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suter [353]
3 years ago
10

(1) [6pts] Let R be the relation {(0, 1), (1, 1), (1, 2), (2, 0), (2, 2), (3, 0)} defined on the set {0, 1, 2, 3}. Find the foll

owing: 1. [3pts] Reflexive closure of R 2. [3pts] Symmetric closure of R
Mathematics
1 answer:
goldenfox [79]3 years ago
8 0

Answer:

Following are the solution to the given points:

Step-by-step explanation:

In point 1:

The Reflexive closure:  

Relationship R reflexive closure becomes achieved with both the addition(a,a) to R Therefore, (a,a) is  (0,0),(1,1),(2,2) \ and \ (3,3)

Thus, the reflexive closure: R={(0,0),(0,1),(1,1),(1,2),(2,0),(2,2),(3,0), (3,3)}

In point 2:

The Symmetric closure:

R relation symmetrically closes by adding(b,a) to R for each (a,b) of R  Therefore, here (b,a) is:   (0,1),(0,2)\ and \ (0,3)

Thus, the Symmetrical closure:

R={(0,1),(0,2),(0,3)(1,0),(1,1)(1,2),(2,0),(2,2),(3,0), (3,3)}

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Answer:

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Step-by-step explanation:

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By applying Pythagoras theorem in ΔDBE,

DE² = DB² + BE²

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By applying Pythagoras theorem in ΔABC,

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AC = 8.49 units

Perimeter of ADEC = 3 + 4.24 + 3 + 8.49

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Area of ADEC = Area of ΔABC - Area of ΔBDE

Area of ΔABC = \frac{1}{2}(AB)(BC)

                       = \frac{1}{2}(6)(6)

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Area of ΔBDE = \frac{1}{2}(BD)(BE)

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