I NEED HELP! WHO EVER ANSWERS FIRST I WILL MARK BRAINLIEST

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

Please help need answer for a quiz

torisob [31]

Answer:

first one with -1...21

Step-by-step explanation:

only one that follows quadratic formula format

5 0

3 years ago

PLEASE HELP!! Compare & Contrast how Completing the Square is used to Convert a quadratic function to vertex form with how i

Likurg_2 [28]

Answer:

Step-by-step explanation:

Given a general quadratic formula given as ax²bx+c = 0

To generate the general formula to solve the quadratic equation, we can use the completing the square method as shown;

Step 1:

Bringing c to the other side

ax²+bx = -c

Dividing through by coefficient of x² which is 'a' will give:

x²+(b/a)x = -c/a

- Completing the square at the left hand side of the equation by adding the square of half the coefficient x i.e (b/2a)² and adding it to both sides of the equation we have:

x²+(b/a)x+(b/2a)² = -c/a+(b/2a)²

(x+b/2a)² = -c/a+(b/2a)²

(x+b/2a)² = -c/a + b²/4a²

- Taking the square root of both sides

√(x+b/2a)² = ±√-c/a + b²/√4a²

x+b/2a = ±√(-4ac+b²)/√4a²

x+b/2a =±√b²-4ac/2a

- Taking b/2a to the other side

x = -b/2a±√√b²-4ac/2a

Taking the LCM:

x = {-b±√b²-4ac}/2a

This gives the vertex form with how it is used to Solve a quadratic equation.

7 0

3 years ago

Please help ill give you points

UkoKoshka [18]

Yes, it is a function

4 0

3 years ago

Read 2 more answers

PLEASE HELP!!<br> Factor <br> 6x^2-42x-54<br><br> Please show work!

skad [1K]

first, rewrite 54 as 6•9

next, rewrite 42 as 6•7

then, that gets you 6x^2 -6•7x- 6•9

lastly, factor out the common term (6)

and then you get 6(x^2-7x-9)

<u>please mark as brainliet <3</u>

6 0

3 years ago