Answer:
1. a late breaking news story = to inform
2. a poetry reading = to entertain
3. an advertisement = to persuade
4. a small group assignment = to collaborate
Explanation:
1. A "breaking news" tells people of what is happening in the society. It <em>informs </em>them of the occurrence of an important event such as the plane crash of Kobe Bryant.
2. Poetry reading is meant to touch the attention of listeners. It tries to entertain them through the poem's interesting verses.
3. An advertisement is being shown/displayed in order to convince people to buy a particular product or service.
4. A group assignment allows the members of the group to contribute their ideas together. Such situation is known as "collaboration." They try to brainstorm together towards a common goal.
Answer:
RJ-11 connector.
Explanation:
RJ-11 (Registered Jack-11) also known as the phone line or modem port is a network cable having twisted wire pairs and a modular jack with two, four or six contacts. It is generally used with telephones and modem.
Hence, the other connector type that can be used with CAT 3 installations is the RJ-11 connector.
Answer: Harry should check that font he has used are readable on every page and element of his website
Harry should check that all images have alt texts
Explanation: edmentum
The simulation, player 2 will always play according to the same strategy.
Method getPlayer2Move below is completed by assigning the correct value to result to be returned.
Explanation:
- You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.
#include <bits/stdc++.h>
using namespace std;
bool getplayer2move(int x, int y, int n)
{
int dp[n + 1];
dp[0] = false;
dp[1] = true;
for (int i = 2; i <= n; i++) {
if (i - 1 >= 0 and !dp[i - 1])
dp[i] = true;
else if (i - x >= 0 and !dp[i - x])
dp[i] = true;
else if (i - y >= 0 and !dp[i - y])
dp[i] = true;
else
dp[i] = false;
}
return dp[n];
}
int main()
{
int x = 3, y = 4, n = 5;
if (findWinner(x, y, n))
cout << 'A';
else
cout << 'B';
return 0;
}
