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Olenka [21]
3 years ago
9

Yeah uh I have no idea how to do this.....

Mathematics
2 answers:
lilavasa [31]3 years ago
6 0

Answer:

\frac{9b}{a^{2}  }\sqrt{\frac{1}{a}  } should be the answer of the problem that you posted

if you try to solve using \sqrt{x}    as as x^{\frac{1}{2}  }

then (a^{-5)} ^{\frac{1}{2}  } \\ =  (a^{-5/2)} = \frac{1}{a^{\frac{5}{2}  } }

\frac{1}{a^{\frac{5}{2}  } } * 9b = \frac{9b}{a^{\frac{5}{2 }  }  }

Step-by-step explanation:

9 \sqrt{(a^-5 b^2)}

x^{-1}  =  \frac{1}{x}

a^{-5}  = \frac{1}{a^{5}  }

\sqrt{b^{2}  } = b

a^{4} a's can be removed from the radicle (one will be left in because you have a^5)

\frac{9b}{a^{2}  }\sqrt{\frac{1}{a}  }

Liula [17]3 years ago
4 0

Answer:

9b /a^5/2

Step-by-step explanation:

9√(a^-5 b^2)

9 sqrt( a^ -5 b^2)

Rewriting sqrt as ^1/2

9 ( a^ -5 b^2)^1/2

9 ( a^ -5) ^1/2 ( b^2)^1/2

we know that  an exponent to an exponent means multiply

9 a^ (-5*1/2)  b^(2*1/2)

9 a^-5/2 b^1

9b a^ -5/2

We know that x^-y = 1/x^y

9b /a^5/2

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