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riadik2000 [5.3K]
3 years ago
9

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

Mathematics
2 answers:
Katen [24]3 years ago
5 0
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
stich3 [128]3 years ago
4 0
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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