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1 year ago

Does the graph above have an Hamiltonian circuit, Hamiltonian path, or neither? Paths and Circuits Help

Mathematics

1 answer:

Hatshy [7]1 year ago

6 0

a. Only a Hamiltonian path

One such path is

1 → 2 → 0 → 4 → 3

which satisfies the requirement that each vertex is visited exactly once.

There is no Hamiltonian circuit, however, since it is impossible for any Hamiltonian path on this graph to visit vertex 0 exactly once.

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What is the distance between the points (–3, 4) and (–7, 4)?

uysha [10]

Answer:

$d = \sqrt{( - 7 + 3) {}^{2} + (4 - 4) {}^{2} } \\ d = \sqrt{16} \\ d = 4$

Good luck!

Intelligent Muslim,

From Uzbekistan.

6 0

3 years ago

How do I solve this equation?

Gnom [1K]

$2x(4-x)^{-1/2}-3(4-x)^{1/2}=(4-x)^{-1/2}\bigg(2x-3(4-x)\bigg)$

We effectively rewrite the equation as

$\dfrac{5x-12}{(4-x)^{1/2}}=0$

In order for the LHS to be defined, we need to restrict $4-x>0$ , or $x$ . Now, the LHS will vanish when the numerator is 0, which happens for

$5x-12=0\implies5x=12\implies x=\dfrac{12}5$

This value is indeed smaller than 4, so the solution is $x=\dfrac{12}5$ .

8 0

2 years ago

What is the common difference between successive terms in the sequence

Lelechka [254]

9,2.5,-4,-10.5,-17

Let's find the difference between two consecutive terms

2.5 - 9 = -6.5

-4 - 2.5 = -6.5

-10.5 - (-4) = -6.5

-17 -(-10.5) = -6.5

We can see that the common difference is -6.5

-6.5 is the common difference between successive terms in the sequence 9,2.5,-4,-10.5,-17

the common difference is negative because we subtract 6.5 to get the next term.

4 0

3 years ago

2y⁴-18x²y²-48xy²-32y²<br><br>

Goryan [66]

Answer:

Step-by-step explanation:

6 0

2 years ago

which of the fallowing functions has a slope 3/2 and contains the midpoint segment between (6, 3) and (-2, 11)?

kolbaska11 [484]

Well, we know the slope is 3/2, what's the midpoint of those anyway?

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 6}}\quad ,&{{ 3}})\quad 
% (c,d)
&({{ -2}}\quad ,&{{ 11}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{-2+6}{2}~~,~~\cfrac{11+3}{2} \right)\implies (2,7)$

so, what's the equation of a line whose slope is 3/2 and runs through 2,7?

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 2}}\quad ,&{{ 7}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{3}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-7=\cfrac{3}{2}(x-2)
\\\\\\
y-7=\cfrac{3}{2}x-3\implies y=\cfrac{3}{2}x+4$

4 0

2 years ago