answer:To solve the problem, we start by understanding the given conditions and translating them into mathematical equations. Given: 1. "There are 5 large containers and 1 small container with a total capacity of 3 hu." This can be translated to the equation 5x + y = 3. 2. "There is 1 large container and 5 small containers with a total capacity of 2 hu." This translates to the equation x + 5y = 2. Therefore, the system of equations that represents the problem is: [ left{begin{array}{l} 5x + y = 3 x + 5y = 2 end{array}right. ] Comparing this system of equations with the given choices: A: left{begin{array}{l}{x+5y=3}{5x+y=2}end{array}right. B: left{begin{array}{l}{5x+y=3}{x+5y=2}end{array}right. C: left{begin{array}{l}{5x=y+3}{x=5y+2}end{array}right. D: left{begin{array}{l}{5x=y+2}{x=5y+3}end{array}right. It is clear that the correct system of equations matches with choice B. Therefore, the correct answer is boxed{B}.

answer:1. The total parts of the work are given by the sum of the ratio numbers: 3 (Allen) + 5 (Ben) + 2 (Charlie) = 10 parts. 2. Each part corresponds to frac{300}{10} = 30 square feet of the wall. 3. Charlie is responsible for 2 parts of the work, so he will paint 30 cdot 2 = 60 square feet of the wall. Conclusion: Charlie will paint boxed{60} square feet of the wall.

answer:To solve the equation dfrac{2x+1}{3}-dfrac{x-1}{6}=2, we start by eliminating the denominators to simplify the equation. We can do this by finding a common denominator, which in this case is 6, and then multiplying each term by this common denominator to get rid of the fractions: [ 6 times left( dfrac{2x+1}{3} right) - 6 times left( dfrac{x-1}{6} right) = 6 times 2 ] This simplifies to: [ 2(2x+1) - (x-1) = 12 ] Expanding the terms on the left side of the equation: [ 4x + 2 - x + 1 = 12 ] Combining like terms (the x terms and the constant terms): [ 3x + 3 = 12 ] Subtracting 3 from both sides to isolate the term with x: [ 3x = 9 ] Finally, dividing both sides by 3 to solve for x: [ x = dfrac{9}{3} ] This gives us: [ x = 3 ] Therefore, the solution to the equation is boxed{x = 3}.

answer:1. **Determine the bounds**: - For the lower bound, frac{5}{7}, observe that it lies between 0 and 1 since frac{5}{7} < 1. [ 0 < frac{5}{7} < 1 ] - For the upper bound, frac{28}{3}, convert it to a mixed number to understand where it lies between the integers: [ frac{28}{3} = 9 frac{1}{3} Rightarrow 9 + frac{1}{3} ] Therefore, frac{28}{3} lies between 9 and 10: [ 9 < frac{28}{3} < 10 ] 2. **Identify the integer range**: - The integers greater than frac{5}{7} must be greater than 0. - The integers less than frac{28}{3} must be less than 10. The integers greater than frac{5}{7} and less than frac{28}{3} therefore are: [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ] 3. **Count the integers**: - There are 9 integers listed in step 2. [ 1, 2, 3, 4, 5, 6, 7, 8, 9 quad text{(Total = 9)} ] # Conclusion: [ boxed{9} ]