Let's solve each part step-by-step, finding the **x** and **y** components of each vector. ### (a) Given: - Vector \( r = 23 \) km - Angle from x-axis = \( 25^\circ \) (above x-axis) The components are: - \( r_x = r \cos(25^\circ) \) - \( r_y = r \sin(25^\circ) \) **Calculations:** - \( r_x = 23 \times \cos(25^\circ) \) - \( r_y = 23 \times \sin(25^\circ) \) Now, using a calculator: - \( \cos(25^\circ) \approx 0.9063 \) - \( \sin(25^\circ) \approx 0.4226 \) So, - \( r_x = 23 \times 0.9063 \) - \( r_x \approx 20.85 \) km - \( r_y = 23 \times 0.4226 \) - \( r_y \approx 9.72 \) km **Final answers (a):** | | x | y | |---|-------|-------| | | 20.85 | 9.72 | --- ### (b) Given: - Vector \( v = 17.6 \) m/s - Angle below x-axis = \( 35^\circ \) (below the x-axis, so y-component is negative) The components are: - \( v_x = v \cos(35^\circ) \) - \( v_y = -v \sin(35^\circ) \) (negative because it is below the axis) **Calculations:** - \( v_x = 17.6 \times \cos(35^\circ) \) - \( v_y = -17.6 \times \sin(35^\circ) \) Now, using a calculator: - \( \cos(35^\circ) \approx 0.8192 \) - \( \sin(35^\circ) \approx 0.5736 \) So, - \( v_x = 17.6 \times 0.8192 \) - \( v_x \approx 14.42 \) m/s - \( v_y = -17.6 \times 0.5736 \) - \( v_y \approx -10.10 \) m/s **Final answers (b):** | | x | y | |---|-------|--------| | | 14.42 | -10.10 | --- ### (c) Given: - \( a = g = 9.8 \) m/s² - The vector is straight down (negative y-axis) So: - \( a_x = 0 \) - \( a_y = -9.8 \) m/s² **Final answers (c):** | | x | y | |---|---|--------| | | 0 | -9.8 | --- ### **Summary Table** #### (a) | | x (km) | y (km) | |---|---------|--------| | | 20.85 | 9.72 | #### (b) | | x (m/s) | y (m/s) | |---|----------|---------| | | 14.42 | -10.10 | #### (c) | | x (m/s²) | y (m/s²) | |---|----------|----------| | | 0 | -9.8 | Let me know if you need a further breakdown or explanation!

# Basketball Players' Parabolic Paths ## Part A: Finding Intersection Points Given the functions for the paths of the basketballs: - \( g(x) = -2x^2 + 13x + 7 \) (First player) - \( h(x) = -x^2 + 4x + 21 \) (Second player) ### Step 1: Set the equations equal to each other To find the distances where the two paths cross, we set \( g(x) = h(x) \): \[ -2x^2 + 13x + 7 = -x^2 + 4x + 21 \] ### Step 2: Rearrange the equation Move all terms to one side: \[ -2x^2 + 13x + 7 + x^2 - 4x - 21 = 0 \] Combine like terms: \[ -x^2 + 9x - 14 = 0 \] ### Step 3: Multiply by -1 for simplicity \[ x^2 - 9x + 14 = 0 \] ### Step 4: Factor the quadratic \[ (x - 7)(x - 2) = 0 \] ### Step 5: Solve for \( x \) Setting each factor to zero gives: \[ x - 7 = 0 \quad \Rightarrow \quad x = 7 \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] ### Final Answer for Part A The distances where the two basketball paths cross are: - \( x = 2 \) - \( x = 7 \) --- ## Part B: Solving for \( f(x) \) Given the function: \[ f(x) = \frac{1}{x} \] ### Step 1: Simplify \( f(x) \) The function is already in its simplest form. ### Step 2: Determine discontinuities A function is discontinuous where its denominator is zero. For \( f(x) \): \[ x = 0 \] ### Final Answer for Part B The function \( f(x) = \frac{1}{x} \) has a discontinuity at: - \( x = 0 \) This means \( f(x) \) is undefined at this point.

