💡Alright, so I have this geometric series to compute: 2 - 4 + 8 - 16 + ... + 2048. Hmm, okay, let me think about how to approach this. I remember that a geometric series is one where each term is obtained by multiplying the previous term by a constant ratio. First, I need to identify the first term, the common ratio, and the number of terms in the series. The first term is clearly 2. Now, looking at the series: 2, -4, 8, -16, and so on. Each term is obtained by multiplying the previous term by -2. So, the common ratio (r) is -2. That makes sense because 2 * (-2) = -4, -4 * (-2) = 8, and so on.Next, I need to figure out how many terms there are in this series. It starts at 2 and goes up to 2048. Let me list out the terms to count them:1. 22. -43. 84. -165. 326. -647. 1288. -2569. 51210. -102411. 2048So, there are 11 terms in total. Okay, that's good to know.Now, I recall the formula for the sum of a geometric series: S_n = a * (r^n - 1) / (r - 1), where S_n is the sum of the first n terms, a is the first term, r is the common ratio, and n is the number of terms. Plugging in the values I have:- a = 2- r = -2- n = 11So, the sum S_11 would be:S_11 = 2 * ((-2)^11 - 1) / (-2 - 1)Let me compute (-2)^11 first. Since the exponent is odd, the result will be negative. 2^11 is 2048, so (-2)^11 is -2048.Now, substituting back into the formula:S_11 = 2 * (-2048 - 1) / (-3)S_11 = 2 * (-2049) / (-3)Multiplying 2 by -2049 gives -4098. Then, dividing by -3:-4098 / -3 = 1366So, the sum of the series is 1366.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting from the beginning:First term (a) = 2Common ratio (r) = -2Number of terms (n) = 11Sum formula: S_n = a * (r^n - 1) / (r - 1)Calculating r^n: (-2)^11 = -2048So, S_11 = 2 * (-2048 - 1) / (-2 - 1) = 2 * (-2049) / (-3) = (-4098) / (-3) = 1366Yes, that seems correct. The negative signs cancel out, and 4098 divided by 3 is indeed 1366.Just to be thorough, maybe I can add the terms manually to verify:2 - 4 = -2-2 + 8 = 66 - 16 = -10-10 + 32 = 2222 - 64 = -42-42 + 128 = 8686 - 256 = -170-170 + 512 = 342342 - 1024 = -682-682 + 2048 = 1366Wow, adding them step by step also gives me 1366. That matches the result from the formula. So, I'm confident that the sum is correct.I think I've covered all the steps and double-checked my work. It seems like the sum of the geometric series is indeed 1366.

💡Alright, so I'm trying to figure out how long it takes for one runner to complete the entire track. There are three runners moving at equal constant speeds on a circular track. When two runners meet, they instantly turn around and start running in the opposite direction. Here's what I know from the problem:1. At a certain moment, the first runner meets the second runner.2. After 20 minutes, the second runner meets the third runner for the first time.3. Another half hour (which is 30 minutes) later, the third runner meets the first runner for the first time.I need to find out how many minutes it takes for one runner to complete the entire track.Okay, let's break this down step by step.First, all runners are moving at the same constant speed. That means their speeds are equal, so when they meet, they just turn around and continue at the same speed but in the opposite direction. Let me visualize the circular track. Since it's circular, the distance between any two points is the same in both directions. But because the runners are turning around when they meet, their paths are effectively changing directions each time they meet someone.Let's denote the time it takes for one runner to complete the entire track as ( T ) minutes. This is the value we need to find.Now, let's consider the events:1. Runner 1 meets Runner 2 at time ( t = 0 ).2. Runner 2 meets Runner 3 at time ( t = 20 ) minutes.3. Runner 3 meets Runner 1 at time ( t = 20 + 30 = 50 ) minutes.Wait, but the problem says "another half hour later," so it's 20 minutes after the first meeting, then 30 minutes after that, which is 50 minutes from the start. So, the third meeting happens at 50 minutes.Now, I need to relate these meeting times to the total time ( T ) it takes to complete the track.Since all runners have the same speed, the time between meetings should be related to the distance between them on the track. But since they turn around upon meeting, their relative motion changes.Let me think about the meetings:- When Runner 1 and Runner 2 meet at ( t = 0 ), they turn around and start moving in the opposite direction.- Then, Runner 2 meets Runner 3 at ( t = 20 ) minutes. So, Runner 2 has been moving for 20 minutes since the last meeting.- Similarly, Runner 3 meets Runner 1 at ( t = 50 ) minutes, which is 30 minutes after the second meeting.Hmm, maybe I can model their positions over time.Let's assume the track has a circumference ( C ). Since all runners have the same speed ( v ), the time to complete the track is ( T = frac{C}{v} ).But since they turn around upon meeting, their effective speed relative to each other changes.Wait, maybe I can think in terms of relative speed.When two runners are moving towards each other, their relative speed is ( 2v ). When they are moving in the same direction, their relative speed is ( 0 ) because they are moving at the same speed.But in this case, when they meet, they turn around, so their direction changes.This is getting a bit complicated. Maybe I can think about the meetings as events that happen at certain intervals.Let me try to find the time intervals between meetings and relate them to the total time ( T ).First, Runner 1 and Runner 2 meet at ( t = 0 ). Then, Runner 2 and Runner 3 meet at ( t = 20 ) minutes. Then, Runner 3 and Runner 1 meet at ( t = 50 ) minutes.So, the time between the first and second meeting is 20 minutes, and between the second