//7 задача
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
	cout << 7 +(( 5 * 3) / 2) - 4 << '\n';
	cout << (12 % 3) + (3 *3) – (4 /5) << '\n';
	cout << (4 * 5 * (5 + (6 * 2 / (6)))) ;
	return 0;

}
// 8 задача
#include <iostream>
#include <cmath>
#include <math.h>
#define PI 3.14

using namespace std;
int main()
{
	double a, b, c, d, e, f, g, h;

	a = 2.0 / 7.0;
	b = 1.0 / 3.0 - 0.1;
	c = 5.0 * 4.0 * 3.0 * 2.0;
	d = 74.0;
	e = PI;
	f = -34.208 * pow(10, -4);
	g = (int)pow(-2, 4);
	h = sqrt(2);

	return 0;
}

//9 задача

#include <iostream>
#include <cmath>
#include <math.h>

using namespace std;

int main() {

	double a = 1;
	double b = 2;
	double c = 3;
	double d = 4;
	double x = 5;
	double e = 6;
	double y = 7;

	double exampleOne = pow(a, 2) + pow(b, 2) + pow(c, 2);
	double exampleTwo = (a + b) / (c + d);
	double exampleThree = ((a + b) / c) * d;
	double exampleFour = (a + b) / (c * d);
	double exampleFive = (1 + (2 * 1) + (3 * 2 * 1)) / (1 + (1 / (2 * 1) + (1 / (3 * 2 * 1))));
	double exampleSix = (sqrt(2) + sqrt(3) + sqrt(4)) / (sqrt(5) + sqrt(6) + sqrt(7));
	double exampleSeven = sin(x) + cos(x) - (pow(tan(x) + (1 / (tan(x))), 3) / log(2 + pow(x, 4)));
	double exampleEight = log(5) / log(2) + log(7) / log(3) + log(9) / log(5);
	double exampleNinth = pow(log(0) * abs(x) + pow(e, x - 1.0), 3.0) / log(2.0 + pow(e, (x + y) / 2));

	cout << exampleOne << endl;
	cout << exampleTwo << endl;
	cout << exampleThree << endl;
	cout << exampleFour<< endl;
	cout << exampleFive << endl;
	cout << exampleSix << endl;
	cout << exampleSeven << endl;
	cout << exampleEight << endl;
       cout << exampleNine << endl;
}

10 задача

//a
#include <iostream>
#include <cmath>
#include <math.h>

using namespace std;


int main()
{

	int a;
	cin >> a;

	bool isDevidable = true;
	bool isNot = false;

	if (a % 3 == 0 && a % 5 == 0 || a % 2 == 0 && a % 7 == 0)
	{
		cout << isDevidable << endl;
	}
	else
	{
		cout << isNot << endl;
	}

return 0;









}
	//й
#include <iostream>
#include <cmath>
#include <math.h>

using namespace std;


int main()
{

	//й
	int a, b, c;
	cin >> a;
	cin >> b;
	cin >> c;

	bool Positive = true;
	bool Negative = false;

	if (a > 0 && b > 0 && c > 0)
	{
		cout << Positive << endl;

	}
	else
	{
		cout << Negative << endl;
	}

	return 0;
}

11 задача

#include <iostream>
using namespace std;

int main() {
	float radius, area_circle;


	cout << "Enter the radius of circle: ";
	cin >> radius;

	area_circle = 3.14 * radius * radius;
	cout << "Area of circle: " << area_circle << endl;

	return 0;
}

12 задача

#include <iostream>
#include <math.h>
using namespace std;

int main() {
	float a, peri, area;
	cout << "Enter value of a : ";
	cin >> a;

	peri = a  * 3;
	cout << "Perimeter of Triangle : " << peri << endl;

	area = (pow(a, 2) * sqrt(3)) / 4;
	cout << "The area of the triangle is: " << area<< endl;

	return 0;
}

13 задача

#include <iostream>
#include <cmath>
#include <math.h>
#define PI 3.14

using namespace std;

int main()
{
	double a, b, c;
	double r, C, S, p;

	double Xa = 1;
	double Xb = 2;
	double Xc = 3;
	double Ya = 1;
	double Yb = 2;
	double Yc = 3;

	c = sqrt((pow(Xa, 2) - (2 * Xa * Xb) + pow(Xb, 2)) + (pow(Ya, 2) - (2 * Ya * Yb) + pow(Yb, 2)));
	b = sqrt((pow(Xc, 2) - (2 * Xa * Xc) + pow(Xa, 2)) + (pow(Ya, 2) - (2 * Ya * Yc) + pow(Yc, 2)));
	a = sqrt((pow(Xc, 2) - (2 * Xc * Xb) + pow(Xb, 2)) + (pow(Yc, 2) - (2 * Yc * Yb) + pow(Yb, 2)));

	p = (a + b + c) / 2;

	S = sqrt(p * (p - a) * (p - b) * (p - c));
	r = (a * b * c) / (4 * S);

	C = 2 * PI * r;
	S = PI * pow(r, 2);

	return 0;
}

\\14 задача а
#include <iostream>
#include <cmath>
#include <math.h>

int main()
{
	double a, b, c, V, B, p;
	double h = 5;
	double Xa = 1;
	double Ya = 1;
	double Xb = 2;
	double Yb = 2;
	double Xc = 3;
	double Yc = 3;

	c = sqrt((pow(Xa, 2) - (2 * Xa * Xb) + pow(Xb, 2)) + (pow(Ya, 2) - (2 * Ya * Yb) + pow(Yb, 2)));
	b = sqrt((pow(Xc, 2) - (2 * Xa * Xc) + pow(Xa, 2)) + (pow(Ya, 2) - (2 * Ya * Yc) + pow(Yc, 2)));
	a = sqrt((pow(Xc, 2) - (2 * Xc * Xb) + pow(Xb, 2)) + (pow(Yc, 2) - (2 * Yc * Yb) + pow(Yb, 2)));

	p = (a + b + c) / 2;
	B = sqrt(p * (p - a) * (p - b) * (p - c));

	V = h * B;


	return 0;
}