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Alex787 [66]
3 years ago
7

Determine whether each of these sets is the power set of a set, where a and b are distinct elements.

Mathematics
1 answer:
Schach [20]3 years ago
4 0
A. \varnothing cannot be the power set of any set. Consult Cantor's theorem, which says that the cardinality of the power set of any set (even the empty set) is strictly greater than the cardinality of the set.

(No part b?)

c. Also not the power set of any set, because any power set must have 2^n elements, where n is the cardinality of the original set. The cardinality of this set is 3, but there is no integer n such that 2^n=3. This set would be a power set if \{\varnothing\} (that is, the set containing the empty set) were a member of it.
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m_a_m_a [10]

Answer:

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Step-by-step explanation:

okay well 0.05 is 5 hundredths

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but simplified

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6 0
3 years ago
If you multiplied 637 by 912, what would be the units digit of the answer
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3 0
3 years ago
Help me answer these questions
Reil [10]

Answer:

See below.

Step-by-step explanation:

1.  x = e^(x/y)   Taking logs:

log x =  x/y

x = y log x

Differentiating:

1 = dy/dx log x +  y * 1/x

dy/dx log x = (1 - y/x)

dy/dx =  (1 - y/x) / log x

dy/dx =  ((x - y)/ x) / log x

dy/dx = (x - y) / x log x)

2.  y^x = e ^(y - x)

Taking logs:

x log y = y - x   --------------(A)

y = x + x logy    --------------(B)

dy/dx = 1 + x * 1/y * dy/dx + 1*logy

dy/dx - dy/dx * x/y = 1 + log y

dy/dx ( (x - y) y))  = 1 + log y

dy/dx = y(1 + log y) / (y - x)

Using (A) and (B) :-

dy/dx =  x(1 + logy)(1 + logy) / x logy          The x's cancel, so:

dy/dx = (1 + logy)^2 / log y

Sorry I have to go right now..

The rest of the answers are  on the re-post of these questions

8 0
3 years ago
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