Explain why there can't be a simple graph with the following sequence of vertex degrees: (a) 5,1,1,1

Mathematics

1 answer:

Contact [7]3 years ago

7 0

Answer:

a cant exist because 3 out of the 4 vertices must have only one degree vertex and the remaining one must have 5.

b) This example can be a single graph because of the handshaking lemma. The sum of the odd degrees vertex must be an even number: 3+3+5=11

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I need help, i don’t understand how to do it

3241004551 [841]

Hai , you can use Pythagoras Theorem to find the length which is hypotenuse.

It will be 7^2 + 3^2 = 58

square root 58 and you will get 7.62

Hope this helps and i am sorry if my answer was wrong .

3 0

2 years ago

Secant RM intersects secant RN at point R. Find the length of RQ. If necessary, round to the hundredths place.

QveST [7]

Answer:

3 Units

Step-by-step explanation:

Secant RM intersects secant RN at point R.

The secants intersects the circle at points P and Q respectively as seen in the diagram.

To determine the length of RQ, we use the Theorem of Intersecting Secants.

Applying this on the diagram, we have:

RP x RN=RQ X RM

4(4+5)=RQ(RQ+9)

Let the length of RQ=x

$4*9=x^2+9x\\36=x^2+9x\\x^2+9x-36=0\\x^2+12x-3x-36=0\\x(x+12)-3(x+12)=0\\(x+12)(x-3)=0\\x+12=0$ x-3=0\\x=-12 or x=3\\Since x cannot be negative$\\x=|RQ|=3$