//=indicating you to do the programming part on your own relating to the description provided against. This done because different programming languages require different coding for that.
n=integer value
n1=dummy storage for n
r=variable used to do the function
{
int n,r,n1,rev=0;
//do the coding here for storing the integer in the variable n
n1=n;
while(n>0){
r=n%10;
rev=(rev*10)+r;
n=n/10;
}
//now add a command for displaying the value of rev
}
this is just a logic i used for java
done.
Answer:
public class Main
{
public static void main(String[] args) {
Main m=new Main();
System.out.println(m.mymath(true,5,2)); // calling the function mymath
}
public int mymath(boolean a,int b,int c) // mymath function definition
{
if(a==true)
{
int d=b+c;
return d;
}
else{int e=b-c;
return e;
}
}
}
Explanation:
because they only function with that one the gates also prevent it from being able to use it
Technological improvements allow me to have greater access to goods around the world. I can buy instantly and communicate instantly with producers. Technology can also help me to monitor economic trends, both in my country and in my own life. In summary, technology helps to give me more freedom to make economic choices.
On e2020
Complete Question:
Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 4096-byte cache using 32-bit memory addresses, 8-byte cache blocks and a 8-way associative design. The cache has :
Cache size = 1024 bytes, sets t = 26.8, tag bits, s = 3.2, set index bit =2
Answer:
Check below for explanations
Explanation:
Cache size = 4096 bytes = 2¹² bytes
Memory address bit = 32
Block size = 8 bytes = 2³ bytes
Cache line = (cache size)/(Block size)
Cache line = 
Cache line = 2⁹
Block offset = 3 (From 2³)
Tag = (Memory address bit - block offset - Cache line bit)
Tag = (32 - 3 - 9)
Tag = 20
Total number of sets = 2⁹ = 512
