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JulijaS [17]
2 years ago
6

Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈

R² | a ≥ b}, the "greater than or equal to" relation, R₃ = {(a, b) ∈ R² | a < b}, the "less than" relation, R₄ = {(a, b) ∈ R² | a ≤ b}, the "less than or equal to" relation, R₅ = {(a, b) ∈ R² | a = b}, the "equal to" relation, R₆ = {(a, b) ∈ R² | a ≠ b}, the "unequal to" relation. Find
a)R1◦R1.b)R1◦R2.c)R1◦R3.d)R1◦R4.e)R1◦R5.f)R1◦R6.g)R2◦R3.h)R3◦R3.
Mathematics
1 answer:
uranmaximum [27]2 years ago
5 0

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = \{ (a,b) \in R^2 \}

d) R1ºR4 = \{ (a,b) \in R^2 \}

e) R1ºR5 = R1

f) R1ºR6 = \{ (a,b) \in R^2 \}

g) R2ºR3 = \{ (a,b) \in R^2 \}

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(<em>a,c</em>) is in R1ºR1 if there exists <em>b</em> such that (<em>a,b</em>) is in R1 and (<em>b,c</em>) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2  

This case is similar to the previous one. (<em>a,c</em>) is in R1ºR2 if there exists <em>b</em> such that (<em>a,b</em>) is in R2 and (<em>b,c</em>) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a < b and b > c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that a<b and b >c. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

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Answer:

Step-by-step explanation:

45 times 7= 315

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Solve for x in the equation x squared + 11 x + StartFraction 121 Over 4 EndFraction = StartFraction 125 Over 4 EndFraction.
Ira Lisetskai [31]

Answer:

Below

Step-by-step explanation:

● x^2 + 11x + 121/4 = 125/4

Substract 125/4 from both sides:

● x^2 + 11x + 121/4-125/4= 125/4 -125/4

● x^2 + 11x - (-4/4) = 0

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● c = 1

● b^2-4ac = 11^2-4*1*1 = 117

So this equation has two solutions:

● x = (-b -/+ √(b^2-4ac) ) / 2a

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There are 135 people in a sport centre. 77 people use the gym. 62 people use the swimming pool. 65 people use the track. 27 peop
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Answer:

The correct answer is \frac{2}{9}.

Step-by-step explanation:

There are 135 people in a sport centre.

Let the experiment be to pick a person from the sport centre who only uses exactly one of the facilities.

Total number of possible outcomes = 135

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77 people use the gym.

27 people use the gym and the pool.

31 people use the gym and the track.

4 people use all three facilities.

Total number of people using only the gym are 77 - ( 27 + 31 + 4) = 15

For the swimming pool:

62 people use the swimming pool.

27 people use the gym and the pool.

23 people use the pool and the track.

4 people use all three facilities.

Total number of people using only the swimming pool are 62 - ( 27 + 23 + 4) = 8

For the tracks:

65 people use the track.

23 people use the pool and the track.

31 people use the gym and the track.

4 people use all three facilities.

Total number of people using only the tracks are 65 - ( 23 + 31 + 4) = 7

A person is selected at random.

Favorable outcomes = 7 + 8 + 15 = 30

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5 0
2 years ago
A random draw is being designed for 210 participants. A single winner is to be chosen, and all the participants must have an equ
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Answer: The correct number of balls is (b) 4.

Step-by-step explanation:  Given that a single winner is to be chosen in a random draw designed for 210 participants. Also, there is an equal probability of winning for each participant.

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Let 'r' represents the number of balls to be picked up.

Since we are choosing from 10 balls, so we must have

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If r = 2, then

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Thus, (b) is the correct option.

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2 years ago
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