The answer for the first question is a. In this problem, we are trying to figure out any numbers that belong to both set X and Y. Set X, has numbers that are less than 10. {etc... -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9} This can go on forever, because there is no specified limitation, besides the numbers having to be lower than 10. Set Y, has numbers that are even. {etc... -4, -2, 0, 2, 4, 6, 8, 10, 12, etc...} This also can go on forever, because there is no specified limitation, besides the numbers having to be even. The only numbers that belong in both set X and Y are {0, 2, 4, 6, 8}. Therefore, answer to the problem is a: {0, 2, 4, 6, 8}.
The answer for the second question is a. You need to replace z with 0, 1, 2, or 3 and solve the equation on the right side of the inequality sign. The answer to the equation must be less than 6. If you replace z with 0: 0 (10-0) = 0. 0 is less than 6. If you replace z with 1: 1 (10-1) = 9. 9 is greater than 6. If you replace z with 2: 2 (10-2) = 16. 16 is greater than 6. If you replace z with 3: 3 (10-3) = 21. 21 is greater than 6. Therefore, 0 is the answer.
when they are combined, there are two taken away, so we can just say that there are two added in total, and the number of grey hexagons is n, and it would be 4n because there are 2 taken away like i said earlier if that makes sense :)
Something goes wrong with the question: " segment AB is parallel to segment DC <span>and segment BC is parallel to segment AD" that is impossible when figure is built.</span>
