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4 years ago

Find x(g) for the graph.

Mathematics

1 answer:

Serjik [45]4 years ago

3 0

1. The x(g) is just the minimum number of colors needed to color the graph without a color being repeated. The answer is 2.

2. The minimum number needed to color this without repeating a color is 3.

3. The minimum number of colors required so that no two adjacent states have the same color is 3.

4. Remember, no colors can be repeated so options like A & B or D & E cannot be choices. The answer is is red (A, C, F); green (B, D, G); blue (E, H); yellow (I).

5. True. No adjacent vertex has the same color.

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Help please (see below)

Elden [556K]

Answer:

The quotient is 10x + 16

The remainder is $28x^2 + 10x +22$

4 0

3 years ago

Help Me Solve This

olasank [31]

Let us assume the number of cd's you can buy = x
Cost of an used CD = $4.25
Total amount of money that you have = $1500
Then
<span>x ≤ 15 − 4.25
</span>x ≤ 10.75

I hope the procedure is clear enough for you to understand. I also hope that this is the answer that you were looking for and the answer has actually come to your desired help.

3 0

3 years ago

51 less than the 8 times an unknown number is equal to 53.

Anton [14]

Let, Unknown number = x

8x-51 = 53

8x = 53+51

x = 104/8

x = 13

8 0

3 years ago

Which equation can be used to find the volume of this solid?

bogdanovich [222]

Answer:

V= 2 times 4 times 5

Step-by-step explanation:

In order to calculate the volume of a figure u must calculate lwh or length times width times height. In this case length is 5 width is 4 and height is 2. This can be plugged in and becomes 2 times 4 times 5.

6 0

3 years ago

Read 2 more answers

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

3 years ago