💡First, we know that the slope of line ( l_1 ) is ( m_1 = 1 ).Since ( l_1 ) is perpendicular to ( l_2 ), their slopes are negative reciprocals of each other. Therefore, the slope of line ( l_2 ) is ( m_2 = -frac{1}{m_1} = -1 ).To find the angle of inclination of line ( l_2 ), we use the formula ( theta = tan^{-1}(m_2) ).Thus, the angle of inclination of line ( l_2 ) is ( theta = tan^{-1}(-1) = 135^{circ} ).Therefore, the answer is (boxed{B: 135^{circ}}).

💡First, I need to determine the areas of triangles ABC and XYZ. For triangle ABC, the coordinates are A(2, 0), B(8, 12), and C(14, 0). I'll use the formula for the area of a triangle with vertices at ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)):[text{Area} = frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|]Plugging in the coordinates for triangle ABC:[text{Area}_{ABC} = frac{1}{2} |2(12 - 0) + 8(0 - 0) + 14(0 - 12)| = frac{1}{2} |24 + 0 - 168| = frac{1}{2} | -144 | = 72]Next, for triangle XYZ with coordinates X(6, 0), Y(8, 4), and Z(10, 0), I'll apply the same formula:[text{Area}_{XYZ} = frac{1}{2} |6(4 - 0) + 8(0 - 0) + 10(0 - 4)| = frac{1}{2} |24 + 0 - 40| = frac{1}{2} | -16 | = 8]Finally, to find the fraction ( d ) of the area of triangle ABC that is the area of triangle XYZ:[d = frac{text{Area}_{XYZ}}{text{Area}_{ABC}} = frac{8}{72} = frac{1}{9}]

💡First, I need to reflect the center of the circle across the x-axis. Reflecting a point across the x-axis changes the sign of its y-coordinate. So, the center (4, -3) becomes (4, 3).Next, I will translate the reflected point 5 units to the right. Translating to the right increases the x-coordinate by 5, so (4, 3) becomes (9, 3).Finally, I will translate the point 3 units up. Translating up increases the y-coordinate by 3, so (9, 3) becomes (9, 6).Therefore, the coordinates of the image of the center of circle R after the reflection and translation are (9, 6).

