CMYK does not include a white color because it is assumed that it will be printed on a white paper and depending on the percentage of each color that is used, the white from the paper will be used to fill the space, therefore making the shades appear lighter. If it's something that will only be seen digitally, use RGB.
There are alot of parts in a computer system. You have to be a little specific
Answer:
Option C. The path from start to finish that passes through all the tasks that are critical to completing the project in the shortest amount of time
is the correct answer.
Explanation:
- Critical path is the term used in Project management.
- It can be known by determining the longest stretch for dependent activities and the time required for them to complete.
- A method known as "critical path method" is used for Project modelling.
- All types of projects including aerospace, software projects, engineering or plant maintenance need Critical method analysis for the modeling of project and estimation.
I hope it will help you!
Answer:
#include <iostream>
using namespace std;
class DogLicense{
public:
void SetYear(int yearRegistered);
void CreateLicenseNum(int customID);
int GetLicenseNum() const;
private:
int licenseYear;
int licenseNum;
};
void DogLicense::SetYear(int yearRegistered) {
licenseYear = yearRegistered;
}
void DogLicense::CreateLicenseNum(int customID) {
licenseNum = (100000 * customID) + licenseYear;
}
int DogLicense::GetLicenseNum() const {
return licenseNum;
}
int main() {
DogLicense dog1;
dog1.SetYear(2014);
dog1.CreateLicenseNum(777);
cout << "Dog license: " << dog1.GetLicenseNum() << endl;
return 0;
}
Explanation:
You can see the whole code above, but let me explain the fixed function.
void DogLicense::CreateLicenseNum(int customID) {
licenseNum = (100000 * customID) + licenseYear;
}
The function header is already declared in the class. It takes <em>customID</em> as a parameter. To find out the <em>lisenseNum</em>, we need to apply the given formula <em><u>(100000 * customID) + licenseYear</u></em>.
