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.
True, because if you need money on the go because you dont know where you are going, so you might need the money.
Explanation:
<h2>
<em><u>Edge computing with 5G creates tremendous opportunities in every industry. It brings computation and data storage closer to where data is generated, enabling better data control, reduced costs, faster insights and actions, and continuous operations. In fact, by 2025, 75% of enterprise data will be processed at the edge, compared to only 10% today.¹</u></em></h2><h2 /><h2>
<em><u>Edge computing with 5G creates tremendous opportunities in every industry. It brings computation and data storage closer to where data is generated, enabling better data control, reduced costs, faster insights and actions, and continuous operations. In fact, by 2025, 75% of enterprise data will be processed at the edge, compared to only 10% today.¹IBM provides an autonomous management offering that addresses the scale, variability and rate of change in edge environments. IBM also offers solutions to help communications companies modernize their networks and deliver new services at the edge.</u></em></h2>
Hope this answer is helpful
Here you go,
import java.util.Scanner;
import java.util.Random;
public class OrderCalculator{
public static void main(String[] args){
float x, y, z, semi, area;
Scanner in = new Scanner(System.in);
System.out.print("Enter the 3 sides: ");
x = in.nextFloat();
y = in.nextFloat();
z = in.nextFloat();
semi = (float) ((x + y + z) / 2.0);
area = (float) Math.sqrt(semi * (semi - x) * (semi - y) * (semi - z));
System.out.printf("The area is: %.3f\n", area);
}
}
To debug a code, means that we locate and fix the errors in a code.
The issue with your code is that:
<em>You did not convert num_owls_A and num_owls_B to integers, when adding them together.</em>
The fix to this is that:
<em>You need to convert num_owls_A and num_owls_B to integers, when adding them together.</em>
The fix is as follows:
<em>total_owls = int(num_owls_A) + int(num_owls_B)</em>
The updated code is as follows:
<em>total_owls = 0
</em>
<em>num_owls_A = input()
</em>
<em>num_owls_B = input()
</em>
<em>total_owls = int(num_owls_A) + int(num_owls_B)
</em>
<em>print('Number of owls:', total_owls)</em>
<em />
<em>The above code will perform addition operations for all inputs</em>
Read more about Python programs at:
brainly.com/question/13246781