💡First, let's denote the original cost of the air-conditioning unit as ( P ).In December, there was a 16% discount, so the price was reduced to ( P - 0.16P = 0.84P ).Six months later, the price was raised by 12%. The new price would be ( 0.84P + 0.12 cdot 0.84P = 0.84P + 0.1008P = 0.9408P ).We are given that the price in November, after the price increase, is 442.18. So we can set up the equation:( 0.9408P = 442.18 )Now we can solve for ( P ):( P = frac{442.18}{0.9408} )( P approx frac{442.18}{0.9408} )( P approx 469.99 )So the original cost of the air-conditioning unit was approximately boxed{469.99}.

💡Okay, so I have this problem about a convex hexagon where each side is at most 1, and I need to prove that at least one of the diagonals AD, BE, or CF is at most 2. Hmm, convex hexagons... I remember they have all their interior angles less than 180 degrees, and all the vertices point outward. First, maybe I should recall some properties of convex polygons, especially hexagons. The sum of the interior angles of a hexagon is (6-2)*180 = 720 degrees. That might come in handy somewhere. Also, since it's convex, any diagonal lies entirely inside the hexagon.Now, the sides are all at most 1. So each side AB, BC, CD, DE, EF, FA is ≤1. I need to look at the diagonals AD, BE, and CF. These are the diagonals that skip three vertices each, right? So AD connects A to D, BE connects B to E, and CF connects C to F.I think the key here might be to use the triangle inequality or some geometric argument. Maybe if I can bound the lengths of these diagonals in terms of the sides. Since the sides are all bounded, perhaps the diagonals can't be too long.Alternatively, maybe I can use the pigeonhole principle. If I consider the three diagonals, maybe at least one of them has to be short because the sides are short. But I'm not sure how to apply that directly.Wait, another idea: in a convex polygon, the length of a diagonal can be related to the sum of certain sides. Maybe if I can express AD, BE, or CF in terms of the sides and some angles, I can find an upper bound.Let me try to visualize the hexagon. Let's label the vertices A, B, C, D, E, F in order. So, AD is the diagonal from A to D, skipping B and C. Similarly, BE skips C and D, and CF skips D and E.Since the hexagon is convex, the diagonals AD, BE, and CF will intersect each other inside the hexagon. Maybe I can use some properties of intersecting diagonals or something related to the areas they form.But I'm not sure about that. Maybe I should think about coordinates. If I assign coordinates to the vertices, I might be able to compute the lengths of the diagonals. But that might get complicated.Wait, another thought: maybe I can use the fact that in a convex polygon, the length of a diagonal is less than the sum of the lengths of the sides it skips. But in this case, AD skips two sides, BC and CD, but each of those is at most 1, so AD would be less than BC + CD, which is at most 2. Wait, is that correct?Hold on, no. The triangle inequality says that the length of a side is less than the sum of the other two sides, but for diagonals, it's a bit different. In a convex polygon, a diagonal is less than the sum of the non-adjacent sides? Hmm, I'm not sure.Wait, maybe I can triangulate the hexagon. If I divide the hexagon into triangles, maybe I can find some relationships between the sides and the diagonals. For example, if I draw diagonals from A, I can split the hexagon into triangles ABC, ACD, ADE, and AEF. But I'm not sure if that