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stich3 [128]
4 years ago
13

Where is the bios in the soft ware or hard ware

Computers and Technology
2 answers:
Natalija [7]4 years ago
7 0

Answer:

Software

Explanation:

Mandarinka [93]4 years ago
6 0

Answer:

The BIOS is special software that interfaces the major hardware components of your computer with the operating system. It is usually stored on a Flash memory chip on the motherboard, but sometimes the chip is another type of ROM.

So software

Hope that helps!!!

You might be interested in
Please help
sergey [27]

Answer:

a = 6, b = 0

Explanation:

The loop ran 3 times before b == 0. The statement "while ((b != 0)" is essentially saying: 'While b is not equal to 0, do what's in my loop'. Same general thing with "&& ((a / b) >= 0)". The "&&" is specifying that there should be another loop condition. The final part of the while loop states: 'as long as a ÷ b is greater than 0, do what's in my loop'. If all of these conditions are met, the loop will run. It will continue to run until at least one of the conditions are not met.

<em />

<em>Side note: I can't help but notice you posted the same question a while ago, so I just copied and pasted my previous response with some tweaking here and there. Hope this helps you! :)</em>

8 0
3 years ago
Create an algorithm that will convert dog years to human years using the assumption that 1 dog year = 7 human years. Prompt the
matrenka [14]

Answer:

Algorithm:

1. Declare and initialize variable one_dog_year=7.

2.Ask user to give dog age.

   2.1 Read the dog age and assign it to variable "dog_age".

3.Create a variable "human_age" to store the equivalent human age.

   3.1 Calculate equivalent human age as "human_age=one_dog_year*dog_age".

4.Print the equivalent human age of dog age.

7. End the program.

// here is algorithm implemented in c++

#include <bits/stdc++.h>

using namespace std;

int main()

{

   // initialize one dog year

   int one_dog_year=7;

   int dog_age;

   int equi_h_age;

   cout<<"Enter the dog age:";

   // read the dog age

   cin>>dog_age;

   

   //calculate the equivalent human age

   equi_h_age=dog_age*one_dog_year;

   cout<<"equivalent human age is : "<<equi_h_age<<endl;

return 0;

}

Output:

Enter the dog age:2                                                                                                                                            

equivalent human age is : 14

4 0
3 years ago
Can someone help me please ??
zavuch27 [327]

Answer:

x=27-the value of 0.1 x all sides

Explanation:

subscribe to me at chrisgot ha x

6 0
3 years ago
Design a class called NumDays. The class’s purpose is to store a value that represents a number of work hours and convert it to
pickupchik [31]

Answer:

Check the explanation

Explanation:

#include <iostream>

#include <string>

using namespace std;

//class declaration

class NumDays{

private:

   double hours;

   double days;

public:

   //constructor

   NumDays(double h = 0){

       hours = h;

       days = h/(8.00);

   }

   //getter functions

   double getHours(){

       return hours;

   }

   double getDays(){

       return days;

   }

   //setter functions

   void setHours(double h){

       hours = h;

       days = h/(8.00);

   }

   void setDays(double d){

       days = d;

       hours = d*(8.00);

   }

   //overload + operator

   double operator+ (const NumDays &right){

       return hours+right.hours;

   }

   //overload - operator

   double operator- (const NumDays &right){

       //check if subtraction will give negative value

       if(hours < right.hours){

           cout << "ERROR! Cannot subtract! Now terminating!\n";

           exit(0);

       }

       return hours-right.hours;

   }

   //overload prefix ++ operator

   NumDays operator++(){

       //pre-increment hours member

       ++hours;

       //update days member

       days = hours/(8.00);

       //return modified calling object

       return *this;

   }

   //overload postfix ++ operator

   NumDays operator++(int){

       //post-increment hours member

       hours++;

       //update days member

       days = hours/(8.00);

       //return modified calling object

       return *this;

   }

   //overload prefix -- operator

   NumDays operator--(){

       //pre-decrement hours member

       --hours;

       //update days member

       days = hours/(8.00);

       //return modified calling object

       return *this;

   }

   //overload postfix -- operator

   NumDays operator--(int){

       //post-decrement hours member

       hours--;

       //update days member

       days = hours/(8.00);

       //return modified calling object

       return *this;

   }

};

int main()

{

   //create first object

   cout << "Creating object with 12 hours...\n";

   NumDays obj1(12);

   cout << obj1.getHours() << " hours = " <<obj1.getDays() << " days.\n";

   //create second object

   cout << "\nCreating object with 18 hours...\n";

   NumDays obj2(18);

   cout << obj2.getHours() << " hours = " <<obj2.getDays() << " days.\n";

   //test overloaded + operator

   cout << endl << "Adding hours... " << obj1 + obj2 << " hours.\n";

   //test overloaded - operator

   cout << endl << "Subtracting hours... " << obj2 - obj1 << " hours.\n\n";

   //test overloaded ++ operators

   cout << "Pre- and post-incrementing first object...\n";

   ++obj1;

   cout << obj1.getHours() << " hours = " <<obj1.getDays() << " days.\n";

   obj1++;

   cout << obj1.getHours() << " hours = " <<obj1.getDays() << " days.\n";

   //test overloaded -- operators

   cout << "\nPre- and post-decrementing second object...\n";

   --obj2;

   cout << obj2.getHours() << " hours = " <<obj2.getDays() << " days.\n";

   obj2--;

   cout << obj2.getHours() << " hours = " <<obj2.getDays() << " days.\n";

   return 0;

}

7 0
4 years ago
Read 2 more answers
A manufacturer wishes to design a hard disk with a capacity of 30 GB or more (using the standard definition of 1 GB = 2^30 bytes
never [62]

Answer:

4 Platters

Explanation:

The capacity of one platter

= 1024 x 2048 x 4096

<em>1 platter capacity= 8GB </em>

For a 30GB hard disk, we need

30/8  = 3.75  platters.

<em>we round off the answer  so 4 platters will be required.</em>

7 0
3 years ago
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