From the following Venn diagram, write the elements of the set A and B and verify n(AUB) + n(A∩B) = n(A) + n(B)

n(A)+n(B)

Step-by-step explanation:

Step-by-step explanation:

From the given Venn diagram ,

A = { a,c,d,f,n }

Number of elements in the set A = 5

=> n(A) = 5

B = { a,b,d,e,g,n }

Number of elements in the set B = 56

=> n(B) = 6

AUB = { a,b,c,d,e,f,g,n}

Number of elements in the set AUB = 8

=> n(AUB) = 8

AUB = { a,d,n}

Number of elements in the set AnB = 3

=> n(AnB) = 3

Now

n(AUB)+n(AnB)

=> 8+3

=> 11

n(AUB)+n(AnB) = 11 (1)

On taking n(A)+n(B)

=> 5+6

=> 11

n(A)+n(B) = 11 (2)

From (1)&(2)

n(AUB) + n(AnB) = n(A)+n(B)

→ Fundamental Theorem on sets

n(AUB) = n(A)+n(B)- n(AnB)

→ The set of all elements in either A or in B or in both A and B is AUB.

→ The set of all common elements in both A and B is AnB.

→ The number of elements in A is denoted by n(A).

Hope this helps.

d≈7.07in

step-by-step explanation:

a.) list the factors of 40.

1/2/4/5/8/10/20/40

b.) list the factors of 40 that represent the number of cans that could be on each shelf.

5

c.) how many different arrangements of cans are possible?

just one because if you put more on one shelf you will go way over 40 and the cabinet will probably break.

d.) what are the possible arrangements of the cans?

put 5 cans on each self to reach 40 cans in total.

e.) which arrangement uses all the shelves?

5 because you can put 5 cans on each shelf, and since there are 8 shelfs you are equally going to get 40 cans.