💡First, I need to determine the number of times Jessy initially plans to read each day. Let's denote this number as ( x ).Jessy reads 6 pages each time she reads. Therefore, the total number of pages she plans to read each day is ( 6x ).Over a week, which has 7 days, the total number of pages she plans to read is ( 7 times 6x = 42x ).However, Jessy needs to read 2 more pages per day to achieve her goal of reading 140 pages in a week. This means she should actually read ( 6x + 2 ) pages each day.Over the week, the total number of pages she needs to read is ( 7 times (6x + 2) = 42x + 14 ).Setting up the equation to reflect her goal:[42x + 14 = 140]Subtracting 14 from both sides:[42x = 126]Dividing both sides by 42:[x = 3]Therefore, Jessy initially plans to read 3 times daily.

💡Okay, so I have this problem about two sequences, an arithmetic sequence {a_n} and a geometric sequence {b_n}. Both start with the same first term, which is 2. Then there are some conditions given: a_3 + a_5 equals 22, and for the geometric sequence, b_2 multiplied by b_4 equals b_6. Part I asks for the general formulas of both sequences. Let me start by recalling what arithmetic and geometric sequences are. An arithmetic sequence has a common difference between consecutive terms. So, if the first term is a_1, then the nth term is a_n = a_1 + (n-1)d, where d is the common difference. A geometric sequence, on the other hand, has a common ratio between consecutive terms. So, the nth term is b_n = b_1 * r^(n-1), where r is the common ratio.Given that a_1 = b_1 = 2, so both sequences start at 2. First, let's tackle the arithmetic sequence. We know that a_3 + a_5 = 22. Let me write expressions for a_3 and a_5 using the formula for arithmetic sequences.a_3 = a_1 + 2d = 2 + 2da_5 = a_1 + 4d = 2 + 4dSo, adding them together: a_3 + a_5 = (2 + 2d) + (2 + 4d) = 4 + 6dWe are told that this equals 22, so:4 + 6d = 22Subtract 4 from both sides: 6d = 18Divide both sides by 6: d = 3So, the common difference d is 3. Therefore, the general formula for {a_n} is:a_n = a_1 + (n-1)d = 2 + (n-1)*3Let me simplify that: 2 + 3n - 3 = 3n - 1So, a_n = 3n - 1Okay, that seems straightforward. Now, moving on to the geometric sequence {b_n}. We know that b_2 * b_4 = b_6.Let me write expressions for b_2, b_4, and b_6 using the geometric sequence formula.b_2 = b_1 * r = 2rb_4 = b_1 * r^3 = 2r^3b_6 = b_1 * r^5 = 2r^5So, according to the given condition:b_2 * b_4 = b_6Substituting the expressions:(2r) * (2r^3) = 2r^5Let me compute the left side: 2r * 2r^3 = 4r^4So, 4r^4 = 2r^5Hmm, let's solve for r. Let me divide both sides by 2r^4 (assuming r ≠ 0):(4r^4) / (2r^4) = (2r^5) / (2r^4)Simplify:2 = rSo, the common ratio r is 2. Therefore, the general formula for {b_n} is:b_n = b_1 * r^(n-1) = 2 * 2^(n-1) = 2^nWait, let me check that. 2 * 2^(n-1) is equal to 2^(1 + n -1) = 2^n. Yes, that's correct.So, b_n = 2^nAlright, so part I is done. The general formulas are a_n = 3n - 1 and b_n = 2^n.Now, part II asks to let c_n = a_n - b_n and find the sum of the first n terms of {c_n}.So, first, let's write c_n:c_n = a_n - b_n = (3n - 1) - 2^nTherefore, the sum of the first n terms, let's denote it as S_n, is:S_n = sum_{k=1}^n c_k = sum_{k=1}^n (3k - 1 - 2^k)I can split this sum into three separate sums:S_n = sum_{k=1}^n (3k) - sum_{k=1}^n 1 - sum_{k=1}^n 2^kCompute each sum separately.First sum: sum_{k=1}^n 3k = 3 * sum_{k=1}^n kI know that sum_{k=1}^n k = n(n + 1)/2, so:3 * sum_{k=1}^n k = 3 * [n(n + 1)/2] = (3n(n + 1))/2Second sum: sum_{k=1}^n 1 = nThird sum: sum_{k=1}^n 2^kThis is a geometric series with first term 2 and common ratio 2. The sum of the first n terms of a geometric series is given by:sum_{k=1}^n ar^(k-1) = a*(r^n - 1)/(r - 1)But here, the series is sum_{k=1}^n 2^k, which is 2 + 4 + 8 + ... + 2^nThis can be written as 2*(2^n - 1)/(2 - 1) = 2*(2^n - 1)/1 = 2^(n+1) - 2So, sum_{k=1}^n 2^k = 2^(n+1) - 2Putting it all together:S_n = (3n(n + 1))/2 - n - (2^(n+1) - 2)Simplify term by term:First term: (3n(n + 1))/2Second term: -nThird term: -2^(n+1) + 2So, let's combine the first and second terms:(3n(n + 1))/2 - n = (3n(n + 1) - 2n)/2 = [3n^2 + 3n - 2n]/2 = (3n^2 + n)/2Then, subtract the third term:(3n^2 + n)/2 - 2^(n+1) + 2So, S_n = (3n^2 + n)/2 - 2^(n+1) + 2Alternatively, we can write this as:S_n = (3n^2 + n)/2 - 2^{n+1} + 2I think that's the final expression for the sum of the first n terms of {c_n}.Let me double-check my steps to make sure I didn't make any mistakes.1. For the arithmetic sequence, I found d = 3, so a_n = 3n - 1. That seems correct because a_3 = 8 and a_5 = 14, which add up to 22.2. For the geometric sequence, I found r = 2, so b_n = 2^n. Let's check the condition: b_2 = 4, b_4 = 16, so b_2 * b_4 = 64. b_6 = 64, which matches. So that's correct.3. For c_n = a_n - b_n, I correctly substituted the expressions.4. When summing c_n, I split it into three sums: 3k, -1, and -2^k. That's correct.5. Calculated each sum:- Sum of 3k is 3*(n(n+1)/2) = (3n(n+1))/2- Sum of 1's is n- Sum of 2^k is 2^(n+1) - 26. Combined all terms:(3n(n+1))/2 - n - (2^(n+1) - 2) = (3n^2 + n)/2 - 2^(n+1) + 2Yes, that seems right.So, I think my answer is correct.**Final Answer**Ⅰ. The general formulas are boxed{a_n = 3n - 1} and boxed{b_n = 2^n}.Ⅱ. The sum of the first n terms of {c_n} is boxed{dfrac{3n^2 + n}{2} - 2^{n+1} + 2}.

💡Okay, so I'm trying to figure out whether Baron Munchausen's claim is possible. He says he drew a polygon and a point inside it such that any line passing through this point divides the polygon into three polygons. Hmm, that sounds pretty interesting. Let me break it down step by step.First, I need to understand what it means for a line to divide a polygon into three polygons. Normally, when you draw a line through a polygon, it can split it into two parts. But here, it's supposed to split it into three. That seems unusual because a straight line can only intersect a polygon in at most two points, right? So, how can a single line create three separate regions?Maybe I'm missing something. Let me think about the structure of the polygon. If the polygon is convex, then any line through a point inside it will intersect the boundary at two points, dividing it into two parts. But if the polygon is concave or has some kind of star shape, maybe the line can intersect more than twice? Wait, no, even in a concave polygon, a straight line can only intersect the boundary at two points. So, that doesn't help.Hold on, maybe the point inside the polygon is special in some way. If the point is such that any line through it intersects the polygon's boundary in three points, then that could create three regions. But how is that possible? A straight line can only cross a polygon's boundary an even number of times, right? It enters and exits, so two points. Unless... unless the polygon is somehow self-intersecting or has multiple boundaries.Wait, but a polygon is a simple closed curve, meaning it doesn't intersect itself. So, it can't have multiple boundaries. Hmm, this is confusing. Maybe the polygon is not simple? But no, by definition, a polygon is a simple closed curve.Let me try to visualize this. Imagine a polygon with a point inside it. If I draw a line through that point, it should intersect the polygon's boundary at two points, dividing it into two parts. To get three parts, the line would have to somehow intersect the boundary three times, but that's impossible with a straight line.Unless... unless the polygon is designed in such a way that the line passes through the point and somehow creates three separate regions without intersecting the boundary more than twice. Maybe the point is at a vertex or something? But if the point is inside, it can't be a vertex.Wait, maybe the polygon has a hole in it. If the polygon is a multiply-connected region, like a donut shape, then a line through the center could intersect the outer boundary and the inner boundary, creating three regions: the outer part, the inner part, and the hole. But is that considered a single polygon? I think a polygon with a hole is still considered a single polygon, but it's a non-simple polygon.But the problem didn't specify whether the polygon is simple or not. If it's allowed to have holes, then maybe it's possible. Let me think about that. If the polygon has an inner boundary (a hole), then a line passing through the center could intersect both the outer and inner boundaries, creating three regions: the part of the polygon on one side of the line, the part on the other side, and the hole. But wait, the hole is already a separate region, so does that count as three polygons?I'm not sure. The original polygon is one, and the hole is another, but if the line intersects both boundaries, maybe it creates three separate regions. But I'm not entirely certain about this interpretation.Alternatively, maybe the polygon is designed in such a way that the line passes through the point and creates three separate polygons by intersecting the boundary at three points. But as I thought earlier, a straight line can only intersect a simple polygon at two points. So, unless the polygon is self-intersecting or has multiple boundaries, which it can't be if it's a simple polygon.Wait, maybe the polygon is star-shaped with respect to the point. In a star-shaped polygon, any line from the point to the boundary doesn't intersect the polygon again. But that still doesn't help because a straight line through the point would only intersect the boundary twice.I'm stuck here. Maybe I need to think differently. Perhaps the polygon is designed in such a way that the line through the point intersects the boundary at two points, but the way the polygon is shaped, it creates three separate regions. For example, if the polygon has a narrow neck or something, and the line cuts through it in a way that creates three parts.But I'm not sure how that would work. Let me try to draw it mentally. Imagine a polygon that's like a triangle with a smaller triangle attached to one side, creating a sort of bowtie shape. If I place a point in the middle, a line through that point could intersect both triangles, creating three regions. But wait, that's a self-intersecting polygon, which is not a simple polygon.So, if we allow self-intersecting polygons, maybe it's possible. But I think the problem assumes a simple polygon because it just says "a polygon." So, maybe the answer is no, it's not possible.But then again, the problem is posed by Baron Munchausen, who is known for his tall tales. So, maybe it's a trick question, and the answer is yes, but in a non-traditional sense.Wait, maybe the polygon is a regular polygon with an even number of sides, and the point is the center. Then, any line through the center would divide it into two congruent polygons. But that's only two, not three.Unless... unless the polygon is designed with multiple layers or something. But I don't think that's possible with a simple polygon.I'm going in circles here. Maybe I need to look up some properties of polygons and lines intersecting them. But since I'm trying to think through this, let me try one more approach.Suppose the polygon is such that the point is where three edges meet, but that would make it a vertex, not an interior point. So, that's not it.Alternatively, maybe the polygon is designed with a sort of tunnel or indentation that allows a line through the point to intersect three times. But again, a straight line can't intersect a simple polygon more than twice.Wait, unless the polygon is not convex. If it's concave, a line through the point could intersect the boundary twice, but the concave part might create an additional region. Hmm, maybe.Imagine a concave polygon where the line through the point intersects the boundary at two points, but due to the concave indentation, it creates three regions. Is that possible?Let me visualize a concave polygon, like a rectangle with a bite taken out of it. If I place a point inside and draw a line through it, it intersects the boundary at two points. Depending on the shape of the concave part, maybe it creates three regions. But I'm not sure if that's the case.I think I need to draw it out, but since I can't, I'll try to imagine it. If the concave part is such that the line passes through the concave area, maybe it creates an additional region. But I'm not certain.Alternatively, maybe the polygon is designed with multiple concave parts, so that a line through the point intersects three times. But again, a straight line can't intersect a simple polygon more than twice.Wait, unless the polygon is designed in such a way that the line passes through the point and intersects the boundary at two points, but the polygon's shape causes the line to create three separate regions. Maybe the polygon has a sort of hourglass shape, but that's self-intersecting.I'm really stuck here. Maybe the answer is no, it's not possible because a straight line can only intersect a simple polygon at two points, thus