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Elenna [48]
3 years ago
11

Each vertex in a graph of n bertices can be the origin of at most ____edges

Computers and Technology
1 answer:
EastWind [94]3 years ago
6 0

Answer:

B. n-1

Explanation:

If there are n vertices then that vertex can be origin of at most  n-1 edges.Suppose that you have a graph with 8 vertices you can select a vertex from these 8 vertices now you have 7 other vertices.So the vertex you selected can have at most 7 edges or it can be origin of at most 7 edges.So we conclude that the answer is n-1.

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Frequent cleaning with an air duster can would do the job
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9. These particular machines can be decentralized.
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A, B

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Find the inverse function of f(x)= 1+squareroot of 1+2x
Svetllana [295]

Answer:

Therefore the inverse function of  f(x)=1+\sqrt{1+2x} is \frac{x^2-2x}{2}

Explanation:

We need to find the inverse of function f(x)=1+\sqrt{1+2x}

Function Inverse definition :

\mathrm{If\:a\:function\:f\left(x\right)\:is\:mapping\:x\:to\:y,\:then\:the\:inverse\:functionof\:f\left(x\right)\:maps\:y\:back\:to\:x.}

y=1+\sqrt{1+2x}\mathrm{Interchange\:the\:variables}\:x\:\mathrm{and}\:y

x=1+\sqrt{1+2y}

\mathrm{Solve}\:x=1+\sqrt{1+2y}\:\mathrm{for}\:y

\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}

1+\sqrt{1+2y}-1=x-1

Simplify

\sqrt{1+2y}=x-1

\mathrm{Square\:both\:sides}

\left(\sqrt{1+2y}\right)^2=\left(x-1\right)^2

\mathrm{Expand\:}\left(\sqrt{1+2y}\right)^2:\quad 1+2y

\mathrm{Expand\:}\left(x-1\right)^2:\quad x^2-2x+1

1+2y=x^2-2x+1

\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}

1+2y-1=x^2-2x+1-1

\mathrm{Simplify}

2y=x^2-2x

\mathrm{Divide\:both\:sides\:by\:}2

\frac{2y}{2}=\frac{x^2}{2}-\frac{2x}{2}

\mathrm{Simplify}

y=\frac{x^2-2x}{2}

Therefore the inverse function of f(x)=1+\sqrt{1+2x} is \frac{x^2-2x}{2}

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3 years ago
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