Answer:
B repeat a chunk of code until the condition is true im 88% sure
1.)
<span>((i <= n) && (a[i] == 0)) || (((i >= n) && (a[i-1] == 0))) </span>
<span>The expression will be true IF the first part is true, or if the first part is false and the second part is true. This is because || uses "short circuit" evaluation. If the first term is true, then the second term is *never even evaluated*. </span>
<span>For || the expression is true if *either* part is true, and for && the expression is true only if *both* parts are true. </span>
<span>a.) (i <= n) || (i >= n) </span>
<span>This means that either, or both, of these terms is true. This isn't sufficient to make the original term true. </span>
<span>b.) (a[i] == 0) && (a[i-1] == 0) </span>
<span>This means that both of these terms are true. We substitute. </span>
<span>((i <= n) && true) || (((i >= n) && true)) </span>
<span>Remember that && is true only if both parts are true. So if you have x && true, then the truth depends entirely on x. Thus x && true is the same as just x. The above predicate reduces to: </span>
<span>(i <= n) || (i >= n) </span>
<span>This is clearly always true. </span>
Answer:
- import java.util.Arrays;
- public class Main {
-
- public static void main(String[] args) {
- String [] first = {"David", "Mike", "Katie", "Lucy"};
- String [] middle = {"A", "B", "C", "D"};
- String [] names = makeNames(first, middle);
-
- System.out.println(Arrays.toString(names));
- }
-
- public static String [] makeNames(String [] array1, String [] array2){
-
- if(array1.length == 0){
- return array1;
- }
-
- if(array2.length == 0){
- return array2;
- }
-
- String [] newNames = new String[array1.length];
-
- for(int i=0; i < array1.length; i++){
- newNames[i] = array1[i] + " " + array2[i];
- }
-
- return newNames;
- }
- }
Explanation:
The solution code is written in Java.
Firstly, create the makeNames method by following the method signature as required by the question (Line 12). Check if any one the input string array is with size 0, return the another string array (Line 14 - 20). Else, create a string array, newNames (Line 22). Use a for loop to repeatedly concatenate the string from array1 with a single space " " and followed with the string from array2 and set it as item of the newNames array (Line 24-26). Lastly, return the newNames array (Line 28).
In the main program create two string array, first and middle, and pass the two arrays to the makeNames methods as arguments (Line 5-6). The returned array is assigned to names array (Line 7). Display the names array to terminal (Line 9) and we shall get the sample output: [David A, Mike B, Katie C, Lucy D]
Answer:
In a bumper-to-bumper traffic, when the engine starts overheating the situation can be handled by tapping the accelator which will revive the engine.
Explanation:
Overheating of engine can be due to many reasons. But one should know what to do when an engine overheats in a traffic. Bumper-to-bumper traffic is when the cars are so close in traffic that they touch each other. This usually happens when there's a traffic for a long time or on very busy lane. During summer times, it is important to keep checking the engine temperature to avoid any problem.
When one is stuck in bumper-to-bumper traffic with overheating engine, then there are some meausres that one can take. They are:
- To put the car on park or neutral mode of driving and tap the accelator which will revive the engine.
- The heat can be disperse by rolling down the window and turn the heater up. It will disperse the heat.
