Answer:
23
Step-by-step explanation:
Brackets go first, 5 (y) - 4 (x) = 1. Now, 3 x 1 = 3, you do this because of the 3 next to the brackets. Now add them together.
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:
3t+2
Step-by-step explanation:
First, we focus on the part with the "t" variable. Then we subtract the values with variables which gets us 3t. Finally, we can't do anything with the +2, so our solution is 3t+2.
Answer:
even I think
Step-by-step explanation:
