answer:To solve this problem, we follow the process of reducing the concentration of ammonia nitrogen in the wastewater through multiple cycles of filtration and calculate the minimum number of cycles required to meet the safety standard. Let's break down the solution step by step: 1. **Initial Concentration**: The initial concentration of ammonia nitrogen in the wastewater is 225mL/L. 2. **Reduction per Cycle**: Each cycle of filtration reduces the ammonia nitrogen content by frac{1}{3}. Therefore, after n cycles, the concentration becomes 225 times left(1 - frac{1}{3}right)^n. 3. **Safety Requirement**: The safety requirement is that the concentration does not exceed half of the national emission standard, which is frac{15mL/L}{2} = 7.5mL/L. 4. **Formulating the Inequality**: To meet the safety requirement, we set up the inequality: [225 times left(1 - frac{1}{3}right)^n leq 7.5] 5. **Simplifying the Inequality**: Simplifying the inequality gives us: [left(frac{2}{3}right)^n geq frac{7.5}{225}] [left(frac{2}{3}right)^n geq frac{1}{30}] 6. **Inverting the Fraction**: Since frac{2}{3} is less than 1, we can invert both sides of the inequality to get: [left(frac{3}{2}right)^n geq 30] 7. **Taking Logarithms**: Taking the logarithm (base 10) of both sides: [n cdot lgleft(frac{3}{2}right) geq lg 30] [n geq frac{lg 30}{lgleft(frac{3}{2}right)}] 8. **Using Logarithm Properties**: We know that lg 30 = lg (10 times 3) = lg 10 + lg 3, and lgleft(frac{3}{2}right) = lg 3 - lg 2. Substituting these values: [n geq frac{lg 10 + lg 3}{lg 3 - lg 2}] 9. **Substituting Approximate Log Values**: Given lg 2 approx 0.3010 and lg 3 approx 0.4771: [n geq frac{1 + 0.4771}{0.4771 - 0.3010}] [n geq frac{1.4771}{0.1761}] [n geq 8.39] 10. **Determining the Minimum Number of Cycles**: Since n must be an integer and n geq 8.39, the minimum number of cycles required is 9. Therefore, the minimum number of cycles required before emission to ensure safety is boxed{9} cycles, which corresponds to choice boxed{D}.

answer:1. **Identify the series type and formula**: The given series is an arithmetic series. The sum of an arithmetic series where a_1 is the first term, a_n is the last term, and n is the number of terms can be calculated using the formula: [ S = frac{n(a_1 + a_n)}{2} ] 2. **Prove the formula for the sum of an arithmetic series**: - Let d be the common difference between consecutive terms. - The n-th term of the series can be expressed as a_n = a_1 + (n-1)d. - Consider the sum S of the series: [ S = a_1 + (a_1 + d) + (a_1 + 2d) + cdots + (a_1 + (n-1)d) ] - Writing the sum in reverse order and adding to the original sum: [ S = (a_1 + (n-1)d) + (a_1 + (n-2)d) + cdots + a_1 ] - Adding these two expressions for S: [ 2S = n(a_1 + a_n) ] - Solving for S gives: [ S = frac{n(a_1 + a_n)}{2} ] 3. **Apply the formula to the given series**: - The first term a_1 = 20. - The last term a_n = 40. - The common difference d = frac{1}{5}. - To find the number of terms n, use the formula for the n-th term: [ 40 = 20 + (n-1)frac{1}{5} ] - Solving for n: [ 40 - 20 = (n-1)frac{1}{5} implies 20 = (n-1)frac{1}{5} implies n-1 = 100 implies n = 101 ] 4. **Calculate the sum**: - Using the sum formula: [ S = frac{101(20 + 40)}{2} = frac{101 times 60}{2} = 101 times 30 = 3030 ] 5. **Conclusion**: - The sum of the arithmetic series is boxed{textbf{(B)} 3030}.

answer:1. **Identify the midpoint of the arc:** - Let ( k ) be a circle with center ( O ) and let ( A ) be a point on ( k ). - Let ( B ) and ( C ) be two other points on ( k ). - Let ( X ) be the intersection of the angle bisector of ( angle ABC ) with ( k ). 2. **Determine the position of ( X ):** - Since ( X ) lies on the angle bisector of ( angle ABC ) and on the circle ( k ), ( X ) is the midpoint of the arc ( AC ) that does not contain ( B ). This is because the angle bisector of an angle inscribed in a circle intersects the circle at the midpoint of the arc opposite the angle. 3. **Reflect ( A ) with respect to ( X ):** - Let ( Y ) be the reflection of ( A ) with respect to ( X ). By definition of reflection, ( X ) is the midpoint of ( AY ), so ( AX = XY ). 4. **Analyze the collinearity and distances:** - Since ( X ) is the midpoint of the arc ( AC ), we have ( AX = XC ). - Given ( AX = XY ) and ( A, X, Y ) are collinear, it follows that ( XY = XC ). 5. **Determine the angle ( angle ACY ):** - Since ( AX = XC ) and ( X ) is the midpoint of the arc ( AC ), the angle ( angle ACX ) is a right angle, i.e., ( angle ACX = frac{pi}{2} ). - Because ( XY = XC ) and ( X ) is the midpoint, ( angle ACY = frac{pi}{2} ). 6. **Find the intersection point ( D ):** - Let ( D ) be the intersection of the line ( YC ) with the circle ( k ). - Since ( angle ACY = frac{pi}{2} ), ( D ) must be the diametral opposite of ( A ) on the circle ( k ). This is because the only point on the circle that forms a right angle with ( A ) and ( C ) is the point diametrically opposite ( A ). 7. **Conclusion:** - Since ( D ) is the diametral opposite of ( A ) on the circle ( k ), it is independent of the choice of points ( B ) and ( C ) on the circle ( k ). (blacksquare)

answer:To calculate the total miles Rocky ran in the first three days of training, we follow the given pattern: - **Day 1**: Rocky runs 4 miles. - **Day 2**: Rocky doubles the miles from day 1, so he runs 2 times 4 = 8 miles. - **Day 3**: Rocky triples the miles from day 2, so he runs 3 times 8 = 24 miles. To find the total, we sum the miles from all three days: [4 + 8 + 24 = 36 , text{miles}] Therefore, the total miles Rocky ran in the first three days of training is boxed{36} miles.