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
πAnswer:
Therefore, the area of a sector of a circle would be 13.5π square units.
Step-by-step explanation:
Given
mi
The area of a sector of a circle is:
A = π r² Ф/360
A = π (9)² 60/360
A = π 81 * 1/6
A = 13.5π square units
Therefore, the area of a sector of a circle would be 13.5π square units.
Answer:
The answer is the forth one , Y-8+2⩽-2+2
Answer:whatever the first number was times 2
Step-by-step explanation: