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LenaWriter [7]
1 year ago
6

A firm will break even (no profit and no loss) as long as revenue just equals cost. The value of x (the number of items

Mathematics
1 answer:
Mamont248 [21]1 year ago
5 0

Step-by-step explanation:

(a)

the cost function is

C(x) = 300 + 20x

(b)

the revenue function is

R(x) = 35x

(e)

the profit function

P(x) = R(x) - C(x) = 35x - 300 - 20x = 15x - 300

(d)

break-even is the point where C(x) and R(x) are equal :

C(x) = R(x)

300 + 20x = 35x

300 = 15x

x = 300/15 = 20

so, the break-even is only achieved when 20 units are produced and sold. and with more than 20 sold units the business would then make a profit.

per our restrictions only max. 19 units can be sold, so, the business will always be in the negative numbers (unless they cheat and sell virtual units that were never produced).

therefore, the product should not be produced at all.

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The radius of a circle is 4 inches. What is the circle's circumference?
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Answer:

8π inches or approximately 25.12 inches

Step-by-step explanation:

What's the circumference of a circle?

 ❖ The circumference of a circle is the distance around a circle.

How do we find the circumference of a circle?

 ❖ The formula to find the circumference of a circle is 2πr, where r =    

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<u>Solving</u>

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<u>or</u>

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3 years ago
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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));

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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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