💡First, I recognize that an isosceles triangle has two equal sides and two equal angles, known as the base angles.Given that there's an exterior angle of 110°, I recall that an exterior angle is supplementary to its adjacent interior angle. Therefore, the adjacent interior angle is 180° - 110° = 70°.Next, I consider two scenarios:1. **If the 70° angle is one of the base angles**, then both base angles are 70°, and the apex angle can be calculated as 180° - 2 × 70° = 40°.2. **If the 70° angle is the apex angle**, then the base angles are equal, and each base angle is (180° - 70°) / 2 = 55°.Thus, the base angle can be either 55° or 70°, depending on whether the 70° angle is the apex or a base angle.

💡First, I need to simplify the given congruence ( 13(3x - 2) equiv 30 pmod{12} ). Since ( 13 equiv 1 pmod{12} ) and ( 30 equiv 6 pmod{12} ), the congruence simplifies to:[3x - 2 equiv 6 pmod{12}]Adding 2 to both sides gives:[3x equiv 8 pmod{12}]To solve for ( x ), I need to find a multiplier that will isolate ( x ). Multiplying both sides by 4 (since ( 4 times 3 equiv 0 pmod{12} )) results in:[12x equiv 32 pmod{12}]However, since ( 12x equiv 0 pmod{12} ), this approach doesn't directly work. Instead, I'll try multiplying by 5:[15x equiv 40 pmod{12}]Simplifying, ( 15 equiv 3 pmod{12} ) and ( 40 equiv 4 pmod{12} ), so:[3x equiv 4 pmod{12}]Next, multiplying by 9 (since ( 9 times 3 equiv 3 pmod{12} )) gives:[27x equiv 36 pmod{12}]Simplifying further, ( 27 equiv 3 pmod{12} ) and ( 36 equiv 0 pmod{12} ), so:[3x equiv 0 pmod{12}]This implies:[x equiv 0 pmod{4}]Therefore, the general solution is ( x = 4k ) for integer values of ( k ). Considering the range ( 0 < x leq 30 ), the solutions are ( 4, 8, 12, 16, 20, 24, 28 ). Summing these solutions:[4 + 8 + 12 + 16 + 20 + 24 + 28 = 112]

💡Alright, so I've got this problem about four volunteers—A, B, C, and D—who are monitoring a garbage point. Each of them has two time slots they can work on a certain day. The goal is to figure out the maximum total monitoring time all of them can contribute, given that at any time, there can be at most two volunteers on duty. Also, each volunteer must work at least one time slot per day.First, I need to understand the time slots each volunteer can work:- **A** can work from 6:00-8:00 and 16:00-18:00.- **B** can work from 6:30-7:30 and 17:00-20:00.- **C** can work from 8:00-11:00 and 18:00-19:00.- **D** can work from 7:00-10:00 and 17:30-18:30.Each volunteer must work at least one time slot, so I need to make sure that everyone is scheduled for at least one of their available slots. Also, at any given time, no more than two volunteers can be on duty. So, I need to schedule their shifts in such a way that these constraints are satisfied, and the total monitoring time is maximized.I think the first step is to list out all the time slots and see where overlaps occur. Overlaps are important because they determine when more than one volunteer is working simultaneously, which is limited to two volunteers at any time.Let's list out the time slots:- **A1:** 6:00-8:00- **A2:** 16:00-18:00- **B1:** 6:30-7:30- **B2:** 17:00-20:00- **C1:** 8:00-11:00- **C2:** 18:00-19:00- **D1:** 7:00-10:00- **D2:** 17:30-18:30Now, let's see where these time slots overlap:1. **6:00-8:00 (A1) overlaps with:** - **B1 (6:30-7:30):** Overlap from 6:30-7:30 - **D1 (7:00-10:00):** Overlap from 7:00-8:002. **16:00-18:00 (A2) overlaps with:** - **B2 (17:00-20:00):** Overlap from 17:00-18:00 - **D2 (17:30-18:30):** Overlap from 17:30-18:003. **8:00-11:00 (C1) overlaps with:** - **D1 (7:00-10:00):** Overlap from 8:00-10:004. **18:00-19:00 (C2) overlaps with:** - **B2 (17:00-20:00):** Overlap from 18:00-19:00 - **D2 (17:30-18:30):** Overlap from 18:00-18:305. **7:00-10:00 (D1) overlaps with:** - **A1 (6:00-8:00):** Overlap from 7:00-8:00 - **C1 (8:00-11:00):** Overlap from 8:00-10:006. **17:30-18:30 (D2) overlaps with:** - **A2 (16:00-18:00):** Overlap from 17:30-18:00 - **B2 (17:00-20:00):** Overlap from 17:30-18:30 - **C2 (18:00-19:00):** Overlap from 18:00-18:30Okay, so there are several overlapping periods where more than one volunteer is available. The challenge is to schedule the volunteers in such a way that no more than two are on duty at any time, while maximizing the total monitoring time.Since each volunteer must work at least one time slot, I need to ensure that each volunteer is assigned to at least one of their available slots. However, to maximize the total monitoring time, I might need to have some volunteers work both of their available slots if possible, without violating the two-volunteer maximum at any time.Let's consider the total possible monitoring time if all volunteers worked both of their slots:- **A:** 2 hours + 2 hours = 4 hours- **B:** 1 hour + 3 hours = 4 hours- **C:** 3 hours + 1 hour = 4 hours- **D:** 3 hours + 1 hour = 4 hoursTotal = 4 + 4 + 4 + 4 = 16 hoursBut, we need to check if this is possible without having more than two volunteers on duty at any time.Looking at the overlaps, especially during the period from 17:30-18:00, we have A2, B2, and D2 overlapping. That's three volunteers, which violates the maximum of two. Similarly, during 18:00-18:30, we have B2, C2, and D2 overlapping—again three volunteers.So, we need to adjust the schedule to avoid having three volunteers on duty during these periods.One way to do this is to have one of the volunteers not work their second slot during these overlapping periods. For example, we could have D not work their D2 slot (17:30-18:30), which would reduce the overlap during 17:30-18:30 to just A2 and B2.But then, D would only be working their D1 slot (7:00-10:00), which is fine since they need to work at least one slot. However, this would reduce the total monitoring time by 1 hour (since D is not working their second slot).Similarly, we might need to adjust other overlaps. For instance, during 6:30-7:30, A1 and B1 overlap. That's two volunteers, which is acceptable. During 7:00-8:00, A1 and D1 overlap—also two volunteers. During 8:00-10:00, C1 and D1 overlap—again two volunteers. During 17:00-18:00, A2 and B2 overlap—two volunteers. During 18:00-19:00, B2 and C2 overlap—two volunteers.So, the problematic overlaps are only during 17:30-18:00 and 18:00-18:30, where three volunteers would be on duty if all worked their second slots.Therefore, to avoid having three volunteers on duty, we need to have one volunteer not work their second slot during these times. As I thought earlier, having D not work their D2 slot would solve the issue during 17:30-18:30.But then, D would only be working their D1 slot, which is 3 hours. However, D could also work their A2 slot if possible, but A is already working A2. Wait, no, A is A, B is B, etc. So, D can't work A's slot.Alternatively, maybe B could not work their B2 slot, but B's B2 slot is from 17:00-20:00, which is quite long. If B doesn't work B2, then B would only work B1 (6:30-7:30), which is 1 hour. That might not be ideal because we want to maximize the total monitoring time.Similarly, C could not work their C2 slot, but C2 is only 1 hour. So, C would still be working C1 (8:00-11:00), which is 3 hours.Alternatively, maybe A could not work their A2 slot, but A2 is 2 hours, and A1 is also 2 hours. If A doesn't work A2, then A would only work A1, which is fine, but we lose 2 hours of monitoring time.So, to minimize the loss of total monitoring time, it's better to have D not work their D2 slot, which is only 1 hour, rather than having someone else not work a longer slot.Therefore, by having D not work their D2 slot, we reduce the total monitoring time by 1 hour, making the total monitoring time 15 hours instead of 16.But wait, let's check if there are other overlaps that might require further adjustments.During 17:30-18:00, if D is not working D2, then only A2 and B2 are overlapping, which is two volunteers—acceptable.During 18:00-18:30, B2 and C2 are overlapping, which