💡First, I identify the three numbers provided: 10, 11, and 12.Next, I determine the smallest number, which is 10.Then, I identify the second largest number, which is 11.Finally, I calculate the product of the smallest number (10) and the second largest number (11), resulting in 110.

💡Alright, so I have this problem here about identifying random events among five different scenarios. Let me try to understand each one step by step. First, what exactly is a random event? From what I remember, a random event is something that has an uncertain outcome; it can happen or not happen, and we can't predict it with certainty. So, events that are certain to happen or certain not to happen aren't random. Got it.Alright, let's look at each event one by one.**Event ①: Tossing the same dice twice in a row, and getting a 2 both times.**Okay, so this is about rolling a die twice and getting a 2 each time. I know that when you roll a die, there are six possible outcomes, and each outcome is equally likely if the die is fair. So, getting a 2 on the first roll is one possibility out of six. Now, if I roll it again, the chance of getting another 2 is also one out of six. Since these are independent events, the combined probability of both happening is (1/6) * (1/6) = 1/36. Hmm, so there's a chance it could happen, but it's not certain. It's also not impossible because it can happen. So, this seems like a random event. I think ① is random.**Event ②: It raining tomorrow.**This one is about the weather. I know that weather can be unpredictable, and even with forecasts, there's always a chance it might rain or not. So, whether it rains tomorrow isn't something we can be 100% sure about. It could happen, or it might not. Therefore, this is also a random event. So, ② is random.**Event ③: Someone winning the lottery.**Lottery wins are based on chance, right? Each ticket has a very low probability of winning, but it's not impossible. So, someone could win, or they might not. Therefore, this is a random event. So, ③ is random.**Event ④: Selecting two elements from the set {1, 2, 3}, and their sum being greater than 2.**Alright, let's think about this. The set is {1, 2, 3}, and we're selecting two elements. Let's list all possible pairs:1. (1, 2) - sum is 32. (1, 3) - sum is 43. (2, 3) - sum is 5Wait, so all possible sums are 3, 4, and 5, which are all greater than 2. So, no matter which two elements I pick, their sum will always be greater than 2. That means this event is certain to happen. It's not random because there's no uncertainty involved. So, ④ is not a random event.**Event ⑤: Water boiling when heated to 90°C under standard atmospheric pressure.**I remember that water boils at 100°C under standard atmospheric pressure. So, if it's heated to 90°C, it shouldn't boil. Therefore, this event is impossible. Since it can't happen, it's not a random event either. So, ⑤ is not random.Okay, so summarizing:- ①: Random- ②: Random- ③: Random- ④: Certain (not random)- ⑤: Impossible (not random)So, out of the five events, three are random. Therefore, the answer should be 3, which is option C.Wait, let me double-check to make sure I didn't miss anything.For event ④, I considered all possible pairs and their sums, and indeed, all sums are greater than 2. So, it's certain. For event ⑤, I'm pretty sure water boils at 100°C, so 90°C won't make it boil. So, that's impossible. The other three events have uncertain outcomes, so they are random.Yeah, I think that's correct. So, the number of random events is 3.

💡Okay, so I have this problem here about finding the contrapositive of a proposition. The proposition is: "If ( x^2 leqslant 1 ), then ( -1 leqslant x leqslant 1 )." And I need to choose the correct contrapositive from the options A to D. First, I remember that the contrapositive of a statement "If P, then Q" is "If not Q, then not P." So, I need to identify what P and Q are in the original statement. In this case, P is ( x^2 leqslant 1 ) and Q is ( -1 leqslant x leqslant 1 ). So, the contrapositive should be "If not Q, then not P." Let me write that out. Not Q would be the negation of ( -1 leqslant x leqslant 1 ). The negation of ( -1 leqslant x leqslant 1 ) is ( x < -1 ) or ( x > 1 ). So, not Q is ( x < -1 ) or ( x > 1 ). Now, not P is the negation of ( x^2 leqslant 1 ), which is ( x^2 > 1 ). So, putting it together, the contrapositive should be: "If ( x < -1 ) or ( x > 1 ), then ( x^2 > 1 )." Looking at the options, let's see which one matches this. Option A: "If ( x^2 geqslant 1 ), then ( x geqslant 1 ) or ( x leqslant -1 )." Hmm, that seems related but not exactly the contrapositive. It's more like the converse or something else. Option B: "If ( -1 < x < 1 ), then ( x^2 < 1 )." That doesn't look like the contrapositive at all. It seems like a different statement, maybe the inverse or something else. Option C: "If ( x geqslant 1 ) or ( x leqslant -1 ), then ( x^2 geqslant 1 )." This is close, but not quite the contrapositive. The contrapositive should have the negation of Q leading to the negation of P, which is ( x < -1 ) or ( x > 1 ) leading to ( x^2 > 1 ). This option uses ( x geqslant 1 ) or ( x leqslant -1 ), which includes equality, whereas the contrapositive should be strict inequalities. Option D: "If ( x > 1 ) or ( x < -1 ), then ( x^2 > 1 )." This matches exactly what I derived earlier. The contrapositive is "If ( x < -1 ) or ( x > 1 ), then ( x^2 > 1 )." So, Option D is correct. Wait, let me double-check. The original statement is "If ( x^2 leqslant 1 ), then ( -1 leqslant x leqslant 1 )." The contrapositive swaps and negates both parts. So, not Q is ( x < -1 ) or ( x > 1 ), and not P is ( x^2 > 1 ). So yes, Option D is indeed the contrapositive. I think I got it right. Option D is the correct contrapositive.

