Let A = {\bullet ,\square, \bigotimes} and B = {\square,\ominus, \bullet}. (a) List the elements of A X B and B X A. The parentheses and comma in an ordered pair are not necessary in cases such as this where the elements of each set are individual symbol.
(b) Identifiy the intersection of A X B and B X A, for the case above, and then guess at a general rule for the intersection of A µ B and B µ A. where A and B are any two sets.

(a) With A = {•, □, ⊗} and B = {□, ⊖, •}, we have

A × B = {(•, □), (•, ⊖), (•, •), (□, □), (□, ⊖), (□, •), (⊗, □), (⊗, ⊖), (⊗, •)}

and

B × A = {(□, •), (□, □), (□, ⊗), (⊖, •), (⊖, □), (⊖, ⊗), (•, •), (•, □), (•, ⊗)}

(b) The intersection of the two sets above is

(A × B) ∩ (B × A) = {(•, •), (•, □), (□, •), (□, □)}

Not sure what µ is supposed to represent, but I suppose you meant to again write × as in the Cartesian product. By definition, for any two sets A and B, we have

A × B = {(a, b) | a ∈ A and b ∈ B}

Then

(A × B) ∩ (B × A) = {(a, b) | a ∈ A ∩ B and b ∈ A ∩ B}

In the product found above, notice that • and □ are both elements of A and B, while ⊗ and ⊖ are exclusive to either set.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

the 4th choice is the correct one

step-by-step explanation: