💡Okay, so I have this problem about triangle ABC where the sides opposite angles A, B, and C are a, b, and c respectively. There's an inequality given: (x^{2}cos C + 2xsin C + frac{3}{2} geqslant 0) which holds for all real numbers x. Part (1) asks for the range of values for (cos C). Hmm, okay. So I need to figure out what constraints this inequality puts on (cos C). First, I remember that for a quadratic inequality (ax^2 + bx + c geq 0) to hold for all real x, the quadratic must be always non-negative. That means two things: the leading coefficient a must be positive, and the discriminant must be less than or equal to zero. So, applying this to the given inequality:The quadratic is (x^{2}cos C + 2xsin C + frac{3}{2}). So, comparing to (ax^2 + bx + c), here:- (a = cos C)- (b = 2sin C)- (c = frac{3}{2})So, for the quadratic to be non-negative for all x, two conditions must be satisfied:1. (a > 0) which means (cos C > 0)2. The discriminant (b^2 - 4ac leq 0)Let me compute the discriminant:(b^2 - 4ac = (2sin C)^2 - 4 cdot cos C cdot frac{3}{2})Simplify that:(4sin^2 C - 6cos C leq 0)So, (4sin^2 C - 6cos C leq 0). Hmm, I can rewrite (sin^2 C) in terms of (cos C) using the identity (sin^2 C = 1 - cos^2 C).Substituting that in:(4(1 - cos^2 C) - 6cos C leq 0)Expanding:(4 - 4cos^2 C - 6cos C leq 0)Let me rearrange this:(-4cos^2 C - 6cos C + 4 leq 0)Multiply both sides by -1 (remember to reverse the inequality sign):(4cos^2 C + 6cos C - 4 geq 0)So, now we have a quadratic inequality in terms of (cos C):(4cos^2 C + 6cos C - 4 geq 0)Let me solve this quadratic inequality. First, find the roots of the quadratic equation (4cos^2 C + 6cos C - 4 = 0).Using the quadratic formula:(cos C = frac{-6 pm sqrt{36 + 64}}{8}) because the quadratic is (4cos^2 C + 6cos C - 4), so a=4, b=6, c=-4.Calculating the discriminant:(36 + 64 = 100), so square root is 10.Thus:(cos C = frac{-6 pm 10}{8})So, two solutions:1. (cos C = frac{-6 + 10}{8} = frac{4}{8} = frac{1}{2})2. (cos C = frac{-6 - 10}{8} = frac{-16}{8} = -2)But (cos C) can't be -2 because the cosine function only takes values between -1 and 1. So, the only valid root is (frac{1}{2}).Now, the quadratic (4cos^2 C + 6cos C - 4) opens upwards (since the coefficient of (cos^2 C) is positive). So, the inequality (4cos^2 C + 6cos C - 4 geq 0) is satisfied when (cos C leq -2) or (cos C geq frac{1}{2}). But since (cos C) can't be less than -1, the only valid interval is (cos C geq frac{1}{2}).But wait, earlier we had the condition that (cos C > 0). So combining these, we have (cos C geq frac{1}{2}).But in a triangle, angle C is between 0 and π, so (cos C) is between -1 and 1. But since (cos C geq frac{1}{2}), angle C must be less than or equal to π/3 (since cosine decreases as the angle increases from 0 to π). So, the range of (cos C) is (frac{1}{2} leq cos C < 1). Wait, why is it less than 1? Because if (cos C = 1), then angle C would be 0, which isn't possible in a triangle. So, yeah, (cos C) is between 1/2 and 1, not including 1.So, part (1) is done. The range is (frac{1}{2} leq cos C < 1).Moving on to part (2). It says when angle C is at its maximum value, and the perimeter of triangle ABC is 9, find the maximum area of triangle ABC.First, from part (1), we know that (cos C) is at least 1/2, so the maximum value of angle C is when (cos C) is minimized, which is 1/2. So, angle C is π/3 (60 degrees) because (cos(pi/3) = 1/2.So, angle C is π/3. Now, the perimeter is 9, so a + b + c = 9.We need to maximize the area of triangle ABC. The area can be given by (frac{1}{2}absin C). Since angle C is π/3, (sin C = sqrt{3}/2). So, area = (frac{sqrt{3}}{4}ab).So, to maximize the area, we need to maximize ab, given that a + b + c = 9 and angle C is π/3.But we need to express c in terms of a and b. Using the