answer:# Solution 1: Detailed Steps To find y, we start by calculating the projection of mathbf{v} onto mathbf{w}, denoted as text{proj}_{mathbf{w}} mathbf{v}. The formula for the projection is given by: [ text{proj}_{mathbf{w}} mathbf{v} = frac{mathbf{v} cdot mathbf{w}}{mathbf{w} cdot mathbf{w}} mathbf{w}. ] Substituting the given vectors mathbf{v} = begin{pmatrix} 1 y end{pmatrix} and mathbf{w} = begin{pmatrix} 9 3 end{pmatrix} into the formula, we get: begin{align*} text{proj}_{mathbf{w}} mathbf{v} &= frac{begin{pmatrix} 1 y end{pmatrix} cdot begin{pmatrix} 9 3 end{pmatrix}}{begin{pmatrix} 9 3 end{pmatrix} cdot begin{pmatrix} 9 3 end{pmatrix}} begin{pmatrix} 9 3 end{pmatrix} &= frac{9 + 3y}{9^2 + 3^2} begin{pmatrix} 9 3 end{pmatrix} &= frac{9 + 3y}{81 + 9} begin{pmatrix} 9 3 end{pmatrix} &= frac{3 + y}{30} begin{pmatrix} 9 3 end{pmatrix}. end{align*} Given that text{proj}_{mathbf{w}} mathbf{v} = begin{pmatrix} -6 -2 end{pmatrix}, we equate the two expressions: [ frac{3 + y}{30} begin{pmatrix} 9 3 end{pmatrix} = begin{pmatrix} -6 -2 end{pmatrix}. ] This implies that the scalar multiplier of begin{pmatrix} 9 3 end{pmatrix} to get begin{pmatrix} -6 -2 end{pmatrix} is -frac{2}{3}. Therefore, we set up the equation: [ frac{3 + y}{30} = -frac{2}{3}. ] Solving for y, we multiply both sides by 30 and simplify: begin{align*} 3 + y &= -20 y &= -23. end{align*} Thus, the value of y is boxed{-23}. # Solution 2: Geometric Approach Given that mathbf{v} - text{proj}_{mathbf{w}} mathbf{v} and mathbf{w} are orthogonal, their dot product equals zero: [ (mathbf{v} - text{proj}_{mathbf{w}} mathbf{v}) cdot mathbf{w} = 0. ] Substituting mathbf{v} = begin{pmatrix} 1 y end{pmatrix}, text{proj}_{mathbf{w}} mathbf{v} = begin{pmatrix} -6 -2 end{pmatrix}, and mathbf{w} = begin{pmatrix} 9 3 end{pmatrix}, we get: [ begin{pmatrix} 1 - (-6) y - (-2) end{pmatrix} cdot begin{pmatrix} 9 3 end{pmatrix} = 0, ] which simplifies to: [ begin{pmatrix} 7 y + 2 end{pmatrix} cdot begin{pmatrix} 9 3 end{pmatrix} = 0. ] Expanding the dot product, we have: [ 7 cdot 9 + (y + 2) cdot 3 = 0. ] Solving for y, we find: begin{align*} 63 + 3y + 6 &= 0 3y &= -69 y &= -23. end{align*} Therefore, the value of y is boxed{-23}.