and third meeting is 30 minutes.I need to find a relationship between these times and the total time ( T ).Since the runners are moving at the same speed, the distance between them when they meet should be related to the time intervals.Let me consider the distance covered by Runner 2 between ( t = 0 ) and ( t = 20 ) minutes. Since Runner 2 turns around at ( t = 0 ), they are moving in the opposite direction for 20 minutes until they meet Runner 3.Similarly, Runner 3, after meeting Runner 2 at ( t = 20 ), turns around and moves in the opposite direction until they meet Runner 1 at ( t = 50 ).So, Runner 3 is moving for 30 minutes from ( t = 20 ) to ( t = 50 ).Similarly, Runner 1, after meeting Runner 2 at ( t = 0 ), turns around and moves in the opposite direction until they meet Runner 3 at ( t = 50 ).So, Runner 1 is moving for 50 minutes from ( t = 0 ) to ( t = 50 ).Wait, but Runner 1 is moving for 50 minutes, but the total time to complete the track is ( T ). So, if Runner 1 has covered the entire track in ( T ) minutes, then in 50 minutes, they would have covered ( frac{50}{T} ) of the track.Similarly, Runner 2, moving for 20 minutes, covers ( frac{20}{T} ) of the track.Runner 3, moving for 30 minutes, covers ( frac{30}{T} ) of the track.But how do these distances relate to each other?Since they meet at certain points, the distances they cover should add up to the circumference ( C ) or some multiple of it.Wait, let's think about the first meeting between Runner 1 and Runner 2 at ( t = 0 ). They are at the same point, then turn around.Then, Runner 2 moves for 20 minutes and meets Runner 3. So, Runner 2 has covered ( frac{20}{T} ) of the track in that time.Similarly, Runner 3, who was moving in the opposite direction, has covered ( frac{20}{T} ) of the track as well, but in the opposite direction.Wait, but since they meet, the sum of the distances they cover should be equal to the circumference ( C ).So, ( frac{20}{T} + frac{20}{T} = 1 ) (since ( C = 1 ) in terms of fractions).Wait, that would mean ( frac{40}{T} = 1 ), so ( T = 40 ) minutes.But that contradicts the next meeting at ( t = 50 ) minutes.Wait, maybe I'm oversimplifying.Let me think again.When Runner 2 and Runner 3 meet at ( t = 20 ), the total distance covered by both since their last meeting should be equal to the circumference ( C ).But Runner 2 has been moving for 20 minutes, and Runner 3 has been moving for 20 minutes as well, but in the opposite direction.So, the distance covered by Runner 2 is ( frac{20}{T} ), and the distance covered by Runner 3 is ( frac{20}{T} ).Since they are moving towards each other, the sum of their distances should be equal to the circumference ( C ).So, ( frac{20}{T} + frac{20}{T} = 1 ), which gives ( T = 40 ) minutes.But then, at ( t = 50 ) minutes, Runner 3 meets Runner 1.So, Runner 3 has been moving for 30 minutes since ( t = 20 ), covering ( frac{30}{T} ) of the track.Similarly, Runner 1 has been moving for 50 minutes since ( t = 0 ), covering ( frac{50}{T} ) of the track.But since they meet, the sum of their distances should be equal to the circumference ( C ).So, ( frac{30}{T} + frac{50}{T} = 1 ), which gives ( frac{80}{T} = 1 ), so ( T = 80 ) minutes.But this contradicts the earlier result of ( T = 40 ) minutes.Hmm, so there's inconsistency here. Maybe my assumption that the sum of their distances equals the circumference is incorrect.Wait, perhaps I need to consider the direction changes.When Runner 1 and Runner 2 meet at ( t = 0 ), they turn around. So, Runner 1 is now moving in the opposite direction, and Runner 2 is also moving in the opposite direction.Then, Runner 2 meets Runner 3 at ( t = 20 ). So, Runner 2 has been moving for 20 minutes in the opposite direction, and Runner 3 has been moving for 20 minutes in the original direction.Wait, no, Runner 3 was moving in the original direction until ( t = 20 ), when they meet Runner 2 and turn around.So, Runner 3 was moving in the original direction for 20 minutes, covering ( frac{20}{T} ) of the track.Similarly, Runner 2 was moving in the opposite direction for 20 minutes, covering ( frac{20}{T} ) of the track.Since they meet, the sum of their distances should be equal to the circumference ( C ).So, ( frac{20}{T} + frac{20}{T} = 1 ), which again gives ( T = 40 ) minutes.But then, at ( t = 50 ), Runner 3 meets Runner 1.Runner 3 has been moving in the opposite direction since ( t = 20 ), so for 30 minutes, covering ( frac{30}{T} ) of the track.Runner 1 has been moving in the opposite direction since ( t = 0 ), so for 50 minutes, covering ( frac{50}{T} ) of the track.Since they meet, the sum of their distances should be equal to the circumference ( C ).So, ( frac{30}{T} + frac{50}{T} = 1 ), which gives ( frac{80}{T} = 1 ), so ( T = 80 ) minutes.But this contradicts ( T = 40 ) minutes from the earlier calculation.This inconsistency suggests that my approach is flawed.Maybe I need to consider the relative speeds and the fact that the runners are turning around, which changes their direction and thus their relative motion.Let me try a different approach.Let's assume that the track has circumference ( C ), and each runner has speed ( v ). So, the time to complete the track is ( T = frac{C}{v} ).When two runners meet, they turn around. So, their direction changes, but their speed remains the same.Let's consider the first meeting at ( t = 0 ): Runner 1 and Runner 2 meet and turn around.Then, Runner 2 starts moving in the opposite direction and meets Runner 3 at ( t = 20 ) minutes.Similarly, Runner 3, after meeting Runner 2, turns around and meets Runner 1 at ( t = 50 ) minutes.Let me try to model the positions of the runners over time.Let's set ( t = 0 ) as the time when Runner 1 and Runner 2 meet at position ( A ).At ( t = 0 ), Runner 1 is at ( A ) moving clockwise, and Runner 2 is at ( A ) moving counterclockwise.Wait, no, they meet at ( A ), and then turn around. So, Runner 1 was moving clockwise and now moves counterclockwise, and Runner 2 was moving counterclockwise and now moves clockwise.Wait, actually, the problem doesn't specify their initial directions, only that they turn around upon meeting.So, perhaps it's better to assume that all runners are moving in the same direction initially, say clockwise.Then, when two runners meet, they turn around and start moving counterclockwise.But this might complicate things.Alternatively, maybe it's better to consider that upon meeting, they reverse their direction.So, if Runner 1 and Runner 2 meet at ( t = 0 ), they both reverse direction.Then, Runner 2, who was moving in one direction, now moves in the opposite direction, and Runner 1 also reverses.Then, Runner 2 meets Runner 3 at ( t = 20 ), so Runner 2 has been moving in the opposite direction for 20 minutes.Similarly, Runner 3, who was moving in the original direction, meets Runner 2 at ( t = 20 ), so Runner 3 has been moving for 20 minutes in the original direction.Then, Runner 3 reverses direction and meets Runner 1 at ( t = 50 ), so Runner 3 has been moving in the opposite direction for 30 minutes.Similarly, Runner 1, after reversing at ( t = 0 ), has been moving in the opposite direction for 50 minutes.So, let's model their positions.Let me denote the positions as fractions of the track, where ( 0 ) is the starting point ( A ), and the track is a circle, so positions wrap around modulo 1.Let me assign:- Runner 1: starts at ( 0 ), moving clockwise, then reverses at ( t = 0 ) to move counterclockwise.- Runner 2: starts at ( 0 ), moving counterclockwise, then reverses at ( t = 0 ) to move clockwise.- Runner 3: starts at some position ( x ), moving clockwise.Wait, but the problem doesn't specify where Runner 3 starts. Hmm, this complicates things.Alternatively, maybe all runners start at the same point, but that would mean they all meet at ( t = 0 ), which contradicts the problem statement.Wait, the problem says "at a certain moment, the first runner met the second runner." So, it's not necessarily at the starting point, but at some point on the track.So, perhaps Runner 1 and Runner 2 meet at position ( A ) at ( t = 0 ), and Runner 3 is somewhere else on the track.This is getting too vague. Maybe I need to use relative motion.Let me consider the relative speed between runners.When two runners are moving towards each other, their relative speed is ( 2v ). When moving in the same direction, their relative speed is ( 0 ) because they have the same speed.But since they turn around upon meeting, their direction changes, which affects their relative motion.Let me try to model the meetings.First meeting: Runner 1 and Runner 2 meet at ( t = 0 ). They turn around.Second meeting: Runner 2 and Runner 3 meet at ( t = 20 ).Third meeting: Runner 3 and Runner 1 meet at ( t = 50 ).I need to find ( T ), the time to complete the track.Let me denote the positions where they meet as fractions of the track.Let me assume that the track is a circle with circumference 1 unit (so ( C = 1 )), and the speed ( v = frac{1}{T} ) units per minute.So, in ( t ) minutes, a runner covers ( frac{t}{T} ) units.Now, let's consider the first meeting at ( t = 0 ): Runner 1 and Runner 2 meet at position ( A ).Then, Runner 2 moves in the opposite direction and meets Runner 3 at ( t = 20 ).So, Runner 2 has moved ( frac{20}{T} ) units in the opposite direction.Similarly, Runner 3 has moved ( frac{20}{T} ) units in the original direction.Since they meet, the sum of the distances they have covered since their last meeting should be equal to the circumference, which is 1.Wait, but they are moving towards each other, so the distance between them when Runner 2 started moving was ( d ), and they covered ( d ) together in 20 minutes.But I don't know the initial distance between Runner 2 and Runner 3.This is getting too abstract. Maybe I need to set up equations based on the meetings.Let me denote:- ( T ): time to complete the track.- ( v = frac{1}{T} ): speed of each runner.- ( t_1 = 0 ): time when Runner 1 and Runner 2 meet.- ( t_2 = 20 ): time when Runner 2 and Runner 3 meet.- ( t_3 = 50 ): time when Runner 3 and Runner 1 meet.Let me consider the positions of the runners at these times.At ( t = 0 ):- Runner 1 is at position ( A ), moving clockwise.- Runner 2 is at position ( A ), moving counterclockwise.- Runner 3 is at position ( B ), moving clockwise.Wait, but the problem doesn't specify where Runner 3 is initially. Hmm.Alternatively, maybe all runners are evenly spaced on the track. If there are three runners, they could be spaced at 120-degree intervals, which is ( frac{1}{3} ) of the track apart.But the problem doesn't specify their initial positions, so I might need to make an assumption.Alternatively, maybe I can consider the track as a unit circle, and assign positions based on the meetings.Let me try to model the positions.Let me assume that at ( t = 0 ), Runner 1 and Runner 2 meet at position ( 0 ).Then, Runner 1 turns around and starts moving counterclockwise, and Runner 2 turns around and starts moving clockwise.Runner 3 is somewhere else on the track, say at position ( x ), moving clockwise.Now, Runner 2 is moving clockwise from ( 0 ) at speed ( v ), and Runner 3 is moving clockwise from ( x ) at speed ( v ).They meet at ( t = 20 ).So, the distance Runner 2 covers in 20 minutes is ( frac{20}{T} ).The distance Runner 3 covers in 20 minutes is ( frac{20}{T} ).Since they are moving in the same direction, the distance between them initially was ( x ), and they meet when Runner 2 has caught up to Runner 3.But since they are