Answer:
Big Oh notation is used to asymptotically bound the growth of running time above and below the constant factor.
Big Oh notation is used to describe time complexity, execution time of an algorithm.
Big Oh describes the worst case to describe time complexity.
For the equation; T(N) = 10000*N + 0.00001*N^3.
To calculate first of all discard all th constants.
And therefore; worst case is the O(N^3).
<span>BCD only goes from digit 0 (0000) to digit 9 (1001), because for 10 you need two digits, so all you've got to do is make a function that produces high for numbers from 10 (1010) to 15 (1111) as follows:
A3 A2 A1 A0 F
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
...........................
1 0 0 0 0
1 0 1 0 1
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
Then simplify the function:
F = A3*A2 + A3*A1
Finally just draw or connect the circuit using NAND</span>
Answer:
False
Explanation:
The private member of a class is not accessible by using the Dot notation ,however the private member are those which are not accessible inside the class they are accessible outside the class .The public member are accessible inside the class so they are accessible by using the dot operator .
<u>Following are the example is given below in C++ Language </u>
#include<iostream> // header file
using namespace std;
class Rectangle
{
private:
double r; // private member
public:
double area()
{ return 3.14*r*r;
}
};
int main()
{
Rectangle r1;// creating the object
r1.r = 3.5;
double t= r1.area(); // calling
cout<<" Area is:"<<t;
return 0;
}
Output:
compile time error is generated
<u>The correct program to access the private member of class is given below </u>
#include<iostream> // header file
using namespace std;
class Rectangle
{
private:
double r; // private member
public:
double area()
{
r1=r;
double t2=3.14*r2*r2;
return(t2); // return the value
}
};
int main()
{
Rectangle r1;// creating the object
r1.r = 1.5;
double t= r1.area(); // calling
cout<<" Area is:"<<t;
return 0;
}
Therefore the given statement is False
Answer:
Hardware.
Explanation:
All electronics devices like computer servers, DVDs, smartphones and cell phones, wrist watches etc, comes with two main components, they are hardware and software.
The hardware components are the physical part of the device, while the software component is the written instructions configured to the hardware for executing tasks.
Cleaning the motherboard in a computer system, fixing a DVD, servicing other internal components of devices are all hardware maintenance and repair services.
Answer:g
public static int addOddMinusEven(int start, int end){
int odd =0;
int even = 0;
for(int i =start; i<end; i++){
if(i%2==0){
even = even+i;
}
else{
odd = odd+i;
}
}
return odd-even;
}
}
Explanation:
Using Java programming language:
- The method addOddMinusEven() is created to accept two parameters of ints start and end
- Using a for loop statement we iterate from start to end but not including end
- Using a modulos operator we check for even and odds
- The method then returns odd-even
- See below a complete method with a call to the method addOddMinusEven()
public class num13 {
public static void main(String[] args) {
int start = 2;
int stop = 10;
System.out.println(addOddMinusEven(start,stop));
}
public static int addOddMinusEven(int start, int end){
int odd =0;
int even = 0;
for(int i =start; i<end; i++){
if(i%2==0){
even = even+i;
}
else{
odd = odd+i;
}
}
return odd-even;
}
}