💡Alright, so I'm trying to figure out this problem about propositions involving lines and planes. Let me read it again to make sure I understand.We have three propositions:- **Proposition A**: Intersecting lines ( l ) and ( m ) are both within plane ( alpha ) and neither are within plane ( beta ).- **Proposition B**: At least one of lines ( l ) and ( m ) intersects with plane ( beta ).- **Proposition C**: Plane ( alpha ) intersects with plane ( beta ).The question is asking, when Proposition A is true, which of the following is correct regarding the relationship between Propositions B and C. The options are about whether B is a sufficient condition, a necessary condition, both, or neither for C.First, I need to recall what sufficient and necessary conditions mean. A sufficient condition means that if B is true, then C must be true. A necessary condition means that if C is true, then B must be true. So, I need to see if B implies C, and if C implies B.Starting with Proposition A: Lines ( l ) and ( m ) intersect and are both in plane ( alpha ). Neither of them is in plane ( beta ). So, ( l ) and ( m ) are in ( alpha ), but not in ( beta ). Since ( l ) and ( m ) intersect, they must intersect at a point, say ( P ). Since both lines are in ( alpha ), their intersection point ( P ) is also in ( alpha ).Now, Proposition B says that at least one of ( l ) or ( m ) intersects with plane ( beta ). So, either ( l ) intersects ( beta ), or ( m ) intersects ( beta ), or both.Proposition C is that plane ( alpha ) intersects plane ( beta ). Planes intersect if they have at least one point in common. Since both ( l ) and ( m ) are in ( alpha ), if either ( l ) or ( m ) intersects ( beta ), then ( alpha ) and ( beta ) share that intersection point, meaning they intersect.So, if B is true (at least one line intersects ( beta )), then C must be true (planes intersect). Therefore, B is a sufficient condition for C.Now, is B a necessary condition for C? That is, if C is true (planes intersect), does it necessarily mean that B is true (at least one line intersects ( beta ))?Suppose that plane ( alpha ) intersects plane ( beta ). Since lines ( l ) and ( m ) are in ( alpha ), their intersection point ( P ) is in ( alpha ). If ( P ) is also in ( beta ), then both ( l ) and ( m ) pass through ( P ), which is in ( beta ). Therefore, both lines would intersect ( beta ) at ( P ). But wait, Proposition A says neither ( l ) nor ( m ) is within ( beta ). So, if ( P ) is in ( beta ), then both lines pass through ( P ), which is in ( beta ), but they are not entirely within ( beta ). So, they intersect ( beta ) at ( P ).Therefore, if ( alpha ) and ( beta ) intersect, then their intersection must be a line. Since ( l ) and ( m ) are in ( alpha ) and intersect at ( P ), if ( alpha ) and ( beta ) intersect, then ( P ) must lie on the intersection line of ( alpha ) and ( beta ). Therefore, both ( l ) and ( m ) intersect ( beta ) at ( P ). So, in this case, B is necessarily true because both lines intersect ( beta ).Wait, but the question is about whether B is a necessary condition for C. So, if C is true, does B have to be true? From the above, yes, because if ( alpha ) and ( beta ) intersect, then their intersection line must contain the point ( P ), which is on both ( l ) and ( m ). Therefore, both lines intersect ( beta ) at ( P ), making B true.But hold on, is it possible for ( alpha ) and ( beta ) to intersect without either ( l ) or ( m ) intersecting ( beta )? Given that ( l ) and ( m ) are in ( alpha ) and intersect at ( P ), if ( alpha ) and ( beta ) intersect, their intersection must be a line that contains ( P ) if ( P ) is in both planes. But since ( P ) is in ( alpha ), if ( alpha ) and ( beta ) intersect, ( P ) could be in ( beta ) or not.Wait, no. If ( alpha ) and ( beta ) intersect, their intersection is a line. If ( P ) is not on that line, then ( P ) is not in ( beta ). But ( l ) and ( m ) are in ( alpha ), so if ( alpha ) and ( beta ) intersect along a line that doesn't pass through ( P ), then neither ( l ) nor ( m ) would intersect ( beta ), right?But wait, if ( l ) and ( m ) are in ( alpha ), and ( alpha ) intersects ( beta ) along a line, then ( l ) and ( m ) could be parallel to that intersection line, meaning they don't intersect ( beta ). But since ( l ) and ( m ) intersect at ( P ), which is in ( alpha ), if ( alpha ) and ( beta ) intersect along a different line, then ( P ) is not in ( beta ), so neither ( l ) nor ( m ) would intersect ( beta ).But wait, if ( alpha ) and ( beta ) intersect, then