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yKpoI14uk [10]
3 years ago
8

Write a complete two column proof for the following information.

Mathematics
1 answer:
andrey2020 [161]3 years ago
4 0
We are asked to show a proof that m∠1 is equal to 62° and the line l is parallel to line m which has an angle of 180°. If two lines are parallel the summation of angles are 180 degrees and they are also considered as a supplementary angle. Hence, by this we have formed the equation and solution below:
m∠l + m∠m = 180° 
replacing the value of two angles, we have:
62° + 118° = 180°
180° = 180°

Therefore, the two lines are parallel.
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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;

}

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);

       }    

   }

   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;

}

8 0
3 years ago
I NEED HELP ASAP!!!!
Degger [83]
I hope this helps you!

4 0
3 years ago
Give the coordinates of Point P without using any new variables. <br><br> (Please show your work.)
poizon [28]

Answer:

(a-b, c)

Step-by-step explanation:

The midpoints of the two diagonals are the same, so we have ...

(P + (-a, 0))/2 = (O +(-b, c))/2

Multiplying by 2 and subtracting (-a, 0), we get ...

P = (0, 0) +(-b, c) -(-a, 0)

P = (a-b, c)

7 0
3 years ago
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