answer:To find the reciprocal of a number, we divide 1 by that number. For -2022, the reciprocal is calculated as follows: [ text{Reciprocal of } -2022 = frac{1}{-2022} = -frac{1}{2022} ] Therefore, the correct answer is boxed{text{B}}.

answer:1. **Domain Analysis**: For 0^circ < x < 90^circ, sin x ranges from 0 to 1 (excluding 0 but including 1). 2. **Function Composition**: Given sin x is between 0 and 1, then sqrt{sin x} is also between 0 and 1. Specifically, since the square root function is increasing and sin x never hits 0 in the open interval (0^circ, 90^circ), sqrt{sin x} ranges from 0 (exclusively) to 1 (inclusively). 3. **Logarithmic Function**: The logarithm base 3 of a number between 0 and 1 is negative. Hence, as sqrt{sin x} ranges from just above 0 up to 1, log_3 (sqrt{sin x}) ranges from a large negative number (tending towards -infty as sqrt{sin x} approaches 0) up to 0 (as sqrt{sin x} equals 1). Conclusion: The range of the function y=log_3 (sqrt{sin x}) for 0^circ < x < 90^circ is boxed{(-infty, 0]}.

answer:Let's denote the two integers as ( a ) and ( b ). We are given that the product of these integers is ( a times b = 19.999999999999996 ). Since the product is very close to 20 and we are dealing with integers, it is reasonable to assume that the two integers are ( a = 20 ) and ( b = 1 ) (or vice versa), as ( 20 times 1 = 20 ), which is extremely close to the given product. Now, we need to find the sum of the cubes of these integers, which is ( a^3 + b^3 ). For ( a = 20 ) and ( b = 1 ), the sum of the cubes is: [ 20^3 + 1^3 = 8000 + 1 = 8001 ] Therefore, the certain number, which is the sum of the cubes of the two integers, is ( boxed{8001} ).

answer:First, we find the derivative of f(x), which is f′(x)=a(x-2)(3x-2). (1) When a > 0, from f′(x) > 0, we get x < dfrac {2}{3} or x > 2; from f′(x) < 0, we get dfrac {2}{3} < x < 2. Therefore, f(x) is increasing on (-∞, dfrac {2}{3}) and (2,+∞), and decreasing on (dfrac {2}{3},2). At this time, when x= dfrac {2}{3}, f(x) reaches its maximum value f( dfrac {2}{3})= dfrac {2}{3}a( dfrac {2}{3}-2)^{2}= dfrac {16}{9}. Solving this, we find a= dfrac {3}{2}. (2) When a < 0, from f′(x) < 0, we get x < dfrac {2}{3} or x > 2; from f′(x) > 0, we get dfrac {2}{3} < x < 2. Therefore, f(x) is decreasing on (-∞, dfrac {2}{3}) and (2,+∞), and increasing on (dfrac {2}{3},2). At this time, when x=2, f(x) reaches its maximum value f(2)=2a(2-2)^{2}= dfrac {16}{9}, which has no solution. In summary, the sought value of a is boxed{dfrac {3}{2}}. Therefore, the correct choice is B. By using the derivative to divide into cases a > 0 and a < 0, we find the maximum value of f(x) to be dfrac {16}{9}. Solving this equation gives the value of a. This problem examines the use of derivatives to find the extreme values of a function, which is a basic question. Mastering the conditions under which a function reaches an extreme value at a certain point is key to solving such problems.