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Inga [223]
2 years ago
6

Urgent plz help me!!!!!plzzzzzz

Mathematics
2 answers:
IRISSAK [1]2 years ago
8 0

Answer:

x is equal to 80, and y is equal to 90.

Step-by-step explanation:

This is because the shape looks odd, due to the weird circle, but the opposite angles are congruent.

#teamtrees #WAP (Water And Plant)

kap26 [50]2 years ago
8 0
X is equal to 80 and y is equal to 90
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Draw a quadrilateral that is similar (but not congruent) to the one given below. Measure and label all sides and angles of quadr
Tanya [424]

Answer:

draw the same shape but multiply all the sizes to be bigger or smaller

5 0
2 years ago
In the graph, if the number of manufactured units falls below 6.25, then the business will be operating in what state?
AURORKA [14]
It would be in a state of loss because the units manufactured neither reaches nor exceeds the break even point.
6 0
3 years ago
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:

<h3>(a)</h3>

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.

<h3>(b)</h3>

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
2 years ago
HELP PLSSSSS PLSSSS I NEED HELPPPPPPP ASAPPPPP
Hitman42 [59]
I would say A (7) but I’m not 100% sure.
5 0
2 years ago
Which conic's equation has 2 squares with the same signs and different leading coefficients?
Romashka [77]

We want to see which is the conic equation that has 2 squares with the same sign and different leading coefficients.

We will see that it is the equation of the ellipse.

Now let's see why that is the correct answer.

The general conic equation is of the form:

A*(x - a)^2 + B*(x - b)^2 = R^2

Where A and B are the leading coefficients, (a, b) is the center of the figure, and R is the average radius of the figure.

For example, if A = B, this would be the equation of a circle, but we must have two different leading coefficients.

If we write:

A = 1/C

B = 1/K

(both are positive, because "it has two squares with the same signs").

Then we get the equation:

\frac{ (x - a)^2}{C} + \frac{(x - b)^2}{K} = R^2

We get the equation we wanted.

The same sign in both square parts and different leading coefficients.

The above equation is the general equation of an ellipse.

If you want to learn more, you can read:

brainly.com/question/10311514

7 0
2 years ago
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