Answer:
See answer below
Step-by-step explanation:
The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.
i) x∈AnB if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.
ii) (I will abbreviate "if and only if" as "iff")
x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.
iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).
iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).
2 kids cross, one brings the boat back, one parent crosses, the other kid brings the boat back, both kids cross again, one kid brings the boat back, other parent crosses, the other kid brings the boat back, both kids cross again, one kid brings the boat back, the fisherman crosses, the other kid brings the boat back, both kids cross again. the fisher man is on the side where the kids and parents are. he can take the boat. total trips 11.
Answer:
6
Step-by-step explanation:
Answer:
it took her 54 mins to finish her walk.
Step-by-step explanation:
Given;
total distance walked by Katie, d = ⁵/₆
time taken to walk ¹/₆ mile = 10 mins
break time = 1 min
There are five ¹/₆ mile in the total distance which is ⁵/₆ miles.
Time for the first ¹/₆ mile = 10 mins +
break time = 1 min
Time for the second ¹/₆ mile = 10 mins +
break time = 1 min
Time for the third ¹/₆ mile = 10 mins +
break time = 1 min
Time for the fourth ¹/₆ mile = 10 mins +
break time = 1 min
Time for the fifth ¹/₆ mile = 10 mins
Total time = 5 (10 mins) + 4 (1 min)
= 50 mins + 4mins
= 54 mins
Therefore, it took her 54 mins to finish her walk.
