answer:Let's denote the speed of the slower train as ( v ) (in meters per second). Then the speed of the faster train would be ( 2v ). When the two trains cross each other, they cover a combined distance equal to the sum of their lengths. Since both trains are 150 m long, the total distance covered when they pass each other is ( 150 + 150 = 300 ) meters. The time taken for them to cross each other is given as 8 seconds. The relative speed at which the two trains cross each other is the sum of their individual speeds because they are moving in opposite directions. Therefore, the relative speed is ( v + 2v = 3v ). We can use the formula for speed, which is distance divided by time, to find the relative speed: [ text{Relative Speed} = frac{text{Total Distance}}{text{Time}} ] Substituting the given values, we get: [ 3v = frac{300 text{ m}}{8 text{ s}} ] [ 3v = 37.5 text{ m/s} ] Now, we solve for ( v ): [ v = frac{37.5 text{ m/s}}{3} ] [ v = 12.5 text{ m/s} ] This is the speed of the slower train. The faster train is moving twice as fast, so its speed is: [ text{Speed of the faster train} = 2v ] [ text{Speed of the faster train} = 2 times 12.5 text{ m/s} ] [ text{Speed of the faster train} = 25 text{ m/s} ] Therefore, the speed of the faster train is boxed{25} meters per second.

answer:To extract the probability that Xiao Tao and Xiao Le together grab red packets totaling no less than ¥5, we enumerate the exhaustive number of base events. The red packets can be combined in C_{5}^{2} = 10 different ways, as each pair of red packets constitutes a base event. Considering that Xiao Tao and Xiao Le will grab two distinct red packets, we must list all combinations of red packets that sum to at least ¥5: 1. (¥2.63, ¥3.26) 2. (¥3.26, ¥1.95) 3. (¥3.26, ¥1.77) There are 3 such combinations. Hence, the probability p that Xiao Tao and Xiao Le together grab red packets totaling no less than ¥5 is given by: p = frac{3}{10}. The solution involves using elementary probability theory and the method of listing or enumeration to determine the number of favorable outcomes. By determining the number of ways Xiao Tao and Xiao Le can grab red packets that sum to at least ¥5 and relating this to the total number of outcomes, we can calculate the desired probability. Therefore, the correct answer is boxed{B left( frac{3}{10} right)}.

answer:1. From the given information, we have begin{cases} e=dfrac{c}{a}=dfrac{sqrt{2}}{2} 2b=2 a^2=b^2+c^2 end{cases}, solving these equations gives a^2=2, b^2=1, Thus, the standard equation of the ellipse C is dfrac{x^2}{2} + y^2 = 1; 2. Let A(x_1,y_1), B(x_2,y_2), by combining the equations begin{cases} overset{y=kx+m}{dfrac{x^2}{2}+y^2=1} end{cases}, Eliminating y gives (1+2k^2)x^2+4kmx+2m^2-2=0. When triangle = 8(2k^2-m^2+1) > 0, i.e., 2k^2 > m^2-1, we have x_1+x_2 = dfrac{-4km}{1+2k^2}, x_1 cdot x_2 = dfrac{2m^2-2}{1+2k^2}. therefore dfrac{x_1+x_2}{2} = dfrac{-2km}{1+2k^2}, dfrac{y_1+y_2}{2} = dfrac{m}{1+2k^2}. Since the perpendicular bisector of segment AB passes through (0, -dfrac{1}{2}), we have dfrac{dfrac{y_1+y_2}{2}-(-dfrac{1}{2})}{dfrac{x_1+x_2}{2}-0} = -dfrac{1}{k}, Simplifying gives 2k^2+1=2m. From begin{cases} overset{2k^2+1=2m}{2k^2+1 > m^2} end{cases}, we get 0 < m < 2. Also, the distance from the origin O to the line AB is d= dfrac{|m|}{sqrt{1+k^2}}. |AB| = sqrt{1+k^2}|x_1-x_2| = 2sqrt{1+k^2}dfrac{sqrt{4k^2-2m^2+2}}{1+2k^2}, therefore S_{triangle AOB} = dfrac{1}{2}|AB| cdot d = dfrac{|m|sqrt{4k^2-2m^2+2}}{1+2k^2}, Since 2k^2+1=2m and 0 < m < 2, S_{triangle AOB} = dfrac{1}{2}sqrt{4m-2m^2}, 0 < m < 2. therefore When m=1, i.e., k^2= dfrac{1}{2}, S_{triangle AOB} reaches its maximum value dfrac{sqrt{2}}{2}. In summary, the maximum value of S_{triangle AOB} is boxed{dfrac{sqrt{2}}{2}}, At this time, the equation of the line l is y= dfrac{sqrt{2}}{2}x+1 or y= -dfrac{sqrt{2}}{2}x+1.

answer:Since |z| = 1, z = e^{itheta} for some angle theta. Then for the first condition: [ left| frac{z}{overline{z}} - frac{overline{z}}{z} right| = left| e^{2itheta} - e^{-2itheta} right| = |2i sin 2theta| = 2|sin 2theta|. ] Thus, |sin 2theta| = 1, which implies sin 2theta = pm 1. The solutions to sin 2theta = pm 1 between 0 and 2pi are theta = frac{pi}{4}, frac{3pi}{4}, frac{5pi}{4}, frac{7pi}{4}. For the second condition: [ left| frac{z^2}{overline{z}^2} - frac{overline{z}^2}{z^2} right| = left| e^{4itheta} - e^{-4itheta} right| = |2i sin 4theta| = 2|sin 4theta|. ] Thus, |sin 4theta| = frac{1}{2}. The solutions to sin 4theta = pm frac{1}{2} between 0 and 2pi are theta = frac{pi}{8}, frac{3pi}{8}, frac{5pi}{8}, frac{7pi}{8}, frac{9pi}{8}, frac{11pi}{8}, frac{13pi}{8}, frac{15pi}{8}. The intersection of these solutions (sin 2theta = pm 1 and sin 4theta = pm frac{1}{2}) needs to be calculated. Intersection points are theta = frac{3pi}{8}, frac{11pi}{8}. Therefore, there are boxed{2} solutions for z. Conclusion: Both conditions are necessary and result in a consistent solution set. Every step of the calculations is supported by trigonometric identities and properties of complex numbers.