💡To determine the number of different orders in which Robin can stack the five different ice cream scoops on her cone, we can use the concept of permutations. Since each scoop must be a different flavor and the order matters, we are essentially looking for the number of ways to arrange 5 distinct items.1. **First Scoop (Bottom):** There are 5 possible flavors to choose from.2. **Second Scoop:** After selecting the first flavor, there are 4 remaining flavors.3. **Third Scoop:** After selecting the first two flavors, there are 3 remaining flavors.4. **Fourth Scoop:** After selecting the first three flavors, there are 2 remaining flavors.5. **Fifth Scoop (Top):** Finally, there is only 1 flavor left.Using the multiplication principle, the total number of ways to stack the scoops is:[ 5 times 4 times 3 times 2 times 1 = 120 ]

💡Alright, so I have this probability problem about a regular hexagon. Each of its 6 sides and 9 diagonals are colored either red or blue randomly and independently with equal probability. I need to find the probability that there will be a triangle (with vertices among the hexagon's vertices) where all three sides are the same color. The options are (A) 0.99, (B) 0.75, (C) 0.995, and (D) 1.0.Hmm, okay. Let me break this down. First, a regular hexagon has 6 sides and 9 diagonals, making a total of 15 edges. Each of these edges is colored either red or blue with a 50-50 chance. So, each edge is like a coin flip—either red or blue.Now, the question is about the probability of having at least one monochromatic triangle. A monochromatic triangle is one where all three sides are the same color, either all red or all blue. So, I need to figure out how likely it is that among all the possible triangles formed by the vertices of the hexagon, at least one of them has all its sides the same color.First, let me figure out how many triangles there are in a hexagon. Since a triangle is defined by three vertices, the number of triangles is the number of ways to choose 3 vertices out of 6. That's a combination problem. The formula for combinations is C(n, k) = n! / (k! (n - k)!). So, C(6, 3) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20. So, there are 20 possible triangles.Each triangle has three edges. For each triangle, the probability that all three edges are red is (1/2)^3 = 1/8. Similarly, the probability that all three edges are blue is also 1/8. So, the probability that a specific triangle is monochromatic (either all red or all blue) is 1/8 + 1/8 = 1/4.Therefore, the probability that a specific triangle is not monochromatic is 1 - 1/4 = 3/4.Now, if I consider all 20 triangles, I might be tempted to calculate the probability that none of them are monochromatic and then subtract that from 1 to get the probability that at least one is monochromatic. That would be:Probability(at least one monochromatic triangle) = 1 - Probability(no monochromatic triangles)Assuming that the events of each triangle being monochromatic are independent, which might not be the case, but let's go with that for a moment. Then, the probability that none of the 20 triangles are monochromatic would be (3/4)^20.Calculating (3/4)^20: Let's see, 3/4 is 0.75. 0.75^20 is a very small number. Let me compute that:0.75^2 = 0.56250.75^4 = (0.5625)^2 ≈ 0.31640.75^8 ≈ (0.3164)^2 ≈ 0.09990.75^16 ≈ (0.0999)^2 ≈ 0.00998Then, 0.75^20 = 0.75^16 * 0.75^4 ≈ 0.00998 * 0.3164 ≈ 0.00316So, approximately 0.00316 is the probability that none of the triangles are monochromatic. Therefore, the probability that at least one triangle is monochromatic is 1 - 0.00316 ≈ 0.99684.Looking at the answer choices, 0.99684 is closest to option (C) 0.995. However, this is an approximation because I assumed independence between the triangles, which isn't actually true. In reality, the triangles share edges, so the coloring of one edge affects multiple triangles. This means that the events are not independent, and my calculation might be slightly off.But given