moving in the same direction, the relative speed is ( 0 ), so they won't meet unless Runner 2 is behind Runner 3.Wait, but Runner 2 is moving clockwise from ( 0 ), and Runner 3 is moving clockwise from ( x ). If ( x ) is ahead of ( 0 ), Runner 2 needs to catch up.But since they meet at ( t = 20 ), Runner 2 must have covered the distance ( x ) plus the distance Runner 3 has moved in 20 minutes.Wait, no, if they are moving in the same direction, the relative speed is ( 0 ), so they won't meet unless they are at the same position.Wait, this is confusing.Alternatively, maybe Runner 2 is moving clockwise, and Runner 3 is moving counterclockwise.But no, the problem says when two runners meet, they turn around. So, Runner 3 was moving in some direction before meeting Runner 2.Wait, maybe Runner 3 was moving counterclockwise, and Runner 2, after meeting Runner 1, is moving clockwise.So, Runner 2 is moving clockwise, and Runner 3 is moving counterclockwise.Therefore, they are moving towards each other.So, their relative speed is ( 2v ).The distance between them at ( t = 0 ) is ( x ), which is the position of Runner 3.They meet at ( t = 20 ), so the time taken to meet is ( t = 20 ).The distance covered by both is ( x ), so:( 2v times 20 = x )But ( v = frac{1}{T} ), so:( 2 times frac{1}{T} times 20 = x )( frac{40}{T} = x )So, ( x = frac{40}{T} )Now, after meeting at ( t = 20 ), Runner 2 and Runner 3 turn around.Runner 2, who was moving clockwise, now moves counterclockwise.Runner 3, who was moving counterclockwise, now moves clockwise.Now, Runner 3 and Runner 1 will meet at ( t = 50 ).Let's see where Runner 1 is at ( t = 50 ).Runner 1 was at position ( 0 ) at ( t = 0 ), moving counterclockwise.From ( t = 0 ) to ( t = 50 ), Runner 1 has been moving counterclockwise for 50 minutes.So, Runner 1's position at ( t = 50 ) is ( frac{50}{T} ) counterclockwise from ( 0 ).Similarly, Runner 3 was at position ( x = frac{40}{T} ) at ( t = 0 ), moving counterclockwise until ( t = 20 ), then turns around and moves clockwise.From ( t = 20 ) to ( t = 50 ), Runner 3 moves clockwise for 30 minutes.So, Runner 3's position at ( t = 50 ) is:Starting from ( x = frac{40}{T} ), moving counterclockwise for 20 minutes to meet Runner 2 at position ( 0 + frac{20}{T} ) (since Runner 2 was moving clockwise from ( 0 )).Wait, no, Runner 2 was moving clockwise from ( 0 ), so Runner 2's position at ( t = 20 ) is ( frac{20}{T} ).But Runner 3 was moving counterclockwise from ( x = frac{40}{T} ), so Runner 3's position at ( t = 20 ) is ( frac{40}{T} - frac{20}{T} = frac{20}{T} ).So, they meet at position ( frac{20}{T} ).Then, Runner 3 turns around and starts moving clockwise.From ( t = 20 ) to ( t = 50 ), Runner 3 moves clockwise for 30 minutes, covering ( frac{30}{T} ).So, Runner 3's position at ( t = 50 ) is ( frac{20}{T} + frac{30}{T} = frac{50}{T} ).Similarly, Runner 1's position at ( t = 50 ) is ( frac{50}{T} ) counterclockwise from ( 0 ), which is the same as ( 1 - frac{50}{T} ) clockwise from ( 0 ).Wait, but Runner 3 is at ( frac{50}{T} ) clockwise from ( 0 ), and Runner 1 is at ( frac{50}{T} ) counterclockwise from ( 0 ).So, the distance between them is ( frac{50}{T} + frac{50}{T} = frac{100}{T} ).But since they meet at ( t = 50 ), the distance between them should be ( 0 ), meaning ( frac{100}{T} = 1 ), so ( T = 100 ) minutes.Wait, that makes sense.Let me verify.If ( T = 100 ) minutes, then:- Runner 1's speed is ( frac{1}{100} ) per minute.- Runner 2's speed is ( frac{1}{100} ) per minute.- Runner 3's speed is ( frac{1}{100} ) per minute.At ( t = 0 ), Runner 1 and Runner 2 meet at position ( 0 ).Runner 1 turns around and moves counterclockwise.Runner 2 turns around and moves clockwise.Runner 3 is at position ( x = frac{40}{100} = 0.4 ) on the track.Runner 3 is moving counterclockwise until ( t = 20 ).At ( t = 20 ):- Runner 2 has moved ( frac{20}{100} = 0.2 ) clockwise from ( 0 ), so Runner 2 is at position ( 0.2 ).- Runner 3 has moved ( frac{20}{100} = 0.2 ) counterclockwise from ( 0.4 ), so Runner 3 is at position ( 0.4 - 0.2 = 0.2 ).So, they meet at position ( 0.2 ).Then, Runner 2 turns around and moves counterclockwise.Runner 3 turns around and moves clockwise.At ( t = 50 ):- Runner 1 has moved ( frac{50}{100} = 0.5 ) counterclockwise from ( 0 ), so Runner 1 is at position ( 0.5 ).- Runner 3 has moved ( frac{30}{100} = 0.3 ) clockwise from ( 0.2 ), so Runner 3 is at position ( 0.2 + 0.3 = 0.5 ).So, they meet at position ( 0.5 ).This checks out.Therefore, the time it takes for one runner to complete the entire track is ( T = 100 ) minutes.

💡Okay, so I have this problem about a rhombus ABCD with an area of 50 square units. Points E and F are the midpoints of sides AB and CD respectively, and I need to find the area of triangle AEC. Hmm, let me try to visualize this first.Alright, a rhombus is a four-sided shape where all sides are equal in length. Also, the opposite sides are parallel, and the opposite angles are equal. The diagonals of a rhombus bisect each other at right angles. That's an important property. So, if I draw the diagonals AC and BD, they intersect at point O, which is the midpoint of both diagonals. And since they bisect each other at right angles, they form four right-angled triangles inside the rhombus.Given that the area of the rhombus is 50 square units, I remember that the area of a rhombus can be calculated in a couple of ways. One common formula is base times height, similar to a parallelogram. Another formula involves the lengths of the diagonals: area = (diagonal1 * diagonal2) / 2. Maybe I can use this formula to find the lengths of the diagonals, but I'm not sure yet.Wait, the problem mentions points E and F are midpoints of AB and CD. So, E is halfway along AB, and F is halfway along CD. Since AB and CD are opposite sides of the rhombus, and in a rhombus opposite sides are equal, so AE = EB and CF = FD.Now, the problem asks for the area of triangle AEC. Let me sketch this out mentally. Points A, E, and C form triangle AEC. So, A is one vertex, E is the midpoint of AB, and C is the opposite vertex from A. Hmm, I need to find the area of this triangle.Since E is the midpoint of AB, maybe I can use some properties of midsegments in triangles or something related to areas in triangles. But first, let me recall that in a rhombus, the diagonals bisect each other at right angles. So, if I consider diagonal AC, it splits the rhombus into two congruent triangles: triangle ABC and triangle ADC. Each of these triangles has an area of half the rhombus, so 25 square units each.Now, triangle AEC is part of triangle ABC. Since E is the midpoint of AB, maybe I can find the area of triangle AEC by considering the ratio of areas in triangle ABC. If I can find how triangle AEC relates to triangle ABC, I can find its area.Let me think about the base and height of triangle ABC. The base could be AB, and the height would be the distance from point C to the line AB. Similarly, for triangle AEC, the base is AE, which is half of AB, and the height remains the same because E is on AB. So, if the base is halved but the height stays the same, the area should also be halved.Wait, that makes sense. If the base is halved and the height remains the same, the area of the triangle would be half of the original. So, since triangle ABC has an area of 25 square units, triangle AEC should have half of that, which is 12.5 square units.But let me double-check to make sure I didn't make a mistake. Another way to think about it is using coordinates. Maybe assigning coordinates to the rhombus and calculating the area that way.Let's place the rhombus in a coordinate system. Let me assume point A is at (0, 0). Since it's a rhombus, all sides are equal, and the diagonals bisect each other at right angles. Let me denote the diagonals AC and BD intersecting at point O. Let me denote the length of diagonal AC as 2p and BD as 2q. Then, the area of the rhombus is (2p * 2q)/2 = 2pq. Given that the area is 50, so 2pq = 50, which means pq = 25.Now, since E is the midpoint of AB, let me find the coordinates of E. If A is at (0, 0), and B is at (a, b), then E would be at ((0 + a)/2, (0 + b)/2) = (a/2, b/2). Similarly, point C would be at (2p, 0) if I align AC along the x-axis, but wait, that might complicate things because the diagonals intersect at O, which would be at (p, q) if BD is along the y-axis.Wait, maybe I should align the rhombus such that diagonal AC is along the x-axis for simplicity. So, point A is at (-p, 0), point C is at (p, 0), and the diagonals intersect at O, which is the origin (0, 0). Then, points B and D would be at (0, q) and (0, -q) respectively. Hmm, but in a rhombus, all sides are equal, so the distance from A to B should be equal to the distance from B to C, and so on.Let me calculate the coordinates properly. If A is at (-p, 0), C is at (p, 0), and O is at (0, 0). Then, points B and D are at (0, q) and (0, -q). Now, the length of AB can be calculated using the distance formula. AB is from (-p, 0) to (0, q), so the distance is sqrt[(0 - (-p))² + (q - 0)²] = sqrt[p² + q²]. Similarly, BC is from (0, q) to (p, 0), which is also sqrt[(p - 0)² + (0 - q)²] = sqrt[p² + q²]. So, all sides are equal, which is consistent with a rhombus.Now, the area of the rhombus is given by (diagonal1 * diagonal2)/2 = (2p * 2q)/2 = 2pq. We know this is 50, so 2pq = 50, which simplifies to pq = 25.Now, point E is the midpoint of AB. So, coordinates of A are (-p, 0), and coordinates of B are (0, q). The midpoint E would be at [(-p + 0)/2, (0 + q)/2] = (-p/2, q/2). Similarly, point F is the midpoint of CD, which is from (p, 0) to (0, -q), so midpoint F is at [(p + 0)/2, (0 + (-q))/2] = (p/2, -q/2). But I don't think F is needed for this problem since we're only asked about triangle AEC.So, triangle AEC has points A(-p, 0), E(-p/2, q/2), and C(p, 0). To find the area of triangle AEC, I can use the shoelace formula or calculate the base and height.Let me try the shoelace formula. The coordinates are A(-p, 0), E(-p/2, q/2), and C(p, 0). Shoelace formula is:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|Plugging in the coordinates:x1 = -p, y1 = 0x2 = -p/2, y2 = q/2x3 = p, y3 = 0So,Area = |[(-p)(q/2 - 0) + (-p/2)(0 - 0) + p(0 - q/2)] / 2|Simplify each term:First term: (-p)(q/2 - 0) = (-p)(q/2) = -pq/2Second term: (-p/2)(0 - 0) = (-p/2)(0) = 0Third term: p(0 - q/2) = p(-q/2) = -pq/2So, adding them up:-pq/2 + 0 - pq/2 = -pqTake absolute value and divide by 2:| -pq | / 2 = pq / 2But we know that pq = 25, so:Area = 25 / 2 = 12.5So, the area of triangle AEC is 12.5 square units.Wait, that matches my earlier reasoning. So, whether I think about it in terms of the base being halved or use coordinate geometry, I get the same result. That makes me more confident that 12.5 is the correct answer.Just to recap, the key steps were:1. Recognize that in a rhombus, the diagonals bisect each other at right angles and split the rhombus into four congruent right triangles.2. The area of the rhombus is 50, so each of the two larger triangles (ABC and ADC) has an area of 25.3. Since E is the midpoint of AB, triangle AEC has half the base of triangle ABC but the same height, so its area is half of 25, which is 12.5.Alternatively, using coordinate geometry, assigning coordinates to the rhombus, finding the coordinates of E, and applying the shoelace formula also led to the same result.I think that's thorough enough. I don't see any mistakes in my reasoning, so I'm pretty sure the area of triangle AEC is 12.5 square units.

💡Okay, so I have this inequality: ( mx^2 + 4mx - 4 < 0 ). The question says that this inequality is always true for any real number ( x ). I need to find the range of ( m ) that satisfies this condition. The options are A, B, C, D, with different intervals for ( m ).First, I remember that for a quadratic inequality ( ax^2 + bx + c < 0 ) to hold for all real ( x ), the quadratic must open downwards (so the coefficient ( a ) must be negative) and it should never touch or cross the x-axis, meaning it has no real roots. This happens when the discriminant is negative.So, applying this to the given inequality ( mx^2 + 4mx - 4 < 0 ):1. **Coefficient Check**: The coefficient of ( x^2 ) is ( m ). For the quadratic to open downwards, ( m ) must be less than 0. So, ( m < 0 ).2. **Discriminant Check**: The discriminant ( D ) of a quadratic ( ax^2 + bx + c ) is ( D = b^2 - 4ac ). For no real roots, ( D < 0 ).Let's compute the discriminant for our quadratic:( D = (4m)^2 - 4(m)(-4) )( D = 16m^2 + 16m )We need ( D < 0 ):( 16m^2 + 16m < 0 )Factor out 16m:( 16m(m + 1) < 0 )So, the inequality simplifies to ( m(m + 1) < 0 ).To solve this inequality, let's find the critical points where the expression equals zero:( m = 0 ) and ( m = -1 ).These points divide the number line into intervals. We can test each interval to see where the product is negative.- **Interval 1**: ( m < -1 ) - Let's pick ( m = -2 ): - ( (-2)(-2 + 1) = (-2)(-1) = 2 ) which is positive. So, this interval doesn't satisfy the inequality.- **Interval 2**: ( -1 < m < 0 ) - Let's pick ( m = -0.5 ): - ( (-0.5)(-0.5 + 1) = (-0.5)(0.5) = -0.25 ) which is negative. So, this interval satisfies the inequality.- **Interval 3**: ( m > 0 ) - Let's pick ( m = 1 ): - ( (1)(1 + 1) = 2 ) which is positive. So, this interval doesn't satisfy the inequality.Therefore, the solution to ( m(m + 1) < 0 ) is ( -1 < m < 0 ).But wait, earlier we had the condition that ( m < 0 ). So, combining both conditions:- From the coefficient check: ( m < 0 )- From the discriminant check: ( -1 < m < 0 )So, the range of ( m ) that satisfies both conditions is ( -1 < m < 0 ).But hold on, let's check the case when ( m = 0 ). If ( m = 0 ), the original inequality becomes ( 0 + 0 - 4 < 0 ), which simplifies to ( -4 < 0 ). That's always true. So, ( m = 0 ) is also a valid solution.Wait, but earlier, from the discriminant condition, ( m ) had to be greater than ( -1 ) and less than 0. But when ( m = 0 ), the quadratic becomes linear, and the inequality still holds. So, should ( m = 0 ) be included?Looking back, when ( m = 0 ), the quadratic term disappears, and we have a linear inequality. Since the inequality is still true for all ( x ), ( m = 0 ) is acceptable.But in the discriminant condition, ( m ) had to be between ( -1 ) and ( 0 ). So, combining both, ( m ) can be from ( -1 ) to ( 0 ), but when ( m = 0 ), it's still valid. So, the range should be ( -1 < m leq 0 ).But wait, let me double-check. If ( m = -1 ), what happens?Plugging ( m = -1 ) into the discriminant:( D = 16(-1)^2 + 16(-1) = 16(1) - 16 = 0 )So, the discriminant is zero, which means the quadratic touches the x-axis at one point. Therefore, the inequality ( mx^2 + 4mx - 4 < 0 ) would not hold for all ( x ) because at the vertex, the quadratic equals zero, which is not less than zero. Hence, ( m = -1 ) is not allowed.Similarly, when ( m = 0 ), as we saw, the inequality holds for all ( x ). So, ( m = 0 ) is allowed, but ( m = -1 ) is not.Therefore, the range of ( m ) is ( -1 < m leq 0 ).Looking back at the options:A: ( -1 leq m leq 0 )B: ( -1 leq m < 0 )C: ( -1 < m < 0 )D: ( -1 < m leq 0 )So, the correct option is D: ( -1 < m leq 0 ).I think that makes sense. Let me just recap:- For the quadratic to be always negative, it must open downward (( m < 0 )) and have no real roots (( D < 0 )).- Solving ( D < 0 ) gives ( -1 < m < 0 ).- Additionally, ( m = 0 ) is allowed because the inequality still holds.- Therefore, combining these, ( m ) can be from just above ( -1 ) up to and including ( 0 ).Yeah, that seems right. So, the answer should be D.

💡Alright, so I have this problem here with two sets, M and N, and I need to find their intersection. Let me try to understand what each set represents.First, set M is defined as {x | |x - 1| < 1}. Okay, that looks like an absolute value inequality. I remember that |x - a| < b means that x is within b units of a. So in this case, |x - 1| < 1 means that x is within 1 unit of 1. That should translate to x being between 0 and 2. Let me write that down:M = {x | 0 < x < 2}Okay, that seems straightforward. Now, set N is defined as {y | y = log₂(x² + 2x + 3)}. Hmm, this one is a bit trickier. It's a logarithmic function, and I need to figure out what values y can take based on x. But wait, the problem is asking for the intersection M ∩ N, which means I need to find the x values that are in both sets. But N is defined in terms of y, which is a function of x. So maybe I need to find the domain of N in terms of x and then see where it overlaps with M.Let me think. For the logarithm to be defined, the argument inside the log must be positive. So x² + 2x + 3 > 0. Let me check if this is always true. The quadratic x² + 2x + 3 can be rewritten as (x + 1)² + 2. Since (x + 1)² is always non-negative, adding 2 makes it always positive. So x² + 2x + 3 is always greater than 0 for all real x. That means the domain of N is all real numbers.But wait, N is defined as {y | y = log₂(x² + 2x + 3)}, so y can take any value depending on x. But I need to find the intersection of M and N, which are both sets of x values. So maybe I need to find the range of y and then see how that relates back to x in M.Let me find the range of y. Since x² + 2x + 3 = (x + 1)² + 2, the minimum value of this expression is 2, which occurs when x = -1. So the argument of the logarithm is always at least 2. Therefore, y = log₂(x² + 2x + 3) is always at least log₂(2) = 1. So y ≥ 1.But how does this relate back to x? I need to find the x values in M such that y ≥ 1. Since y = log₂(x² + 2x + 3) ≥ 1, that implies x² + 2x + 3 ≥ 2. Let's solve this inequality:x² + 2x + 3 ≥ 2x² + 2x + 1 ≥ 0(x + 1)² ≥ 0Well, (x + 1)² is always greater than or equal to 0, so this inequality holds for all real x. That means y is always greater than or equal to 1, regardless of x. But wait, that doesn't help me narrow down x any further because it's always true.Hmm, maybe I'm approaching this wrong. The intersection M ∩ N would be the set of x values that are in both M and N. Since M is {x | 0 < x < 2} and N is defined in terms of y, which is a function of x, I need to see for which x in M does y satisfy the condition