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#1
inputstr1=input('Enter elements separated by ,(comma) for set1:')
inputstr2=input('Enter elements separated by ,(comma) for set2:')
inputt1=inputstr1.split(',')
inputt2=inputstr2.split(',')
inset1=set(inputt1)
inset2=set(inputt2)
print('Symmetric_difference of set1 and set2:',list(inset1.symmetric_difference(inset2)))
print('set1 ^ set2:',list(inset1^inset2))
inset1.symmetric_difference_update(inset2)
print('Sorted order of set1 after symmetric_difference_update:',sorted(list(inset1)))
print('Sorted order of set2 after symmetric_difference_update:',sorted(list(inset2)))
inputstr3=input('Enter elements separated by ,(comma) for set3:')
inputstr4=input('Enter elements separated by ,(comma) for set4:')
inputt3=inputstr3.split(',')
inputt4=inputstr4.split(',')
inset3=set(inputt3)
inset4=set(inputt4)
inset3^=inset4
print('Sorted order of set3 after (set3 ^= set4):',sorted(list(inset3)))
print('Sorted order of set4 after (set3 ^= set4):’,sorted(list(inset4)))
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Go to https://quillbot.com/ .here You can get the paraphaser of your required para with same meaning containing different words.
# remember that only for 150 words, you can get the paraphrase to work.
# because of subscription.Hope you make use of it.
a) Increased morbidity and mortality: Without adequate healthcare services, individuals are at a higher risk of developing preventable illnesses, experiencing complications from existing conditions, or even facing premature death. Lack of access to healthcare disproportionately affects vulnerable populations, such as low-income communities, marginalized groups, and those living in remote areas.
b) Limited disease prevention and early detection: Regular check-ups, vaccinations, and screenings are essential for preventing diseases and detecting potential health issues at early stages when treatment is more effective. A lack of access to healthcare hampers disease prevention efforts and delays diagnoses, resulting in more advanced diseases and poorer health outcomes.
c) Economic burden: When individuals lack access to healthcare, they often postpone or forgo necessary medical treatment. This can lead to increased healthcare costs in the long run due to the need for more extensive interventions or emergency care. Moreover, the economic burden falls not only on individuals but also on society as a whole.
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0/1000
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#include <stdio.h>
int main() {
int n, reversed = 0, remainder, original;
printf("Enter an integer: ");
scanf("%d", &n);
original = n;
// reversed integer is stored in reversed variable
while (n != 0) {
remainder = n % 10;
reversed = reversed * 10 + remainder;
n /= 10;
}
// palindrome if orignal and reversed are equal
if (original == reversed)
printf("%d is a palindrome.", original);
else
printf("%d is not a palindrome.", original);
return 0;
}
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void addLast(struct node **head, int val)
{
//create a new node
struct node *newNode = malloc(sizeof(struct node));
newNode->data = val;
newNode->next = NULL;
}
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#include <iostream>
#include <cmath>
using namespace std;
int main() {
// Declare R. P
double r;
double p = 3.14;
// Read R
cout << "Enter the Radius: ";
cin >> radius;
// Solve the Radius of a circle
// using radius formula in circle
radius = pi * pow(radius, 2);
double circle = 2 * pi * radius;
// Display R and C
cout << "Radius of Circle = " << radius << endl;
cout << "Circle Area = " << circle << endl;
return 0;
}
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#include <iostream>
#include <cmath>
using namespace std;
int main() {
// Declare R. P
double r;
double p = 3.14;
// Read R
cout << "Enter the Radius: ";
cin >> radius;
// Solve the Radius of a circle
// using radius formula in circle
radius = pi * pow(radius, 2);
double circle = 2 * pi * radius;
// Display R and C
cout << "Radius of Circle = " << radius << endl;
cout << "Circle Area = " << circle << endl;
return 0;
}
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#include <iostream>
