121 spaces on a Chinese checkers board
private void jButton1ActionPerformed(java.awt.event.ActionEvent evt) { int a,b,c,d,g=0,s=0; a=Integer.parseInt(jTextField1.getText()); b=Integer.parseInt(jTextField2.getText()); c=Integer.parseInt(jTextField3.getText()); d=Integer.parseInt(jTextField3.getText()); if(a>b){ if(b>c){g=a;s=c;} else{g=a;s=b;} } else if(b>c){ if(c>a){g=b;s=a;} else{g=b;s=c;} } else if(c>a){ if(b>a){g=c;s=a;} else {g=c;s=c;} } switch(d){ case 1:JOptionPane.showMessageDialog(this, "the greates no. is"+g); break; case 2:JOptionPane.showMessageDialog(this, "the smallest no. is"+s); break; default:JoptionPane.showMessageDialog(this," please enter the choice"+"correctly 1 for greatest and 2 for smallest"); }
//Written in C++. ALFRED OMONDI ORIMBO (orimbo@gmail.com)# include# includeusing namespace std;void main(){double s,a,b,c,area;couta;coutb;coutc;s=(a+b+c)/2;area = sqrt(s*(s-a)*(s-b)*(s-c));cout
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
1^2 = 111^2 = 121111^2 = 123211111^2 = 123432111111^2 = 123454321#include#include#includevoid main(){clrscr(); int n;//soullessgodcoutn;int a=1,b,s=1;for(;a=1;b--){cout
asdfghjkl;' s-sum and c'-carry see for half adder s=a(xor)b and c'=ab for full adder s=a(xor)b(xor)c and c=ab+bc+ac or ab+c(a(xor)b) we can convert two half adder to full adder with help of and or gate. . . ! we got two half adder * for first half adder input is a and b therefore. . .s=a(xor)b and c'=ab * for second half adder input is a(xor)b and c therefore. . .s=a(xor)b(xor)c and c' is (a(xor)b)c note: now connect the c' of first half adder and second half adder to 'or' gate resulting is ab+c(a(xor)b)
It is sqrt{s*(s-a)*(s-b)*(s-c)} where the lengths of the three sides are a, b and c units and s = (a+b+c)/2.
Suppose the sides are a, b and c units. Calculate s= (a+b+c)/2 Then Area = sqrt[s*(s-a)*(s-b)*(s-c)] square units
Use the Hero's formula: Let s = (a + b + c)/2. Then the area of the triangle equals√[s(s - a)(s - b)(s - c)], where a, b, and c denote the sides of the triangle.
Let the sides be a, b, c Area = sq rt [s(s-a)(s-b)(s-c)] where s= 1/2 (a+b+c)
120 sq metres. To see how you get this answer, read on: If the sides are a, b and c, then calculate s = 0.5*(a+b+c) Then the area is sqrt[s*(s-a)*(s-b)*(s-c)]
The answer depends on the information that you do have. Suppose you know all the edge lengths: the three sides of the triangle are a, b and c and the length of the prism is d. Let s = (a + b + c)/2 Then the area of the triangular cross section is sqrt[s*(s-a)*(s-b)*(s-c)] square units. So, surface area = 2*sqrt[s*(s-a)*(s-b)*(s-c)] + d*(a+b+c) square units. Volume = sqrt[s*(s-a)*(s-b)*(s-c)]*d cubic units.
The information depends on what information is available. For example, for a triangle in which all three sides are known (a, b and c) then calculate s = (a+b+c)/2 And the area = sqrt[s*(s-a)*(s-b)*(s-c)] If a, b and the angle between then, C, is known then area = 0.5*a*b*sin(C). There are other formulae for other circumstances.
If A is the area of a triangle with side length a, b and c and s is (a+b+c)/2 ( known as the semiperimeter of the triangle) Then, Heron's formula tells us that A=square root of [s(s-a)(s-b)(s-c)].
One possible form is sqrt{s*(s-a)*(s-b)*(s-c)} square units where the lengths of the three sides of the triangle are a, b and c units and s = (a+b+c)/2.
One possible form is sqrt{s*(s-a)*(s-b)*(s-c)} square units where the lengths of the three sides of the triangle are a, b and c units and s = (a+b+c)/2.
#include<stdio.h> #include<math.h> main() { float A,a,b,c,s; printf("entr the value of a,b and c"); scanf("%f %f %f",&a,&b,&c); A=sqrt(s*s-a*s-b*s-c) printf("area of the triangle %f\n"A); getch(); }
Let A,B,C,D Four different sides with different lengths. Apply the following formula. A = Sqrt ( ( s-a) ( s-b) (s-c) (s-c) ) Where S = ( a+b+c+d) / 2.