💡Okay, so I have this problem about a tetrahedron PQRS. A tetrahedron has four triangular faces, right? Each face is a triangle, so all the edges must satisfy the triangle inequality. The edges given are 8, 15, 21, 29, 35, and 45 units. They tell me that the length of edge PQ is 45 units, and I need to find the length of edge RS. The options are A) 8, B) 15, C) 21, D) 29, E) 35.Alright, so first, let me visualize the tetrahedron. It has four vertices: P, Q, R, S. Each pair of vertices is connected by an edge, so there are six edges in total. We know all six edge lengths: 8, 15, 21, 29, 35, 45. PQ is given as 45, which is the longest edge. So, PQ = 45.Now, I need to assign the remaining lengths to the other edges: PR, PS, QR, QS, and RS. The goal is to figure out which of these is RS.Since PQ is 45, which is the longest edge, the other edges connected to P and Q must be such that the triangle inequality holds for each face.Let me list the remaining edge lengths: 8, 15, 21, 29, 35.So, we have to assign these to PR, PS, QR, QS, RS.Let me think about the triangle inequality. For any triangle, the sum of any two sides must be greater than the third side.First, let's consider triangle PQR. It has edges PQ, PR, and QR. Since PQ is 45, the sum of PR and QR must be greater than 45. So, PR + QR > 45.Looking at the remaining edges: 8, 15, 21, 29, 35.We need two edges that add up to more than 45. Let's see:- 8 + 35 = 43, which is less than 45. So that doesn't work.- 15 + 35 = 50, which is greater than 45. That works.- 21 + 29 = 50, which is also greater than 45. That works.- 21 + 35 = 56, which is greater than 45.- 29 + 35 = 64, which is also greater than 45.So, possible pairs for PR and QR are (15,35), (21,29), (21,35), (29,35). But we can't use all of them because we have to assign the remaining edges to PS, QS, and RS.Let me try assigning the largest possible remaining edges to PR and QR first because they need to satisfy the triangle inequality with PQ=45.If I take PR=35 and QR=29, then PR + QR = 64, which is greater than 45. Alternatively, PR=35 and QR=21 would give 56, which is also good. Similarly, PR=29 and QR=21 gives 50.But let's see if we can find a consistent assignment.Suppose I assign PR=21 and QR=29. Then, PR + QR = 50 > 45, which is good. Then, the remaining edges are 8, 15, 35.These remaining edges need to be assigned to PS, QS, and RS.Now, let's look at triangle PQS. It has edges PQ=45, PS, and QS. So, PS + QS must be greater than 45.Looking at the remaining edges: 8, 15, 35.Possible pairs:- 8 + 35 = 43 < 45, which doesn't work.- 15 + 35 = 50 > 45, which works.So, if I assign PS=15 and QS=35, then PS + QS = 50 > 45, satisfying the triangle inequality.That leaves RS=8.Now, let's check the other triangles to make sure everything works.First, triangle PRS: It has edges PR=21, PS=15, and RS=8.Check triangle inequalities:- 21 + 15 > 8: 36 > 8, which is true.- 21 + 8 > 15: 29 > 15, which is true.- 15 + 8 > 21: 23 > 21, which is true.Good.Next, triangle QRS: It has edges QR=29, QS=35, and RS=8.Check triangle inequalities:- 29 + 35 > 8: 64 > 8, true.- 29 + 8 > 35: 37 > 35, true.- 35 + 8 > 29: 43 > 29, true.Good.Lastly, triangle PQS: PQ=45, PS=15, QS=35.Check:- 45 + 15 > 35: 60 > 35, true.