they share at least one point. If that point is not ( P ), then neither ( l ) nor ( m ) would intersect ( beta ), because they only intersect at ( P ). But that would mean that ( alpha ) and ( beta ) intersect at some other point, but ( l ) and ( m ) don't pass through that point.But hold on, if ( alpha ) and ( beta ) intersect, their intersection is a line. If ( l ) and ( m ) are in ( alpha ) and intersect at ( P ), then if ( P ) is not on the intersection line of ( alpha ) and ( beta ), then ( l ) and ( m ) don't intersect ( beta ). But if ( P ) is on the intersection line, then both ( l ) and ( m ) intersect ( beta ) at ( P ).So, it's possible for ( alpha ) and ( beta ) to intersect without either ( l ) or ( m ) intersecting ( beta ), right? Because if their intersection line doesn't pass through ( P ), then ( l ) and ( m ) don't intersect ( beta ).But wait, if ( alpha ) and ( beta ) intersect, then they share infinitely many points along a line. If ( l ) and ( m ) are in ( alpha ), and they intersect at ( P ), then ( P ) is in ( alpha ). If ( P ) is not in ( beta ), then ( l ) and ( m ) don't intersect ( beta ). But if ( P ) is in ( beta ), then both ( l ) and ( m ) intersect ( beta ) at ( P ).So, in the case where ( alpha ) and ( beta ) intersect, it's possible for B to be true or false, depending on whether ( P ) is in ( beta ) or not. Therefore, B is not necessarily true when C is true.Wait, but Proposition A says that neither ( l ) nor ( m ) is within ( beta ). So, if ( P ) is in ( beta ), then both ( l ) and ( m ) intersect ( beta ) at ( P ), but they are not entirely within ( beta ). So, in that case, B is true.But if ( alpha ) and ( beta ) intersect along a different line that doesn't pass through ( P ), then ( l ) and ( m ) don't intersect ( beta ), so B is false. But is that possible?Wait, if ( alpha ) and ( beta ) intersect, their intersection is a line. If ( l ) and ( m ) are in ( alpha ), and they intersect at ( P ), then ( P ) is in ( alpha ). If ( P ) is also in ( beta ), then ( P ) is on the intersection line of ( alpha ) and ( beta ). If ( P ) is not in ( beta ), then ( l ) and ( m ) don't intersect ( beta ), but ( alpha ) and ( beta ) still intersect along some other line.So, in that case, C is true (planes intersect), but B is false (neither line intersects ( beta )). Therefore, B is not a necessary condition for C.Wait, but Proposition A says that neither ( l ) nor ( m ) is within ( beta ). So, if ( alpha ) and ( beta ) intersect along a line that doesn't pass through ( P ), then ( l ) and ( m ) don't intersect ( beta ), so B is false. But C is true because the planes intersect.Therefore, B is not a necessary condition for C, because C can be true without B being true.But earlier, I thought that if C is true, then B must be true, but now I'm seeing that it's not necessarily the case.Wait, let me clarify. If ( alpha ) and ( beta ) intersect, their intersection is a line. If ( l ) and ( m ) are in ( alpha ) and intersect at ( P ), then ( P ) is in ( alpha ). If ( P ) is also in ( beta ), then both ( l ) and ( m ) intersect ( beta ) at ( P ), so B is true. If ( P ) is not in ( beta ), then ( l ) and ( m ) don't intersect ( beta ), so B is false. Therefore, C can be true without B being true, meaning B is not a necessary condition for C.But earlier, I thought that if C is true, then B must be true, but that's only if ( P ) is in ( beta ). If ( P ) is not in ( beta ), then C is true, but B is false.Therefore, B is not a necessary condition for C.But wait, in Proposition A, it says that neither ( l ) nor ( m ) is within ( beta ). So, if ( l ) or ( m ) were within ( beta ), they would be entirely in ( beta ), but Proposition A says they are not. So, if ( l ) or ( m ) intersects ( beta ), it's only at a single point, not the entire line.So, going back, if ( alpha ) and ( beta ) intersect along a line that doesn't pass through ( P ), then ( l ) and ( m ) don't intersect ( beta ), so B is false, but C is true. Therefore, B is not necessary for C.But if B is true, meaning at least one line intersects ( beta ), then ( alpha ) and ( beta ) must intersect at that point, so C is true. Therefore, B is a sufficient condition for C.So, in summary, B is sufficient for C, but not necessary. Therefore, the correct answer is A: Proposition B is a sufficient but not necessary condition for Proposition C.Wait, but in the initial thought process, I thought B was necessary, but after reconsidering, it's not necessary because C can be true without B being true.So, the correct answer should be A.

