(start) /a=0 c=0\ \b=0 / /input a/ /input b/ /input c/ /a>b\ no /b>c\ yes /display b/ -> (a) \ / \ / yes no /a>c\ no /display c/ -> (a) \ / yes /display a/ <- (a) (end)
max = a > b ? a : b; max = max > c ? max : c;
dim a,b,c a=cint(inputbox("enter value for a")) b=cint(inputbox("enter value for b")) c=cint(inputbox("enter value for c")) if((a>b)and(a>c)) then msgbox "greatest number is a="&a else if ((b>a)and(b>c)) then msgbox "greatest number is b="&b else msgbox "greatest number is c="&c end if end if
If you know the lengths of two sides, a and b, and the included angle, C, then area = 0.5*a*b*sin(C)
The answer is 4! (4 factorial), the same as 4x3x2x1, which equals 24 combinations. The answer is 24 and this is how: A b c d A b d c A c d b A c b d A d c b A d b c B c d a B c a d B d a c B d c a B a c d B a d c C d a b C d b a C a b d C a d b C b d a C b a d D a b c D a c b D b c a D b a c D c a b D c b a
(A+B+C)' = A'B'C' by using truth table
If a + b = cThen a = c -b
a=a^b; b=a^b; a=a^b;
biggest3 (a,b,c) = biggest2 (a, biggest2 (b,c))
A)Testing and debugging difficulty B)Deadlocks C)Data races D)B and C E)A, B and C Any one?
(start) /a=0 c=0\ \b=0 / /input a/ /input b/ /input c/ /a>b\ no /b>c\ yes /display b/ -> (a) \ / \ / yes no /a>c\ no /display c/ -> (a) \ / yes /display a/ <- (a) (end)
(a+b).c=(a.b)+(b.c) (a+b).c=(a.b)+(b.c)
You use a person's name to spell it.
int min (int a, int b, int c) {if (a
A = A xor B B = A xor B A = A xor B in C... A^=B; B^=A; A^=B;
mga bogo man mo ///// c HAPAY NI ,,
R = (A > B && A > C) ? A : (B > C) ? B : C; // parentheses not necessary - for clarity only