num1 = float(input("Enter the first number: "))
num2 = float(input("Enter the second number: "))
operation = input("Which operation are you performing? (a/s/m/d) ")
if operation == "a":
print("{} + {} = {}".format(num1, num2, num1+num2))
elif operation == "s":
print("{} - {} = {}".format(num1, num2, num1-num2))
elif operation == "m":
print("{} * {} = {}".format(num1, num2, num1*num2))
elif operation == "d":
print("{} / {} = {}".format(num1, num2, num1/num2))
I hope this helps!
Answer:
Explanation:
260 cost units, Big O(n) complexity for a push
Answer:
The solution code is written in Java
- public static void checkCommonValues(int arr1[], int arr2[]){
- if(arr1.length < arr2.length){
- for(int i = 0; i < arr1.length; i++){
- for(int j = 0; j < arr2.length; j++){
- if(arr1[i] == arr2[j]){
- System.out.print(arr1[i] + " ");
- }
- }
- }
- }
- else{
- for(int i = 0; i < arr2.length; i++){
- for(int j = 0; j < arr1.length; j++){
- if(arr2[i] == arr1[j]){
- System.out.print(arr2[i] + " ");
- }
- }
- }
- }
- }
Explanation:
The key idea of this method is to repeated get a value from the shorter array to check against the all the values from a longer array. If any comparison result in True, the program shall display the integer.
Based on this idea, an if-else condition is defined (Line 2). Outer loop will traverse through the shorter array (Line 3, 12) and the inner loop will traverse the longer array (Line 4, 13). Within the inner loop, there is another if condition to check if the current value is equal to any value in the longer array, if so, print the common value (Line 5-7, 14-16).
By looking at your code, it seems like you're trying to let the user enter positive numbers until the user enters a negative number. To achieve this, you need to indent int(input("Enter a number, negative to stop")) inside the loop.
The second answer choice is correct.
Answer:
yes they are: Central Processor Unit (CPU)
Memory (RAM)
Input (keyboard, mouse, etc)
Output (monitor, printer, etc)
Explanation:
True
