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Muhammad al-xorazmiy nomidagi toshkent axborot texnologiyalari universiteti samarqand filiali kompyuter injiniringi fakulteti
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| Sana | 12.11.2023 | | Hajmi | 0.71 Mb. | | #97602 |
Bog'liq E2 Dasturlash 5 SINF2-mak.Axmedova.S, 2-MA\'RUZA TAQDIMOTI, sharti-oldindan-beriluvchi-sharti, namunaviy-dastur, 2-21 guruhi 1-simmestr, sherzod YUMB, YOQUBJON PAPKASI, Бошлангич-таълим-1-булим-туплам (1), ASOSIY kalit, ISH rejasining yuzi Norqizilov Mansur (2), 14-dars, TOBE TEST, o\'zbek tili, 1,2,3
O‘ZBEKISTON RESPUBLIKASIAXBOROT TEXNOLOGIYALARI VA KOMMUNIKATSIYALARINI RIVOJLANTIRISH VAZIRLIGI
MUHAMMAD AL-XORAZMIY NOMIDAGI
TOSHKENT AXBOROT TEXNOLOGIYALARI UNIVERSITETI SAMARQAND FILIALI
KOMPYUTER INJINIRINGI FAKULTETI
21-04-guruh
Nematova Elnura
2- MUSTAQIL ISHI
Tekshirdi: Mahkamova D
Bajardi: Nematova E
SAMARQAND 2023
#include
#include
using namespace std;
int main() {
double x = 0; // boshlang'ich qiymat
double y1, y2; // funksiya qiymatlari
y1 = sin(x); // funksiya qiymati x da
y2 = sin(x+1); // funksiya qiymati x+1 da
if (y1 == y2) { // ildizga ega bo'lsa
cout << "Funksiya X, X+1 oraliqda ildizga ega" << endl;
} else { // ildizga ega emas
cout << "Funksiya X, X+1 oraliqda ildizga ega emas" << endl;
}
return 0;
}
#include
using namespace std;
int main() {
int speed;
cout << "Tezlikni kiriting: ";
cin >> speed;
if (speed <= 30) {
cout << "You are following the traffic rules.\n";
} else {
cout << "You have violated the traffic rules.\n";
}
return 0;
}
#include
#include // sqrt() funksiyasi uchun kerak
using namespace std;
int main() {
int x1, y1, x2, y2, x3, y3;
cout << " birinchi nuqtaning koordinatalarini kiriting: ";
cin >> x1 >> y1;
cout << "ikkinchi nuqtaning koordinatalarini kiriting: ";
cin >> x2 >> y2;
cout << " uchunchi nuqtaning koordinatalarini kiriting: ";
cin >> x3 >> y3;
// Uchburchakning to'g'ri chiziqqa tushirilgan burchagi
double a = sqrt(pow(x2 - x1, 2) + pow(y2 - y1, 2));
double b = sqrt(pow(x3 - x2, 2) + pow(y3 - y2, 2));
double c = sqrt(pow(x1 - x3, 2) + pow(y1 - y3, 2));
// Perpendikulyar bo'lgan kesmalar uzunliklari
double d1, d2, d3;
if (a > b && a > c) {
d1 = b;
d2 = c;
d3 = a;
} else if (b > a && b > c) {
d1 = a;
d2 = c;
d3 = b;
} else {
d1 = a;
d2 = b;
d3 = c;
}
// Perpendikulyar bo'lganligini tekshirish
if (pow(d1, 2) + pow(d2, 2) == pow(d3, 2)) {
cout << "The line segment connecting the first and third points is perpendicular to the line segment connecting the second and third points.\n";
} else {
cout << "The line segment connecting the first and third points is not perpendicular to the line segment connecting the second and third points.\n";
}
return 0;
}
#include
using namespace std;
int main() {
double fuelEfficiency, distanceToGasStation, gasTankCapacity, fuelRemaining;
cout << " avtomobilning yoqilg'i samaradorligini kiriting (gallon uchun milda): ";
cin >> fuelEfficiency;
cout << " avtomobilning yoqilg'i samaradorligini kiriting (gallon uchun milda): ";
cin >> distanceToGasStation;
cout << " avtomobilning yoqilg'i samaradorligini kiriting (gallon uchun milda): ";
cin >> gasTankCapacity;
cout << " avtomobilning yoqilg'i samaradorligini kiriting (gallon uchun milda): ";
cin >> fuelRemaining;
double distancePossible = fuelEfficiency * gasTankCapacity; // Masofa, qancha benzin borligiga qarab
double distanceRemaining = distancePossible - distanceToGasStation; // Qolgan masofa, shaxobchaga yetish uchun
if (distanceRemaining <= 0) {
cout << "You will not be able to reach the gas station.\n";
} else {
if (distanceRemaining <= fuelRemaining * fuelEfficiency) {
cout << "You have enough fuel to reach the gas station.\n";
} else {
cout << "You do not have enough fuel to reach the gas station.\n";
}
}
return 0;
}
#include
#include
using namespace std;
int main() {
double a, b, c, d;
cout << "yon tomonning uzunligini kiriting a: ";
cin >> a;
cout << " yon tomonning uzunligini kiriting b: ";
cin >> b;
cout << " yon tomonning uzunligini kiriting c: ";
cin >> c;
cout << " yon tomonning uzunligini kiriting d: ";
cin >> d;
double diagonal1 = sqrt(pow(a, 2) + pow(b, 2)); // Birinchi diagonal uzunligi
double diagonal2 = sqrt(pow(c, 2) + pow(d, 2)); // Ikkinchi diagonal uzunligi
if (diagonal1 <= c && diagonal2 <= a) { // Agar birinchi diagonal ikkinchi tomonidan kichik yoki teng bo'lsa va ikkinchi diagonal birinchi tomonidan kichik yoki teng bo'lsa
cout << "The quadrilateral can be inscribed in a circle.\n";
} else {
cout << "The quadrilateral cannot be inscribed in a circle.\n";
}
return 0;
}
double S1, S2;
cout << "doira maydonini kirting: ";
cin >> S1;
cout << "kvadrat maydonini kiriting: ";
cin >> S2;
double a = sqrt(S2); // Kvadratning bir tomoni
double diagonal = sqrt((4 * S1) / M_PI); // Doira uzunligi
double r = diagonal / 2; // Doiraning radiusi
if (a <= 2 * r) { // Agar kvadratning har bir tomoni doiraning ichiga sigsa
cout << "Kvadrat doira ichiga yozilishi mumkin
.\n";
} else {
cout << "Kvadratni aylanaga yozib bo'lmaydi
.\n";
}
#include
#include
using namespace std;
int main() {
double x1, y1, x2, y2;
cout << "nuqtaning koordinatalarini kiriting (x1, y1
): ";
cin >> x1 >> y1;
cout << " nuqtaning koordinatalarini kiritingB (x2, y2): ";
cin >> x2 >> y2;
double distanceA = sqrt(pow(x1, 2) + pow(y1, 2)); // distance of point A from the origin
double distanceB = sqrt(pow(x2, 2) + pow(y2, 2)); // distance of point B from the origin
if (distanceA > distanceB) {
cout << "A nuqtasi boshlang'ichdan uzoqroq
.";
} else if (distanceB > distanceA) {
cout << "Point B is farther from the origin.";
} else {
cout << "Both points are equidistant from the origin.";
}
return 0;
}
#include
#include
using namespace std;
int main() {
double a, b, c;
cout << "tomonining uzunligini kiriting
a: ";
cin >> a;
cout << " tomonining uzunligini kiriting b: ";
cin >> b;
cout << " tomonining uzunligini kiriting c: ";
cin >> c;
// check if the sides can form a triangle
if (a + b > c && a + c > b && b + c > a) {
double perimeter = a + b + c;
double s = perimeter / 2; // half of the perimeter
double area = sqrt(s * (s - a) * (s - b) * (s - c)); // Heron's formula
cout << "Perimeter of the triangle is: " << perimeter << endl;
cout << "Area of the triangle is: " << area << endl;
} else {
cout << "These sides cannot form a triangle." << endl;
}
return 0;
}
#include
#include
using namespace std;
int main() {
int n;
cout << "o’zgaruvchilar sonini kiriting: ";
cin >> n;
vector> A(n, vector(n)); // matrix A
vector b(n); // vector b
// input matrix A
cout << "Enter the elements of matrix A: " << endl;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
cin >> A[i][j];
}
}
// input vector b
cout << "Enter the elements of vector b: " << endl;
for (int i = 0; i < n; i++) {
cin >> b[i];
}
// solve the system of equations using Gauss-Jordan elimination
for (int i = 0; i < n; i++) {
// pivot element
double pivot = A[i][i];
// divide the current row by the pivot element
for (int j = i; j < n; j++) {
A[i][j] /= pivot;
}
b[i] /= pivot;
// eliminate the elements below the pivot element
for (int k = i + 1; k < n; k++) {
double factor = A[k][i];
for (int j = i; j < n; j++) {
A[k][j] -= factor * A[i][j];
}
b[k] -= factor * b[i];
}
}
// back substitution to obtain the solution
vector x(n);
for (int i = n - 1; i >= 0; i--) {
x[i] = b[i];
for (int j = i + 1; j < n; j++) {
x[i] -= A[i][j] * x[j];
}
}
// output the solution
cout << "The solution of the system of equations is: ";
for (int i = 0; i < n; i++) {
cout << x[i] << " ";
}
cout << endl;
return 0;
}
#include
using namespace std;
int main() {
int age;
cout << "yoshingizni kiriting: ";
cin >> age;
if (age < 7) {
cout << "siz bolalar bog'chasidasiz." << endl;
} else if (age >= 7 && age < 18) {
cout << "siz talabasiz." << endl;
} else if (age >= 18 && age < 60) {
cout << "siz ishchisiz." << endl;
} else {
cout << "siz pensionersiz." << endl;
}
return 0;
}
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Muhammad al-xorazmiy nomidagi toshkent axborot texnologiyalari universiteti samarqand filiali kompyuter injiniringi fakulteti
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