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Question:
Write one for loop to print out each element of the list several_things. Then, write another for loop to print out the TYPE of each element of the list several_things.
Answer:
The solution in python is as follows:
for element in several_things:
print(element)
for element in several_things:
print(type(element))
Explanation:
The solution assumes that the list several_things has already been initialized.
So, the rest of the code is explained as follows:
This line iterates through the list, several_things
for element in several_things:
This line prints each element
print(element)
This line iterates through the list, several_things for the second time
for element in several_things:
This line prints the type of each element
print(type(element))
Answer:
Explanation:
The following code is written in Java. It creates the raiseToPower method that takes in two int parameters. It then uses recursion to calculate the value of the first parameter raised to the power of the second parameter. Three test cases have been provided in the main method of the program and the output can be seen in the attached image below.
class Brainly {
public static void main(String[] args) {
System.out.println("Base 5, Exponent 3: " + raiseToPower(5,3));
System.out.println("Base 2, Exponent 7: " + raiseToPower(2,7));
System.out.println("Base 5, Exponent 9: " + raiseToPower(5,9));
}
public static int raiseToPower(int base, int exponent) {
if (exponent == 0) {
return 1;
} else if (exponent == 1) {
return base;
} else {
return (base * raiseToPower(base, exponent-1));
}
}
}
Answer:
B. {1, 2, 2, 3, 3, 4, 5}
Explanation:
Given
The above code segment
Required
Determine which list does not work
The list that didn't work is 
Considering options (A) to (E), we notice that only list B has consecutive duplicate numbers i.e. 2,2 and 3,3
All other list do not have consecutive duplicate numbers
Option B can be represented as:
![nums[0] = 1](https://tex.z-dn.net/?f=nums%5B0%5D%20%3D%201)
![nums[1] = 2](https://tex.z-dn.net/?f=nums%5B1%5D%20%3D%202)
![nums[2] = 2](https://tex.z-dn.net/?f=nums%5B2%5D%20%3D%202)
![nums[3] = 3](https://tex.z-dn.net/?f=nums%5B3%5D%20%3D%203)
![nums[4] = 3](https://tex.z-dn.net/?f=nums%5B4%5D%20%3D%203)
![nums[5] = 4](https://tex.z-dn.net/?f=nums%5B5%5D%20%3D%204)
![nums[6] = 5](https://tex.z-dn.net/?f=nums%5B6%5D%20%3D%205)
if (nums.get(j).equals(nums.get(j + 1)))
The above if condition checks for duplicate numbers.
In (B), when the elements at index 1 and 2 (i.e. 2 and 2) are compared, one of the 2's is removed and the Arraylist becomes:
![nums[0] = 1](https://tex.z-dn.net/?f=nums%5B0%5D%20%3D%201)
![nums[1] = 2](https://tex.z-dn.net/?f=nums%5B1%5D%20%3D%202)
![nums[2] = 3](https://tex.z-dn.net/?f=nums%5B2%5D%20%3D%203)
![nums[3] = 3](https://tex.z-dn.net/?f=nums%5B3%5D%20%3D%203)
![nums[4] = 4](https://tex.z-dn.net/?f=nums%5B4%5D%20%3D%204)
![nums[5] = 5](https://tex.z-dn.net/?f=nums%5B5%5D%20%3D%205)
The next comparison is: index 3 and 4. Meaning that comparison of index 2 and 3 has been skipped.
<em>This is so because of the way the if statement is constructed.</em>
DDR memory is 64-bit so that means that it is 64