dividing it into at most two regions. Therefore, Baron Munchausen must be incorrect.But then again, maybe there's a special polygon where this is possible. I'm not sure. I think I need to conclude that it's not possible because of the limitations of straight lines intersecting polygons.Wait, but the problem says "any line passing through this point divides the polygon into three polygons." So, it's not just one line, but any line. That makes it even more impossible because if it's true for any line, then the polygon would have to be designed in a way that every possible line through the point intersects the boundary three times, which is impossible for a simple polygon.Therefore, I think the answer is no, Baron Munchausen cannot be correct because it's impossible for any line through a point inside a simple polygon to divide it into three polygons.

💡Okay, so I have this equation: (x^6 + 64 = 0). I need to find all the solutions, both real and complex. Hmm, let me think about how to approach this.First, I remember that solving equations like this often involves factoring or using complex numbers. Since it's a sixth-degree polynomial, there should be six solutions, right? According to the Fundamental Theorem of Algebra, a degree n polynomial has n roots in the complex plane, counting multiplicities. So, I should expect six solutions here.Looking at the equation (x^6 + 64 = 0), I can rewrite it as (x^6 = -64). That seems like a good start. Now, I need to find all the sixth roots of -64. I think complex numbers can help here because -64 is a real number, but it's negative, so its roots won't all be real.I remember that complex numbers can be expressed in polar form, which might make taking roots easier. The general form is (r(cos theta + i sin theta)), or (re^{itheta}) using Euler's formula. So, maybe I can write -64 in polar form.Let me recall that any complex number can be written as (re^{itheta}), where (r) is the modulus (or absolute value) and (theta) is the argument (or angle). For -64, the modulus is 64 because it's 64 units away from the origin on the real axis. But since it's negative, it's located on the negative real axis. The angle for the positive real axis is 0, so for the negative real axis, the angle should be (pi) radians, right? Because that's 180 degrees, pointing directly to the left on the complex plane.So, -64 can be written as (64e^{ipi}). But wait, angles in complex numbers are periodic with (2pi), so technically, it's also (64e^{i(pi + 2kpi)}) for any integer (k). That might be important when finding all the roots.Now, I need to find all the sixth roots of (64e^{ipi}). I remember that to find the nth roots of a complex number, you take the nth root of the modulus and divide the angle by n, then add increments of (2pi/n) to get all the distinct roots.So, let's apply that here. The modulus of -64 is 64, so the sixth root of 64 is... let me calculate that. 64 is (2^6), so the sixth root of 64 is 2. That's straightforward.Now, for the angles. The original angle is (pi), so dividing that by 6 gives (pi/6). But since angles are periodic, I also need to add increments of (2pi/6) or (pi/3) to get all six distinct roots. So, the general formula for the k-th root is:(x_k = 2e^{i(pi/6 + 2kpi/6)}) for (k = 0, 1, 2, 3, 4, 5).Simplifying the exponent, that becomes:(x_k = 2e^{i(pi/6 + kpi/3)}).So, plugging in the values of k from 0 to 5, I can find all six roots.Let me list them out:1. For (k = 0): (x_0 = 2e^{ipi/6})2. For (k = 1): (x_1 = 2e^{i(pi/6 + pi/3)} = 2e^{ipi/2})3. For (k = 2): (x_2 = 2e^{i(pi/6 + 2pi/3)} = 2e^{i5pi/6})4. For (k = 3): (x_3 = 2e^{i(pi/6 + 3pi/3)} = 2e^{i7pi/6})5. For (k = 4): (x_4 = 2e^{i(pi/6 + 4pi/3)} = 2e^{i3pi/2})6. For (k = 5): (x_5 = 2e^{i(pi/6 + 5pi/3)} = 2e^{i11pi/6})Okay, so these are the six roots in polar form. But the question