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 relations from {1, 2, 3} to {1, 2, 3, 4}. Find a. R1 ∪ R2 b. R1 ∩ R2 c. R1 – R2 d. R2 – R1

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)}

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation:

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)