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question:Given frac{x}{y}=frac{4}{3}, then frac{x-y}{y}=______.
answer:Given frac{x}{y}=frac{4}{3}, we aim to find the value of frac{x-y}{y}.Starting with the given equation:frac{x}{y}=frac{4}{3}From this, we can express x in terms of y:x=frac{4}{3}yNow, we substitute x=frac{4}{3}y into the expression frac{x-y}{y}:begin{align*}frac{x-y}{y} &= frac{frac{4}{3}y - y}{y} &= frac{frac{4}{3}y - frac{3}{3}y}{y} &= frac{frac{1}{3}y}{y} &= frac{1}{3}end{align*}Therefore, the answer is boxed{frac{1}{3}}.
question:A market survey shows that the sales volume (in pieces) and price (in yuan) of a certain small commodity in a city in the past 20 days are functions of time t (in days). The sales volume approximately satisfies g(t) = 80 - 2t (pieces), and the price approximately satisfies f(t) = 20 - |t - 10| (yuan).1. Write the function expression of the daily sales y of this commodity with time t (0 ≤ t ≤ 20);2. Find the maximum and minimum values of the daily sales y of this commodity.
answer:1. y = g(t) ⋅ f(t) = (80 - 2t) ⋅ (20 - |t - 10|) = begin{cases} (80 - 2t)(10 + t), & 0 leq t < 10 (80 - 2t)(30 - t), & 10 leq t leq 20 end{cases} 2. When 0 ≤ t < 10, y = -2t² + 60t + 800 is monotonically increasing on [0, 10), and the range of y is [800, 1200). When 10 ≤ t ≤ 20, y = 2t² - 140t + 2400 is monotonically decreasing on [10, 20], and the range of y is [1200, 400]. At t = 20, y takes the minimum value of boxed{400}. At t = 10, y takes the maximum value of boxed{1200}. Therefore, on the 10th day, the daily sales y takes the maximum value of boxed{1200} yuan; On the 20th day, the daily sales y takes the minimum value of boxed{400} yuan.
question:The graph of the function y=sin ( frac {1}{2}x+ frac {π}{3}) is translated to the right by frac {π}{3}. The expression corresponding to the resulting graph is ( )A: y=sin frac {1}{2}xB: y=sin ( frac {1}{2}x+ frac {π}{6})C: y=sin ( frac {1}{2}x- frac {π}{3})D: y=sin ( frac {1}{2}x- frac {2π}{3})
answer:The graph of the function y=sin ( frac {1}{2}x+ frac {π}{3}) is translated to the right by frac {π}{3}, resulting in the expression y=sin [ frac {1}{2}(x- frac {π}{3})+ frac {π}{3}].This simplifies to y=sin ( frac {1}{2}x+ frac {π}{6}),Hence, the answer is boxed{B}.This problem primarily tests the understanding of the transformation rules of the graph of the function y=Asin (ωx+φ), which is a basic problem.
question:## Task 1 - V00501The fencing of a square garden is being renewed. It costs 992.00 DM. One meter of fencing is charged at 4.00 DM.Calculate the area of this garden and convert the result into hectares.
answer:The perimeter u of a square with side length a is u=4a. Since frac{992}{4}=248 text{~m} of fencing was installed, a=62 text{~m}.For the area, A=a^{2}=62^{2}=3844 text{~m}^{2}, i.e., approximately 0.38 ha.
question:1. If a>0, b>0, and a+2b=6, then the maximum value of ln a+2 ln b is qquad
answer:ln a+2 ln b=ln left(a b^{2}right) leqslant ln left(frac{a+2 b}{3}right)^{3}=3 ln 2, equality holds when a=b=2. Therefore, the maximum value of ln a+2 ln b is 3 ln 2.
question:10. The graph of the quadratic function f(x)=a x^{2}+(3 a-1) x-(10 a+3)(a neq 0) always passes through two fixed points. The coordinates of these two points are qquad .
answer:10. (2,-5) and (-5,2).y=left(x^{2}+3 x-10right) a-(x+y+3), if the fixed point is independent of a, then left{begin{array}{l}x^{2}+3 x-10=0 x+y+3=0,end{array}right. we get left{begin{array}{l}x=2 y=-5end{array}right. or left{begin{array}{l}x=-5 y=2end{array}right.