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aksik [14]
3 years ago
10

Please help i need the answer quick :(

Mathematics
1 answer:
nevsk [136]3 years ago
3 0

Answer:

a

Step-by-step explanation:

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Let N be the universal set and A, B, C, D be its subsets given by A={x:xisaevennaturalnumber},B={x:xNandxisamultipleof3} C={x:x
Svet_ta [14]

Answer:

A^{'} = {1,3,5,7,9,....}

B^{'} = {1,2,4,5,7,8,10,11,13,14......}

C^{'} = {1,2,3,4,5}

D^{'} = {10,11,12,13,14,15,......}

Step-by-step explanation:

Solution:

It is given that, Universal set is the set of all Natural Numbers.

and

Set A, Set B, Set  C, Set D are subsets of Universal set.

A = {x:x is even natural numbers} = {2,4,6,8,10,.....}

B = {x:x ∈ N and x is a multiple of 3} = {3,6,9,12,15,....}

C = {x:x∈ N and x < 5} = {6,7,8,9,10,.....}

D = {x:x ∈ N and x > 10} = {1,2,3,4,5,6,7,8,9}

U = {1,2,3,4,5,6,7,8,9,10,11,........}

Complements:

A^{'} = U-A = {1,2,3,4,5,6,7,8,9,10,11,........} - {2,4,6,8,10,.....}

A^{'} = {1,3,5,7,9,....}

B^{'} = U-B =  {1,2,3,4,5,6,7,8,9,10,11,........} - {3,6,9,12,15,....}

B^{'} = {1,2,4,5,7,8,10,11,13,14......}

C^{'} = U-C = {1,2,3,4,5,6,7,8,9,10,11,........} - {6,7,8,9,10,.....}

C^{'} = {1,2,3,4,5}

D^{'} = {1,2,3,4,5,6,7,8,9,10,11,........} - {1,2,3,4,5,6,7,8,9}

D^{'} = {10,11,12,13,14,15,......}

8 0
3 years ago
The two perpendicular lines of a coordinate plane that intersect at the origin (singular: axis)
Svetach [21]
3. Axes
That’s what it should be
3 0
2 years ago
Read 2 more answers
Beatrice withdrew 28 dollars from her checking account. What is the integer?
Bingel [31]
An integer is a whole number so 28 is the answer
8 0
3 years ago
Read 2 more answers
11to the power of 3 multiply by 21 to the power of three
julia-pushkina [17]

Answer:

12326391

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A^2(64a^2-3)/3=27-4a^2/4
frutty [35]
The answer is a = 3/4 = 0.75

First get rid of the paranthesis,
{a}^{2} (64 {a}^{2}  - 3) = 64 {a}^{4} -  3 {a}^{2}
Then set the denominators equal:
\frac{ 256 {a}^{4}  - 12 {a}^{2}}{12}  =  \frac{81 - 12 {a}^{2} }{12}
Then remove the denominators and solve:
256 {a}^{4}  - 12 {a}^{2}  = 81 - 12 {a}^{2}
Eliminate -12a^2 by adding 12a^2 to both sides:
256 {a}^{4}  = 81
Take the fourth root of them or take the square root twice:
\sqrt[4]{256 {a}^{4} }  =  \sqrt[4]{81}   \\  4a = 3
Divide both sides by 4:
a =  \frac{3}{4}  = 0.75
7 0
3 years ago
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