Law of Cosines:(c^2 = a^2 + b^2 - 2abcos C)Since C is π/3, (cos C = 1/2), so:(c^2 = a^2 + b^2 - ab)So, c = (sqrt{a^2 + b^2 - ab})So, the perimeter is a + b + (sqrt{a^2 + b^2 - ab}) = 9.We need to maximize ab with this constraint.Hmm, this seems a bit complicated. Maybe we can use some inequality here. Let me think.Let me denote s = a + b. Then, the perimeter is s + c = 9, so c = 9 - s.But c is also equal to (sqrt{a^2 + b^2 - ab}). So,(sqrt{a^2 + b^2 - ab} = 9 - s)Squaring both sides:(a^2 + b^2 - ab = (9 - s)^2 = 81 - 18s + s^2)But s = a + b, so s^2 = a^2 + 2ab + b^2. So,Left side: a^2 + b^2 - abRight side: 81 - 18s + a^2 + 2ab + b^2Subtract left side from both sides:0 = 81 - 18s + a^2 + 2ab + b^2 - (a^2 + b^2 - ab)Simplify:0 = 81 - 18s + 3abSo,3ab = 18s - 81Divide both sides by 3:ab = 6s - 27So, ab = 6s - 27.But we need to express ab in terms of s, and then perhaps find the maximum.But we also know that in a triangle, the sides must satisfy the triangle inequality. So, a + b > c, which is 9 - s. So,s > 9 - sWhich implies 2s > 9, so s > 4.5.Also, since c > 0, 9 - s > 0, so s < 9.So, s is between 4.5 and 9.But we also have the expression ab = 6s - 27. So, to maximize ab, we need to maximize s, since ab increases as s increases.But s is less than 9, so the maximum ab would be when s approaches 9, but s can't be 9 because then c would be 0, which isn't possible.Wait, but maybe there's another constraint. Let's think about the relationship between a and b.We have ab = 6s - 27, and s = a + b.So, ab = 6(a + b) - 27Let me write this as ab - 6a - 6b + 27 = 0Hmm, maybe factor this.ab - 6a - 6b + 27 = 0Add 36 to both sides:ab - 6a - 6b + 63 = 36Hmm, not sure. Maybe try to express it as (a - 6)(b - 6) = something.Let me see:ab - 6a - 6b + 36 = 9So, (a - 6)(b - 6) = 9Yes, because:(a - 6)(b - 6) = ab - 6a - 6b + 36 = 9So, (a - 6)(b - 6) = 9Interesting. So, the product of (a - 6) and (b - 6) is 9.Now, since a and b are sides of a triangle, they must be positive. Also, since s = a + b is between 4.5 and 9, a and b are each less than 9.But let's see, (a - 6) and (b - 6) multiply to 9. So, both (a - 6) and (b - 6) are positive or both negative.But since a + b = s < 9, and a and b are positive, each of a and b is less than 9. So, a - 6 and b - 6 could be positive or negative.But if (a - 6)(b - 6) = 9, which is positive, so both factors are positive or both negative.Case 1: Both (a - 6) and (b - 6) are positive.Then, a > 6 and b > 6. But since a + b < 9, each of a and b is less than 9, but greater than 6. So, 6 < a, b < 9, but a + b < 9. Wait, but if a > 6 and b > 6, then a + b > 12, which contradicts a + b < 9. So, this case is impossible.Case 2: Both (a - 6) and (b - 6) are negative.So, a < 6 and b < 6.Then, (a - 6)(b - 6) = 9Let me denote x = 6 - a and y = 6 - b, so x and y are positive.Then, ( -x)( -y) = xy = 9So, xy = 9.Also, s = a + b = (6 - x) + (6 - y) = 12 - (x + y)But s must be less than 9, so 12 - (x + y) < 9 => x + y > 3Also, since a and b are positive, 6 - x > 0 => x < 6, similarly y < 6.So, x and y are positive numbers less than 6, with x + y > 3 and xy = 9.We need to find the maximum value of ab.But ab = (6 - x)(6 - y) = 36 - 6x - 6y + xy = 36 - 6(x + y) + 9 = 45 - 6(x + y)So, ab = 45 - 6(x + y)We need to maximize ab, which is equivalent to minimizing (x + y).Given that xy = 9, and x + y > 3, with x, y < 6.We know from AM ≥ GM that x + y ≥ 2√(xy) = 2√9 = 6.But x + y must be greater than 3, but the minimum of x + y is 6 (when x = y = 3). Wait, but if x + y = 6, then ab = 45 - 6*6 = 45 - 36 = 9.But if x + y is greater than 6, ab becomes less than 9. So, the maximum ab is 9 when x + y is minimized at 6.But wait, x + y = 6 is achievable when x = y = 3, so a = 6 - 3 = 3, b = 6 - 3 = 3.So, a = b = 3.So, when a = b = 3, then c = 9 - a - b = 9 - 6 = 3.Wait, so triangle ABC would be equilateral? Because all sides are 3.But angle C was given as π/3, which is 60 degrees, which is consistent with an equilateral triangle.So, in this case, the area would be (frac{sqrt{3}}{4} times 3 times 3 = frac{9sqrt{3}}{4}).Is this the maximum area? Let me check.If a and b are not equal, say a = 4, then b would be such that (a - 6)(b - 6) = 9.If a = 4, then (4 - 6)(b - 6) = (-2)(b - 6) = 9 => b - 6 = -4.5 => b = 1.5Then, a = 4, b = 1.5, c = 9 - 4 - 1.5 = 3.5Check if this satisfies the triangle inequality: 4 + 1.5 > 3.5? 5.5 > 3.5, yes. 4 + 3.5 > 1.5? 7.5 > 1.5, yes. 1.5 + 3.5 > 4? 5 > 4, yes. So, it's a valid triangle.Compute ab = 4 * 1.5 = 6, which is less than 9. So, the area would be (frac{sqrt{3}}{4} * 6 = frac{3sqrt{3}}{2}), which is less than (frac{9sqrt{3}}{4}).Similarly, if a = 5, then (5 - 6)(b - 6) = (-1)(b - 6) = 9 => b - 6 = -9 => b = -3, which is invalid because side lengths can't be negative.So, the maximum ab occurs when a = b = 3, giving ab = 9, and area (frac{9sqrt{3}}{4}).Therefore, the maximum area is (frac{9sqrt{3}}{4}).**Final Answer**(1) The range of (cos C) is (boxed{left[ dfrac{1}{2}, 1 right)}).(2) The maximum area of (triangle ABC) is (boxed{dfrac{9sqrt{3}}{4}}).

💡Okay, so I have this permutation problem here, and I need to represent it as a product of disjoint cycles. Let me see... The set A is {1, 2, 3, 4, 5, 6, 7, 8, 9}, and the permutation f is given in a two-row notation. The top row is the original positions, and the bottom row is where each element gets mapped to. So, f is:f = 1 2 3 4 5 6 7 8 93 4 7 6 9 2 1 8 5Alright, so I need to figure out the cycles. I remember that to find cycles, I start with an element and follow where it maps until I come back to the starting element. Then, I move to the next element that hasn't been included in a cycle yet and repeat the process.Let me start with 1. So, f(1) is 3. Then, I need to see where 3 maps to. Looking at the permutation, f(3) is 7. Next, where does 7 go? f(7) is 1. So, we've come back to 1. That means we've completed a cycle: 1 → 3 → 7 → 1. So, that's the cycle (1 3 7).Now, let's move to the next element not in the first cycle, which is 2. f(2) is 4. Then, f(4) is 6. Next, f(6) is 2. So, we've completed another cycle: 2 → 4 → 6 → 2. That's the cycle (2 4 6).Next, the next element not in the previous cycles is 5. f(5) is 9. Then, f(9) is 5. So, that's a cycle of length 2: 5 → 9 → 5, which is (5 9).Finally, the last element is 8. f(8) is 8, so it maps to itself. That's a trivial cycle, which we can write as (8), but sometimes it's omitted since it doesn't change anything.So, putting it all together, the permutation f can be written as the product of disjoint cycles: (1 3 7)(2 4 6)(5 9)(8). Since (8) is a trivial cycle, it might be left out, but it's still technically part of the permutation.Let me double-check to make sure these cycles are disjoint and cover all elements from 1 to 9. The first cycle includes 1, 3, 7. The second includes 2, 4, 6. The third includes 5, 9. And the last one is 8. Yep, all elements are accounted for, and none overlap. So, I think that's correct.I also remember that the order of cycles doesn't matter in the product, as long as they are disjoint. So, writing them in any order is fine. But usually, people write them starting from the smallest element or in the order they were found. I think starting from 1 makes sense here.Just to be thorough, let me verify each cycle:- For (1 3 7): f(1)=3, f(3)=7, f(7)=1. Correct.- For (2 4 6): f(2)=4, f(4)=6, f(6)=2. Correct.- For (5 9): f(5)=9, f(9)=5. Correct.- For (8): f(8)=8. Correct.Everything checks out. So, I'm confident that the permutation f is correctly represented as the product of these disjoint cycles.