of N.But N is just the set of all y values, which are log₂(x² + 2x + 3). Since y is defined for all x, and M restricts x to be between 0 and 2, the intersection should just be the x values in M that also satisfy the condition for N. But since N is defined for all x, the intersection is just M itself.Wait, that doesn't make sense because the options don't include M itself. Let me look back at the problem. Maybe I misinterpreted N. It says N = {y | y = log₂(x² + 2x + 3)}. So N is a set of y values, not x values. But M is a set of x values. So how do we find the intersection of M and N? They are sets of different variables.Oh, I see now. Maybe the problem is actually asking for the intersection in terms of x, where both conditions are satisfied. So M is about x, and N is about y, but y is a function of x. So perhaps we need to find x such that x is in M and y is in N. But since y is determined by x, it's more about finding x in M such that y satisfies the condition for N.But N is just the set of all y values from the function, so y is always defined and greater than or equal to 1. So maybe the intersection is just M, but since M is 0 < x < 2 and N requires y ≥ 1, which translates to x² + 2x + 3 ≥ 2, which is always true, so the intersection is still M.But the options don't have M as an answer. The options are:A: {x | 1 ≤ x < 2}B: {x | 0 < x < 2}C: {x | 1 < x < 2}D: ∅So B is M itself. But I thought the intersection would be M, but maybe I'm missing something. Let me double-check.Wait, N is {y | y = log₂(x² + 2x + 3)}, which is a set of y values. M is a set of x values. So their intersection would be the set of x values that are in both M and N. But N is a set of y values, not x values. So unless there's a misunderstanding, maybe the problem is miswritten.Alternatively, perhaps N is intended to be a set of x values such that y is defined or something. Maybe it's a typo, and N should be {x | y = log₂(x² + 2x + 3)}. That would make more sense because then both M and N would be sets of x values, and their intersection would be meaningful.Assuming that, let's proceed. So N would be {x | y = log₂(x² + 2x + 3)}. But y is defined for all x, so N would still be all real numbers. But that can't be right because the options suggest a specific interval.Wait, maybe N is supposed to be the range of y, but then it's not about x. I'm confused. Let me try another approach.Perhaps the intersection is considering x values where y is in N, but N is defined as y = log₂(x² + 2x + 3). So for each x in M, y is determined, and we need to see if y is in N. But N is just the set of all possible y values from that function, which is y ≥ 1. So for x in M, y is always ≥ 1, so the intersection would be all x in M, which is 0 < x < 2.But option B is {x | 0 < x < 2}, which is M itself. But the answer choices include A, C, and D as well. Maybe I'm still missing something.Wait, perhaps the intersection is considering the x values where y is in N, but N is defined as y = log₂(x² + 2x + 3). So for each x, y is determined, and we need to see if y is in N. But N is just the set of all y values from that function, so y is always in N. Therefore, the intersection would be all x in M, which is 0 < x < 2.But again, that's option B. However, the correct answer might be different. Let me think again.Maybe the problem is asking for the intersection of the domains of M and N. M is defined for x, and N is defined for y, but y is a function of x. So perhaps the intersection is the set of x values where both M and N are defined. But N is defined for all x, so the intersection is just M.But the options include A: {x | 1 ≤ x < 2}, which is a subset of M. Maybe there's a restriction I'm missing. Let me check the function y = log₂(x² + 2x + 3). The minimum value of the argument is 2, so y ≥ 1. But does that impose any restriction on x? No, because the argument is always ≥ 2, regardless of x. So y is always ≥ 1, but that doesn't restrict x any further.Wait, maybe the problem is asking for the intersection of M and N as sets, but since N is a set of y values and M is a set of x values, their intersection would be empty because they are sets of different variables. That would make the answer D: ∅.But that seems too straightforward. The problem probably expects an intersection in terms of x, considering that y is a function of x. So maybe the intersection is the set of x values in M where y is in N. But N is all y values from the function, so y is always in N. Therefore, the intersection is M itself, which is option B.But I'm still unsure because the options include A and C, which are subsets of M. Maybe I need to consider the range of y and see how it affects x. Since y ≥ 1, and y = log₂(x² + 2x + 3), we can write:log₂(x² + 2x + 3) ≥ 1x² + 2x + 3 ≥ 2x² + 2x + 1 ≥ 0(x + 1)² ≥ 0Which is always true. So this doesn't restrict x any further. Therefore, the intersection is still M, which is 0 < x < 2, option B.But wait, maybe the problem is considering the intersection of the domains and ranges. M is a domain of x, and N is a range of y. So their intersection would be empty because they are different types of sets. That would make the answer D: ∅.I'm confused now. I think the problem is a bit ambiguous. If we consider M and N as sets of x values, then N is all real numbers, so the intersection is M. If we consider them as sets of different variables, the intersection is empty. But since the options include specific x intervals, I think the intended answer is B: {x | 0 < x < 2}.However, looking back at the original problem, it says "determine the intersection M ∩ N". Since M is a set of x values and N is a set of y values, their intersection would be empty because they are sets of different variables. Therefore, the answer should be D: ∅.Wait, but that contradicts the earlier reasoning. Maybe I need to think differently. Perhaps N is intended to be a set of x values such that y is defined. But y is defined for all x, so N would be all real numbers, making the intersection M ∩ N = M, which is option B.I think the confusion arises from whether N is a set of y values or x values. If N is a set of y values, then M ∩ N is empty. If N is a set of x values, then M ∩ N is M. Since the options include M as option B, I think the intended answer is B.But I'm still not entirely sure. Maybe the problem expects us to consider the x values where y is in N, but since N is all y values from the function, which is y ≥ 1, and y is determined by x, we need to find x such that y ≥ 1, which is always true. Therefore, the intersection is M, option B.Yes, I think that's the correct approach. So the answer is B: {x | 0 < x < 2}.

