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3 years ago

What is the degree of vertex B and G?

Mathematics

1 answer:

Marat540 [252]3 years ago

6 0

Answer:

degree of vertex B = 2

degree of vertex g = 4

Step-by-step explanation:

Using given picture we need to find about what is the degree of vertex B and G.

In graph theory, we know that the degree (or valency) of a vertex of a graph is the number of edges incident to the vertex.

So we just need to count how many edges are incindent on vertex B and G.

From picture we see that number of edges incident on vertex B = 2

Hence degree of vertex B = 2

From picture we see that number of edges incident on vertex G = 4

Hence degree of vertex g = 4

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If you graph the system of inequalities given below, which points would lie in the solution set? Select all the correct answers.

vivado [14]

Its just a matter of subbing...but keep in mind, for it to be a solution, it has to satisfy all 3 inequalities

x - y > = -4
2x - y < = 5
2y + x > 1

solutions are : (-1,3) and (3,5)

5 0

3 years ago

Read 2 more answers

What is the value of t?

frutty [35]

Answer:

-39 2/3

Step-by-step explanation:

t - 122= 2t - 3

subtract 't' from each side to get:

-122 = 3t - 3

add 3 to each side to get:

-119 = 3t

divide each side by 3 to get:

t = -39 2/3

5 0

2 years ago

The difference of a number and five is negative one. Find the number.

Veronika [31]

Answer:

The number is 4

Step-by-step explanation:

Write algebraically:

n-5=-1

n=4

The number is 4

8 0

3 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Andre45 [30]

For this case, we must indicate which of the given functions is not defined for $x = 0$

By definition, we know that:

$f (x) = \sqrt {x}$ has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$ , your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

$y = \sqrt [3] {x-2}$ with x = 0: $y = \sqrt [3] {- 2}$ is defined.