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3 years ago

6

A Euler circuit has ___ ODD vertices

1 answer:

xenn [34]3 years ago

4 0

Answer:

A Euler circuit has TWO ODD vertices

Step-by-step explanation:

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Determine the number of solutions for the linear equation shown below?<br> 4(3x+8)−9=2(6x−8)−15

natta225 [31]

Answer:

0 = -54

Step-by-step explanation:

4(3x + 8) − 9 = 2(6x − 8) − 15

12x + 32 - 9 = 12x - 16 - 15

12x + 23 = 12x - 31

12x = 12x - 54

0 = -54

No solutions.

Best of Luck!

5 0

3 years ago

How much does 6 goes into 4,296

Dominik [7]

If you divide 4,296 by 6 it would equal 716.

Answer: 716

4 0

3 years ago

Read 2 more answers

Plz help been stuck on this for awhile

Kazeer [188]

Answer:

<em>d</em> = $\sqrt{74}$ or 8.60

Step-by-step explanation:

(5, 8) (-2, 3)

Let 5 = $x_{1}$

8 = $y_{1}$

-2 = $x_{2}$

3 = $y_{2}$

Plug in the values into the equation to get the distance:

<em>d = </em> $\sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} }$ <em />

<em>d </em>= $\sqrt{(-2-5)^{2} + (3-8)^{2} }$

<em>d = </em> $\sqrt{(-7)^{2} + (5)^{2} }$

<em>d</em> = $\sqrt{49 + 25}$

<em>d = </em> $\sqrt{74}$ or 8.60

7 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

3 years ago

F(x)=5x^5+ 24x^4+25x^3-86x^2-224x+48

Serjik [45]

I believe you graph this.. please let me know if wrong and I’ll help solve otherwise! The picture is the answer on how to graph

4 0

3 years ago