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.
Using the two parallel line theorems we proved that ∠8 ≅ ∠4.
In the given question,
Given: f || g
Prove: ∠8 ≅ ∠4
We using given diagram in proving that ∠8 ≅ ∠4
Since f || g, by the Corresponding Angles Postulate which states that "When a transversal divides two parallel lines, the resulting angles are congruent." So
∠8≅∠6
Then by the Vertical Angles Theorem which states that "When two straight lines collide, two sets of linear pairs with identical angles are created."
∠6≅∠4
Then, by the Transitive Property of Congruence which states that "All shapes are congruent to one another if two shapes are congruent to the third shape."
∠8 ≅ ∠4
Hence, we proved that ∠8 ≅ ∠4.
To learn more about parallel linetheorems link is here
