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malfutka [58]
3 years ago
9

Consider the following repeating decimal. 0.619

Mathematics
1 answer:
slega [8]3 years ago
6 0

Answer:

a.  0.6\overline {19} = 0.6 + \ \sum \limits ^{\infty}_{n=0} \  0.019 \ ( \dfrac{1}{100})^n

b. 0.6\overline {19}  = \mathbf{\dfrac{613}{990}}  

Step-by-step explanation:

Consider the following repeating decimal. 0.6\overline {19}

a) Write the repeating decimal as a geometric series.

0.6\overline {19} is being expressed as 0.6191919...

0.6\overline {19} = 0.6 + 0.019 + 0.00019+ 0.0000019 + ...

0.6\overline {19} =0.6 +  \dfrac{19}{1000}+ \dfrac{19}{100000}+ \dfrac{19}{10000000}+ ...

0.6\overline {19} = 0.6+ \dfrac{19}{1000} \begin {bmatrix} 1 + \dfrac{1}{100} + \dfrac{1}{10000}+ ... \end  {bmatrix}

0.6\overline {19} = 0.6 + \ \sum \limits ^{\infty}_{n=0} \  0.019 \ ( \dfrac{1}{100})^n

(b) Write the sum of the series as the ratio of two integers

0.6\overline {19}  = 0.6 + 0.019  ( \dfrac{1}{1-0.01})

0.6\overline {19}  =0.6 +  \dfrac{19}{1000}\times  \dfrac{100}{99}

0.6\overline {19}  = 0.6 +  \dfrac{19}{990}

0.6\overline {19}  = \dfrac{594+19}{990}

0.6\overline {19}  = \mathbf{\dfrac{613}{990}}  

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How many integers in the set {n ∈ Z | 1 ≤ n ≤ 700} are divisible by 2 or 7?
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\large\begin{array}{l}\\\\ \textsf{This question gives us a set}\\\\ \mathsf{S=\{n \in\mathbb{Z}:~1\le n\le 700\}}\\\\ \mathsf{S=\{1,\,2,\,3,\,\ldots,\,699,\,700\}}\\\\\\ \bullet~~\textsf{Set of integers that are divible by 2 (even integers):}\\\\ \mathsf{A=\{n\in \mathbb{Z}:~n=2k,\,k\in\mathbb{Z}\}}\\\\ \mathsf{A=\{\ldots,\,-4,\,-2,\,0,\,2,\,4,\,\ldots\}}\\\\\\ \bullet~~\textsf{Set of integers that are divible by 7:}\\\\ \mathsf{B=\{n\in \mathbb{Z}:~n=7k,\,k\in\mathbb{Z}\}}\\\\ \mathsf{A=\{\ldots,\,-14,\,-7,\,0,\,7,\,14,\,\ldots\}} \end{array}

___________


\large\begin{array}{l}\\\\ \textsf{We want to know how many elements there are in the}\\\textsf{following set:}\\\\ \mathsf{S\cap (A\cup B)=(S\cap A)\cup(S\cap B)\qquad(i)} \end{array}

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\large\begin{array}{l}\\\\ \bullet~~\mathsf{S\cap A=\{n\in\mathbb{N}:~n=2k~~and~~1\le n\le 700,\,k\in\mathbb{Z}\}}\\\\ \mathsf{S\cap A=\{2,\,4,\,6,\,\ldots,\,698,\,700\}}\\\\ \mathsf{S\cap A=\{1\cdot 2,\,2\cdot 2,\,3\cdot 2,\,\ldots,\,349\cdot 2,\,350\cdot 2\}}\\\\\\ \textsf{So, there are 350 elements in }\mathsf{S\cap A:}\\\\ \mathsf{\#(S\cap A)=350.} \\\\\\ \bullet~~\mathsf{S\cap B=\{n\in\mathbb{N}:~n=7k~~and~~1\le n\le 700,\,k\in\mathbb{Z}\}}\\\\ \mathsf{S\cap B=\{7,\,14,\,21,\,\ldots,\,693,\,700\}}\\\\ \mathsf{S\cap B=\{1\cdot 7,\,2\cdot 7,\,3\cdot 7,\,\ldots,\,99\cdot 7,\,100\cdot 7\}} \\\\\\ \textsf{So, there are 100 elements in }\mathsf{S\cap B:}\\\\ \mathsf{\#(S\cap B)=100.} \end{array}

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\large\begin{array}{l}\\\\ \textsf{Therefore,}\\\\ \mathsf{\#\big[S\cap (A\cup B)\big]}\\\\ =\mathsf{\#\big[(S\cap A)\cup(S\cap B)\big]}\\\\ =\mathsf{\#(S\cap A)+\#(S\cap B)-\#\big[(S\cap A)\cap(S\cap B)\big]}\\\\ =\mathsf{350+100-50}\\\\ =\mathsf{450-50}\\\\ =\mathsf{400~elements.}\\\\\\ \textsf{There are 400 integers in S that are divisible by 2 or 7.} \end{array}


If you're having problems understanding the answer, try to see it through your browser: brainly.com/question/2105863


\large\begin{array}{l}\\\\ \textsf{Any doubts? Please, comment below.}\\\\\\ \textsf{Best wishes! :-)} \end{array}


Tags: <em>set theory divibilility divisible integers union intersection</em>

3 0
3 years ago
The volume of
BARSIC [14]
0.1

1*100 = 100
100/1000
=0.1
8 0
3 years ago
Read 2 more answers
PLEASE HURRY THIS IS DUE IN 10 MINUTES AND I AM REALLY CONFUSED!!
Harrizon [31]

Answer:

A. 8 beads per bracelet.

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GaryK [48]

Answer:

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Step-by-step explanation:

Define listed price = C

Based on the term, to have the balance on both plan, the equation is 30+90%C = 80+80%C

  1. 30+0.9C=80+0.8C
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