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Brums [2.3K]
3 years ago
10

A wheelchair access ramp has an angle of elevation of 20°. If the ramp reaches to the top of a 33 inch high porch, how long is t

he ramp?
Mathematics
2 answers:
sashaice [31]3 years ago
8 0
Sinα=height/length

length=height/sinα, we are given that α=20° and h=33 in

length=33/sin20° in

length≈96.49 in (to nearest hundredth of an inch)
ella [17]3 years ago
3 0
This is a sin operation, Sin is the relationship between the opposite (height) and the hypotenuse (length of ramp). Plugging in: Sin(20deg) = 33/x

Sin(20) = 33/x

x*Sin(20) = 33

x = 33/Sin(20)

x = 96.49in long.

*This answer is the length of the SLOPE of the ramp.
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Answer:

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Step-by-step explanation:

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3 years ago
Which relation describes a function?
Kisachek [45]

Answer:

{(-2, -3), (-3, -2), (2,3), (3, 2)}

Step-by-step explanation:

A function is a set of ordered pairs in which <u>no two pairs have the same first number</u>.

In other words, an <em>x</em> cannot be paired with two <em>y</em>'s.

A function takes an <em>x</em> and pairs it with one and only one <em>y</em>.

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{(2, 2), (2, 3), (3, 2), (3, 3)} is not a function because it pairs 2 with both 2 and 3.

8 0
3 years ago
Read 2 more answers
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;

}

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