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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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leonid [27]

Answer: 250 mi

Step-by-step explanation:

Here we can think in a triangle rectangle:

The distance from Birmingham to Atlanta is roughly 150 mi, and this is one of the cathetus.

And the distance from Birmingham to Nashville is roughly 200 mi, this is the other cathetus of the triangle.

Now, the distance from Atlanta to Nashville will be the hypotenuse of this triangle rectangle.

Now we can apply the Pythagorean's theorem:

A^2 + B^2 = H^2

Where A and B are the cathetus, and H is the hypotenuse:

Then:

H = √(A^2 + B^2)

H = √(150^2 + 200^2) mi = √(62,500) mi = 250 mi

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6 0
2 years ago
Mrs. Gomes found that 40% of students at her high school take chemistry. She randomly surveys 12 students. What is the probabili
zlopas [31]

Answer:

The correct answer to the following question will be "0.438".

Step-by-step explanation:

Just because Mrs. Gomes finds around 40% of students in herself high school are studying chemistry.  

Although each student becomes independent of one another, we may conclude:  

"x" number of the students taking chemistry seems to be binomial to p = constant probability = 0.40

Given:

Number of surveys

= 12

Exactly 4 students have taken chemistry:

=P(X\leq4)

=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)

=\Sigma_{0}^{4} C_{r}(0.4)^r(0.6)^{12-r}

On substituting the above equation, we get the probability of approximately "0.438".

So that the above would be the appropriate answer.

5 0
2 years ago
Solve for<br> a. 7a-2b = 5a b
Nookie1986 [14]

For this case we have the following expression:

7a-2b = 5ab

From here, we must clear the value of a.

We then have the following steps:

Place the terms that depend on a on the same side of the equation:

7a - 5ab = 2b

Do common factor "a":

a (7 - 5b) = 2b

Clear the value of "a" by dividing the factor within the parenthesis:

a =\frac{2b}{7-5b}

Answer:

The clear expression for "a" is given by:

a =\frac{2b}{7-5b}

8 0
1 year ago
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Answer: y = -9

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