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kompoz [17]
3 years ago
13

computer hardware without software is useless while computer software without hardware is meaningless. explain what do you under

stand by this statement ​
Computers and Technology
1 answer:
Rainbow [258]3 years ago
6 0

Answer:

That the software is useless

Explanation:

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3. Which swim_backwards method is called when sammy.swim_backwards() is executed? Why?
Georgia [21]
Classes called child classes or subclasses inherit methods and variables
4 0
2 years ago
Implementations <br> in c++ programming when finding lcm and gcd in c++​
vazorg [7]

Answer:

#include<iostream>

using namespace std;

int lcm(int m, int n) {

  int a;

  a = (m > n) ? m: n;

  while (true) {

     if (a % m == 0 && a % n == 0)

        return a;

        ++a;

  }

}

int gcd(int m, int n) {

  int r = 0, a, b;

  a = (m > n) ? m : n;

  b = (m < n) ? m : n;

  r = b;

  while (a % b != 0) {

     r = a % b;

     a = b;

     b = r;

  }

  return r;

}

int main(int argc, char **argv) {

  cout << "Enter the two numbers: ";

  int m, n;

  cin >> m >> n;

  cout << "The LCM of two numbers is: " << lcm(m, n) << endl;

  cout << "The GCD of two numbers is: " << gcd(m, n) << endl;  

  return 0;

}

Explanation:

7 0
2 years ago
Consider a classful IPv4 address 200.200.200.200? What is its implied subnet mask? If we take the same address and append the CI
topjm [15]

Answer:

a. 255.255.255.0 (class C)

b. 255.255.255.224

Explanation:

Here, we want to give the implied subnet mask of the given classful IPV4 address

We proceed as follows;

Given IPv4 address: 200.200.200.200

Classes

Class A : 0.0.0.0 to 127.255.255.255

Class B: 128.0.0.0 to 191.255.255.255

Class C: 192.0.0.0 to 223.255.255.255

Class D: 224.0.0.0 to 239.255.255.255

so 200.200.200.200 belongs to Class C and it's subnet mask is 255.255.255.0

In CIDR(Classless Inter Domain Routing)

subnet /27 bits means 27 1s and 5 0s. so subnet

i.e 11111111.11111111.11111111.11100000 which in dotted decimal format is 255.255.255.224 .

4 0
3 years ago
How many bits does it take to store a 3-minute song using an audio encoding method that samples at the rate of 40,000 bits/secon
blondinia [14]

Answer:

115200000 bits

Explanation:

Given Data:

Total Time = 3 minute = 3 x 60 sec = 180 sec

Sampling rate = 40,000 bits / sample

Each sample contain bits = 16 bits /sample

Total bits of song = ?

Solution

Total bits of song = Total Time x Sampling rate x Each sample contain bits

                             = 180 sec x 40,000 bits / sec x 16 bits /sample

                             = 115200000 bits

                             =  115200000/8 bits

                             = 14400000 bytes

                             = 144 MB

There are total samples in one second are 40000. Total time of the song is 180 seconds and one sample contains 16 bits. so the total bits in the song are 144 MB.

5 0
3 years ago
I’m joining the robotics team in my school, but my knowledge is pretty basic. What should I do to improve myself?
Aleonysh [2.5K]
Start learning the functions of a computer and really put ur heart into it
8 0
3 years ago
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