Please Help! On federal income tax returns, self employed people can depreciate the value of business equipment. Suppose a computer valued at $1200 depreciates at a rate of 25% a year. In 2005, the average retirement salary for Colorado teachers was $47,000. The contract promised a 2% raise every year. How much money would that teacher be making in the year 2010? Round to the nearest $100.
I'm not sure the first part is a complete question. For the second part, you multiply 47000 by 1.02, 5 times. Once for each year between 2005 and 2010.
\[a_n=a_{n-1}+(.02)a_{n-1}\] \[a_n=(1+.02)a_{n-1}\] \[a_n=(1.02)a_{n-1}\] \[a_n=(1.02)(1.02)a_{n-2}\] \[a_n=(1.02)(1.02)(1.02)a_{n-3}\] \[a_n=(1.02)^4a_{n-4}\] \[a_n=(1.02)^ra_{n-r}\] \[a_n=(1.02)^na_{n-n}\] \[a_n=(1.02)^na_{0}\]
since a_0 = 47000 we have a closed form of an equation that can tell us the value at the nth year; 2005 2010 -2005 -2005 ------------- 0 5 \[a_5=(1.02)^5\ *47000\]
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On federal income tax returns, self employed people can depreciate the value of business equipment. Suppose a computer valued at $1200 depreciates at a rate of 25% a year. In 2005, the average retirement salary for Colorado teachers was $47,000. The contract promised a 2% raise every year. In what year would that teacher earn $60000?
so 1.02^x * 47000 = 60000. Solve for x. Divide by 47000, take the log of both sides.
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