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Mathematics 17 Online
OpenStudy (anonymous):

need help? 18KRx (over) = E F^2

OpenStudy (anonymous):

solving for x

OpenStudy (anonymous):

\[{18kRx}= F^2E\] \[x = \frac{F^2E}{18kR}\]

OpenStudy (anonymous):

x=F^2E/18kR

OpenStudy (anonymous):

thanks !!

OpenStudy (anonymous):

my pleasure

OpenStudy (anonymous):

np : )

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