answer:Starting from the inequality x^{2}+(3-a)x+3-2a^{2} < ke^{x}, we rewrite it as frac {x^{2}+(3-a)x+3-2a^{2}}{e^{x}} < k We define a function f(x) as follows: f(x) = frac {x^{2}+(3-a)x+3-2a^{2}}{e^{x}} To determine the behavior of f(x), we find its derivative: f′(x) = frac {-x^{2}-(1-a)x+a(2a-1)}{e^{x}} = frac {(-x+2a-1)(x+a)}{e^{x}} Setting f′(x) = 0, we get the critical points x_{1} = 2a-1 and x_{2} = -a. Since a in (-infty,0), we conclude that x_{1} = 2a-1 < 0 and x_{2} = -a > 0. Now since x in (0, +infty), the function f(x) will be monotonic increasing on (0, -a) because f′(x) > 0 and monotonic decreasing on (-a, +infty) because f′(x) < 0. When x = -a, f(x) reaches its maximum value: f(-a) = frac {3-3a}{e^{-a}} Since a is negative, this can be further simplified as f(-a) = frac {3a-3}{e^{a}} Because a in (-infty,0), we have f(a) < f(0). Therefore, f(a) < f(0) leqslant k From the evaluating f(0), we see that k geqslant 3 Hence, the correct option is boxed{text{B: }[3, +infty)}. We used the derivative to investigate the monotonicity of the function and determine the value which constitutes the lower bound for k. The problem mainly examines the solution to the persistent function problem, applies classification discussion, transformation thinking, and uses the maximum and minimum values of a quadratic function on a closed interval in conjunction with monotonicity.

answer:Let's start by determining David's current age. If David was 7 years old 7 years ago, then David is now 7 + 7 = 14 years old. Now, let's find Richard's current age. Since Richard is 6 years older than David, Richard is 14 + 6 = 20 years old. Next, let's find Scott's current age. Since David is 8 years older than Scott, Scott is 14 - 8 = 6 years old. In 8 years, Richard will be 20 + 8 = 28 years old, and Scott will be 6 + 8 = 14 years old. The problem states that in 8 years, Richard's age will be some multiple of Scott's age. We can see that 28 is indeed a multiple of 14 (28 = 14 * 2). The ratio of Richard's age to Scott's age in 8 years will be 28:14, which simplifies to 2:1. Therefore, the ratio of Richard's age to Scott's age in 8 years is boxed{2:1} .

answer:1. **Calculate the mass of water in cubic meter of air**: Given: - Mass of one droplet, ( m_{text{drop}} = 5 text{ mg} = 5 times 10^{-6} text{ kg} ) - Number of droplets per cubic meter, ( n = 400 ) The total mass of water in one cubic meter of air can be calculated as: [ m_{text{water}} = n times m_{text{drop}} = 400 times 5 times 10^{-6} = 0.002 text{ kg} ] 2. **Calculate the mass of water falling onto one square meter surface in one second**: Given the velocity of the droplets, ( v = 2 text{ m/s} ), in one second, the number of droplets falling onto a surface of 1 square meter (1 m²) will be equivalent to the droplets in a 2 m³ column of air. Therefore: [ m_{text{water per sec per m}^2} = 0.002 text{ kg/m}^3 times 2 text{ m} = 0.004 text{ kg} ] 3. **Calculate the mass of water falling onto 10 cm² surface in one second**: Surface area of the cylinder's base: [ A = 10 text{ cm}^2 = 10 times 10^{-4} text{ m}^2 = 10^{-3} text{ m}^2 ] Mass of water per second falling onto this area: [ m_{text{water per sec per 10 cm}^2} = 0.004 text{ kg/m}^2 times 10^{-3} text{ m}^2 = 4 times 10^{-6} text{ kg} ] 4. **Calculate the total mass of water needed to fill the cylinder**: Given: - Density of water, (rho_{B} = 1000 text{ kg/m}^3) - Height of the cylinder, ( h = 20 text{ cm} = 0.2 text{ m} ) - Base area of the cylinder, ( A = 10 text{ cm}^2 = 10^{-3} text{ m}^2 ) Volume of the cylinder: [ V = A times h = 10^{-3} text{ m}^2 times 0.2 text{ m} = 2 times 10^{-4} text{ m}^3 ] Total mass of water to fill the cylinder: [ m_{text{required}} = rho_{B} times V = 1000 text{ kg/m}^3 times 2 times 10^{-4} text{ m}^3 = 0.2 text{ kg} ] 5. **Calculate the time needed to fill the cylinder completely**: Using the mass flow rate calculated in step 3: [ t = frac{m_{text{required}}}{m_{text{water per sec per 10 cm}^2}} = frac{0.2 text{ kg}}{4 times 10^{-6} text{ kg/sec}} = 5 times 10^4 text{ sec} ] 6. **Convert seconds to minutes and hours**: [ t = frac{5 times 10^4 text{ sec}}{60 text{ sec/min}} approx 833.33 text{ min} ] [ t = frac{833.33 text{ min}}{60 text{ min/hr}} approx 13.89 text{ hours} ] # Conclusion: The cylindrical vessel will be completely filled with water in approximately: [ boxed{13.9 text{ hours}} ]

