Let's write each of the given sets in roster form:
(i) [tex]\(A_1 = \{x: 2x + 3 = 11\}\)[/tex]
To find [tex]\(x\)[/tex], solve the equation [tex]\(2x + 3 = 11\)[/tex]:
[tex]\[2x + 3 = 11 \Rightarrow 2x = 8 \Rightarrow x = 4\][/tex]
Thus, the set in roster form is:
[tex]\[A_1 = \{4\}\][/tex]
(ii) [tex]\(A_2 = \{x: x^2 - 4x - 5 = 0\}\)[/tex]
First, solve the quadratic equation [tex]\(x^2 - 4x - 5 = 0\)[/tex]:
Factoring, we have:
[tex]\[(x - 5)(x + 1) = 0 \Rightarrow x = 5 \text{ or } x = -1\][/tex]
Thus, the set in roster form is:
[tex]\[A_2 = \{-1, 5\}\][/tex]
(iii) [tex]\(A_3 = \{x: x \in \mathbb{Z}, -3 \leq x < 4\}\)[/tex]
Here, [tex]\(\mathbb{Z}\)[/tex] denotes the set of all integers. We need all integers from [tex]\(-3\)[/tex] to [tex]\(3\)[/tex]:
Thus, the set in roster form is:
[tex]\[A_3 = \{-3, -2, -1, 0, 1, 2, 3\}\][/tex]
(iv) [tex]\(A_4 = \{x: x \text{ is a two-digit number and sum of the digits of } x \text{ is } 7\}\)[/tex]
Find all two-digit numbers whose digits sum to 7:
These numbers are [tex]\(16, 25, 34, 43, 52, 61, 70\)[/tex].
Thus, the set in roster form is:
[tex]\[A_4 = \{16, 25, 34, 43, 52, 61, 70\}\][/tex]
(v) [tex]\(A_5 = \{x: x = 4n, n \in \mathbb{W} \text{ and } n < 4\}\)[/tex]
[tex]\(\mathbb{W}\)[/tex] denotes the set of whole numbers [tex]\(\{0, 1, 2, 3, \ldots\}\)[/tex]. We need [tex]\(n \in \mathbb{W}\)[/tex] such that [tex]\(n < 4\)[/tex]:
[tex]\[n = 0, 1, 2, 3\][/tex]
Thus, [tex]\(x = 4n\)[/tex] gives:
[tex]\[x = 4 \cdot 0 = 0\][/tex]
[tex]\[x = 4 \cdot 1 = 4\][/tex]
[tex]\[x = 4 \cdot 2 = 8\][/tex]
[tex]\[x = 4 \cdot 3 = 12\][/tex]
Thus, the set in roster form is:
[tex]\[A_5 = \{0, 4, 8, 12\}\][/tex]
(vi) [tex]\(A_8 = \left\{x: x = \frac{n}{n+2}; n \in \mathbb{N} \text{ and } n > 5\right\}\)[/tex]
[tex]\(\mathbb{N}\)[/tex] denotes the set of natural numbers [tex]\(\{1, 2, 3, \ldots\}\)[/tex]. We need [tex]\(n \in \mathbb{N}\)[/tex] such that [tex]\(n > 5\)[/tex]:
[tex]\[n = 6, 7, 8, 9, 10 \text{ (Example range to show the pattern)}\][/tex]
Thus, for each [tex]\(n\)[/tex], calculate [tex]\(x = \frac{n}{n+2}\)[/tex]:
[tex]\[n = 6 \Rightarrow x = \frac{6}{6+2} = \frac{6}{8} = 0.75\][/tex]
[tex]\[n = 7 \Rightarrow x = \frac{7}{7+2} = \frac{7}{9} \approx 0.7778\][/tex]
[tex]\[n = 8 \Rightarrow x = \frac{8}{8+2} = \frac{8}{10} = 0.8\][/tex]
[tex]\[n = 9 \Rightarrow x = \frac{9}{9+2} = \frac{9}{11} \approx 0.8182\][/tex]
[tex]\[n = 10 \Rightarrow x = \frac{10}{10+2} = \frac{10}{12} \approx 0.8333\][/tex]
Thus, the set in roster form is:
[tex]\[A_8 = \{0.75, 0.7778, 0.8, 0.8182, 0.8333\}\][/tex]
