a model rocket is launched from a roof into a large field the path of the rocket can be modeled by the equation y equals -0.04 ^ 2 + 8.3 x + 4.3 where X is the horizontal distance in meters from the starting point on the roof and why is the height in meters of the rocket above the ground how far horizontally from its starting point will the rocket land round your answer to the nearest hundredth meter

Answer:

Step-by-step explanation:

on no

Answer:

Option (2)

Step-by-step explanation:

Two points having real and imaginary parts are,

w₁ = 4 + 2i

w₂ = -1 - 3i

By adding these numbers,

w₁ + w₂ = (4 + 2i) + (-1 - 3i)

            = (4 - 1) + (2 - 3)i

            = 3 - (1)i

Since, coordinates of a number on the graph are (x, y)

Here x = Real part

y = Imaginary part

Therefore, sum of these numbers will be (3, -1).

Option (2) will be the correct graph.

Answer:

B on Edge

Step-by-step explanation:

Got 100% on the test.

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;

}

The distance between two points on a plane can be found by Pythagoras's theorem.

The shortest route from her home to her workplace is the route that passes along the block in the middle of the graph ,which is approximately 2.5 blocks shorter, than the apparently next shortest route.

Reason:

The shortest distance between two points is a straight line.

The shortest route to her workplace are the routes that approximate a straight line the most.

Following the dotted lines, there are two routes, that are approximately direct and therefore short routes to her workplace which are;

• The route heading past the block close to the center of the map to her workplace

• The route to the left of the above route

Taking the route that passes close to the center of the map to her workplace ,gives;

Path 1 = √(2² + 4²) = 2·√5 (by Pythagoras's theorem)

Path 2 = √(1² + 2²) = √5

Path 3 = 1 + 1 = 2

Path 4 = √(2² + 3²) = √13

Path 5 = 3

Cumulativedistance = 2·√5 + √5 + 2 + √13 + 3 ≈ 15.3

Taking the route to the left of the above route, gives;

Path 1 = √(2² + 4²) = 2·√5

Path 2 = 7 + 6 = 13

Cumulative distance = 2·√5 + 13 ≈ 17.5

Therefore, the shortest route is the route that heads to her workplace, through, close to the center of the map, which approximately 17.5 blocks - 15.3 blocks  = 2.5 blocks shorter

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