the options, 0.995 is the closest to my approximate value. Also, considering that the exact probability is likely even higher than 0.99684 because of dependencies, but since 0.995 is an option and it's very close, I think that's the intended answer.Wait, but hold on. Another thought: in Ramsey theory, there's a concept that in any coloring of the edges of a complete graph, certain monochromatic subgraphs must exist. For a complete graph with 6 vertices, K6, the Ramsey number R(3,3) is 6, meaning that any 2-coloring of the edges of K6 must contain a monochromatic triangle. So, does that mean the probability is actually 1.0?Hmm, that's conflicting with my earlier calculation. Let me think about that. Ramsey's theorem states that for any given integers, there exists a minimum number such that any coloring will contain a monochromatic complete subgraph of a certain size. Specifically, R(3,3) = 6, which means that any 2-coloring of K6 must contain a monochromatic triangle.So, does that mean that in our case, since we're dealing with K6 (the complete graph on 6 vertices), any coloring must have at least one monochromatic triangle? If that's the case, then the probability should be 1.0, which is option (D).But wait, in the problem statement, it's a regular hexagon, which is a 6-cycle, but when considering all sides and diagonals, it's indeed K6. So, according to Ramsey's theorem, every 2-coloring of K6 must contain a monochromatic triangle. Therefore, the probability is 1.0.But then why did my initial calculation give approximately 0.99684? Because I assumed independence, which isn't the case, and also because Ramsey's theorem tells us it's certain, not just highly probable.So, perhaps the correct answer is (D) 1.0.But I need to verify this. Let me recall Ramsey's theorem. R(m, n) is the smallest number such that any 2-coloring of the edges of K_{R(m,n)} contains a red K_m or a blue K_n. For R(3,3), it's 6, meaning that in any 2-coloring of K6, there must be a red triangle or a blue triangle.Therefore, in our problem, since we're dealing with K6, the probability is indeed 1.0. So, the answer should be (D) 1.0.But wait, the problem says "each of the 6 sides and the 9 diagonals of a regular hexagon are randomly and independently colored red or blue with equal probability." So, it's not just any coloring, but a random coloring. But according to Ramsey's theorem, regardless of the coloring, there must be a monochromatic triangle. So, in every possible coloring, there is at least one monochromatic triangle. Therefore, the probability is 1.0.So, the answer is (D) 1.0.But then why did my initial calculation give a high probability but not exactly 1? Because my calculation was based on independence, which isn't the case, and Ramsey's theorem tells us it's a certainty.Therefore, the correct answer is (D) 1.0.

💡First, I need to determine the volume of one cube. The formula for the volume of a cube is ( V = s^3 ), where ( s ) is the side length of the cube.Given that each cube has an edge length of 5 meters, I can substitute this value into the formula:[V = 5^3 = 125 text{ cubic meters}]Since there are four identical cubes, I will multiply the volume of one cube by 4 to find the total volume:[text{Total volume} = 125 times 4 = 500 text{ cubic meters}]

💡First, I need to determine how many hours Bill worked in total given his total pay of 1200.Bill earns 20 per hour for the first 40 hours. If he works more than 40 hours, he earns double the rate, which is 40 per hour for the additional hours.First, I'll calculate the pay for the initial 40 hours:40 hours × 20/hour = 800.Next, I'll find out how much pay is left after accounting for the first 40 hours:1200 - 800 = 400.This remaining 400 is earned at the double rate of 40 per hour. To find the number of additional hours worked:400 ÷ 40/hour = 10 hours.Finally, I'll add the initial 40 hours to the additional 10 hours to find the total hours worked:40 hours + 10 hours = 50 hours.

