Make S the subject of the relation p s frac sm ... - JAMB Mathematics 2017 Question
Make S the subject of the relation
p = s + \(\frac{sm^2}{nr}\)
p = s + \(\frac{sm^2}{nr}\)
A
s = \(\frac{nrp}{nr + m^2}\)
B
s = nr + \(\frac{m^2}{mrp}\)
C
s = \(\frac{nrp}{mr}\) + m2
D
s = \(\frac{nrp}{nr}\) + m2
correct option: d
p = s + \(\frac{sm^2}{nr}\)
p = s + ( 1 + \(\frac{m^2}{nr}\))
p = s (1 + \(\frac{nr + m^2}{nr}\))
nr × p = s (nr + m2)
s = \(\frac{nrp}{nr + m^2}\)
p = s + ( 1 + \(\frac{m^2}{nr}\))
p = s (1 + \(\frac{nr + m^2}{nr}\))
nr × p = s (nr + m2)
s = \(\frac{nrp}{nr + m^2}\)
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