- 45 + 35 > 15: 80 > 15, true.- 15 + 35 > 45: 50 > 45, true.Everything checks out.So, in this assignment, RS=8.Wait, but let me make sure there isn't another possible assignment. Maybe if I assign PR=35 and QR=21, would that work?Let's try that.Assign PR=35, QR=21. Then, PR + QR = 56 > 45, which is good.Remaining edges: 8, 15, 29.Now, assign PS, QS, RS.Looking at triangle PQS: edges PQ=45, PS, QS.We need PS + QS > 45.Possible pairs:- 8 + 29 = 37 < 45, doesn't work.- 15 + 29 = 44 < 45, doesn't work.- 8 + 15 = 23 < 45, doesn't work.So, none of the pairs add up to more than 45. That means this assignment doesn't work because triangle PQS would violate the triangle inequality.Therefore, PR=35 and QR=21 is not a valid assignment.Similarly, if I try PR=35 and QR=15, then PR + QR = 50 > 45, but then remaining edges are 8, 21, 29.Assigning to PS, QS, RS.Triangle PQS: PS + QS > 45.Possible pairs:- 8 + 29 = 37 < 45- 21 + 29 = 50 > 45So, assign PS=21, QS=29.Then, RS=8.Check triangle PRS: PR=35, PS=21, RS=8.Check:- 35 + 21 > 8: 56 > 8, true.- 35 + 8 > 21: 43 > 21, true.- 21 + 8 > 35: 29 > 35? No, that's false.So, this assignment doesn't work because 21 + 8 is not greater than 35.Therefore, PR=35 and QR=15 is invalid.Next, try PR=29 and QR=21.Then, PR + QR = 50 > 45, good.Remaining edges: 8, 15, 35.Assign to PS, QS, RS.Triangle PQS: PS + QS > 45.Possible pairs:- 15 + 35 = 50 > 45.So, assign PS=15, QS=35, RS=8.Check triangle PRS: PR=29, PS=15, RS=8.Check:- 29 + 15 > 8: 44 > 8, true.- 29 + 8 > 15: 37 > 15, true.- 15 + 8 > 29: 23 > 29? No, false.So, again, this assignment fails because 15 + 8 is not greater than 29.Hmm, so that doesn't work either.Wait, so if I try PR=29 and QR=15, then PR + QR=44 < 45, which violates the triangle inequality for triangle PQR. So that's invalid.Similarly, PR=15 and QR=35 would give PR + QR=50 > 45, but then remaining edges are 8, 21, 29.Assign to PS, QS, RS.Triangle PQS: PS + QS > 45.Possible pairs:- 21 + 29 = 50 > 45.Assign PS=21, QS=29, RS=8.Check triangle PRS: PR=15, PS=21, RS=8.Check:- 15 + 21 > 8: 36 > 8, true.- 15 + 8 > 21: 23 > 21, true.- 21 + 8 > 15: 29 > 15, true.Good.Now, check triangle QRS: QR=35, QS=29, RS=8.Check:- 35 + 29 > 8: 64 > 8, true.- 35 + 8 > 29: 43 > 29, true.- 29 + 8 > 35: 37 > 35, true.Good.Triangle PQS: PQ=45, PS=21, QS=29.Check:- 45 + 21 > 29: 66 > 29, true.- 45 + 29 > 21: 74 > 21, true.- 21 + 29 > 45: 50 > 45, true.Everything works here too.So, in this case, RS=8 as well.Wait, so whether I assign PR=21, QR=29 or PR=15, QR=35, I end up with RS=8.Is there another way to assign?Let me see.If I assign PR=8, QR=35, then PR + QR=43 < 45, which violates the triangle inequality for triangle PQR. So that's invalid.Similarly, PR=8, QR=29: 8 + 29=37 <45, invalid.PR=8, QR=21: 29 <45, invalid.PR=8, QR=15: 23 <45, invalid.So, PR cannot be 8.Therefore, the only valid assignments are when PR and QR are either (21,29) or (15,35), both leading to RS=8.Hence, RS must be 8.I think that's the answer. Let me just recap.Given PQ=45, the other edges must be assigned such that all triangle inequalities are satisfied. The only way to do that without violating any inequalities is to have RS=8.So, the answer is 8, which is option A.

