<span>2. basketball
This is a classic case of overloading in C++. You have 2 functions, both named "printSport", but one of the functions receives an input of type double, and the other receives an input of type int. The specified method call passes a parameter of type int, so the version of printSport is called that receives a parameter of type int. And that version of printSport only prints the word "basketball". The other version of printSport is never called at all.</span>
Answer:
Explanation: The first thing to calculate is what IP addresses are used by the largest LAN. Because the LAN has 100 hosts, 7 bits must be left for host bits.
Explanation:
Explanation: The first thing to calculate is what IP addresses are used by the largest LAN. Because the LAN has 100 hosts, 7 bits must be left for host bits.
Answer:
There is no value of the number variable, for which the loop can be true in any iteration.
Explanation:
- The above loop states the condition that the value should be less than 100 and greater than 500. It is because the loop holds the and condition which gives the true if both conditions will be true.
- The first condition of the while loop states that the value of the number variable is less than the 100.
- The second condition of the while loop state that the value of the number variable is greater than the 500.
- The and condition of the while loop will true if both conditions will true.
- But there is no number which is less than 100 and greater than 500.
- So no number can satisfy the while condition to be true.
I'd go with command-line interface.
A CLI enables users to type commands in a console or a terminal window expressed as a sequence of characters and presses the enter key on the keyboard to execute that command. And in this case, Sarah is typing an “open document 3” command to the command-line interface to open a file in her computer.
Answer:
7 bytes
Explanation:
<u>2 Address Instruction</u>
The 2 address instruction consist 3 components in the format.
One is opcode,other two are addresses of destination and source.
<u>Example-</u>
load b,c | Opcode destination address,source address
add a,d | Opcode destination address,source address
sub c,f | Opcode destination address,source address
Opcode consists of 1 bytes whereas destination address and source address consist of 3 bytes each.
(1+3+3) bytes=7 bytes