Question on: WAEC Mathematics - 2016
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 = \(\frac{nrp}{nr + m^2}\)
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Correct Option: D
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 = \(\frac{nrp}{nr + m^2}\)
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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