Answer:too many words ahhh
Explanation:
(assuming jsx)
function Buttons (props) {
return(
{props.counterValue}
counter
increment
reset
);
}
var counterValue = 1;
function addup(a){
if(counterValue + a <= 20){
counterValue += a;
} else if (counterValue + a > 20){
//do nothing
}
ReactDOM.render(
,
document.getElementById('root')
);
}
function reset() {
counterValue = 1;
ReactDOM.render(
,
document.getElementById('root')
);
}
Answer:
C++.
Explanation:
#include <iostream>
using namespace std;
////////////////////////////////////////////////////////////////
void printPrime(int n) {
if (n > 0) {
cout<<2<<endl;
for (int i=3; i<=n; i++) {
bool isPrime = true;
for (int j=2; j<i; j++) {
if (i % j == 0) {
isPrime = false;
break;
}
}
if (isPrime == true)
cout<<i<<endl;
}
}
else {
cout<<"Invalid input";
}
}
int main() {
int n;
cout<<"Enter positive integer: ";
cin>>n;
printPrime(n);
return 0;
}
Answer:
an electronic device for storing and processing data, typically in binary form, according to instructions given to it in a variable program
Explanation:
Answer:
a divice to play games on, learn on, and other helpful things
Explanation:
B. Degree to which each user or group may interact with other users.
Answer: groups or users with permission to access the folder.
Explanation:
Properties dialog box simply contains the options that allow one to modify a document's look. To do this on the computer, one has to click on the 'Document Properties' which is located in the top-left side of the Document Panel. Then, one will select "Advanced Properties" option and the Properties dialog box will come up on the screen.
In the Properties dialog box for each folder, it should be noted that the top pane shows the groups or users with permission to access the folder while the bottom pane simply shows the degree of interactions of the user or group with the folder and the folder contents.
Answer:
Wireshark.
Explanation:
Wireshark is a free and open source network packet sniffing tool, used for analysing, troubleshooting, software and communication protocol development etc.
It uses various extensions to capture wireless and cabled connections like AirPcap (air packet capture), pcapng ( packet capture next generation.
It uses the device's antenna to capture wireless communication and gives the signal strength, noise ratio and other information of the antenna.
Answer:
#include <iostream>
using namespace std;
struct complex{//structure of comlex number..
double real;
double img;
};
complex add_complex(complex n1,complex n2)//add_complex functrion to add 2 complex numbers..
{
double a,b;//variables to hold the sum of real and imaginary.
complex sum;//result sum complex number.
a=n1.real+n2.real;//adding real parts.
b=n1.img+n2.img;//adding imaginary parts..
sum.real=a;//assinging values to sum.
sum.img=b;
return sum;//returning complex number sum.
}
int main()
{
complex c1,c2,sum;
double a,b;
cout<<"Enter values of c1"<<endl;
cin>>a>>b;
c1.real=a;
c1.img=b;
cout<<"Enter values of c2"<<endl;
cin>>a>>b;
c2.real=a;
c2.img=b;
sum=add_complex(c1,c2);
cout<<"The sum is : "<<sum.real<<" + i"<<sum.img<<endl;
return 0;
}
OUTPUT:-
Enter values of c1
8.5 7.2
Enter values of c2
4.5 3.9
The sum is : 13 + i11.1
Explanation:
First I have created a structure for complex number.Then created a function to add two complex numbers according to the question which returns a complex number.
Answer:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
double side, height;
cout<<"Enter the length of one of the sides of the base: ";
cin>>side;
cout<<"Enter the height: ";
cin>>height;
double area = side * side;
double volume = area * height / 3;
cout<<"The area of the base of the pyramid is: "<<area<<endl;
cout<<"The height of the pyramid is: "<<height<<endl;
cout<<"The volume of the pyramid is: "<<fixed<<setprecision(2)<<volume<<endl;
return 0;
}
Pseudocode:
Declare side, height
Get side
Get height
Set area = side * side
Set volume = area * height / 3
Print area
Print height
Print volume
Explanation:
Include <iomanip> to have two decimal places
Declare the side and height
Ask the user to enter side and height
Calculate the base area, multiply side by side
Calculate the volume using the given formula
Print area, height and volume (use fixed and setprecision(2) to have two decimal places)