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.
a-punch s-block k-kick b-back f-forward d-down u-up Kitana fan lift-b,b,a fan throw-f,f,a fly-b,f,a leg throw-d,d,k fatality-d,d,d,d,a Noob Saibot Shadow Clone-d,d,f,k freeze-d,f,a shuriken throw-d,b,f,a telekick-d,d,k fatality-f,d,f,a Sub Zero freeze-d,f,a slide kick-b,f,k ice breath-b,f,a fatality-f,f,b,d,a Nightwolf arrow shoot-d,b,a shove-f,f,a grab 'n throw-d,f,a fatality-b,f,b,a Kabal energy ball-b,b,a spinnin' throw-d,f,a flash-b,f,k fatality-d,d,b,f,a
Sickbed Bed Sick
bed, bake, yea
There is no English Scrabble word that meets the criteria.
quet
#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;
The most general form is (ax - b)*(cx - d) = k where a, b, c, d and k are constants.