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salantis [7]
3 years ago
14

2v(-4v^2+4v+4) BE FAST! Best gets the Brainliest.

Mathematics
2 answers:
ella [17]3 years ago
5 0

Answer:

8v + 8v² - 8v³

Step-by-step explanation:

According to the Product-to-Power Exponential Rule, whenever you multiply terms with exponents and coefficients, you add the exponents.

2v[-4v² + 4v + 4]

-8v³ + 8v² + 8

The above answer is written in reverse, but it is the exact same result.

I am joyous to assist you anytime.

dezoksy [38]3 years ago
5 0

2v(-4v^2+4v+4)

Multiply the bracket by 2v

(2v)(-4v^2)=-8v^3

(2v)(4v)=8v^2

(2v)(4)=8v

Answer: -8v^3+8v^2+8v

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Wittaler [7]

Step-by-step explanation:

If you need help with how I got my answer, you can ask me.

\frac{2yx {}^{ - 4} }{(x {}^{ - 4}y {}^{4}) {}^{3}  \times 2x {}^{ - 1}y {}^{ - 3}    }

\frac{2yx {}^{ - 4} }{x {}^{ - 12}y {}^{12}   \times 2x {}^{ - 1}y {}^{ - 3}  }

\frac{2y x {}^{ - 4}  }{2x {}^{  - 13} y {}^{ - 9} }

= x {}^{9} y {}^{ - 8}

=  \frac{x {}^{9} }{y {}^{8} }

12.

( \frac{2u {}^{  4} }{ - u {}^{2}v {}^{ - 1}   \times 2uv {}^{ - 4} } ) {}^{ - 1}

( \frac{2u {}^{4} \times 1 }{ - 2 {u}^{3}v {}^{ - 5}  } ) {}^{ - 1}

( - u v {}^{ 5} ) {}^{   - 1}

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-   \frac{2m {}^{4}n {}^{ - 1}  }{( - m {}^{2}n {}^{ - 2}) {}^{ - 1}    \times  - nm {}^{ - 3} }

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( \frac{x {}^{3}y {}^{ - 4}  }{ -  {x}^{2} y {}^{ - 3} \times yx {}^{0}  } ) {}^{3}

( \frac{x {}^{3}y {}^{ - 4}  }{ -  {x}^{ - 2}y {}^{ - 2}  } ) {}^{3}

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16.

\frac{h {}^{7}j  {k}^{4} }{4h {}^{4} }  =  \frac{1}{4} h {}^{3}  =  \frac{h {}^{3} jk {}^{4} }{4}

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