Answer: Even though the hardware is inexpensive the writing of program is not efficient through this method as proper development of program is necessary for the clear execution due to factors like:-
- The facility of writing program even the cost of hardware is less but it is not a free facility.
- It also has a slower processing for the execution of the program
- The construction of the efficient program is necessary for the compilation and execution of it rather than poorly constructed program is worthless and inefficient in working.
Yes very much so! You could learn previous mistakes/bugs/etc and find solutions to fixing them and avoiding getting them. As well as much more. History is always a fun subject for anything, really in my opinion.
It’s a piece of hardware (you can touch it) that changes something on the screen. Examples are a keyboard putting letters on the screen or a mouse moving the cursor
Answer:
The output is 24
Explanation:
Given
The above code segment
Required
The output
We have (on the first line):
On the second line:
Substitute the value of each variable
Solve the inner brackets
8%3 implies that, the remainder when 8 is divided by 3.
The remainder is 2
So:
<em>Hence, the output is 24</em>
Answer:
The answer to this question is given below in the explanation section.
Explanation:
First, we need to convert these hexadecimal numbers into decimal numbers, then we can easily identify which one is the lowest hexadecimal.
The hexadecimal numbers are F2, 81, 3C, and 39.
F2 = (F2)₁₆ = (15 × 16¹) + (2 × 16⁰) = (242)₁₀
81 = (81)₁₆ = (8 × 16¹) + (1 × 16⁰) = (129)₁₀
3C = (3C)₁₆ = (3 × 16¹) + (12 × 16⁰) = (60)₁₀
39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀
The 39 is the lowest hexadecimal number among the given numbers.
Because 39 hex is equal to 57 decimal.
39 = (39)₁₆ = (3 × 16¹) + (9 × 16⁰) = (57)₁₀
