answer:Given a number composed of 100 units, represented as: [ 111ldots 11 quad (text{100 ones}) ] This can be expressed in the form: [ a_0 + a_1 cdot 10^1 + a_2 cdot 10^2 + ldots + a_{99} cdot 10^{99} ] where (a_0, a_1, ldots, a_{99}) are non-negative integers, and the sum of these integers is at most 100. We need to prove that each (a_i) is equal to 1. # Detailed Steps: 1. **Representation of the Number**: The given number (111ldots 11) (with 100 ones) can be written as: [ sum_{i=0}^{99} 10^i ] 2. **Sum of Digits Constraint**: It is given that (a_0 + a_1 + cdots + a_{99} leq 100). 3. **Comparative Sum of Digits**: Note that the sum of the digits of (111ldots 11) is exactly 100 because each digit is `1` and there are 100 digits. 4. **Equating and Understanding the Constraint**: The number (111ldots 11) can indeed be written as: [ 1 cdot 10^0 + 1 cdot 10^1 + 1 cdot 10^2 + ldots + 1 cdot 10^{99} ] This matches exactly our desired sum where each (a_i = 1). 5. **Considering another Possibility**: Suppose there is any (a_i) which is greater than 1 or any (a_i) which is 0. Then the left-hand side (LHS) which is (a_0 + a_1 + cdots + a_{99}) will either exceed 100 or be less than 100. 6. **Sum Constraint Violation**: Any case where any (a_i neq 1) would violate the constraint (a_0 + a_1 + cdots + a_{99} leq 100) while still trying to sum up correctly to the numerical value represented by the infinite series. # Conclusion: Given the constraints and the structure of the problem, we must have each (a_i = 1) for (i=0) to (99). Therefore, all coefficients (a_0, a_1, ldots, a_{99}) must be equal to 1. ( boxed{ text{Each } a_i text{ is 1} } )

answer:Firstly, convert both polar equations to Cartesian coordinates. For r = 6 sin theta, begin{align*} x &= r cos theta = 6 sin theta cos theta = 3 sin 2theta, y &= r sin theta = 6 sin^2 theta = 3 - 3 cos 2theta. end{align*} Thus, the graph of r = 6 sin theta results in: [x^2 + (y-3)^2 = 9.] This is a circle centered at (0,3) with radius 3. For r = 2 sin(2theta + pi), begin{align*} r &= -2 sin(2theta) = -4 sin theta cos theta, x &= r cos theta = -4 sin theta cos theta = -2 sin 2theta, y &= r sin theta = -4 sin^2 theta = -2 + 2 cos 2 theta. end{align*} Thus, the graph of r = 2 sin(2theta + pi) results in: [x^2 + (y+2)^2 = 4.] This is a circle centered at (0,-2) with radius 2. Now, to find intersections: The centers of the circles are at (0,3) and (0,-2), with a distance of 5 between them. The sum of the radii is 5, which is exactly the distance between the centers. Hence, they touch at exactly one point. Conclusion: The graphs r = 6 sin theta and r = 2 sin(2theta + pi) intersect at boxed{1} point.

answer:When x > 0, x geq ax always holds, which means a leq 1. When x = 0, 0 geq a times 0 always holds, which means a in mathbb{R}. When x < 0, -x geq ax always holds, which means a geq -1. If for any x in mathbb{R}, the inequality |x| geq ax always holds, then -1 leq a leq 1. Therefore, the correct choice is boxed{text{B}}.

answer:# Solution: Part 1: - **Number of Athletes**: There are 11 athletes in the group. - **Average Weight Calculation**: The average weight is calculated by summing all the weights and dividing by the number of athletes. [ frac{82.7 + 78.7 + 78.8 + 77.3 + 83.6 + 85.4 + 73 + 80.6 + 83.2 + 71.3 + 74.4}{11} = frac{879}{11} = 79 , text{kg} ] - **Median Weight Calculation**: First, arrange the weights in ascending order: 71.3, 73, 74.4, 77.3, 78.7, 78.8, 80.6, 82.7, 83.2, 83.6, 85.4. Since there are 11 weights, the median weight is the weight in the 6th position, which is 78.8 , text{kg}. Therefore, the answers are: - Number of athletes: boxed{11} - Average weight: boxed{79 , text{kg}} - Median weight: boxed{78.8 , text{kg}} Part 2: - **Total Number of People**: 11 , text{(athletes)} + 4 , text{(ladies)} = 15. Since 15 < 18, the number of people does not exceed the elevator's capacity. - **Total Weight Calculation**: The total weight of the athletes and the ladies is calculated as follows: [ (79 times 11) + (52.3 times 4) = 869 + 209.2 = 1078.2 , text{kg} ] Since 1078.2 , text{kg} < 1100 , text{kg}, the total weight does not exceed the safe load of the elevator. Therefore, they can safely ride the elevator together, which can be encapsulated as boxed{text{Yes}}.