Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\)

  • A s = \(\frac{mrp}{nr + m^2}\)
  • B s = \(\frac{nr + m^2}{mrp}\)
  • C s = \(\frac{nrp}{mr + m^2}\)
  • D s = \(\frac{nrp}{nr + m^2}\)
waec 2016mathematics

The correct answer is D. s = \(\frac{nrp}{nr + m^2}\)

P = S + \(\frac{sm^2}{nr}\)

P = S(1 + \(\frac{m^2}{nr}\))

P = S(1 + \(\frac{nr + m^2}{nr}\))

nrp = S(nr + m2)

S = \(\frac{nrp}{nr + m^2}\)

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