Sets and their representations

The multiples of 7 around the required interval are 28, 35, 42, 49, 56, and 63. The strict inequalities 30 < x < 60 exclude 28 and 63, as well as the endpoints 30 and 60 if they were multiples. The remaining values are 35, 42, 49, and 56. Therefore, option A gives the complete roster form.

The condition −5 < x excludes −5 because the left inequality is strict. The condition x ≤ 1 includes 1 because the right inequality is inclusive. Therefore, the integers beginning at −4 and ending at 1 are −4, −3, −2, −1, 0, and 1. This complete list is option A; neither endpoint should be handled in the same way.

A number that is both a multiple of 10 and divisible by 25 must be a common multiple of 10 and 25. Their least common multiple is 50, so the positive common multiples are 50, 100, 150, and so on. The condition x < 100 excludes 100 and all larger values. Therefore, the only member is 50, making option A correct.

Since x = 12/n must be a natural number and n is natural, n must be a positive divisor of 12. The possible divisors are 1, 2, 3, 4, 6, and 12. Their corresponding x-values are 12, 6, 4, 3, 2, and 1. A set is normally written in increasing order, giving {1, 2, 3, 4, 6, 12}. Option B contains the same members but is not the standard increasing roster form; option C incorrectly includes 5.

The word mathematics has the letters m, a, t, h, e, m, a, t, i, c, and s. In a set, repeated elements are counted only once. Thus the distinct-letter set is {m, a, t, h, e, i, c, s}, which contains 8 elements. Therefore option A is correct. The repeated letters m, a, and t do not increase the cardinality of the set.

To find the members of P₁, solve the condition x² − 10x + 21 = 0. Factoring gives (x − 3)(x − 7) = 0, so x = 3 or x = 7. Both values are natural numbers, so both belong to the set. The roster form is therefore {3, 7}. Negative values are not roots of this equation, 1 and 21 are not solutions, and 0 is not a solution.

The inequality x² < 10 implies −√10 < x < √10. Since √10 is slightly greater than 3, the possible integer values are −3, −2, −1, 0, 1, 2, and 3. Applying the additional condition that x must be even leaves only −2, 0, and 2. Hence the roster form is {−2, 0, 2}; values such as ±4 do not satisfy the inequality.

A two-digit number is at least 10, and the condition x < 20 restricts the search to 10 through 19. Among these numbers, 11, 13, 17, and 19 have no positive divisors other than 1 and themselves, so they are prime. The other candidates are composite: 10, 12, 14, 15, 16, and 18. Therefore option A gives the correct roster form.

The positive natural-number divisors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. The condition says that the divisor must not be even, so we retain only the odd divisors. Among the listed divisors, 1, 3, 5, and 15 are odd, while 2, 6, 10, and 30 are even. Therefore, the roster form of the set is S₁ = {1, 3, 5, 15}, so option A is correct.

Test natural numbers less than 10 against both prime conditions. For x = 3, 10 − x = 7, and both are prime. For x = 5, 10 − x = 5, so both are prime. For x = 7, 10 − x = 3, and both are prime. The other natural numbers below 10 fail at least one condition, so T₁ = {3, 5, 7}.

For 20/x to be a natural number, x must divide 20 exactly. The positive divisors of 20 are 1, 2, 4, 5, 10, and 20. Each of these values makes 20/x a natural number, whereas 3 does not divide 20 and values such as 40 give a non-natural fraction. Hence the roster form is U₁ = {1, 2, 4, 5, 10, 20}.

We examine every number from 1 through 30 and select those whose digits add to 5. The number 5 has digit sum 5, 14 has digit sum 1 + 4 = 5, and 23 has digit sum 2 + 3 = 5. The other listed alternatives contain numbers such as 15, 25, 30, or 32, whose digit sums are not 5 or whose values are outside the interval 1 to 30. Hence V₁ = {5, 14, 23}, making option A correct.

Solve the defining equation x² = 2x by bringing all terms to one side: x² − 2x = 0. Factoring gives x(x − 2) = 0. By the zero-product property, x = 0 or x = 2. Both are integers and therefore satisfy the domain restriction. The set is consequently W₁ = {0, 2}. Dividing by x would incorrectly discard the valid solution x = 0.

The positive divisors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42. The condition x > 6 excludes 1, 2, 3, and 6 because they are not greater than 6. The remaining divisors are therefore 7, 14, 21, and 42. Hence the roster form is X₁ = {7, 14, 21, 42}, making option A correct. Notice that 6 is excluded because the inequality is strict, not x ≥ 6.

List the integers from 1 through 20 and reject every number divisible by 2 or by 3. The numbers divisible by 2 are all even numbers, while the multiples of 3 include 3, 6, 9, 12, 15, and 18. The numbers that belong to neither group are 1, 5, 7, 11, 13, 17, and 19, so option A is correct.

Start with x² + x = 20 and bring all terms to one side: x² + x − 20 = 0. Factoring gives (x + 5)(x − 4) = 0, so the algebraic solutions are x = −5 and x = 4. However, the set specifies x ∈ ℕ, and −5 is not a natural number. Only 4 satisfies both the equation and the domain condition, so Z₁ = {4}. Therefore, option A is correct.

Factor the quadratic expression: x² − 7x + 10 = (x − 2)(x − 5). Setting each factor equal to zero gives x = 2 or x = 5. Both values are natural numbers and satisfy the original equation: 4 − 14 + 10 = 0 and 25 − 35 + 10 = 0. Therefore, the roster form is A = {2, 5}.

Because x must be an integer, list the integers lying strictly greater than −3 and less than or equal to 2. They are −2, −1, 0, 1, and 2. The lower endpoint −3 is excluded because the inequality is strict, while 2 is included because the upper inequality allows equality. Thus C has five elements, so option B is correct.

The governing idea is to apply both restrictions in the definition: a member must be a positive factor of 36 and must be odd. The positive factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Selecting only the odd factors leaves 1, 3, and 9. Thus D = {1, 3, 9}, so option A is correct. Option B omits the valid factor 1; C includes even 6; D includes 12, which is also even.

The inequality x² < 10 is equivalent to |x| < √10. Since √10 is approximately 3.16, the integers satisfying this condition range from −3 to 3. Checking them confirms that (−3)² = 9 < 10, while (−4)² = 16 > 10; the positive side is symmetric. Zero must also be included because 0² = 0 < 10. Hence E = {-3, -2, -1, 0, 1, 2, 3}, making option B correct.