helps directly.Alternatively, maybe I can use vectors or coordinate geometry. Assign coordinates to the vertices and express the diagonals in terms of vectors. But that might be too involved.Wait, another approach: maybe use the fact that in a convex polygon, the maximum distance between any two vertices is less than the sum of all the sides. But since each side is at most 1, the total perimeter is at most 6, but that doesn't directly help with individual diagonals.Hmm, maybe I should think about the diameter of the hexagon. The diameter is the maximum distance between any two vertices. If I can show that the diameter is at most 2, then that would mean all diagonals are at most 2, but the problem only asks for at least one of AD, BE, or CF to be at most 2.Wait, maybe I can argue by contradiction. Suppose all three diagonals AD, BE, and CF are greater than 2. Then, maybe I can derive some contradiction based on the side lengths being at most 1.Let me try that. Assume AD > 2, BE > 2, and CF > 2. Then, perhaps I can find that some sides must be longer than 1, which contradicts the given condition.But how? Let me think. If AD > 2, then the distance from A to D is greater than 2. Similarly, BE > 2 and CF > 2. Maybe I can use the triangle inequality on triangles involving these diagonals.For example, consider triangle ABD. Wait, no, AD is a diagonal, not a side. Maybe consider triangles ABD, BCE, and CAF. Hmm, not sure.Wait, another idea: maybe use the fact that in a convex polygon, the sum of the lengths of any set of non-adjacent sides is greater than the length of a diagonal. But I'm not sure about that.Alternatively, maybe use graph theory. The hexagon can be represented as a graph with vertices and edges. The diagonals AD, BE, and CF form a triangle in the graph. Maybe I can use some graph-theoretic properties.But I'm not sure. Maybe I should go back to the beginning. The hexagon is convex, so all its vertices are vertices of their convex hull. So, the hexagon is its own convex hull.Wait, maybe I can use the fact that in a convex polygon, the length of a diagonal is less than the sum of the lengths of the sides it connects. But that seems similar to the triangle inequality.Wait, let's think about triangle ABD. The sides AB and BD, but BD is another diagonal. Hmm, not helpful.Wait, maybe I can use the law of cosines. If I consider triangle ABD, then AD² = AB² + BD² - 2*AB*BD*cos(theta), where theta is the angle at B. But I don't know theta.Alternatively, maybe consider the entire hexagon and use some averaging argument. Since the total perimeter is at most 6, maybe the average side length is 1, but that doesn't directly help.Wait, another approach: maybe use the fact that in a convex polygon, the length of a diagonal is less than the sum of the lengths of the sides it skips. So, for diagonal AD, it skips sides BC and CD. So, AD < BC + CD. Since BC and CD are each at most 1, AD < 2. Wait, is that correct?Hold on, in a convex polygon, is a diagonal less than the sum of the sides it skips? I think that's a theorem. Let me recall. Yes, in a convex polygon, the length of any diagonal is less than the sum of the lengths of the sides it skips. So, for diagonal AD, which skips sides BC and CD, we have AD < BC + CD. Since BC and CD are each ≤1, then AD < 1 + 1 = 2. So, AD < 2. But the problem says "at most 2," so maybe AD ≤ 2.Wait, but in the problem statement, it says "at least one of the three diagonals AD, BE, or CF is at most 2." So, if I can show that each of these diagonals