Answer: A #3f107f.
Letter B would produce a lighter blue-purple color rather than a darker shade of purple. Letter C would produce a very light purple color. Letter D would produce a Black color, making it very dark and not in the range of being purple.
Letter A would produce a darker shade of purple. As explained in the way that the RGB color hexadecimal uses, the amount of each respective color is 2 digits for each color Red, Green, Blue respectively. By reducing the amount of each color in the RGB mode, the output will become a darker shade as the RGB mode is an ADDITITIVE type of color mode.
Answer:
> we8thereRatings[1:5,]
Food Service Value Atmosphere Overall
1 5 5 5 5 5
2 5 5 5 5 5
5 5 5 4 4 5
11 5 5 5 5 5
12 5 5 5 4 5
Answer:
c. decDiscount = GetDiscount(decSales, decRate)
Explanation:
Option a. is incorrect because it is using Call word which is not a valid way to invoke a function.
Similarly option b. is also incorrect because it uses Call word to invoke function GetDiscount() which is not a valid way to call a function and also it is passing it the contents of three variables decSales, decRate and decDiscount and as mentioned in the question only two parameters are to be passed to GetDiscount() function.
Option c. is correct as it invokes the function GetDiscount() and passes it the contents of two variables decSales and decRate and assigns this to a variable decDiscount. For example if the GetDiscount() method has to calculate the discount using decSales and decRate then the resultant value of this computation is assigned to decDiscount. So whatever this function returns or computers is assigned to and stored in decDiscount variable. So this is a valid way to invoke a method.
Answer:
Online issue tracking system is a computer software solution for managing issues lists. The lists of issues are defined by the organization activity field.The best example of online issue tracking software is a customer support call center. ... A Bug Tracker is also one of possible online issue tracking software
Explanation:
the answer has the explanation needed.
Answer:
for (char outerChar='a'; outerChar<='e'; outerChar++){
for (char innerChar='a'; innerChar<='e'; innerChar++){
cout << outerChar << innerChar << "\n";
}
}
