8
(xM, yM) = [(-3 + 1)/2, (3 + -3)/2] = (-2/2, 0/2) = (-1, 0)
(0 + 1 + 5 + 3 + 3 + 3 + 0 + 2 + 2) / 9 = 2.1 recurring (that is, 2.1111...)
2,2,0,5,1,4,1,3,0,0,1,4,4,0,1,4,3,4,2,1,0
Believe the expression is 10x*2 - 17x + 3 = 0, then we have10x*2 -17x + 3 = 0 or10x*2 - 15x - 2x + 3 = 0 or5x(2x - 3) - 1(2x - 3) = 0 or(5x - 1)(2x - 3) = 0 which says that either(5x - 1 ) = 0 in which case x = 1/5 or(2x - 3 ) = 0 in which case x = 3/2Believe the expression is 10x*2 - 17x + 3 = 0, then we have10x*2 -17x + 3 = 0 or10x*2 - 15x - 2x + 3 = 0 or5x(2x - 3) - 1(2x - 3) = 0 or(5x - 1)(2x - 3) = 0 which says that either(5x - 1 ) = 0 in which case x = 1/5 or(2x - 3 ) = 0 in which case x = 3/2
A huge number. 0 + 1 + 2 = 3 0 + 2 + 1 = 3 1 + 0 + 2 = 3 1 + 2 + 0 = 3 2 + 0 + 1 = 3 2 + 1 + 0 = 3 -0 + 1 + 2 = 3 -0 + 2 + 1 = 3 1 - 0 + 2 = 31 + 2 - 0 = 32 - 0 + 1 = 32 + 1 - 0 = 3 0 - 1 + 3 = 2 0 + 3 - 1 = 2 -1 + 0 + 3 = 2 -1 + 3 + 0 = 2 3 + 0 - 1 = 2 3 - 1 + 0 = 2 -0 - 1 + 3 = 2-0 + 3 - 1 = 2-1 - 0 + 3 = 2-1 + 3 - 0 = 23 - 0 - 1 = 23 - 1 - 0 = 2 0 - 2 + 3 = 1 0 + 3 - 2 = 1 -2 + 0 + 3 = 1 -2 + 3 + 0 = 1 3 + 0 - 2 = 1 3 - 2 + 0 = 1 -0 - 2 + 3 = 1-0 + 3 - 2 = 1-2 - 0 + 3 = 1-2 + 3 - 0 = 13 - 0 - 2 = 13 - 2 - 0 = 1 1 + 2 - 3 = 0 1 - 3 + 2 = 0 2 + 1 - 3 = 0 2 - 3 + 1 = 0 -3 + 1 + 2 = 0 -3 + 2 + 1 = 0 For each of these equations there is a counterpart in which all signs have been switched. For example 0 + 1 + 2 = 3 gives -0 - 1 - 2 = -3and so on. Now, all of the above equations has three numbers on the left and one on the right. Each can be converted to others with two numbers on each side. For example:the equation 0 + 1 + 2 = 3 gives rise to0 + 1 = 3 - 20 + 1 = -2 + 30 + 2 = 3 - 10 + 2 = -1 + 31 + 2 = 3 - 01 + 2 = -0 + 3-0 + 1 = 3 - 2-0 + 1 = -2 + 3-0 + 2 = 3 - 1-0 + 2 = -1 + 31 + 2 = 3 + 01 + 2 = +0 + 3 As you can see, the number of equations is huge!
