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vlabodo [156]
4 years ago
14

1. Let R1 = {(1, 2), (2, 3), (3, 4)} and R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)} be relati

ons from {1, 2, 3} to {1, 2, 3, 4}. Find a. R1 ∪ R2 b. R1 ∩ R2 c. R1 – R2 d. R2 – R1
Mathematics
1 answer:
PilotLPTM [1.2K]4 years ago
7 0

Answer:

a) R1 ∪ R2={(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)}

b)  R1 ∩ R2= {(1, 2), (2, 3), (3, 4)}

c) R1 – R2= ∅

d) R2 – R1= {(1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3)}

Step-by-step explanation:

We know that:

R1 = {(1, 2), (2, 3), (3, 4)}

R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)}

Therefore, we get  

a) R1 ∪ R2={(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4)}

b)  R1 ∩ R2= {(1, 2), (2, 3), (3, 4)}

c) R1 – R2= ∅

d) R2 – R1= {(1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3)}

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2=9k . <em>divi</em><em>ding</em><em> </em><em>throu</em><em>gh</em><em> </em><em>by</em><em> </em><em>9</em><em> </em><em>to</em><em> </em><em>get</em><em> </em><em>the</em><em> </em><em>valu</em><em>e</em><em> </em><em>of</em><em> </em><em>k</em>

<em>\frac{2}{9}  =  \frac{9k}{9}</em>

<em>k =  \frac{2}{9}</em>

<em>pu</em><em>tting</em><em> </em><em>it</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>ge</em><em>neral</em><em> </em><em>expres</em><em>sion</em>

<em>y =  \frac{2}{9} x</em>

<em>the</em><em> </em><em>valu</em><em>e</em><em> </em><em>of</em><em> </em><em>y</em><em> </em><em>whe</em><em>n</em><em> </em><em>x</em><em>=</em><em>6</em>

<em>y =  \frac{2}{9}  \times 6</em>

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<em>4</em><em>/</em><em>3</em>

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