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Elodia [21]
3 years ago
12

Multiply the polynomials.

Mathematics
2 answers:
navik [9.2K]3 years ago
8 0
The answer should be A, no parentheses just multiply within the numbers
velikii [3]3 years ago
7 0

Answer:

2x^3 + 13x^2 + 26x + 16

Step-by-step explanation:

x(2x^2 + 9x + 8) +2(2x^2 + 9x + 8)

2x^3 + 9x^2 + 8x + 4x^2 + 18x + 16

2x^3 + 13x^2 + 26x + 16

  • HOPE IT'S HELPFUL
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Answer:

1.)(I cant draw a slope but here is the equation) 45x+150

2.) 25(45) + 150

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the coding project will cost <u>$1275</u> if the programmer charges $45 and hour, and takes up 25 hours

8 0
4 years ago
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;

}

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)

   {

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       y = p[i].second.second;

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   }

   return minimumCost;

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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;

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   }

   // Sort the edges in the ascending order

   sort(p, p + edges);

   minimumCost = kruskal(p);

   cout << minimumCost << endl;

   return 0;

}

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