A The surface area of the smallercan is 80%of the surface area of the larger can
Answer:
a. 4600
b. 6200
c. 6193
Step-by-step explanation:
Let
the number of elements in A.
Remember, the number of elements in
satisfies

Then,
a) If
, and if 
Since 
So

b) Since the sets are pairwise disjoint

c) Since there are two elements in common to each pair of sets and one element in all three sets, then

The lowest value of sinα is -1.
When x=1, you get:
So the maximum value of the function is 15.
Does it tell you what X represents?