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Lorico [155]
3 years ago
11

a box was measured with a degree of accuracy to the nearest 2cm; 24cm x 24cm x 20cm. what is the largest possible volume of the

box to the nearest cm3
Mathematics
1 answer:
DochEvi [55]3 years ago
5 0
For any value, e.g. a, to the nearest x units, the upper and lower bounds are:
(a + x/2) and (a - x/2)
For your question, there are three dimensions so:
Dealing with the first one, we have 24 cm to the nearest 2 cm so the boundaries are:
Upper boundary: 24 + 2/2 = 25
Lower boundary: 24 - 2/2 = 23
The second dimension is the same as the first in value and is also given to the nearest 2 cm so the boundaries are the same as for the first.
The third dimension is 20 cm to the nearest 2 cm so the boundaries are:
Upper boundary: 20 + 2/2 = 21
Lower boundary: 20 - 2/2 = 19

To get the largest possible area, we take the upper bounds of all the dimensions and multiply them so:
25 * 25 * 21 = 13125 cm³
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Can the set of lengths be the side lengths of a right triangle?
andreyandreev [35.5K]

Answer:

B. No.

Step-by-step explanation:

We have been given 3 side lengths 7 ft, 12 ft, 17 ft. We are asked to determine, whether the given set of lengths can form a right triangle or not.

We will use Pythagoras theorem to solve our given problem, which states that the square of hypotenuse of a right triangle is equal to the sum of squares of two legs of right triangle.

17^2=12^2+7^2

289=144+49

289>193

Since the sum of  squares of both legs is less than square of hypotenuse, therefore, the given set of lengths can not be the side lengths of a right triangle.

4 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;

}

8 0
3 years ago
Please help! Will get brainliest!!!
ASHA 777 [7]
Chart;the answer is b

and it makes sense answer is ; also b
3 0
3 years ago
If the quadrilateral below is a kite, find m
Debora [2.8K]

Answer:

49°

Step-by-step explanation:

7.5x-15+90=180

7.5x+75 =180

7.5x=105

x =14

so NPQ = ( 4×14 -7 ) = 49° this the answer

6 0
2 years ago
Read 2 more answers
Which of these expressions is equivalent to 3x(x-1)-5(x-1)? Select all that apply.
Maru [420]

Answer:

3x(x-1)-5(x-1)

=3x²-3x-5x+5 (we can count it one by one)

=3x²-8x+5 (we can calculate the same variable)

#i'm from indonesia

hope it helps.

6 0
2 years ago
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