Mathematics 48. A truck carries a load of 50 boxes; some are 20 kg boxes and the rest are 25 kg boxes. If the total weight of all boxes is 1175 kg , how many of each type are there? 49. The product of two numbers is 5 . If their sum \( 9 / 2 \), find the numbers. 50. Find the domain and range of the function \( y=\sqrt{3 x-x^{2}} \). 51. Let \( R \) be a relation on the set \( A=\{1,2,3,4,5,6\} \) defined by \( R=\{(a, b): a+b \leq 9\} \) \( \begin{array}{ll}\text { (a) List the elements of } R & \text { (b) Is } R=R^{-1}\end{array} \)

Certainly! Let's tackle each of the problems step by step. --- ### **Problem 48** **A truck carries a load of 50 boxes; some are 20 kg boxes and the rest are 25 kg boxes. If the total weight of all boxes is 1175 kg, how many of each type are there?** **Solution:** Let's define: - \( x \) = number of 20 kg boxes - \( y \) = number of 25 kg boxes We have two equations based on the problem statement: 1. **Total number of boxes:** \[ x + y = 50 \] 2. **Total weight of boxes:** \[ 20x + 25y = 1175 \] **Step 1:** Solve the first equation for \( y \): \[ y = 50 - x \] **Step 2:** Substitute \( y \) into the second equation: \[ 20x + 25(50 - x) = 1175 \] \[ 20x + 1250 - 25x = 1175 \] \[ -5x + 1250 = 1175 \] \[ -5x = -75 \] \[ x = 15 \] **Step 3:** Substitute \( x = 15 \) back into \( y = 50 - x \): \[ y = 50 - 15 = 35 \] **Answer:** - **20 kg boxes:** 15 - **25 kg boxes:** 35 --- ### **Problem 49** **The product of two numbers is 5. If their sum is \( \frac{9}{2} \), find the numbers.** **Solution:** Let the two numbers be \( a \) and \( b \). We have: \[ a \times b = 5 \quad \text{(1)} \] \[ a + b = \frac{9}{2} \quad \text{(2)} \] **Step 1:** Express one variable in terms of the other using equation (2): \[ b = \frac{9}{2} - a \] **Step 2:** Substitute \( b \) into equation (1): \[ a \left( \frac{9}{2} - a \right) = 5 \] \[ \frac{9a}{2} - a^2 = 5 \] \[ - a^2 + \frac{9a}{2} - 5 = 0 \] \[ 2(-a^2) + 9a - 10 = 0 \] \[ -2a^2 + 9a - 10 = 0 \] \[ 2a^2 - 9a + 10 = 0 \] **Step 3:** Solve the quadratic equation: \[ a = \frac{9 \pm \sqrt{81 - 80}}{4} = \frac{9 \pm 1}{4} \] So, \[ a = \frac{10}{4} = 2.5 \quad \text{or} \quad a = \frac{8}{4} = 2 \] **Step 4:** Find \( b \) for each \( a \): - If \( a = 2.5 \), then \( b = \frac{9}{2} - 2.5 = 2 \) - If \( a = 2 \), then \( b = \frac{9}{2} - 2 = 2.5 \) **Answer:** The two numbers are **2** and **2.5**. --- ### **Problem 50** **Find the domain and range of the function \( y = \sqrt{3x - x^2} \).