$A\cup B=A\cup(A-B)\\\\A\ and\ A-B\ are\ disjoint\ that\ means\ there\ is\ no\ common\ element\ in\ these\ two.\\\\n(A\cup B)=n(B\cup(A-B))\\\\n(A\cup B)=n(B)+n(A-B)\ \ \ \ \ \ \ as\ for\ disjoint\ set\ P,Q\ n(P\cup Q)=n(P)+n(Q)\\\\(n(P-Q)=n(P)-n(P\cap Q))\\\\n(AUB)=n(B)+n(A-B)\\\\n(AUB)=n(B)+n(A)-n(A\cap B)\\\\n(AUB)=n(A)+n(B)-n(A\cap B)\\\\Hence\ given\ statement\ is\ true$

4 0

3 years ago

The following stem and leaf plot shows the number of students at the local recreational center each day at 4 PM.

marishachu [46]

Answer:

Step-by-step explanation:

By leaf plot...

4 0

3 years ago

Read 2 more answers

The nth term of a sequence is n^2 -n-2 .find the sum of the first and third terms

Troyanec [42]

Answer:

Step-by-step explanation:

substitute n=1 and n=3 in the given equation

1 st term is (1)^2-1-2 = 1 -1 -2 = -2

3rd term is (3)^2-3-2 = 9-3-2=4

sum of 1st and 3rd term is -2 + 4 = 2

8 0

3 years ago

H(w)=-5x^2+80 to factored form

algol13

Answer:

- 5(x - 4)(x + 4)

Step-by-step explanation:

Given

- 5x² + 80 ← factor out - 5 from each term

= - 5(x² - 16) ← x² - 16 is a difference of squares and factors in general as

a² - b² = (a - b)(a + b), thus

x² - 16 = x² - 4² = (x - 4)(x + 4)

Thus

- 5x² + 80 = - 5(x - 4)(x + 4) ← in factored form

3 0

3 years ago