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Rus_ich [418]
3 years ago
11

A Venn diagram is shown below: a Venn diagram showing two categories, A and B. In the A only circle is 1 and 2, in the B circle

is 5 and 6, in the intersection is 3 and 4, and in outside the circles is U What are the elements of (A ∩ B) ' ? {3, 4} {1, 2, 3, 4} {1, 2, 5, 6} {3, 4, 5, 6}

Mathematics
2 answers:
erica [24]3 years ago
7 0
The correct answer is {1, 2, 5, 6}.

A intersect B would be {3, 4}; the complement of this is everything else.
wel3 years ago
6 0
Let
<span>ξ -----------> is a universal set.
A----------> is a  subsets of ξ
</span>B----------> is a  subsets of ξ

we know that
<span>ξ = {1, 2, 3, 4, 5, 6}
</span><span>A = {1, 2, 3, 4} 
B = {3, 4, 5, 6} 

</span>(A ∩ B)'<span> = {elements of ξ which are not in A ∩ B}

</span>and

A ∩ B<span> = {elements which are common to both A and B}</span>
A ∩ B<span>= {3,4} 
</span>
therefore
the elements of (A ∩ B) ' ={1,2.5,6}

the answer is the option
<span>{1, 2, 5, 6}</span>
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Match each equation with its solution set.
taurus [48]

Answer:

1- The solution of I2x + 5I = 9 is {-7 , 2}

2- The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- The solution of I5 - xI = 6 is {-1 , 11}

4- The solution of I6x - 8I + 7 = 5 is ∅

5- The solution of Ix + 3I = 12 is {-15 , 9}

6- The solution of Ix - 3I = -12 is ∅

Step-by-step explanation:

* At first lets explain the meaning of IxI = a

- If IxI = a ⇒ then x = a or x = -a

- IxI never give a negative answer, because IxI means the

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Ex: I-2I is 2

* Now lets find the solution of each equation

1- ∵ I2x + 5I = 9

∴ 2x + 5 = 9 ⇒ subtract 5 from both sides

∴ 2x = 4 ⇒ divide both sides by 2

∴ x = 2

OR

∴ 2x + 5 = -9 ⇒ subtract 5 from both sides

∴ 2x = -14 ⇒ divide both sides by 2

∴ x = -7

* The solution of I2x + 5I = 9 is {-7 , 2}

2- ∵ I2x + 7I + 2 = 11 ⇒ Subtract 2 from both sides

∴ I2x + 7I = 9

∴ 2x + 7 = 9 ⇒ subtract 7 from both sides

∴ 2x = 2 ⇒ divide both sides by 2

∴ x = 1

OR

∴ 2x + 7 = -9 ⇒ subtract 7 from both sides

∴ 2x = -16 ⇒ divide both sides by 2

∴ x = -8

* The solution of I2x + 7I + 2 = 11 is {-8 , 1}

3- ∵ I5 - xI = 6

∴ 5 -x = 6 ⇒ subtract 5 from both sides

∴ -x = 1 ⇒ divide both sides by -1

∴ x = -1

OR

∴ 5 -x = -6 ⇒ subtract 5 from both sides

∴ -x = -11 ⇒ divide both sides by -1

∴ x = 11

* The solution of I5 - xI = 6 is {-1 , 11}

4- ∵ I6x - 8I + 7 = 5 ⇒ Subtract 7 from both sides

∴ I6x - 8I = -2

- I  I never give negative answer

* The solution of I6x - 8I + 7 = 5 is ∅

5- ∵ Ix + 3I = 12

∴ x + 3 = 12 ⇒ subtract 3 from both sides

∴ x = 9

OR

∴ x + 3 = -12 ⇒ subtract 3 from both sides

∴ x = -15

* The solution of Ix + 3I = 12 is {-15 , 9}

6- ∵ Ix - 3I = -12

- I  I never give negative answer

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