Answer:
is this overdue
Explanation:
it probly is you are smart you can do this
Answer:
See the explanation please
Explanation:
A.
The algorithm tries to subtract the number of credits taken so far and the required credits for the degree before getting the number of credits required for the degree program and the number of credits the student has taken so far. Since the values are not known, calculation cannot be done.
Write the step 3 after step 5
B.
Step 1, 2, 4, 5 require user interaction, since it is asking the user to enter values
Two:
<span>Hiding the complexities of hardware from the user.
Managing between the hardware's resources which include the processors, memory, data storage and I/O devices.
Handling "interrupts" generated by the I/O controllers.
<span>Sharing of I/O between many programs using the CPU.
Three:
</span></span><span>Desktops
icons
menus
windows.</span><span>
Seven:
</span><span>The convention is that ellipses following a menu item usually means that the menu item will open a dialog box with further choices, rather than immediately carrying out an action.
</span>
Eight:
Organize Your Files
Nine:
Use the save as Command
name the File
Ten:
Open the file you want to delete
Select the file you want to delete
select Delete from the file menu
Confirm the deletion
Hope this helps
Answer:
Explanation:
The following code is written in Python. It creates a method for each one of the questions asked and then tests all three with the same test case which can be seen in the picture attached below.
def alternating_list(lst1, lst2):
lst3 = []
for x in range(len(lst1)):
lst3.append(lst1[x])
try:
lst3.append(lst2[x])
except:
pass
if len(lst2) > len(lst1):
lst3.extend(lst2[len(lst1):])
return lst3
def reverse_alternating(lst1, lst2):
lst3 = []
if len(lst1) == len(lst2):
for x in range(len(lst1) - 1, -1, -1):
lst3.append(lst1[x])
lst3.append(lst2[x])
return lst3
def alternating_list_no_extra(lst1, lst2):
lst3 = []
max = 0
if len(lst1) > len(lst2):
max = len(lst2)
else:
max = len(lst1)
for x in range(max):
lst3.append(lst1[x])
try:
lst3.append(lst2[x])
except:
pass
return lst3
Information can be transferred and passed way easier.
