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Mama L [17]
3 years ago
8

Suppose that X is a subset of Y. Let p be the proposition ‘x is an element ofX’ and let q be the proposition ‘x is an element of

Y’. Write down apropositional term which represents the statement ‘x is an element ofY \X ’. Hence or otherwise, establish the following identities for sets:(i) A n B = A\Bc;(ii) A U (B\A) = A U B;(iii) A\(B U C) = (A\B) n (A\C);(iv) A\(B n C) = (A\B) U (A\C).
Mathematics
1 answer:
Genrish500 [490]3 years ago
8 0

Answer:

See answer below

Step-by-step explanation:

The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.

i) x∈AnB  if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB  then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.

ii) (I will abbreviate "if and only if" as "iff")

x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.

iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).

iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).

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7 0
3 years ago
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The school production of Our Town' was a big success. For opening night. 434 tickets were sold. Students paid $4.00 each, while
Alex73 [517]

Answer:

286 students attended

148 non students attended

Step-by-step explanation:

Given

Tickets = 434

Students = \$4.00

Non-Students = \$6.00

Total = \$2032.00

Solving (a): Number of students

Represent students with S and non students with N

So:

S + N = 434 --- (1)

4S + 6N = 2032 --- (2)

Make N the subject of formula in (1)

N = 434 - S

Substitute 434 - S for N in (2)

4S +6(434 - S) = 2032

Open Bracket

4S + 2604 - 6S = 2032

Collect Like Terms

4S - 6S = 2032 - 2604

-2S = -572

Divide through by -2

S = 286

<em>Hence; 286 students attended</em>

Solving (b):

Recall that

N = 434 - S

N = 434 - 286

N = 148

<em>148 non students attended</em>

4 0
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the second and last sentences.

Step-by-step explanation:

Its the most reasonable!

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Answer:

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SOLVE THIS EXTREMELY IMPORTANT EQUATION
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Answer:

<em>Option B; EF = ( About ) 4.47 units, Perimeter of Δ EFG = ( About ) 12.94 units</em>

Step-by-step explanation:

If we were to consider the height of this triangle EFG, it would be 4 units of length, supposedly splitting base GF into two ≅ parts, each 2 units of length. First let us name the point drawn to base GF ⇒ point H, so that the height of Δ EFG ⇒ EH. Now as EH splits GF into two ≅ parts, by Converse to Coincidence Theorem, Δ EFG ⇒ Isosceles Δ;

EH and FH are legs of a right triangle EFH, so that Pythagorean Theorem can be applied to solve for the length of EF and EG, knowing that as Δ EFG ⇒ Isosceles Δ, EF ≅ EG;

( EH )^2 + ( FH )^2 = ( EF )^2,

( 4 )^2 + ( 2 )^2 = ( EF )^2 ⇒ take square of 4 & 2,

16 + 4 = ( EF )^2 ⇒ combine like terms,

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<em>Solution; Option B</em>

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