It’s says the same thing to me too!
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.
Answer:
Explanation:
I will consult the following strategies:
Step 1: Being a discerning consultant I should be knowing what exactly has caused the drive to get corrupted.
Step 2: After examining the current state of the hard drive I will explain the actual problem to the client in a very lucid and eloquent manner whether it is feasible to retrieve the lost data or it is beyond repair.
Step 3: Identify a solution to the problem with things at my disposal(finding a right tool).
Step 4: apply the solution found to retrieve the data from the hard drive.
Step 5: submitting the retrieved data from the corrupt drive to the customer and advocating them the preventive measures for future reference.
The above five steps will work perfectly to resolve the issue faced if followed with great care and being heedful of the situation at hand.
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:
Number of strings = (10, 2)×(8,3) = 2520
Explanation:
The number of possible combinations for taking two 0's is C(10, 2)
It remains 8 Positions
The number of possible combinations for taking three 1's is C(8,3)
So there remains 5 spots
