If these two figures are similar what is the measure of the missing side length J?

Naddika [18.5K]

All The Subsets

For theset {a,b,c}:

The empty set {} is a subset of {a,b,c}And these are subsets: {a}, {b} and {c}And these are also subsets: {a,b}, {a,c} and {b,c}And {a,b,c} is a subset of {a,b,c}

And when we list all the subsets of S={a,b,c} we get the Power Set of {a,b,c}:

P(S) = { {}, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} }

Think of it as all the different ways we can select the items (the order of the items doesn't matter), including selecting none, or all.

Example: The shop has banana, chocolate and lemon ice cream.

What do you order?

Nothing at all: {}Or maybe just banana: {banana}. Or just {chocolate} or just {lemon}Or two together: {banana,chocolate} or {banana,lemon} or {chocolate,lemon}Or all three! {banana, chocolate,lemon}

Question: if the shop also has strawberry flavor what are your options? Solution later.

How Many Subsets

Easy! If the original set has n members, then the Power Set will have 2n members

Example: in the {a,b,c} example above, there are three members (a,b and c).

So, the Power Set should have 23 = 8, which it does!

Notation

The number of members of a set is often written as |S|, so when S has n members we can write:

|P(S)| = 2n

Example: for the set S={1,2,3,4,5} how many members will the power set have?

Well, S has 5 members, so:

|P(S)| = 2n = 25 = 32

You will see in a minute why the number of members is a power of 2

It's Binary!

And here is the most amazing thing. To create the Power Set, write down the sequence of binary numbers (using n digits), and then let "1" mean "put the matching member into this subset".

So "101" is replaced by 1 a, 0 b and 1 c to get us {a,c}

Like this:

abcSubset0000{ }1001{c}2010{b}3011{b,c}4100{a}5101{a,c}6110{a,b}7111{a,b,c}

Well, they are not in a pretty order, but they are all there.

Another ExampleLet's eat! We have four flavors of ice cream: banana, chocolate, lemon, and strawberry. How many different ways can we have them?Let's use letters for the flavors: {b, c, l, s}. Example selections include:{} (nothing, you are on a diet){b, c, l, s} (every flavor){b, c} (banana and chocolate are good together)etcLet's make the table using "binary": bclsSubset00000{}10001{s}20010{l}30011{l,s}...... etc ..... etc ...121100{b,c}131101{b,c,s}141110{b,c,l}151111{b,c,l,s}

And the result is (more neatly arranged):

P = { {}, {b}, {c}, {l}, {s}, {b,c}, {b,l}, {b,s}, {c,l}, {c,s}, {l,s}, {b,c,l}, {b,c,s},
{b,l,s}, {c,l,s}, {b,c,l,s} }

SymmetryIn the table above, did you notice that the first subset is empty and the last has every member?But did you also notice that the second subset has "s", and the second last subset has everything except "s"? In fact when we mirror that table about the middle we see there is a kind of symmetry.This is because the binary numbers (that we used to help us get all those combinations) have a beautiful and elegant pattern.A Prime Example

The Power Set can be useful in unexpected areas.

I wanted to find all factors (not just the prime factors, but all factors) of a number.

I could test all possible numbers: I could check 2, 3, 4, 5, 6, 7, etc...

That took a long time for large numbers.

But could I try to combine the prime factors?

Let me see, the prime factors of 510 are 2×3×5×17 (using prime factor tool).

So, all the factors of 510 are:

2, 3, 5 and 17,2×3, 2×5 and 2×17 as well, and2×3×5 and 2×3×17 and ..... aha! Just like ice cream I needed a Power Set!

And this is what I got:

2,3,5,17SubsetFactors of 51000000{ }110001{17}1720010{5}530011{5,17}5 × 17 = 8540100{3}350101{3,17}3 × 17 = 51 ... etc ...... etc ...... etc ...151111{2,3,5,17}2 × 3 × 5 × 17 = 510

And the result? The factors of 510 are 1, 2, 3, 5, 6, 10, 15, 17, 30, 34, 51, 85, 102, 170, 255 and 510 (and −1, −2, −3, etc as well). See the All Factors Tool.

Automated

I couldn't resist making Power Sets available to you in an automated way.

So, when you need a power set, try Power Set Maker.

6 0

4 years ago

Read 2 more answers

What is the inverse of y=(x-6)^2 +2 ?

drek231 [11]

Answer:

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

If a club charges an initial fee of $99 to join, and a monthly fee of $45. Write an equation that represents the cost of being a

ANEK [815]

Equation: 99 + 45x, with x as how many ever months.

After five months:
99 + 45(5) = 99 + 225 = $324.

8 0

3 years ago

Multiply the polynomials (5x^2+4x-4)(4x^3-2x+6) Explain your steps!!

Lapatulllka [165]

So distribute using distributive property
a(b+c)=ab+ac so

split it up
(5x^2+4x-4)(4x^3-2x+6)=(5x^2)(4x^3-2x+6)+(4x)(4x^3-2x+6)+(-4)(4x^3-2x+6)=[(5x^2)(4x^3)+(5x^2)(-2x)+(5x^2)(6)]+[(4x)(4x^3)+(4x)(-2x)+(4x)(6)]+[(-4)(4x^3)+(-4)(-2x)+(-4)(6)]=(20x^5)+(-10x^3)+(30x^2)+(16x^4)+(-8x^2)+(24x)+(-16x^3)+(8x)+(-24)
group like terms
[20x^5]+[16x^4]+[-10x^3-16x^3]+[30x^2-8x^2]+[24x+8x]+[-24]=20x^5+16x^4-26x^3+22x^2+32x-24

the asnwer is 20x^5+16x^4-26x^3+22x^2+32x-24

4 0

3 years ago

Given the sequence 1/2 ; 4 ; 1/4 ; 7 ; 1/8 ; 10;.. calculate the sum of 50 terms

miv72 [106K]

Hint :-

Break the given sequence into two parts .
Notice the terms at gap of one term beginning from the first term .They are like $\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8}$ . Next term is obtained by multiplying half to the previous term .
Notice the terms beginning from 2nd term , $4,7,10,13$ . Next term is obtained by adding 3 to the previous term .

Solution :- 

We need to find out the sum of 50 terms of the given sequence . After splitting the given sequence ,

$\implies S_1 = \dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8}$ .

We can see that this is in Geometric Progression where 1/2 is the common ratio . Calculating the sum of 25 terms , we have ,

$\implies S_1 = a\dfrac{1-r^n}{1-r} \\\\\implies S_1 = \dfrac{1}{2}\left[ \dfrac{1-\bigg(\dfrac{1}{2}\bigg)^{25}}{1-\dfrac{1}{2}}\right]$

Notice the term $\dfrac{1}{2^{25}}$ will be too small , so we can neglect it and take its approximation as 0 .

$\implies S_1\approx \cancel{ \dfrac{1}{2} } \left[ \dfrac{1-0}{\cancel{\dfrac{1}{2} }}\right]$

$\\\implies \boxed{ S_1 \approx 1 }$

$\rule{200}2$

Now the second sequence is in Arithmetic Progression , with common difference = 3 .

$\implies S_2=\dfrac{n}{2}[2a + (n-1)d]$

Substitute ,

$\implies S_2=\dfrac{25}{2}[2(4) + (25-1)3] =\boxed{ 908}$

Hence sum = 908 + 1 = 909

7 0

3 years ago