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3 years ago

A 6 square units B 8 square units C 10 square units D 16 square units

Mathematics

1 answer:

Dahasolnce [82]3 years ago

5 0

The answer would be D which is 16 square units

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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

What is the inverse of the function? f(x) = 2x - 10?

gulaghasi [49]

Answer:

Step-by-step explanation:

Let $f(x)=y$

y=2x-10

Switch x and y,

x=2y-10

x+10=2y

(x/2)+5=y

So $f^{-1}(x)=\frac{x}{2}+5$

3 0

2 years ago

How do you solve law of cosine?

Effectus [21]

Draw and label a standard Oblique Triangle, as we’ve done in our previous lessons.

Determine the given congruence, either SAS or SSS, and pick the equation that helps you solve for either a missing side or angle.

Plug into your chosen equation and solve.

The "Law of Cosines" can be used to calculate one side of a triangle when the angle opposite and the other two sides are known. The "Law of Cosines" can be expressed as c2 = a2 + b2 - 2 a b cos C (1)

The cosine rule is an extension of this mathematic principal that makes it effective for non-right triangles and states that in regard to a certain angle, the square of the side of the triangle opposite that angle is equal to the squares of the other two sides added together, minus two times both..

4 0

3 years ago

Read 2 more answers

Can y'all help me step by step with number 5

aev [14]

If the wall with 4 ft. of storage area can fit 6 sections of plant and the wall with 7 ft. Can fit 5, that means each section is 3 ft. because 7-4=3, so that extra section on the wall with 4 ft. You do this because there is one extra section on the wall with four. Knowing this, the wall with 7 ft. Would be the 7 ft. + 5 sections of 3 ft. So 7+(5x3) = 22. On the wall with 4 ft., do the same,
4+(6x3) =22. Since they are equal and the walls are supposed to be equal, we can confirm this is the right answer

6 0

3 years ago

1. Storing milk at temperatures colder than 35°F can affect its quality of taste. However, storing milk at temperatures warmer t

Mazyrski [523]

Answer:

Part (A): The required inequality is T < 35 or t >40.

Part (B): The correct option is C) Only x=7.

Part (C) |x+1|+5=2 has no solution; |4x+12|=0 has one solution; |3x|=9 has two solution.

Step-by-step explanation:

Consider the provided information.

Part (A)

Storing milk at temperatures colder than 35°F can affect its quality of taste. However, storing milk at temperatures warmer than 40°F is an unsafe food practice.

The union is written as A∪B or “A or B”.

The intersection of two sets is written as A∩B or “A and B”

We need to determine the inequalities represent the union of these improper storage.

That means we will use A∪B or “A or B”.

The improper storage temperatures is when temperature is less than 35°F or greater than 40°F.

Hence, the required inequality is T < 35 or t >40.

Part (B) 15| x-7|+4=10|x-7|+4

Solve the inequality as shown below:

Subtract 4 from both sides.

$15| x-7|+4-4=10|x-7|+4-4$