answer:Solution: Since a > b > 0, the equation of the ellipse C_{1} is dfrac{x^{2}}{a^{2}} + dfrac{y^{2}}{b^{2}} = 1. The eccentricity of C_{1} is dfrac{sqrt{a^{2}-b^{2}}}{a}. The equation of the hyperbola C_{2} is dfrac{x^{2}}{a^{2}} - dfrac{y^{2}}{b^{2}} = 1. The eccentricity of C_{2} is dfrac{sqrt{a^{2}+b^{2}}}{a}. Since the product of the eccentricities of C_{1} and C_{2} is dfrac{sqrt{3}}{2}, dfrac{sqrt{a^{2}-b^{2}}}{a} cdot dfrac{sqrt{a^{2}+b^{2}}}{a} = dfrac{sqrt{3}}{2}, left(dfrac{b}{a}right)^{2} = dfrac{1}{2}, dfrac{b}{a} = dfrac{sqrt{2}}{2}, The equation of the asymptotes of C_{2} is: y = pm dfrac{sqrt{2}}{2}x, which is equivalent to x pm sqrt{2}y = 0. Therefore, the correct choice is boxed{B}. This problem examines the basic properties of ellipses and hyperbolas, including eccentricity and the equation of asymptotes. Understanding the relationship between the eccentricities of an ellipse and a hyperbola and establishing equations based on this relationship is key to solving this problem.

answer:1. **Base Case Verification**: - a_1 = 2 and a_2 = frac{5}{11} are given as initial terms. 2. **Inductive Hypothesis**: - Assume a_{m-1} = frac{5}{6m-1} and a_m = frac{5}{6m+5} correctly represent the terms for some m geq 2. 3. **Inductive Step**: - For n = m+1, using the recursive formula: [ a_{m+1} = frac{2a_{m-1} cdot a_m}{3a_{m-1} - a_m} ] - Plugging in the hypothesis: [ a_{m+1} = frac{2 cdot frac{5}{6m-1} cdot frac{5}{6m+5}}{3 cdot frac{5}{6m-1} - frac{5}{6m+5}} ] - Simplify the denominator: [ = frac{15}{6m-1} - frac{5}{6m+5} = frac{90m+45 - 30m+5}{(6m-1)(6m+5)} = frac{60m+50}{(6m-1)(6m+5)} ] - Substituting back into a_{m+1}: [ a_{m+1} = frac{frac{50}{(6m-1)(6m+5)}}{frac{60m+50}{(6m-1)(6m+5)}} = frac{50}{60m+50} = frac{5}{6m+11} ] - Thus, a_{m+1} = frac{5}{6(m+1)+5}. 4. **Conclusion**: - By mathematical induction, a_n = frac{5}{6n+5} holds for all n geq 1. - For n = 2023, a_{2023} = frac{5}{6 times 2023 + 5} = frac{5}{12143}. - Here, p = 5 and q = 12143 are coprime. - Thus, p+q = 5 + 12143 = 12148. The final answer is boxed{12148}.

answer:Let's denote the number of associate professors as A and the number of assistant professors as B. We are given that there are 5 people present, so we can write the first equation as: A + B = 5 (Equation 1) We are also given that the total number of pencils brought to the meeting is 10, and the total number of charts is 5. Let's denote the number of pencils each associate professor brings as P (since this is what we want to find out). We know that each assistant professor brings 1 pencil, so we can write the second equation as: P * A + 1 * B = 10 (Equation 2) And since each associate professor brings 1 chart and each assistant professor brings 2 charts, we can write the third equation as: 1 * A + 2 * B = 5 (Equation 3) Now we have a system of three equations with three unknowns (A, B, and P). We can solve this system to find the values of A, B, and P. From Equation 1, we can express B in terms of A: B = 5 - A (Equation 4) Now we can substitute B from Equation 4 into Equations 2 and 3: P * A + 1 * (5 - A) = 10 (Equation 5) 1 * A + 2 * (5 - A) = 5 (Equation 6) Simplifying Equation 5, we get: P * A + 5 - A = 10 P * A - A = 10 - 5 A * (P - 1) = 5 (Equation 7) Simplifying Equation 6, we get: A + 10 - 2A = 5 -A = 5 - 10 -A = -5 A = 5 (Equation 8) Now that we know A = 5, we can substitute this value back into Equation 4 to find B: B = 5 - 5 B = 0 So there are no assistant professors present, only associate professors. Now we can substitute A = 5 into Equation 7 to find P: 5 * (P - 1) = 5 P - 1 = 1 P = 2 Each associate professor brought boxed{2} pencils to the meeting.