answer:To solve the given problem, we start by rewriting the given equation using trigonometric identities: [ tan 2x + tan 3x = sec 3x ] We can express sec 3x in terms of cos 3x and use the identity to combine the terms: [ tan 2x = sec 3x - tan 3x = frac{1}{cos 3x} - frac{sin 3x}{cos 3x} = frac{1 - sin 3x}{cos 3x} ] Next, we recall the half-angle identity for tangent: [ tan frac{theta}{2} = frac{1 - cos theta}{sin theta} ] By comparing the forms, we can rewrite our expression as: [ frac{1 - sin 3x}{cos 3x} = frac{1 - cos (frac{pi}{2} - 3x)}{sin (frac{pi}{2} - 3x)} = tan left( frac{pi}{4} - frac{3x}{2} right) ] Therefore, we have: [ tan 2x = tan left( frac{pi}{4} - frac{3x}{2} right) ] Given that the tangent function has a period of pi, we can equate the arguments of the tangent functions and solve for x: [ 2x - left( frac{pi}{4} - frac{3x}{2} right) = n pi ] Solving for x, we find: [ x = frac{(4n + 1) pi}{14} ] To find the smallest positive solution, we consider the smallest integer value for n, which is n=0. Substituting n=0 into the expression for x, we obtain: [ x = frac{(4(0) + 1) pi}{14} = frac{pi}{14} ] Therefore, the smallest positive solution to the given equation in radians is boxed{frac{pi}{14}}.

answer:The surface area of a sphere is given by the formula A = 4pi r^2. - For Niraek: A = 4 pi times 5^2 = 100pi mm² - For Theo: A = 4 pi times 7^2 = 196pi mm² - For Akshaj: A = 4 pi times 11^2 = 484pi mm² To find when all three workers finish a donut hole at the same time, we compute the LCM of the surface areas: - 100 = 2^2 cdot 5^2 - 196 = 2^2 cdot 7^2 - 484 = 2^2 cdot 11^2 The LCM of 100pi, 196pi, and 484pi is 2^2 cdot 5^2 cdot 7^2 cdot 11^2 pi = 107800pi mm². The number of donut holes Niraek will have covered by this point is frac{107800pi}{100pi} = 1078. Thus, by the first time all three workers finish their current donut hole at the same time, Niraek will have covered boxed{1078} donut holes.

answer:(1) The probability of drawing exactly one yellow ball from a bag containing 1 red ball and 2 yellow balls is frac{2}{3}. (2) The probability that oil will float on the surface when dropped into water is 1. (3) The probability of rolling a fair die and getting a number less than 3 on the top face is frac{1}{3}. (4) The probability of drawing a red card from a deck of cards (excluding jokers) at random is frac{1}{2}. (5) The probability that two line segments can form a triangle is 0. Therefore, the correct answer is boxed{text{D}}.

answer:Part (a) 1. **Initial Condition**: - There are 2002 candidates initially. - A candidate is eliminated if they have the fewest votes in any round and if no one has more than half the votes. - The process continues until only one candidate remains, collecting all votes from eliminated candidates over successive rounds. 2. **Identifying Constraints**: - If Ostap Bender won in the 1002nd round, then for each of the previous 1001 rounds, some candidates were eliminated. - The least possible value for k is 1. The largest possible value for k is when Ostap Bender was just above the candidate who gets eliminated in the first round, meaning Ostap Bender could have been second to last. 3. **Maximum k Calculation**: - In the worst-case scenario, Ostap Bender can be positioned such that he almost always advances but barely avoids elimination. - Over the rounds, Ostap Bender improves his standing by gathering the votes from eliminated candidates one by one. To determine this maximum possible k, consider he needs to stay-in even until the 1002nd round happens. - To simplify, assume all 2002 candidates tied almost equally: If (k = 2001), this means Ostap would be the second-to-last candidate, but since Ostap wins, he must have survived every round with the minimum number of votes to avoid elimination. 4. **Final Vote Distribution**: - To calculate the number of votes, if each candidate initially gets votes in an almost equal distribution of 10^6, 10^6 + 1, ldots 10^6 + 2001, - When it comes to half the rounds, it means Ostap has outlasted 1001 candidates, collecting their votes one by one. [ text{Total votes Ostap could get:} 10^6 + left(10^6 + 1 + ldots + 10^6 + 1001right) = 1002 cdot 10^6 + frac{1001 cdot 1002}{2} = 10^6 + 501501. ] 5. **Conclusion for Part (a)**: - Ostap's maximum k ensures he missed elimination just until gaining the majority: [ boxed{2001} ] Part (b) 1. **Different Condition**: - Ostap Bender was elected in the 1001st round this time. 2. **Feasibility Check**: - Assume Ostap did not hold the first minimum position (k = 1). - For every more than half the total rounds, there were at least 1001 eliminations. - To win in the exact 1001st round, Ostap must avoid elimination from the start until he gathers a sufficient majority just at the transition between the rounds. 3. **Candidates and Votes**: - Naming the candidates who lost as the least (1001) rounds - If starting minimum, Ostap's placement can only remain viable if each round aggregates lost votes exactly matching his progress until the final positioning. 4. **Final Check**: - Ostap thus takes at each point a position ensuring he does not go beneath the threshold, leading to the conclusion: [ boxed{1} ]