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.
First get the factors of 24: 2*2*2*3Then the factors of 9: 3*3 Comparing the factors, the GCF is 3 Then you can rewrite the expression: 3*8 + 3*3 At this point, I'm not sure whether what you mean is really distributive property or not since this case is more of a factoring. 3*(8+3)