asks for all solutions, real and otherwise. I think it's fine to leave them in polar form, but maybe I should convert them to rectangular form (a + bi) for clarity, especially since the final answer is supposed to be boxed.Let me recall how to convert from polar to rectangular form. For a complex number (re^{itheta}), the rectangular form is (rcostheta + irsintheta). So, I can compute the cosine and sine for each angle and multiply by 2.Let me compute each one step by step.1. (x_0 = 2e^{ipi/6}): - (cos(pi/6) = sqrt{3}/2) - (sin(pi/6) = 1/2) - So, (x_0 = 2(sqrt{3}/2 + i(1/2)) = sqrt{3} + i)2. (x_1 = 2e^{ipi/2}): - (cos(pi/2) = 0) - (sin(pi/2) = 1) - So, (x_1 = 2(0 + i(1)) = 2i)3. (x_2 = 2e^{i5pi/6}): - (cos(5pi/6) = -sqrt{3}/2) - (sin(5pi/6) = 1/2) - So, (x_2 = 2(-sqrt{3}/2 + i(1/2)) = -sqrt{3} + i)4. (x_3 = 2e^{i7pi/6}): - (cos(7pi/6) = -sqrt{3}/2) - (sin(7pi/6) = -1/2) - So, (x_3 = 2(-sqrt{3}/2 + i(-1/2)) = -sqrt{3} - i)5. (x_4 = 2e^{i3pi/2}): - (cos(3pi/2) = 0) - (sin(3pi/2) = -1) - So, (x_4 = 2(0 + i(-1)) = -2i)6. (x_5 = 2e^{i11pi/6}): - (cos(11pi/6) = sqrt{3}/2) - (sin(11pi/6) = -1/2) - So, (x_5 = 2(sqrt{3}/2 + i(-1/2)) = sqrt{3} - i)Alright, so converting each root to rectangular form, I get:1. (sqrt{3} + i)2. (2i)3. (-sqrt{3} + i)4. (-sqrt{3} - i)5. (-2i)6. (sqrt{3} - i)Let me double-check my calculations to make sure I didn't make any mistakes.Starting with (x_0): (pi/6) is 30 degrees. Cosine of 30 degrees is indeed (sqrt{3}/2), and sine is 1/2. Multiplying by 2 gives (sqrt{3} + i). That seems correct.(x_1): (pi/2) is 90 degrees. Cosine is 0, sine is 1. So, 2i is correct.(x_2): (5pi/6) is 150 degrees. Cosine is negative (sqrt{3}/2), sine is 1/2. So, (-sqrt{3} + i) is correct.(x_3): (7pi/6) is 210 degrees. Cosine is negative (sqrt{3}/2), sine is -1/2. So, (-sqrt{3} - i) is correct.(x_4): (3pi/2) is 270 degrees. Cosine is 0, sine is -1. So, -2i is correct.(x_5): (11pi/6) is 330 degrees. Cosine is (sqrt{3}/2), sine is -1/2. So, (sqrt{3} - i) is correct.Okay, all the rectangular forms look correct.Now, just to make sure, let me verify that these are indeed the roots of the equation (x^6 + 64 = 0).Take (x = sqrt{3} + i). Let's compute (x^6).First, find the modulus of (x). The modulus is (sqrt{(sqrt{3})^2 + (1)^2} = sqrt{3 + 1} = 2). So, modulus is 2, which is correct because we expect each root to have modulus 2.The argument of (x) is (pi/6), as we found earlier. So, (x = 2e^{ipi/6}). Then, (x^6 = (2)^6 e^{i6(pi/6)} = 64e^{ipi} = -64). So, (x^6 + 64 = 0), which checks out.Similarly, take (x = 2i). The modulus is 2, argument is (pi/2). So, (x^6 = (2)^6 e^{i6(pi/2)} = 64e^{i3pi} = 64(-1) = -64). So, (x^6 + 64 = 0), which is correct.Let me check one more, say (x = -sqrt{3} - i). The modulus is (sqrt{(-sqrt{3})^2 + (-1)^2} = sqrt{3 + 1} = 2). The argument is (7pi/6). So, (x^6 = (2)^6 e^{i6(7pi/6)} = 64e^{i7pi} = 64(-1) = -64). So, (x^6 + 64 = 0), which is correct.Alright, so all the roots satisfy the equation. Therefore, I can be confident that these are the correct solutions.In summary, the solutions to (x^6 + 64 = 0) are the six complex numbers:1. (sqrt{3} + i)2. (2i)3. (-sqrt{3} + i)4. (-sqrt{3} - i)5. (-2i)6. (sqrt{3} - i)These are all the sixth roots of -64, evenly spaced around the circle of radius 2 in the complex plane, each separated by an angle of (pi/3) radians.I think I've covered all the steps and checked my work, so I feel confident about this solution.**Final Answer**The solutions are (boxed{sqrt{3} + i}), (boxed{2i}), (boxed{-sqrt{3} + i}), (boxed{-sqrt{3} - i}), (boxed{-2i}), and (boxed{sqrt{3} - i}).

