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THARRIS1818 THARRIS1818
  • 18-06-2014
  • Mathematics
Answered

2700=2300(1+r)^5 solve for r Please show all work

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Ryan2
Ryan2 Ryan2
  • 18-06-2014

\(2700=2300(1+r)^5\\ \\ (1+r)^5=\frac{2700}{2300}\\ \\ (1+r)^5=\frac{27}{23}\\ \\ \sqrt[5]{(1+r)^5}= \sqrt[5]{\frac{27}{23}}\\ \\ 1+r= 1.032\\ \\ r=1.032-1\\ \\ \boxed{r=0.032}\)

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kate200468
kate200468 kate200468
  • 19-06-2014

\(2700=2300(1+r)^5\ /:2300\\\\(1+r)^5= \frac{27}{23} \ \ \ \Rightarrow\ \ \ 1+r= \sqrt[5]{ \frac{27}{23}} \ \ \ \Rightarrow\ \ \ r= \sqrt[5]{ \frac{27}{23}} -1\\\\ r= \sqrt[5]{ \frac{27\cdot23^4}{23^5}}-1= \frac{ \sqrt[5]{ 27\cdot23^4} }{23} -1= \frac{ \sqrt[5]{7555707} }{23} -1\)

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