** **Solution:** The function involves a square root, so the expression inside must be non-negative: \[ 3x - x^2 \geq 0 \] \[ x(3 - x) \geq 0 \] **Step 1:** Solve the inequality \( x(3 - x) \geq 0 \): - \( x = 0 \) and \( x = 3 \) are the critical points. - Test intervals: - \( x < 0 \): Negative - \( 0 \leq x \leq 3 \): Positive or zero - \( x > 3 \): Negative **Domain:** \[ x \in [0, 3] \] **Step 2:** Find the range by determining the possible values of \( y \): \[ y = \sqrt{3x - x^2} \] The maximum value of \( 3x - x^2 \) occurs at the vertex of the parabola \( y = -x^2 + 3x \): \[ x = \frac{-b}{2a} = \frac{-3}{2(-1)} = 1.5 \] \[ \text{Maximum } y = \sqrt{3(1.5) - (1.5)^2} = \sqrt{4.5 - 2.25} = \sqrt{2.25} = 1.5 \] **Range:** \[ y \in [0, 1.5] \] **Answer:** - **Domain:** \( [0, 3] \) - **Range:** \( [0, 1.5] \) --- ### **Problem 51** **Let \( R \) be a relation on the set \( A = \{1, 2, 3, 4, 5, 6\} \) defined by \( R = \{(a, b) : a + b \leq 9\} \).** #### **(a) List the elements of \( R \).** **Solution:** We need to list all ordered pairs \( (a, b) \) where \( a, b \in A \) and \( a + b \leq 9 \). **List of Pairs:** - **When \( a = 1 \):** \[ (1,1),\ (1,2),\ (1,3),\ (1,4),\ (1,5),\ (1,6) \quad [1+6 = 7 \leq 9] \] - **When \( a = 2 \):** \[ (2,1),\ (2,2),\ (2,3),\ (2,4),\ (2,5),\ (2,6) \quad [2+6 = 8 \leq 9] \] - **When \( a = 3 \):** \[ (3,1),\ (3,2),\ (3,3),\ (3,4),\ (3,5),\ (3,6) \quad [3+6 = 9 \leq 9] \] - **When \( a = 4 \):** \[ (4,1),\ (4,2),\ (4,3),\ (4,4),\ (4,5) \quad [4+5 = 9 \leq 9] \] - **When \( a = 5 \):** \[ (5,1),\ (5,2),\ (5,3),\ (5,4) \quad [5+4 = 9 \leq 9] \] - **When \( a = 6 \):** \[ (6,1),\ (6,2),\ (6,3) \quad [6+3 = 9 \leq 9] \] **Complete List of Elements in \( R \):** \[ \begin{aligned} R = \{ & (1,1),\ (1,2),\ (1,3),\ (1,4),\ (1,5),\ (1,6), \\ & (2,1),\ (2,2),\ (2,3),\ (2,4),\ (2,5),\ (2,6), \\ & (3,1),\ (3,2),\ (3,3),\ (3,4),\ (3,5),\ (3,6), \\ & (4,1),\ (4,2),\ (4,3),\ (4,4),\ (4,5), \\ & (5,1),\ (5,2),\ (5,3),\ (5,4), \\ & (6,1),\ (6,2),\ (6,3) \} \end{aligned} \] #### **(b) Is \( R = R^{-1} \)?** **Solution:** \( R^{-1} \) is the inverse relation, consisting of all pairs \( (b, a) \) where \( (a, b) \in R \). To determine if \( R = R^{-1} \), we need to check if for every \( (a, b) \) in \( R \), the pair \( (b, a) \) is also in \( R \). **Analysis:** - Since \( a + b = b + a \), if \( a + b \leq 9 \), then \( b + a \leq 9 \) as well. - This means that if \( (a, b) \) is in \( R \), then \( (b, a) \) is automatically in \( R \). **Conclusion:** Yes, \( R \) is equal to its inverse relation \( R^{-1} \). **Answer:** - **(a)** All ordered pairs \( (a, b) \) where \( a, b \in \{1,2,3,4,5,6\} \) and \( a + b \leq 9 \) are: \[ \begin{aligned} R = \{ & (1,1),\ (1,2),\ (1,3),\ (1,4),\ (1,5),\ (1,6), \\ & (2,1),\ (2,2),\ (2,3),\ (2,4),\ (2,5),\ (2,6), \\ & (3,1),\ (3,2),\ (3,3),\ (3,4),\ (3,5),\ (3,6), \\ & (4,1),\ (4,2),\ (4,3),\ (4,4),\ (4,5), \\ & (5,1),\ (5,2),\ (5,3),\ (5,4), \\ & (6,1),\ (6,2),\ (6,3) \} \end{aligned} \] - **(b)** Yes, \( R = R^{-1} \) because for every \( (a, b) \) in \( R \), the pair \( (b, a) \) is also in \( R \).