fomengchangoxyx1e fomengchangoxyx1e
  • 17-10-2017
  • Mathematics
contestada

19 ) 23a ³b ² / 69ab ³

pakai cara ya guys

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Аноним Аноним
  • 17-10-2017
[tex] \frac{23a^3b^2}{69ab^3} [/tex]

You simplify out...69 and 23 (divide each for 3), a^3 and a (since you divide them, you have to subtract the exponents...3-1 = 2), b^2 and b^3 (3-2 = 1)

[tex] \frac{a^2}{3b} [/tex] (final result)

So a²/3b
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