💡Okay, so I have this geometry problem here: "The relationship between a line connecting a point inside a plane and a point outside the plane, and a line within this plane is __________." Hmm, let me try to visualize this. First, I need to understand what it's asking. There's a plane, let's say it's a flat surface like a tabletop. Now, there's a point inside this plane, which means it's somewhere on the tabletop. Then there's another point outside this plane, so maybe it's somewhere above or below the tabletop. The line connecting these two points would go from the tabletop up to the point above or below.Now, there's also a line that's within this plane, so it's lying entirely on the tabletop. The question is asking about the relationship between this line connecting the two points (one inside and one outside the plane) and the line that's entirely on the plane.I remember that in geometry, lines can have different relationships: they can be parallel, intersecting, or skew. But wait, skew lines are only in three-dimensional space, right? So, if both lines are in the same plane, they can either be parallel or intersecting. But in this case, one line is in the plane and the other is connecting a point inside the plane to a point outside. So, the second line isn't entirely in the plane.Let me think about this. If I have a line that's entirely on the plane, and another line that starts on the plane and goes out, can they ever intersect? Well, if they intersect, the point of intersection would have to be on the plane because that's where the line that's entirely on the plane is. But the other line starts on the plane and goes out, so it could potentially intersect the line on the plane at that starting point. So, is that possible?Wait, but the line connecting the two points (inside and outside the plane) starts at the point inside the plane and goes to the point outside. So, if the line within the plane passes through that starting point, then they would intersect at that point. But if the line within the plane doesn't pass through that starting point, then they wouldn't intersect. So, in some cases, they could intersect, and in other cases, they might not.But what about skew lines? Skew lines are lines that don't intersect and aren't parallel. They exist in different planes. So, in this case, the line within the plane is in one plane, and the line connecting the two points is in a different plane because it goes outside. So, could they be skew?Wait, but the line connecting the two points starts in the plane, so it's not entirely in a different plane. It's partially in the plane and partially outside. So, does that mean it's not entirely in a different plane? Hmm, I'm a bit confused here.Maybe I should draw a diagram. Imagine the plane as a horizontal line. Point A is on the plane, and point B is above the plane. So, the line AB connects A to B. Now, there's another line CD that's entirely on the plane. If CD passes through A, then AB and CD intersect at A. If CD doesn't pass through A, then AB and CD don't intersect. But are they parallel? Well, CD is on the plane, and AB starts on the plane but goes up. So, unless CD is parallel to AB's direction, they might not be parallel.Wait, but AB is not entirely in the plane, so its direction isn't confined to the plane's orientation. So, it's possible that AB and CD are neither intersecting nor parallel, which would make them skew. But earlier, I thought skew lines have to be in different planes entirely. But AB starts in the plane, so part of it is in the plane, and part is outside.Is there a different way to think about this? Maybe in terms of projections. If I project AB onto the plane, it would just be the point A. So, the projection of AB is a single point, while CD is a line. So, unless CD passes through A, they don't intersect. But in terms of being skew, I think that applies only to lines that are entirely in different planes. Since AB is partially in the plane, maybe it's not considered skew.Wait, I'm getting conflicting thoughts here. Let me check the definitions again. Skew lines are non-parallel, non-intersecting lines that are not coplanar. So, if two lines are not in the same plane and don't intersect, they are skew. In this case, AB is not entirely in the plane, but CD is entirely in the plane. So, are AB and CD coplanar? If they were, they would either intersect or be parallel. But since AB is not entirely in the plane, they might not be coplanar.Wait, but AB starts in the plane, so at least one point is shared with the plane. So, if CD is in the plane and AB starts in the plane, could they be coplanar? If CD passes through A, then AB and CD would intersect at A, making them coplanar at that point. But if CD doesn't pass through A, then AB and CD don't intersect, and since AB is not entirely in the plane, they might not be coplanar. So, in that case, they would be skew.But I'm not entirely sure. Maybe I should think about specific examples. Imagine the plane is the floor, point A is on the floor, and point B is somewhere above the floor. Line AB goes from the floor to above. Now, line CD is on the floor. If CD passes through A, then AB and CD intersect at A. If CD doesn't pass through A, then AB and CD don't intersect and aren't parallel, so they must be skew.Wait, but AB is not entirely in the plane, so it's not confined to the floor's orientation. So, even if CD is on the floor, AB could be going in a different direction, making them skew.But I'm still a bit confused because AB starts in the plane, so it's not entirely outside. Does that affect whether they can be skew? I think the key is that skew lines don't have to be entirely in different planes; they just can't be in the same plane. Since AB is not entirely in the plane, and CD is entirely in the plane, they might not be in the same plane, making them skew if they don't intersect.So, putting it all together, the relationship between the line connecting a point inside the plane and a point outside the plane, and a line within the plane can either be intersecting (if they meet at the point inside the plane) or skew (if they don't intersect and aren't parallel). Therefore, the answer should be that they are either skew or intersecting.

💡Okay, so I have this problem here about a sequence {a_n} where the sum of the first n terms is denoted as S_n. The problem says that for any natural numbers n and m, the sum S_n plus S_m equals S_{n+m}. Also, they give me that a_1 is 5, and I need to find the value of a_8. The options are A: 40, B: 35, C: 12, D: 5.Alright, let me break this down. First, I know that S_n is the sum of the first n terms of the sequence. So, S_n = a_1 + a_2 + ... + a_n. Similarly, S_m is the sum of the first m terms. The condition given is that S_n + S_m = S_{n+m}. That seems interesting because normally, if you add two sums, S_n and S_m, you would expect it to be equal to S_{n+m} only if the sequence has some special property.Let me think about what kind of sequences have this property. If I take n=1 and m=1, then S_1 + S_1 should equal S_2. Since S_1 is just a_1, which is 5, then 5 + 5 = S_2. So, S_2 is 10. But S_2 is also a_1 + a_2, which is 5 + a_2. Therefore, 5 + a_2 = 10, so a_2 must be 5.Wait, so a_2 is 5? That's the same as a_1. Let me check with another value. Let's take n=2 and m=1. Then S_2 + S_1 should equal S_3. S_2 is 10, S_1 is 5, so 10 + 5 = 15, which should be S_3. But S_3 is a_1 + a_2 + a_3, which is 5 + 5 + a_3. So, 5 + 5 + a_3 = 15, which means a_3 is 5 again.Hmm, so a_3 is also 5. This seems like a pattern. Maybe all the terms are 5? Let me test this further. Let's take n=3 and m=1. Then S_3 + S_1 should equal S_4. S_3 is 15, S_1 is 5, so 15 + 5 = 20, which should be S_4. S_4 is a_1 + a_2 + a_3 + a_4, which is 5 + 5 + 5 + a_4. So, 15 + a_4 = 20, meaning a_4 is 5.Okay, so a_4 is also 5. It seems like every term is 5. Let me generalize this. Suppose for any k, a_k = 5. Then S_n would be 5n, because it's the sum of n terms each equal to 5. Let's see if this satisfies the given condition S_n + S_m = S_{n+m}.If S_n = 5n, then S_n + S_m = 5n + 5m = 5(n + m) = S_{n+m}, which is exactly what the condition says. So, yes, this works. Therefore, the sequence {a_n} is a constant sequence where each term is 5.Therefore, a_8 is 5. Looking back at the options, D is 5, so the answer should be D.Wait, but let me make sure I didn't miss anything. The problem didn't specify whether the sequence is arithmetic or geometric or something else. But from the condition given, it seems that the only way S_n + S_m = S_{n+m} is if each term is constant. Because if the terms weren't constant, adding S_n and S_m wouldn't just give you S_{n+m}; there would be some overlap or something else.For example, if the sequence was arithmetic with a common difference, then S_n would be a quadratic function of n, and adding two quadratics wouldn't generally give another quadratic unless the coefficients are zero, which would make it a linear function, implying a constant sequence.Similarly, if it was geometric, S_n would be exponential, and adding two exponentials wouldn't give another exponential unless the ratio is 1, which again makes it a constant sequence.So, yeah, it seems that the only possibility is that all terms are equal to 5. Therefore, a_8 is 5, which is option D.

💡Okay, so I have this probability problem here. Let me read it again to make sure I understand it correctly. It says: From the set of positive integers {1,2,3,…,9}, let event A be "the sum of the two chosen numbers is even," and event B be "both chosen numbers are even." I need to calculate the conditional probability P(B|A). The options are A: 1/8, B: 1/4, C: 3/8, D: 1/2.Alright, so first, I need to recall what conditional probability means. The formula for conditional probability is P(B|A) = P(B ∩ A) / P(A). That is, the probability of both B and A happening divided by the probability of A happening. So, I need to figure out both P(A) and P(B ∩ A).Let me start by understanding the sample space. The set is {1,2,3,4,5,6,7,8,9}. We are choosing two numbers from this set. So, the total number of possible ways to choose two numbers is the combination of 9 numbers taken 2 at a time, which is denoted as C(9,2). I remember that the formula for combinations is C(n,k) = n! / (k!(n - k)!). So, plugging in the numbers, C(9,2) = 9! / (2! * 7!) = (9 * 8) / (2 * 1) = 36. So, there are 36 possible pairs.Now, let's figure out event A: the sum of the two chosen numbers is even. When is a sum even? Well, in mathematics, the sum of two numbers is even if both numbers are even or both are odd. So, that gives me two scenarios to consider for event A.First scenario: both numbers are even. In the set {1,2,3,4,5,6,7,8,9}, the even numbers are {2,4,6,8}. So, there are 4 even numbers. The number of ways to choose two even numbers is C(4,2). Calculating that: C(4,2) = 4! / (2! * 2!) = (4 * 3) / (2 * 1) = 6. So, 6 ways.Second scenario: both numbers are odd. The odd numbers in the set are {1,3,5,7,9}, which are 5 numbers. The number of ways to choose two odd numbers is C(5,2). Calculating that: C(5,2) = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10. So, 10 ways.Therefore, the total number of favorable outcomes for event A is 6 (both even) + 10 (both odd) = 16. So, P(A) = number of favorable outcomes / total number of possible outcomes = 16 / 36. Simplifying that, 16 divided by 36 is 4/9. Wait, 16 divided by 36 is actually 4/9? Let me check: 16 divided by 4 is 4, and 36 divided by 4 is 9. Yes, so P(A) = 4/9.Now, moving on to event B: both chosen numbers are even. We already calculated the number of ways to choose two even numbers, which is 6. So, the probability of event B is 6 / 36, which simplifies to 1/6. But wait, in the formula for conditional probability, I need P(B ∩ A), which is the probability that both B and A occur. But in this case, if both numbers are even, their sum is definitely even. So, event B is a subset of event A. That means P(B ∩ A) is just P(B). So, P(B ∩ A) = 6 / 36 = 1/6.Now, plugging these into the conditional probability formula: P(B|A) = P(B ∩ A) / P(A) = (1/6) / (4/9). Dividing fractions is the same as multiplying by the reciprocal, so (1/6) * (9/4) = (9/24) = 3/8. So, P(B|A) is 3/8.Let me double-check my calculations to make sure I didn't make any mistakes. Total number of ways is 36, correct. For event A, both even: 6, both odd: 10, total 16. So, 16/36 is 4/9, that's correct. For event B, it's 6/36, which is 1/6. Then, 1/6 divided by 4/9 is indeed 3/8. So, that seems right.Wait, just to make sure, let me think about it another way. If I know that the sum is even, what's the chance both numbers are even? So, out of the 16 favorable pairs for A, how many of them are pairs where both numbers are even? That would be 6. So, 6 out of 16 is 3/8. Yep, that's another way to see it. So, 6/16 simplifies to 3/8.So, yeah, I think 3/8 is the correct answer. Looking back at the options, that's option C.**Final Answer**boxed{C}

💡Okay, so I have this problem here: "If the ratio of (3x - 2y) to (2x + 3y) is (frac{3}{4}), what is the ratio of (x) to (y)?"Hmm, let's see. I need to find the ratio of (x) to (y), which means I need to express (x) in terms of (y) or vice versa. The given ratio is between two expressions involving (x) and (y), so I can set up an equation and solve for the ratio.First, the ratio of (3x - 2y) to (2x + 3y) is (frac{3}{4}). So, I can write this as:[frac{3x - 2y}{2x + 3y} = frac{3}{4}]Alright, now I need to solve for (frac{x}{y}). To do this, I can cross-multiply to eliminate the fractions. Let's do that:[4(3x - 2y) = 3(2x + 3y)]Expanding both sides:Left side: (4 times 3x = 12x) and (4 times (-2y) = -8y), so left side is (12x - 8y).Right side: (3 times 2x = 6x) and (3 times 3y = 9y), so right side is (6x + 9y).Now, the equation becomes:[12x - 8y = 6x + 9y]I need to get all the (x) terms on one side and the (y) terms on the other. Let me subtract (6x) from both sides:[12x - 6x - 8y = 9y]Which simplifies to:[6x - 8y = 9y]Now, I'll add (8y) to both sides to get all the (y) terms on the right:[6x = 9y + 8y]Simplifying the right side:[6x = 17y]Okay, so now I have (6x = 17y). To find the ratio (frac{x}{y}), I can divide both sides by (6y):[frac{x}{y} = frac{17}{6}]So, the ratio of (x) to (y) is (frac{17}{6}).Let me just double-check my steps to make sure I didn't make any mistakes. Starting from the ratio, cross-multiplied correctly, expanded both sides, moved all (x) terms to one side and (y) terms to the other, simplified, and then solved for (frac{x}{y}). Everything seems to add up. I think that's the correct answer.

💡Okay, so I have this probability problem here about two people, A and B, shooting at a target. The problem has two parts, and I need to figure out both. Let me start with part (I).Part (I) says: If both A and B shoot once, find the probability that A hits the target while B does not. Hmm, okay. So, A and B each take one shot. I know the probability that A hits the target is 3/4, and for B, it's 4/5. They also mention that each shot is independent, so the outcome of A's shot doesn't affect B's shot and vice versa.So, I need the probability that A hits and B misses. Since these are independent events, I can multiply their individual probabilities. The probability that A hits is 3/4, and the probability that B does not hit is 1 minus the probability that B hits. So, that would be 1 - 4/5, which is 1/5.Therefore, the probability that A hits and B misses is (3/4) * (1/5). Let me calculate that: 3/4 is 0.75, and 1/5 is 0.2. Multiplying them together gives 0.15, which is 3/20. So, that should be the probability for part (I). I think that's straightforward.Now, moving on to part (II). It says: If both A and B shoot twice, find the probability that they hit the target an equal number of times. Hmm, okay. So, both A and B are shooting twice, and I need the probability that they each hit the target the same number of times. That could mean both hitting 0 times, both hitting 1 time, or both hitting 2 times. So, I need to calculate the probabilities for each of these scenarios and then add them up.First, let me recall that each shot is independent, so the number of hits follows a binomial distribution. For A, the probability of hitting k times out of 2 shots is given by the binomial formula: C(2, k) * (3/4)^k * (1/4)^(2 - k). Similarly, for B, it's C(2, l) * (4/5)^l * (1/5)^(2 - l), where l is the number of hits.So, I need to compute the probability that both A and B have 0 hits, both have 1 hit, and both have 2 hits, then sum these probabilities.Let me start with both having 0 hits. For A, the probability of 0 hits is C(2, 0) * (3/4)^0 * (1/4)^2. C(2, 0) is 1, (3/4)^0 is 1, and (1/4)^2 is 1/16. So, that's 1 * 1 * 1/16 = 1/16.For B, the probability of 0 hits is C(2, 0) * (4/5)^0 * (1/5)^2. Similarly, that's 1 * 1 * 1/25 = 1/25.Since these are independent, the joint probability that both A and B have 0 hits is (1/16) * (1/25) = 1/400.Next, both having 1 hit. For A, the probability of exactly 1 hit is C(2, 1) * (3/4)^1 * (1/4)^1. C(2, 1) is 2, so that's 2 * (3/4) * (1/4) = 2 * 3/16 = 6/16, which simplifies to 3/8.For B, the probability of exactly 1 hit is C(2, 1) * (4/5)^1 * (1/5)^1. That's 2 * (4/5) * (1/5) = 8/25.So, the joint probability that both A and B have exactly 1 hit is (3/8) * (8/25). Let me compute that: 3/8 * 8/25 = (3 * 8) / (8 * 25) = 24/200. Simplifying that, divide numerator and denominator by 8: 3/25. Wait, no, 24 divided by 8 is 3, and 200 divided by 8 is 25. So, 3/25. Hmm, that seems right.Wait, but let me double-check. 3/8 * 8/25: the 8s cancel out, so it's 3/25. Yeah, that's correct.Now, both having 2 hits. For A, the probability of 2 hits is C(2, 2) * (3/4)^2 * (1/4)^0. C(2, 2) is 1, (3/4)^2 is 9/16, and (1/4)^0 is 1. So, that's 9/16.For B, the probability of 2 hits is C(2, 2) * (4/5)^2 * (1/5)^0. That's 1 * 16/25 * 1 = 16/25.So, the joint probability that both A and B have 2 hits is (9/16) * (16/25). Let me compute that: 9/16 * 16/25. The 16s cancel out, so it's 9/25.Now, adding up all these probabilities: the probability that both have 0 hits is 1/400, both have 1 hit is 3/25, and both have 2 hits is 9/25. So, total probability is 1/400 + 3/25 + 9/25.First, let me convert all fractions to have the same denominator to add them up. The denominators are 400, 25, and 25. Let me convert everything to 400.1/400 is already in 400. 3/25 is equal to (3 * 16)/400 = 48/400. Similarly, 9/25 is (9 * 16)/400 = 144/400.So, adding them up: 1/400 + 48/400 + 144/400 = (1 + 48 + 144)/400 = (1 + 48 is 49, 49 + 144 is 193)/400.Wait, that's 193/400. Let me see if that can be simplified. 193 is a prime number, I think, because it's not divisible by 2, 3, 5, 7, 11, 13. Let me check: 193 divided by 2 is 96.5, not integer. 193 divided by 3 is about 64.333, not integer. Divided by 5 is 38.6, nope. 7? 193/7 is about 27.571, not integer. 11? 17.545, nope. 13? 14.846, nope. So, 193 is prime, so 193/400 is the simplified fraction.But wait, the initial calculation: 1/400 + 3/25 + 9/25. 3/25 is 0.12, 9/25 is 0.36, so adding them up: 0.12 + 0.36 = 0.48, plus 1/400 which is 0.0025, so total is 0.4825. 0.4825 is equal to 193/400 because 193 divided by 400 is 0.4825. So, that's correct.Alternatively, 193 divided by 400 is 0.4825, which is 48.25%. So, the probability is 193/400 or 0.4825.Wait, but let me double-check my calculations because 1/400 is 0.0025, 3/25 is 0.12, and 9/25 is 0.36. Adding them: 0.0025 + 0.12 = 0.1225, plus 0.36 is 0.4825. So, yes, that's correct.But wait, in the initial step, when I calculated the joint probabilities, I think I might have made a mistake. Let me check again.For both having 0 hits: A has 0 hits with probability (1/4)^2 = 1/16, and B has 0 hits with probability (1/5)^2 = 1/25. So, 1/16 * 1/25 = 1/400. That's correct.For both having 1 hit: A's probability is C(2,1)*(3/4)*(1/4) = 2*(3/4)*(1/4) = 6/16 = 3/8. B's probability is C(2,1)*(4/5)*(1/5) = 2*(4/5)*(1/5) = 8/25. So, 3/8 * 8/25 = 24/200 = 3/25. That's correct.For both having 2 hits: A's probability is (3/4)^2 = 9/16, and B's probability is (4/5)^2 = 16/25. So, 9/16 * 16/25 = 9/25. That's correct.So, adding up: 1/400 + 3/25 + 9/25. Converting to 400 denominator: 1/400 + 48/400 + 144/400 = 193/400, which is 0.4825.Wait, but in the initial problem statement, part (II) says "they hit the target an equal number of times." So, that includes 0, 1, or 2 hits. So, I think my approach is correct.Alternatively, maybe I can think of it as the sum over k=0 to 2 of P(A=k) * P(B=k). So, that's exactly what I did.Wait, but let me check if I did the calculations correctly. For A=0, B=0: 1/16 * 1/25 = 1/400.For A=1, B=1: 3/8 * 8/25 = 3/25.For A=2, B=2: 9/16 * 16/25 = 9/25.So, total is 1/400 + 3/25 + 9/25. Let me compute 3/25 + 9/25 first: that's 12/25. Then, 12/25 is 0.48, plus 1/400 which is 0.0025, so total is 0.4825, which is 193/400.Wait, but 12/25 is 48/100, and 1/400 is 0.25/100, so total is 48.25/100, which is 193/400. Yeah, that's correct.So, the probability is 193/400, which is 0.4825, or 48.25%.Wait, but the problem says "find the probability that they hit the target an equal number of times." So, that's exactly what I calculated. So, I think that's the answer.But let me think again: is there another way to approach this? Maybe using generating functions or something else? Hmm, but for two shots, it's manageable with the binomial approach.Alternatively, I can think of all possible outcomes for A and B and count the cases where the number of hits is equal.For A, possible hits: 0, 1, 2.For B, possible hits: 0, 1, 2.So, the total number of possible outcomes is (number of outcomes for A) * (number of outcomes for B). But since each shot is independent, the number of outcomes is 3 (for A) * 3 (for B) = 9 possible combinations. But actually, the probabilities are not uniform, so I can't just count them equally. So, the approach I took earlier is correct.So, I think my answer is correct: 193/400, which is 0.4825.But wait, let me check if 193/400 simplifies. 193 is a prime number, as I thought earlier, so it can't be simplified further. So, 193/400 is the simplest form.Alternatively, if I want to write it as a decimal, it's 0.4825, which is 48.25%.So, I think that's the answer for part (II).Wait, but in the initial problem, part (II) says "they hit the target an equal number of times." So, that's exactly what I calculated. So, I think that's correct.But just to make sure, let me recast the problem in terms of joint probabilities.For A and B, each shooting twice, the joint probability distribution is the product of their individual distributions because the shots are independent.So, the probability that A has k hits and B has l hits is P(A=k) * P(B=l).Therefore, the probability that they have the same number of hits is the sum over k=0 to 2 of P(A=k) * P(B=k).Which is exactly what I did.So, I think my answer is correct.Therefore, the answers are:(I) 3/20(II) 193/400 or 0.4825Wait, but in the initial problem, part (II) says "they hit the target an equal number of times." So, that's exactly what I calculated. So, I think that's correct.But just to make sure, let me recast the problem in terms of joint probabilities.For A and B, each shooting twice, the joint probability distribution is the product of their individual distributions because the shots are independent.So, the probability that A has k hits and B has l hits is P(A=k) * P(B=l).Therefore, the probability that they have the same number of hits is the sum over k=0 to 2 of P(A=k) * P(B=k).Which is exactly what I did.So, I think my answer is correct.Therefore, the answers are:(I) 3/20(II) 193/400 or 0.4825Wait, but 193/400 is 0.4825, which is 48.25%. That seems reasonable.Alternatively, if I want to write it as a fraction, 193/400 is the simplest form.So, I think that's the answer.**Final Answer**(I) boxed{dfrac{3}{20}}(II) boxed{dfrac{193}{400}}

💡Okay, so I have this function p(x) = x⁴ + 6x² + 9, and I need to find its range given that the domain is [0, ∞). Hmm, let me think about how to approach this. First, I remember that the range of a function is the set of all possible output values (y-values) it can produce. Since the domain is [0, ∞), x can be any non-negative real number. So, I need to figure out what values p(x) can take when x is in that domain.Looking at the function p(x) = x⁴ + 6x² + 9, it seems like a polynomial function. Polynomials are continuous everywhere, so that might help. Also, since all the exponents are even, the function is going to behave similarly for positive and negative x, but since the domain is only [0, ∞), I don't have to worry about negative x values.Wait, actually, the function is x⁴ + 6x² + 9. I wonder if this can be factored or rewritten in a simpler form. Let me try to factor it. Looking at x⁴ + 6x² + 9, I notice that x⁴ is (x²)², and 9 is 3². So, maybe this is a perfect square trinomial. Let me check:(x² + a)² = x⁴ + 2a x² + a². Comparing this to x⁴ + 6x² + 9, I see that 2a should be 6, so a = 3. Then, a² is 9, which matches. So, yes, p(x) can be written as (x² + 3)². That simplifies things a bit.So, p(x) = (x² + 3)². Now, since x is in [0, ∞), x² is also in [0, ∞). Therefore, x² + 3 is in [3, ∞). Then, squaring that, (x² + 3)² will be in [9, ∞). Wait, let me make sure. If x² is 0, then x² + 3 is 3, and squaring that gives 9. As x increases, x² increases, so x² + 3 increases, and squaring it makes it increase even more. So, the smallest value p(x) can take is 9, and it goes to infinity as x increases. But just to double-check, let me consider the derivative to see if there's a minimum or maximum. The derivative of p(x) with respect to x is p'(x) = 4x³ + 12x. Setting this equal to zero to find critical points:4x³ + 12x = 0 x(4x² + 12) = 0 So, x = 0 or 4x² + 12 = 0. But 4x² + 12 = 0 has no real solutions because x² is always non-negative, and 4x² + 12 is always positive. So, the only critical point is at x = 0. Now, evaluating p(x) at x = 0: p(0) = 0 + 0 + 9 = 9. Since this is the only critical point and the function tends to infinity as x increases, the function must have a minimum at x = 0, which is 9, and it increases without bound as x increases. Therefore, the range of p(x) is [9, ∞). I think that makes sense. The function starts at 9 when x is 0 and keeps increasing as x gets larger. There's no upper limit, so the range should be all real numbers from 9 onwards.**Final Answer**The range of the function is boxed{[9, infty)}.