💡Okay, so I have this chess tournament problem where there are grandmasters and masters participating. Each participant scored exactly half of their points in matches against masters. I need to prove that the total number of participants is a perfect square. Hmm, let's try to break this down step by step.First, let me understand the setup. In a chess tournament, each participant plays against every other participant exactly once. The scoring is straightforward: 1 point for a win, 0.5 points for a draw, and 0 points for a loss. So, if there are, say, n participants, each participant plays n-1 games.Now, the problem states that each participant scored exactly half of their points in matches against masters. That means if a participant is a master, half of their total points came from games against other masters, and the other half came from games against grandmasters. Similarly, if a participant is a grandmaster, half of their points came from games against masters, and the other half from games against other grandmasters.Let me denote the number of masters as m and the number of grandmasters as g. So, the total number of participants is m + g, which is the number we need to prove is a perfect square.Each master plays m - 1 games against other masters and g games against grandmasters. Similarly, each grandmaster plays g - 1 games against other grandmasters and m games against masters.Now, the total points scored by all participants in the tournament can be calculated in two ways. One way is to consider all the games played and sum up the points. Since each game contributes exactly 1 point in total (either 1 for a win and 0 for a loss, or 0.5 each for a draw), the total points scored by everyone is C(m + g, 2) = (m + g)(m + g - 1)/2.Another way to calculate the total points is to consider the points scored by masters and grandmasters separately. Let's denote the total points scored by all masters as P_m and the total points scored by all grandmasters as P_g. Then, P_m + P_g = (m + g)(m + g - 1)/2.Now, according to the problem, each participant scored exactly half of their points in matches against masters. So, for each master, half of their points came from games against other masters, and half came from games against grandmasters. Similarly, for each grandmaster, half of their points came from games against masters, and half came from games against other grandmasters.Let me denote the total points scored by masters in games against other masters as P_mm. Similarly, the total points scored by grandmasters in games against other grandmasters is P_gg. Then, the points scored by masters against grandmasters would be P_mg, and the points scored by grandmasters against masters would be P_gm.But since each game between a master and a grandmaster contributes to both P_mg and P_gm, we have P_mg + P_gm = m * g. Because each of the m masters plays each of the g grandmasters once, so there are m * g games in total, each contributing 1 point, hence m * g points in total.Now, since each master's points are half from masters and half from grandmasters, the total points scored by masters, P_m, is equal to 2 * P_mm. Similarly, the total points scored by grandmasters, P_g, is equal to 2 * P_gg.But wait, let me think again. If each master scores half their points against masters, then P_mm = (1/2) * P_m. Similarly, P_gg = (1/2) * P_g. So, P_m = 2 * P_mm and P_g = 2 * P_gg.Now, the total points scored in games between masters is P_mm, which is the sum of all points scored by masters in games against other masters. Similarly, the total points scored in games between grandmasters is P_gg.But in reality, the total points scored in games between masters is C(m, 2) = m(m - 1)/2, because each game between two masters contributes 1 point in total. Similarly, the total points scored in games