Consider sets r, s, and t, defined as follows: r = { x ϵ z | x is divisible by 2 }s = { y ϵ z | y is divisible by 3 }t = { z ϵ z | z is divisible by 6 }a) is r ⊆ t? explain. b) is t ⊆ r? explain. c) is t ⊆ s? explain.

a) No b)Yes c) No

Step-by-step explanation:

a) No, R is not a subset of T that is NOT ALL the elements of R can be found in T. For R ⊆ T, it means that ALL the element of R can be found in T which is false in this case.

b) Yes, T is a subset of R that is ALL the element of T can be found in T since all the elements in both sets are all even. For T ⊆ R, it means that ALL the element of T can be found in R

c) No, T is not a subset of S that is NOT ALL the elements of T can be found in S. For T ⊆ S, it means that ALL the element of T can be found in S which is false in this case.

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

it would open into a square depending on the type of envelope i’m not sure about the second part