Make L the subjects of the formula if d = \(\sqrt{\frac{42w}{5l}}\)
- \(\sqrt{\frac{42w}{5d}}\)
-
\(\frac{42W}{5d^2}\)
- \(\frac{42}{5dW}\)
- \(\frac{1}{d}\sqrt{\frac{42w}{5}}\)
The correct answer is: B
Explanation
d = \(\sqrt{\frac{42w}{5l}}\)
square both side of the equation
\(d^2 = \left(\sqrt{\frac{42W}{5l}}\right)^2\\
d^2 = \frac{42W}{5l}\\
5ld^2=42W\\
l = \frac{42W}{5d^2}\)
There is an explanation video available .