is less than 2, then all of them are less than 2, which would satisfy the condition. But I'm not sure if that's the case.Wait, no, the problem says "at least one," so maybe not all of them, but at least one. So, perhaps not all of them are necessarily less than 2, but at least one must be.But earlier, I thought that in a convex polygon, a diagonal is less than the sum of the sides it skips. So, for AD, it skips BC and CD, so AD < BC + CD ≤ 2. Similarly, BE skips CD and DE, so BE < CD + DE ≤ 2, and CF skips DE and EF, so CF < DE + EF ≤ 2. So, actually, all three diagonals are less than 2, which would mean that all of them are at most 2, which is stronger than the problem's requirement.But wait, the problem says "at least one," so maybe my initial thought is correct, but perhaps the problem is expecting a different approach. Maybe the user wants me to think about it differently.Alternatively, maybe the user is thinking about the hexagon being convex and using some geometric properties. Maybe if I consider the hexagon inscribed in a circle or something like that.Wait, another idea: maybe use the fact that in a convex polygon, the diameter is at least the length of the longest side. But since all sides are at most 1, the diameter is at least 1, but that doesn't help.Wait, maybe use the pigeonhole principle. Since the hexagon has six sides, and the diagonals AD, BE, CF are three in number, maybe at least one of them has to be short.But I'm not sure how to apply the pigeonhole principle here. Maybe if I can partition the hexagon into regions and argue that one of the diagonals must lie within a certain region.Alternatively, maybe use the fact that in a convex hexagon, the three main diagonals AD, BE, CF intersect at a common point, and use properties of that intersection.Wait, I think in a convex hexagon, the three main diagonals do intersect at a common point, called the centroid or something like that. Maybe I can use properties of that point to bound the lengths of the diagonals.But I'm not sure. Maybe I should look for a simpler approach.Wait, going back to the earlier idea: in a convex polygon, a diagonal is less than the sum of the sides it skips. So, for AD, it skips BC and CD, so AD < BC + CD ≤ 2. Similarly for BE and CF. So, all three diagonals are less than 2, which means they are at most 2. Therefore, at least one of them is at most 2.But wait, the problem says "at least one," but if all three are at most 2, then certainly at least one is. So, maybe that's the proof.But I'm not sure if the theorem I recalled is correct. Let me double-check. In a convex polygon, is a diagonal less than the sum of the sides it skips?Yes, I think that's a standard result. In a convex polygon, the length of any diagonal is less than the sum of the lengths of the sides it skips. So, for diagonal AD in a convex hexagon, it skips sides BC and CD, so AD < BC + CD. Since BC and CD are each ≤1, AD < 2.Similarly, BE skips CD and DE, so BE < CD + DE ≤ 2, and CF skips DE and EF, so CF < DE + EF ≤ 2.Therefore, all three diagonals AD, BE, and CF are less than 2, which means they are at most 2. Hence, at least one of them is at most 2.Wait, but the problem says "at least one," so maybe the user is expecting a different approach where only one of them is necessarily ≤2, but the others could be longer. But according to this, all three are necessarily <2, so all are ≤2.But maybe the problem is considering that in some cases, one of them could be exactly 2, while the others are less. But in any case, the conclusion holds.So, I think this is the way to go. Use the theorem that in a convex polygon, a diagonal is less than the sum of the sides it skips. Since each side is at most 1, the sum is at most 2, so each diagonal is less than 2, hence at most 2. Therefore, at least one of them is at most 2.I think that's the proof.

