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snow_tiger [21]
4 years ago
14

Prove the set of subsets of a finite set has cardinality 2^x

Mathematics
1 answer:
Nata [24]4 years ago
8 0

Let's take the simple case of a set with four elements: the letters a-d

\{a,b,c,d\}

Two subsets that this - and any other - set contains are the empty set ∅ and the set itself. Now, if we wanted, we could construct the rest of the subsets by picking elements from the original set at random - {a, b, c}, {a, c}, and {c, d} to name a few - but this process is incredibly inefficient, and there's a good chance you'll miss a few subsets this way.

There's a part in that last paragraph that's extremely important: we're <em>picking</em> elements from the original set to put in our subsets, and this selection process boils down to a single yes or no question: <em>do we want to add this element to our subset? </em>This is where that 2 emerges in the original question - we're asking a question with 2 possible outcomes, and we're asking it x times, where x is the number of elements in our set.

For instance, with the set {a, b, c, d}, constructing subsets consists of four questions:

- Should we add a to the subset? Yes/No

- Should we add b? Yes/No

- Should we add c? Yes/No

- Should we add d? Yes/No

The space of possible outcomes, and consequently possible subsets, these questions produce is the same as the space of possible outcomes for 4 yes-or-no questions: 2^4=16

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Answer:

-18=2x-3(-1/3x+3)

-18=2x+x-9

-18=3x-9

-18+9=3x

3x=-9

x=-9

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3 years ago
Factor the polynomial 4x4 – 20x2 – 3x2 + 15 by grouping. What is the resulting expression?
siniylev [52]

Factoring the polynomial 4x^4-20x^2-3x^2 + 15 by grouping means separation of the polynomial in two addends, each of them consists of two addends:

4x^4-20x^2-3x^2 + 15=(4x^4-20x^2)+(-3x^2+15).

In first brackets the common factor is 4x^2, in second brackets the common factor is -3.

Then

(4x^4-20x^2)+(-3x^2+15)=4x^2(x^2-5)-3(x^2-5)=(x^2-5)(4x^2-3).

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7 0
4 years ago
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[90° is given)]

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MrRissso [65]

Answer:

Step-by-step explanation:

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const2013 [10]

Answer:

2^7

Step-by-step explanation:

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