is two volunteers—acceptable.So, by having D not work their D2 slot, we avoid the overlap of three volunteers, and the total monitoring time becomes 15 hours.But let's see if there's a way to have more than 15 hours without violating the two-volunteer rule.Is there another volunteer whose second slot we could have them not work, but in a way that doesn't reduce the total monitoring time as much?For example, if B doesn't work their B2 slot, which is 3 hours, then B would only work B1 (1 hour). That would reduce the total monitoring time by 2 hours (from 4 to 2 hours for B), which is worse than just having D not work their D2 slot.Similarly, if C doesn't work their C2 slot, which is 1 hour, then C would only work C1 (3 hours). That would reduce the total monitoring time by 1 hour, same as having D not work D2.But if both C and D don't work their second slots, that would reduce the total monitoring time by 2 hours, which is worse.Alternatively, what if we have someone work only their first slot and someone else work only their second slot in a way that avoids the overlap?For example, if A works only A1 (6:00-8:00) and A2 (16:00-18:00), which is 4 hours, and B works only B1 (6:30-7:30) and B2 (17:00-20:00), which is 4 hours, and C works only C1 (8:00-11:00) and C2 (18:00-19:00), which is 4 hours, and D works only D1 (7:00-10:00) and D2 (17:30-18:30), which is 4 hours.But as we saw earlier, this causes overlaps of three volunteers during 17:30-18:00 and 18:00-18:30.So, to avoid that, we need to have one of the volunteers not work their second slot.If we have D not work D2, then D works only D1 (3 hours), and the total monitoring time is 15 hours.Alternatively, if we have C not work C2, then C works only C1 (3 hours), and the total monitoring time is also 15 hours.Either way, we lose 1 hour of monitoring time.Is there a way to have both D and C not work their second slots, but have someone else work an extra slot to compensate?Wait, no, because each volunteer must work at least one slot, but they can't work more than their available slots. So, we can't have someone work more than their available slots to compensate.Therefore, the maximum total monitoring time we can achieve without violating the two-volunteer rule is 15 hours.But wait, let's double-check.If D doesn't work D2, then D works only D1 (7:00-10:00), which is 3 hours.A works A1 and A2: 6:00-8:00 and 16:00-18:00, total 4 hours.B works B1 and B2: 6:30-7:30 and 17:00-20:00, total 4 hours.C works C1 and C2: 8:00-11:00 and 18:00-19:00, total 4 hours.But wait, if C works C2 (18:00-19:00), and B works B2 (17:00-20:00), then during 18:00-19:00, B and C are both on duty, which is fine—only two volunteers.Similarly, during 17:00-18:00, A2 and B2 are overlapping, which is two volunteers.During 17:30-18:00, A2 and B2 are overlapping, which is two volunteers.So, by having D not work D2, we avoid the overlap of three volunteers, and the total monitoring time is:A: 4 hoursB: 4 hoursC: 4 hoursD: 3 hoursTotal = 4 + 4 + 4 + 3 = 15 hoursBut the options given are 12, 14, 16, and 18 hours. 15 isn't an option. So, perhaps I made a mistake in my reasoning.Wait, maybe I need to consider that some volunteers can work only one slot, and others can work both, but in a way that the overlaps are managed.Let me try a different approach.Let's try to schedule the volunteers in such a way that during the overlapping periods, only two are on duty, and see what the maximum total monitoring time can be.First, let's look at the morning shifts:- A1: 6:00-8:00- B1: 6:30-7:30- D1: 7:00-10:00So, during 6:00-6:30, only A is on duty.During 6:30-7:00, A and B are on duty.During 7:00-7:30, A, B, and D are on duty. That's three volunteers, which is not allowed.So, to avoid three volunteers during 7:00-7:30, we need to have one of them not work during that time.Options:1. Have A not work A1: But A must work at least one slot, so if A doesn't work A1, they have to work A2.2. Have B not