💡First, I need to determine the truck's speed in feet per second. The truck travels (frac{b}{4}) feet every (t) seconds, so its speed is (frac{b}{4t}) feet per second.Next, I convert the total travel time from minutes to seconds. Since there are 60 seconds in a minute, 4 minutes is (4 times 60 = 240) seconds.Now, I calculate the total distance traveled in feet by multiplying the speed by the time:[text{Distance in feet} = frac{b}{4t} times 240 = frac{240b}{4t} = frac{60b}{t}]Finally, I convert the distance from feet to yards. There are 3 feet in a yard, so:[text{Distance in yards} = frac{60b}{t} div 3 = frac{20b}{t}]

💡First, I need to find the coordinates of point ( D ), which is the midpoint of segment ( overline{AB} ). Given that ( A ) is at ( (0, 8) ) and ( B ) is at ( (0, 0) ), I can use the midpoint formula:[D = left( frac{0 + 0}{2}, frac{8 + 0}{2} right) = (0, 4)]Next, I have the coordinates of points ( C ) at ( (10, 0) ) and ( D ) at ( (0, 4) ). To find the slope of the line passing through these two points, I use the slope formula:[text{slope} = frac{4 - 0}{0 - 10} = frac{4}{-10} = -0.4]The ( y )-intercept of the line is the ( y )-coordinate of point ( D ), which is ( 4 ).Finally, I add the slope and the ( y )-intercept:[-0.4 + 4 = 3.6]

💡First, I need to determine the effective speed of the boat when it is moving downstream. Since the boat is moving with the current, the effective speed is the sum of the boat's speed in still water and the speed of the current.The boat's speed in still water is 20 km/hr, and the current's speed is 5 km/hr. Adding these together gives an effective downstream speed of 25 km/hr.Next, I need to calculate the distance traveled downstream. The boat traveled for 12 minutes, but the speed is given in hours. To match the units, I'll convert 12 minutes to hours by dividing by 60, which results in 0.2 hours.Finally, I'll use the formula for distance, which is speed multiplied by time. Multiplying the effective downstream speed of 25 km/hr by the time of 0.2 hours gives the distance traveled downstream.

💡Okay, so I have this question about geometric sequences, and I need to figure out which of the given statements is always true. Let me start by recalling 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 times r^{n-1} ]where ( a_1 ) is the first term.Now, the question gives four options, labeled A to D, each stating that certain terms form a geometric sequence. I need to check each one to see if they always form a geometric sequence, regardless of the original sequence's terms or common ratio.Let me go through each option one by one.**Option A: ( a_1, a_3, a_9 ) form a geometric sequence.**First, let's write out what each term is in terms of ( a_1 ) and r.- ( a_1 = a_1 )- ( a_3 = a_1 times r^{2} )- ( a_9 = a_1 times r^{8} )Now, to check if these three terms form a geometric sequence, the ratio between consecutive terms should be constant. So, let's compute the ratio between ( a_3 ) and ( a_1 ), and then between ( a_9 ) and ( a_3 ).- Ratio between ( a_3 ) and ( a_1 ): ( frac{a_3}{a_1} = frac{a_1 r^2}{a_1} = r^2 )- Ratio between ( a_9 ) and ( a_3 ): ( frac{a_9}{a_3} = frac{a_1 r^8}{a_1 r^2} = r^{6} )For these to form a geometric sequence, these two ratios should be equal. So, we have:[ r^2 = r^6 ]This simplifies to:[ r^6 - r^2 = 0 ][ r^2 (r^4 - 1) = 0 ][ r^2 = 0 quad text{or} quad r^4 = 1 ]So, either ( r = 0 ) or ( r = pm 1 ). However, in a geometric sequence, if ( r = 0 ), all terms after the first would be zero, which is a trivial case. If ( r = 1 ), all terms are equal, which is also a specific case. But for other values of r, this equality doesn't hold. For example, if ( r = 2 ), then ( r^2 = 4 ) and ( r^6 = 64 ), which are not equal. Therefore, option A is not always true.