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question:Calculate:<br/>(1)2^2+(frac{1}{2})^4+(3π)^0;<br/>(2){5}^{2022}×(frac{1}{5})^{2023}.
answer:For the given problems, let's break them down step by step: # Problem 1: Calculate (1)-2^2+(-frac{1}{2})^4+(3-π)^0. Solution: We start by evaluating each term individually: - The first term is -2^2. Remember, the negative sign affects only the 2, not the squaring, so we have: -2^2 = -(2cdot2) = -4. - The second term is (-frac{1}{2})^4. Raising a fraction to the fourth power means multiplying it by itself four times: (-frac{1}{2})^4 = left(-frac{1}{2}right)cdotleft(-frac{1}{2}right)cdotleft(-frac{1}{2}right)cdotleft(-frac{1}{2}right) = frac{1}{16}. - The third term is (3-π)^0. Any non-zero number raised to the power of 0 equals 1: (3-π)^0 = 1. Adding these results together: -4 + frac{1}{16} + 1 = -4 + frac{1}{16} + frac{16}{16} = -frac{64}{16} + frac{1}{16} + frac{16}{16} = -frac{47}{16}. Therefore, the final answer for problem (1) is boxed{-frac{47}{16}}. # Problem 2: Calculate {5}^{2022}times(-frac{1}{5})^{2023}. Solution: We can break down the expression by separating the powers of 5 and -frac{1}{5}: - First, notice that {5}^{2022}times(-frac{1}{5})^{2023} can be rewritten as: 5^{2022}times (-frac{1}{5})^{2022}times (-frac{1}{5}). - We can then combine the powers of 5 and -frac{1}{5}, since multiplying them together for the same exponent gives: (-5times frac{1}{5})^{2022}times (-frac{1}{5}) = (-1)^{2022}times (-frac{1}{5}). - Since (-1)^{2022} is an even power of -1, it equals 1: left(-1right)^{2022}times (-frac{1}{5}) = 1times (-frac{1}{5}) = -frac{1}{5}. Thus, the final answer for problem (2) is boxed{-frac{1}{5}}.
question:We define a function g(x) such that g(12)=37, and if there exists an integer a such that g(a)=b, then g(b) is defined and follows these rules: 1. g(b)=3b+1 if b is odd 2. g(b)=frac{b}{2} if b is even. What is the smallest possible number of integers in the domain of g?
answer:Since g(12)=37, we know that g(37) is defined, and it must equal 112. Following this rule: begin{align*} g(112) &= 56 g(56) &= 28 g(28) &= 14 g(14) &= 7 g(7) &= 22 g(22) &= 11 g(11) &= 34 g(34) &= 17 g(17) &= 52 g(52) &= 26 g(26) &= 13 g(13) &= 40 g(40) &= 20 g(20) &= 10 g(10) &= 5 g(5) &= 16 g(16) &= 8 g(8) &= 4 g(4) &= 2 g(2) &= 1 g(1) &= 4 end{align*} The sequence from g(1) enters the previously known cycle 1 rightarrow 4 rightarrow 2 rightarrow 1, etc. Since there are no integers left that need defining, counting all defined integers, we find: {12, 37, 112, 56, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} Thus the minimum number of integers we can define within this domain is boxed{23}.
question:The slope of the tangent line to the curve y= frac {1}{3}x^{3}2 at the point (1, frac {5}{3}) is ( ) A: sqrt {3} B: 1 C: 1 D: sqrt {3}
answer:The derivative of y= frac {1}{3}x^{3}-2 is: y'=x^{2}, Substituting the x-coordinate of the point (1,- frac {5}{3}), we get the slope as: k=1. Hence, the answer is: boxed{B}. To find the slope of the tangent line to a curve at a specific point, we need to find the derivative of the function at that point. This is done by first finding the derivative function, and then substituting the coordinates of the point into the derivative function. This question tests the geometric meaning of derivatives, which connects the derivative of a function with the tangent line of the curve, making the derivative an important carrier for the intersection of function knowledge and analytic geometry knowledge. This is a basic question.
question:f alpha is an obtuse angle, then the terminal side of frac{alpha}{2} lies in underline{hspace{1cm}}. A: Quadrant I B: Quadrant II C: Quadrant III D: Quadrant I or III
answer:Given that alpha is an obtuse angle, therefore frac{pi}{2} < alpha < pi Dividing by 2, we get frac{pi}{4} < frac{alpha}{2} < frac{pi}{2} Since frac{alpha}{2} lies in the first quadrant, Hence, the answer is: boxed{A}. To solve this problem, we determine the range of frac{alpha}{2} based on the given range of alpha. This problem primarily tests the understanding of quadrant angles, with the key being to identify the terminal side's position. It is a relatively basic question.