where:
![3r^{4}=\frac{48}{625} \\r^{4}= \frac{48}{625*3} \\r^{4} = \frac{16}{625} \\r=\sqrt[4]{\frac{16}{625} } \\r = \frac{2}{5}](/tpl/images/0670/2508/90f91.png)
Find the three positive geometric means between 3 and 48/625
A. 9/5, 27/25, 81/125
...
Mathematics, 29.05.2020 15:57 lilbaby1433
Find the three positive geometric means between 3 and 48/625
A. 9/5, 27/25, 81/125
B. 6/5, 12/25, 24/125
C. 3/5, 3/25, 3/125
D. 1, 1/3, 1/9
Answers: 1
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