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mafiozo [28]
4 years ago
7

A = {1, 3, 5, 7, 9} B = {2, 4, 6, 8, 10} C = {1, 5, 6, 7, 9} A ∩ (B ∪ C) =

Mathematics
1 answer:
vovikov84 [41]4 years ago
6 0
A = {1, 3, 5, 7, 9}
B = {2, 4, 6, 8, 10}
C = {1, 5, 6, 7, 9}

(B ∪ C) = {1, 2, 4, 5, 6, 7, 8, 9, 10}
so
A ∩ (B ∪ C) = {1, 5, 7 , 9}
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Remember, adjectives in Spanish agree with the gender of the noun they describe.
Zanzabum

Answer:

1) A serious girl - <em>Una chica seria/Una joven seria</em>

2) The talented boy - <em>El chico talentoso/El joven talentoso</em>

3) An outgoing girl - <em>Una chica extrovertida/Una joven extrovertida</em>

4) The tasty ice cream - <em>El helado delicioso/El helado sabroso</em>

5) A funny book - <em>Un libro divertido/Un libro entretenido</em>

Step-by-step explanation:

Now, we present a possible set of solution, as there are solutions as many synonyms exist in Spanish. Then:

1) A serious girl - <em>Una chica seria/Una joven seria</em>

2) The talented boy - <em>El chico talentoso/El joven talentoso</em>

3) An outgoing girl - <em>Una chica extrovertida/Una joven extrovertida</em>

4) The tasty ice cream - <em>El helado delicioso/El helado sabroso</em>

5) A funny book - <em>Un libro divertido/Un libro entretenido</em>

4 0
3 years ago
What is the sum of the probabilities in a uniform probability distribution? A) 0 B) infinite C) 1 D)2
lubasha [3.4K]

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Probability.

Basically we have to know that, when there is a specific event for which a probability has to be done, there are always two possibilities that is,

Event favourable and Not Favourable.

the Sum of both always end up as 1.

thus sum of the probability distribution is 1.

8 0
3 years ago
Evaluate each logarithm. Do not use a calculator. ln ^3 square root e^4
Alona [7]

Answer:

\large\boxed{\ln\sqrt[3]{e^4}=\dfrac{4}{3}}

Step-by-step explanation:

\text{Use}\\\\\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\ln a^n=n\ln a\\\\\ln e=1\\-----------\\\\\ln\sqrt[3]{e^4}=\ln e^\frac{4}{3}=\dfrac{4}{3}\ln e=\dfrac{4}{3}\cdot1=\dfrac{4}{3}

6 0
3 years ago
Here is 3 &amp; 4 I NEED DONE ASAP!
Marrrta [24]

Given matrices are from here

brainly.com/question/18267865

=====================================================

Problem 3

a)

B - C = DNE

We cannot subtract matrices of different sizes. Matrix B is 2x2 while C is 3x2. Both matrices must have the same number of rows, and they must also have the same number of columns. The matrices don't have to be square.

------------------------------

b)

A+B = \begin{bmatrix}-5 & 6\\7 & 4\end{bmatrix}

You add the corresponding elements. For instance, in the top left corner we have -1+(-4) = -5. The other entries are treated in a similar manner.

------------------------------

c)

-2E = \begin{bmatrix}6 & -4 & 8\\-12 & 14 & -16\\-10 & -18 & 20\end{bmatrix}

You get this from multiplying each entry in matrix E by -2. Eg: top left corner has -2*(-3) = 6

------------------------------

d)

CD = \begin{bmatrix}4 & 7 & 8\\ 1 & 3 & 4\\14 & 17 & 16\end{bmatrix}

Matrix C has 3 rows and D has 3 columns. The final answer will be size 3x3

To generate each value in the answer matrix, you'll highlight rows of C to pair with columns of D. Then you'll multiply out the corresponding values, after which you add those products. This is done for every entry in the answer shown above.

For example, the first row of C is highlighted and the second column of D is highlighted. Those values pair up and multiply getting -1*(-1) + 2*3 = 1+6 = 7, which goes in the first row and second column of the answer matrix. The other entries are handled in a similar fashion.

------------------------------

e)

\det(B) = -46

The 2x2 matrix determinant formula is \begin{vmatrix}a & b\\c & d\end{vmatrix} = a*d - b*c

In this case, a = -4, b = 6, c = 5, d = 4.

------------------------------

f)

B^{-1} = \begin{bmatrix}-2/23 & 3/23\\ 5/46 & 2/23\end{bmatrix}

Swap the top left and bottom right corners of matrix B. Change the sign of the other two corner values. Then multiply each entry by 1/d where d is the determinant found back in part (e) above. Be sure to reduce any fraction as much as possible.

=====================================================

Problem 4

Answers: x = 2  and y = -10

-----------------

Work Shown:

\begin{bmatrix}2x & 4\\-5 & -2\end{bmatrix}+\begin{bmatrix}3 & -7\\11 & y-1\end{bmatrix} = \begin{bmatrix}7 & -3\\6 & -13\end{bmatrix}\\\\\\\begin{bmatrix}2x+3 & 4+(-7)\\-5+11 & -2+(y-1)\end{bmatrix} = \begin{bmatrix}7 & -3\\6 & -13\end{bmatrix}\\\\\\\begin{bmatrix}2x+3 & -3\\6 & y-3\end{bmatrix} = \begin{bmatrix}7 & -3\\6 & -13\end{bmatrix}\\\\\\

In the top left corners of each matrix, in line 3, we have 2x+3 = 7 which solves to...

2x+3 = 7

2x = 7-3

2x = 4

x = 4/2

x = 2

In the bottom right corners, we have y-3 = -13 which solves to....

y-3 = -13

y = -13+3

y = -10

4 0
3 years ago
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harkovskaia [24]
-3isqrt11!!! (The last answer!!)
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1 year ago
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