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Vlad [161]
2 years ago
14

A system of equations can never have

Mathematics
2 answers:
Oduvanchick [21]2 years ago
6 0

Answer:

This shows that a system of equations may have one solution (a specific x,y-point), no solution at all,

explanation: solution (being all the solutions to the equation). You will never have a system with two or three solutions; it will always be one, none, or infinitely-many.

miskamm [114]2 years ago
4 0
More variables than equations
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PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA POINTS*.. <br> IM GIVING 40 POINTS !! DONT SKIP :((.
steposvetlana [31]

Answer:

Step-by-step explanation:

(29, 35) (43, 21)

(21 - 35)/(43 - 29) = -14/14 = -1

y - 35 = -(x - 29)

y - 35 = -x + 29

y = -x + 64

3 0
3 years ago
Anyone, please help me please answer my question because I have to pass this tomorrow morning:(
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} }

15.

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

-  \frac{ - y {}^{4} x {}^{4} z {}^{ - 4} }{ {z}^{4} y {}^{8} }  =  - y {}^{4} x {}^{4} z {}^{ - 8}  =  -  \frac{(xy) {}^{4} }{ {z}^{8} }

16.

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

6 0
2 years ago
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Nat2105 [25]

Answer:

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6.5 = 6.5 = 66 1/2

Step-by-step explanation:

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6 0
3 years ago
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