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zimovet [89]
3 years ago
9

What is the slope of the line through point (2/3, 4/7) and (2/3, 11/7)

Mathematics
1 answer:
mart [117]3 years ago
6 0
\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ \frac{2}{3} &,& \frac{4}{7}~) 
%  (c,d)
&&(~ \frac{2}{3} &,& \frac{11}{7}~)
\end{array}
\\\\\\
% slope  = m
slope \implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{\frac{11}{7}-\frac{4}{7}}{\frac{2}{3}-\frac{2}{3}}\implies \cfrac{\frac{7}{7}}{0}\implies \cfrac{1}{0}\impliedby un de fined
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adelina 88 [10]

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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}

=  - u {}^{ - 1} v {}^{ - 5}  =   - \frac{ 1}{uv {}^{5} }

13.

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

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

-  \frac{2m {}^{4} n {}^{ - 1} }{m {}^{ - 5}n {}^{3}  }  =  - 2m {}^{9} n {}^{ - 4}

=   - \frac{2m {}^{9} }{n {}^{4} }

14.

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

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

( -  {x}^{5} y {}^{ - 2} ) {}^{3}  =  -  x {}^{15} y {}^{ - 6}

-  \frac{x {}^{15} }{ {y}^{6} }

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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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Alex17521 [72]

C.


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Therefore, it's C

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