💡Okay, so Vitya has five math lessons a week, from Monday to Friday. He knows that there's a 50% chance the teacher won't check his homework at all during the week, and a 50% chance that the teacher will check it, but only on one day, with each day being equally likely.Now, by the end of Thursday, Vitya realizes that his homework hasn't been checked yet. He wants to know the probability that the homework will be checked on Friday.Alright, let's break this down. First, there are two scenarios:1. The teacher doesn't check the homework at all during the week. The probability of this is 1/2.2. The teacher does check the homework, but only on one day, with each day (Monday to Friday) being equally likely. The probability of this is also 1/2.Given that Vitya hasn't had his homework checked by Thursday, we need to find the probability that it will be checked on Friday.So, we're dealing with conditional probability here. The probability that the homework is checked on Friday given that it hasn't been checked from Monday to Thursday.Let me define the events:- Let A be the event that the homework has not been checked from Monday to Thursday.- Let B be the event that the homework is checked on Friday.We need to find P(B|A), the probability that the homework is checked on Friday given that it hasn't been checked from Monday to Thursday.Using the formula for conditional probability:P(B|A) = P(A ∩ B) / P(A)So, we need to find P(A ∩ B) and P(A).First, let's find P(A ∩ B). This is the probability that the homework is checked on Friday and not checked from Monday to Thursday. Since the teacher checks the homework on only one day if they do check it, this is the same as the probability that the homework is checked on Friday.Given that the teacher checks the homework on one day with equal probability, the chance it's on Friday is 1/5. But we also have to consider the probability that the teacher decides to check the homework at all, which is 1/2.So, P(A ∩ B) = P(check on Friday) = (1/2) * (1/5) = 1/10.Next, let's find P(A), the probability that the homework hasn't been checked from Monday to Thursday.There are two scenarios contributing to this probability:1. The teacher doesn't check the homework at all during the week. The probability of this is 1/2.2. The teacher checks the homework, but only on Friday. The probability of this is (1/2) * (1/5) = 1/10.So, P(A) = P(no check) + P(check on Friday) = 1/2 + 1/10 = 5/10 + 1/10 = 6/10 = 3/5.Now, plugging these into the conditional probability formula:P(B|A) = (1/10) / (3/5) = (1/10) * (5/3) = 1/6.Wait, that doesn't seem right. Let me double-check.Wait, no, I think I made a mistake here. The P(A ∩ B) is the probability that the homework is checked on Friday, which is 1/5, but considering that the teacher might not check at all, it's actually (1/2) * (1/5) = 1/10.And P(A) is the probability that the homework hasn't been checked from Monday to Thursday, which includes both the case where the teacher doesn't check at all and the case where the teacher checks on Friday.So, P(A) = P(no check) + P(check on Friday) = 1/2 + (1/2)*(1/5) = 1/2 + 1/10 = 6/10 = 3/5.Therefore, P(B|A) = (1/10) / (3/5) = (1/10) * (5/3) = 1/6.Hmm, that seems low. Let me think again.Alternatively, maybe I should consider the conditional probability differently. Since we know that the homework hasn't been checked from Monday to Thursday, we can think of the possible days left for the check.If the teacher is going to check the homework, it must be on Friday. So, given that it hasn't been checked from Monday to Thursday, the only remaining day is Friday.But we also have to consider the probability that the teacher doesn't check at all.So, the total probability space is now reduced to two possibilities:1. The teacher doesn't check the homework at all.2. The teacher checks the homework on Friday.These are the only two possibilities given that it hasn't been checked from Monday to Thursday.So, the probability that the homework is checked on Friday is the probability that the teacher does check it, which is 1/2, divided by the total probability of either not checking or checking on Friday.Wait, that might not be the right way to think about it.Alternatively, let's model it as follows:There are two possibilities:- The teacher doesn't check the homework: probability 1/2.- The teacher checks the homework on one day: probability 1/2, with each day equally likely.Given that the homework hasn't been checked from Monday to Thursday, we can update our probabilities.So, the probability that the homework is checked on Friday is the probability that the teacher checks it on Friday divided by the probability that the homework hasn't been checked from Monday to Thursday.So, P(B|A) = P(check on Friday) / P(not checked from Monday to Thursday).P(check on Friday) is (1/2)*(1/5) = 1/10.P(not checked from Monday to Thursday) is P(no check) + P(check on Friday) = 1/2 + 1/10 = 6/10 = 3/5.So, P(B|A) = (1/10) / (3/5) = 1/6.Wait, that seems consistent with my earlier calculation.But intuitively, if the teacher is going to check the homework, and it hasn't been checked from Monday to Thursday, then the probability it's checked on Friday should be higher than 1/6.Maybe I'm missing something.Alternatively, perhaps I should consider the problem as a conditional probability where we have two possibilities: either the teacher doesn't check at all, or the teacher checks on one day, with each day equally likely.Given that the homework hasn't been checked from Monday to Thursday, we can update the probabilities.So, the probability that the teacher doesn't check at all is 1/2.The probability that the teacher checks on Friday is (1/2)*(1/5) = 1/10.The probability that the teacher checks on any of Monday to Thursday is (1/2)*(4/5) = 2/5.But since we know that the homework hasn't been checked from Monday to Thursday, we can eliminate the possibility that the teacher checks on Monday to Thursday.So, the total probability is now P(no check) + P(check on Friday) = 1/2 + 1/10 = 6/10 = 3/5.Therefore, the probability that the homework is checked on Friday given that it hasn't been checked from Monday to Thursday is P(check on Friday) / P(no check from Monday to Thursday) = (1/10) / (3/5) = 1/6.Hmm, that still seems low. Maybe I'm overcomplicating it.Alternatively, let's think of it as a reduced sample space.Originally, there are two possibilities:1. No check: probability 1/2.2. Check on one day: probability 1/2, with each day having probability 1/5.Given that the homework hasn't been checked from Monday to Thursday, we can eliminate the possibility of the check being on Monday to Thursday.So, the remaining possibilities are:- No check: probability 1/2.- Check on Friday: probability (1/2)*(1/5) = 1/10.So, the total probability is 1/2 + 1/10 = 6/10 = 3/5.Therefore, the probability that the homework is checked on Friday is (1/10) / (3/5) = 1/6.Wait, that seems consistent.But intuitively, if the teacher is going to check the homework, and it hasn't been checked from Monday to Thursday, then the probability it's checked on Friday should be higher than 1/6.Wait, maybe I'm not accounting for the fact that the teacher might not check at all.So, the probability that the teacher checks on Friday is 1/10, and the probability that the teacher doesn't check at all is 1/2.So, the probability that the homework is checked on Friday given that it hasn't been checked from Monday to Thursday is 1/10 divided by (1/2 + 1/10) = 1/10 / 6/10 = 1/6.Yes, that seems correct.Alternatively, if we think of it as a conditional probability where we have two possibilities: either the teacher doesn't check at all, or the teacher checks on Friday.Given that the homework hasn't been checked from Monday to Thursday, the probability that it's checked on Friday is 1/6.So, the final answer is 1/6.

💡Okay, so I have this geometry problem here, and I need to figure out the minimum value of the dot product of vectors CA and CB. Let me try to break it down step by step.First, let me understand the given information. There's a circle O with radius 2. Points P and Q are on this circle, and the angle between them, ∠POQ, is 60 degrees. AB is a diameter of the circle, so points A and B are at opposite ends of the circle, each 2 units away from O. Now, point C is defined by the vector equation: OC = (λ - 1) OP + λ OQ, where λ is a real number. I need to find the minimum value of the dot product CA · CB.Hmm, okay. So, let me recall some vector properties. The dot product of two vectors CA and CB can be expressed in terms of their magnitudes and the cosine of the angle between them. But since I don't know the angle, maybe it's better to express everything in terms of vectors from the origin O.Let me write down the expressions for vectors CA and CB. Vector CA = OA - OCVector CB = OB - OCSo, the dot product CA · CB is (OA - OC) · (OB - OC). Let me expand this:CA · CB = OA · OB - OA · OC - OB · OC + OC · OCSimplify that:= OA · OB - OC · (OA + OB) + |OC|²Now, since AB is a diameter, OA and OB are in opposite directions. So, OA + OB = 0. That simplifies things a bit.So, CA · CB = OA · OB + |OC|²Wait, OA · OB is the dot product of two vectors of length 2 with an angle of 180 degrees between them because AB is a diameter. So, OA · OB = |OA||OB|cos(180°) = 2*2*(-1) = -4.So, CA · CB = -4 + |OC|²Therefore, to minimize CA · CB, I need to minimize |OC|² because -4 is a constant. So, the problem reduces to finding the minimum value of |OC|².Given that OC = (λ - 1) OP + λ OQ. Let me write that as OC = (λ - 1) OP + λ OQ. Maybe I can factor this expression:OC = λ (OP + OQ) - OPHmm, that might not be immediately helpful. Alternatively, I can write OC as a linear combination of OP and OQ. Since OP and OQ are vectors from the origin to points P and Q on the circle, each with magnitude 2, and the angle between them is 60 degrees.Let me denote OP as vector p and OQ as vector q. So, |p| = |q| = 2, and the angle between p and q is 60 degrees. Therefore, p · q = |p||q|cos(60°) = 2*2*(1/2) = 2.So, OC = (λ - 1)p + λ q = λ(p + q) - pLet me compute |OC|²:|OC|² = |λ(p + q) - p|²Let me expand this:= [λ(p + q) - p] · [λ(p + q) - p]= λ²(p + q)·(p + q) - 2λ(p + q)·p + p·pCompute each term:First term: λ²(p + q)·(p + q) = λ²(|p|² + 2p·q + |q|²) = λ²(4 + 4 + 4) = λ²(12) because p·q = 2, so 2*2=4, but wait, p·q is 2, so 2p·q is 4.Wait, let me recast that:(p + q)·(p + q) = |p|² + 2p·q + |q|² = 4 + 2*2 + 4 = 4 + 4 + 4 = 12. So, first term is 12λ².Second term: -2λ(p + q)·p = -2λ(p·p + q·p) = -2λ(4 + 2) = -2λ(6) = -12λ.Third term: p·p = |p|² = 4.So, putting it all together:|OC|² = 12λ² - 12λ + 4So, |OC|² is a quadratic function in terms of λ: 12λ² - 12λ + 4.To find the minimum value, since the coefficient of λ² is positive, the parabola opens upwards, so the minimum occurs at the vertex.The vertex of a quadratic aλ² + bλ + c is at λ = -b/(2a). Here, a = 12, b = -12.So, λ = -(-12)/(2*12) = 12/24 = 0.5.So, the minimum occurs at λ = 0.5.Now, let's compute |OC|² at λ = 0.5:|OC|² = 12*(0.5)² - 12*(0.5) + 4 = 12*(0.25) - 6 + 4 = 3 - 6 + 4 = 1.So, the minimum value of |OC|² is 1.Therefore, going back to CA · CB = -4 + |OC|² = -4 + 1 = -3.So, the minimum value of CA · CB is -3.Wait, but let me double-check my steps to make sure I didn't make a mistake.Starting from OC = (λ - 1)p + λ q.Expressed as OC = λ(p + q) - p.Then, |OC|² = |λ(p + q) - p|².Expanding that, we have:= [λ(p + q) - p] · [λ(p + q) - p]= λ²(p + q)·(p + q) - 2λ(p + q)·p + p·pWe found that (p + q)·(p + q) = 12, (p + q)·p = 6, and p·p = 4.So, substituting:= 12λ² - 12λ + 4Yes, that seems correct.Then, the quadratic 12λ² - 12λ + 4 has its minimum at λ = 0.5, and plugging that in gives |OC|² = 1.Therefore, CA · CB = -4 + 1 = -3.So, the minimum value is -3, which corresponds to option C.I think that's correct. I don't see any errors in my calculations.

💡Okay, so I need to find the equation of a parabola. The problem says that the vertex is at the origin, and the focus is on the y-axis. The directrix is given by the equation y = 3. Hmm, let me think about how to approach this.First, I remember that a parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. Since the vertex is at the origin, that should be the midpoint between the focus and the directrix. Given that the focus is on the y-axis, it must have coordinates (0, p) where p is some value. The directrix is given as y = 3, which is a horizontal line. Since the vertex is at the origin, it should be halfway between the focus and the directrix. So, the distance from the vertex to the focus should be equal to the distance from the vertex to the directrix.Let me visualize this. The vertex is at (0,0). The focus is at (0, p), and the directrix is the line y = 3. The distance from the vertex to the focus is |p - 0| = |p|. The distance from the vertex to the directrix is |3 - 0| = 3. Since these distances should be equal, I can set up the equation |p| = 3. But wait, since the focus is on the y-axis and the directrix is above the vertex, the focus should be below the vertex, right? Because the parabola opens downward if the focus is below the vertex and the directrix is above it. So, p should be negative. Therefore, p = -3.Now, I recall that the standard form of a parabola that opens upward or downward with vertex at the origin is x² = 4py. Since p is negative, the parabola opens downward. Plugging p = -3 into the equation, we get x² = 4*(-3)y, which simplifies to x² = -12y.Wait, but let me double-check. The standard form is x² = 4py, where p is the distance from the vertex to the focus. If p is negative, the parabola opens downward. So, yes, x² = -12y should be the correct equation.But hold on, the problem says the focus is on the y-axis, which it is at (0, -3), and the directrix is y = 3, which is 3 units above the vertex. The distance from the vertex to the focus is 3 units downward, which matches the distance to the directrix. So, everything seems consistent.I think I got it. The equation of the parabola is x² = -12y.

💡Okay, so I have this probability problem here about seven envelopes and seven people. Each envelope has a letter for a different person, and the mail is being randomly distributed. I need to find the probability that exactly four people get the right letter. Hmm, okay, let me try to break this down.First, I know that when dealing with probabilities involving permutations, especially where things are being distributed randomly, it often involves combinations and derangements. A derangement is a permutation where none of the objects appear in their original position. I remember that concept from somewhere, but I'm a bit fuzzy on the details. Maybe I should look that up or think through it carefully.So, the problem is about exactly four people getting their correct letters. That means four people get the right letter, and the remaining three do not. So, I need to calculate the number of ways this can happen and then divide that by the total number of possible ways the letters can be distributed.Let me start by figuring out the total number of ways to distribute the seven letters. Since each person gets one letter, and there are seven letters, this is simply 7 factorial, which is 7! So, 7! equals 5040. That's the total number of possible distributions.Now, for the number of favorable outcomes, where exactly four people get the correct letter. To find this, I think I need to use combinations and derangements. First, I need to choose which four people out of the seven will receive the correct letters. The number of ways to choose four people out of seven is given by the combination formula, which is "7 choose 4," written as C(7,4).Calculating C(7,4), that's 7! divided by (4! times (7-4)!), which simplifies to 35. So, there are 35 ways to choose the four people who get the correct letters.Now, the remaining three people must not get their correct letters. This is where derangements come into play. A derangement of n objects is the number of permutations where none of the objects appear in their original position. The formula for derangements, often denoted as !n, is n! times the sum from k=0 to n of (-1)^k divided by k factorial. For n=3, this would be !3.Let me calculate !3. Using the formula, it's 3! times (1 - 1/1! + 1/2! - 1/3!). Calculating that step by step: 3! is 6. Then, inside the parentheses, 1 minus 1 is 0, plus 1/2 is 0.5, minus 1/6 is approximately 0.1667. So, 0.5 minus 0.1667 is about 0.3333. Multiplying 6 by 0.3333 gives approximately 2. So, !3 is 2.Therefore, the number of derangements for the remaining three letters is 2. So, for each of the 35 ways to choose the four correct recipients, there are 2 ways to distribute the remaining three letters such that none of them get their correct letter.Multiplying these together, the number of favorable outcomes is 35 times 2, which is 70.Now, to find the probability, I take the number of favorable outcomes divided by the total number of possible outcomes. That's 70 divided by 5040. Simplifying that fraction, 70 divided by 5040 is the same as 1 divided by 72. So, the probability is 1/72.Wait, let me double-check that. So, 7! is 5040, and 70 divided by 5040 is indeed 1/72. That seems right. I think I did everything correctly. I used combinations to choose the four correct people, then derangements for the remaining three, multiplied them together, and then divided by the total number of permutations. Yeah, that makes sense.I guess another way to think about it is that if you fix four letters correctly, the remaining three have to be all wrong, which is a derangement. And since derangements for three items are only two, it's a small number compared to the total permutations, which is why the probability is so low, just 1/72.I don't think I made any mistakes here. Maybe I should verify the derangement calculation again. For n=3, !3 is 2. Let's list them out to be sure. Suppose the three letters are A, B, and C, and their correct recipients are 1, 2, and 3 respectively. The derangements would be:1. A goes to 2, B goes to 3, C goes to 1.2. A goes to 3, B goes to 1, C goes to 2.Yes, that's two derangements. So, !3 is indeed 2. Okay, that checks out.So, putting it all together, the probability is 1/72. I think that's the correct answer.

💡Okay, so I have this problem where I'm given two logarithmic expressions, a and b, defined as follows:a = log base 5 of (3^x + 4^x)b = log base 4 of (5^x - 3^x)And I'm told that a is greater than or equal to b. I need to compare the sizes of a and x. Hmm, interesting. Let me try to unpack this step by step.First, let's write down what a and b represent. a is the logarithm base 5 of the sum of 3^x and 4^x. So, if I rewrite that in exponential form, it would be 5^a = 3^x + 4^x.Similarly, b is the logarithm base 4 of (5^x - 3^x). So, in exponential form, that would be 4^b = 5^x - 3^x.Okay, so now I have two equations:1. 5^a = 3^x + 4^x2. 4^b = 5^x - 3^xAnd I know that a >= b. I need to figure out how a and x relate in terms of size.Let me see. Maybe I can add these two equations together? Let's try that.Adding equation 1 and equation 2:5^a + 4^b = (3^x + 4^x) + (5^x - 3^x)Simplify the right-hand side:3^x + 4^x + 5^x - 3^x = 4^x + 5^xSo, 5^a + 4^b = 4^x + 5^xHmm, that's interesting. So, 5^a + 4^b equals 4^x + 5^x.Now, since a >= b, and both a and b are exponents, maybe I can find a relationship between 5^a and 4^b.Wait, 5^a is equal to 3^x + 4^x, and 4^b is equal to 5^x - 3^x. So, if I plug those back into the equation, I get:(3^x + 4^x) + (5^x - 3^x) = 4^x + 5^xWhich simplifies to 4^x + 5^x = 4^x + 5^x, which is just an identity. That doesn't help me much.Maybe I need to approach this differently. Let's think about the functions involved.Consider the function f(t) = 4^t + 5^t. This function is strictly increasing because both 4^t and 5^t are exponential functions with bases greater than 1, so they are increasing. Therefore, f(t) is also increasing.From equation 1 and 2, we have:5^a = 3^x + 4^x4^b = 5^x - 3^xIf I add these two equations, I get:5^a + 4^b = 4^x + 5^xBut since f(t) = 4^t + 5^t is increasing, and we know that a >= b, can I relate this to x?Wait, let's see. If a >= b, then 5^a >= 5^b because 5^t is increasing. Similarly, 4^b <= 4^a because 4^t is increasing.So, 5^a + 4^b <= 5^a + 4^aBut from earlier, 5^a + 4^b = 4^x + 5^xTherefore, 4^x + 5^x <= 5^a + 4^aBut since f(t) is increasing, if 4^x + 5^x <= 4^a + 5^a, then x <= a.Is that correct? Let me verify.Yes, because f(t) is strictly increasing, so if f(x) <= f(a), then x <= a.Therefore, x <= a.Wait, but let me make sure I didn't make a leap there. So, f(t) = 4^t + 5^t is strictly increasing, so if f(x) <= f(a), then x <= a. That makes sense because for increasing functions, if the function value at x is less than or equal to the function value at a, then x must be less than or equal to a.So, putting it all together:Given a >= b,We have 5^a + 4^b = 4^x + 5^xSince a >= b, 4^b <= 4^a and 5^a >= 5^bTherefore, 5^a + 4^b <= 5^a + 4^aBut 5^a + 4^b = 4^x + 5^xSo, 4^x + 5^x <= 5^a + 4^aSince f(t) = 4^t + 5^t is strictly increasing, this implies x <= a.Therefore, the conclusion is that x is less than or equal to a.I think that makes sense. Let me just recap to make sure I didn't miss anything.1. Start with definitions of a and b in exponential form.2. Add the two equations to get 5^a + 4^b = 4^x + 5^x.3. Recognize that f(t) = 4^t + 5^t is strictly increasing.4. Use the fact that a >= b to establish that 5^a + 4^b <= 5^a + 4^a.5. Therefore, 4^x + 5^x <= 5^a + 4^a.6. Since f is increasing, this implies x <= a.Yes, that seems solid. I don't see any gaps in the logic. So, the answer should be that x is less than or equal to a.

