Let Ebe the set of all even positive integers in the universe Zof integers, and XE : Z R be the characteristic function of E. Then
1. XE(2) =
2. XP(-2) =
3. {z 62: XE() = 1} =

Step-by-step explanation:

Let E be the set of all even positive integers in the universe Z of integers,

i.e

E = {2,4,6,8,10 ....∞}

be the characteristic function of E.

∴

For XE(2)

 since x is an element of E (i.e the set of all even numbers)

For XE(-2)

  since  - 2 is less than 0 , and -2 is not an element of E

For { x ∈ Z: XE(x) = 1}

This can be read as:

x which is and element of Z such that X is also an element of x which is equal to 1.

∴

E = {2,4,6,8,10 ....∞}

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!