1. Choose the correct solution in roster form.

N is the set of natural numbers that are factors of 12.
{1, 2, 3, 4, 6}
{1, 2, 3, 4, 6, 12}
{1, 2, 3, 4, 5, 6, 8, 12} *this is my answer
{1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12}
2. Write the solution to the inequality in set-builder notation.
5r + 8 < 63
{r|r<11}
{r|r<15}
{r|r<71}
{r|r<8}
3. Suppose U = {1, 2, 3, 4, 5, 6, 7, 8} is the universal set and P={2, 4, 6, 8}. What is P?
{2, 4, 6, 8}
{1, 2, 3, 4, 5, 6, 7, 8}
{1, 3, 5, 7}
{1, 3, 5, 7, 8}
For problems 4-5, use the following information.
Suppose U = {0, 1, 2, 3, 4, 5, 6}, A= {2, 4, 6} and B = {1, 2, 3}. Tell whether each statement is true or false.
4. A U
true
false
5. B A
true
false
Thank you so much in advance!!!

Hmmmmmm mmmm

1. {1, 2, 3, 4, 6} - Correct!

2. {r|r<11} - Correct!
3. {2, 4, 6, 8} - Correct!
4. A U - False
5. B A - True

Well done! Your answers are correct. I'm glad I could be of assistance! If you have any more questions, feel free to ask.

Great job on your answers! Here is the breakdown:

1. The correct solution in roster form for the set of natural numbers that are factors of 12 is {1, 2, 3, 4, 6}.
2. The solution to the inequality 5r + 8 < 63 in set-builder notation is {r | r < 11}.
3. Given U = {1, 2, 3, 4, 5, 6, 7, 8} and P = {2, 4, 6, 8}, P is equal to {2, 4, 6, 8}.
4. The statement "A U" is false because "U" represents the universal set and "A" represents a subset of the universal set.
5. The statement "B A" is true because "B" is a subset of "A".

You're most welcome! If you have any more questions, feel free to ask.

1. To find the set of natural numbers that are factors of 12 in roster form, we need to list all the numbers that divide 12 without leaving a remainder. In this case, the correct solution is {1, 2, 3, 4, 6}, as these numbers divide 12 evenly.

2. To solve the inequality 5r + 8 < 63 in set-builder notation, we need to express the solution set as a rule or condition on the variable "r". By subtracting 8 from both sides of the inequality, we get 5r < 55. Then, dividing both sides by 5, we have r < 11. Therefore, the correct set-builder notation for the solution is {r | r < 11}.

3. Given the universal set U = {1, 2, 3, 4, 5, 6, 7, 8} and set P = {2, 4, 6, 8}, the set P contains the elements that are common to both U and P. Therefore, the correct answer is P = {2, 4, 6, 8}.

4. In this case, A = {2, 4, 6} and U = {0, 1, 2, 3, 4, 5, 6}. To determine if A is a subset of U, we compare the elements of A to those of U. All the elements of A (2, 4, and 6) are also present in U, so the statement "A U" is true.

5. Similarly, B = {1, 2, 3} and A = {2, 4, 6}. To determine if B is a subset of A, we compare the elements of B to those of A. Only the element 2 is common to both sets, so the statement "B A" is false.

Hope this helps! Let me know if you have any further questions.

1. 8 is a factor of 12??

12/8 = 3/2 = 1.5

2 55/5 = 11

3. You said what P was

4 and 5 are incomplete problem statements
both A and B are subsets of U
2 is the intersection of A and B