+0  
 
0
59
4
avatar+781 

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!!..frown

 Oct 21, 2020
 #1
avatar+10583 
+1

Hi juriemagic,

it is solved correctly.
You can still put the W under the root.

\(L=\frac{\sqrt{Z^2-R^2}}{W}\)

laugh  !

 Oct 21, 2020
 #3
avatar+781 
0

Thank you Asinus...laugh

juriemagic  Oct 21, 2020
 #2
avatar+111602 
+1

I think it looks ok except L can be positive or negative.  So you need the +/- sign

 Oct 21, 2020
 #4
avatar+781 
+1

Thank you Melody....laugh

juriemagic  Oct 21, 2020

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