answer:Let's analyze and illustrate the set of solutions for the inequality: [ left(y^{2}-arcsin ^{2}(sin x)right) cdotleft(y^{2}-arcsin ^{2}(sin (x+pi / 6))right) cdotleft(y^{2}-arcsin ^{2}(sin (x-pi / 6))right)<0 ] 1. **Periodicity and Interval Consideration**: We recognize that the left-hand side of the inequality is periodic with a period of (2pi). This implies that we only need to solve the problem over a single interval of length (2pi), say from (0) to (2pi). 2. **Equality Case Analysis**: We start by analyzing the corresponding equality: [ left(y^{2}-arcsin ^{2}(sin x)right) cdotleft(y^{2}-arcsin ^{2}(sin (x+pi / 6))right) cdotleft(y^{2}-arcsin ^{2}(sin (x-pi / 6))right)=0 ] We decompose each factor as follows: [ y^{2}-arcsin ^{2}(sin x) = (y - arcsin (sin x))(y + arcsin (sin x)) ] Applying this, our product becomes: [ begin{align*} &((y - arcsin (sin x))(y + arcsin (sin x))) cdot ((y - arcsin (sin (x + pi / 6)))(y + arcsin (sin (x + pi / 6)))) cdot &((y - arcsin (sin (x - pi / 6)))(y + arcsin (sin (x - pi / 6)))) = 0 end{align*} ] This product will equal zero when (y = pm arcsin (sin x)), (y = pm arcsin (sin (x + pi / 6))), or (y = pm arcsin (sin (x - pi / 6))). 3. **Graph of (y = arcsin (sin x))**: We analyze the graph of (y = arcsin (sin x)) on ([- pi, pi]). Note that: - For ( - frac{pi}{2} leq x leq frac{pi}{2} ), (arcsin (sin x) = x) - For ( frac{pi}{2} < x < frac{3 pi }{2} ), ( arcsin (sin x) = pi - x ) Extending this periodic behavior, we effectively see the piecewise linear pattern described. 4. **Vertical Shifts and Reflections**: The additional terms (x + pi / 6) and (x - pi / 6) just translate the original (arcsin (sin x)) graph horizontally by ( pm pi / 6), respectively. 5. **Inequality Analysis**: To find where the product of the terms is negative, we consider intervals where exactly one of the squared terms is positive and the others are negative. 6. **Solution Visualization**: By solving ( (y = pm arcsin (sin x)) ), ( (y = pm arcsin (sin (x + pi / 6))) ), and ( (y = pm arcsin (sin (x - pi / 6))) ), we look for regions in the ((x, y))-plane where each term changes sign, leading to a negative product of the three factors. The regions satisfying the inequality are situated between curves defined by (y = pm arcsin(sin x)) and similarly for (y = pm arcsin(sin (x pm pi/6))). The resulting shaded regions (as shown in the figures and consistent with the described calculations) provide the required solution set. Conclusion: [ boxed{begin{aligned} & text{The solution regions in the } (x, y) text{ plane will be} & text{the areas between the graphs of } y = pm arcsin(sin x) text{, } y = pm arcsin(sin (x + pi / 6))text{, and} & y = pm arcsin(sin (x - pi / 6)) text{ where the product of the terms is negative.} end{aligned}} ]

answer:1. **Finding the Coordinates Change from A to B:** The change in x-coordinates from A to B is 17 - 3 = 14. The change in y-coordinates from A to B is 7 - 1 = 6. 2. **Calculating the Extended Segment from B to C:** Since BC = frac{2}{5} cdot AB, we calculate frac{2}{5} of both the x and y differences: [ frac{2}{5} times 14 = frac{28}{5} = 5.6 ] [ frac{2}{5} times 6 = frac{12}{5} = 2.4 ] 3. **Finding Coordinates of C:** Adding these to the coordinates of B, the x-coordinate of C is 17 + 5.6 = 22.6. The y-coordinate of C is 7 + 2.4 = 9.4. Thus, the coordinates for C are: [ boxed{(22.6, 9.4)} ]

answer:Let's denote the first term of the arithmetic progression as ( a ) and the common difference as ( d ). The ( n )th term of an arithmetic progression can be given by the formula: [ T_n = a + (n - 1)d ] According to the problem, the sum of the 5th and 21st term is equal to the sum of the 8th, 15th, and 13th term. So we can write this as: [ T_5 + T_{21} = T_8 + T_{15} + T_{13} ] Using the formula for the ( n )th term, we can express this as: [ (a + 4d) + (a + 20d) = (a + 7d) + (a + 14d) + (a + 12d) ] Simplifying both sides, we get: [ 2a + 24d = 3a + 33d ] Now, let's solve for ( a ) in terms of ( d ): [ 3a - 2a = 33d - 24d ] [ a = 9d ] Now, we want to find out which term of the arithmetic progression is 0. For a term ( T_n ) to be 0, we have: [ T_n = a + (n - 1)d = 0 ] Substitute ( a ) with ( 9d ): [ 9d + (n - 1)d = 0 ] [ 9d + nd - d = 0 ] [ nd + 8d = 0 ] [ d(n + 8) = 0 ] Since ( d ) cannot be 0 (otherwise, it wouldn't be a progression), we must have: [ n + 8 = 0 ] [ n = -8 ] This means the term that would be 0 is the ( -8 )th term, which doesn't make sense in the context of a standard arithmetic progression where we start counting from the first term. However, if we consider the sequence extending infinitely in both directions, the ( -8 )th term would be 0. In a typical arithmetic progression starting from the first term, there would be no term that is exactly 0 with the given conditions, unless the first term ( a ) is boxed{0} and the common difference ( d ) is also 0, which is not the case here.

answer:**Answer**: First, factorize 1995 to get 1995 = 3 times 5 times 7 times 19. There are 4 types of equilateral triangles. For isosceles triangles with unequal bases and legs, there are the following 8 types: (3, 3, 5), (5, 5, 3), (5, 5, 7), (7, 7, 3), (7, 7, 5), (19, 19, 3), (19, 19, 5), and (19, 19, 7). For scalene triangles, there is only 1 type: (3, 5, 7). In total, there are boxed{13} types of triangles. Therefore, the correct answer is boxed{C}.