💡Okay, so I need to figure out Zoe's overall percentage of correct answers based on the information given. Let me try to break this down step by step.First, both Chloe and Zoe have a homework assignment with a total number of problems. They each solved one-third of the problems alone and two-thirds together. So, if I let the total number of problems be, say, ( t ), then each of them solved ( frac{1}{3}t ) alone and ( frac{2}{3}t ) together.Now, Chloe had a 70% correct rate on the problems she solved alone. That means she got ( 0.70 times frac{1}{3}t ) correct alone. Let me calculate that: ( 0.70 times frac{1}{3}t = frac{7}{30}t ).Chloe's overall correct rate was 85%. So, overall, she got ( 0.85t ) problems correct. This includes both the ones she solved alone and the ones she solved together with Zoe.To find out how many problems Chloe got correct when they solved them together, I can subtract the ones she got correct alone from her total correct. That would be ( 0.85t - frac{7}{30}t ). Let me compute that:First, convert ( 0.85t ) to a fraction to make it easier. ( 0.85 = frac{17}{20} ), so ( 0.85t = frac{17}{20}t ).Now, subtract ( frac{7}{30}t ) from ( frac{17}{20}t ). To do this, I need a common denominator, which is 60.( frac{17}{20}t = frac{51}{60}t ) and ( frac{7}{30}t = frac{14}{60}t ).So, ( frac{51}{60}t - frac{14}{60}t = frac{37}{60}t ).This means that Chloe got ( frac{37}{60}t ) problems correct when they solved them together.Since Chloe and Zoe solved the two-thirds of the problems together, the number of problems they solved together is ( frac{2}{3}t ). If Chloe got ( frac{37}{60}t ) correct out of ( frac{2}{3}t ), then the number of correct answers they had together is ( frac{37}{60}t ).Now, let's think about Zoe. Zoe had an 85% correct rate on the problems she solved alone. So, she got ( 0.85 times frac{1}{3}t = frac{17}{60}t ) correct alone.To find Zoe's overall correct answers, we need to add the correct answers she got alone and the correct answers they got together. So, Zoe's total correct answers would be ( frac{17}{60}t + frac{37}{60}t = frac{54}{60}t ).Simplifying ( frac{54}{60}t ) gives ( frac{9}{10}t ), which is 90%.Wait, let me double-check that. If Chloe got ( frac{37}{60}t ) correct on the two-thirds they solved together, does that mean Zoe also got ( frac{37}{60}t ) correct? Or is there a possibility that their correct answers could differ?Hmm, the problem doesn't specify whether they both got the same number of correct answers when working together. It just says they solved the remaining two-thirds together. So, maybe I need to assume that they both got the same number of correct answers on the problems they solved together.But Chloe's total correct is 85%, which includes her correct alone and correct together. Similarly, Zoe's total correct would be her correct alone plus correct together. If we assume that the number of correct answers they got together is the same for both, then Zoe's total correct would be higher because she had a higher correct rate on her alone problems.Wait, but Chloe got 70% on her alone, and Zoe got 85% on her alone. So, if they solved the same number of problems together, but Chloe's overall is 85%, which is higher than her alone rate, that suggests that she did better on the together problems.But Zoe's alone rate is already 85%, so if she also did well on the together problems, her overall would be higher.But the problem doesn't specify whether the together problems were all correct or not. It just says they solved them together. So, perhaps we need to assume that the number of correct answers on the together problems is the same for both Chloe and Zoe.But that might not necessarily be the case. Maybe they worked together but might have different correct rates.Wait, the problem says they solved the remaining two-thirds together, but it doesn't specify whether they both got the same number correct or not. So, perhaps the correct approach is to assume that the number of correct answers on the together problems is the same for both Chloe and Zoe.But then, if Chloe got ( frac{37}{60}t ) correct on the together problems, then Zoe also got ( frac{37}{60}t ) correct. So, Zoe's total correct would be her alone correct plus the together correct, which is ( frac{17}{60}t + frac{37}{60}t = frac{54}{60}t = 0.9t ), which is 90%.Alternatively, maybe the together problems were all correct, meaning that both Chloe and Zoe got all the together problems correct. But Chloe's overall correct is 85%, which is higher than her alone rate, so that suggests that she got more correct on the together problems.But if the together problems were all correct, then Chloe's total correct would be her alone correct plus all the together problems. Let's see:If the together problems were all correct, then Chloe's total correct would be ( frac{7}{30}t + frac{2}{3}t = frac{7}{30}t + frac{20}{30}t = frac{27}{30}t = 0.9t ), which is 90%. But the problem says Chloe's overall correct is 85%, so that can't be the case.Therefore, the together problems weren't all correct. So, Chloe got some number of correct answers on the together problems, which we calculated as ( frac{37}{60}t ).Therefore, Zoe's total correct would be her alone correct plus the same number of correct on the together problems, which is ( frac{17}{60}t + frac{37}{60}t = frac{54}{60}t = 0.9t ), which is 90%.So, Zoe's overall percentage of correct answers is 90%.Wait, but let me think again. If Chloe and Zoe solved the together problems together, does that mean they both got the same number correct? Or could they have different correct rates?The problem doesn't specify, so perhaps we need to assume that the number of correct answers on the together problems is the same for both. Therefore, if Chloe got ( frac{37}{60}t ) correct on the together problems, then Zoe also got ( frac{37}{60}t ) correct.Therefore, Zoe's total correct is ( frac{17}{60}t + frac{37}{60}t = frac{54}{60}t = 0.9t ), which is 90%.Alternatively, if the together problems had a certain correct rate, say ( p ), then both Chloe and Zoe would have ( p times frac{2}{3}t ) correct on the together problems. But since Chloe's overall correct is 85%, we can solve for ( p ).Let me try that approach.Let ( p ) be the correct rate on the together problems. Then, Chloe's total correct is:( 0.70 times frac{1}{3}t + p times frac{2}{3}t = 0.85t )So,( frac{7}{30}t + frac{2}{3}pt = frac{17}{20}t )Multiply both sides by 60 to eliminate denominators:( 14t + 40pt = 51t )So,( 40pt = 51t - 14t = 37t )Therefore,( p = frac{37t}{40t} = frac{37}{40} = 0.925 )So, the correct rate on the together problems is 92.5%.Therefore, Zoe's total correct is:( 0.85 times frac{1}{3}t + 0.925 times frac{2}{3}t )Let me compute that:First, ( 0.85 times frac{1}{3}t = frac{17}{60}t )Second, ( 0.925 times frac{2}{3}t = frac{37}{40} times frac{2}{3}t = frac{74}{120}t = frac{37}{60}t )Adding them together:( frac{17}{60}t + frac{37}{60}t = frac{54}{60}t = 0.9t = 90% )So, Zoe's overall percentage of correct answers is 90%.Yes, that makes sense. So, the correct answer is 90%.