💡Okay, so I have this problem where I need to define a function f(x) as the minimum of three linear functions: 3x + 1, -4/3 x + 3, and 1/3 x + 9. Then, I need to find the maximum value of this function f(x). Hmm, that sounds a bit tricky, but I think I can figure it out step by step.First, let me visualize what's going on here. Each of these three functions is a straight line on a graph. Since f(x) is the minimum of these three at each x, the graph of f(x) will be the lower envelope of these three lines. That means, for each x, f(x) will be the lowest y-value among the three lines. So, to find the maximum value of f(x), I need to find the highest point on this lower envelope.I remember that the maximum of the minimum function will occur at one of the intersection points of these lines because that's where the minimum switches from one line to another. So, my plan is to find the intersection points of each pair of lines and then evaluate f(x) at those points to see which one gives the highest value.Alright, let's start by finding the intersection points.First, I'll find where 3x + 1 intersects with -4/3 x + 3. To do this, I set them equal to each other:3x + 1 = -4/3 x + 3Hmm, let me solve for x. I'll multiply both sides by 3 to eliminate the fraction:3*(3x) + 3*1 = 3*(-4/3 x) + 3*3Which simplifies to:9x + 3 = -4x + 9Now, I'll bring all the x terms to one side and constants to the other:9x + 4x = 9 - 313x = 6So, x = 6/13Now, plug this back into one of the original equations to find y. Let's use 3x + 1:y = 3*(6/13) + 1 = 18/13 + 13/13 = 31/13So, the intersection point is (6/13, 31/13).Next, I'll find where -4/3 x + 3 intersects with 1/3 x + 9. Setting them equal:-4/3 x + 3 = 1/3 x + 9Again, I'll multiply both sides by 3 to eliminate fractions:3*(-4/3 x) + 3*3 = 3*(1/3 x) + 3*9Which simplifies to:-4x + 9 = x + 27Bringing x terms to one side and constants to the other:-4x - x = 27 - 9-5x = 18x = -18/5Now, plug this back into one of the equations to find y. Let's use -4/3 x + 3:y = -4/3*(-18/5) + 3 = (72/15) + 3 = (72/15) + (45/15) = 117/15 = 39/5So, the intersection point is (-18/5, 39/5).Lastly, I should also check where 3x + 1 intersects with 1/3 x + 9. Setting them equal:3x + 1 = 1/3 x + 9Multiply both sides by 3:9x + 3 = x + 27Bringing x terms to one side:9x - x = 27 - 38x = 24x = 3Plugging back into 3x + 1:y = 3*3 + 1 = 9 + 1 = 10So, the intersection point is (3, 10).Now, I have three intersection points: (6/13, 31/13), (-18/5, 39/5), and (3, 10). But wait, I need to make sure that at each intersection point, the two lines involved are actually the ones contributing to the minimum at that x-value.Let me think about this. For f(x) to switch from one line to another at an intersection point, that point must lie on the lower envelope. So, I need to check whether each intersection point is indeed a point where the minimum switches.Let me analyze each intersection:1. (6/13, 31/13): This is where 3x + 1 and -4/3 x + 3 intersect. I need to check whether at this x, both lines are indeed the minimum.Let me pick a point just before 6/13, say x = 0. At x=0, 3x +1 =1, -4/3 x +3=3, and 1/3 x +9=9. So, the minimum is 1, which is from 3x +1.At x=6/13, the value is 31/13 ≈2.38. Let's pick a point just after 6/13, say x=1. At x=1, 3x +1=4, -4/3 x +3≈1.666, and 1/3 x +9≈9.333. So, the minimum is ≈1.666, which is from -4/3 x +3.Therefore, at x=6/13, the minimum switches from 3x +1 to -4/3 x +3. So, this point is indeed on the lower envelope.2. (-18/5, 39/5): This is where -4/3 x +3 and 1/3 x +9 intersect. Let's check the behavior around x=-18/5.Take x=-4, which is less than -18/5 (-3.6). At x=-4, 3x +1=-11, -4/3 x +3≈6.333, and 1/3 x +9≈7. So, the minimum is -11 from 3x +1.At x=-18/5 (-3.6), the value is 39/5=7.8. Let's take x=-3, which is greater than -18/5. At x=-3, 3x +1=-8, -4/3 x +3=7, and 1/3 x +9=8. So, the minimum is -8 from 3x +1.Wait, that's strange. At x=-18/5, both -4/3 x +3 and 1/3 x +9 equal 7.8, but at x=-4 and x=-3, the minimum is still from 3x +1. So, does this mean that the point (-18/5, 39/5) is not on the lower envelope?Hmm, maybe because 3x +1 is still lower than both -4/3 x +3 and 1/3 x +9 around that x-value. So, perhaps this intersection point is not part of the lower envelope.Let me verify. Let's take x=-3.6, which is exactly -18/5. At x=-3.6, 3x +1= -10.8 +1= -9.8, which is less than 7.8. So, actually, at x=-3.6, the minimum is still -9.8 from 3x +1, not 7.8. Therefore, the intersection point (-18/5, 39/5) is not on the lower envelope because 3x +1 is still lower there.So, that means this intersection doesn't contribute to the lower envelope. Therefore, f(x) doesn't switch here.3. (3, 10): This is where 3x +1 and 1/3 x +9 intersect. Let's check around x=3.Take x=2: 3x +1=7, -4/3 x +3≈-1.666 +3=1.333, and 1/3 x +9≈6.333 +9=15.333. So, the minimum is 1.333 from -4/3 x +3.At x=3, both 3x +1 and 1/3 x +9 equal 10. Let's take x=4: 3x +1=13, -4/3 x +3≈-5.333 +3≈-2.333, and 1/3 x +9≈12 +9=21. So, the minimum is -2.333 from -4/3 x +3.Wait, so at x=3, the value is 10, but around x=3, the minimum is still from -4/3 x +3. So, does that mean that the intersection at (3,10) is not on the lower envelope?Wait, let's check x=3. At x=3, 3x +1=10, -4/3 x +3= -4 +3=-1, and 1/3 x +9=3 +9=12. So, the minimum is -1, not 10. So, actually, the point (3,10) is not the minimum at x=3. Therefore, this intersection is also not on the lower envelope.Hmm, so only the first intersection point (6/13, 31/13) is on the lower envelope. The other two intersections are not because 3x +1 is still lower in their vicinity.Therefore, the lower envelope is made up of three segments:1. From negative infinity up to x=6/13, the minimum is 3x +1.2. From x=6/13 to some point, the minimum is -4/3 x +3.3. After that point, the minimum is 1/3 x +9.Wait, but earlier, I saw that at x=3, the minimum is still -4/3 x +3, which is -1, but 1/3 x +9 at x=3 is 12, so 1/3 x +9 is higher. So, when does 1/3 x +9 become the minimum?Let me check where -4/3 x +3 and 1/3 x +9 cross each other, but we saw that at their intersection point, 3x +1 is still lower. So, perhaps 1/3 x +9 never becomes the minimum? Or maybe it does beyond a certain point.Wait, let's analyze the behavior as x approaches infinity.As x becomes very large, 3x +1 and 1/3 x +9 both go to infinity, but 3x +1 grows faster. Meanwhile, -4/3 x +3 goes to negative infinity. So, as x increases, -4/3 x +3 will dominate and will be the minimum.Wait, but that contradicts my earlier thought. Maybe I made a mistake.Wait, let's think about the slopes:- 3x +1 has a positive slope of 3.- -4/3 x +3 has a negative slope of -4/3.