**Option B: ( a_2, a_3, a_6 ) form a geometric sequence.**Again, let's express each term:- ( a_2 = a_1 times r^{1} )- ( a_3 = a_1 times r^{2} )- ( a_6 = a_1 times r^{5} )Now, check the ratios:- Ratio between ( a_3 ) and ( a_2 ): ( frac{a_3}{a_2} = frac{a_1 r^2}{a_1 r} = r )- Ratio between ( a_6 ) and ( a_3 ): ( frac{a_6}{a_3} = frac{a_1 r^5}{a_1 r^2} = r^{3} )For these to form a geometric sequence, the ratios must be equal:[ r = r^3 ][ r^3 - r = 0 ][ r(r^2 - 1) = 0 ][ r = 0 quad text{or} quad r = pm 1 ]Again, similar to option A, this only holds for specific values of r. For example, if ( r = 2 ), then the first ratio is 2 and the second is 8, which are not equal. Therefore, option B is not always true.**Option C: ( a_2, a_4, a_8 ) form a geometric sequence.**Expressing each term:- ( a_2 = a_1 times r^{1} )- ( a_4 = a_1 times r^{3} )- ( a_8 = a_1 times r^{7} )Checking the ratios:- Ratio between ( a_4 ) and ( a_2 ): ( frac{a_4}{a_2} = frac{a_1 r^3}{a_1 r} = r^{2} )- Ratio between ( a_8 ) and ( a_4 ): ( frac{a_8}{a_4} = frac{a_1 r^7}{a_1 r^3} = r^{4} )Setting these equal:[ r^2 = r^4 ][ r^4 - r^2 = 0 ][ r^2 (r^2 - 1) = 0 ][ r^2 = 0 quad text{or} quad r^2 = 1 ][ r = 0 quad text{or} quad r = pm 1 ]Same issue as before. For other values of r, like ( r = 2 ), the ratios would be 4 and 16, which aren't equal. So, option C isn't always true.**Option D: ( a_3, a_6, a_9 ) form a geometric sequence.**Let's write out the terms:- ( a_3 = a_1 times r^{2} )- ( a_6 = a_1 times r^{5} )- ( a_9 = a_1 times r^{8} )Now, compute the ratios:- Ratio between ( a_6 ) and ( a_3 ): ( frac{a_6}{a_3} = frac{a_1 r^5}{a_1 r^2} = r^{3} )- Ratio between ( a_9 ) and ( a_6 ): ( frac{a_9}{a_6} = frac{a_1 r^8}{a_1 r^5} = r^{3} )Here, both ratios are equal to ( r^3 ). Therefore, regardless of the value of r (as long as it's not zero, which would make the sequence trivial), the ratios are the same. This means ( a_3, a_6, a_9 ) will always form a geometric sequence.Just to double-check, let me plug in a specific example. Suppose ( a_1 = 1 ) and ( r = 2 ).- ( a_3 = 1 times 2^{2} = 4 )- ( a_6 = 1 times 2^{5} = 32 )- ( a_9 = 1 times 2^{8} = 256 )Now, check the ratios:- ( 32 / 4 = 8 )- ( 256 / 32 = 8 )They are equal, so it works for r = 2. Let's try another ratio, say r = 3.- ( a_3 = 1 times 3^{2} = 9 )- ( a_6 = 1 times 3^{5} = 243 )- ( a_9 = 1 times 3^{8} = 6561 )Compute the ratios:- ( 243 / 9 = 27 )- ( 6561 / 243 = 27 )Again, equal ratios. What about a fractional ratio, like r = 1/2.- ( a_3 = 1 times (1/2)^2 = 1/4 )- ( a_6 = 1 times (1/2)^5 = 1/32 )- ( a_9 = 1 times (1/2)^8 = 1/256 )Compute the ratios:- ( (1/32) / (1/4) = (1/32) * (4/1) = 1/8 )- ( (1/256) / (1/32) = (1/256) * (32/1) = 1/8 )Still equal. Even if r is negative, say r = -2.- ( a_3 = 1 times (-2)^2 = 4 )- ( a_6 = 1 times (-2)^5 = -32 )- ( a_9 = 1 times (-2)^8 = 256 )Compute the ratios:- ( (-32) / 4 = -8 )- ( 256 / (-32) = -8 )Equal ratios again. So, regardless of the common ratio, positive, negative, fractional, or even 1, the ratios between ( a_3, a_6, a_9 ) are equal. Therefore, option D is always true.I think I've checked all the options and only D holds true in all cases. So, the correct answer should be D.**Final Answer**boxed{D}