💡Alright, so I've got this problem here about a function and a triangle. Let me try to tackle it step by step. First, the function is given as ( f(x) = sin x cosleft(x + frac{pi}{6}right) + 1 ). I need to find where this function is monotonically decreasing. Hmm, okay. Monotonicity usually involves looking at the derivative. If the derivative is negative, the function is decreasing. So, I think I need to find the derivative of ( f(x) ) and then determine where it's negative.But before I jump into taking the derivative, maybe I can simplify the function a bit. The term ( sin x cosleft(x + frac{pi}{6}right) ) looks a bit complicated. Maybe I can use a trigonometric identity to simplify it. I remember that ( sin A cos B = frac{1}{2} [sin(A + B) + sin(A - B)] ). Let me try that.So, applying the identity:[sin x cosleft(x + frac{pi}{6}right) = frac{1}{2} left[ sinleft(x + x + frac{pi}{6}right) + sinleft(x - left(x + frac{pi}{6}right)right) right]]Simplifying the arguments inside the sine functions:[= frac{1}{2} left[ sinleft(2x + frac{pi}{6}right) + sinleft(-frac{pi}{6}right) right]]I know that ( sin(-theta) = -sintheta ), so:[= frac{1}{2} left[ sinleft(2x + frac{pi}{6}right) - sinleft(frac{pi}{6}right) right]]And since ( sinleft(frac{pi}{6}right) = frac{1}{2} ):[= frac{1}{2} sinleft(2x + frac{pi}{6}right) - frac{1}{4}]So, plugging this back into the original function:[f(x) = frac{1}{2} sinleft(2x + frac{pi}{6}right) - frac{1}{4} + 1 = frac{1}{2} sinleft(2x + frac{pi}{6}right) + frac{3}{4}]Alright, that's a simpler form. Now, to find where it's decreasing, I need the derivative. Let's compute ( f'(x) ):[f'(x) = frac{d}{dx} left[ frac{1}{2} sinleft(2x + frac{pi}{6}right) + frac{3}{4} right] = frac{1}{2} cdot 2 cosleft(2x + frac{pi}{6}right) = cosleft(2x + frac{pi}{6}right)]So, ( f'(x) = cosleft(2x + frac{pi}{6}right) ). For the function to be decreasing, ( f'(x) < 0 ). Therefore:[cosleft(2x + frac{pi}{6}right) < 0]When is cosine negative? Cosine is negative in the second and third quadrants, i.e., when the angle is between ( frac{pi}{2} ) and ( frac{3pi}{2} ) plus any multiple of ( 2pi ). So, we can write:[frac{pi}{2} + 2kpi < 2x + frac{pi}{6} < frac{3pi}{2} + 2kpi quad text{for some integer } k]Let me solve for ( x ):Subtract ( frac{pi}{6} ) from all parts:[frac{pi}{2} - frac{pi}{6} + 2kpi < 2x < frac{3pi}{2} - frac{pi}{6} + 2kpi]Simplify the angles:[frac{3pi}{6} - frac{pi}{6} = frac{2pi}{6} = frac{pi}{3}][frac{9pi}{6} - frac{pi}{6} = frac{8pi}{6} = frac{4pi}{3}]So:[frac{pi}{3} + 2kpi < 2x < frac{4pi}{3} + 2kpi]Divide all parts by 2:[frac{pi}{6} + kpi < x < frac{2pi}{3} + kpi]Therefore, the function ( f(x) ) is monotonically decreasing on the intervals ( left( frac{pi}{6} + kpi, frac{2pi}{3} + kpi right) ) for all integers ( k ).Wait, the question says "interval(s)", so I guess I need to express this as intervals. Since ( k ) can be any integer, these intervals repeat every ( pi ). So, for each integer ( k ), there's an interval where the function is decreasing.But maybe they just want the general form? I think that's acceptable.Okay, moving on to part (2). It's about triangle ABC. Let me read it again.In triangle ABC, sides ( a ), ( b ), ( c ) are opposite angles ( A ), ( B ), ( C ) respectively. Given ( f(C) = frac{5}{4} ), ( b = 4 ), and ( overrightarrow{AC} cdot overrightarrow{BC} = 12 ). Find ( c ).Alright, let's break this down.First, ( f(C) = frac{5}{4} ). From part (1), we have ( f(x) = frac{1}{2} sinleft(2x + frac{pi}{6}right) + frac{3}{4} ). So, plugging in ( x = C ):[frac{1}{2} sinleft(2C + frac{pi}{6}right) + frac{3}{4} = frac{5}{4}]Subtract ( frac{3}{4} ) from both sides:[frac{1}{2} sinleft(2C + frac{pi}{6}right) = frac{5}{4} - frac{3}{4} = frac{2}{4} = frac{1}{2}]Multiply both sides by 2:[sinleft(2C + frac{pi}{6}right) = 1]When is sine equal to 1? At ( frac{pi}{2} + 2kpi ). So:[2C + frac{pi}{6} = frac{pi}{2} + 2kpi]Solving for ( C ):[2C = frac{pi}{2} - frac{pi}{6} + 2kpi = frac{3pi}{6} - frac{pi}{6} + 2kpi = frac{2pi}{6} + 2kpi = frac{pi}{3} + 2kpi][C = frac{pi}{6} + kpi]But in a triangle, angles are