A server can have many definitions, but in this particular case, the answer is A.
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
D. white with a hue
HOPE THIS HELPS
Answer:
The answer to this question is given below in the explanation section.
Explanation:
The value stored by a variable can be changed after it is assigned(true).
The value of a variable can be changed after it is assigned, for example:
int a=10;
and we can change the value of variable a in letter program such as:
a=15;
Variables are a name for a spot in the computer's memory (true).
it is true, because the variables value stored in the computer's memory and we can access theses values by their name (variable name). so Variables are a name for a spot in the computer's memory.
Variable names can be words: such as temperature or height (true).
Yes, the variable name can be words such as height, width, temperature etc.
The value stored by a variable cannot be changed after it is assigned (false).
It is noted that the value stored by a variable can be changed after it is assigned. However, it is noted that is some programming language, you can't change the value of static variable.
Answer:
If all the character pairs match after processing both strings, one string in stack and the other in queue, then this means one string is the reverse of the other.
Explanation:
Lets take an example of two strings abc and cba which are reverse of each other.
string1 = abc
string2 = cba
Now push the characters of string1 in stack. Stack is a LIFO (last in first out) data structure which means the character pushed in the last in stack is popped first.
Push abc each character on a stack in the following order.
c
b
a
Now add each character of string2 in queue. Queue is a FIFO (first in first out) data structure which means the character inserted first is removed first.
Insert cba each character on a stack in the following order.
a b c
First c is added to queue then b and then a.
Now lets pop one character from the stack and remove one character from queue and compare each pair of characters of both the strings to each other.
First from stack c is popped as per LIFO and c is removed from queue as per FIFO. Then these two characters are compared. They both match
c=c. Next b is popped from stack and b is removed from queue and these characters match too. At the end a is popped from the stack and a is removed from queue and they both are compared. They too match which shows that string1 and string2 which are reverse of each other are matched.
