answer:To solve for 3^{-1} pmod{29}, we need to find an integer x such that: [3 cdot x equiv 1 pmod{29}] We start by checking multiples of 3 modulo 29: - 3 cdot 1 = 3 - 3 cdot 2 = 6 - 3 cdot 3 = 9 - 3 cdot 4 = 12 - 3 cdot 5 = 15 - 3 cdot 6 = 18 - 3 cdot 7 = 21 - 3 cdot 8 = 24 - 3 cdot 9 = 27 - 3 cdot 10 = 30 equiv 1 pmod{29} Since 3 cdot 10 equiv 1 pmod{29}, we find that: [3^{-1} equiv boxed{10} pmod{29}]

answer:Given that x > 2, we have x-2 > 0. The function y= frac {4}{x-2}+x can be rewritten as y= frac {4}{x-2}+(x-2)+2. According to the Arithmetic Mean-Geometric Mean Inequality, the arithmetic mean of two non-negative real numbers is always greater than or equal to the geometric mean. Therefore, we have: frac{frac{4}{x-2} + (x-2)}{2} geq sqrt{frac{4}{x-2} cdot (x-2)}. Multiplying both sides by 2 gives: frac {4}{x-2}+(x-2) geq 2sqrt{frac{4}{x-2} cdot (x-2)} = 4. Adding 2 to both sides of the inequality, we obtain: y = frac {4}{x-2}+(x-2)+2 geq 4 + 2 = 6. The equality holds if and only if frac{4}{x-2} = x-2, which is true when x=4. Thus, the minimum value of the function is boxed{6}.

answer:To solve this problem, we start by defining the variables based on the information given. Let's denote the average speed of the bus as x km/h. Given that the car travels at 1.5 times the speed of the bus, the average speed of the car can be represented as 1.5x km/h. The distance to the museum is 50 kilometers for both the bus and the car, and they both arrive at the same time. However, the teacher (driving the car) starts frac{1}{3} hour later than the bus. The time taken by the bus to reach the museum is the distance divided by its speed, frac{50}{x} hours. Similarly, the time taken by the car is frac{50}{1.5x} hours. Since the car starts frac{1}{3} hour later, the difference in their travel times is frac{1}{3} hour, leading to the equation: [ frac{50}{x} - frac{50}{1.5x} = frac{1}{3} ] To solve for x, we first simplify the equation: [ frac{50}{x} - frac{50}{1.5x} = frac{1}{3} Rightarrow frac{50}{x} - frac{33.33}{x} = frac{1}{3} ] [ Rightarrow frac{16.67}{x} = frac{1}{3} ] Multiplying both sides by 3x gives: [ 16.67 = x ] However, upon reviewing the calculation, it seems there was a mistake in the simplification process. Let's correct that and solve the equation properly: [ frac{50}{x} - frac{50}{1.5x} = frac{1}{3} ] Multiplying both sides by 3x cdot 1.5x to clear the denominators, we get: [ 3 cdot 1.5 cdot 50 - 3 cdot 50 = x cdot 1.5x ] [ 150 - 100 = 0.5x^2 ] [ 50 = 0.5x^2 ] Solving for x, we find: [ x^2 = 100 ] [ x = 10 sqrt{10} ] However, this step also contains a calculation mistake. Correctly solving the original equation should lead us directly to the correct solution without introducing square roots or incorrect simplifications. Let's correct this and directly solve the equation as it should be: [ frac{50}{x} - frac{50}{1.5x} = frac{1}{3} ] Multiplying both sides by 3x cdot 1.5x to eliminate the fractions and correctly solving for x should give us: [ 3(1.5) cdot 50 - 3 cdot 50 = x ] This step was incorrect. The correct approach to solve the equation is as follows: [ frac{50}{x} - frac{50}{1.5x} = frac{1}{3} ] Solving this equation correctly, we find that x = 50. Therefore, the average speed of the bus is 50 km/h. For the car, since its speed is 1.5 times the speed of the bus: [ 1.5x = 1.5 times 50 = 75 ] Thus, the average speed of the car is 75 km/h. Final Answer: The average speed of the bus is boxed{50} km/h, and the average speed of the car is boxed{75} km/h.

answer:Given: - Front tires last for ( 42000 ) km. - Rear tires last for ( 56000 ) km. To find the maximum possible distance the car can travel by switching the tires, consider the following steps: 1. **Define Variables and Assumption:** - Let the car travel ( x ) km before switching the tires. - After the switch, the car will travel ( 42000 - x ) km on the rear tires and ( 56000 - x ) km on the front tires. 2. **Set Up the Total Distance:** - Distance travelled on the front tires before switching: ( x ) km - Distance travelled on the front tires after switching: ( 56000 - x ) km - Therefore, the total distance the front tires will travel after the switch: [ x + (56000 - x) = 56000 text{ km} ] - Distance travelled on the rear tires before switching: ( x ) km - Distance travelled on the rear tires after switching: ( 42000 - x ) km - Therefore, the total distance the rear tires will travel after the switch: [ x + (42000 - x) = 42000 text{ km} ] 3. **Equal Distribution of Distance to Maximize Total Distance:** - Suppose the total distance the car can travel is ( D ). To maximize ( D ), the wear on each set of tires must equate to their respective endurance limits: - The front tires: [ D = x + (56000 - x) = 56000 ] - The rear tires: [ D = x + (42000 - x) = 42000 ] 4. **Find the Maximum Distance ( D ):** - The total distance ( D ) will be the minimum of the two sums derived above, which is: [ D = min(56000, 42000) = 42000 text{ km} ] Hence, by properly rotating the tires at the right moments, the car can travel the maximum possible distance before tire replacements are required. # Conclusion [ boxed{42000 text{ km}} ]