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soldier1979 [14.2K]
3 years ago
7

Computer can do work very___​

Computers and Technology
2 answers:
Andrei [34K]3 years ago
6 0

Answer:

Explanation:

quickly      

omeli [17]3 years ago
4 0

Answer:

fast......................

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Assume that you have an ArrayList variable named a containing 4 elements, and an object named element that is the correct type t
jonny [76]

Answer:

Option (4) is the correct answer.

Explanation:

In Java programming language ,array collection starts from 0 index location and ends in a size-1 index location. So to access the last elements the user needs to use a[Size-1] statement. so to modify the value of the last location of the array the user needs to use "a[size-1]= element;".

But when the user wants to add some new value to the end of the array list collection then he needs to use the statement--

a.add(element); //where add is a function, element is a value and a is a array list object.

Another option is invalid because--

  • Option 1 is not the correct because "a[3]=element;" modify the value of the 3rd element of the array.
  • Option 2 gives a compile-time error because add functions bracts are not closed.
  • Option 3 gives the error because a[4] gives the location of the 5th element of the array but the above question says that a is defined with 4 elements.
7 0
3 years ago
You and your friend play a video game where a superbird has to find and eat radiation leaks at nuclear power plants before the p
koban [17]

Answer:

Bird with superpowers

Explanation:

It is the main idea

7 0
4 years ago
Read 2 more answers
Implement the following logic in C++, Use appropriate data types. Data types are represented as either numeric (num) or string.
ICE Princess25 [194]

Answer:

Follows are the modified code in c++ language:

#include<iostream>//header file

#include<string>//header file

using namespace std;

int main()//main method

{

string name,address, MSG_YES, MSG_NO;//defining string variable

int item,quantity,size=6, i=0;//defining integer variable

double price;//defining double variable  

int VALID_ITEM[]={106,108,307,405,457,688};//defining integer array and assign value

double VALID_ITEM_PRICE[]={0.59,0.99,4.50,15.99,17.50,39.00};//defining double array and assign value

bool foundIt=false;//defining bool variable  

MSG_YES="Item available";//use string variable to assign value

MSG_NO = "Item not found"; //use string variable to assign value

cout<<"Input name: ";//print message

cin>>name;//input value in string variable

cout<<"Input Address: ";//print message

cin>>address;//input value in string variable

cout<<"Input Item: "<<endl;//print message

cin>>item;//input value in string variable

cout<<"Input Quantity: "<<endl;//print message

cin>>quantity;//input value in string variable

while(i <size)//defining while that checks i less then size  

{

if (item ==VALID_ITEM[i]) //use if block to match item in double array

{

foundIt = true;//change bool variable value

price = VALID_ITEM_PRICE[i];//hold item price value in price variable  

}

i++;//increment the value of i

}

if (foundIt == true)//use if to check bool variable value equal to true

{

cout<<MSG_YES<<endl;//print value

cout<<"Quantity "<<quantity<<" at "<<"Price "<<price<<"each"<<endl;//print value

cout<<"Total"<<quantity*price;//calculate the total value

}

else//else block

cout<<MSG_NO;//print message  

}

Output:

please find the attached file.

Explanation:

In the above given C++ language modified code, four-string variable " name, address, MSG_YES, and MSG_NO", four integer variable "item, quantity, size, and i", and "integer and double" array is defined, that holds values.

In the string and integer variable "name, address, and item, quantity", we input value from the user-end and use the while loop, which uses the if block to check the "item and quantity" value from the user end and print its respective value.

6 0
3 years ago
The disadvantage of this topology is that a problem with one node will crash the entire network.
DanielleElmas [232]

Answer:

Bus Topology

Explanation:

In local area networks (LANs), Bus topology is used to connect all the computers on the single cable. This cable is called backbone of the network.  In this type of network when a computer sends data, it will be received on entire network.

For Example

In water line system, all the taps that are connected to single pipe. If the pipe or a single tap broken this will affect the whole system.

Entire system can be crash if any node or wire will break.

6 0
3 years ago
The compare_strings function is supposed to compare just the alphanumeric content of two strings, ignoring upper vs lower case a
Korolek [52]

Answer:

There is a problem in the given code in the following statement:

Problem:

punctuation = r"[.?!,;:-']"

This produces the following error:

Error:

bad character range

Fix:

The hyphen - should be placed at the start or end of punctuation characters. Here the role of hyphen is to determine the range of characters. Another way is to escape the hyphen - using using backslash \ symbol.

So the above statement becomes:

punctuation = r"[-.?!,;:']"  

You can also do this:

punctuation = r"[.?!,;:'-]"  

You can also change this statement as:

punctuation = r"[.?!,;:\-']"

Explanation:

The complete program is as follows. I have added a print statement print('string1:',string1,'\nstring2:',string2) that prints the string1 and string2 followed by return string1 == string2  which either returns true or false. However you can omit this print('string1:',string1,'\nstring2:',string2) statement and the output will just display either true or false

import re  #to use regular expressions

def compare_strings(string1, string2):  #function compare_strings that takes two strings as argument and compares them

   string1 = string1.lower().strip()  # converts the string1 characters to lowercase using lower() method and removes trailing blanks

   string2 = string2.lower().strip()  # converts the string1 characters to lowercase using lower() method and removes trailing blanks

   punctuation = r"[-.?!,;:']"  #regular expression for punctuation characters

   string1 = re.sub(punctuation, r"", string1)  # specifies RE pattern i.e. punctuation in the 1st argument, new string r in 2nd argument, and a string to be handle i.e. string1 in the 3rd argument

   string2 = re.sub(punctuation, r"", string2)  # same as above statement but works on string2 as 3rd argument

   print('string1:',string1,'\nstring2:',string2)  #prints both the strings separated with a new line

   return string1 == string2  # compares strings and returns true if they matched else false

#function calls to test the working of the above function compare_strings

print(compare_strings("Have a Great Day!","Have a great day?")) # True

print(compare_strings("It's raining again.","its raining, again")) # True

print(compare_strings("Learn to count: 1, 2, 3.","Learn to count: one, two, three.")) # False

print(compare_strings("They found some body.","They found somebody.")) # False

The screenshot of the program along with its output is attached.

4 0
4 years ago
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