Answer:
Distribution factor P = =38.33
V = 7.826 ml
Explanation:
given details:
BOD =230 mg/l
DO inital = 8.0mg/l
DO final = 2.0mg/l
we know
BOD = [DO inital -DO final] * distribution factor
230 = [8 - 2] D.F
Distribution factor P 
Distribution factor P = =38.33
THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is
distribution factor 
V = 7.826 ml
Answer:
See explaination
Explanation:
#include <iostream>
#include<string.h>
using namespace std;
bool isPalindrome(string str, int lower, int upper){
if(str.length() == 0 || lower>=upper){
return true;
}
else{
if(str.at(lower) == str.at(upper)){
return isPalindrome(str,lower+1,upper-1);
}
else{
return false;
}
}
}
int main(){
string input;
cout<<"Enter string: ";
cin>>input;
if(isPalindrome(input,0,input.length()-1)){
cout<<input<<" is a palindrome"<<endl;
}
else{
cout<<input<<" is NOT a palindrome"<<endl;
}
return 0;
}
Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