work B1: Similarly, B must work at least one slot, so if B doesn't work B1, they have to work B2.3. Have D not work D1: D must work at least one slot, so if D doesn't work D1, they have to work D2.So, to minimize the loss of monitoring time, let's see which option causes the least reduction.Option 1: A not working A1 (6:00-8:00). Then A works A2 (16:00-18:00), which is 2 hours. So, A's total monitoring time is 2 hours instead of 4.Option 2: B not working B1 (6:30-7:30). Then B works B2 (17:00-20:00), which is 3 hours. So, B's total monitoring time is 3 hours instead of 4.Option 3: D not working D1 (7:00-10:00). Then D works D2 (17:30-18:30), which is 1 hour. So, D's total monitoring time is 1 hour instead of 4.Clearly, Option 3 causes the biggest reduction in total monitoring time (from 4 to 1 hour for D), so it's not ideal. Option 1 reduces A's monitoring time by 2 hours, and Option 2 reduces B's monitoring time by 1 hour.So, Option 2 is better because it only reduces B's monitoring time by 1 hour.Therefore, let's choose Option 2: Have B not work B1, so B only works B2 (17:00-20:00).Now, during 7:00-7:30, only A and D are on duty—two volunteers, which is acceptable.Now, let's see the rest of the schedule.Morning shifts:- A1: 6:00-8:00- D1: 7:00-10:00So, during 6:00-6:30: A6:30-7:00: A and B (but B is not working B1, so only A)Wait, no, B is not working B1, so during 6:30-7:00, only A is on duty.Then, during 7:00-8:00: A and DDuring 8:00-10:00: DAfternoon shifts:- A2: 16:00-18:00- B2: 17:00-20:00- C1: 8:00-11:00- C2: 18:00-19:00- D2: 17:30-18:30So, let's see the overlaps:From 16:00-17:00: Only A2From 17:00-17:30: A2 and B2From 17:30-18:00: A2, B2, and D2—three volunteers, which is not allowed.So, we need to adjust this overlap.Again, we have three volunteers overlapping during 17:30-18:00.To fix this, we need to have one of them not work during that time.Options:1. Have A not work A2: But A must work at least one slot, so if A doesn't work A2, they have to work A1, which they are already doing.But A is already working A1 (6:00-8:00), so they can't work A2 as well.Wait, no, A can work both A1 and A2, but during 17:30-18:00, if A is working A2, then we have A, B, and D overlapping.So, to avoid that, we need to have one of them not work during that time.Options:1. Have A not work A2: But A is already working A1, so they can't work A2. Wait, no, A can work both A1 and A2, but during 17:30-18:00, if A is working A2, then we have three volunteers. So, to avoid that, A could stop working A2 before 17:30, but A2 is from 16:00-18:00, so they can't stop early.Alternatively, have D not work D2: D would only work D1 (7:00-10:00), which is 3 hours.Or have B not work B2: B would only work B1 (6:30-7:30), but we already decided to have B not work B1, so B would have to work B2.Wait, no, we had B not work B1, so B is working B2 (17:00-20:00). So, if we have B not work B2, they wouldn't be working any slot, which violates the requirement that each volunteer must work at least one slot.Therefore, we can't have B not work B2.So, the only options are to have A not work A2 or D not work D2.But A is already working A1, and A2 is their second slot. If A doesn't work A2, they still meet the requirement of working at least one slot.Similarly, D is working D1, so if D doesn't work D2, they still meet the requirement.So, which option causes less reduction in total monitoring time?If A doesn't work A2, A's total monitoring time is 2 hours (only A1).If D doesn't work D2, D's total monitoring time is 3 hours (only D1).So, having D not work D2 causes less reduction in total monitoring time.Therefore, let's have D not work D2.Now, during 17:30-18:00, only A2 and B2 are overlapping—two volunteers, which is acceptable.Now, let's tally up the total monitoring time:- A: A1 (6:00-8:00) and A2 (16:00-18:00) = 4 hours- B: B2 (17:00-20:00) = 3 hours- C: C1 (8:00-11:00) and C2 (18:00-19:00) = 4 hours- D: D1 (7:00-10:00) = 3 hoursTotal = 4 + 3 + 