💡Okay, so I have this problem where I need to find the smallest value of ( t ) such that when I plot the polar equation ( r = sin theta ) from ( 0 ) to ( t ), it forms the entire circle. Hmm, I remember that polar graphs can sometimes trace over themselves, so I need to figure out when the graph completes a full circle without retracing.First, I recall that ( r = sin theta ) is a circle in polar coordinates. I think it's a circle with a diameter along the y-axis, passing through the origin. To confirm, maybe I can convert it to rectangular coordinates. Let me try that.Starting with ( r = sin theta ), I know that ( r = sqrt{x^2 + y^2} ) and ( sin theta = frac{y}{r} ). So substituting, I get:[r = frac{y}{r}]Multiplying both sides by ( r ):[r^2 = y]Which translates to:[x^2 + y^2 = y]To make this look like a standard circle equation, I can complete the square for ( y ):[x^2 + y^2 - y = 0][x^2 + left(y^2 - y + frac{1}{4}right) = frac{1}{4}][x^2 + left(y - frac{1}{2}right)^2 = left(frac{1}{2}right)^2]So yes, it's a circle with center at ( (0, frac{1}{2}) ) and radius ( frac{1}{2} ). That makes sense.Now, going back to the polar equation ( r = sin theta ). I know that as ( theta ) increases, ( r ) changes. Let me think about how ( r ) behaves as ( theta ) goes from 0 to ( 2pi ).At ( theta = 0 ), ( r = 0 ). So the point is at the origin.As ( theta ) increases to ( frac{pi}{2} ), ( r ) increases to 1. So the point moves up along the positive y-axis to ( (0, 1) ).Then, as ( theta ) goes from ( frac{pi}{2} ) to ( pi ), ( r ) decreases back to 0. So the point moves back towards the origin along the negative x-axis.Wait, but in polar coordinates, when ( r ) becomes negative, it plots the point in the opposite direction. So when ( theta ) is between ( pi ) and ( 2pi ), ( r ) becomes negative, which would plot points in the lower half-plane.But does the graph complete the circle before ( 2pi )? Let me think.At ( theta = pi ), ( r = sin pi = 0 ). So we're back at the origin.As ( theta ) increases beyond ( pi ), ( r ) becomes negative. For example, at ( theta = frac{3pi}{2} ), ( r = sin frac{3pi}{2} = -1 ). So in polar coordinates, ( r = -1 ) at ( theta = frac{3pi}{2} ) is equivalent to ( r = 1 ) at ( theta = frac{3pi}{2} + pi = frac{5pi}{2} ). But ( frac{5pi}{2} ) is more than ( 2pi ), so maybe it's just plotted in the opposite direction.Wait, actually, in polar coordinates, ( (r, theta) ) is the same as ( (-r, theta + pi) ). So when ( r ) is negative, it's like reflecting the point across the origin.So, when ( theta ) is between ( pi ) and ( 2pi ), ( r ) is negative, which effectively plots the lower half of the circle.But I need to see if the entire circle is traced out before ( 2pi ). Let me consider the points:- At ( theta = 0 ), ( r = 0 ).- At ( theta = frac{pi}{2} ), ( r = 1 ).- At ( theta = pi ), ( r = 0 ).- At ( theta = frac{3pi}{2} ), ( r = -1 ).- At ( theta = 2pi ), ( r = 0 ).So, from ( 0 ) to ( frac{pi}{2} ), it goes from the origin up to ( (0,1) ).From ( frac{pi}{2} ) to ( pi ), it goes back to the origin, but along the negative x-axis.From ( pi ) to ( frac{3pi}{2} ), ( r ) is negative, so it's plotting the lower half of the circle, going from the origin down to ( (0,-1) ).From ( frac{3pi}{2} ) to ( 2pi ), it goes back to the origin, but along the positive x-axis.Wait, but does it actually trace the entire circle? Because when ( r ) is negative, it's plotting points in the opposite direction, so maybe it's overlapping.Let me think about the parametric equations.In polar coordinates, ( x = r cos theta ) and ( y = r sin theta ).So for ( r = sin theta ), we have:[x = sin theta cos theta][y = sin^2 theta]Using the double-angle identity, ( x = frac{1}{2} sin 2theta ), and ( y = frac{1 - cos 2theta}{2} ).So, parametrically, it's:[x = frac{1}{2} sin 2theta][y = frac{1}{2} - frac{1}{2} cos 