answer:First, let's calculate how many hours Peter will take to read one book. If Peter reads one book in 12 hours, then Kristin, who reads three times slower, will take 3 times longer to read one book. So, Kristin will take 12 hours * 3 = 36 hours to read one book. Now, let's calculate the total number of hours Peter and Kristin will take to read 20 books each. For Peter: 20 books * 12 hours/book = 240 hours For Kristin: 20 books * 36 hours/book = 720 hours Next, we need to consider the time constraints. They can only read between 6 am and 10 pm each day, which is a 16-hour reading window per day. Let's calculate how many days it will take for each of them to finish reading their books. For Peter: 240 hours / 16 hours/day = 15 days For Kristin: 720 hours / 16 hours/day = 45 days Since they must finish reading their books at the same time, we need to adjust the reading time to fit within the 45-day window, which is the time Kristin will take to read her books. Now, let's calculate the actual number of days they have available to read, considering the 16-hour reading window per day. For Kristin: 45 days * 16 hours/day = 720 hours (which matches the total hours she needs) For Peter: To finish at the same time as Kristin, Peter will also have 45 days to read. However, since he reads faster, he doesn't need to use the full 16-hour window each day. Let's calculate how many hours per day Peter needs to read to finish in 45 days. 240 hours / 45 days = 5.33 hours/day So, Peter needs to read approximately 5.33 hours each day to finish at the same time as Kristin. In conclusion, Kristin will take 36 hours to read one book and will finish reading her 20 books in 45 days, using the full 16-hour reading window each day. Peter will take 12 hours to read one book and will finish reading his 20 books in 45 days as well, but he only needs to read about 5.33 hours each day. They will both finish reading their books at the same time, which is boxed{45} days after they start.

answer:Let's denote the percentage of students who attend the chess class as P%. Since 50% of the students in the chess class are also enrolled for swimming, this means that half of the students in the chess class will attend the swimming class. We know that 100 students will attend the swimming class. Since this is 50% of the students in the chess class, we can set up the following equation to find the number of students in the chess class: 0.5 * (P% of 2000 students) = 100 students Let's solve for P%: 0.5 * (P/100 * 2000) = 100 0.5 * (P * 20) = 100 P * 10 = 100 P = 100 / 10 P = 10% Therefore, boxed{10%} of the students attend the chess class.

answer:1. Begin with the given equation: [ frac{frac{1}{x} + frac{1}{y}}{frac{1}{x} + 2frac{1}{y}} = 4 ] 2. Multiply both sides by ((frac{1}{x} + 2frac{1}{y})): [ frac{1}{x} + frac{1}{y} = 4left(frac{1}{x} + 2frac{1}{y}right) ] 3. Expand and simplify the right-hand side: [ frac{1}{x} + frac{1}{y} = 4cdotfrac{1}{x} + 8cdotfrac{1}{y} ] 4. Rearrange all terms: [ frac{1}{x} - 4cdotfrac{1}{x} = 8cdotfrac{1}{y} - frac{1}{y} ] 5. Simplify both sides: [ -3cdotfrac{1}{x} = 7cdotfrac{1}{y} ] 6. Solve for (frac{1}{x}) in terms of (frac{1}{y}): [ frac{1}{x} = -frac{7}{3} cdot frac{1}{y} ] 7. Substitute ( x = -frac{3}{7}y ) based on relationship between ( x ) and ( y ): [ frac{x+y}{x+2y} = frac{-frac{3}{7}y + y}{-frac{3}{7}y + 2y} ] 8. Simplify the fraction: [ frac{y(frac{4}{7})}{y(frac{11}{7})} = frac{4}{11} ] Conclusion with the answer boxed: [ frac{4{11}} ] The final answer is B. boxed{frac{4}{11}}

answer:1. Let's denote ( x + 3a ) by ( t ). This substitution does not change the number of solutions of the equation: [ 3^{x^2 + 6ax + 9a^2} = ax^2 + 6a^2 x + 9a^3 + a^2 - 4a + 4. ] Therefore, the equation in terms of ( t ) becomes: [ 3^{t^2} = a t^2 + a^2 - 4a + 4. ] 2. Observe that both sides of the equation remain unchanged when ( t ) is replaced by ( -t ). Therefore, for the equation to have exactly one solution, ( t = 0 ) must be a solution: [ 3^{0} = a cdot 0 + a^2 - 4a + 4. ] This simplifies to: [ 1 = a^2 - 4a + 4. ] 3. Solving the quadratic equation ( a^2 - 4a + 3 = 0 ): [ a^2 - 4a + 3 = 0 Rightarrow (a - 3)(a - 1) = 0. ] So the possible values for ( a ) are ( a = 3 ) and ( a = 1 ). We need to check if these values satisfy the condition of having exactly one solution. 4. When ( a = 1 ): The equation becomes: [ 3^{t^2} = t^2 + 1. ] Let's analyze whether this equation has more than one solution. Note that for ( t neq 0 ): [ 3^{t^2} > t^2 ln 3 + 1. ] This inequality can be shown using calculus, specifically by considering the derivatives of both sides: - The left-hand side derivative ( (3^{t^2})' = 2t cdot 3^{t^2} ln 3 ). - The right-hand side derivative ( (t^2 + 1)' = 2t ). Comparing these derivatives, for ( t > 0 ) or ( t < 0 ): [ 2t cdot 3^{t^2} ln 3 > 2t. ] Because ( ln 3 > 1 ), thus: [ 3^{t^2} > t^2 + 1 quad text{for} quad t neq 0. ] Therefore, the only solution is ( t = 0 ). 5. When ( a = 3 ): The equation becomes: [ 3^{t^2} = 3t^2 + 1. ] To check the number of solutions, we evaluate specific points: - For ( t = 1 ): [ 3^{1^2} = 3 quad text{and} quad 3 cdot 1^2 + 1 = 4. ] - For ( t = 2 ): [ 3^{2^2} = 81 quad text{and} quad 3 cdot 2^2 + 1 = 13. ] Since: [ 3 < 4 quad text{and} quad 81 > 13, ] By the Intermediate Value Theorem, there must be some value ( t ) in ( (1, 2) ) where the equation has another solution. Thus ( a = 3 ) does not result in exactly one solution. 6. Conclusion: The only value for ( a ) that results in exactly one solution is: [ boxed{1} ]