a. speed + 12 - miles * 2 = 10 + 12 - 5 * 2. With order of operations, we do the multiplication first so the equation is now 10 + 12 - 10 = 22 - 10 = 12
b. speed + miles * 3 = 10 + 5 * 3 and again, order of operations gives us 10 + 15 = 25
c. (speed + miles) * 3 = (10 + 5) * 3 = 15 * 3 = 45
d. speed + speed * miles + miles = 10 + 10 * 5 + 5 = 10 + 50 + 5 = 60 + 5 = 65
e. (10 – speed) + miles / miles = (10 - 10) + 5 / 5 = 0 + 5 / 5 = 5 / 5 = 1
Local
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Answer:
#include <iostream>
using namespace std;
class Digits
{
public:
int num;
int read() //method to read num from user
{
cout<<"Enter number(>0)\n";
cin>>num;
return num;
}
int digit_count(int num) //method to count number of digits of num
{
int count=0;
while(num>0) //loop till num>0
{
num/=10;
count++; //counter which counts number of digits
}
return count;
}
int countDigits(int num) //method to return remainder
{
int c=digit_count(num); //calls method inside method
return num%c;
}
};
int main()
{
Digits d; //object of class Digits is created
int number=d.read(); //num is read from user
cout<<"\nRemainder is : "<<d.countDigits(number); //used to find remainder
return 0;
}
Output :
Enter number(>0)
343
Remainder is : 1
Explanation:
As program is missing to find errors , a logically write program is written to find the remainder when a number is divided by its number of digits. A class Digits is constructed which has public variable num and methods read(), digit_count(), countDigits().
- read() - This method reads value of num from the user and return num.
- digit_count() - This method takes a integer as parameter and counts the number of digits of a number passed as argument. while loop is used to increement the counter until num<0. This returns the value of count.
- countDigits() - This method takes a integer as a parameter and returns remainder when the argument is divided by number of digits of argument. Number of digits is calculated by using method digit_count().
At last in main method , object of Digits class is created and its methods are used to find the output.
Answer:
Following are the code to the given question:
int power(int x, int n)//defining a method power that accepts two integer parameters
{
if (n == 0)//defining if block to check n equal to 0
{
return 1; //return value 1
}
else//defining else block
{
x = x * power(x, --n); //use x variable to call method recursively
}
return x; //return x value
}
Explanation:
In the above-given code, a method power is defined that accepts two integer variable in its parameter, in the method a conditional statement is used which can be defined as follows:
- In the if block, it checks "n" value, which is equal to 0. if the condition is true it will return value 1.
- In the else block, an integer variable x is defined that calls the method recursively and return x value.