2theta]This is a circle with center at ( (0, frac{1}{2}) ) and radius ( frac{1}{2} ), as we saw earlier.Now, to see when the entire circle is traced, we can look at the parametric equations.The parametric equations for a circle can be written as:[x = r cos phi][y = r sin phi]Where ( phi ) is the parameter. In our case, the parametric equations are:[x = frac{1}{2} sin 2theta][y = frac{1}{2} - frac{1}{2} cos 2theta]Let me see if I can express this in terms of a single parameter.Let me set ( phi = 2theta ). Then:[x = frac{1}{2} sin phi][y = frac{1}{2} - frac{1}{2} cos phi]This is the parametric equation of a circle with radius ( frac{1}{2} ), centered at ( (0, frac{1}{2}) ), parameterized by ( phi ).To trace the entire circle, ( phi ) needs to go from 0 to ( 2pi ). So, ( phi = 2theta ) implies that ( theta ) needs to go from 0 to ( pi ).Wait, that's interesting. So if ( theta ) goes from 0 to ( pi ), then ( phi ) goes from 0 to ( 2pi ), completing the circle.But earlier, I thought that ( r ) becomes negative beyond ( pi ), so maybe the graph retraces.Wait, let's check.If ( theta ) goes from 0 to ( pi ), then ( phi = 2theta ) goes from 0 to ( 2pi ), so the parametric equations trace the entire circle once.But in polar coordinates, when ( r ) becomes negative, does it plot the same points again?Wait, let's see.At ( theta = pi ), ( r = sin pi = 0 ).At ( theta = frac{3pi}{2} ), ( r = sin frac{3pi}{2} = -1 ). So in polar coordinates, ( (-1, frac{3pi}{2}) ) is the same as ( (1, frac{3pi}{2} + pi) = (1, frac{5pi}{2}) ), which is equivalent to ( (1, frac{pi}{2}) ) because ( frac{5pi}{2} ) is coterminal with ( frac{pi}{2} ).But ( (1, frac{pi}{2}) ) is the point ( (0,1) ), which is already plotted at ( theta = frac{pi}{2} ).Similarly, at ( theta = frac{3pi}{2} ), we're just retracing the point ( (0,1) ).Wait, so does that mean that the graph completes the circle by ( theta = pi )?But when I think about the polar plot, from ( theta = 0 ) to ( theta = pi ), ( r ) goes from 0 to 1 and back to 0, but in the upper half-plane.Then, from ( theta = pi ) to ( 2pi ), ( r ) becomes negative, which would plot the lower half-plane.But if I only go up to ( theta = pi ), I only get the upper half of the circle.Wait, but according to the parametric equations, if ( theta ) goes from 0 to ( pi ), ( phi ) goes from 0 to ( 2pi ), so the entire circle is traced.But in polar coordinates, when ( r ) is negative, it's plotting points in the opposite direction, so maybe it's overlapping.I'm getting conflicting information here.Let me try to plot some points.At ( theta = 0 ): ( r = 0 ) → (0,0)At ( theta = frac{pi}{4} ): ( r = sin frac{pi}{4} = frac{sqrt{2}}{2} ) → ( x = frac{sqrt{2}}{2} cos frac{pi}{4} = frac{sqrt{2}}{2} cdot frac{sqrt{2}}{2} = frac{1}{2} ), ( y = frac{sqrt{2}}{2} sin frac{pi}{4} = frac{1}{2} ) → (0.5, 0.5)At ( theta = frac{pi}{2} ): ( r = 1 ) → (0,1)At ( theta = frac{3pi}{4} ): ( r = sin frac{3pi}{4} = frac{sqrt{2}}{2} ) → ( x = frac{sqrt{2}}{2} cos frac{3pi}{4} = frac{sqrt{2}}{2} cdot (-frac{sqrt{2}}{2}) = -frac{1}{2} ), ( y = frac{sqrt{2}}{2} sin frac{3pi}{4} = frac{sqrt{2}}{2} cdot frac{sqrt{2}}{2} = frac{1}{2} ) → (-0.5, 0.5)At ( theta = pi ): ( r = 0 ) → (0,0)So from ( 0 ) to ( pi ), we've gone from the origin up to (0,1), then to (-0.5, 0.5), and back to the origin. That's only the upper half of the circle.Wait, but according to the parametric equations, it should have traced the entire circle. Maybe I'm missing something.Wait, let me think again about the parametric equations.We have:[x = frac{1}{2} sin 2theta][y = frac{1}{2} - frac{1}{2} cos 2theta]So, as ( theta ) goes from 0 to ( pi ), ( 2theta ) goes from 0 to ( 2pi ), which should trace the entire circle once.But when I plot points from ( theta = 0 ) to ( theta = pi ), I only see the upper half.Wait, maybe I'm