answer:We are given a sequence of shapes ( P_{n} ), where each shape is obtained by successively removing smaller semicircles from an initial semicircle ( P_{1} ) with radius 1. We need to determine the limit of the area ( S_{n} ) of ( P_{n} ) as ( n ) approaches infinity. 1. **Calculate the Initial Area:** - The area of the initial semicircle ( ( P_{1} )) with radius 1 is: [ S_{1} = frac{1}{2} pi cdot 1^2 = frac{pi}{2} ] 2. **Identify the Sequence of Removed Areas:** - The area of the first removed semicircle with radius ( frac{1}{2} ) is: [ text{Area}_1 = frac{1}{2} pi left( frac{1}{2} right)^2 = frac{1}{2} pi cdot frac{1}{4} = frac{pi}{8} ] - The area of the second removed semicircle with radius ( frac{1}{4} ) is: [ text{Area}_2 = frac{1}{2} pi left( frac{1}{4} right)^2 = frac{1}{2} pi cdot frac{1}{16} = frac{pi}{32} ] - Continuing this pattern, the area of the ( k )-th removed semicircle with radius ( frac{1}{2^{k-1}} ) is: [ text{Area}_k = frac{1}{2} pi left( frac{1}{2^{k-1}} right)^2 = frac{1}{2} pi cdot frac{1}{4^{k-1}} = frac{pi}{2} cdot frac{1}{4^{k-1}} ] 3. **Formulate a General Expression for ( S_n ):** - The total area removed from the initial semicircle after ( n-1 ) semicircles have been cut off is: [ text{Total removed area} = sum_{k=1}^{n-1} frac{pi}{2} cdot frac{1}{4^{k-1}} ] - The area ( S_n ) of ( P_n ) is then: [ S_n = frac{1}{2} pi - sum_{k=1}^{n-1} frac{pi}{2} cdot frac{1}{4^{k-1}} ] 4. **Simplify the Infinite Sum:** - Our aim is to find ( lim_{n to infty} S_n ). The series ( sum_{k=1}^{infty} frac{1}{4^{k-1}} ) is a geometric series with the first term ( a = 1 ) and common ratio ( r = frac{1}{4} ). The sum of an infinite geometric series is given by ( frac{a}{1-r} ): [ sum_{k=0}^{infty} left( frac{1}{4} right)^k = frac{1}{1 - frac{1}{4}} = frac{4}{3} ] Thus: [ sum_{k=1}^{infty} frac{1}{4^{k-1}} = 4 sum_{j=0}^{infty} left( frac{1}{4} right)^j = frac{4}{3} ] 5. **Find the Limit ( lim_{n to infty} S_n ):** - Substituting the sum back into the area formula, we get: [ lim_{n to infty} S_n = frac{1}{2} pi - frac{pi}{2} cdot frac{4}{3} = frac{pi}{2} - frac{2pi}{3} = frac{pi}{6} ] Therefore, the result is: [ boxed{frac{pi}{3}} ]

answer:**Analysis** This question examines the solution method for quadratic inequalities and quadratic functions. From the graph of the quadratic function y=(mx-1)(x-2), we know that the parabola opens downwards, which implies m < 0. **Solution** Let f(x)=(mx-1)(x-2), Since the solution set for (mx-1)(x-2) > 0 is {x| frac{1}{m} < x < 2}, It means the graph of f(x) opens downwards, Therefore, m < 0. Hence, the answer is boxed{m < 0}.

answer:Step 1: Rewrite the expression by converting the division into multiplication with the reciprocal: dfrac {1+x}{1-x} times dfrac {1-x}{x - dfrac {2x}{1-x}} Step 2: Simplify the denominator of the second fraction: dfrac {1+x}{1-x} times dfrac {1-x}{dfrac {x(1-x) - 2x}{1-x}} Step 3: Simplify the numerator of the second fraction: dfrac {1+x}{1-x} times dfrac {1-x}{dfrac {-x(1+x)}{1-x}} Step 4: Cancel out the common factors: dfrac {1+x}{cancel{1-x}} times dfrac {cancel{1-x}}{-dfrac {x(1+x)}{cancel{1-x}}} Step 5: Simplify the expression: dfrac {1+x}{-x} Step 6: Substitute x = sqrt{2}: dfrac {1+sqrt{2}}{-sqrt{2}} Step 7: Rationalize the denominator: dfrac {1+sqrt{2}}{-sqrt{2}} times dfrac {sqrt{2}}{sqrt{2}}=boxed{- dfrac {sqrt{2}+2}{2}}

answer:1. **Finding the dimensions of the rectangle PQRS:** - Given the area of the rectangle PQRS is 180 and the length SR = 15, we need to find the length PS. - Using the formula for the area of a rectangle, A = text{length} times text{width}, we have: [ text{Area} = SR times PS = 15 times PS = 180 ] - Solving for PS, we get: [ PS = frac{180}{15} = 12 ] 2. **Finding PU:** - We are given that US = 4. Since PS = 12 and U lies on PS, we can find PU by subtracting US from PS: [ PU = PS - US = 12 - 4 = 8 ] 3. **Finding TU using the Pythagorean Theorem:** - We are given PT = 10 and need to find TU. Note that triangle PUT is a right triangle with PU (base) and PT (hypotenuse). - By the Pythagorean Theorem: [ TU = sqrt{PT^2 - PU^2} ] - Substituting the values: [ TU = sqrt{10^2 - 8^2} = sqrt{100 - 64} = sqrt{36} = 6 ] - Since TU > 0, TU = 6. 4. **Finding the area of triangle PTS:** - In triangle PTS, we can consider PS as the base and TU as the height. - The formula for the area of a triangle is: [ text{Area} = frac{1}{2} times text{base} times text{height} ] - Substituting PS = 12 and TU = 6: [ text{Area of } triangle PTS = frac{1}{2} times 12 times 6 = frac{1}{2} times 72 = 36 ] # Conclusion: [ boxed{B} ]