Asked by Tyalysia
Sue invested $1000 at 6% per year compounded yearly.Find the values of joe's investment at the end of each of the first five years
Answers
Answered by
Steve
after the nth year, it is worth
1000(1+.06)^n
1000(1+.06)^n
Answered by
Reiny
Why are you calling this trig ?
assuming his first deposit is NOW
end of 1st year: 1000(1.06) + 1000 = 2060
end of 2nd year: 2060(1.06) + 1000 = 3183.60
...
end of 5th year: .......... = 6637.09
fill in the rest using your calculator
assuming his first deposit is NOW
end of 1st year: 1000(1.06) + 1000 = 2060
end of 2nd year: 2060(1.06) + 1000 = 3183.60
...
end of 5th year: .......... = 6637.09
fill in the rest using your calculator
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