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Gekata [30.6K]
3 years ago
5

Can anyone walk me step by step through this?

Mathematics
1 answer:
Jlenok [28]3 years ago
5 0

I assume the set A referred to among all the answer choices is actually supposed to be B, or the other way around so that

A = {1, 2, 3, {4, 5, 6, {7}}, 8}

A has <u>5</u> elements:

• 1

• 2

• 3

• {4, 5, 6, {7}}

• 8

Any non-empty subset of A is a set containing at least one of these elements.

The fourth element listed above, {4, 5, 6, {7}}, is itself a set with 4 elements (4, 5, 6, and yet another set {7} that contains just the 1 element 7).

Now,

• { } ⊆ A is true - the empty set is subset of every set

• {2, 3} ⊆ A is true - both 2 and 3 are elements of A, so {2, 3} is a subset of A

• {4, 5, 6, {7}} ⊆ A is false - the set {4, 5, 6, {7}} is an element of A, but not a subset

• {{2, 3}} ⊆ A is false - 2 and 3 are elements of A, but the set {2, 3} is not an element of A, so {{2, 3}} is not a subset

• {{4, 5, 6}, 8} ⊆ A is false - the set {4, 5, 6} is not an element of A

• {{4, 5, 6, {7}}} ⊆ A is true - the set {4, 5, 6, {7}} is an element of A

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y=n^2+3n+4

Step-by-step explanation:

Sequence,S: 8,14,22,32 ,44, 58, 74...

First differences of S: 6,8,10,12,14,16,...

Second differences of S: 2,2,2,2,2,2,....

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So the 1st term is 8. If the 0th term was listes it would be 4 since 8-4=4 and 6-4=2.

This means our quadratic it is in the form

y=an^2+bn+c where c=4.

So we have y=an^2+bn+4.

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(1,8) gives us 8=a(1)^2+b(1)+4

Simplifying gives 8=a+b+4

Subtracting 4 on both sides gives: 4=a+b

(2,14) gives us 14=a(2)^2+b(2)+4

Simplifying gives 14=4a+2b+4

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So we have the following system to solve:

4=a+b

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I'm going to solve using elimination/linear combination style.

Dividing second equation by 2 gives the system as:

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Subtract the 2nd equation from the first gives:

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This implies a=1 since -1=-1(1) is true.

If a+b=4 and a=1, then b=3. Because 3+1=4.

So the equation is y=1n^2+3n+4 or y=n^2+3n+4.

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List the range pls help
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Answer:

D(1, -2)

E(3, -4)

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