💡Okay, so I have this problem about a geometric sequence. Let me try to figure it out step by step. The problem says that in the geometric sequence {a_n}, it's known that a_5 = -2, and we need to find the product of the first 9 terms of this sequence. The options are A: 512, B: -512, C: 256, D: -256.First, let me recall what a geometric sequence is. A geometric sequence is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as r. So, the nth term of a geometric sequence can be written as a_n = a_1 * r^(n-1), where a_1 is the first term.Given that a_5 = -2, I can write that as:a_5 = a_1 * r^(5-1) = a_1 * r^4 = -2.Now, the question is asking for the product of the first 9 terms. So, I need to find a_1 * a_2 * a_3 * a_4 * a_5 * a_6 * a_7 * a_8 * a_9.Let me write each term in terms of a_1 and r:a_1 = a_1a_2 = a_1 * ra_3 = a_1 * r^2a_4 = a_1 * r^3a_5 = a_1 * r^4 = -2a_6 = a_1 * r^5a_7 = a_1 * r^6a_8 = a_1 * r^7a_9 = a_1 * r^8So, the product P = a_1 * a_2 * a_3 * a_4 * a_5 * a_6 * a_7 * a_8 * a_9.Let me substitute each term with the expressions in terms of a_1 and r:P = (a_1) * (a_1 * r) * (a_1 * r^2) * (a_1 * r^3) * (a_1 * r^4) * (a_1 * r^5) * (a_1 * r^6) * (a_1 * r^7) * (a_1 * r^8)Let me count how many a_1 terms there are. Each term from a_1 to a_9 has one a_1, so that's 9 a_1s. So, that's a_1^9.Now, for the r terms, let's add up the exponents:0 (from a_1) + 1 (from a_2) + 2 (from a_3) + 3 (from a_4) + 4 (from a_5) + 5 (from a_6) + 6 (from a_7) + 7 (from a_8) + 8 (from a_9).So, the exponents of r are 0 through 8. The sum of exponents is 0+1+2+3+4+5+6+7+8.I can calculate that: 0+1 is 1, 1+2 is 3, 3+3 is 6, 6+4 is 10, 10+5 is 15, 15+6 is 21, 21+7 is 28, 28+8 is 36.So, the total exponent for r is 36. Therefore, P = a_1^9 * r^36.Hmm, that seems a bit complicated. Maybe there's a smarter way to approach this.Wait, I remember that in a geometric sequence, the product of terms equidistant from the ends is equal. For example, a_1 * a_9 = a_2 * a_8 = a_3 * a_7 = a_4 * a_6 = (a_5)^2.Is that true? Let me check. Since it's a geometric sequence, a_k * a_{n+1 -k} = (a_1 * r^{k-1}) * (a_1 * r^{n -k}) ) = a_1^2 * r^{n -1}.But in this case, n is 9, so a_k * a_{10 -k} = a_1^2 * r^{9 -1} = a_1^2 * r^8.Wait, but a_5 is the middle term, so a_5 = a_1 * r^4. So, (a_5)^2 = (a_1 * r^4)^2 = a_1^2 * r^8.Which is the same as a_k * a_{10 -k} for k=1 to 4.So, indeed, each pair multiplies to (a_5)^2.Therefore, the product of all 9 terms can be thought of as the product of these 4 pairs and the middle term a_5.So, P = (a_1 * a_9) * (a_2 * a_8) * (a_3 * a_7) * (a_4 * a_6) * a_5.Since each pair is (a_5)^2, and there are 4 pairs, so that's (a_5)^8, and then multiplied by a_5, so total is (a_5)^9.Therefore, P = (a_5)^9.Given that a_5 = -2, so P = (-2)^9.Calculating that, (-2)^9 is -512 because 2^9 is 512 and since the exponent is odd, it's negative.So, the product is -512, which is option B.Wait, let me double-check my reasoning. I used the property that in a geometric sequence, the product of terms equidistant from the ends is equal, which is (a_5)^2 for each pair. Then, since there are 4 such pairs and the middle term a_5, the total product is (a_5)^9. That seems correct.Alternatively, if I go back to the initial approach where P = a_1^9 * r^36, and since a_5 = a_1 * r^4 = -2, then a_1 = (-2) / r^4.Substituting into P, we get [(-2)/r^4]^9 * r^36 = (-2)^9 / r^{36} * r^{36} = (-2)^9 = -512.So, both methods give the same result, which is reassuring.Therefore, the answer is B: -512.**Final Answer**boxed{B}