Answer:
The external business environment consists of economic, political and legal, demographic, social, competitive, global, and technological sectors.
The most appropriate change in this situation would be to change the message "User input deemed invalid" to "You made a mistake, try again, you can".
<h3>What should we change to improve the children's experience?</h3>
To improve children's experience with this new calculator program, we must adapt all the features of the program for children's users. Therefore, the buttons, the messages and everything related to this program must be suitable for their motor and brain development.
Therefore, it is considered that the message "User input deemed invalid" is not suitable for children because they may not understand this message or misinterpret it and desist from using the new program.
So, the most appropriate message to replace the previous message would be:
- "You made a mistake, try again, you can"
Because this message is suitable for the friendly language that children use in their daily lives and they would better understand the message without giving up using the new program or the calculator.
Learn more about programs in: brainly.com/question/13264074
#SPJ1
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
Answer:
Explanation:
The following code is written in Java. It is a method that calculates the square root of a number as requested. The method first checks with an IF statement if the parameter value is a positive number and then calculates the square root and prints it to the screen. Otherwise, it prints Number must not be negative. A test case has been provided in the main method and the output can be seen in the attached image below.
import java.util.Scanner;
class Brainly {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.print("Enter a number of type double to calculate square root:");
double num = in.nextDouble();
rootPositive(num);
}
public static void rootPositive(double num) {
if (num > 0) {
System.out.println(Math.sqrt(num));
} else {
System.out.println("Number must not be negative.");
}
}
}
