answer:Given the vertex of angle alpha is at the origin, and the initial side coincides with the non-negative half-axis of the x-axis, with point P(4m, -3m), where m < 0, on the terminal side of angle alpha, we can find the values of sinalpha and cosalpha using the coordinates of point P. First, we calculate the radius (r) of the circle that point P lies on, which is the distance from the origin to point P. This is found using the Pythagorean theorem: begin{align*} r &= sqrt{(4m)^2 + (-3m)^2} &= sqrt{16m^2 + 9m^2} &= sqrt{25m^2} &= 5|m| &= -5m quad text{(since m < 0)}. end{align*} Next, we find sinalpha and cosalpha using the definitions sinalpha = frac{text{opposite}}{text{hypotenuse}} and cosalpha = frac{text{adjacent}}{text{hypotenuse}}: begin{align*} sinalpha &= frac{-3m}{-5m} = frac{3}{5}, cosalpha &= frac{4m}{-5m} = -frac{4}{5}. end{align*} Now, we can calculate frac{2sinalpha + cosalpha}{sinalpha - cosalpha}: begin{align*} frac{2sinalpha + cosalpha}{sinalpha - cosalpha} &= frac{2left(frac{3}{5}right) - frac{4}{5}}{frac{3}{5} - left(-frac{4}{5}right)} &= frac{frac{6}{5} - frac{4}{5}}{frac{3}{5} + frac{4}{5}} &= frac{frac{2}{5}}{frac{7}{5}} &= frac{2}{7}. end{align*} Therefore, the answer is boxed{frac{2}{7}}.

answer:1. The lateral surface of the cylinder is viewed as a rectangle when "unrolled," with dimensions where the length is the circumference (10 inches) and the height is 24 inches. 2. The stripe spirals twice, implying it covers twice the height of the cylinder in its path. Thus, the effective height it covers is (2 times 24 = 48) inches. 3. The stripe forms the hypotenuse of a right triangle with base 10 inches (one complete wrap around the circumference) and height 48 inches. Using the Pythagorean theorem: [ text{Length of stripe} = sqrt{10^2 + 48^2} = sqrt{100 + 2304} = sqrt{2404} = 49 text{ inches} ] Hence, the length of the stripe is boxed{49} inches.

answer:Assume for contradiction that ( f(x) ) can be factored in (mathbf{Z}[x]) as ( f(x) = g(x) h(x) ), where ( g(x) ) and ( h(x) ) are polynomials with integer coefficients. 1. Note that: [ f(b) = k p quad text{with} quad p quad text{a prime number}. ] Therefore: [ k p = |f(b)| = |g(b) h(b)|, ] implying ( p ) divides either (|g(b)|) or (|h(b)|). 2. Without loss of generality, assume: [ p mid h(b). ] 3. Since ( |g(b)| leq k ), let: [ g(x) = b_0 (x - r_1)(x - r_2) cdots (x - r_j), ] where ( b_0 in mathbf{Z} ) and ( |b - r_i| > sqrt{k} ) for ( 1 leq i leq j ). 4. Evaluate (|g(b)|): [ |g(b)| = |b_0| cdot |b - r_1| cdot |b - r_2| cdots |b - r_j|. ] 5. Given ( |b - r_i| > sqrt{k} ), then: [ |g(b)| > |b_0| cdot (sqrt{k})^j. ] 6. Since ( |g(b)| leq k ), it follows that: [ k geq |g(b)| > |b_0| cdot k^{frac{j}{2}}. ] 7. By simplifying the inequality: [ k^{1 - frac{j}{2}} geq |b_0|. ] 8. Given ( b_0 ) is a nonzero integer and assuming ( b_0 > 0): - If ( b_0 < 0 ), use ( -g(x) ) instead. 9. Consider ( j = 1 ): [ g(x) = b_0 x + b_1. ] By substituting, obtain: [ k geq |g(b)| = |b_0 b + b_1|. ] 10. Since ( b_1 ) is a non-negative integer to avoid ( f(x) ) having a real root, by condition ( (1) ): [ b_1 geq 0. ] - If higher reasons conclude ( b_1 < 0 ), a contradiction emerges with non-negative coefficients leading to: [ b > k. ] 11. Ultimately: [ k geq |b_0| b + b_1. ] Revealing ( g(b) geq b ), asserts a contradiction ( b > k geq b ). Therefore, from the derived contradiction, it follows that ( f(x) ) in (mathbf{Z}[x]) is irreducible. blacksquare

answer:**Analysis** This question tests the calculation of derivatives. First, we find the derivative of the function, substitute x=e to get the value of f'(e), and then substitute it back into the function's expression to find the solution. **Solution** Given f(x) = 2x f''(e) + ln x, we can derive that f'(x) = 2f''(e) + dfrac{1}{x}. Therefore, f'(e) = 2f''(e) + dfrac{1}{e}. Thus, f''(e) = -dfrac{1}{e}. Hence, the correct answer is boxed{text{D}}.