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sergejj [24]
2 years ago
12

Determine the number of subsets and list down all of them.

Mathematics
2 answers:
Setler [38]2 years ago
7 0

We have to use Powerset .All of the elements are subsets of the set.

  • If n(A)=n

\boxed{\sf P(A)=2^n}

#1

  • {a,b,c}
  • It has 2^3=8subsets

Now

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

#2

  • {2,4,6,8}
  • It has 2^4=16 subsets

Now

P(A)={{2},{4},{6},{8},{2,4},{2,8},{2,8},{4,8},{4,6},{6,8},{2,4,6},{2,4,8},{2,6,8},{4,6,8},{2,4,6,8}}

#3

It has 2^5=32 subsets

Now

P(A)={{1},{2},{3},{4},{5},{1,2},{1,3},{1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5},{1,2,3},{1,2,4},{1,2,5}...{1,2,3,4,5}}

Sauron [17]2 years ago
5 0
<h2>✏️ <u>SUBSETS </u></h2>

\purple{\underline {\bold{\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}

\bold \purple{PROBLEM:}

  • Determine the number of subsets and list down all of them.

  • » 1. { a,b,c }
  • » 2.{ 2,4,6,8 }
  • » 3.{ 1,2,3,4,5,}
  • » 4.{d,e,f,g,h,i}

\bold \purple{FORMULA:}

\underline{ \boxed{ \green{ \rm{ \wp =  2^{n}}}}}

\bold \purple{ANSWER:}

» \rm \: {1.) \:  { a,b,c }}

  • \rm{ \wp =  2^{3} =   \green{\boxed{8}}}

  • {},{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}

» \rm \: {2.)\:  { 2,4,6,8 }}

  • \rm{ \wp =  2^{4} =   \green{\boxed{16}}}

  • {},{2},{4},{6},{8},{2,4},{2,8},{2,8},{4,8},{4,6},{6,8},{2,4,6},{2,4,8},{2,6,8},{4,6,8},{2,4,6,8}

» \rm \: {3. )\:  { 1,2,3,4,5  }}

  • \rm{ \wp =  2^{5} =   \green{\boxed{32}}}

  • {1},{2},{3},{4,}{5},{1,2},{1,3},{1,4},{1,5},{1,2,3},{1,2,4},{1,2,5},{1,3,4},{1,3,5},{1,4,5},{1,2,3,4},{1,2,3,5},{1,2,4,5},{1,3,4,5},{1,2,3,4,5},{2,3},{2,4},{2,5},{2,3,4},{2,3,5},{2,4,5},{2,3,4,5},{3,4}{3,5}{3,4,5}{4,5},{1,2,3,4,5}.

» \rm \: {4. )\:  { d,e,f,g,h,i }}

  • \rm{ \wp =  2^{6} =   \green{\boxed{64}}}

  • {d},{e},{f},{g},{h},{i},{d,e},{d,f},{d,g},{d,h},{d,i},{ e,f },{ e,g },{ e,h },{ e,i },{ f,g },{f,h },{ f,i },{ g,h},{ g,i },{ h,i },{ d,e,f },{ d,e,g },{d,e,h },{ d,e,i },{ d,f,g },{ d,f,h },{ d,f,i },{d,g,h },{ d,g,i },{ d,h,i },{ e,f,g },{ e,f,h },{e,f,i },{e,g,h },{ e,g,i },{ e,h,i },{ f,g,h },{ f,g,i },{g,h,i },{d,e,f,g },{d,e,f,h},{ d,e,f,i },{ d,f,g,h },{d,f,g,i},{d,g,h,i},{e,f,g,h},{e,f,g,i},{f,g,h,i},{d,e,f,g,h},{d,e,f,g,i},{e,f,g,h,i},{d,e,f,g,h,i}.

\purple{\underline {\bold{\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}

\(^.^\)

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A worker was paid a salary of $10,500 in 1985. Each year, a salary increase of 6% of the previous year's salary was awarded. How
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Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.

Here is how much the worker earned each year:


In the year 1985 the worker earned <span>$10,500. 

</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

                                            $10,500(1+0.06).



In the year 1987 the worker earned

$10,500(1+0.06) + 0.06[$10,500(1+0.06)].

Now we can factorize $10,500(1+0.06) and write the earnings as:

$10,500(1+0.06) [1+0.06]=$10,500(1.06)^2.


Similarly we can check that in the year 1987 the worker earned $10,500(1.06)^3, which makes the pattern clear. 


We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

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Factorizing, we have

=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]

We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r}) (where a is the first term and r is the common ratio) we have:

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The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating (1.06)^{26}.


Answer: D



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