Given that the universal set. U = {X : 30 ≤ X ≤ 40. X is an even number}.
P = {X : X is a multiple of 4}
and Q = {X : X is a number such that the sum of its digits is an odd number}.
find n(P∩Q')

A. 1
B. 2
C. 3
D. None of the above​

Given that

U = {X : 30 ≤ X ≤ 40. X is an even number}.

so Mathematically

U = {30,32,34,36,38 40}.

P = {X : X is a multiple of 4}

P = { 4,8,12,16,20,24,28,32,36,40,44,}

and Q = {X : X is a number such that the sum of its digits is an odd number}.

Q = {12,14,16,18,21,23,25,27,29,30,32,34,36,38,41,43,45,}.

(Q')= U-Q  ={40}

(P∩Q')= {40}

n(P∩Q')= 1

(Q') is Q complement which means a set containing those elements of the universal set which are not element of the set Q.

(P∩Q') is a set containing those elements of set P and Q complement which are common to both sets.

n(P∩Q') denotes the number of elements in a set containing those elements of set P and Q complement which are common to both sets which  has only 1 element which is 40.

So the best answer is part A = 1.

a) 0

step-by-step explanation:

i^413 x i^0 = 0 as any number multiplied by a zero is

happy to

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hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb