Make 'n' the subject of the formula if w = \(\frac{v(2 + cn)}{1 - cn}\)
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Correct Answer: Option A
Explanation:
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 = \(\frac{1}{c}\frac{w - 2v}{v + w}\)
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 = \(\frac{1}{c}\frac{w - 2v}{v + w}\)