answer:We are asked to prove the following inequality for ( 0 < x < frac{pi}{2} ): [ 0 < frac{x - sin x}{tan x - sin x} < frac{1}{3} . ] Let's break down the proof into detailed steps: 1. **Establishing basic inequalities**: - For ( 0 < x < frac{pi}{2} ), we know that: [ tan x > x > sin x . ] - This is because (tan x) is greater than (x), and for small (x), (x) is greater than (sin x). 2. **Define a function ( f(x) ) and show its properties**: - Define ( f(x) = x - sin x ). - Calculate the derivative of ( f(x) ): [ f'(x) = 1 - cos x . ] - Since ( cos x < 1 ), it follows that: [ f'(x) = 1 - cos x > 0 text{ for } 0 < x < frac{pi}{2} . ] - Therefore, ( f(x) ) is an increasing function in the interval ( left(0, frac{pi}{2}right) ). 3. **Conclusion from ( f(x) )'s properties**: - Because ( f(x) ) is increasing and ( f(0) = 0 ), we have: [ f(x) > f(0) = 0 quad text{for} quad 0 < x < frac{pi}{2} . ] - This means ( x > sin x ). 4. **Another basic inequality using ( tan x )**: - Similarly, we know ( tan x > x ), and we can use this to define another function: [ g(x) = tan x - sin x - 3(x - sin x) . ] 5. **Analyzing ( g(x) )**: - Calculate the derivative of ( g(x) ): [ g'(x) = frac{1}{cos^2 x} + 2 cos x - 3 . ] - To analyze ( g'(x) ), we use the Arithmetic Mean-Geometric Mean (AM-GM) Inequality: [ frac{1}{cos^2 x} + 2 cos x geq 3 sqrt[3]{frac{1}{cos^2 x} cdot cos x cdot cos x} . ] Simplifying this, we get: [ frac{1}{cos^2 x} + 2 cos x geq 3 . ] - Hence, we have: [ g'(x) geq 3 - 3 = 0 . ] 6. **Conclusion about the monotonicity of ( g(x) )**: - Since ( g'(x) geq 0 ), ( g(x) ) is a non-decreasing function in ( left(0, frac{pi}{2}right) ). - Evaluate ( g(x) ) at ( x = 0 ): [ g(0) = tan 0 - sin 0 - 3(0 - sin 0) = 0 . ] - Since ( g(x) ) is non-decreasing and ( g(0) = 0 ), it follows that: [ g(x) > 0 quad forall quad 0 < x < frac{pi}{2} . ] 7. **Final verification**: - If ( g(x) > 0 ), then: [ tan x - sin x > 3(x - sin x) . ] - Rearranging the inequality, we get: [ frac{x - sin x}{tan x - sin x} < frac{1}{3} . ] - Also, since ( x > sin x ), we have ( 0 < frac{x - sin x}{tan x - sin x} ). # Conclusion: Thus, the given inequality is satisfied: [ boxed{0 < frac{x - sin x}{tan x - sin x} < frac{1}{3}} . ]