In: Computer Science
c program
i ) 1 && (30%10 >=0) && (30 %10<=0) will return 1.
30%10 = 0 return Trrue
30 %10 =3 which is less than 3 , return true
ii) no paranthases required,it will return 1
proof for i And ii
iii)
method 1
-> (1 + (30%10 >=0))' = 0
-> 0 && ((30 %10<=3) ) ' = 0 + 0 = 0
then
compliment of 1 && (30%10 >=0) && (30 %10<=0) = 0
demorgans law
(A+B+C )' =A' * B' * C'
(1 && (30%10 >=0) && (30 %10<=0)) ' = 1 ' * (30%10 >=0)' * (30 %10<=0)'
=0 * (30%10 >=0)' * (30 %10<=0)'
= 0