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Eine Kleine Nachtmusik, K.525Piano Sonata No. 16, K.545Andante for Clavier in C, K.1Horn Concerto No. 2, K.417Concerto for Flute and Harp, K.299Rondo alla Turca, K.331Lullaby, K.350Flute Quartet No. 4, K.298Symphony No. 41, "Jupiter", K.551Requiem in D minor, K.626
FAEKCDHGB
fried D+I+C+K+S on a stick
fried D+I+C+K+S on a stick
Yes. To show the conditions on a, b, c and d given that if a/b = c/d then a+b = c+d. Suppose b != d (and that both b and d are non-zero) then: d = kb for some number k (!= 0), so c/d = c/kb = (c/k)/b so a/b = (c/k)/b => a = c/k => c = ka Thus: c + d = ka + kb = k(a + b) Which means that c + d = a + b only if k = 1. Thus if a/b = c/d then a + b = c + d only if a = c and b = d. The condition on b and d both being non-zero prevents the possibility of division by zero. If either is zero, a division by zero will occur and at least one of the fractions is infinite.
#include<stdio.h> #include<conio.h> void main() { int i,j,k,l,m; clrscr(); for(i=0;i<4;i++) { for(j=i+1;j<=4;j++) { printf("%d",j); } for(k=1;k<=i;k++) { printf("%d",k); } printf("\n"); for(k=i;k>=1;k--) { printf("%d",k); } for(j=4;j>=i+1;j--) { printf("%d",j); } getch(); }
#include<stdio.h> #include<conio.h> main() { float a,b,c,real,imag,r1,r2,d; int k; printf("Enter the values of a,b,c:"); scanf("f%f",&a,&b,&c); d=b*b-4*a*c; if(d<0) k=1; else if(d==0) k=2; else k=3; switch(k) { case 1: printf("Roots are imaginary\n"); real=-b/(2*a); d=-d;
if u have a big d i c k dont search up this question if ur a girl get a d i c k -_-;
C K C. Metz has written: 'Time and cost control for R and D Manapers'
I LOve d-i-c-k-s
my d. I. C. K
a d i c k
D-U-C-K
false there are five, class a, b, c, d, and k.