Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Sure, let's solve each part step by step.
### Part A: Cartesian Products of Sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex]
Step 1: Define Set [tex]\( A \)[/tex]
Given:
[tex]\[ A = \{ x \leq 4 \mid x \in \mathbb{N} \} \][/tex]
This means that [tex]\( A \)[/tex] contains all natural numbers [tex]\( x \)[/tex] such that [tex]\( x \leq 4 \)[/tex]. Thus:
[tex]\[ A = \{ 0, 1, 2, 3, 4 \} \][/tex]
Step 2: Define Set [tex]\( B \)[/tex]
Given:
[tex]\[ B = \{ x \mid x^2 - 5x - 6 = 0 \} \][/tex]
To solve for [tex]\( B \)[/tex], we must find the roots of the quadratic equation [tex]\( x^2 - 5x - 6 = 0 \)[/tex]. Factoring the quadratic equation, we get:
[tex]\[ (x - 6)(x + 1) = 0 \][/tex]
Hence, the solutions are:
[tex]\[ x = 3 \text{ and } x = -2 \][/tex]
So,
[tex]\[ B = \{ 3, -2 \} \][/tex]
Step 3: Find [tex]\( A \times B \)[/tex]
The Cartesian product [tex]\( A \times B \)[/tex] is the set of all possible ordered pairs [tex]\( (a, b) \)[/tex] where [tex]\( a \in A \)[/tex] and [tex]\( b \in B \)[/tex]. Thus:
[tex]\[ A \times B = \{ (a, b) \mid a \in A \text{ and } b \in B \} \][/tex]
Calculating the pairs:
[tex]\[ A \times B = \{ (0, 3), (0, -2), (1, 3), (1, -2), (2, 3), (2, -2), (3, 3), (3, -2), (4, 3), (4, -2) \} \][/tex]
Step 4: Find [tex]\( B \times A \)[/tex]
The Cartesian product [tex]\( B \times A \)[/tex] is the set of all possible ordered pairs [tex]\( (b, a) \)[/tex] where [tex]\( b \in B \)[/tex] and [tex]\( a \in A \)[/tex]. Thus:
[tex]\[ B \times A = \{ (b, a) \mid b \in B \text{ and } a \in A \} \][/tex]
Calculating the pairs:
[tex]\[ B \times A = \{ (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (-2, 0), (-2, 1), (-2, 2), (-2, 3), (-2, 4) \} \][/tex]
### Part B: Cartesian Products of Sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]
Step 1: Define Set [tex]\( P \)[/tex]
Given:
[tex]\[ P = \{ x \mid 2 < x < 7 \text{ and } x \in \mathbb{N} \} \][/tex]
This means that [tex]\( P \)[/tex] contains all natural numbers [tex]\( x \)[/tex] such that [tex]\( 2 < x < 7 \)[/tex]. Thus:
[tex]\[ P = \{ 3, 4, 5, 6 \} \][/tex]
Step 2: Define Set [tex]\( Q \)[/tex]
Given:
[tex]\[ Q = \{ x \mid x^2 = 3x \} \][/tex]
To solve for [tex]\( Q \)[/tex], we need to find the roots of the equation [tex]\( x^2 - 3x = 0 \)[/tex]. Factoring the equation we get:
[tex]\[ x(x - 3) = 0 \][/tex]
Hence, the solutions are:
[tex]\[ x = 0 \text{ and } x = 3 \][/tex]
So,
[tex]\[ Q = \{ 0, 3 \} \][/tex]
Step 3: Find [tex]\( P \times Q \)[/tex]
The Cartesian product [tex]\( P \times Q \)[/tex] is the set of all possible ordered pairs [tex]\( (p, q) \)[/tex] where [tex]\( p \in P \)[/tex] and [tex]\( q \in Q \)[/tex]. Thus:
[tex]\[ P \times Q = \{ (p, q) \mid p \in P \text{ and } q \in Q \} \][/tex]
Calculating the pairs:
[tex]\[ P \times Q = \{ (3, 0), (3, 3), (4, 0), (4, 3), (5, 0), (5, 3), (6, 0), (6, 3) \} \][/tex]
Step 4: Find [tex]\( Q \times P \)[/tex]
The Cartesian product [tex]\( Q \times P \)[/tex] is the set of all possible ordered pairs [tex]\( (q, p) \)[/tex] where [tex]\( q \in Q \)[/tex] and [tex]\( p \in P \)[/tex]. Thus:
[tex]\[ Q \times P = \{ (q, p) \mid q \in Q \text{ and } p \in P \} \][/tex]
Calculating the pairs:
[tex]\[ Q \times P = \{ (0, 3), (0, 4), (0, 5), (0, 6), (3, 3), (3, 4), (3, 5), (3, 6) \} \][/tex]
### Summary of Results
- [tex]\( A \times B = \{ (0, 3), (0, -2), (1, 3), (1, -2), (2, 3), (2, -2), (3, 3), (3, -2), (4, 3), (4, -2) \} \)[/tex]
- [tex]\( B \times A = \{ (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (-2, 0), (-2, 1), (-2, 2), (-2, 3), (-2, 4) \} \)[/tex]
- [tex]\( P \times Q = \{ (3, 0), (3, 3), (4, 0), (4, 3), (5, 0), (5, 3), (6, 0), (6, 3) \} \)[/tex]
- [tex]\( Q \times P = \{ (0, 3), (0, 4), (0, 5), (0, 6), (3, 3), (3, 4), (3, 5), (3, 6) \} \)[/tex]
### Part A: Cartesian Products of Sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex]
Step 1: Define Set [tex]\( A \)[/tex]
Given:
[tex]\[ A = \{ x \leq 4 \mid x \in \mathbb{N} \} \][/tex]
This means that [tex]\( A \)[/tex] contains all natural numbers [tex]\( x \)[/tex] such that [tex]\( x \leq 4 \)[/tex]. Thus:
[tex]\[ A = \{ 0, 1, 2, 3, 4 \} \][/tex]
Step 2: Define Set [tex]\( B \)[/tex]
Given:
[tex]\[ B = \{ x \mid x^2 - 5x - 6 = 0 \} \][/tex]
To solve for [tex]\( B \)[/tex], we must find the roots of the quadratic equation [tex]\( x^2 - 5x - 6 = 0 \)[/tex]. Factoring the quadratic equation, we get:
[tex]\[ (x - 6)(x + 1) = 0 \][/tex]
Hence, the solutions are:
[tex]\[ x = 3 \text{ and } x = -2 \][/tex]
So,
[tex]\[ B = \{ 3, -2 \} \][/tex]
Step 3: Find [tex]\( A \times B \)[/tex]
The Cartesian product [tex]\( A \times B \)[/tex] is the set of all possible ordered pairs [tex]\( (a, b) \)[/tex] where [tex]\( a \in A \)[/tex] and [tex]\( b \in B \)[/tex]. Thus:
[tex]\[ A \times B = \{ (a, b) \mid a \in A \text{ and } b \in B \} \][/tex]
Calculating the pairs:
[tex]\[ A \times B = \{ (0, 3), (0, -2), (1, 3), (1, -2), (2, 3), (2, -2), (3, 3), (3, -2), (4, 3), (4, -2) \} \][/tex]
Step 4: Find [tex]\( B \times A \)[/tex]
The Cartesian product [tex]\( B \times A \)[/tex] is the set of all possible ordered pairs [tex]\( (b, a) \)[/tex] where [tex]\( b \in B \)[/tex] and [tex]\( a \in A \)[/tex]. Thus:
[tex]\[ B \times A = \{ (b, a) \mid b \in B \text{ and } a \in A \} \][/tex]
Calculating the pairs:
[tex]\[ B \times A = \{ (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (-2, 0), (-2, 1), (-2, 2), (-2, 3), (-2, 4) \} \][/tex]
### Part B: Cartesian Products of Sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]
Step 1: Define Set [tex]\( P \)[/tex]
Given:
[tex]\[ P = \{ x \mid 2 < x < 7 \text{ and } x \in \mathbb{N} \} \][/tex]
This means that [tex]\( P \)[/tex] contains all natural numbers [tex]\( x \)[/tex] such that [tex]\( 2 < x < 7 \)[/tex]. Thus:
[tex]\[ P = \{ 3, 4, 5, 6 \} \][/tex]
Step 2: Define Set [tex]\( Q \)[/tex]
Given:
[tex]\[ Q = \{ x \mid x^2 = 3x \} \][/tex]
To solve for [tex]\( Q \)[/tex], we need to find the roots of the equation [tex]\( x^2 - 3x = 0 \)[/tex]. Factoring the equation we get:
[tex]\[ x(x - 3) = 0 \][/tex]
Hence, the solutions are:
[tex]\[ x = 0 \text{ and } x = 3 \][/tex]
So,
[tex]\[ Q = \{ 0, 3 \} \][/tex]
Step 3: Find [tex]\( P \times Q \)[/tex]
The Cartesian product [tex]\( P \times Q \)[/tex] is the set of all possible ordered pairs [tex]\( (p, q) \)[/tex] where [tex]\( p \in P \)[/tex] and [tex]\( q \in Q \)[/tex]. Thus:
[tex]\[ P \times Q = \{ (p, q) \mid p \in P \text{ and } q \in Q \} \][/tex]
Calculating the pairs:
[tex]\[ P \times Q = \{ (3, 0), (3, 3), (4, 0), (4, 3), (5, 0), (5, 3), (6, 0), (6, 3) \} \][/tex]
Step 4: Find [tex]\( Q \times P \)[/tex]
The Cartesian product [tex]\( Q \times P \)[/tex] is the set of all possible ordered pairs [tex]\( (q, p) \)[/tex] where [tex]\( q \in Q \)[/tex] and [tex]\( p \in P \)[/tex]. Thus:
[tex]\[ Q \times P = \{ (q, p) \mid q \in Q \text{ and } p \in P \} \][/tex]
Calculating the pairs:
[tex]\[ Q \times P = \{ (0, 3), (0, 4), (0, 5), (0, 6), (3, 3), (3, 4), (3, 5), (3, 6) \} \][/tex]
### Summary of Results
- [tex]\( A \times B = \{ (0, 3), (0, -2), (1, 3), (1, -2), (2, 3), (2, -2), (3, 3), (3, -2), (4, 3), (4, -2) \} \)[/tex]
- [tex]\( B \times A = \{ (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (-2, 0), (-2, 1), (-2, 2), (-2, 3), (-2, 4) \} \)[/tex]
- [tex]\( P \times Q = \{ (3, 0), (3, 3), (4, 0), (4, 3), (5, 0), (5, 3), (6, 0), (6, 3) \} \)[/tex]
- [tex]\( Q \times P = \{ (0, 3), (0, 4), (0, 5), (0, 6), (3, 3), (3, 4), (3, 5), (3, 6) \} \)[/tex]
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