# Basketball Players' Parabolic Paths ## Part A: Finding Intersection Points Given the functions for the paths of the basketballs: - \( g(x) = -2x^2 + 13x + 7 \) (First player) - \( h(x) = -x^2 + 4x + 21 \) (Second player) ### Step 1: Set the equations equal to each other To find the distances where the two paths cross, we set \( g(x) = h(x) \): \[ -2x^2 + 13x + 7 = -x^2 + 4x + 21 \] ### Step 2: Rearrange the equation Move all terms to one side: \[ -2x^2 + 13x + 7 + x^2 - 4x - 21 = 0 \] Combine like terms: \[ -x^2 + 9x - 14 = 0 \] ### Step 3: Multiply by -1 for simplicity \[ x^2 - 9x + 14 = 0 \] ### Step 4: Factor the quadratic \[ (x - 7)(x - 2) = 0 \] ### Step 5: Solve for \( x \) Setting each factor to zero gives: \[ x - 7 = 0 \quad \Rightarrow \quad x = 7 \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] ### Final Answer for Part A The distances where the two basketball paths cross are: - \( x = 2 \) - \( x = 7 \) --- ## Part B: Solving for \( f(x) \) Given the function: \[ f(x) = \frac{g(x)}{h(x)} = \frac{-2x^2 + 13x + 7}{-x^2 + 4x + 21} \] ### Step 1: Simplify \( f(x) \) To simplify, we will factor both \( g(x) \) and \( h(x) \). **Factoring \( g(x) \)**: - We already have \( g(x) = -2(x - 7)(x - 2) \). **Factoring \( h(x) \)**: - We need to factor \( -x^2 + 4x + 21 \): Using the quadratic formula or factoring method: \[ h(x) = - (x^2 - 4x - 21) \] Now, we need to find factors of \( -21 \) that sum to \( -4 \), which are \( -7 \) and \( 3 \): \[ h(x) = - (x - 7)(x + 3) \] ### Step 2: Combine into \( f(x) \) Now substituting back into \( f(x) \): \[ f(x) = \frac{-2(x - 7)(x - 2)}{-(x - 7)(x + 3)} \] Cancelling the common factor \( -(x - 7) \) (valid for \( x \neq 7 \)): \[ f(x) = \frac{2(x - 2)}{x + 3} \] ### Step 3: Determine discontinuities A function is discontinuous where its denominator is zero. For \( f(x) \): Set the denominator equal to zero: \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] ### Final Answer for Part B The simplified form of \( f(x) \) is: \[ f(x) = \frac{2(x - 2)}{x + 3} \quad (x \neq 7) \] The function \( f(x) \) has a discontinuity at: - \( x = -3 \) Additionally, \( f(x) \) is undefined at \( x = 7 \) due to cancellation. Thus, the discontinuities are: - \( x = -3 \) - \( x = 7 \)

# Calculation of CD Maturity Value To calculate the amount the theater will receive when cashing in the Certificate of Deposit (CD), we will use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: - \( A \) = the amount of money accumulated after n years, including interest. - \( P \) = the principal amount (the initial amount of money). - \( r \) = annual interest rate (decimal). - \( n \) = number of times that interest is compounded per unit \( t \). - \( t \) = the time the money is invested for in years. ### Given Values - \( P = 1600 \) (the initial investment) - \( r = 3.5\% = 0.035 \) (annual interest rate) - \( n = 12 \) (compounded monthly) - \( t = \frac{18}{12} = 1.5 \) years (18 months) ### Step 1: Plug Values into the Formula Substituting the values into the formula: \[ A = 1600 \left(1 + \frac{0.035}{12}\right)^{12 \times 1.5} \] ### Step 2: Calculate the values 1. Calculate \( \frac{r}{n} \): \[ \frac{0.035}{12} \approx 0.00291667 \] 2. Calculate \( nt \): \[ 12 \times 1.5 = 18 \] 3. Now substitute back into the formula: \[ A = 1600 \left(1 + 0.00291667\right)^{18} \] \[ A = 1600 \left(1.00291667\right)^{18} \] 4. Calculate \( (1.00291667)^{18} \): Using a calculator, \( (1.00291667)^{18} \approx 1.055273 \). 5. Finally, calculate \( A \): \[ A \approx 1600 \times 1.055273 \approx 1688.43 \] ### Final Answer The theater will receive **$1688.43** when it cashes in the CD.