#include <cmath>
using namespace std;
int main() {
// Declare R. P
double r;
double p = 3.14;
// Read R
cout << "Enter the Radius: ";
cin >> radius;
// Solve the Radius of a circle
// using radius formula in circle
radius = pi * pow(radius, 2);
double circle = 2 * pi * radius;
// Display R and C
cout << "Radius of Circle = " << radius << endl;
cout << "Circle Area = " << circle << endl;
return 0;
}
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start:
MOV R0,#0x4000
MOV R1,#0x00
MOV R2,#0x100
loopa:
LDRB R3,[R0]
TST R3,#0x01 //Check if the number is odd or even
BEQ next //If it is even, it will go to next
ADD R1,R1,#1 //If it is odd, R1+=1
next:
ADD R0,R0,#1 //Increment the memory address
SUBS R2,R2,#1 //Decrement the memory address
BNE loopa
MOV R0,#0x2000
STRB R1,[R0] //Save the calculated data
loop:
B loopb
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public class City {
public String name;
public City(String n) {
name = n;
}
}
public class Nation {
public String name;
public City capital;
public int population;
public Nation(String name, City capital, int population) {
this.name = name;
this.capital = capital;
this.population = population;
}
}
public class Question2Main {
public static void main(String args[]) {
City city = new City("Rome");
Nation nation = new Nation("Italy", city, 60000000);
System.out.println(nation.name + "␣has␣" + city.name + "␣as␣its␣capital"
);
}
}
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%1.Parameters
promptEf='Value of Ef=';
promptvf='Value of vf=';
promptEm='Value of Em=';
promptvm='Value of vm=';
promptVf='Value of Vf=';
promptVm='Value of Vm=';
promptGf='Value of Gf=';
promptGm='Value of Gm=';
Ef=input(promptEf);
vf=input(promptvf);
Em=input(promptEm);
vm=input(promptvm);
Vf=input(promptVf);
Vm=input(promptVm);
Gf=input(promptGf);
Gm=input(promptGm);
E11=Ef*Vf+Em*Vm;
v12=vf*Vf+vm*Vm;
E22=Ef*Em/(Ef*Vm+Em*Vf);
v21=E22*v12/E11;
G12=Gf*Gm/(Gf*Vm+Gm*Vf);
%2&3-Number of layers,Thickness
prompt_thickness='layer thickness=';
t=input(prompt_thickness);
promptN = 'Number of layers=';
N=input(promptN);
orientation=zeros(1,N);
fprintf('Enter values of layer orientation from bottom to top\n')
%Orientation of fibers
for i=1:N
fprintf('Angle of fibers in layers %i\n',i);
prompt_theta='Theta=';
theta=input(prompt_theta);
orientation(1,i)=theta;
end
disp(orientation);
%4.Qij matrix
Q11=E11/(1-v12*v21);
Q22=E22/(1-v12*v21);
Q12=v12*E22/(1-v12*v21);
Q21=v21*E11/(1-v12*v21);
Q66=G12;
Q=[Q11 Q12 0;Q21 Q22 0;0 0 Q66];
fprintf('Matrix Q=\n');
disp(Q);
R=[1 0 0;0 1 0;0 0 2];
Rinv=inv(R);
%loop starting for different layers
m=cell(N,1);
for n=1:N
c=cosd(orientation(1,n));%degree
s=sind(orientation(1,n));%degree
%Transformation matrix
a11=c.^2; a12=s.^2; a13=2*c*s;
a21=s.^2; a22=c.^2; a23=(-2*c*s);
a31=(-c*s); a32=c*s; a33=c.^2-s.^2;
T=[a11 a12 a13;a21 a22 a23;a31 a32 a33];
Tinv=inv(T);
%5.Qij-bar matrix
Q_bar=Tinv*Q*R*T*Rinv;
%Q_bar=R*inv(T)*Rinv*Q*T;
fprintf('Q_bar Matrix=\n');
disp(Q_bar);
m{n}=Q_bar;
end
%6.z values
z=zeros(1,(N+1));
h=N*t;
for j=1:N+1
z(1,j)=h/2-t*(j-1);
end
%disp(z);
aa=0;
bb=0;
dd=0;
A=zeros(3,3);
B=zeros(3,3);
D=zeros(3,3);
for yy=1:3
for zz=1:3
for xx=1:n
aa=aa+(m{xx}(yy,zz))*(z(1,(xx))-z(1,xx+1));
bb=bb+(m{xx}(yy,zz))*(z(1,(xx))^2-(z(1,xx+1))^2);
dd=dd+(m{xx}(yy,zz))*(z(1,(xx))^3-(z(1,xx+1))^3);
end
A(zz,yy)=aa;
B(zz,yy)=0.5*bb;
D(zz,yy)=(1/3)*dd;
aa=0;
bb=0;
dd=0;
end
end
%7. ABD Matrix
ABBD=[A(1,1) A(1,2) A(1,3) B(1,1) B(1,2) B(1,3);
A(2,1) A(2,2) A(2,3) B(2,1) B(2,2) B(2,3);
A(3,1) A(3,2) A(3,3) B(3,1) B(3,2) B(3,3);
B(1,1) B(1,2) B(1,3) D(1,1) D(1,2) D(1,3);
B(2,1) B(2,2) B(2,3) D(2,1) D(2,2) D(2,3);
B(3,1) B(3,2) B(3,3) D(3,1) D(3,2) D(3,3)];
fprintf('Matrix ABBD= \n');
disp(ABBD);
A = [A(1,1) A(1,2) A(1,3);A(2,1) A(2,2) A(2,3);A(3,1) A(3,2) A(3,3)];
B = [B(1,1) B(1,2) B(1,3);B(2,1) B(2,2) B(2,3);B(3,1) B(3,2) B(3,3)];
D = [D(1,1) D(1,2) D(1,3); D(2,1) D(2,2) D(2,3);D(3,1) D(3,2) D(3,3)];
Ainv = inv(A);
Bx = (B)*(Ainv)*(B);
Dstar = (D)- Bx;
D1 = inv(Dstar);
B1 = -(Ainv)*(B)*(D1);
C1 = -(D1)*(B)*(Ainv);
A1 = (Ainv)+((Ainv)*(B)*(Dstar)*(B)*(Ainv));
fprintf ('matrix A1= \n');
disp (A1);
fprintf ('matrix B1= \n');
disp (B1);
fprintf ('matrix C1= \n');
disp (C1);
fprintf ('matrix D1= \n');
disp (D1);