If ξ={1 , 2 , 3 , . . . , 8 , 9 , 10} A={x : x is multiples of 3},
B={x : x is a prime numbers} and
C={x : x<6},
then what is (A⋃B)ʹ⋂C ?

{1,4}

Step-by-step explanation:

A={3,6,9}

B={2,3,5,7}

C={1,2,3,4,5}

AUB={2,3,5,6,7,9}

(AUB)'={1,4,8,10}

(AUB)'⋂C={1,4}

Step-by-step explanation:

A = {3 , 6 , 9}

B = {2,3,5,7}

A ∪ B = { 2,3,5,6,7,9}

(A∪B)' - is set of all elements in universal set  that are not in (A∪B)

(A∪B)'  = {1,4,8,10}

C = {1,2,3,4,5}

(A∪B)' ∩C = {1 , 4}

y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice