💡Okay, so I have this problem where I need to find the intersection of two sets, A and B. Let me first understand what each set represents.Set A is defined as all x such that x + 1/2 is greater than or equal to 3/2. Hmm, let me write that down:A = {x | x + 1/2 ≥ 3/2}I can solve this inequality for x. Subtracting 1/2 from both sides gives:x ≥ 3/2 - 1/2x ≥ 1So, set A is all real numbers x where x is greater than or equal to 1. But wait, is that all? Let me double-check. The original inequality was x + 1/2 ≥ 3/2. If I subtract 1/2, I definitely get x ≥ 1. So yes, A is [1, ∞).Now, set B is defined as all x such that x² + x < 6. Let me write that down:B = {x | x² + x < 6}This is a quadratic inequality. To solve it, I can rearrange it to:x² + x - 6 < 0Now, I need to factor this quadratic or find its roots. Let me try factoring. Looking for two numbers that multiply to -6 and add to 1. Hmm, 3 and -2 work because 3 * (-2) = -6 and 3 + (-2) = 1. So, the quadratic factors as:(x + 3)(x - 2) < 0Now, to solve this inequality, I can use a sign chart. The roots are x = -3 and x = 2. These divide the number line into three intervals:1. x < -32. -3 < x < 23. x > 2I need to test each interval to see where the product (x + 3)(x - 2) is negative.- For x < -3, let's pick x = -4: (-4 + 3)(-4 - 2) = (-1)(-6) = 6, which is positive.- For -3 < x < 2, let's pick x = 0: (0 + 3)(0 - 2) = (3)(-2) = -6, which is negative.- For x > 2, let's pick x = 3: (3 + 3)(3 - 2) = (6)(1) = 6, which is positive.So, the inequality (x + 3)(x - 2) < 0 holds true for -3 < x < 2. Therefore, set B is (-3, 2).Now, I need to find the intersection of sets A and B, which is A ∩ B.Set A is [1, ∞) and set B is (-3, 2). The intersection will be the overlap of these two intervals.Visualizing the number line:- Set A starts at 1 and goes to infinity.- Set B starts just above -3 and goes up to 2.The overlap between these two sets will be from 1 up to 2. But I need to be careful about the endpoints.Set A includes 1, so the lower bound is 1, inclusive. Set B does not include 2, so the upper bound is 2, exclusive. Therefore, the intersection is [1, 2).Wait, but hold on. Set A also includes numbers less than or equal to -2, right? Because when I solved x + 1/2 ≥ 3/2, I only got x ≥ 1. Did I miss something?Wait, maybe I misinterpreted the original set A. Let me go back.Set A is {x | x + 1/2 ≥ 3/2}. Solving for x, as I did before, gives x ≥ 1. But is there another condition? Because sometimes inequalities can have two cases, especially if they involve absolute values or something else. But in this case, it's a simple linear inequality. So, I think my initial solution is correct: A is [1, ∞).But looking at the options provided, some of them include intervals like (-3, -2] and [1, 2). Maybe I need to reconsider.Wait, perhaps I made a mistake in interpreting set A. Let me check again.The original set A is {x | x + 1/2 ≥ 3/2}. Solving this:x + 1/2 ≥ 3/2Subtract 1/2: x ≥ 1So, A is [1, ∞). There's no other interval here because it's a simple linear inequality. So, why do the options include negative numbers?Wait, maybe I misread the problem. Let me look again.Oh, wait! The problem says:"A = {x | x + 1/2 ≥ 3/2}" and "B = {x | x² + x < 6}"So, A is indeed [1, ∞). But in the options, there are negative intervals. That suggests that maybe A has another part? Or perhaps I need to consider the inequality differently.Wait, maybe the inequality x + 1/2 ≥ 3/2 can also be interpreted as x + 1/2 ≤ -3/2? Because sometimes with inequalities involving absolute values, you have two cases. But in this case, it's not an absolute value; it's just a linear inequality.Wait, no, that's not correct. The inequality is x + 1/2 ≥ 3/2, which only gives x ≥ 1. There's no other case here. So, A is [1, ∞).But then, why do the options include negative numbers? Maybe I need to consider that the original inequality might have been miswritten or I misread it.Wait, let me double-check the original problem:"A = {x | x + 1/2 ≥ 3/2}" and "B = {x | x² + x < 6}"No, it's definitely x + 1/2 ≥ 3/2, which simplifies to x ≥ 1.So, perhaps the options are incorrect, or maybe I'm missing something.Wait, no, the options are given, and I need to choose from them. So, maybe I need to consider that set A is not just [1, ∞), but also includes another interval.Wait, let me think differently. Maybe the original inequality was supposed to be |x + 1/2| ≥ 3/2, which would give two intervals: x + 1/2 ≥ 3/2 or x + 1/2 ≤ -3/2. That would make sense because then A would have two parts: x ≥ 1 and x ≤ -2.But the problem doesn't have absolute value signs. It's just x + 1/2 ≥ 3/2. So, unless it's a typo, I think A is only [1, ∞).But given the options, which include negative intervals, I think maybe the original problem was supposed to have absolute value. Maybe it's a misprint. Alternatively, perhaps I need to consider that x + 1/2 can be both greater than or equal to 3/2 and less than or equal to -3/2, but that would be the case if it's an absolute value inequality.Wait, let me assume for a moment that it's an absolute value inequality, even though it's not written. So, if it were |x + 1/2| ≥ 3/2, then:x + 1/2 ≥ 3/2 or x + 1/2 ≤ -3/2Solving the first inequality:x + 1/2 ≥ 3/2x ≥ 1Solving the second inequality:x + 1/2 ≤ -3/2x ≤ -2So, set A would be (-∞, -2] ∪ [1, ∞).Now, that makes sense because the options include negative intervals. So, maybe the problem was supposed to have absolute value, but it's missing. Alternatively, maybe I need to interpret it differently.But since the problem is written as x + 1/2 ≥ 3/2, I think it's just x ≥ 1. However, given the options, I think the intended answer assumes that A is (-∞, -2] ∪ [1, ∞). So, perhaps I should proceed with that assumption.So, assuming A is (-∞, -2] ∪ [1, ∞), and B is (-3, 2), then the intersection A ∩ B would be:For the first part of A: (-∞, -2] intersected with (-3, 2) is (-3, -2].For the second part of A: [1, ∞) intersected with (-3, 2) is [1, 2).So, combining these, A ∩ B is (-3, -2] ∪ [1, 2).Looking at the options, option C is (-3, -2] ∪ [1, 2), which matches.But wait, I'm confused because initially, I thought A was only [1, ∞). But given the options, it seems like A is supposed to have two intervals. Maybe the problem was miswritten, or perhaps I need to consider that x + 1/2 ≥ 3/2 can also be interpreted as x + 1/2 ≤ -3/2, but that doesn't make sense because it's not an absolute value.Alternatively, maybe the original inequality was supposed to be x + 1/2 ≥ 3/2 or x + 1/2 ≤ -3/2, which would make A have two intervals. But since it's written as x + 1/2 ≥ 3/2, I think it's just x ≥ 1.But given the options, I think the intended answer is C, which is (-3, -2] ∪ [1, 2). So, maybe I should go with that.Wait, but let me double-check. If A is [1, ∞), and B is (-3, 2), then A ∩ B is [1, 2). But none of the options have just [1, 2). Option A is [-3, -2) ∪ (1, 2], which includes negative numbers. Option B is (-3, -2] ∪ (1, ∞), which includes (1, ∞). Option C is (-3, -2] ∪ [1, 2), which includes both negative and positive intervals. Option D is (-∞, -3] ∪ (1, 2], which also includes negative and positive intervals.So, if A is only [1, ∞), then A ∩ B is [1, 2). But none of the options have just [1, 2). The closest is option C, which includes (-3, -2] ∪ [1, 2). But why would there be a negative interval?Wait, maybe I need to consider that set A is not just [1, ∞), but also includes another interval. Let me think again.If A is {x | x + 1/2 ≥ 3/2}, then x ≥ 1. There's no other interval unless it's an absolute value. So, unless the problem is miswritten, I think A is only [1, ∞). Therefore, A ∩ B is [1, 2). But since none of the options have just [1, 2), I think the problem might have intended A to be (-∞, -2] ∪ [1, ∞), which would make the intersection (-3, -2] ∪ [1, 2), which is option C.Alternatively, maybe I'm overcomplicating it. Let me try solving it again without assuming absolute value.Set A: x + 1/2 ≥ 3/2 → x ≥ 1.Set B: x² + x < 6 → x² + x - 6 < 0 → (x + 3)(x - 2) < 0 → -3 < x < 2.So, A ∩ B is [1, 2).But none of the options have just [1, 2). The closest is option C, which includes (-3, -2] ∪ [1, 2). But why is there a (-3, -2] interval?Wait, maybe I made a mistake in solving set A. Let me check again.Set A: x + 1/2 ≥ 3/2 → x ≥ 1.Is there another condition? If it's x + 1/2 ≥ 3/2, then x ≥ 1. There's no other solution. So, A is [1, ∞).But then, why do the options include negative numbers? Maybe the problem is miswritten, or perhaps I need to consider that the inequality is x + 1/2 ≥ 3/2 or x + 1/2 ≤ -3/2, which would give A as (-∞, -2] ∪ [1, ∞). But that would require the original inequality to be an absolute value, which it's not.Alternatively, maybe the problem is written correctly, and the options are incorrect. But since I need to choose from the given options, and the only one that includes [1, 2) is option C, which also includes (-3, -2], I think the intended answer is C.But I'm still confused because, according to the problem, A should only be [1, ∞). Maybe the problem intended A to be |x + 1/2| ≥ 3/2, which would give two intervals. So, perhaps I should proceed with that assumption.Therefore, assuming A is (-∞, -2] ∪ [1, ∞), and B is (-3, 2), then A ∩ B is (-3, -2] ∪ [1, 2), which is option C.So, I think the answer is C.

💡Okay, so I've got this equation to solve: 16 times (x minus 1) squared minus 9 equals zero. Hmm, let me write that down to make it clearer: 16(x - 1)² - 9 = 0. Alright, I need to find the values of x that satisfy this equation.First, I remember that when dealing with equations like this, it's often helpful to isolate the squared term. So, maybe I should start by getting rid of that -9 on the left side. To do that, I can add 9 to both sides of the equation. Let me try that:16(x - 1)² - 9 + 9 = 0 + 9Simplifying that, the -9 and +9 on the left side cancel each other out, so I'm left with:16(x - 1)² = 9Okay, now I have 16 times (x - 1) squared equals 9. I want to solve for (x - 1)², so I should divide both sides of the equation by 16 to get that term by itself. Let's do that:16(x - 1)² / 16 = 9 / 16On the left side, the 16 in the numerator and denominator cancel out, leaving me with:(x - 1)² = 9/16Alright, now I have (x - 1) squared equals 9 over 16. To solve for x, I need to take the square root of both sides. Remember, when you take the square root of both sides of an equation, you have to consider both the positive and negative roots. So, let's do that:√[(x - 1)²] = ±√(9/16)On the left side, the square and the square root cancel each other out, leaving me with:x - 1 = ±3/4Okay, so x minus 1 equals plus or minus 3 over 4. Now, I need to solve for x in both cases. Let's start with the positive case:x - 1 = 3/4To solve for x, I'll add 1 to both sides:x - 1 + 1 = 3/4 + 1Simplifying that, the -1 and +1 on the left side cancel out, so I have:x = 3/4 + 1Now, to add 3/4 and 1, I need to express 1 as a fraction with the same denominator. 1 is the same as 4/4, so:x = 3/4 + 4/4 = 7/4Alright, so one solution is x equals 7 over 4.Now, let's solve the negative case:x - 1 = -3/4Again, I'll add 1 to both sides to solve for x:x - 1 + 1 = -3/4 + 1Simplifying, the -1 and +1 on the left side cancel out, leaving:x = -3/4 + 1Again, I need to express 1 as a fraction with the same denominator to add it to -3/4. So, 1 is 4/4:x = -3/4 + 4/4 = 1/4Okay, so the other solution is x equals 1 over 4.Let me just double-check my work to make sure I didn't make any mistakes. Starting from the original equation:16(x - 1)² - 9 = 0Adding 9 to both sides:16(x - 1)² = 9Dividing both sides by 16:(x - 1)² = 9/16Taking the square root of both sides:x - 1 = ±3/4Solving for x:x = 1 ± 3/4Which gives x = 7/4 and x = 1/4.Yep, that all checks out. So, the solutions are x equals 7 over 4 and x equals 1 over 4.

💡First, I'll define the variables for the number of red, blue, and green gumballs as ( R ), ( B ), and ( G ) respectively.From the problem, I know that the number of blue gumballs is half the number of red gumballs:[B = frac{R}{2}]Next, the number of green gumballs is four times the number of blue gumballs:[G = 4B]The total number of gumballs is 56, so:[R + B + G = 56]Substituting ( B ) and ( G ) in terms of ( R ) into the total equation:[R + frac{R}{2} + 4left(frac{R}{2}right) = 56]Simplifying the equation:[R + frac{R}{2} + 2R = 56]Combining like terms:[frac{7R}{2} = 56]Solving for ( R ):[7R = 112][R = 16]Therefore, there are 16 red gumballs in the machine.

💡Alright, so I have this probability problem here: Seven distinct integers are picked at random from the set {1,2,3,…,12}. I need to find the probability that, among those selected, the third smallest is 4. Hmm, okay. Let me try to break this down step by step.First, I know that when dealing with probability, it's often helpful to figure out the total number of possible outcomes and then the number of favorable outcomes. The probability will then be the ratio of favorable to total outcomes.So, the total number of ways to choose 7 distinct integers from 12 is given by the combination formula, which is C(12,7). Let me calculate that. C(12,7) is equal to 12! / (7! * (12-7)!) = 792. Okay, so there are 792 possible ways to pick any 7 numbers from the set.Now, I need to find how many of these combinations have 4 as the third smallest number. Let me think about what that means. If 4 is the third smallest number in the selected set, that implies that there are exactly two numbers smaller than 4 and four numbers larger than 4 in the selected set.Wait, hold on. Let me clarify that. If 4 is the third smallest, then there must be two numbers that are smaller than 4 and the remaining four numbers must be larger than 4. Because when you order the selected numbers from smallest to largest, the third one is 4, so the first two are smaller, and the next four are larger.So, to find the number of favorable outcomes, I need to calculate how many ways I can choose two numbers from the numbers smaller than 4 and four numbers from the numbers larger than 4.Looking back at the original set {1,2,3,…,12}, the numbers smaller than 4 are {1,2,3}, and the numbers larger than 4 are {5,6,7,8,9,10,11,12}. So, there are 3 numbers smaller than 4 and 8 numbers larger than 4.Therefore, the number of ways to choose two numbers from the smaller set is C(3,2), and the number of ways to choose four numbers from the larger set is C(8,4). Then, the total number of favorable outcomes is the product of these two combinations.Let me compute these. C(3,2) is 3, because there are three ways to choose two numbers from three. And C(8,4) is 70, since 8! / (4! * 4!) = 70. So, multiplying these together, 3 * 70 = 210. Hmm, so there are 210 favorable outcomes.Wait, but hold on a second. The problem says seven distinct integers, and I just calculated the number of ways to choose two small numbers and four large numbers, which adds up to six numbers. But we need seven numbers. Oh! I see, I forgot to include the number 4 itself in the count. So, actually, the total number of numbers selected should be 2 (smaller) + 1 (the number 4) + 4 (larger) = 7. So, I need to adjust my calculation.Therefore, the number of favorable outcomes is C(3,2) * C(8,4). Wait, but that's still 3 * 70 = 210. But wait, does that include the number 4? Or do I need to account for it separately?Hmm, actually, when I choose two numbers from {1,2,3} and four numbers from {5,6,7,8,9,10,11,12}, the number 4 is not included in either of these selections. So, to have 4 as the third smallest, I need to include 4 in the selected set. Therefore, the total number of numbers selected is 2 (from below 4) + 1 (which is 4) + 4 (from above 4) = 7.So, the number of favorable outcomes is indeed C(3,2) * C(8,4) = 3 * 70 = 210. Wait, but earlier I thought it was 168. Hmm, maybe I made a mistake in my initial calculation.Wait, let me double-check. The total number of ways to choose 7 numbers is 792. The number of favorable outcomes is 210. So, the probability would be 210 / 792. Let me simplify that. Dividing numerator and denominator by 6, we get 35 / 132. Hmm, that's approximately 0.265.But wait, in my initial thought process, I thought of 168 / 792, which simplifies to 7 / 33, approximately 0.212. So, which one is correct?Wait, maybe I confused the number of numbers to choose. Let me go back. If 4 is the third smallest, then there are two numbers less than 4 and four numbers greater than 4. So, the total numbers selected are 2 + 1 + 4 = 7. So, the number of ways is C(3,2) * C(8,4) = 3 * 70 = 210.But in the initial problem, the user had a different approach. They considered the third smallest as 4, so they chose 2 numbers below 4 and 5 numbers above 4. Wait, that would make the total numbers selected as 2 + 1 + 5 = 8, which is more than 7. That can't be right.Wait, no. Let me clarify. If 4 is the third smallest, then in the selected set of 7 numbers, the first two are below 4, the third is 4, and the remaining four are above 4. So, the number of numbers above 4 should be 4, not 5. Therefore, the number of ways is C(3,2) * C(8,4) = 3 * 70 = 210.But in the initial problem, the user had C(8,5) = 56, which would mean selecting 5 numbers above 4, making the total numbers selected as 2 + 1 + 5 = 8, which is incorrect because we need only 7 numbers. So, that was a mistake in the initial approach.Therefore, the correct number of favorable outcomes is 210, leading to a probability of 210 / 792. Simplifying that, divide numerator and denominator by 6: 35 / 132. Can we simplify further? 35 and 132 share a common factor of... 35 is 5*7, 132 is 4*33, which is 4*3*11. No common factors, so 35/132 is the simplified form.Wait, but 35/132 can be simplified by dividing numerator and denominator by... Let me check. 35 divided by 5 is 7, 132 divided by 5 is 26.4, which is not an integer. So, no, 35/132 is the simplest form.But wait, in the initial problem, the user had 168 / 792 = 7/33. So, why the discrepancy? Because in their approach, they considered selecting 5 numbers above 4, which would make the total numbers selected as 2 + 1 + 5 = 8, which is incorrect. Therefore, their count was wrong.So, the correct count is 210, leading to 35/132. Hmm, but let me double-check my reasoning.Alternatively, maybe I can think of it as fixing 4 as the third smallest, which means that in the selected 7 numbers, exactly two are less than 4, and four are greater than 4. So, the number of ways is C(3,2) * C(8,4) = 3 * 70 = 210.Yes, that seems correct. So, the probability is 210 / 792. Simplifying, divide numerator and denominator by 6: 35 / 132. So, 35/132 is the probability.Wait, but 35/132 is approximately 0.265, which seems reasonable. Alternatively, 7/33 is approximately 0.212, which is less. So, which one is correct?Let me think differently. Maybe I can use the concept of order statistics. The probability that the third order statistic is 4.In general, the probability that the k-th order statistic is equal to a particular value can be calculated using combinations.The formula is: C(m - 1, k - 1) * C(N - m, n - k) / C(N, n), where N is the total number of elements, n is the number of selected elements, m is the value we are interested in, and k is the order statistic.In this case, N = 12, n = 7, m = 4, k = 3.So, plugging into the formula: C(4 - 1, 3 - 1) * C(12 - 4, 7 - 3) / C(12,7).That is, C(3,2) * C(8,4) / C(12,7).Which is exactly what I did earlier: 3 * 70 / 792 = 210 / 792 = 35 / 132.So, that confirms it. Therefore, the correct probability is 35/132.Wait, but in the initial problem, the user had 168 / 792 = 7/33. So, why the difference? Because in their approach, they considered selecting 5 numbers above 4, which would make the total numbers selected as 2 + 1 + 5 = 8, which is incorrect because we need only 7 numbers. Therefore, their count was wrong.So, the correct count is 210, leading to 35/132.But wait, let me check the formula again. The formula is C(m - 1, k - 1) * C(N - m, n - k). So, m = 4, k = 3, N = 12, n = 7.C(3,2) * C(8,4) = 3 * 70 = 210.Yes, that's correct.Alternatively, another way to think about it is: To have 4 as the third smallest, we need exactly two numbers less than 4 and four numbers greater than 4. So, the number of ways is C(3,2) * C(8,4) = 3 * 70 = 210.Therefore, the probability is 210 / 792. Simplifying, divide numerator and denominator by 6: 35 / 132.Yes, that seems correct.Wait, but 35/132 can be simplified further? Let me check. 35 is 5*7, 132 is 4*33, which is 4*3*11. No common factors, so 35/132 is the simplest form.Alternatively, if I divide numerator and denominator by 3, 35/132 becomes approximately 11.666/44, which is not helpful. So, 35/132 is the simplest form.Therefore, the probability is 35/132.Wait, but in the initial problem, the user had 168 / 792 = 7/33. So, why the discrepancy? Because in their approach, they considered selecting 5 numbers above 4, which would make the total numbers selected as 2 + 1 + 5 = 8, which is incorrect because we need only 7 numbers. Therefore, their count was wrong.So, the correct count is 210, leading to 35/132.Alternatively, maybe I made a mistake in interpreting the problem. Let me read it again: "Seven distinct integers are picked at random from {1,2,3,…,12}. What is the probability that, among those selected, the third smallest is 4?"Yes, so we need exactly two numbers less than 4, and four numbers greater than 4, plus the number 4 itself, making a total of 7 numbers.Therefore, the number of favorable outcomes is C(3,2) * C(8,4) = 3 * 70 = 210.Total outcomes: C(12,7) = 792.Probability: 210 / 792 = 35 / 132.So, I think that's the correct answer.Wait, but let me check with another approach. Suppose I list all possible combinations where 4 is the third smallest. So, the first two numbers are from {1,2,3}, and the next four are from {5,6,7,8,9,10,11,12}.So, the number of ways is C(3,2) * C(8,4) = 3 * 70 = 210.Yes, that's consistent.Alternatively, if I think about the position of 4 in the ordered set. For 4 to be the third smallest, there must be exactly two numbers before it and four after it. So, the number of ways is C(3,2) * C(8,4) = 210.Therefore, the probability is 210 / 792 = 35 / 132.Yes, that seems correct.Wait, but 35/132 is approximately 0.265, which is about 26.5%. That seems reasonable.Alternatively, if I think about the probability step by step. The probability that 4 is included in the selected set is C(11,6) / C(12,7) = 462 / 792 = 7/12. Then, given that 4 is included, the probability that exactly two numbers less than 4 are also included.So, given that 4 is included, we have 6 more numbers to choose from the remaining 11 numbers. But actually, since 4 is already included, we have to choose 6 more numbers from the remaining 11, which are {1,2,3,5,6,7,8,9,10,11,12}.Out of these, we need exactly two numbers from {1,2,3} and four numbers from {5,6,7,8,9,10,11,12}.So, the number of favorable outcomes is C(3,2) * C(8,4) = 3 * 70 = 210.The total number of ways to choose 6 numbers from 11 is C(11,6) = 462.Therefore, the conditional probability is 210 / 462 = 5/11.Then, the overall probability is the probability that 4 is included times the conditional probability, which is (7/12) * (5/11) = 35 / 132.Yes, that's the same result as before.So, that confirms it again. The probability is 35/132.Therefore, I think the correct answer is 35/132.But wait, in the initial problem, the user had 168 / 792 = 7/33. So, why the difference? Because in their approach, they considered selecting 5 numbers above 4, which would make the total numbers selected as 2 + 1 + 5 = 8, which is incorrect because we need only 7 numbers. Therefore, their count was wrong.So, the correct count is 210, leading to 35/132.Alternatively, maybe I can think of it as fixing 4 as the third smallest, which means that in the selected 7 numbers, exactly two are less than 4, and four are greater than 4. So, the number of ways is C(3,2) * C(8,4) = 3 * 70 = 210.Yes, that seems correct.Therefore, the probability is 210 / 792. Simplifying, divide numerator and denominator by 6: 35 / 132.So, the final answer is 35/132.

💡Okay, so I have this geometry problem here, and I need to figure out the correct answer. Let me read it again carefully."If there is a point on a line that is outside a known plane, then the points on the line are (    )A: All outside the plane.B: There are infinitely many points outside the plane.C: There are infinitely many points inside the plane.D: There is at least one point inside the plane."Alright, so the question is about the relationship between a line and a plane in space. It says that if there's at least one point on the line that's outside the known plane, what can we say about the points on the line?Let me visualize this. Imagine a plane, like a flat tabletop, and a line somewhere in space. If the line has at least one point that's not on the tabletop, what does that tell me about the rest of the line?First, I know that a line can either lie entirely on a plane, intersect the plane at one point, or be parallel to the plane without intersecting it at all. Since the problem states that there's a point on the line outside the plane, the line can't be entirely on the plane. So, that rules out the possibility that all points are on the plane.Now, considering the options:A: All outside the plane. Hmm, is that necessarily true? If the line is parallel to the plane, then yes, all points would be outside. But if the line intersects the plane at one point, then only that one point is on the plane, and the rest are outside. So, it's not necessarily true that all points are outside; it depends on whether the line is parallel or intersects the plane.B: There are infinitely many points outside the plane. Well, if the line is parallel to the plane, then all points are outside, which means infinitely many. If the line intersects the plane, then all points except one are outside, which is still infinitely many. So, regardless of whether the line is parallel or intersects the plane, there are infinitely many points outside. This seems promising.C: There are infinitely many points inside the plane. Wait, if the line intersects the plane, only one point is on the plane. If it's parallel, none are on the plane. So, there can't be infinitely many points inside the plane. This seems incorrect.D: There is at least one point inside the plane. If the line is parallel to the plane, then there are no points on the plane. So, this isn't necessarily true either.So, putting it all together, the correct answer should be B: There are infinitely many points outside the plane.

💡Okay, so I have this function f(x) = (1 - x)e^x, and I need to find its monotonic increasing interval. Hmm, I remember that to find where a function is increasing or decreasing, I need to look at its derivative. If the derivative is positive, the function is increasing, and if it's negative, the function is decreasing. Alright, let me start by finding the derivative of f(x). The function is a product of two functions: u(x) = (1 - x) and v(x) = e^x. So, I should use the product rule for differentiation. The product rule says that the derivative of u*v is u’*v + u*v’. First, let me find u’(x). Since u(x) = (1 - x), the derivative u’(x) should be -1, right? Because the derivative of 1 is 0, and the derivative of -x is -1. Next, v(x) is e^x, and I remember that the derivative of e^x is still e^x. So, v’(x) = e^x. Now, applying the product rule: f’(x) = u’(x)*v(x) + u(x)*v’(x). Plugging in the values I found, that would be f’(x) = (-1)*e^x + (1 - x)*e^x. Let me simplify that. So, f’(x) = -e^x + (1 - x)e^x. I can factor out e^x from both terms, which gives me f’(x) = e^x*(-1 + 1 - x). Wait, that simplifies to e^x*(-x). So, f’(x) = -x e^x. Hmm, that seems right. Let me double-check my steps. I took the derivative of (1 - x), got -1, multiplied by e^x. Then, took (1 - x) multiplied by the derivative of e^x, which is e^x. So, f’(x) = -e^x + (1 - x)e^x. Factoring out e^x gives e^x*(-1 + 1 - x) which is indeed e^x*(-x). So, f’(x) = -x e^x. Now, I need to find where this derivative is positive because that's where the function is increasing. So, I set f’(x) > 0, which is -x e^x > 0. I know that e^x is always positive for any real number x because the exponential function is always positive. So, e^x > 0 for all x. That means the sign of f’(x) depends solely on the term -x. So, if -x is positive, then f’(x) is positive. Let's solve -x > 0. Multiplying both sides by -1 (and remembering to flip the inequality sign when multiplying by a negative), we get x < 0. Therefore, f’(x) > 0 when x < 0. That means the function f(x) is increasing on the interval from negative infinity to 0. Let me just visualize this to make sure. The function f(x) = (1 - x)e^x. As x approaches negative infinity, (1 - x) becomes very large positive because x is negative and subtracting a negative makes it positive. e^x approaches 0 because as x is very negative, e^x tends to 0. So, the function tends to 0 from the positive side. At x = 0, f(0) = (1 - 0)e^0 = 1*1 = 1. So, the function starts at 0 when x is very negative, increases to 1 at x = 0, and then beyond x = 0, what happens? Let me plug in a positive x, say x = 1. f(1) = (1 - 1)e^1 = 0*e = 0. So, it decreases back to 0 at x = 1. So, the function increases from negative infinity up to x = 0, reaching a maximum at x = 0, and then decreases afterward. That makes sense with the derivative we found. Just to be thorough, let me test a value in the interval x < 0, say x = -1. f’(-1) = -(-1)e^{-1} = 1/e, which is positive. So, the function is indeed increasing there. And if I test a value greater than 0, say x = 1, f’(1) = -1*e^1 = -e, which is negative, confirming that the function is decreasing there. So, putting it all together, the function f(x) = (1 - x)e^x is increasing on the interval (-∞, 0). That should be the answer.