# Solution Using Gaussian-Jordan Elimination We will solve the following system of linear equations: \[ \begin{align*} 1. & \quad 4x + y + z = 5 \\ 2. & \quad -2x - 2y + z = 5 \\ 3. & \quad 2x + 3y - z = -7 \end{align*} \] ### Step 1: Write in Augmented Matrix Form The augmented matrix corresponding to the system is: \[ \begin{bmatrix} 4 & 1 & 1 & | & 5 \\ -2 & -2 & 1 & | & 5 \\ 2 & 3 & -1 & | & -7 \end{bmatrix} \] ### Step 2: Row Reduction **Step 2.1: Make the leading coefficient of the first row to be 1** We can divide the first row by 4: \[ R_1 = \frac{1}{4}R_1 \Rightarrow \begin{bmatrix} 1 & \frac{1}{4} & \frac{1}{4} & | & \frac{5}{4} \\ -2 & -2 & 1 & | & 5 \\ 2 & 3 & -1 & | & -7 \end{bmatrix} \] **Step 2.2: Eliminate the x-term from the second and third rows** \[ R_2 = R_2 + 2R_1 \quad \text{and} \quad R_3 = R_3 - 2R_1 \] Calculating these: \[ R_2: -2 + 2(1) = 0, \quad -2 + 2\left(\frac{1}{4}\right) = -2 + \frac{1}{2} = -\frac{3}{2}, \quad 1 + 2\left(\frac{1}{4}\right) = 1 + \frac{1}{2} = \frac{3}{2}, \quad 5 + 2\left(\frac{5}{4}\right) = 5 + \frac{10}{4} = \frac{20}{4} + \frac{10}{4} = \frac{30}{4} = \frac{15}{2} \] \[ R_3: 2 - 2(1) = 0, \quad 3 - 2\left(\frac{1}{4}\right) = 3 - \frac{1}{2} = \frac{6}{2} - \frac{1}{2} = \frac{5}{2}, \quad -1 - 2\left(\frac{1}{4}\right) = -1 - \frac{1}{2} = -\frac{3}{2}, \quad -7 - 2\left(\frac{5}{4}\right) = -7 - \frac{10}{4} = -\frac{28}{4} - \frac{10}{4} = -\frac{38}{4} = -\frac{19}{2} \] The augmented matrix now looks like: \[ \begin{bmatrix} 1 & \frac{1}{4} & \frac{1}{4} & | & \frac{5}{4} \\ 0 & -\frac{3}{2} & \frac{3}{2} & | & \frac{15}{2} \\ 0 & \frac{5}{2} & -\frac{3}{2} & | & -\frac{19}{2} \end{bmatrix} \] **Step 2.3: Make the leading coefficient of the second row to be 1** Multiply the second row by \(-\frac{2}{3}\): \[ R_2 = -\frac{2}{3}R_2 \Rightarrow \begin{bmatrix} 1 & \frac{1}{4} & \frac{1}{4} & | & \frac{5}{4} \\ 0 & 1 & -1 & | & -5 \\ 0 & \frac{5}{2} & -\frac{3}{2} & | & -\frac{19}{2} \end{bmatrix} \] **Step 2.4: Eliminate the y-term from the third row** \[ R_3 = R_3 - \frac{5}{2}R_2 \] Calculating this: For \(R_3\): - \( 0 \) remains \( 0 \) - \( \frac{5}{2} - \frac{5}{2}(1) = 0 \) - \(-\frac{3}{2} + \frac{5}{2}(1) = -\frac{3}{2} + \frac{5}{2} = 1\) - \(-\frac{19}{2} - \frac{5}{2}(-5) = -\frac{19}{2} + \frac{25}{2} = 3\) The augmented matrix now looks like: \[ \begin{bmatrix} 1 & \frac{1}{4} & \frac{1}{4} & | & \frac{5}{4} \\ 0 & 1 & -1 & | & -5 \\ 0 & 0 & 1 & | & 3 \end{bmatrix} \] ### Step 3: Back Substitution **From \( R_3 \)**: \[ z = 3 \] **Substituting into \( R_2 \)**: \[ y - z = -5 \Rightarrow y - 3 = -5 \Rightarrow y = -2 \] **Substituting into \( R_1 \)**: \[ 4x + y + z = 5 \Rightarrow 4x - 2 + 3 = 5 \Rightarrow 4x + 1 = 5 \Rightarrow 4x = 4 \Rightarrow x = 1 \] ### Final Solution The solution to the system of equations is: \[ \begin{align*} x &= 1 \\ y &= -2 \\ z &= 3 \end{align*} \]

# Solution Using Gaussian-Jordan Elimination We will solve the following system of linear equations: \[ \begin{align*} 1. & \quad 4x + y + z = 5 \\ 2. & \quad -2x - 2y + z = 5 \\ 3. & \quad 2x + 3y - z = -7 \end{align*} \] ## Step 1: Write in Augmented Matrix Form The augmented matrix corresponding to the system is: \[ \begin{bmatrix} 4 & 1 & 1 & | & 5 \\ -2 & -2 & 1 & | & 5 \\ 2 & 3 & -1 & | & -7 \end{bmatrix} \] ## Step 2: Row Reduction ### Step 2.1: Make the leading coefficient of the first row to be 1 We can divide the first row by 4: \[ R_1 = \frac{1}{4}R_1 \Rightarrow \begin{bmatrix} 1 & \frac{1}{4} & \frac{1}{4} & | & \frac{5}{4} \\ -2 & -2 & 1 & | & 5 \\ 2 & 3 & -1 & | & -7 \end{bmatrix} \] ### Step 2.2: Eliminate the x-term from the second and third rows \[ R_2 = R_2 + 2R_1 \quad \text{and} \quad R_3 = R_3 - 2R_1 \] Calculating these: - For \( R_2 \): \[ R_2: -2 + 2(1) = 0, \quad -2 + 2\left(\frac{1}{4}\right) = -2 + \frac{1}{2} = -\frac{3}{2}, \quad 1 + 2\left(\frac{1}{4}\right) = 1 + \frac{1}{2} = \frac{3}{2}, \quad 5 + 2\left(\frac{5}{4}\right) = 5 + \frac{10}{4} = \frac{30}{4} = \frac{15}{2} \] - For \( R_3 \): \[ R_3: 2 - 2(1) = 0, \quad 3 - 2\left(\frac{1}{4}\right) = 3 - \frac{1}{2} = \frac{5}{2}, \quad -1 - 2\left(\frac{1}{4}\right) = -1 - \frac{1}{2} = -\frac{3}{2}, \quad -7 - 2\left(\frac{5}{4}\right) = -7 - \frac{10}{4} = -\frac{38}{4} = -\frac{19}{2} \] The augmented matrix now looks like: \[ \begin{bmatrix} 1 & \frac{1}{4} & \frac{1}{4} & | & \frac{5}{4} \\ 0 & -\frac{3}{2} & \frac{3}{2} & | & \frac{15}{2} \\ 0 & \frac{5}{2} & -\frac{3}{2} & | & -\frac{19}{2} \end{bmatrix} \] ### Step 2.3: Make the leading coefficient of the second row to be 1 Multiply the second row by \(-\frac{2}{3}\): \[ R_2 = -\frac{2}{3}R_2 \Rightarrow \begin{bmatrix} 1 & \frac{1}{4} & \frac{1}{4} & | & \frac{5}{4} \\ 0 & 1 & -1 & | & -5 \\ 0 & \frac{5}{2} & -\frac{3}{2} & | & -\frac{19}{2} \end{bmatrix} \] ### Step 2.4: Eliminate the y-term from the third row \[ R_3 = R_3 - \frac{5}{2}R_2 \] Calculating this: - For \( R_3 \): - \( 0 \) remains \( 0 \) - \( \frac{5}{2} - \frac{5}{2}(1) = 0 \) - \(-\frac{3}{2} + \frac{5}{2}(1) = -\frac{3}{2} + \frac{5}{2} = 1\) - \(-\frac{19}{2} - \frac{5}{2}(-5) = -\frac{19}{2} + \frac{25}{2} = 3\) The augmented matrix now looks like: \[ \begin{bmatrix} 1 & \frac{1}{4} & \frac{1}{4} & | & \frac{5}{4} \\ 0 & 1 & -1 & | & -5 \\ 0 & 0 & 1 & | & 3 \end{bmatrix} \] ## Step 3: Back Substitution **From \( R_3 \)**: \[ z = 3 \] **Substituting into \( R_2 \)**: \[ y - z = -5 \Rightarrow y - 3 = -5 \Rightarrow y = -2 \] **Substituting into \( R_1 \)**: \[ 4x + y + z = 5 \Rightarrow 4x - 2 + 3 = 5 \Rightarrow 4x + 1 = 5 \Rightarrow 4x = 4 \Rightarrow x = 1 \] ## Final Solution The solution to the system of equations is: \[ \begin{align*} x &= 1 \\ y &= -2 \\ z &= 3 \end{align*} \]