Sets and their representations

The positive divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. We now retain only the odd divisors, meaning those not divisible by 2. These are 1, 3, and 9. Hence the roster form is R = {1, 3, 9}. The number 1 must be included because 1 divides every positive integer and 1 is odd.

First solve the inequality: 2x − 1 ≤ 11. Adding 1 to both sides gives 2x ≤ 12, and dividing by the positive number 2 gives x ≤ 6. Because x belongs to N, the admissible natural-number values are 1, 2, 3, 4, 5, and 6, using the convention N = {1, 2, 3, ...}. Therefore, S = {1, 2, 3, 4, 5, 6}.

The conditions translate to the strict double inequality 5 < x < 11, with x restricted to the integers. Since the inequalities are strict, neither endpoint 5 nor endpoint 11 belongs to the set. The integers strictly between them are 6, 7, 8, 9, and 10. Therefore, the roster form is T = {6, 7, 8, 9, 10}; this is a finite set of five integers.

A prime number is a natural number greater than 1 with exactly two positive divisors: 1 and itself. The prime numbers less than 10 are 2, 3, 5, and 7. The number 1 is not prime, and 9 is composite because it has divisors 1, 3, and 9. Therefore the correct roster form is option A.

For x = 24/n to be a natural number, n must be a positive divisor of 24. The even positive divisors of 24 are 2, 4, 6, 8, 12, and 24. Substitution gives x = 12, 6, 4, 3, 2, and 1, respectively. Therefore V = {12, 6, 4, 3, 2, 1}; the order of elements does not matter in a set.

Since x is restricted to integers, inspect the integers near zero. The values −1, 0, and 1 have squares 1, 0, and 1, all of which are less than 2. The next integers, −2 and 2, have square 4 and therefore fail the strict inequality. All integers farther from zero have even larger squares. Thus the complete roster form is {−1, 0, 1}, so option A is correct.

The variable x is restricted to integers. The inequality x² < 10 allows the integers -3 through 3, because 3² = 9 is less than 10 but 4² = 16 is not. The additional condition x > -3 excludes -3 itself, while -2, -1, 0, 1, 2, and 3 satisfy both conditions. Hence A = {-2, -1, 0, 1, 2, 3}.

The listed elements are precisely the even positive integers from 2 through 10, inclusive. Option A states all three necessary facts: x is positive, x is even, and its value lies between 2 and 10. Option B includes infinitely many extra even numbers, while C and D include odd numbers. Therefore A is the exact set-builder description.

The inequality x² ≤ 16 gives -4 ≤ x ≤ 4, so the integer candidates are -4, -3, -2, -1, 0, 1, 2, 3, and 4. Among these, the only prime numbers are 2 and 3. Removing them leaves -4, -3, -2, -1, 0, 1, and 4. Negative integers, zero, and one are not prime by the definition of a prime number.

A member of P must be a natural-number divisor of both 18 and 24. The positive divisors of 18 are 1, 2, 3, 6, 9, and 18. The positive divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The common elements in these two lists are 1, 2, 3, and 6. Therefore, P = {1, 2, 3, 6}.

The listed elements are consecutive squares: 1 = 1², 4 = 2², 9 = 3², 16 = 4², and 25 = 5². Thus every element has the form n², where n is a natural number from 1 through 5. The correct set-builder description is S = {n² : n ∈ ℕ, 1 ≤ n ≤ 5}. The other options describe consecutive natural numbers, even numbers, or cubes.

The positive multiples of 6 below 40 are 6, 12, 18, 24, 30, and 36. The positive multiples of 9 below 40 are 9, 18, 27, and 36. Taking the union gives {6, 9, 12, 18, 24, 27, 30, 36}, which has 8 elements. The common multiples 18 and 36 are counted only once because a set has no repeated elements.

The roster form lists every element that satisfies both conditions. The positive divisors of 12 are 1, 2, 3, 4, 6, and 12. Among these, only 1 and 3 are odd; 2, 4, 6, and 12 are even and must be excluded. Therefore, C = {1, 3}, so option A is correct. Option B lists all positive divisors without applying the oddness condition, while option D includes 5, which is not a divisor of 12.

Under the stated convention, N = {1, 2, 3, ...}, so no natural number is less than 1. Therefore the set in option A has no elements and is empty. Option B contains 0, option C contains 1, and option D also contains the integer 0. Specifying the convention for N removes any ambiguity about whether zero is included.

The condition |x − 2| ≤ 3 means that x is at a distance of at most 3 from 2. Therefore, subtracting and adding 3 gives −1 ≤ x ≤ 5. Since x must be an integer, we list every integer in this closed interval: −1, 0, 1, 2, 3, 4, and 5. Hence the roster form is D = {−1, 0, 1, 2, 3, 4, 5}, so option A is correct.

To find the members of T, solve the condition x² − 5x + 6 = 0. Factoring gives (x − 2)(x − 3) = 0, so x = 2 or x = 3. Both values are natural numbers, so both satisfy the restriction x ∈ ℕ. Therefore, the set contains exactly these two elements: T = {2, 3}. Thus option A is correct; the other options either contain incorrect roots or include numbers that are not solutions.

For x² = 9, the integer solutions are x = −3 and x = 3. For x² = 16, the integer solutions are x = −4 and x = 4. These four values are different, so E = {−4, −3, 3, 4}. Consequently, the cardinality of E is 4, making option C correct. Both positive and negative square roots must be included.

The roster form lists exactly the five English vowels: a, e, i, o, and u. A correct descriptive or set-builder form must include every one of these letters and no other letter. Option A describes precisely the vowels of the English alphabet. The other choices either describe consonants, include letters beyond the set, or include all alphabet letters.

Start with the inequality 3x + 1 < 16. Subtracting 1 from both sides gives 3x < 15, and dividing by the positive number 3 gives x < 5. Because x belongs to ℕ, using the usual school convention ℕ = {1, 2, 3, …}, the possible natural numbers less than 5 are 1, 2, 3, and 4. Therefore G = {1, 2, 3, 4}, making option A correct.

The inequality has two different boundary conditions. The sign −2 < x means that −2 is excluded, while x ≤ 3 means that 3 is included. The integers strictly greater than −2 and less than or equal to 3 are −1, 0, 1, 2, and 3. Consequently, H = {−1, 0, 1, 2, 3}, so option A is the only correct answer.