Answer:
Step-by-step explanation: A or the top answer choice. The highes exponent needs to be 2.
1) 2
2) 3
3) 4
4) 7
Hope this helps ^-^
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:
have you found salena yet?
Step-by-step explanation:
try looking behind your nearest Mcdonalds
Answer:
1. 30j+12+6j
2. 36j+12
3. 6(6j+2)
Step-by-step explanation:
1. You just multiply so... 6x5j ... 6x2 ... 6xj
2. you multiply and then simplify so first you have 30j+12+6j, so you add like terms so... 30j+6j= 36j and then +12
3. you do what's in the parentheses first so 5j+j=6j so .... 6(6j+2)