E|----------------|-------0-1-0----|----------------|--------1-0-----|--| B|--0-1-----0-1--|--0-1--------1-|------0-1-------|------1----1---|--| G|0-------0------|0--------------2|0-------------0-|-----2-------2-|0-| D|----------------|-----------------|----------------|3--3------------|--| A|----------------|-----------------|----------------|-----------------|--| E|----------------|-----------------|----------------|-----------------|--| E|----3-1---------|-----------------0-|1---------------| B|--------1-------|-----------3-1-----|----------------| G|----------0-----|0-------0----------|----------------| D|------------2-3|----2-3------------|----------------| A|----------------|--------------------|----------------| E|----------------|--------------------|----------------| E|----------------|--------0-1-0----|----------------|-------1-0-----|--| B|--0-1-----0-1-|---0-1-------1--|------0-1-------|------1----1---|--| G|0-------0------|-0--------------2|0------------0-|-----2-------2-|0-| D|----------------|------------------|----------------|3--3-----------|--| A|----------------|------------------|----------------|----------------|--| E|----------------|------------------|----------------|----------------|--|
+4 +3 +3 -4 -3 -2 0 -1 +3 +1+1 0 +1
I was only able to get 13 but I think there can be more, this is what i got: 0+0+4=4 4+0+0=4 0+4+0=4 0+1+3=4 0+3+1=4 0+2+2=4 1+2+1=4 1+1+2=4 1+3+0=4 2+2+0=4 2+0+2=4 3+1+0=4 3+0+1=4
|------------------------------------------------0--1--3------E |---------------------------------------0--1--3---------------A |---------------------------------0--2------------------------D |------------------------0--2--3------------------------------G |---------------0--2--3---------------------------------------B |------0--1--3------------------------------------------------e
8
from: http://local.wasp.uwa.edu.au/~pbourke/other/determinant/ /* Recursive definition of determinate using expansion by minors. */ double Determinant(double **a,int n) { int i,j,j1,j2; double det = 0; double **m = NULL; if (n < 1) { /* Error */ } else if (n j1) continue; m[i-1][j2] = a[i][j]; j2++; } } det += pow(-1.0,1.0+j1+1.0) * a[0][j1] * Determinant(m,n-1); for (i=0;i<n-1;i++) free(m[i]); free(m); } } return(det); } //New Answer By Shaikh SOHIAL Hussain form PAKISTAN #include<stdio.h> #include<conio.h> void main () { clrscr(); int a[10][10],row,i=0,j=0,result,w,x,y,z; printf("This Program made 4 solve matrix determintae\n"); scanf("%d",&row); while(i<row) {printf("\n"); for(j=0;j<row;j++) { printf("Enter the value of %d%d\n",i,j); scanf("%d",&a[i][j]); } i++; } i=0; while(i<row) { printf("\n"); j=0; while( j<row) {printf("%d",a[i][j]); printf(" ");printf(" "); j++;} i++; } if(row==2) { result=(a[0][0]*a[1][1])-(a[0][1]*a[1][0]); printf("The answer is\t %d",result); } else if(row==3) { result=(a[0][0]*((a[1][1]*a[2][2])-(a[1][2]*a[2][1]))) - (a[0][1]*((a[1][0]*a[2][2])-(a[1][2]*a[2][0]))) + (a[0][2]*((a[1][0]*a[2][1])-(a[1][1]*a[2][0]))); printf("\nThe answer is \t %d",result); } else if(row==4) { w=a[0][0]*(a[1][1]*(a[2][2]*a[3][3]-a[2][3]*a[3][2])-a[1][2]*(a[2][1]*a[3][3]-a[2][3]*a[3][1])+a[1][3]*(a[2][1]*a[3][2]-a[2][2]*a[3][1])); x=a[0][1]*(a[1][0]*(a[2][2]*a[3][3]-a[2][3]*a[3][2])-a[1][2]*(a[2][0]*a[3][3]-a[2][3]*a[3][0])+a[1][3]*(a[2][0]*a[3][2]-a[2][2]*a[3][0])); y=a[0][2]*(a[1][0]*(a[2][1]*a[3][3]-a[2][3]*a[3][1])-a[1][1]*(a[2][0]*a[3][3]-a[2][3]*a[3][0])+a[1][3]*(a[2][0]*a[3][1]-a[2][1]*a[3][0])); z=a[0][3]*(a[1][0]*(a[2][1]*a[3][2]-a[2][2]*a[3][1])-a[1][1]*(a[2][0]*a[3][2]-a[2][2]*a[3][0])+a[1][2]*(a[2][0]*a[3][1]-a[2][1]*a[3][0])); result=w-x+y-z; printf("\n The answer is \t %d", result); } else{ printf("Determinate Limit is 4*4 max\n"); getch(); }
(xM, yM) = [(-3 + 1)/2, (3 + -3)/2] = (-2/2, 0/2) = (-1, 0)
(0 + 1 + 5 + 3 + 3 + 3 + 0 + 2 + 2) / 9 = 2.1 recurring (that is, 2.1111...)
d-0(2) d-0(2) d-1 d-3 d-2 d-1 a-0(2) a-0(2) a-0 a-1 d-2 d-3 d-1(2) d-1(2) d-1 d-3 d-2 d-1 d-0 a-3 a-2 a-1 a-0 d-3 d-2 d-1 d-0(2) d-0(2) d-1 d-3 d-2 d-1 a-0(2) a-0(2) a-0 a-1 d-2 d-3 d-1(2) d-1(2) d-1 d-3 d-2 d-0 a-0 d-1 d-2 d-0 a-3(2) (2)=2 beats [half note] the rest are 1 beat [quarter note] d= the string on the violin (g,d,a,e) -0=how many fingers you put on that string hope this helps!!!!!! >_< hehe...... =P
10 different combinations can be made. Raisins Peanuts Pretzels 3 0 0 2 1 1 2 0 1 1 2 0 1 0 2 1 1 1 0 0 3 0 3 0 0 1 2 0 2 1
1 (0+1) 3 (0+1+2) 6 (0+1+2+3) 10 (0+1+2+3+4) 15 (0+1+2+3+4+5) 21 (0+1+2+3+4+5+6) Notice these are the numbers you can arrange into equalateral triangles.