54 equals 9 more than a number t .

Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is $\{1,2,3,4\}$ .

a) We need to find a relation R reflexive and transitive that contain the relation $R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}$

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

1R4 and 4R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
4R1 and 1R2, then 4 must be related with 2.

Therefore $\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\}$ is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

since 1R2, then 2 must be related with 1.
since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

1R2 and 2R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
2R1 and 1R4, then 2 must be related with 4.
4R1 and 1R2, then 4 must be related with 2.
2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

1 must be related with 1,

2 must be related with 2,

3 must be related with 3,

4 must be related with 4

For be symmetric

since 1R2, 2 must be related with 1,
since 1R4, 4 must be related with 1.

For be transitive

Since 4R1 and 1R2, 4 must be related with 2,
since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

5 0

2 years ago

In the trapezoid ABCD ( AB ∥ CD ) point M∈ AD , so that AM:MD=3:5. Line l ∥ AB and going trough point M intersects diagonal AC a

geniusboy [140]

Answer:

BC:BN=8:3

Step-by-step explanation:

ABCD is a trapezoid and there is a point m which belongs to AD such that AM:MD=3:5.Line "l" parallel to AB intersects the diagonal AC at p and BD at N.

Now, we know that the parallel lines divide the transversal into the segments with equal ratio, therefore, BN:NC=AM:MD

But, BC= BN+NC

Therefore, BC:BN=(BN+NC):BN

⇒BC:BN=(3+5):3

⇒BC:BN=8:3

4 0

2 years ago

A manager associated each employee's name with a number on one ball in a container, then drew balls without looking to select a

RUDIKE [14]

Answer: This type of sampling is Simple Random Sampling

7 0

2 years ago

Read 2 more answers

Minato drove 390 miles. Part of the drive was along local roads, where his average speed was 20 mph, and the rest was along a hi

pashok25 [27]

Answer:

45 miles.

Step-by-step explanation:

Given that the:

Total distance covered = 390 miles

Total time = 8 hours

Let the distance covered along the local way = L

And the distance covered along the highway = H

Along with local way,

Speed = distance/ time

20 = L / T

T = L /20 .... (1)

Along the highway,

Distance covered H = 390 - L

Let the time = t

Speed = distance/time

60 = (390 - L)/t

t = ( 390 - L)/60

But total time = T + t

That is

8 = L/20 + (390 - L)/60

The LCM at right hand side will be 60

8 = ( 3L + 390 - L )/60

Cross multiply

480 = 2L + 390

Collect the like terms

2L = 480 - 390

2L = 90

L = 90/2

L = 45 miles.

Therefore, the distance Minato drive along local roads is 45 miles

4 0

2 years ago

Quincy feeds his dog 2.75 cups food each day each cup of dog food costs $1.25. He multiplies the numbers together to determine h

Flura [38]

2times1 because 2.75 take the 75 off and 1.25 take the 25 off so you multiply 2 and 1 so 2times1 equals 2

6 0

2 years ago