answer:In stratified sampling, the ratio of each stratum in the total population is the same as that in the sample. Let the number of 9th-grade students in the sample be n. According to the given information, we have frac{240}{600} = frac{n}{560}, solving this equation gives n=224.Therefore, the correct answer is boxed{D}.This problem mainly examines the definition and method of stratified sampling, utilizing the principle that the ratio of the number of individuals in each stratum of the total population is equal to the ratio of the corresponding stratum's sample size in the sample. It is a basic question.

answer:Solution: Let P be a good point. Let l, m, n be respetively the number of parts the sides B C, C A, A B are divided by the rays starting from P. Note that a ray must pass through each of the vertices the triangle A B C; otherwise we get some quadrilaterals.Let h_{1} be the distance of P from B C. Then h_{1} is the height for all the triangles with their bases on B C. Equality of areas implies that all these bases have equal length. If we denote this by x, we get l x=a. Similarly, taking y and z as the lengths of the bases of triangles on C A and A B respectively, we get m y=b and n z=c. Let h_{2} and h_{3} be the distances of P from C A and A B respectively. Thenh_{1} x=h_{2} y=h_{3} z=frac{2 Delta}{27}where Delta denotes the area of the triangle A B C. These lead toh_{1}=frac{2 Delta}{27} frac{l}{a}, quad h_{1}=frac{2 Delta}{27} frac{m}{b}, quad h_{1}=frac{2 Delta}{27} frac{n}{c}Butfrac{2 Delta}{a}=h_{a}, quad frac{2 Delta}{b}=h_{b}, quad frac{2 Delta}{c}=h_{c}Thus we getfrac{h_{1}}{h_{a}}=frac{l}{27}, quad frac{h_{2}}{h_{b}}=frac{m}{27}, quad frac{h_{3}}{h_{c}}=frac{n}{27}However, we also havefrac{h_{1}}{h_{a}}=frac{[P B C]}{Delta}, quad frac{h_{2}}{h_{b}}=frac{[P C A]}{Delta}, quad frac{h_{3}}{h_{c}}=frac{[P A B]}{Delta}Adding these three relations,frac{h_{1}}{h_{a}}+frac{h_{2}}{h_{b}}+frac{h_{3}}{h_{c}}=1Thusfrac{l}{27}+frac{m}{27}+frac{n}{27}=frac{h_{1}}{h_{a}}+frac{h_{2}}{h_{b}}+frac{h_{3}}{h_{c}}=1We conclude that l+m+n=27. Thus every good point P determines a partition (l, m, n) of 27 such that there are l, m, n equal segments respectively on B C, C A, A B.Conversely, take any partition (l, m, n) of 27 . Divide B C, C A, A B respectively in to l, m, n equal parts. Defineh_{1}=frac{2 l Delta}{27 a}, quad h_{2}=frac{2 m Delta}{27 b}Draw a line parallel to B C at a distance h_{1} from B C; draw another line parallel to C A at a distance h_{2} from C A. Both lines are drawn such that they intersect at a point P inside the triangle A B C. Then[P B C]=frac{1}{2} a h_{1}=frac{l Delta}{27}, quad[P C A]=frac{m Delta}{27}Hence[P A B]=frac{n Delta}{27}This shows that the distance of P from A B ish_{3}=frac{2 n Delta}{27 c}Therefore each traingle with base on C A has area frac{Delta}{27}. We conclude that all the triangles which partitions A B C have equal areas. Hence P is a good point.Thus the number of good points is equal to the number of positive integral solutions of the equation l+m+n=27. This is equal tobinom{26}{2}=325

answer:First, we find the vectors overrightarrow{a}+2overrightarrow{b} and 2overrightarrow{a}-overrightarrow{b}:overrightarrow{a}+2overrightarrow{b}=(-1,2)+2(m,1)=(2m-1,4),2overrightarrow{a}-overrightarrow{b}=2(-1,2)-(m,1)=(-2-m,3).Since the vectors overrightarrow{a}+2overrightarrow{b} and 2overrightarrow{a}-overrightarrow{b} are parallel, their cross product is zero:(2m-1,4)times(-2-m,3)=0.Expanding the cross product, we get:3(2m-1)-4(-2-m)=0.Solving for m, we obtain m=-frac{1}{2}.Now, we can find the dot product of overrightarrow{a} and overrightarrow{b}:overrightarrow{a}cdotoverrightarrow{b}=(-1)cdot(-frac{1}{2})+2cdot1=frac{5}{2}.Therefore, the answer is boxed{frac{5}{2}}.

answer:When x=0, f(x)=0 ; x>0 Rightarrow p^{prime}(x) is monotonically increasing, and p^{prime}(0)=0 Rightarrow p^{prime}(x)>0 Rightarrow g^{prime}(x)>0 Rightarrow g(x) is monotonically increasing, thus g(x)<lim _{x rightarrow 0} frac{1-mathrm{e}^{-x}}{x^{2}-2 x}=lim _{x rightarrow 0} frac{mathrm{e}^{-x}}{2 x-2}=-frac{1}{2}.In summary, the range of real number A is left[-frac{1}{2},+inftyright).

answer:First, we break down the integral into two parts based on the piecewise function: int_{-1}^{2}f(x)dx= int_{-1}^{1} sqrt {1-x^{2}}dx+ int_{1}^{2} e^{x}dx.The first integral, int_{-1}^{1} sqrt {1-x^{2}}dx, represents the area of the upper half of the circle defined by x^{2}+y^{2}=1(ygeqslant 0). Therefore, int_{-1}^{1} sqrt {1-x^{2}}dx= frac{pi}{2}.The second integral, int_{1}^{2} e^{x}dx, can be solved using the fundamental theorem of calculus: int_{1}^{2} e^{x}dx= e^{x} |_{1}^{2} = e^{2}-e.Combining these results, we get int_{-1}^{2}f(x)dx= int_{-1}^{1} sqrt {1-x^{2}}dx+ int_{1}^{2} e^{x}dx = frac{pi}{2}+e^{2}-e.Therefore, the final answer is boxed{frac{pi}{2}+e^{2}-e}.

answer:By the Law of Sines, B X / sin B D X = B D / sin B X D = 2 R sin psi / sin B X D. Moreover, sin B D X = sin B D C = sin varphi; the measure of angle B X D is easily calculated: if points C and D lie on the same side of A B, then angle B X D = pi - varphi - psi, and if on opposite sides, then angle B X D = |varphi - psi|. Therefore, B X = 2 R sin varphi sin psi / sin |varphi pm psi|.