R = { (x,y): 3x-y=0 }
The condition is 3x=y so that's not going to be any of these things.
R is reflexive if (x,x)∈R for all x. Let's check.
3x - y = 3x - x = 2x ≠ 0 necessarily. NOT REFLEXIVE
R is symmetric if (x,y)∈R → (y,x)∈R. Let's check.
(x,y)∈R so
3x-y = 0
y = 3x
Is (y,x)∈R. That would be true if 3y-x=0
3y - x = 3(3x) - x = 8x ≠ 0 necessarily NOT SYMMETRIC
R is transitive if (x,y)∈R and (y,z)∈R → (x,z)∈R. Let's check.
3x-y = 0 so y=3x
3y-z = 0 so z=3y = 9x
3x - z = 3x - 9x = -6x ≠ 0 necessarily NOT TRANSITIVE
Answer:
by the way what grade are you in? If you are more than 7th grade can you help me? brainly.com/question/16233847
35 mph
Step-by-step explanation:
7 times 5
Answer:
No
Step-by-step explanation:
Multiplying the 2 and the 5 by 3 will get you 6 and 14. 14 is not a factor of 5 making it incorrect.
Answer:
70.2
Step-by-step explanation:
Answer:
Description
Step-by-step explanation:
a. 2 solutions
b. 2 imaginary solutions
c. If the discriminant is positive, then it will have 2 real solutions as the square root of a positive number always equals a positive number. If the discriminant is negative, the quadratic equation will have 2 imaginary solutions, as the square root of a negative number is always imaginary. If the discriminant equals 0, it will have only 1 real solution.
