the number of elements in the union of the A sets is:5(30)−rAwhere r is the number of repeats.Likewise the number of elements in the B sets is:3n−rB
Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x. Likewise on the B side, 8x of those elements should be thrown away. so now we have:150−9x=3n−8x⟺150−x=3n⟺50−x3=n
Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set. if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times. This means that:150 /10=15is the number of elements in the the A's without repeats counted (same for the Bs as well).So now we have:50−15 /3=n⟺n=45
I don't understand the question
Answer:
It is a unit of radius that is radius of 1. Thus, the distant to the middle to any edge is always 1.
Step-by-step explanation:
F(x) = g(x)
3(2) - 4 = (2)² - 2
6 - 4 = 4 - 2
2 = 2
I just substituted all the values in the domain to see which one would work. That's not really the quickest way but that's the only way I know
