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Artist 52 [7]
3 years ago
8

Mathematics with applications in the management, Natural, and Social Sciences Twelfth edition

Mathematics
1 answer:
Monica [59]3 years ago
4 0

Answer:

In 2015 the both populations were the same and from that year the population of millennials surpassed the population of boomers

Step-by-step explanation:

\left \{ {{10x+13y = 1125} \atop {-2x+7y = 495}} \right.

x=14

Boomer: 10(14)+13y=1125

               140+13y=1125

                       13y=1125-140

                       y= 985/13

                       y= 75.77 (75.77 millions of boomers in 2014)

Millenials: -2(14) +7y = 495

                  -28 +7y = 495

7y= 495+28

y= 523/7

y=74.71  (74.71 millios on millenials in 2014)

In 2014 the population of boombers were still greater than the population of millennials

The solution of the system of equations will give us the point where the populations were equalized, and from that point the population of boombers will be less than that of the millennials.

Boomers: 10x+13y = 1125

                 y= (-10x +1125)/13

Millenials: -2x+7y = 495

                 y= (2x+495)/7

We match both expressions of "y"

(-10x +1125)/13 =(2x+495)/7

cross multiply:

(-10x +1125)*7 =(2x+495)*13

-70x + 7875 = 26x + 6435

we group similar terms:

7875 -6435 = 26x+70x

1440 = 96x

x= 1440/96

x= 15

In 2015 the both populations were the same and from that year the population of millennials surpassed the population of boomers

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Step-by-step explanation:

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Using the Breadth-First Search Algorithm, determine the minimum number of edges that it would require to reach
jekas [21]

Answer:

The algorithm is given below.

#include <iostream>

#include <vector>

#include <utility>

#include <algorithm>

using namespace std;

const int MAX = 1e4 + 5;

int id[MAX], nodes, edges;

pair <long long, pair<int, int> > p[MAX];

void initialize()

{

   for(int i = 0;i < MAX;++i)

       id[i] = i;

}

int root(int x)

{

   while(id[x] != x)

   {

       id[x] = id[id[x]];

       x = id[x];

   }

   return x;

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void union1(int x, int y)

{

   int p = root(x);

   int q = root(y);

   id[p] = id[q];

}

long long kruskal(pair<long long, pair<int, int> > p[])

{

   int x, y;

   long long cost, minimumCost = 0;

   for(int i = 0;i < edges;++i)

   {

       // Selecting edges one by one in increasing order from the beginning

       x = p[i].second.first;

       y = p[i].second.second;

       cost = p[i].first;

       // Check if the selected edge is creating a cycle or not

       if(root(x) != root(y))

       {

           minimumCost += cost;

           union1(x, y);

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   }

   return minimumCost;

}

int main()

{

   int x, y;

   long long weight, cost, minimumCost;

   initialize();

   cin >> nodes >> edges;

   for(int i = 0;i < edges;++i)

   {

       cin >> x >> y >> weight;

       p[i] = make_pair(weight, make_pair(x, y));

   }

   // Sort the edges in the ascending order

   sort(p, p + edges);

   minimumCost = kruskal(p);

   cout << minimumCost << endl;

   return 0;

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8 0
3 years ago
The width of a rectangle is (x+1) and the length is (x-6). What is the length and width of the rectangle if the area is 30 squar
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Answer:

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The width of the rectangle (W) =  10 feet

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the width of the rectangle  = (x+1) feet</em>

<em>Given that the length of the rectangle = ( x-6) feet</em>

<em>The area of the rectangle = 30 square feet</em>

<u><em>Step(ii):-</em></u>

We know that the area of the rectangle

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⇒ x² - 5 x - 36 =0

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<u><em>Step(iii):-</em></u>

we have to choose x =9

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      The width of the rectangle (W) =  x+1 =  9 +1 = 10

<u><em>Final answer:-</em></u>

The length of the rectangle (l) = 3 feet

The width of the rectangle (W) =  10 feet

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both are correct!

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