- 1/3 x +9 has a positive slope of 1/3.So, as x increases, 3x +1 and 1/3 x +9 both increase, but 3x +1 increases much faster. Meanwhile, -4/3 x +3 decreases as x increases.Therefore, as x increases, -4/3 x +3 will eventually become the minimum because it's decreasing, while the other two are increasing.Wait, but at x=3, -4/3 x +3 is -1, which is lower than both 3x +1=10 and 1/3 x +9=12. So, as x increases beyond 3, -4/3 x +3 becomes even more negative, so it's definitely the minimum.But wait, when does 1/3 x +9 become the minimum? Let me check for very large negative x.As x approaches negative infinity, 3x +1 goes to negative infinity, -4/3 x +3 goes to positive infinity, and 1/3 x +9 goes to negative infinity. So, which one is lower?3x +1 has a steeper negative slope than 1/3 x +9. So, for very large negative x, 3x +1 will be lower than 1/3 x +9.Therefore, 1/3 x +9 will never be the minimum because 3x +1 is always lower for negative x, and for positive x, -4/3 x +3 is lower after a certain point.Wait, so does that mean that 1/3 x +9 is never the minimum? Or is there a point where it becomes the minimum?Wait, let's check at x=0: 3x +1=1, -4/3 x +3=3, 1/3 x +9=9. So, 3x +1 is the minimum.At x=6/13≈0.46, the minimum switches to -4/3 x +3.As x increases beyond 6/13, -4/3 x +3 decreases, while 3x +1 and 1/3 x +9 increase. So, -4/3 x +3 remains the minimum until it intersects with another line.But we saw that when -4/3 x +3 intersects with 1/3 x +9 at x=-18/5, which is negative, and at that point, 3x +1 is still lower.Wait, so maybe 1/3 x +9 is never the minimum? Because for x < 6/13, 3x +1 is the minimum, and for x > 6/13, -4/3 x +3 is the minimum, which continues to decrease as x increases.Wait, but 1/3 x +9 is above both 3x +1 and -4/3 x +3 for all x? Let me check at x=6.At x=6: 3x +1=19, -4/3 x +3= -8 +3=-5, 1/3 x +9=2 +9=11. So, minimum is -5.At x=12: 3x +1=37, -4/3 x +3= -16 +3=-13, 1/3 x +9=4 +9=13. Minimum is -13.So, yeah, 1/3 x +9 is always above the other two lines for positive x.Wait, but what about for negative x? Let's take x=-6.At x=-6: 3x +1=-18 +1=-17, -4/3 x +3=8 +3=11, 1/3 x +9=-2 +9=7. So, minimum is -17 from 3x +1.At x=-3: 3x +1=-9 +1=-8, -4/3 x +3=4 +3=7, 1/3 x +9=-1 +9=8. Minimum is -8.At x=-2: 3x +1=-6 +1=-5, -4/3 x +3≈2.666 +3=5.666, 1/3 x +9≈-0.666 +9=8.333. Minimum is -5.At x=-1: 3x +1=-3 +1=-2, -4/3 x +3≈1.333 +3=4.333, 1/3 x +9≈-0.333 +9=8.666. Minimum is -2.At x=0: 1, 3, 9. Minimum is 1.So, in all these cases, 3x +1 is the minimum for x < 6/13, and -4/3 x +3 is the minimum for x > 6/13.Therefore, 1/3 x +9 is never the minimum. So, the lower envelope is made up of two lines: 3x +1 for x ≤ 6/13, and -4/3 x +3 for x ≥ 6/13.Therefore, f(x) is a piecewise function:f(x) = 3x +1, when x ≤ 6/13f(x) = -4/3 x +3, when x ≥ 6/13Now, to find the maximum value of f(x), we need to find the highest point on this lower envelope.Since f(x) is increasing for x ≤ 6/13 (because the slope is positive) and decreasing for x ≥ 6/13 (because the slope is negative), the maximum must occur at the point where these two segments meet, which is at x=6/13.Therefore, the maximum value of f(x) is f(6/13) = 3*(6/13) +1 = 18/13 +13/13=31/13.So, the maximum value is 31/13.Wait, let me just double-check that.At x=6/13, f(x)=31/13≈2.38.For x <6/13, f(x)=3x +1, which is less than 31/13 because as x increases, f(x) increases.For x >6/13, f(x)=-4/3 x +3, which is less than 31/13 because as x increases, f(x) decreases.Therefore, yes, 31/13 is indeed the maximum value of f(x).So, the answer is 31/13.**Final Answer**The maximum value of ( f(x) ) is (boxed{dfrac{31}{13}}).

