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Vlada [557]
4 years ago
14

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]4 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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Please solve the combination of equations here.
klemol [59]

Answer:

\left \{ {{x=5} \atop {y=7}} \right.

Step-by-step explanation:

\left \{ {{x+y=12} \atop {3y = 4x+1}} \right.

\left \{ {{x=12-y} \atop {3y=4x+1}} \right.

We are going to do a replacement of x for a second equation

So instead

3y=4x+1

in the place of x we are putting 12-y from the first equation, so

3y = 4(12-y)+1

To leave the brackets we need to do multiplication

3y = 4*12-4*y+1

Which is

3y = 48-4y+1

3y = -4y+49 /+4y\\7y = 49 /:7\\y = 7

How to determine x? From the first equation:

If the x +y = 12 then the x = 12-y so x = 12-7 = 5

We should write solution as a pair:

\left \{ {{x=5} \atop {y=7}} \right.

7 0
3 years ago
How many groups of two are in six
GuDViN [60]

Answer:

3 groups

Step-by-step explanation:

6 / 2 = 3 groups

plz mark me brainliest

4 0
3 years ago
Read 2 more answers
Will Give Brainliest Please Help!
BlackZzzverrR [31]

The given angle theorems and triangle theorems with the description that matches each is shown in the image attached below.

<h3>Angle Theorems and Triangle Theorems?</h3>
  • Angle theorems are theorems that states the relationship between angle pairs that are formed when two parallel lines are intersected by a transversal line.
  • Triangle theorems, such as midsegment theorem and angle sum theorem, relates specifically to triangles and its measures.

Thus, the given angle theorems and triangle theorems with the description that matches each is shown in the image attached below.

Learn more about angle and triangle theorems on:

brainly.com/question/19805547

6 0
3 years ago
Mark wants to use a grid like the ones in exercises 1 and 2 to model the percent equivalent of the fraction 2/3.How many grid sq
Marrrta [24]

Answer:

67 squares or 66.66 squares.

2/3 turned into a decimal is 66.66 or rounded, 67 squares.

Hope this helped! (plz mark me brainliest!)

4 0
3 years ago
Evaluate the surface integral. S xz dS S is the boundary of the region enclosed by the cylinder y2 + z2 = 16 and the planes x =
bagirrra123 [75]

If you project S onto the (x,y)-plane, it casts a "shadow" corresponding to the trapezoidal region

T = {(x,y) : 0 ≤ x ≤ 10 - y and -4 ≤ y ≤ 4}

Let z = f(x, y) = √(16 - y²) and z = g(x, y) = -√(16 - y²), each referring to one half of the cylinder to either side of the plane z = 0.

The surface element for the "positive" half is

dS = √(1 + (∂f/∂x)² + (∂f/dy)²) dx dy

dS = √(1 + 0 + 4y²/(16 - y²)) dx dy

dS = √((16 + 3y²)/(16 - y²)) dx dy

The the surface integral along this half is

\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16-y^2} \sqrt{\frac{16+3y^2}{16-y^2}} \, dx \, dy

\displaystyle \iint_T xz \,dS = \int_{-4}^4 \int_0^{10-y} x \sqrt{16+3y^2}\, dx \, dy

\displaystyle \iint_T xz \,dS = \frac12 \int_{-4}^4 (10-y)^2 \sqrt{16+3y^2} \, dy

\displaystyle \iint_T xz \,dS = 416\pi

You'll find that the integral over the "negative" half has the same value, but multiplied by -1. Then the overall surface integral is 0.

8 0
3 years ago
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