This is true. As it is the same number for A and B so taking C from one would be the same as taking C from the other.
No- this is not true in general. Counterexample: Let a = {1,2}, b = {1} and c ={2}. a union c = [1,2} and b union c = {1,2} but a does not equal b. The statement be made true by putting additional restrictions on the sets.
No. Suppose A = {1,2}, B = {1,2,3,4,5,6} and C = {1,2,3,5,7,11}. The intersection of A with B is {1,2}, the intersection of A with C is also {1,2}, but B is not equal to C.
a - b = c -(a - b) = -c b - a = -c
2a. (a, b and c are all equal.)
A.
False : Cos B = 16.67/24 = 0.6946 : Therefore angle B = 46° (not 26°).
True : Sin B = 13.5/28.9 = 0.46713 : Therefore Angle B = 27.8 (1dp)
a b means return true if the value of a is equal to the value of b, otherwise return false. a = b means assign the value of b to the variable a.
No; it is false. The sum of all the angles of a quadrilateral always equals 360o.
Yes. If all the question's parts are true, then the answer is true. If all the question's parts are false, then the answer is false. If one of the question's parts is false and the rest true, then the answer is false. Logically, this is illustrated below using: A = True, B = True, C = True, D = False, E = False, F = False A and B and C = True D and E and F = False A and B and D = False If you add NOT, it's a bit more complicated. A and NOT(D) = True and True = True NOT(D) and D = True and False = False NOT(A) and NOT(B) = False and False = False Using OR adds another layer of complexity. A OR NOT(E) = True OR True = True NOT(D) OR D = True OR False = False NOT(A) OR NOT(B) = False OR False = False Logic is easy once you understand the rules.
No- this is not true in general. Counterexample: Let a = {1,2}, b = {1} and c ={2}. a union c = [1,2} and b union c = {1,2} but a does not equal b. The statement be made true by putting additional restrictions on the sets.
No. Suppose A = {1,2}, B = {1,2,3,4,5,6} and C = {1,2,3,5,7,11}. The intersection of A with B is {1,2}, the intersection of A with C is also {1,2}, but B is not equal to C.
F B D A False False False Not Given True C D B
It is true (allowing for rounding).
There is nothing to "solve". You can evaluate the expression when each of a, b and c are TRUE or FALSE. But that is not solving.
a - b = c -(a - b) = -c b - a = -c
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)