between 0 and ( pi ). So, ( C = frac{pi}{6} ) or ( C = frac{pi}{6} + pi = frac{7pi}{6} ). But ( frac{7pi}{6} ) is more than ( pi ), which isn't possible for a triangle. So, ( C = frac{pi}{6} ).Alright, so angle ( C ) is ( 30^circ ).Next, we're given ( b = 4 ). So, side ( b ) opposite angle ( B ) is 4 units.Then, ( overrightarrow{AC} cdot overrightarrow{BC} = 12 ). Hmm, vectors. Let me recall that the dot product of two vectors is equal to the product of their magnitudes times the cosine of the angle between them.So, ( overrightarrow{AC} cdot overrightarrow{BC} = |AC||BC| cos theta ), where ( theta ) is the angle between vectors ( overrightarrow{AC} ) and ( overrightarrow{BC} ).Wait, but in triangle ABC, points A, B, C. So, vector ( overrightarrow{AC} ) goes from A to C, and vector ( overrightarrow{BC} ) goes from B to C. So, the angle between them is the angle at point C, right? Because both vectors originate from C.Wait, no. Vector ( overrightarrow{AC} ) is from A to C, and vector ( overrightarrow{BC} ) is from B to C. So, they both end at C, but originate from A and B respectively. So, the angle between them is actually the angle at C, which is angle ( C ).Wait, but in the triangle, angle at C is between sides AC and BC. So, yes, the angle between vectors ( overrightarrow{AC} ) and ( overrightarrow{BC} ) is angle ( C ). So, the dot product is:[overrightarrow{AC} cdot overrightarrow{BC} = |AC||BC| cos C]Given that ( overrightarrow{AC} cdot overrightarrow{BC} = 12 ), and ( |AC| = b = 4 ), ( |BC| = a ) (since in triangle ABC, side opposite angle A is a, opposite B is b, opposite C is c). Wait, hold on.Wait, in triangle ABC, side a is opposite angle A, which is BC. Side b is opposite angle B, which is AC. Side c is opposite angle C, which is AB.So, ( |AC| = b = 4 ), ( |BC| = a ), and angle between them is ( C = frac{pi}{6} ).So, the dot product is:[overrightarrow{AC} cdot overrightarrow{BC} = |AC||BC| cos C = b cdot a cdot cos C = 4a cos left( frac{pi}{6} right) = 12]We know ( cos left( frac{pi}{6} right) = frac{sqrt{3}}{2} ), so:[4a cdot frac{sqrt{3}}{2} = 12][2a sqrt{3} = 12][a sqrt{3} = 6][a = frac{6}{sqrt{3}} = 2sqrt{3}]So, side ( a = 2sqrt{3} ).Now, we need to find side ( c ). We have sides ( a = 2sqrt{3} ), ( b = 4 ), and angle ( C = frac{pi}{6} ). So, perhaps we can use the Law of Cosines to find ( c ).Law of Cosines states:[c^2 = a^2 + b^2 - 2ab cos C]Plugging in the known values:[c^2 = (2sqrt{3})^2 + 4^2 - 2 cdot 2sqrt{3} cdot 4 cdot cos left( frac{pi}{6} right)]Calculate each term:[(2sqrt{3})^2 = 4 cdot 3 = 12][4^2 = 16][2 cdot 2sqrt{3} cdot 4 = 16sqrt{3}][cos left( frac{pi}{6} right) = frac{sqrt{3}}{2}]So, putting it all together:[c^2 = 12 + 16 - 16sqrt{3} cdot frac{sqrt{3}}{2}]Simplify the last term:[16sqrt{3} cdot frac{sqrt{3}}{2} = 16 cdot frac{3}{2} = 24]So:[c^2 = 12 + 16 - 24 = 28 - 24 = 4]Therefore, ( c = sqrt{4} = 2 ).Wait, that seems straightforward. Let me double-check my steps.1. Simplified ( f(x) ) correctly using trigonometric identities.2. Took the derivative correctly, found where it's negative.3. For part (2), solved ( f(C) = frac{5}{4} ) correctly, found ( C = frac{pi}{6} ).4. Calculated the dot product correctly, found ( a = 2sqrt{3} ).5. Applied Law of Cosines correctly, ended up with ( c = 2 ).Everything seems to check out. I think that's the correct answer.**Final Answer**The value of ( c ) is boxed{2}.

💡First, I need to determine the unknown number when 3/8 of it equals 141. To do this, I'll set up the equation (3/8) * number = 141 and solve for the number.Next, I'll calculate 32.08% of the found number. This involves converting the percentage to a decimal by dividing by 100, resulting in 0.3208.Finally, I'll multiply the number by 0.3208 to find the exact value and then round it to the nearest whole number to get the approximate value.