answer:The series representation in (2) can be used to evaluate integrals of the form int f(x)tan(1/x),dx over an interval, where f(x) is a function that can be expanded in a power series. By substituting the power series expansion of f(x) into the integral and using the series representation of inttan(1/x),dx, we can obtain a series representation for the integral int f(x)tan(1/x),dx. This can be useful for approximating the integral or for studying its properties.

answer:To multiply these polynomials, we can use the distributive property twice. First, distribute p(x) over q(x): (9sqrt{2}x^2)(5sqrt{2}x^2) + (9sqrt{2}x^2)(3sqrt{2}x) + (9sqrt{2}x^2)(-7sqrt{2}) + (9sqrt{2}x)(5sqrt{2}x^2) + (9sqrt{2}x)(3sqrt{2}x) + (9sqrt{2}x)(-7sqrt{2}) Now, simplify each term: 90x^4 + 54x^3 - 126x^2 + 45x^3 + 27x^2 - 63x Combine like terms: 90x^4 + (54x^3 + 45x^3) - (126x^2 - 27x^2) - 63x 90x^4 + 99x^3 - 99x^2 - 63x The expanded form is 90x^4 + 99x^3 - 99x^2 - 63x.

answer:The characteristic polynomial of the given matrix is calculated as follows: 1. Let the matrix be denoted by A, with its elements a_{ij}. 2. The characteristic polynomial, p(lambda), is given by the determinant of A - lambda I, where lambda is the variable and I is the identity matrix of the same size as A. 3. For the 3x3 matrix provided, we have: A - lambda I = begin{pmatrix} -frac{86}{9} - lambda & frac{26}{3} & frac{17}{3} -frac{70}{9} & -frac{55}{9} - lambda & frac{53}{9} frac{74}{9} & -frac{53}{9} & -frac{14}{9} - lambda end{pmatrix} 4. Compute the determinant of A - lambda I. 5. The characteristic polynomial is found to be: p(lambda) = left( -frac{86}{9} - lambda right) left( left(-frac{55}{9} - lambda right) left(-frac{14}{9} - lambda right) - left(frac{53}{9}right) left(-frac{70}{9}right) right) - left(frac{26}{3}right) left( left(-frac{70}{9}right) left(-frac{14}{9} - lambda right) - left(frac{17}{3}right) left(-frac{53}{9}right) right) 6. Simplify the above expression to obtain the characteristic polynomial: p(lambda) = lambda^3 + frac{155lambda^2}{9} + frac{3733lambda}{27} - frac{106154}{243} Thus, the characteristic polynomial of the given matrix is: lambda^3 + frac{155lambda^2}{9} + frac{3733lambda}{27} - frac{106154}{243}

answer:To find the distance from a point to a line, we can use the formula: d = frac{|ax_0 + by_0 + c|}{sqrt{a^2 + b^2}} where (x_0, y_0) is the point, and ax + by + c = 0 is the equation of the line. Plugging in the given values, we get: d = frac{left|frac{141}{32}left(frac{85}{32}right) + frac{75}{16}left(frac{13}{16}right) - frac{89}{32}right|}{sqrt{left(frac{141}{32}right)^2 + left(frac{75}{16}right)^2}} Simplifying, we get: d = frac{left|frac{141(85)}{32^2} + frac{75(13)}{16^2} - frac{89}{32}right|}{sqrt{frac{141^2}{32^2} + frac{75^2}{16^2}}} d = frac{left|frac{11985}{1024} + frac{975}{256} - frac{89}{32}right|}{sqrt{frac{19881}{1024} + frac{5625}{256}}} d = frac{left|frac{11985 + 975 - 2848}{1024}right|}{sqrt{frac{19881 + 5625}{1024}}} d = frac{left|frac{9112}{1024}right|}{sqrt{frac{25506}{1024}}} d = frac{9112}{1024} cdot frac{1024}{159.65} d = frac{9112}{159.65} d approx 56.82 Therefore, the distance from the point left(frac{85}{32}, frac{13}{16}right) to the line frac{141 x}{32}+frac{75 y}{16}-frac{89}{32}=0 is approximately 56.82. The answer is frac{13037}{96 sqrt{4709}}