💡Okay, so I have this problem here: I need to find the value of 'a' such that the minimum value of the function f(x) = cos(2x) - 2a(1 + cosx) is -1/2. Hmm, okay. Let me try to break this down step by step.First, I remember that cos(2x) can be expressed using a double-angle formula. I think it's cos(2x) = 2cos²x - 1. Let me verify that. Yeah, that seems right. So, substituting that into the function, I get:f(x) = 2cos²x - 1 - 2a(1 + cosx)Let me expand that out:f(x) = 2cos²x - 1 - 2a - 2a cosxHmm, so that simplifies to:f(x) = 2cos²x - 2a cosx - (1 + 2a)Okay, so now I have a quadratic in terms of cosx. Let me denote y = cosx to make it easier. So, the function becomes:f(y) = 2y² - 2a y - (1 + 2a)Now, since y = cosx, the range of y is between -1 and 1. So, y ∈ [-1, 1].Now, I need to find the minimum value of this quadratic function f(y) over the interval y ∈ [-1, 1]. The minimum value is given as -1/2, so I need to find 'a' such that the minimum of f(y) is -1/2.Quadratic functions have their extrema either at the vertex or at the endpoints of the interval. So, I should check both possibilities.First, let's find the vertex of the quadratic. The general form of a quadratic is f(y) = Ay² + By + C, and the vertex occurs at y = -B/(2A). In our case, A = 2, B = -2a. So, the vertex is at:y = -(-2a)/(2*2) = (2a)/4 = a/2So, the vertex is at y = a/2. Now, I need to check if this vertex lies within the interval [-1, 1]. If it does, then the minimum (since the parabola opens upwards, as A = 2 > 0) will be at the vertex. If it doesn't, then the minimum will be at one of the endpoints.So, let's consider different cases based on the value of 'a'.Case 1: a/2 is within [-1, 1], which means a ∈ [-2, 2]. In this case, the minimum occurs at y = a/2.Case 2: a/2 < -1, which implies a < -2. Then, the minimum occurs at y = -1.Case 3: a/2 > 1, which implies a > 2. Then, the minimum occurs at y = 1.Now, let's compute the minimum value in each case.Starting with Case 1: a ∈ [-2, 2], minimum at y = a/2.Plugging y = a/2 into f(y):f(a/2) = 2*(a/2)² - 2a*(a/2) - (1 + 2a)= 2*(a²/4) - a² - 1 - 2a= (a²/2) - a² - 1 - 2a= (-a²/2) - 2a - 1So, the minimum value is (-a²/2) - 2a - 1. We are told this equals -1/2.So, set up the equation:(-a²/2) - 2a - 1 = -1/2Let's solve for 'a':Multiply both sides by 2 to eliminate the fraction:- a² - 4a - 2 = -1Bring all terms to one side:- a² - 4a - 2 + 1 = 0- a² - 4a - 1 = 0Multiply both sides by -1 to make it standard:a² + 4a + 1 = 0Now, solve this quadratic equation using the quadratic formula:a = [-b ± sqrt(b² - 4ac)] / (2a)Here, a = 1, b = 4, c = 1.So,a = [-4 ± sqrt(16 - 4*1*1)] / 2= [-4 ± sqrt(12)] / 2= [-4 ± 2*sqrt(3)] / 2= -2 ± sqrt(3)So, the solutions are a = -2 + sqrt(3) and a = -2 - sqrt(3).Now, since we are in Case 1 where a ∈ [-2, 2], let's check if these solutions fall within this interval.Compute -2 + sqrt(3):sqrt(3) ≈ 1.732, so -2 + 1.732 ≈ -0.268, which is within [-2, 2].Compute -2 - sqrt(3):-2 - 1.732 ≈ -3.732, which is less than -2, so it falls outside the interval for Case 1.Therefore, in Case 1, the valid solution is a = -2 + sqrt(3).Now, let's check the other cases to ensure there are no other solutions.Case 2: a < -2, minimum at y = -1.Compute f(-1):f(-1) = 2*(-1)² - 2a*(-1) - (1 + 2a)= 2*1 + 2a - 1 - 2a= 2 + 2a - 1 - 2a= 1So, the minimum value is 1, which is greater than -1/2. Therefore, this case does not satisfy the condition that the minimum is -1/2.Case 3: a > 2, minimum at y = 1.Compute f(1):f(1) = 2*(1)² - 2a*(1) - (1 + 2a)= 2 - 2a - 1 - 2a= 1 - 4aWe are told the minimum is -1/2, so set 1 - 4a = -1/2.Solving for 'a':1 - 4a = -1/2Subtract 1 from both sides:-4a = -3/2Divide both sides by -4:a = (-3/2)/(-4) = (3/2)/4 = 3/8But wait, in Case 3, we assumed a > 2, but 3/8 is approximately 0.375, which is less than 2. Therefore, this solution does not satisfy the condition a > 2. Hence, no solution in this case.Therefore, the only valid solution is a = -2 + sqrt(3).Let me just double-check my calculations to make sure I didn't make any mistakes.Starting from f(x) = cos(2x) - 2a(1 + cosx). Using the double-angle formula, I correctly transformed it into 2cos²x - 1 - 2a - 2a cosx. Then, substituting y = cosx, I got f(y) = 2y² - 2a y - (1 + 2a). That seems correct.Finding the vertex at y = a/2, and considering the cases for a. Then, computing f(a/2) led me to (-a²/2) - 2a - 1. Setting that equal to -1/2, solving the quadratic, and getting a = -2 ± sqrt(3). Checking the interval, only a = -2 + sqrt(3) is valid.In Cases 2 and 3, the minimum values were 1 and 1 - 4a respectively, which didn't satisfy the required minimum of -1/2. So, yeah, I think that's correct.Just to be thorough, let me plug a = -2 + sqrt(3) back into the original function and see if the minimum is indeed -1/2.Compute f(x) with a = -2 + sqrt(3):f(x) = cos(2x) - 2*(-2 + sqrt(3))(1 + cosx)= cos(2x) + 4(1 + cosx) - 2*sqrt(3)(1 + cosx)= cos(2x) + 4 + 4cosx - 2*sqrt(3) - 2*sqrt(3)cosxHmm, that seems a bit messy. Maybe it's better to compute the minimum value directly using the expression we had earlier.We had f(y) = 2y² - 2a y - (1 + 2a), and at y = a/2, the minimum is (-a²/2) - 2a - 1.Plugging a = -2 + sqrt(3):First, compute a²:a = -2 + sqrt(3)a² = (-2 + sqrt(3))² = 4 - 4*sqrt(3) + 3 = 7 - 4*sqrt(3)Now, (-a²/2) - 2a - 1:= (- (7 - 4*sqrt(3))/2 ) - 2*(-2 + sqrt(3)) - 1= (-7/2 + 2*sqrt(3)) + 4 - 2*sqrt(3) - 1= (-7/2 + 4 - 1) + (2*sqrt(3) - 2*sqrt(3))= (-7/2 + 3) + 0= (-7/2 + 6/2)= (-1/2)Yes, that checks out. So, the minimum value is indeed -1/2 when a = -2 + sqrt(3).Therefore, I'm confident that the correct value of 'a' is -2 + sqrt(3).

💡To determine the value of ( p ) for which the equation ( frac{x-3}{x-4} = frac{x-p}{x-8} ) has no solution, follow these steps:1. **Identify the domain restrictions**: The equation is undefined for ( x = 4 ) and ( x = 8 ) because these values make the denominators zero.2. **Cross-multiply to eliminate the fractions**: [ (x - 3)(x - 8) = (x - p)(x - 4) ]3. **Expand both sides**: [ x^2 - 11x + 24 = x^2 - (p + 4)x + 4p ]4. **Simplify the equation by subtracting ( x^2 ) from both sides**: [ -11x + 24 = -(p + 4)x + 4p ]5. **Rearrange the terms to isolate ( x )**: [ -11x + (p + 4)x = 4p - 24 ] [ (p - 7)x = 4p - 24 ]6. **Determine the value of ( p ) that leads to no solution**: - For the equation to have no solution, the coefficient of ( x ) must be zero, and the constant term must be non-zero. - Set ( p - 7 = 0 ) to make the coefficient of ( x ) zero: [ p = 7 ] - Substitute ( p = 7 ) into the constant term: [ 4(7) - 24 = 28 - 24 = 4 ] - This results in the equation ( 0x = 4 ), which is a contradiction and has no solution.Therefore, the value of ( p ) that makes the equation have no solution is ( 7 ).

💡First, I need to determine the total number of fishes caught by each team member over the 5-day competition.Jackson catches 6 fishes per day. Over 5 days, he catches:6 fishes/day * 5 days = 30 fishesJonah catches 4 fishes per day. Over 5 days, he catches:4 fishes/day * 5 days = 20 fishesGeorge catches 8 fishes per day. Over 5 days, he catches:8 fishes/day * 5 days = 40 fishesNow, I will add the number of fishes caught by each team member to find the total number of fishes caught by the team:30 fishes (Jackson) + 20 fishes (Jonah) + 40 fishes (George) = 90 fishesTherefore, the team caught a total of 90 fishes throughout the competition.