If not all tiles are necessary, try "bonks".
bodkins
Bilked
becked
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.
Sickbed Bed Sick
quet
There is no English Scrabble word that meets the criteria.
bed, bake, yea
#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(); }
t-r-i-s-k-a-i-d-e-k-a-p-h-o-b-i-a.
#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;
betake
The most general form is (ax - b)*(cx - d) = k where a, b, c, d and k are constants.