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
$13.68 because you do [.18+.06=.24] [18•.24=4.32] [18-4.32=13.68]
Answer:
Step-by-step explanation:
To solve this problem is multiply 2 by 2 and then 2 which gives you 8.
Then you will add 2+2+2 and get six. For when multiply exponents you are adding.
Answer:
The cat should receive 1/5 of the can of food at each meal.
Step-by-step explanation:
Given that a cat is to receive 90 kcal of a canned food at each meal, and the food has a caloric density of 450 kcal per can, to determine what fraction of a can it should receive at each meal, the following calculation must be performed:
1 / (450/90) = X
1/5 = X
Thus, the cat should receive 1/5 of the can of food at each meal.
So easy , two plus two equals four ahh
