answer:(1) Since cos alpha = -frac{4}{5} and alpha is in the third quadrant, both the sine and cosine functions are negative. Therefore, sin alpha = -sqrt{1 - cos^2 alpha} = -sqrt{1 - left(-frac{4}{5}right)^2} = -sqrt{1 - frac{16}{25}} = -sqrt{frac{9}{25}} = -frac{3}{5}. Hence, sin alpha = -frac{3}{5}. (2) Knowing that tan alpha = -3, we substitute tan alpha and simplify the given expression: begin{align*} frac{4sin alpha - 2cos alpha}{5cos alpha + 3sin alpha} &= frac{4tan alpha - 2}{5 + 3tan alpha} &= frac{4(-3) - 2}{5 + 3(-3)} &= frac{-12 - 2}{5 - 9} &= frac{-14}{-4} &= frac{7}{2}. end{align*} Therefore, the value of the expression is boxed{frac{7}{2}}.

answer:1. **Substitute a with kb:** Given a = kb, substitute into (a-b)^n: [ (a-b)^n = (kb - b)^n = (b(k-1))^n ] 2. **Apply the Binomial Theorem:** The Binomial Theorem states that [ (x+y)^n = sum_{i=0}^n binom{n}{i} x^{n-i} y^i ] Applying this to (b(k-1))^n, we get: [ (b(k-1))^n = sum_{i=0}^n binom{n}{i} b^{n-i} (k-1)^i ] 3. **Identify the second and third terms:** The second term (where i=1) and the third term (where i=2) are: [ text{Second term: } binom{n}{1} b^{n-1} (k-1)^1 = nb^{n-1}(k-1) ] [ text{Third term: } binom{n}{2} b^{n-2} (k-1)^2 = frac{n(n-1)}{2} b^{n-2}(k-1)^2 ] 4. **Set the sum of these terms to zero:** According to the problem, the sum of these terms is zero: [ nb^{n-1}(k-1) + frac{n(n-1)}{2} b^{n-2}(k-1)^2 = 0 ] 5. **Factor out common terms:** Factor out b^{n-2}(k-1)n: [ b^{n-2}(k-1)n left(b(k-1) + frac{n-1}{2}(k-1)right) = 0 ] 6. **Simplify the expression inside the brackets:** Simplify and solve for n: [ b(k-1) + frac{n-1}{2}(k-1) = 0 ] [ (k-1)left(b + frac{n-1}{2}right) = 0 ] Since b neq 0 and k neq 1, we focus on: [ b + frac{n-1}{2} = 0 quad text{which is not possible as } b neq 0 ] Re-examine the factorization: [ n(-2k + n - 1) = 0 ] Since n neq 0, solve -2k + n - 1 = 0: [ n = 2k + 1 ] 7. **Conclude with the value of n:** The correct value of n that satisfies the condition is 2k + 1. Thus, the answer is boxed{text{E}}.

answer:1. **Orientation and dimensions**: Use the 90 m side along the concrete wall, fencing the two 45 m sides and one additional 90 m side. 2. **Number of posts for the 90 m side not along the wall (Length = 90 m)**: Posts are placed every 15 meters, creating frac{90}{15} = 6 intervals. Including both endpoints, this side requires 6 + 1 = 7 posts. 3. **Number of posts for each 45 m side (Length = 45 m)**: Each 45 m side will have frac{45}{15} = 3 intervals. Normally, this would require 3 + 1 = 4 posts per side. However, the posts at the ends (meeting the 90 m side and concrete wall) are shared. Thus, we only need 4 - 1 = 3 additional posts for each 45 m side. 4. **Total posts required**: The total is computed as follows: - 7 posts for the standalone 90 m side. - 3 posts for each of the two 45 m sides: 3 + 3 = 6 posts. - Combined total: 7 + 6 = 13 posts. 5. **Conclusion**: The minimal number of posts required to fence the park, using the concrete wall as one boundary, is 13. The final answer is The correct answer, given the choices, is boxed{text{(B)} 13}.

answer:Since 2cos left(x- frac {pi}{4}right)= sqrt {2}, it follows that cos left(x- frac {pi}{4}right)= frac { sqrt {2}}{2}. Therefore, x- frac {pi}{4}=kpipm frac {pi}{4}(kinmathbb{Z}). Thus, x=kpi+ frac {pi}{2} or kpi(kinmathbb{Z}). Since xin(0,pi), it follows that x= frac {pi}{2}. Hence, the answer is: boxed{left{ frac {pi}{2}right}}. First, we find the cosine value of x- frac {pi}{4}, then determine the value of x- frac {pi}{4} using the properties of the cosine function, and finally determine the value of x based on its range. The answer must be presented in the form of a set. This question mainly tests the problem of finding x given the value of a trigonometric function. It requires students to be proficient in understanding the graphs and properties of sine and cosine functions to solve the problem efficiently.