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If P and Q are two sets then n( P-Q) means ?

Lapatulllka [165]

$\huge{\mathbb{ \tt{ANSWER}}}$

I shall use the symbol U for union

^ for intersection

~ for complement

The property of set subtraction states that for any sets X and Y, X-Y = X ^ Y~

That is in words, X throw away Y is the same as the intersection of X and the complement of Y

or X and anything NOT in Y

At the SET level, the proof looks like this:

(P-Q) U (Q-P) = « given

(P^Q~) U (Q ^ P~) = « property of set subtraction

(P U Q) ^ (P U P~) ^ (Q~ U Q) ^ (Q~ U P~) = « distributive property

(P u Q) ^ (Q~ U P~) = « Union of any set with it's complement is the universal

set which gets dropped by any intersection of a non-universal set

(P u Q ) ^ (Q ^ P)~ = « De Morgan's Law

(P u Q) - (Q ^ P) = « property of set subtraction

(P u Q) - (P ^ Q) « commutative property

At the ELEMENT level, the proof looks like this:

» Proves left side is subset of right side

If x is in (P-Q) U (Q-P) then x is in (P-Q) OR x is in (Q-P)

If x is in P-Q then x is in P but not in Q.

Then X is in P U Q.

Since x is not in Q, x is NOT in P ^ Q

Hence x is in P U Q but NOT in P^Q, which

means it is an element if (P U Q) - (P ^ Q)

Therefore x is an element of the set defined by the right side, which means (P-Q) U (Q-P) is a subset of (P U Q) - (P ^ Q)

« proves right side is subset of the left side

If x is an element of (P U Q) - (P ^ Q) then

x is in (P U Q) but NOT in (P^Q)

Since x is in P U Q, x is in P OR x is in Q.

Since x is NOT in (P^Q) then x is NOT in BOTH P and Q.

Hence x is in either P or Q but not both.

If x is in P, then x is not in Q, so x is in P-Q.

Therefore x is in (P-Q) U (Q-P).if x is in Q, then x is not in P, so x is in Q-P.Therefore x is in (Q-P) U (P-Q).There x is an element of set defined by the left side, which menas (P u Q)-(P ^Q) is a subset of (P-Q) U (Q-P).

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8 0

3 years ago

In a basketball tournament, team A scored 6 more points than 3 times as many points as team B scored. Team C scored 45 more poin

Sophie [7]

A + B + C = 476
A = 3B + 6
C = B + 45

now its just a matter of subbing..
A + B + C = 476
(3B + 6) + B + (B + 45) = 476...combine like terms
5B + 51 = 476
5B = 476 - 51
5B = 425
B = 425/5
B = 85 <== team B scored 85

A = 3B + 6
A = 3(85) + 6
A = 255 + 6
A = 261 <=== team A scored 261

C = B + 45
C = 85 + 45
C = 130 <=== team C scored 130

6 0

3 years ago

Read 3 more answers

Helppppppppppppppppp

Softa [21]

Answer:

CA = 32

Step-by-step explanation:

By intersecting chords theorem:

$12(4x + 1) = 14(3 + 3x) \\ \\ 48x + 12 = 42 + 42x \\ \\ 48x - 42x = 42 - 12 \\ \\ 6x = 30 \\ \\ x = \frac{30}{6} \\ \\ x = 5$