💡Alright, so I have this function here: y equals negative five cosine of (2x plus pi over three) plus four. I need to figure out the amplitude, phase shift, and vertical shift of this graph. Hmm, okay, let's break this down step by step.First, I remember that the general form of a cosine function is y equals A cosine(Bx minus C) plus D. In this form, A is the amplitude, which tells me how tall the wave is from the middle line. B affects the period, which is how long it takes to complete one full cycle. C is related to the phase shift, which is how much the graph is shifted horizontally, and D is the vertical shift, which moves the graph up or down.Looking at my function, y equals negative five cosine of (2x plus pi over three) plus four, I can try to match this to the general form. Let me rewrite it to make it clearer:y = -5 cos(2x + π/3) + 4So, comparing this to y = A cos(Bx - C) + D, I can see that:- A is -5- B is 2- Inside the cosine, it's 2x plus π/3, which is like Bx + C, so C must be π/3- D is 4Wait, hold on. In the general form, it's Bx minus C, but in my function, it's Bx plus C. Does that mean C is negative? Let me think. If I have cos(Bx + C), that's the same as cos(B(x + C/B)). So, actually, the phase shift is -C/B, right? Because it's shifted to the left by C/B if it's plus, and to the right by C/B if it's minus.So, in my case, since it's 2x + π/3, that's like 2(x + π/6). So, the phase shift is -π/6, which means a shift to the left by π/6. But wait, in the general form, it's B(x - C), so if I have B(x + π/6), that's equivalent to B(x - (-π/6)). So, the phase shift is -π/6, which is a shift to the left by π/6.But sometimes, people prefer to express phase shifts as positive numbers indicating the direction. So, a shift to the left by π/6 can also be written as a phase shift of π/6 to the left.Okay, so amplitude is the absolute value of A, which is |-5| = 5. That's straightforward.The vertical shift is D, which is 4. So, the graph is shifted up by 4 units.Now, about the phase shift. I think I got that it's π/6 to the left. But let me double-check. If I have cos(2x + π/3), I can factor out the 2:cos(2(x + π/6))So, this is a horizontal shift. Since it's x plus π/6 inside the function, it's a shift to the left by π/6. So, yes, the phase shift is π/6 to the left.Wait, but sometimes people write phase shift as a positive number with a direction. So, if it's shifted to the left, it's negative, and to the right, it's positive. So, in terms of the formula, phase shift is -π/6.But when reporting it, I think it's more common to say the phase shift is π/6 to the left, rather than negative π/6.Let me confirm with an example. Suppose I have cos(x + π/2). That's the same as cos(x - (-π/2)), so the phase shift is -π/2, which means a shift to the left by π/2. So, yes, in this case, the phase shift is π/6 to the left.Okay, so putting it all together:- Amplitude: 5- Phase shift: π/6 to the left- Vertical shift: 4 upWait, but in the general form, the phase shift is C/B, right? So, in my function, C is π/3, and B is 2, so phase shift is (π/3)/2 = π/6. But since it's a plus inside the cosine, it's a shift to the left, so phase shift is -π/6.But when expressing it, should I write it as π/6 to the left or -π/6? I think both are correct, but maybe in the answer, they prefer the direction specified.Also, the amplitude is always positive, so even though A is -5, the amplitude is 5.And the vertical shift is straightforward, it's just D, which is 4.So, to summarize:- Amplitude: 5- Phase shift: π/6 to the left- Vertical shift: 4 upI think that's it. Let me just visualize the graph to make sure. The cosine wave is normally centered around the x-axis, but here it's shifted up by 4, so the midline is y=4. The amplitude is 5, so the maximum is 4 + 5 = 9, and the minimum is 4 - 5 = -1. The phase shift is π/6 to the left, so the starting point of the cosine wave is shifted π/6 units to the left compared to the standard cosine function.Yeah, that makes sense. I think I've got it right.

💡First, I need to determine how many candies Josh gave to his siblings. Since he has 3 siblings and gave each 35 candies, the total given to siblings is 3 multiplied by 35, which equals 105 candies.Next, I'll subtract the candies given to his siblings from the total number of candies Josh initially had. So, 500 minus 105 equals 395 candies remaining.Then, Josh gave half of the remaining candies to his best friend. Half of 395 is 197.5, but since we can't have half a candy, I'll assume he gave his friend 197 candies. Subtracting this from the remaining candies, 395 minus 197 equals 198 candies left.After that, Josh gave 25% of the remaining candies to his cousin. 25% of 198 is 49.5, but again, assuming we can't have half a candy, I'll consider he gave his cousin 49 candies. Subtracting this from the remaining candies, 198 minus 49 equals 149 candies remaining.Josh then kept 50 candies for himself. Subtracting this from the remaining candies, 149 minus 50 equals 99 candies left.Finally, Josh gave his neighbor 3 times the amount he kept for himself, which is 3 multiplied by 50, equaling 150 candies. However, Josh only has 99 candies left at this point, which means he doesn't have enough to give his neighbor 150 candies. This indicates an inconsistency in the problem as stated.If we ignore the neighbor's part, Josh would be left with 99 candies after keeping some for himself. However, if we consider the neighbor's part, Josh would not have enough candies to give, and we would need to adjust the number of candies given to the neighbor or reconsider the previous steps.

