Make 'n' the subject of the formula if w = \(\frac{v(2 + cn)}{1 - cn}\)
index-0
\(\frac{1}{c}(\frac{w - 2v}{v + w})\)
index-1
\(\frac{1}{c}(\frac{w - 2v}{v - w})\)
index-2
\(\frac{1}{c}(\frac{w + 2v}{v - w})\)
index-3
\(\frac{1}{c}(\frac{w + 2v}{v + w})\)
Correct Answer:
A
w = \(\frac{v(2 + cn)}{1 - cn}\)
2v + cnv = w(1 - cn)
2v + cnv = w - cnw
2v - w = -cnv - cnw
Multiply through by negative sign
-2v + w = cnv + cnw
-2v + w = n(cv + cw)
n = \(\frac{-2v + w}{cv + cw}\)
n = \(\frac{1}{c}\frac{-2v + w}{v + w}\)
Re-arrange...
n = \(\frac{1}{c}\frac{w - 2v}{v + w}\)