"% or \"0.30\".\r
\n" ); document.write( "\n" ); document.write( "Each year, the boat is worth \"100-30=70\"% of its value the previous year, so after \"3\"years, its value will be\r
\n" ); document.write( "\n" ); document.write( "$\"15000%2A+0.30+=4500\" will cost less than year before; so price firs year after is $\"15000-4500=10500\"\r
\n" ); document.write( "\n" ); document.write( "next year value \"10500\" depreciates again for \"30\"%; so new value will be
\n" ); document.write( "\"10500%2A0.30=3150\"; so price second year after is $\"10500-3150=7350\"\r
\n" ); document.write( "\n" ); document.write( "and finally, \r
\n" ); document.write( "\n" ); document.write( "$\"7350%2A0.30=2205\"; so price in three years is \r
\n" ); document.write( "\n" ); document.write( "$\"7350-2205=5145\"\r
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\n" ); document.write( "\n" ); document.write( "or use exponential growth/decay formula:\r
\n" ); document.write( "\n" ); document.write( "Since the decay equation takes on the form \"P=P%5B0%5D%281-k%29%5Et\", we can take note that \"k=.30\", \"P%5B0%5D=15000\" and \"t=3\".
\n" ); document.write( "Plugging this into the equation, we have \r
\n" ); document.write( "\n" ); document.write( "\"P=15000%281-.3%29%5E3\"\r
\n" ); document.write( "\n" ); document.write( "\"P=5145\"\r
\n" ); document.write( "\n" ); document.write( "So after 3 years, the car will be worth \"5145\"\r
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