💡Okay, so I have this problem about a geometric sequence. It says it's increasing, which probably means each term is bigger than the one before. The first term is a₁, and the fifth term is a₅. They tell me that a₁ plus a₅ equals 17, and the product of a₂ and a₄ is 16. I need to find the common ratio q.Hmm, let's recall what a geometric sequence is. Each term is the previous term multiplied by a common ratio q. So, a₂ is a₁ times q, a₃ is a₂ times q, which is a₁ times q squared, and so on. So, a₅ would be a₁ times q to the fourth power.They gave me two equations: a₁ + a₅ = 17 and a₂ * a₄ = 16. Let me write those out in terms of a₁ and q.First equation: a₁ + a₅ = 17. Since a₅ is a₁ * q⁴, that becomes a₁ + a₁ * q⁴ = 17. I can factor out a₁: a₁(1 + q⁴) = 17.Second equation: a₂ * a₄ = 16. Let's express a₂ and a₄ in terms of a₁ and q. a₂ is a₁ * q, and a₄ is a₁ * q³. So, multiplying them together: (a₁ * q) * (a₁ * q³) = a₁² * q⁴ = 16.So now I have two equations:1. a₁(1 + q⁴) = 172. a₁² * q⁴ = 16I need to solve for q. Maybe I can express a₁ from the first equation and substitute into the second.From the first equation: a₁ = 17 / (1 + q⁴)Plugging that into the second equation: (17 / (1 + q⁴))² * q⁴ = 16That looks a bit complicated, but let's try to simplify it.Let me denote q⁴ as x for simplicity. So, x = q⁴.Then, the equation becomes:(17 / (1 + x))² * x = 16Let me write that out:(289 / (1 + x)²) * x = 16Multiply both sides by (1 + x)² to eliminate the denominator:289x = 16(1 + x)²Expand the right side:289x = 16(1 + 2x + x²)289x = 16 + 32x + 16x²Bring all terms to one side:16x² + 32x + 16 - 289x = 0Combine like terms:16x² - 257x + 16 = 0Hmm, that's a quadratic in terms of x. Let me see if I can solve this.Using the quadratic formula: x = [257 ± sqrt(257² - 4*16*16)] / (2*16)Calculate the discriminant:257² = 660494*16*16 = 1024So, sqrt(66049 - 1024) = sqrt(65025) = 255So, x = [257 ± 255] / 32Calculate both possibilities:First solution: (257 + 255)/32 = 512/32 = 16Second solution: (257 - 255)/32 = 2/32 = 1/16So, x = 16 or x = 1/16But x is q⁴, so q⁴ = 16 or q⁴ = 1/16Taking the fourth root:If q⁴ = 16, then q = ±2If q⁴ = 1/16, then q = ±1/2But the sequence is increasing, so q must be positive and greater than 1, right? Because if q is less than 1, the terms would decrease.Wait, actually, if q is positive and greater than 1, the sequence increases. If q is between 0 and 1, it decreases. If q is negative, the sequence alternates in sign, but since it's increasing, it can't alternate. So q must be positive.So, q must be 2 or 1/2. But 1/2 is less than 1, so the sequence would decrease. Therefore, q must be 2.Let me check if that makes sense.If q = 2, then a₁ = 17 / (1 + 16) = 17/17 = 1So, the sequence would be 1, 2, 4, 8, 16, which is increasing. a₁ + a₅ = 1 + 16 = 17, which matches. a₂ * a₄ = 2 * 8 = 16, which also matches.So, q = 2 is the correct answer.Wait, but the options are A: -4, B:4, C:-2, D:2.So, D:2 is the answer.I think that's solid. I don't see any mistakes in my reasoning.