3 rivers in York
3 faults for a refusal in show jumping
Type your answer here... 8 miracles of Jesus
5 Brothers in the Jackson Five
-44 Assassination of Julius Caesar (-44 means 44 BC)
The answer to this Ditloid is somewhat ridiculous, and very hard indeed. The 6 Heroines of Jane Austin Novels. The answer to this Ditloid is somewhat ridiculous, and very hard indeed. The 6 Heroines of Jane Austin Novels.
f = 54
Frank is 24. Sue is 8 and John is 9. F = 3S F = J + 15 S = J - 1 Where F is Frank's age, S is Sue's age, and J is John's age. You can substitute J - 1 for S in the first equation: F = 3(J - 1) Then replace F with the expression from the second equation: J + 15 = 3(J - 1) 2J = 18 J = 9 S = 8 F = 24 --- You could also put both expressions in terms of the smallest value, S F = 3S F = 15 + (S +1) 3S = 16 + S 2S = 16 S = 8 F = 24
One possibility is : 2 Thieves Crucified with Jesus.
s, o, n (the months of the year)
It's the designer's initial - J. E. Fraser. It's not a mint mark.
that its from France
J.-F Reymond has written: 'J.-F. Reymond' -- subject(s): Exhibitions
#include"stdio.h" #include"conio.h" #include"math.h" void main() { int i,j; float a[3][4],b[3][4],c[4][4]; float x,y,z,p,q,r; m: printf("\nEnter the coefficients: "); for(i=0;i<3;i++) for(j=0;j<3;j++) scanf("%f",&a[i][j]); printf("\nEnter the constants: "); for(i=0;i<3;i++) scanf("%f",&a[i][3]); if(a[0][0]!=0.0) { p=a[1][0]; q=a[0][0]; r=a[2][0]; for(j=0;j<=3;j++) { b[0][j]=-(p/q)*a[0][j]; a[1][j]+=b[0][j]; c[0][j]=-(r/q)*a[0][j]; a[2][j]+=c[0][j]; } p=a[2][1]; q=a[1][1]; for(j=0;j<=3;j++) { b[1][j]=-(p/q)*a[1][j]; a[2][j]+=b[1][j]; } printf("\n\nThe matrix becomes\n"); for(i=0;i<3;i++) { for(j=0;j<4;j++) { printf("%.4f\t",a[i][j] } printf("\n"); } z=a[2][3]/a[2][2]; y=(a[1][3]-a[1][2]*z)/a[1][1]; x=(a[0][3]-a[0][2]*z-a[0][1]*y)/a[0][0]; printf("\nThe solution is"); printf("\nX=%f, Y=%f , Z=%f",x,y,z); } else { printf("\nThe first cofficient must not be zero,Enter again"); goto m; } }