#1**+1 **

I'm assuming that we are solving for *v* in the equation \(v^2=\frac{25}{81}\). Let's do that!

\(v^2=\frac{25}{81}\) | Take the square root of both sides to isolate v. Of course, the square root always results in 2 answers: The positive and negative. |

\(v=\pm\sqrt{\frac{25}{81}}\) | "Distribute" the square to both the numerator and denominator. |

\(v=\pm\frac{\sqrt{25}}{\sqrt{81}}\) | Now, simplify both the numerator and denominator. |

\(v=\pm\frac{5}{9}=\pm0.\overline{55}\) | |

TheXSquaredFactor
Aug 25, 2017

#1**+1 **

Best Answer

I'm assuming that we are solving for *v* in the equation \(v^2=\frac{25}{81}\). Let's do that!

\(v^2=\frac{25}{81}\) | Take the square root of both sides to isolate v. Of course, the square root always results in 2 answers: The positive and negative. |

\(v=\pm\sqrt{\frac{25}{81}}\) | "Distribute" the square to both the numerator and denominator. |

\(v=\pm\frac{\sqrt{25}}{\sqrt{81}}\) | Now, simplify both the numerator and denominator. |

\(v=\pm\frac{5}{9}=\pm0.\overline{55}\) | |

TheXSquaredFactor
Aug 25, 2017