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faltersainse [42]
2 years ago
9

Here are the first five terms of a number sequence .

Mathematics
1 answer:
dybincka [34]2 years ago
4 0

Answer:

t=4+6n

Step-by-step explanation:

a = 10

d=6

in general

tn = a + (n-1)*d

tn = 10 + (n-1)*6

tn = 10 + 6n - 6

tn = 4 + 6n

So if n = 5 then

t5= 4+6*5

t5= 4 + 30

t5 = 34 which agrees with your givens.

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3 0
3 years ago
Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9}={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} A={1, 3, 4, 5, 7,
blagie [28]

Answer:

a) A' = [0,2,6,8]

b) (AUB)' = [0,2,6]

c) (AUB')' = [9]

d) A∩B′= [3,5,9]

Step-by-step explanation:

Assuming this problem: "Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} , and B={1, 4, 7, 8} . List the elemetns of the following sets in the increasing order: a) A′=  b) (A∪B)′={ , , }} c) (A∪B′)′={ }} d) A∩B′={ , , }}"

Part a

For this case we just need to find the elements in the universal set that are not in A. And we see that:

A' = [0,2,6,8]

And that represent the complement for A

Part b

For this case we need to find first the Union AUB who are the elements on A or B without repetition and we got:

AUB = [1,3,4,5,7,8,9]

And now the complement for (AUB)' are the elements that are not in AUB but are on the universal set and we got:

(AUB)' = [0,2,6]

Part c

For this case we need to find B' who are the elements on the universal set that are not in B

B' = [0,2,3,5,6,9]

Then we can find the union between AUB' and we got:

AUB' = [0,1,2,3,4,5,6,7,9]

And then the complment is just:

(AUB')' = [9]

Part d

For this case we just need to see the elements in common between A and B' and we got:

A∩B′= [3,5,9]

6 0
4 years ago
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