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GalinKa [24]
3 years ago
13

Let A, B, C and D be sets. Prove that A \ B and C \ D are disjoint if and only if A ∩ C ⊆ B ∪ D

Mathematics
1 answer:
ANEK [815]3 years ago
7 0

Step-by-step explanation:

We have to prove both implications of the affirmation.

1) Let's assume that A \ B and C \ D are disjoint, we have to prove that A ∩ C ⊆ B ∪ D.

We'll prove it by reducing to absurd.

Let's suppose that A ∩ C ⊄ B ∪ D. That means that there is an element x that belongs to A ∩ C but not to B ∪ D.

As x belongs to A ∩ C, x ∈ A and x ∈ C.

As x doesn't belong to B ∪ D, x ∉ B and x ∉ D.

With this, we can say that x ∈ A \ B and x ∈ C \ D.

Therefore, x ∈ (A \ B) ∩ (C \ D), absurd!

It's absurd because we were assuming that A \ B and C \ D were disjoint, therefore their intersection must be empty.

The absurd came from assuming that A ∩ C ⊄ B ∪ D.

That proves that A ∩ C ⊆ B ∪ D.

2) Let's assume that A ∩ C ⊆ B ∪ D, we have to prove that A \ B and C \ D are disjoint (i.e.  A \ B ∩ C \ D is empty)

We'll prove it again by reducing to absurd.

Let's suppose that  A \ B ∩ C \ D is not empty. That means there is an element x that belongs to  A \ B ∩ C \ D. Therefore, x ∈ A \ B and x ∈ C \ D.

As x ∈ A \ B, x belongs to A but x doesn't belong to B.  

As x ∈ C \ D, x belongs to C but x doesn't belong to D.

With this, we can say that x ∈ A ∩ C and x ∉ B ∪ D.

So, there is an element that belongs to A ∩ C but not to B∪D, absurd!

It's absurd because we were assuming that A ∩ C ⊆ B ∪ D, therefore every element of A ∩ C must belong to B ∪ D.

The absurd came from assuming that A \ B ∩ C \ D is not empty.

That proves that A \ B ∩ C \ D is empty, i.e. A \ B and C \ D are disjoint.

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Tbh I forgot :)


3 0
3 years ago
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-8+5w^4=28 what is w=
Murljashka [212]

Answer:

w = ±\sqrt[4]{36/5}

Step-by-step explanation:

-8+5w^4=28

You can regroup it:

5w^4 - 8 = 28

Add 8 to both side:

5w^4 = 28 + 8

5w^4 = 36

Divide both side by 5:

5w^4/5 = 36/5

w^4 = 36/5

Take the 4 root of both sides:

w = ±\sqrt[4]{36/5}

Hope this help you :3

3 0
3 years ago
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Consider triangle ABC where AB=X+5, BC=X-2, area is 30cm squared. Find X by solving X^2+3X-70=0, and find the perimeter.
klasskru [66]

Answer:

What the hell AC is not given so you can't find the Perimeter

Step-by-step explanation:

  • x^2+3x-70=0
  • (x-7)(x+10)=0
  • x-7=0

       x+10=0

  • x=7

       x=-10

AB=x+5

  • AB=12

        AB=-5

BC=x-2

  • BC=5

        BC=-12

Area of ABC = 30cm2

<u>Perimeter of ABC = AB+BC+</u><u>AC</u>

<u />

8 0
3 years ago
Solve<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3x%20%2B%204%7D%7B2%7D%20%20%3D%209.5" id="TexFormula1" title=" \frac{3x
ELEN [110]
Answer: 5

Step by step:

(3x+4)/2 = 9.5
3x+4 = 9.5*2
3x+4 = 19
3x = 19 -4
3x = 15
x = 5

7 0
3 years ago
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The endpoints of CD are C(–8, 4) and D(6, –6). What are the coordinates of point P on CD such that P is the length of the line s
GalinKa [24]
If you graph the end points C and D then graph the 4 points at the end it is difficult to tell which points are on CD without a line.  
Using the endpoints find the slope (change in y/ change in x) then substitute a point in to find the intercept.  
Slope = (-6-4)/(6- -8) = -5/7
Intercept equation (-6) = -5/7 (6) + b
b = -1.71428571429
Graphing the line shows only 2 points on the line (–2.75, 0.25) and <span>(0.75, –2.25)
I am confused by the part, "</span><span>P is the length of the line segment from D".  Were you given a length P to help you determine which point.  Using the distance formula to find the length from each point to D doesn't help determine which one is best with the information you have given.  The image shows the distances I calculated and the graphed points. 
I hope this helps!</span>

5 0
3 years ago
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