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-BARSIC- [3]
2 years ago
9

How can you rotate and resize your drawing?​

Computers and Technology
2 answers:
jonny [76]2 years ago
5 0

Answer:

By using the PICTURE application built into your computer, you can set a custom image editing type. It goes very easily from there

Explanation:

...

tekilochka [14]2 years ago
4 0

Answer:

Do one of the following:

To rotate the shape counterclockwise, on the Home tab, in the Arrange group, click Position, point to Rotate Shapes, and select Rotate Left 90º.

To rotate the shape clockwise, on the Home tab, in the Arrange group, click Position, point to Rotate Shapes, and select Rotate Right 90º

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Energy requirements of analog and digital signals.
jek_recluse [69]
Analog signals require more energy.

Analog=Constant flowing modulated

Digital is bursts of on/off binary data.
8 0
3 years ago
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Which statement best describes one reason why assembly language is easier
viktelen [127]

Answer:

machine language uses binary code and assembly language uses mnemonic codes to write a program.

Explanation:

In a nutshell, machine language uses binary code, which is almost impossible for humans to decipher, whereas assembly language uses mnemonic codes to write a program. Mnemonic codes make it simpler for humans to understand or remember something, and so make the language a bit easier for humans to use than machine code.

6 0
2 years ago
Alexi is writing a program which prompts users to enter their race times without decimals. Which function should she use?
Juli2301 [7.4K]

Answer:

It's either B or C

Explanation:

I saw this on another brainly but they said it wasn't print so id say b is most likely

6 0
3 years ago
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There are n poor college students who are renting two houses. For every pair of students pi and pj , the function d(pi , pj ) ou
Nuetrik [128]

Answer:

Here the given problem is modeled as a Graph problem.

Explanation:

Input:-  n, k and the function d(pi,pj) which outputs an integer between 1 and n2

Algorithm:-We model each student as a node. So, there would be n nodes. We make a foothold between two nodes u and v (where u and v denote the scholars pu and pv respectively) iff d(pu,pv) > k. Now, Let's call the graph G(V, E) where V is that the vertex set of the graph ( total vertices = n which is that the number of students), and E is that the edge set of the graph ( where two nodes have edges between them if and only the drama between them is bigger than k).

We now need to partition the nodes of the graph into two sets S1 and S2 such each node belongs to precisely one set and there's no edge between the nodes within the same set (if there's a foothold between any two nodes within the same set then meaning that the drama between them exceeds k which isn't allowed). S1 and S2 correspond to the partition of scholars into two buses.

The above formulation is akin to finding out if the graph G(V,E) is a bipartite graph. If the Graph G(V, E) is bipartite then we have a partition of the students into sets such that the total drama <= k else such a partition doesn't exist.

Now, finding whether a graph is bipartite or not is often found using BFS (Breadth First algorithm) in O(V+E) time. Since V = n and E = O(n2) , the worst-case time complexity of the BFS algorithm is O(n2). The pseudo-code is given as

PseudoCode:

// Input = n,k and a function d(pi,pj)

// Edges of a graph are represented as an adjacency list

1. Make V as a vertex set of n nodes.

2. for each vertex  u ∈ V

\rightarrow  for each vertex v ∈ V

\rightarrow\rightarrowif( d(pu, pj) > k )

\rightarrow\rightarrow\rightarrow add vertex u to Adj[v]   // Adj[v] represents adjacency list of v

\rightarrow\rightarrow\rightarrow add vertex v to Adj[u] // Adj[u] represents adjacency list of u

3.  bool visited[n] // visited[i] = true if the vertex i has been visited during BFS else false

4. for each vertex u ∈ V

\rightarrowvisited[u] = false

5. color[n] // color[i] is binary number used for 2-coloring the graph  

6. for each vertex u ∈ V  

\rightarrow if ( visited[u] == false)

\rightarrow\rightarrow color[u] = 0;

\rightarrow\rightarrow isbipartite = BFS(G,u,color,visited)  // if the vertices reachable from u form a bipartite graph, it returns true

\rightarrow\rightarrow if (isbipartite == false)

\rightarrow\rightarrow\rightarrow print " No solution exists "

\rightarrow\rightarrow\rightarrow exit(0)

7.  for each vertex u ∈V

\rightarrow if (color[u] == 0 )

\rightarrow\rightarrow print " Student u is assigned Bus 1"

\rightarrowelse

\rightarrow\rightarrow print " Student v is assigned Bus 2"

BFS(G,s,color,visited)  

1. color[s] = 0

2. visited[s] = true

3. Q = Ф // Q is a priority Queue

4. Q.push(s)

5. while Q != Ф {

\rightarrow u = Q.pop()

\rightarrow for each vertex v ∈ Adj[u]

\rightarrow\rightarrow if (visited[v] == false)

\rightarrow\rightarrow\rightarrow color[v] = (color[u] + 1) % 2

\rightarrow\rightarrow\rightarrow visited[v] = true

\rightarrow\rightarrow\rightarrow Q.push(v)

\rightarrow\rightarrow else

\rightarrow\rightarrow\rightarrow if (color[u] == color[v])

\rightarrow\rightarrow\rightarrow\rightarrow return false // vertex u and v had been assigned the same color so the graph is not bipartite

}

6. return true

3 0
3 years ago
I will give brainliest !!
OverLord2011 [107]
Do these questions have given answers already?
3 0
2 years ago
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