Answer:
i think G
Step-by-step explanation:
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
What types of problems can be solved using the greatest common factor? What types of problems can be solved using the least common multiple? Complete the explanation.
<span>*** Use the words 'same' and 'different' to complete the following sentences.*** </span>
<span>Problems in which two different amounts must be split into (the same) number of groups can be solved using the GCF. Problems with events that occur on (different) schedules can be solved using the LCM.</span>
Answer:
8v^2+24-48x^3
Distribute the -8 to each term to get your answer.
Hope I helped!
