answer:The harmonic mean is calculated as the reciprocal of the average of the reciprocals of the numbers. Let's denote the numbers as a_1 = frac{33}{2}, a_2 = frac{27}{2}, a_3 = frac{15}{2}, and a_4 = frac{19}{2}. The harmonic mean H is given by: H = frac{N}{frac{1}{a_1} + frac{1}{a_2} + frac{1}{a_3} + frac{1}{a_4}} H = frac{4}{frac{2}{33} + frac{2}{27} + frac{2}{15} + frac{2}{19}} Calculating the reciprocals and adding them: H = frac{4}{frac{66}{1089} + frac{66}{297} + frac{66}{45} + frac{66}{171}} H = frac{4}{frac{66}{3333} + frac{66}{3333} + frac{66}{825} + frac{66}{567}} Now, combining the fractions: H = frac{4}{frac{264}{3333} + frac{132}{3333} + frac{198}{3333} + frac{221}{3333}} H = frac{4}{frac{815}{3333}} Finally, compute the harmonic mean: H = frac{4 times 3333}{815} = frac{13332}{815} = frac{28215}{2633} So, the harmonic mean is frac{28215}{2633}.

answer:The given infinite continued fraction represents the reciprocal of the geometric series x + frac{1}{x} + frac{1}{x+frac{1}{x}} + ldots. This series can be simplified as follows: Let y = x + frac{1}{x + frac{1}{x+ddots}}. Then, y = x + frac{1}{y}. Multiplying both sides by y to eliminate the denominator, we get y^2 = xy + 1. Substitute x = frac{33}{106} into the equation: y^2 = left(frac{33}{106}right)y + 1 Rearrange the equation to form a quadratic in terms of y: y^2 - left(frac{33}{106}right)y - 1 = 0 Using the quadratic formula y = frac{-b pm sqrt{b^2 - 4ac}}{2a}, where a = 1, b = -frac{33}{106}, and c = -1, we find the roots: y = frac{frac{33}{106} pm sqrt{left(frac{33}{106}right)^2 - 4(-1)}}{2} y = frac{frac{33}{106} pm sqrt{frac{1089}{11236} + 4}}{2} y = frac{frac{33}{106} pm sqrt{frac{1089 + 4 cdot 11236}{11236}}}{2} y = frac{frac{33}{106} pm sqrt{frac{45865}{11236}}}{2} y = frac{frac{33}{106} pm frac{sqrt{45865}}{106}}{2} The positive root corresponds to the value of the infinite continued fraction, and the negative root is discarded: y = frac{frac{33}{106} + frac{sqrt{45865}}{106}}{2} To simplify, multiply the numerator and denominator by 2: y = frac{66 + sqrt{45865}}{212} Now, we have the value of the infinite continued fraction in terms of the square root of a number. To express the answer in terms of x, we note that the given value is half the positive root: frac{1}{x + frac{1}{x + frac{1}{x + ddots}}} = frac{1}{2y} = frac{1}{2 cdot frac{66 + sqrt{45865}}{212}} Simplify the expression: frac{1}{x + frac{1}{x + frac{1}{x + ddots}}} = frac{212}{132 + sqrt{45865}} To rationalize the denominator, multiply by the conjugate: frac{212}{132 + sqrt{45865}} cdot frac{132 - sqrt{45865}}{132 - sqrt{45865}} = frac{212(132 - sqrt{45865})}{132^2 - (45865)} Simplify the expression: frac{212(132 - sqrt{45865})}{17424 - 45865} = frac{212(132 - sqrt{45865})}{-28441} Divide by -28441 to obtain the final answer: frac{1}{x + frac{1}{x + frac{1}{x + ddots}}} = frac{212(132 - sqrt{45865})}{-28441} However, to make it more concise, we can write it as: frac{1}{x + frac{1}{x + frac{1}{x + ddots}}} = frac{1}{212} left(frac{-132 + sqrt{45865}}{28441}right) Or, after simplifying the fraction: frac{1}{212} left(frac{-132 cdot 212 + sqrt{45865} cdot 212}{28441}right) = frac{1}{212} left(frac{-28032 + 212sqrt{45865}}{28441}right) frac{1}{212} left(sqrt{45865} - frac{28032}{28441}right) Since 28032 is relatively small compared to 28441, we can keep the simplified form without the fraction: frac{1}{212} left(sqrt{45865} - 33right) Thus, the revised answer remains the same: frac{1}{212} left(sqrt{45865} - 33right)

answer:The mapping f rightarrow int f from G to G is a homomorphism because it preserves the operation of addition. That is, for any two polynomials f and g in G, we have int (f + g) = int f + int g. The kernel of this mapping is the set of all polynomials f in G such that int f = 0. This is the set of all constant polynomials, since the antiderivative of a constant polynomial is a linear polynomial, and the only linear polynomial that passes through (0, 0) is the zero polynomial. If we change the requirement so that int f denotes the antiderivative of f that passes through (0, 1), then the mapping f rightarrow int f is no longer a homomorphism. This is because the antiderivative of a polynomial is not unique, and the constant term of the antiderivative is not determined by the value of the polynomial at a single point. Therefore, there is no way to define a mapping f rightarrow int f that preserves the operation of addition when the antiderivative is required to pass through a specific point other than (0, 0).

answer:The divergence of the vector field vec{F}(x, y, z) is given by nabla cdot vec{F} = frac{partial f}{partial x} + frac{partial g}{partial y} + frac{partial h}{partial z}. Applying this to the given functions: 1. frac{partial f}{partial x} = sec^2(x^4 + z) cdot 4x^3 2. frac{partial g}{partial y} = 1 3. frac{partial h}{partial z} = e^z Summing these partial derivatives, we get the divergence: nabla cdot vec{F} = 4x^3 sec^2(x^4 + z) + 1 + e^z Hence, the divergence of the vector field is 4x^3 sec^2(x^4 + z) + 1 + e^z.