💡Okay, so I have this problem here about finding the domain of the inverse function of ( f(x) = 3^x ) where the domain of ( f(x) ) is given as ( 0 < x leq 2 ). The options are A: ( (0, +infty) ), B: ( (1, 9] ), C: ( (0, 1) ), and D: ( [9, +infty) ). Hmm, let me try to figure this out step by step.First, I remember that for a function and its inverse, the domain of the inverse function is the range of the original function. So, if I can find the range of ( f(x) = 3^x ) when ( x ) is between 0 and 2, that should give me the domain of the inverse function.Alright, let's think about ( f(x) = 3^x ). I know that exponential functions like this are always increasing because the base, 3 in this case, is greater than 1. So, as ( x ) increases, ( f(x) ) increases as well. That means the smallest value of ( f(x) ) will be when ( x ) is smallest, and the largest value will be when ( x ) is largest.Given that ( x ) is in the interval ( (0, 2] ), let's plug in the endpoints to find the range. When ( x = 0 ), ( f(0) = 3^0 = 1 ). But wait, the domain is ( 0 < x leq 2 ), so ( x = 0 ) isn't actually included. That means the function doesn't actually reach 1; it just approaches it. So, the lower bound of the range is just above 1.Next, when ( x = 2 ), ( f(2) = 3^2 = 9 ). Since ( x = 2 ) is included in the domain, the function does reach 9. Therefore, the upper bound of the range is 9.Putting this together, the range of ( f(x) ) is all real numbers greater than 1 and up to and including 9. In interval notation, that's ( (1, 9] ).So, since the domain of the inverse function is the range of the original function, the domain of ( f^{-1}(x) ) is ( (1, 9] ). Looking back at the options, that's option B.Wait, let me double-check. If I consider the inverse function, it should take the outputs of ( f(x) ) as its inputs. Since ( f(x) ) outputs values from just above 1 up to 9, the inverse function must accept inputs in that same interval. Yeah, that makes sense. So, the domain of the inverse is ( (1, 9] ).I don't think I made any mistakes here. The function is strictly increasing, so the range is straightforward. The endpoints were handled correctly, considering whether they're included or not based on the original domain. So, I feel confident that option B is the right answer.**Final Answer**boxed{B}

💡Okay, so I need to find the coefficient of ( x^2 ) in the expansion of ( (1 - x)^5 ). Hmm, I remember that when expanding binomials like this, the binomial theorem is really useful. The binomial theorem says that ( (a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k ). In this case, ( a ) is 1 and ( b ) is ( -x ), and ( n ) is 5. So, applying the binomial theorem, the expansion of ( (1 - x)^5 ) would be ( sum_{k=0}^{5} binom{5}{k} (1)^{5 - k} (-x)^k ). Simplifying that, since ( (1)^{5 - k} ) is just 1 for any ( k ), the expansion becomes ( sum_{k=0}^{5} binom{5}{k} (-x)^k ). Now, I need to find the term where the power of ( x ) is 2. That means I need to look at the term where ( k = 2 ). So, let's calculate that term. First, the binomial coefficient ( binom{5}{2} ). I remember that ( binom{n}{k} = frac{n!}{k!(n - k)!} ). Plugging in the values, ( binom{5}{2} = frac{5!}{2!3!} ). Calculating that, ( 5! = 120 ), ( 2! = 2 ), and ( 3! = 6 ). So, ( binom{5}{2} = frac{120}{2 times 6} = frac{120}{12} = 10 ).Next, the term also includes ( (-x)^2 ). Since ( (-x)^2 = (-1)^2 x^2 = 1 times x^2 = x^2 ). Putting it all together, the term for ( k = 2 ) is ( 10 times x^2 ). Therefore, the coefficient of ( x^2 ) is 10.Wait, let me double-check that. If I expand ( (1 - x)^5 ) manually, would the coefficient of ( x^2 ) indeed be 10? Let's see:( (1 - x)^5 = 1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5 ). Yes, the coefficient of ( x^2 ) is 10. So, I think I did it correctly.

💡Okay, so I have this problem with two parts, both involving inequalities and logical conditions. Let me try to break it down step by step.Starting with part (1): We are given two inequalities, p and q. p is defined as x² - 7x + 10 < 0, and q is (x - m)(x - 3m) < 0, where m is 4. We need to find the range of x where both p and q are true, meaning their conjunction p ∧ q is true.First, I'll tackle the inequality for p: x² - 7x + 10 < 0. To solve this quadratic inequality, I should factor the quadratic expression. Let me see, the quadratic is x² - 7x + 10. I need two numbers that multiply to 10 and add up to -7. Hmm, -2 and -5. So, the factored form is (x - 2)(x - 5) < 0.Now, to find where this product is less than zero, I can use a sign chart. The roots are x = 2 and x = 5. These divide the number line into three intervals: (-∞, 2), (2, 5), and (5, ∞). I'll test each interval:1. For x < 2, say x = 1: (1 - 2)(1 - 5) = (-1)(-4) = 4 > 0.2. For 2 < x < 5, say x = 3: (3 - 2)(3 - 5) = (1)(-2) = -2 < 0.3. For x > 5, say x = 6: (6 - 2)(6 - 5) = (4)(1) = 4 > 0.So, the inequality (x - 2)(x - 5) < 0 holds true for 2 < x < 5. Therefore, p is true when x is between 2 and 5.Next, let's look at q: (x - m)(x - 3m) < 0 with m = 4. Substituting m = 4, we get (x - 4)(x - 12) < 0. Again, this is a quadratic inequality. The roots are x = 4 and x = 12. These divide the number line into three intervals: (-∞, 4), (4, 12), and (12, ∞). Testing each interval:1. For x < 4, say x = 0: (0 - 4)(0 - 12) = (-4)(-12) = 48 > 0.2. For 4 < x < 12, say x = 8: (8 - 4)(8 - 12) = (4)(-4) = -16 < 0.3. For x > 12, say x = 13: (13 - 4)(13 - 12) = (9)(1) = 9 > 0.So, the inequality (x - 4)(x - 12) < 0 holds true for 4 < x < 12. Therefore, q is true when x is between 4 and 12.Now, since we need both p and q to be true, we have to find the intersection of the intervals where p is true and where q is true. The interval for p is (2, 5) and for q is (4, 12). The overlap between these two intervals is from 4 to 5. So, the range of x where both p and q are true is 4 < x < 5.Moving on to part (2): We are told that ¬q is a sufficient and necessary condition for ¬p. This means that ¬q if and only if ¬p. In logical terms, this is equivalent to saying that p if and only if q. So, p and q must have the same truth values for all x.First, let's recall what p and q represent. From part (1), we know that p is true for 2 < x < 5. For q, we have (x - m)(x - 3m) < 0. Since m > 0, the roots of q are x = m and x = 3m. The interval where q is true depends on the values of m.If m is positive, then m < 3m because m is positive. So, the inequality (x - m)(x - 3m) < 0 will hold true for m < x < 3m.Now, since ¬q is a necessary and sufficient condition for ¬p, it means that whenever ¬p is true, ¬q is also true, and vice versa. In other words, the intervals where p is false must exactly match the intervals where q is false.Let's find where ¬p is true. p is true for 2 < x < 5, so ¬p is true for x ≤ 2 or x ≥ 5.Similarly, q is true for m < x < 3m, so ¬q is true for x ≤ m or x ≥ 3m.Since ¬q must be equivalent to ¬p, their intervals must be the same. Therefore, the intervals where ¬q is true (x ≤ m or x ≥ 3m) must be exactly the same as where ¬p is true (x ≤ 2 or x ≥ 5).This gives us two conditions:1. The lower bound of ¬q must match the lower bound of ¬p: m must be equal to 2.2. The upper bound of ¬q must match the upper bound of ¬p: 3m must be equal to 5.Wait, that would mean m = 2 and 3m = 5. But if m = 