Then my last one for the day...
Make L the subject
\(Z= {\sqrt{R^2+W^2L^2}}\)
I did this:
\(R^2+W^2L^2=Z^2 \)
\(W^2L^2=Z^2-R^2\)
\(L^2={Z^2-R^2 \over W^2}\)
\(L= \sqrt{{Z^2-R^2} \over W^2}\)
mmmmmm...I'm not so sure!!..
Hi juriemagic,
it is solved correctly.You can still put the W under the root.
\(L=\frac{\sqrt{Z^2-R^2}}{W}\)
!
Thank you Asinus...
I think it looks ok except L can be positive or negative. So you need the +/- sign
Thank you Melody....
+1124 
!
