answer:To find the eigenvalues and their multiplicities of the Laplace operator Delta = text{div grad} on a sphere of radius R in Euclidean space of dimension n, we need to consider the properties of the Laplace-Beltrami operator on the sphere S^{n-1}. 1. **Laplace-Beltrami Operator on the Sphere:** The Laplace-Beltrami operator Delta on the sphere S^{n-1} is a generalization of the Laplace operator to curved surfaces. For a function f on S^{n-1}, the Laplace-Beltrami operator is given by: [ Delta f = frac{1}{sqrt{g}} frac{partial}{partial x^i} left( sqrt{g} g^{ij} frac{partial f}{partial x^j} right) ] where g is the determinant of the metric tensor g_{ij} of the sphere. 2. **Eigenvalues of the Laplace-Beltrami Operator:** The eigenvalues of the Laplace-Beltrami operator on the sphere S^{n-1} are well-known and given by: [ lambda_k = -k(k+n-2) ] where k is a non-negative integer. 3. **Eigenfunctions:** The eigenfunctions corresponding to these eigenvalues are the spherical harmonics Y_{k,m}, where m is a multi-index that labels the different spherical harmonics of degree k. 4. **Multiplicity of Eigenvalues:** The multiplicity of the eigenvalue lambda_k = -k(k+n-2) is given by the dimension of the space of spherical harmonics of degree k on S^{n-1}. This dimension is: [ text{dim}(mathcal{H}_k) = binom{n+k-1}{k} - binom{n+k-3}{k-2} ] where mathcal{H}_k denotes the space of spherical harmonics of degree k. 5. **Radius of the Sphere:** Since the sphere has radius R, the Laplace-Beltrami operator on the sphere of radius R is related to the operator on the unit sphere by a scaling factor. Specifically, the eigenvalues on the sphere of radius R are scaled by a factor of 1/R^2. Therefore, the eigenvalues on the sphere of radius R are: [ lambda_k = -frac{k(k+n-2)}{R^2} ] Conclusion: The eigenvalues of the Laplace-Beltrami operator on a sphere of radius R in Euclidean space of dimension n are: [ lambda_k = -frac{k(k+n-2)}{R^2} ] with multiplicities: [ text{dim}(mathcal{H}_k) = binom{n+k-1}{k} - binom{n+k-3}{k-2} ] The final answer is ( boxed{ lambda_k = -frac{k(k+n-2)}{R^2} } ) with multiplicities (text{dim}(mathcal{H}_k) = binom{n+k-1}{k} - binom{n+k-3}{k-2}).

answer:If Ajay can ride 50 km in 1 hour, then in 30 hours he can ride: 50 km/hour * 30 hours = 1500 km So, the distance he rides in 30 hours is boxed{1500} km.

answer:Let A(x_{1}, y_{1}) and C(x_{2}, y_{2}). According to the problem, points A and B are the intersection points of a line passing through the origin and the hyperbola frac{x^{2}}{a^{2}}-frac{y^{2}}{b^{2}}=1. Due to the symmetry of the hyperbola, points A and B are symmetric about the origin. Therefore, B(-x_{1}, -y_{1}). Then, k_{1}k_{2}=frac{y_{2}-y_{1}}{x_{2}-x_{1}}cdotfrac{y_{2}+y_{1}}{x_{2}+x_{1}}=frac{y_{2}^{2}-y_{1}^{2}}{x_{2}^{2}-x_{1}^{2}}. Since points A and C both lie on the hyperbola, we have frac{x_{1}^{2}}{a^{2}}-frac{y_{1}^{2}}{b^{2}}=1 and frac{x_{2}^{2}}{a^{2}}-frac{y_{2}^{2}}{b^{2}}=1. Subtracting these two equations, we obtain k_{1}k_{2}=frac{b^{2}}{a^{2}} > 0. Consider the function y=frac{2}{x}+ln x (x > 0). By calculating its derivative, y'=-frac{2}{x^{2}}+frac{1}{x}, we find the critical points x=0 (rejected as it's not in the domain) and x=2. The function is increasing when x > 2 and decreasing when 0 < x < 2. Therefore, the function attains its minimum value when x=2. Hence, when frac{2}{k_{1}k_{2}}+ln(k_{1}k_{2}) is minimized, k_{1}k_{2}=frac{a^{2}}{b^{2}}=2. Thus, the eccentricity of the hyperbola is e=sqrt{1+frac{b^{2}}{a^{2}}}=boxed{sqrt{3}}.

answer:Let's begin by factoring the numerator and the denominator of the right-hand side: 1. Factorize the quadratic expression x^2 + 120x + 1575. We look for two numbers that multiply to 1575 and add up to 120. Factoring 1575 we find x^2 + 120x + 1575 = (x + 25)(x + 45). 2. Factorize the quadratic expression x^2 - 144x + 1050. We look for two numbers that multiply to 1050 and add up to -144. Observing potential factors, x^2 - 144x + 1050 = (x - 30)(x - 70). The simplified form of the right-hand side fraction is then: [ frac{(x + 25)(x + 45)}{(x - 30)(x - 70)} = frac{x + 25}{x - 30} text{ for } x neq -45 text{ and } x neq 70. ] For the left-hand side frac{x-alpha}{x+beta} to match this, we set alpha = -25 and beta = 30. Therefore, alpha + beta = -25 + 30 = boxed{5}. Conclusion: Both the polynomial on the numerator and the denominator simplifies correctly after proper factoring, suggesting alpha = -25 and beta = 30. The solution is consistent with the given transformation.