2, then 3m = 6, which is not equal to 5. That's a contradiction. Hmm, maybe I need to approach this differently.Perhaps instead of setting m = 2 and 3m = 5, I need to ensure that the intervals x ≤ m and x ≥ 3m cover exactly x ≤ 2 and x ≥ 5. So, for ¬q to be equivalent to ¬p, the intervals where ¬q is true must be subsets of the intervals where ¬p is true and vice versa.This means that m must be less than or equal to 2, so that x ≤ m is within x ≤ 2. Similarly, 3m must be greater than or equal to 5, so that x ≥ 3m is within x ≥ 5.So, we have two inequalities:1. m ≤ 22. 3m ≥ 5Solving the second inequality: 3m ≥ 5 ⇒ m ≥ 5/3 ≈ 1.6667.So, combining both inequalities, m must be between 5/3 and 2. Therefore, the range of m is [5/3, 2].Let me double-check this. If m is 5/3, then 3m is 5. So, ¬q would be x ≤ 5/3 or x ≥ 5. But ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 5/3 is a subset of x ≤ 2, and x ≥ 5 is the same. So, ¬q is a subset of ¬p, but not exactly equal. Wait, that might not satisfy the necessary and sufficient condition.Wait, actually, for ¬q to be equivalent to ¬p, the intervals must be exactly the same. So, if m = 5/3, then ¬q is x ≤ 5/3 or x ≥ 5. But ¬p is x ≤ 2 or x ≥ 5. These are not exactly the same because x ≤ 5/3 is a subset of x ≤ 2. So, to make them exactly the same, m must be equal to 2 and 3m must be equal to 5. But as I saw earlier, that's impossible because m can't be both 2 and 5/3 at the same time.Hmm, maybe I need to adjust my approach. Perhaps the intervals where ¬q is true must cover exactly the intervals where ¬p is true. That is, the lower bound of ¬q (m) must be equal to the lower bound of ¬p (2), and the upper bound of ¬q (3m) must be equal to the upper bound of ¬p (5). But as I saw, that would require m = 2 and 3m = 5, which is impossible.Alternatively, maybe the intervals where ¬q is true must be exactly the same as ¬p, meaning that m must be 2 and 3m must be 5. But again, that's impossible because m can't satisfy both m = 2 and 3m = 5 simultaneously.Wait, perhaps I'm overcomplicating this. Let's think about it differently. If ¬q is a necessary and sufficient condition for ¬p, then the truth sets of ¬q and ¬p must be identical. That is, the set of x where ¬q is true is exactly the same as the set of x where ¬p is true.So, ¬q is x ≤ m or x ≥ 3m, and ¬p is x ≤ 2 or x ≥ 5. For these to be identical, the intervals must match exactly. Therefore, m must be 2 and 3m must be 5. But as before, m can't be both 2 and 5/3 at the same time.This seems like a contradiction, which suggests that there might be no solution. But that can't be right because the problem asks for the range of m. So, perhaps I'm missing something.Wait, maybe the intervals don't have to match exactly, but rather, the condition is that whenever ¬q is true, ¬p is true, and whenever ¬p is true, ¬q is true. So, it's a bi-conditional: ¬q ↔ ¬p.This means that the intervals where ¬q is true must be exactly the same as where ¬p is true. Therefore, the lower bound of ¬q (m) must be equal to the lower bound of ¬p (2), and the upper bound of ¬q (3m) must be equal to the upper bound of ¬p (5). So, m = 2 and 3m = 5. But as before, m can't be both 2 and 5/3.Wait, maybe m can be such that the intervals where ¬q is true are exactly the same as where ¬p is true. So, if m = 2, then ¬q is x ≤ 2 or x ≥ 6. But ¬p is x ≤ 2 or x ≥ 5. These are not the same because x ≥ 6 is a subset of x ≥ 5, but x ≥ 5 includes more values than x ≥ 6.Similarly, if m = 5/3, then ¬q is x ≤ 5/3 or x ≥ 5. But ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 5/3 is a subset of x ≤ 2, but x ≤ 2 includes more values than x ≤ 5/3.Therefore, to make ¬q exactly equal to ¬p, we need m = 2 and 3m = 5, which is impossible. So, perhaps there is no solution? But the problem states that m > 0, so maybe I'm missing something.Wait, maybe I need to consider that the intervals where ¬q is true must cover exactly the intervals where ¬p is true. That is, the lower bound of ¬q (m) must be greater than or equal to the lower bound of ¬p (2), and the upper bound of ¬q (3m) must be less than or equal to the upper bound of ¬p (5). But that would mean m ≥ 2 and 3m ≤ 5, which implies m ≥ 2 and m ≤ 5/3. But 2 > 5/3, so there's no overlap. That can't be right either.I'm getting confused here. Let me try to visualize this.¬p is x ≤ 2 or x ≥ 5.¬q is x ≤ m or x ≥ 3m.For ¬q to be equivalent to ¬p, the intervals must be the same. So, x ≤ m must be x ≤ 2, and x ≥ 3m must be x ≥ 5.Therefore, m must be 2, and 3m must be 5. But m can't be both 2 and 5/3. So, this is impossible. Therefore, there is no such m that satisfies this condition.But the problem says "find the range of real number m," implying that there is a solution. So, perhaps I'm misunderstanding the condition.Wait, maybe the condition is that ¬q is a sufficient and necessary condition for ¬p, which means that ¬q implies ¬p and ¬p implies ¬q. So, they are equivalent.Therefore, the sets where ¬q is true must be exactly the same as where ¬p is true. So, x ≤ m or x ≥ 3m must be exactly x ≤ 2 or x ≥ 5.This requires that m = 2 and 3m = 5. But as before, m can't be both 2 and 5/3. So, perhaps the only way this can happen is if m is such that the intervals overlap in a way that they effectively cover the same regions.Wait, maybe m is between 5/3 and 2. Let me test m = 5/3. Then, ¬q is x ≤ 5/3 or x ≥ 5. But ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 5/3 is a subset of x ≤ 2, and x ≥ 5 is the same. So, ¬q is a subset of ¬p. But for equivalence, they need to be exactly the same. So, m needs to be such that x ≤ m is x ≤ 2 and x ≥ 3m is x ≥ 5. Therefore, m = 2 and 3m = 5. But again, m can't be both.Wait, maybe m can be between 5/3 and 2. Let's say m is 1.8, which is between 5/3 (~1.6667) and 2. Then, 3m = 5.4. So, ¬q is x ≤ 1.8 or x ≥ 5.4. But ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 1.8 is a subset of x ≤ 2, and x ≥ 5.4 is a subset of x ≥ 5. Therefore, ¬q is a subset of ¬p, but not the other way around. So, ¬q implies ¬p, but ¬p does not imply ¬q because there are values of x where ¬p is true but ¬q is false, like x = 1.9 or x = 5.2.Therefore, to have ¬q equivalent to ¬p, we need that whenever ¬q is true, ¬p is true, and whenever ¬p is true, ¬q is true. So, the intervals must match exactly.Therefore, m must be such that x ≤ m is exactly x ≤ 2, and x ≥ 3m is exactly x ≥ 5. So, m must be 2 and 3m must be 5. But m can't be both 2 and 5/3. Therefore, there is no such m that satisfies both conditions simultaneously.But the problem states that m > 0, so perhaps I'm missing something. Maybe the intervals don't have to match exactly, but rather, the condition is that ¬q is a subset of ¬p and ¬p is a subset of ¬q, which would require that m ≤ 2 and 3m ≥ 5. So, m ≤ 2 and m ≥ 5/3. Therefore, m must be between 5/3 and 2.Wait, that makes sense. If m is between 5/3 and 2, then:- For m = 5/3, ¬q is x ≤ 5/3 or x ≥ 5. But ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 5/3 is a subset of x ≤ 2, and x ≥ 5 is the same. Therefore, ¬q is a subset of ¬p.- For m = 2, ¬q is x ≤ 2 or x ≥ 6. But ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 2 is the same, and x ≥ 6 is a subset of x ≥ 5. Therefore, ¬q is a subset of ¬p.But for m between 5/3 and 2, say m = 1.8, then ¬q is x ≤ 1.8 or x ≥ 5.4. And ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 1.8 is a subset of x ≤ 2, and x ≥ 5.4 is a subset of x ≥ 5. Therefore, ¬q is a subset of ¬p.But for equivalence, we need that ¬q is exactly ¬p. So, to have ¬q exactly equal to ¬p, we need that m = 2 and 3m = 5, which is impossible. Therefore, perhaps the problem is asking for m such that ¬q is equivalent to ¬p, which would require that m = 2 and 3m = 5, but since that's impossible, maybe the problem is asking for m such that ¬q is a subset of ¬p and ¬p is a subset of ¬q, which would require that m ≤ 2 and 3m ≥ 5, leading to m between 5/3 and 2.Yes, that makes sense. So, the range of m is [5/3, 2].Let me verify this. If m is 5/3, then ¬q is x ≤ 5/3 or x ≥ 5. But ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 5/3 is a subset of x ≤ 2, and x ≥ 5 is the same. Therefore, ¬q is a subset of ¬p.If m is 2, then ¬q is x ≤ 2 or x ≥ 6. But ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 2 is the same, and x ≥ 6 is a subset of x ≥ 5. Therefore, ¬q is a subset of ¬p.For m between 5/3 and 2, say m = 1.8, then ¬q is x ≤ 1.8 or x ≥ 5.4. And ¬p is x ≤ 2 or x ≥ 5. So, x ≤ 1.8 is a subset of x ≤ 2, and x ≥ 5.4 is a subset of x ≥ 5. Therefore, ¬q is a subset of ¬p.But for equivalence, we need that ¬q is exactly ¬p. So, perhaps the problem is asking for m such that ¬q is equivalent to ¬p, which would require that m = 2 and 3m = 5, but since that's impossible, the next best thing is that ¬q is a subset of ¬p and ¬p is a subset of ¬q, which would require that m ≤ 2 and 3m ≥ 5, leading to m between 5/3 and 2.Yes, that seems to be the correct approach. Therefore, the range of m is [5/3, 2].

💡To solve the equation ( a^{2} + b^{2} = (a + b)^{2} + text{____} ), I start by expanding the right-hand side.First, expand ( (a + b)^{2} ):[(a + b)^{2} = a^{2} + 2ab + b^{2}]Now, substitute this back into the original equation:[a^{2} + b^{2} = a^{2} + 2ab + b^{2} + text{____}]To isolate the missing term, subtract ( a^{2} + b^{2} ) from both sides:[0 = 2ab + text{____}]Solving for the missing term gives:[text{____} = -2ab]Therefore, the missing term is (-2ab).

💡First, I need to determine the relationship between the cost price (CP), marked price (MP), and selling price (SP) after a discount.The shopkeeper aims for an 8% gain on the cost price, so the selling price should be:[ SP = 1.08 times CP ]A discount of 16.92307692307692% is applied to the marked price, meaning the selling price is 83.07692307692308% of the marked price:[ SP = 0.8307692307692308 times MP ]Equating the two expressions for SP:[ 1.08 times CP = 0.8307692307692308 times MP ]Expressing the marked price as a percentage above the cost price:[ MP = CP times (1 + P%) ]Substituting this into the equation:[ 1.08 times CP = 0.8307692307692308 times CP times (1 + P%) ]Dividing both sides by CP:[ 1.08 = 0.8307692307692308 times (1 + P%) ]Solving for P%:[ frac{1.08}{0.8307692307692308} = 1 + P% ][ 1.2999999999999998 approx 1 + P% ]Subtracting 1 from both sides:[ P% approx 0.2999999999999998 ]Converting to a percentage:[ P% approx 29.99999999999998% ]Therefore, the marked price is fixed at approximately 30% above the cost price.