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question:Simplify left( frac{1}{2k} right)^{2} cdot (k)^3.
answer:To simplify the expression left( frac{1}{2k} right)^{-2} cdot (-k)^3, we follow the steps below: 1. Apply the negative exponent rule, which states that a^{-n} = frac{1}{a^n}, to the first part of the expression. This gives us (2k)^2 because taking the reciprocal of frac{1}{2k} and squaring it results in (2k)^2. 2. The second part of the expression, (-k)^3, remains unchanged as it's already simplified. 3. Now, we multiply the two parts together: (2k)^2 cdot (-k)^3. 4. Simplify (2k)^2 to 4k^2 because squaring 2k results in 4k^2. 5. Then, multiply 4k^2 by (-k^3). This is done by multiplying the coefficients and adding the exponents of k. 6. The multiplication gives 4k^2 cdot (-k^3) = -4k^{2+3}. 7. Simplify the exponent by adding 2+3 to get -4k^5. Therefore, the simplified expression is boxed{-4k^5}.
question:a dance class, there are 3 girls for every 4 boys. If there are 35 students in total in the class, how many boys are there?
answer:Let the number of boys be 4k and the number of girls be 3k. The total number of students in the class is given by: 4k + 3k = 7k. Given there are 35 students in total, we set up the equation: 7k = 35. Solving for k, we get: k = frac{35}{7} = 5. So, the number of boys in the class is: 4k = 4 times 5 = boxed{20}.
question:the fictional country of Mathland, the populations of cities F, G, H, I, and J in years 1990 and 2000 are recorded. The city J had a population of 200,000 in 1990 and had a 50% increase by 2000. For the other cities, the populations are as follows: City F: 120,000 in 1990 and 150,000 in 2000 City G: 150,000 in 1990 and 195,000 in 2000 City H: 80,000 in 1990 and 100,000 in 2000 City I: 200,000 in 1990 and 260,000 in 2000 Which city had the greatest percentage increase in population from 1990 to 2000? textbf{(A)} 25%qquad textbf{(B)} 30%qquad textbf{(C)} 40%qquad textbf{(D)} 50%qquad textbf{(E)} 35%
answer:To resolve this question, we begin by calculating the population of city J in 2000: [ text{Population of J in 2000} = 200,000 + (0.50 times 200,000) = 300,000 ] Next, calculate the percentage increase for each city: - **City F:** [ frac{150,000 - 120,000}{120,000} times 100% = frac{30,000}{120,000} times 100% = 25% ] - **City G:** [ frac{195,000 - 150,000}{150,000} times 100% = frac{45,000}{150,000} times 100% = 30% ] - **City H:** [ frac{100,000 - 80,000}{80,000} times 100% = frac{20,000}{80,000} times 100% = 25% ] - **City I:** [ frac{260,000 - 200,000}{200,000} times 100% = frac{60,000}{200,000} times 100% = 30% ] - **City J:** [ frac{300,000 - 200,000}{200,000} times 100% = 50% ] Given these increases: - City F: 25% - City G: 30% - City H: 25% - City I: 30% - City J: 50% City J has the highest percentage increase. Conclusion: The answer is City J with a percentage increase of 50%. The final answer is boxed{textbf{(D)} 50%}
question:A train running at the speed of 54 km/hr crosses a pole in some seconds. The length of the train is 105 meters. How long does it take for the train to cross the pole?
answer:To find out how long it takes for the train to cross the pole, we need to convert the speed of the train from kilometers per hour (km/hr) to meters per second (m/s) because the length of the train is given in meters. We know that: 1 km = 1000 meters 1 hour = 3600 seconds So, to convert 54 km/hr to m/s, we use the following conversion: [ text{Speed in m/s} = text{Speed in km/hr} times frac{1000 text{ meters}}{1 text{ km}} times frac{1 text{ hour}}{3600 text{ seconds}} ] [ text{Speed in m/s} = 54 times frac{1000}{3600} ] [ text{Speed in m/s} = 54 times frac{5}{18} ] [ text{Speed in m/s} = 15 text{ m/s} ] Now that we have the speed of the train in meters per second, we can calculate the time it takes to cross the pole using the formula: [ text{Time} = frac{text{Distance}}{text{Speed}} ] The distance to be covered is equal to the length of the train, which is 105 meters. The speed is 15 m/s. So the time taken to cross the pole is: [ text{Time} = frac{105 text{ meters}}{15 text{ m/s}} ] [ text{Time} = 7 text{ seconds} ] Therefore, it takes boxed{7} seconds for the train to cross the pole.