between grandmasters is C(g, 2) = g(g - 1)/2.Wait, that's a key point. The total points from games between masters is m(m - 1)/2, and the total points from games between grandmasters is g(g - 1)/2. The remaining points, which are from games between masters and grandmasters, is m * g.But from the problem's condition, we have that P_mm = (1/2) * P_m and P_gg = (1/2) * P_g. So, P_m = 2 * P_mm = 2 * [m(m - 1)/2] = m(m - 1). Similarly, P_g = 2 * P_gg = 2 * [g(g - 1)/2] = g(g - 1).But wait, that can't be right because P_m is the total points scored by all masters, which includes both their points against masters and against grandmasters. So, P_m = P_mm + P_mg. Similarly, P_g = P_gg + P_gm.But from the problem, P_mm = (1/2) * P_m and P_gg = (1/2) * P_g. So, substituting, we get:P_m = 2 * P_mm = 2 * [m(m - 1)/2] = m(m - 1)Similarly, P_g = 2 * P_gg = 2 * [g(g - 1)/2] = g(g - 1)But also, P_m = P_mm + P_mg = m(m - 1)/2 + P_mgSimilarly, P_g = P_gg + P_gm = g(g - 1)/2 + P_gmBut since P_m = m(m - 1), we have:m(m - 1) = m(m - 1)/2 + P_mgSo, P_mg = m(m - 1) - m(m - 1)/2 = m(m - 1)/2Similarly, P_gm = g(g - 1)/2But we also know that P_mg + P_gm = m * g, because each of the m * g games contributes to either P_mg or P_gm.So, m(m - 1)/2 + g(g - 1)/2 = m * gLet me write that equation:[m(m - 1) + g(g - 1)] / 2 = m * gMultiply both sides by 2:m(m - 1) + g(g - 1) = 2 m gLet's expand the left side:m² - m + g² - g = 2 m gBring all terms to one side:m² - m + g² - g - 2 m g = 0Let me rearrange terms:m² - 2 m g + g² - m - g = 0Notice that m² - 2 m g + g² is (m - g)², so:(m - g)² - (m + g) = 0So, (m - g)² = m + gLet me denote t = m + g, which is the total number of participants. Then, we have:(m - g)² = tBut t is the total number of participants, which is m + g. So, we have:(m - g)² = m + gLet me denote d = m - g. Then, d² = m + g.But since d = m - g, we can write m = g + d.Substituting into d² = m + g:d² = (g + d) + g = 2 g + dSo, d² - d = 2 gThus, g = (d² - d)/2Similarly, since m = g + d, m = (d² - d)/2 + d = (d² - d + 2 d)/2 = (d² + d)/2So, both m and g are expressed in terms of d. Now, since m and g must be integers (number of participants can't be fractional), (d² - d) must be even, and (d² + d) must be even as well.Let's check for d:If d is even, say d = 2k, then:g = (4k² - 2k)/2 = 2k² - km = (4k² + 2k)/2 = 2k² + kBoth m and g are integers.If d is odd, say d = 2k + 1, then:g = [(4k² + 4k + 1) - (2k + 1)] / 2 = (4k² + 4k + 1 - 2k - 1)/2 = (4k² + 2k)/2 = 2k² + km = [(4k² + 4k + 1) + (2k + 1)] / 2 = (4k² + 4k + 1 + 2k + 1)/2 = (4k² + 6k + 2)/2 = 2k² + 3k + 1Again, both m and g are integers.So, regardless of whether d is even or odd, m and g are integers as long as d is an integer.Now, since t = m + g = d², from earlier, t must be a perfect square.Therefore, the total number of participants, t, is equal to d², which is a perfect square.So, we've shown that t = d², hence t is a perfect square.I think that makes sense. Let me recap:1. Let m be the number of masters and g the number of grandmasters.2. Each participant scores half their points against masters, leading to equations involving m and g.3. Through these equations, we derive that (m - g)² = m + g.4. Letting t = m + g, we find t = (m - g)², so t is a perfect square.Yes, that seems to cover it. I don't see any gaps in the logic, so I think this proves that the total number of participants is a perfect square.

💡First, I need to find the sum of vectors (overrightarrow{a}) and (overrightarrow{b}). [overrightarrow{a} + overrightarrow{b} = (1 + 3, -2 + 4) = (4, 2)]Next, I will calculate the dot product of ((overrightarrow{a} + overrightarrow{b})) and (overrightarrow{c}).[(overrightarrow{a} + overrightarrow{b}) cdot overrightarrow{c} = (4, 2) cdot (2, -1) = 4 times 2 + 2 times (-1) = 8 - 2 = 6]Therefore, the result of the dot product is 6.