not accounting for the negative ( r ) values correctly.Wait, when ( theta ) is between ( pi ) and ( 2pi ), ( r ) is negative, so the points are plotted in the opposite direction.So, for example, at ( theta = frac{3pi}{2} ), ( r = -1 ), which is the same as ( r = 1 ) at ( theta = frac{3pi}{2} + pi = frac{5pi}{2} ), which is equivalent to ( theta = frac{pi}{2} ).But in reality, ( theta = frac{3pi}{2} ) is pointing downward, so ( r = -1 ) would be plotted at ( theta = frac{3pi}{2} + pi = frac{5pi}{2} ), which is the same as ( theta = frac{pi}{2} ), but that's already plotted.Wait, so maybe the entire circle is traced when ( theta ) goes from 0 to ( pi ), because the negative ( r ) values beyond ( pi ) just retrace the same points.But that contradicts the parametric equations, which suggest that ( theta ) needs to go to ( pi ) to complete the circle.Wait, perhaps I'm overcomplicating this.Let me think about the graph of ( r = sin theta ). It's a circle that is traced twice as ( theta ) goes from 0 to ( 2pi ). So, to trace it once, you only need ( theta ) to go from 0 to ( pi ).But wait, when I plot it, from 0 to ( pi ), I only get the upper half, and from ( pi ) to ( 2pi ), I get the lower half.But according to the parametric equations, it's a full circle when ( theta ) goes from 0 to ( pi ).Hmm, maybe I'm confusing the parameterization.Wait, let's consider the parametric equations again:[x = frac{1}{2} sin 2theta][y = frac{1}{2} - frac{1}{2} cos 2theta]If ( theta ) goes from 0 to ( pi ), then ( 2theta ) goes from 0 to ( 2pi ), so ( x ) and ( y ) trace the entire circle once.But when I plot points from ( theta = 0 ) to ( theta = pi ), I only see the upper half. So maybe the parametric equations are correct, but my point-by-point plotting is incomplete.Wait, no, because when ( theta ) is between ( pi/2 ) and ( pi ), ( r ) is decreasing from 1 to 0, but in the negative x direction.Wait, let me plot ( theta = pi ): ( r = 0 ), so (0,0).At ( theta = pi/2 ): (0,1).At ( theta = pi/4 ): (0.5, 0.5).At ( theta = 3pi/4 ): (-0.5, 0.5).So, from ( 0 ) to ( pi ), the graph goes from (0,0) up to (0,1), then to (-0.5, 0.5), and back to (0,0). That's only the upper half.But according to the parametric equations, it should trace the entire circle. So maybe the parametric equations are not directly mapping to the polar plot.Wait, perhaps the issue is that in polar coordinates, when ( r ) is negative, it's plotting points in the opposite direction, so the lower half is plotted when ( r ) is negative, but that happens when ( theta ) is between ( pi ) and ( 2pi ).But if I only go up to ( theta = pi ), I don't get the lower half.Wait, so maybe the entire circle is traced when ( theta ) goes from 0 to ( pi ), but the lower half is plotted as ( r ) becomes negative beyond ( pi ).Wait, no, because ( r ) becomes negative beyond ( pi ), but in the parametric equations, the entire circle is traced when ( theta ) goes from 0 to ( pi ).I'm confused now.Let me try to think differently.In polar coordinates, the equation ( r = sin theta ) is symmetric about the y-axis because ( sin theta ) is symmetric about ( theta = pi/2 ).So, when ( theta ) goes from 0 to ( pi ), it traces the right half of the circle, and from ( pi ) to ( 2pi ), it traces the left half.But wait, no, because ( r = sin theta ) is positive from 0 to ( pi ), and negative from ( pi ) to ( 2pi ).So, when ( r ) is positive, it's plotting the upper half, and when ( r ) is negative, it's plotting the lower half.Wait, that makes more sense.So, from ( 0 ) to ( pi ), ( r ) is positive, so it's plotting the upper half of the circle.From ( pi ) to ( 2pi ), ( r ) is negative, so it's plotting the lower half of the circle.Therefore, to get the entire circle, ( theta ) needs to go from 0 to ( 2pi ).But wait, earlier, the parametric equations suggested that ( theta ) only needs to go to ( pi ).I think the confusion arises because in the parametric equations, ( phi = 2theta ), so ( theta ) only needs to go to ( pi ) to cover ( phi ) from 0 to ( 2pi ).But in polar coordinates, the negative ( r ) values beyond ( pi ) are necessary to plot the lower half.Wait, maybe the parametric equations are considering the entire circle in one sweep, but in polar coordinates, the negative ( r ) values are necessary to complete the circle.So, perhaps the smallest ( t ) is ( pi ), because the parametric equations cover the entire circle when ( theta ) goes to ( pi ).But when I plot it, I only see the upper half. So, maybe I'm missing something.Wait, let me think about the graph.When ( theta ) is 0, ( r = 0 ).As ( theta ) increases to ( pi/2 ), ( r ) increases to 1, so the point moves up to (0,1).As ( theta ) increases beyond ( pi/2 ), ( r ) starts decreasing, but since ( r ) is still positive, it moves back towards the origin, but in the negative x direction.So, at ( theta = pi ), ( r = 0 ), so it's back at the origin.But in the parametric equations, this should have traced the entire circle.Wait, maybe the parametric equations are correct, and the polar plot is just overlapping.Wait, let me consider the parametric equations again.If ( theta ) goes from 0 to ( pi ), then ( phi = 2theta ) goes from 0 to ( 2pi ), so the parametric equations trace the entire circle once.But in polar coordinates, when ( theta ) goes from 0 to ( pi ), ( r ) is positive, so it's only plotting the upper half.Wait, so perhaps the parametric equations are not directly reflecting the polar plot.I think I need to reconcile these two perspectives.In the parametric equations, the entire circle is traced as ( theta ) goes from 0 to ( pi ), but in the polar plot, the lower half is plotted when ( r ) is negative, which happens when ( theta ) is between ( pi ) and ( 2pi ).Therefore, to get the entire circle in polar coordinates, ( theta ) needs to go from 0 to ( 2pi ).But wait, that contradicts the parametric equations.I think the key is that in the parametric equations, the entire circle is traced as ( theta ) goes from 0 to ( pi ), but in the polar plot, the negative ( r ) values beyond ( pi ) are necessary to plot the lower half.Wait, but in the parametric equations, the entire circle is traced without needing negative ( r ) values because the parameterization inherently covers the entire circle as ( theta ) goes from 0 to ( pi ).So, perhaps in the polar plot, the negative ( r ) values are redundant because the parametric equations already cover the entire circle.Wait, but in reality, when you plot ( r = sin theta ) in polar coordinates, you do need ( theta ) to go from 0 to ( 2pi ) to get the entire circle.Wait, let me check online or in my notes.Wait, I recall that ( r = sin theta ) is a circle that is traced once as ( theta ) goes from 0 to ( pi ), and then retraced as ( theta ) goes from ( pi ) to ( 2pi ).Wait, that makes sense because when ( theta ) is between ( pi ) and ( 2pi ), ( r ) is negative, which reflects the points across the origin, effectively retracing the circle.So, the entire circle is traced when ( theta ) goes from 0 to ( pi ), and then it retraces the same circle from ( pi ) to ( 2pi ).Therefore, the smallest ( t ) needed to plot the entire circle is ( pi ).But wait, earlier, when I plotted points from 0 to ( pi ), I only saw the upper half.Wait, maybe I'm missing that the parametric equations cover the entire circle, but the polar plot only shows the upper half when ( r ) is positive.Wait, no, because when ( r ) is negative, it's plotting the lower half.Wait, let me think again.When ( theta ) is between ( pi ) and ( 2pi ), ( r = sin theta ) is negative, so the points are plotted in the opposite direction.So, for example, at ( theta = frac{3pi}{2} ), ( r = -1 ), which is plotted at ( theta = frac{3pi}{2} + pi = frac{5pi}{2} ), which is the same as ( theta = frac{pi}{2} ), but that's already plotted.Wait, so maybe the lower half is plotted as ( theta ) goes from ( pi ) to ( 2pi ), but it's just overlapping with the upper half.Wait, no, because when ( r ) is negative, it's plotting points in the opposite direction, so the lower half is plotted in the lower half-plane.Wait, let me take specific points.At ( theta = frac{3pi}{2} ), ( r = -1 ), so the point is ( (-1 cdot cos frac{3pi}{2}, -1 cdot sin frac{3pi}{2}) = (0, 1) ).Wait, that's the same as ( theta = frac{pi}{2} ), ( r = 1 ).Wait, so it's the same point.Wait, so when ( r ) is negative, it's plotting the same points as when ( r ) is positive but in the opposite direction.So, in effect, the entire circle is plotted when ( theta ) goes from 0 to ( pi ), and then it retraces the same points from ( pi ) to ( 2pi ).Therefore, the smallest ( t ) needed to plot the entire circle is ( pi ).But earlier, when I plotted points from 0 to ( pi ), I only saw the upper half.Wait, maybe I'm not accounting for the fact that the parametric equations cover the entire circle, but the polar plot only shows the upper half when ( r ) is positive.Wait, no, because when ( r ) is negative, it's plotting points in the lower half-plane.Wait, let me take ( theta = frac{3pi}{2} ), ( r = -1 ). So, in polar coordinates, that's ( (-1, frac{3pi}{2}) ), which is equivalent to ( (1, frac{3pi}{2} + pi) = (1, frac{5pi}{2}) ), which is the same as ( (1, frac{pi}{2}) ), which is (0,1). So, it's the same point as ( theta = frac{pi}{2} ).Wait, so when ( r ) is negative, it's not plotting a new point, but rather the same point as when ( r ) is positive.Therefore, the entire circle is plotted when ( theta ) goes from 0 to ( pi ), and then from ( pi ) to ( 2pi ), it retraces the same points.Therefore, the smallest ( t ) needed is ( pi ).But wait, when I think about the parametric equations, they suggest that the entire circle is traced when ( theta ) goes from 0 to ( pi ), which is consistent with this conclusion.So, I think the answer is ( pi ).But wait, let me check with another approach.The general equation ( r = a sin theta ) is a circle with diameter ( a ) along the y-axis.To plot the entire circle, you need ( theta ) to go from 0 to ( pi ), because beyond ( pi ), it retraces the same points.Therefore, the smallest ( t ) is ( pi ).Wait, but I'm still confused because when I plot points from 0 to ( pi ), I only see the upper half.Wait, maybe I'm not considering that the parametric equations cover the entire circle, but in the polar plot, the lower half is plotted when ( r ) is negative, which happens beyond ( pi ).Wait, but if the parametric equations cover the entire circle when ( theta ) goes to ( pi ), then the polar plot must also cover the entire circle when ( theta ) goes to ( pi ).Wait, maybe I'm overcomplicating it.Let me think about the graph of ( r = sin theta ).It's a circle that is traced once as ( theta ) goes from 0 to ( pi ), and then retraced as ( theta ) goes from ( pi ) to ( 2pi ).Therefore, the smallest ( t ) needed is ( pi ).Yes, that makes sense.So, the answer is ( pi ).

💡First, identify the number of action figures and their respective values. Jimmy has 4 figures worth 15 each and 1 figure worth 20.Next, determine the selling price for each figure by subtracting 5 from their original value. This means the 15 figures will be sold for 10 each, and the 20 figure will be sold for 15.Calculate the total earnings from the 15 figures: 4 figures multiplied by 10 each equals 40.Then, calculate the earnings from the 20 figure: 1 figure multiplied by 15 equals 15.Finally, add the earnings from both sets of figures to find the total amount Jimmy will earn: 40 plus 15 equals 55.