4 + 3 = 14 hoursBut wait, earlier I thought we could get 15 hours by having D not work D2, but now it's 14 hours. Did I make a mistake?Wait, in this scenario, B is only working B2, which is 3 hours, and D is only working D1, which is 3 hours. So, total is 14 hours.But earlier, I thought if D doesn't work D2, and B still works B2, then total would be 15 hours. But now, it's 14 hours.I think the confusion arises because when we have B not work B1, B has to work B2, which is 3 hours, and D not working D2, D works D1, which is 3 hours.So, total monitoring time is 4 (A) + 3 (B) + 4 (C) + 3 (D) = 14 hours.But earlier, I thought if D doesn't work D2, and B still works both B1 and B2, but that would cause three volunteers during 7:00-7:30.Wait, no, because we already decided to have B not work B1 to avoid the overlap during 7:00-7:30.So, B is only working B2, which is 3 hours.Therefore, the total monitoring time is 14 hours.But the options are 12, 14, 16, and 18 hours. So, 14 is an option.But earlier, I thought we could get 15 hours, but that seems to conflict with the requirement that each volunteer must work at least one slot.Wait, no, in the scenario where D doesn't work D2, D is still working D1, which is their first slot, so they meet the requirement.Similarly, B is working B2, which is their second slot, so they meet the requirement.A is working both A1 and A2, which is fine.C is working both C1 and C2, which is fine.So, total monitoring time is 14 hours.But wait, is there a way to have more than 14 hours?What if instead of having B not work B1, we have someone else not work their slot?Let's try having C not work C2.So, C only works C1 (8:00-11:00), which is 3 hours.Then, during 18:00-19:00, only B2 is on duty, since C isn't working C2.But then, during 17:30-18:00, we still have A2, B2, and D2 overlapping, which is three volunteers.So, we still need to adjust that.If we have D not work D2, then during 17:30-18:00, only A2 and B2 are overlapping—two volunteers.So, in this scenario:- A: A1 and A2 = 4 hours- B: B1 and B2 = 4 hours- C: C1 = 3 hours- D: D1 = 3 hoursTotal = 4 + 4 + 3 + 3 = 14 hoursSame as before.Alternatively, if we have C not work C2 and D not work D2, then:- A: 4 hours- B: 4 hours- C: 3 hours- D: 3 hoursTotal = 14 hoursSame result.Alternatively, what if we have A not work A2?Then, A only works A1 (6:00-8:00), which is 2 hours.Then, during 17:30-18:00, only B2 and D2 are overlapping—two volunteers.So, in this scenario:- A: 2 hours- B: 4 hours- C: 4 hours- D: 4 hoursTotal = 2 + 4 + 4 + 4 = 14 hoursAgain, same total.So, regardless of which volunteer we have not work their second slot, the total monitoring time is 14 hours.Therefore, the maximum total monitoring time is 14 hours.But wait, earlier I thought we could get 15 hours, but that seems to require violating the two-volunteer rule or the requirement that each volunteer must work at least one slot.So, 14 hours seems to be the maximum possible without violating the constraints.Therefore, the answer should be 14 hours, which is option B.

💡First, I need to understand the definition of set ( B ). It consists of all elements ( x ) such that ( -x ) is in set ( A ) and ( 1 - x ) is not in set ( A ).Given set ( A = {-2, 0, 1, 3} ), I will identify all possible values of ( x ) that satisfy the condition ( -x in A ). This means ( x ) can be ( 2, 0, -1, -3 ) because:- If ( -x = -2 ), then ( x = 2 )- If ( -x = 0 ), then ( x = 0 )- If ( -x = 1 ), then ( x = -1 )- If ( -x = 3 ), then ( x = -3 )Next, I need to ensure that ( 1 - x ) is not in set ( A ). I will check each candidate value of ( x ):- For ( x = 2 ), ( 1 - 2 = -1 ), which is not in ( A )- For ( x = 0 ), ( 1 - 0 = 1 ), which is in ( A )- For ( x = -1 ), ( 1 - (-1) = 2 ), which is not in ( A )- For ( x = -3 ), ( 1 - (-3) = 4 ), which is not in ( A )After evaluating all candidates, the values of ( x ) that satisfy both conditions are ( 2, -1, ) and ( -3 ). Therefore, set ( B ) is ( {-3, -1, 2} ).

