answer:To swap two numbers, we need to use an intermediate variable. Therefore, the correct answer is boxed{text{B}}.

answer:1. Given that the function (f(x)) satisfies the property (f(x+2) = f(x)), it indicates that the function is periodic with a period of 2. 2. When (x in (0,1)), the function is defined as (f(x) = 2^x). 3. When (x in (-1,0)), since (0 < -x < 1), we consider (f(-x)). By the given function, (f(-x) = 2^{-x}). 4. From the periodic property (f(x+2) = f(x)), it implies that the function (f(x)) can be extended for any (x) using the intervals ((0,1)) and ((-1,0)). 5. To find (fleft(log_{frac{1}{2}}23right)), first we note the properties of logarithms with respect to the change of base. Since (log_{frac{1}{2}}(23) = -log_2(23)) by the change of base formula. 6. Next, we calculate the range for (log_{frac{1}{2}}23). Since (frac{1}{2}^4 < 23 < frac{1}{2}^5) and noting the properties of logarithms, [ 4 < - log_2(23) < 5 ] implying [ -5 < log_{frac{1}{2}}23 < -4. ] 7. To make this more systematic within the interval ((-1,0)), observe that [ log_{frac{1}{2}}23 = -log_2(23) = -(log_2(32) - log_2(frac{32}{23})) = -5 + log_2(frac{32}{23}). ] This simplification might be tricky but helps us understand the inclusion of the periodicity. 8. Now using periodicity: [ log_{frac{1}{2}}23 = x approx 4 + text{(remaining decimal part)}. ] This conversion puts (x) in consideration as (-x) interval results in [ f(x) = f(log_{frac{1}{2}}(23 / 16)). ] 9. Given (f(x) = -2^{-x}), when (x in (-1, 0)) we calculate, [ boxed{f left( log_{frac{1}{2}} frac{23}{16} right) = -frac{23}{16}.} ]

answer:To find a sine function f(x) that satisfies the given conditions, we start by expressing it in the general form of a sine function: f(x) = Asin(omega x + varphi), where A > 0 and omega > 0. 1. **Condition on Maximum Value**: Given that for all x in mathbb{R}, there exists an x_0 such that f(x) leqslant f(x_0) = 2, it implies that the maximum value of f(x) is 2. Therefore, we deduce that A = 2. 2. **Condition on Period**: The smallest positive period of f(x) is given as pi. The period T of a sine function is related to omega by T = frac{2pi}{omega}. Setting T = pi, we solve for omega: [ pi = frac{2pi}{omega} implies omega = 2. ] Thus, the function now takes the form f(x) = 2sin(2x + varphi). 3. **Condition on Monotonicity**: The function is monotonically increasing on the interval [0, frac{pi}{4}]. For a sine function, this means that within this interval, the argument of the sine function, 2x + varphi, must be increasing and within a range where sine itself is increasing. Sine increases in the interval [-frac{pi}{2}, frac{pi}{2}]. Therefore, for x in [0, frac{pi}{4}], the argument 2x + varphi must map into an interval that is a subset of [-frac{pi}{2}, frac{pi}{2}]. Considering the transformation, we have 2x + varphi in [varphi, varphi + frac{pi}{2}]. To ensure f(x) is increasing in the given interval, we need [varphi, varphi + frac{pi}{2}] subseteq [-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpi] for some k in mathbb{Z}. This implies -frac{pi}{2} + 2kpi leqslant varphi leqslant 2kpi. Choosing k = 0 for simplicity, we get -frac{pi}{2} leqslant varphi leqslant 0. Let's select varphi = -frac{pi}{4} to satisfy this condition. Therefore, the function that meets all the given conditions is f(x) = 2sin(2x - frac{pi}{4}). However, it's important to note that the solution is not unique. Any function of the form f(x) = 2sin(2x + varphi), where -frac{pi}{2} + 2kpi leqslant varphi leqslant 2kpi and k in mathbb{Z}, will satisfy the given conditions. Hence, the answer is: boxed{2sin(2x - frac{pi}{4})}.

answer:In linear regression, the coefficient of determination R^2 is equal to the square of the correlation coefficient. Given x_1=1, x_2=2, x_3=3, x_4=4, we have: overline {x}=2.5, y_1=1.4, y_2=2.2, y_3=3, y_4=3.8, we have: overline {y}=2.6, Therefore, the correlation coefficient r is calculated as r= frac { sumlimits_{i=1}^{4}(x_{i}- overline {x})(y_{i}- overline {y})}{ sqrt { sumlimits_{i=1}^{4}(x_{i}- overline {x})^{2}} sqrt { sumlimits_{i=1}^{4}(y_{i}- overline {y})^{2}}} = frac {1.5×1.2+0.5×0.4+0.5×0.4+1.5×1.2}{ sqrt {1.5^{2}+0.5^{2}+1.5^{2}+0.5^{2}} sqrt {1.2^{2}+0.4^{2}+1.2^{2}+0.4^{2}}} = frac {4}{ sqrt {5}× sqrt {3.2}}= frac {4}{4} = 1. Therefore, the correct choice is boxed{C}. Since the two variables are linearly related, the coefficient of determination is equal to the square of the correlation coefficient. Directly substituting into the formula for the correlation coefficient to solve is sufficient. This question tests the understanding of the correlation coefficient, computational skills, and memory of formulas. The key to solving this question is knowing that the coefficient of determination between two linearly related variables is equal to the square of the correlation coefficient. This question is a basic one.