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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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here are several types of objects. For each type of object, estimate how many there are in a stack that is 5 feet high. Be prepa
lorasvet [3.4K]

We need to estimate how tall is, on average, one of these object, and then count how many would be in a 5 feet high stack.

For example, on Google you may see that "1.5 cubic foot boxes are the standard box, manufactured by most companies". So, we assume that a standard cardboard box is 1.5 feet tall.

So, if we set the equation

1.5k = 5 \iff k=\dfrac{5}{1.5}=3.\bar{3}

So, there would be between 3 and 4 cardboard boxes in a 5 feet tall stack.

Similarly, we can see that the average book is 9 inches tall. 9 inches are 0.75

feet, so we have

0.75k=5 \iff k=6.\bar{6}

So, there would be between 6 and 7 books in a 5 feet tall stack.

The average brick is 75 millimeters tall, which means 0.25 feet tall. Again, we have

0.25k=5 \iff k=20

So, there would be 20 bricks in a 5 feet tall stack.

Finally, a coin is about 0.006 feet, which leads to

0.006k=5 \iff k=833.\bar{3}

So, there would be between 833 and 834 coins in a 5 feet tall stack.

5 0
3 years ago
Which statements are true about the area of the
cupoosta [38]

Answer:

A=bh..I don't get for those numbers

7 0
3 years ago
10 POINTS!! Made myself broke with my last two questions!
Svet_ta [14]

Answer:

D

Step-by-step explanation:

The radius r is the distance from the centre to a point on the circle.

Calculate r using the distance formula

r = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (0, 0) and (x₂, y₂ ) = (- 5, - 12)

r = \sqrt{(-5-0)^2+(-12-0)^2}

  = \sqrt{(-5)^2+(-12)^2}

  = \sqrt{25+144}

  = \sqrt{169}

  = 13 → D

4 0
3 years ago
Jessica rides the bus 8% miles each day.how many miles does she ride in 10 days?
liq [111]

Answer:

80 miles

Step-by-step explanation:

3 0
3 years ago
Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?
zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
<span><span>→n</span>=<1,2−2></span>

The equation of the plane parallel to the original one passing through <span>P<span>(<span>x0</span>,<span>y0</span>,<span>z0</span>)</span></span>is:

<span><span>→n</span>⋅< x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span><1,2,−2>⋅<x−<span>x0</span>,y−<span>y0</span>,z−<span>z0</span>>=0</span>
<span>x−<span>x0</span>+2y−2<span>y0</span>−2z+2<span>z0</span>=0</span>
<span>x+2y−2z−<span>x0</span>−2<span>y0</span>+2<span>z0</span>=0</span>

Or

<span>x+2y−2z+d=0</span> [1]
where <span>a=1</span>, <span>b=2</span>, <span>c=−2</span> and <span>d=−<span>x0</span>−2<span>y0</span>+2<span>z0</span></span>

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when <span>x=0</span> and <span>y=0</span>:
<span>x+2y−2z=1</span> => <span>0+2⋅0−2z=1</span> => <span>z=−<span>12</span></span>
<span>→<span>P1</span><span>(0,0,−<span>12</span>)</span></span>

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping <span>D=2</span>, and d as d itself, we get:

<span><span>D=<span><span>|a<span>x1</span>+b<span>y1</span>+c<span>z1</span>+d|</span><span>√<span><span>a2</span>+<span>b2</span>+<span>c2</span></span></span></span></span>
<span>2=<span><span><span>∣∣</span>1⋅0+2⋅0+<span>(−2)</span>⋅<span>(−<span>12</span>)</span>+d<span>∣∣</span></span><span>√<span>1+4+4</span></span></span></span>
<span><span>|d+1|</span>=2⋅3</span> => <span><span>|d+1|</span>=6</span>First solution:
<span>d+1=6</span> => <span>d=5</span>
<span>→x+2y−2z+5=0</span>Second solution:
<span>d+1=−6</span> => <span>d=−7</span>
<span>→x+2y−2z−7=<span>0</span></span></span>
8 0
3 years ago
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