💡Okay, so I need to figure out which of these power functions is decreasing on the interval from negative infinity to zero. The options are A: y equals x to the one-third power, B: y equals x to the one-half power, C: y equals x cubed, and D: y equals x squared. Hmm, let's break this down step by step.First, I remember that power functions are of the form y equals x raised to some exponent. The behavior of these functions can vary depending on whether the exponent is positive or negative, even or odd, and whether it's a fraction or a whole number. So, let's look at each option one by one.Starting with option A: y equals x to the one-third power. I know that the cube root function, which is what this is, is defined for all real numbers, including negative numbers. It's also an odd function, meaning it's symmetric about the origin. Now, is this function increasing or decreasing? Well, if I think about the graph of y equals x cubed, it's an odd function that passes through the origin and increases as x increases. Similarly, the cube root function should also be increasing because as x increases, the cube root of x increases as well. So, on the interval from negative infinity to zero, this function is increasing, not decreasing. So, option A is not the answer.Moving on to option B: y equals x to the one-half power. That's the square root function. Wait a minute, the square root function is only defined for x greater than or equal to zero because you can't take the square root of a negative number in the set of real numbers. So, on the interval from negative infinity to zero, this function isn't even defined. Therefore, it can't be decreasing there because it doesn't exist in that interval. So, option B is out.Next is option C: y equals x cubed. Again, this is a power function with an exponent of three. I recall that for odd exponents, the function is defined for all real numbers and is also an odd function. The graph of y equals x cubed passes through the origin and is increasing throughout its entire domain. That means as x increases, y increases, and as x decreases, y decreases. But wait, if I'm looking at the interval from negative infinity to zero, is it increasing or decreasing? Let me think. If I take two points on the negative side, say x1 = -2 and x2 = -1. Plugging into the function, y1 = (-2)^3 = -8 and y2 = (-1)^3 = -1. So, as x increases from -2 to -1, y increases from -8 to -1. That means the function is increasing on that interval as well. So, option C is also increasing on (-∞, 0), not decreasing.Finally, option D: y equals x squared. This is a quadratic function, and its graph is a parabola opening upwards. I know that for x greater than zero, the function is increasing, and for x less than zero, it's decreasing. Let me verify that. If I take two negative numbers, say x1 = -3 and x2 = -2. Then y1 = (-3)^2 = 9 and y2 = (-2)^2 = 4. So, as x increases from -3 to -2, y decreases from 9 to 4. That confirms that the function is decreasing on the interval from negative infinity to zero. Therefore, option D is the correct answer.Just to make sure I haven't missed anything, let me recap:- Option A: Increasing on (-∞, 0)- Option B: Undefined on (-∞, 0)- Option C: Increasing on (-∞, 0)- Option D: Decreasing on (-∞, 0)Yep, that seems right. So, the function that's decreasing on (-∞, 0) is y equals x squared, which is option D.

💡First, I recognize that there are two types of tickets: adult tickets priced at 12 and student tickets priced at 6. The total number of tickets sold is 1,500, and the total revenue collected is 16,200. Additionally, it's given that 300 student tickets were sold.To find the number of adult tickets sold, I can set up an equation based on the total revenue. Let ( A ) represent the number of adult tickets sold. The revenue from adult tickets would be ( 12A ), and the revenue from student tickets would be ( 6 times 300 = 1800 ) dollars.Adding these together gives the total revenue:[12A + 1800 = 16200]Next, I'll solve for ( A ):[12A = 16200 - 1800][12A = 14400][A = frac{14400}{12}][A = 1200]Therefore, 1,200 adult tickets were sold.