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Find the geometric mean of the pair of numbers. 3 and 6

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same as last one' multiply them , take the square root
\(c^2=a\times b\) c is the geometrical mean of a and b
only this time you do not get a prefect square like the last one

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Other answers:

use that or this\[{a\over c}={c\over b}\]
yeah I did and it came out as 18 then when I got 4.24 but I cant figure out what way to write that out
\[\sqrt{9\times 3}=\sqrt{9}\times \sqrt{3}=3\sqrt{3}\]
or use a calculator if you want a decimal approximation
i did use a calculator I am saying I dont know what way to write it out let me show you the options @satellite73
sorry thats messy
ok use the option i wrote above
with the calculator?
oh damn i messed up
no i wrote the wrong thing lets take is slow
\[\sqrt{3\times 6}=\sqrt{18}=\sqrt{9\times 2}=\sqrt{9}\times \sqrt{2}=3\sqrt{2}\]
final answer is \(3\sqrt{2}\)
okay thanks

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