answer：Alright, let's tackle this problem step by step. So, Charlotte wants to treat herself, her daughter, and her three granddaughters to a spa day, which means she needs to pay for 5 mani/pedis. The regular price for one mani/pedi is 40.00, but there's a Mother's Day special offering 25% off the regular rate. First, I need to figure out what the discounted price per mani/pedi is. To do that, I'll calculate 25% of 40.00. 25% of 40.00 is:[0.25 times 40 = 10]So, the discount per mani/pedi is 10.00. Now, I'll subtract the discount from the original price to find the discounted price:[40 - 10 = 30]That means each mani/pedi will cost 30.00 after the discount.Since Charlotte is getting 5 mani/pedis, I'll multiply the discounted price by 5 to find the total cost:[5 times 30 = 150]Therefore, Charlotte will spend 150.00 in total for the 5 mani/pedis.

answer：First, I need to determine the revenue generated from the first 10 people who received a 40% discount on the 20 tickets. Calculating the discount amount and the price each person paid, I find that each of these 10 people paid 12, resulting in a total revenue of 120.Next, I'll calculate the revenue from the next 20 people who received a 15% discount. The discount amount is 3 per ticket, so each person paid 17. The total revenue from this group is 340.Adding the revenue from the first two groups gives me 460. Since the total revenue was 980, the remaining revenue from full-price tickets is 520. Dividing this by the full price of 20 per ticket, I find that 26 full-price tickets were sold.Finally, adding up all the ticket buyers—10 from the first group, 20 from the second group, and 26 full-price buyers—gives a total of 56 people who bought tickets to the concert.

answer：Alright, so I have this problem about a tetrahedron (ABCD). I need to show that the lines (AA') and (BB') intersect if and only if (AB) is perpendicular to (CD). Then, I also need to determine if they intersect when (AC = AD = BC = BD). Hmm, okay, let's break this down step by step.First, let me recall what a tetrahedron is. It's a three-dimensional shape with four triangular faces, right? So, (ABCD) has four vertices, and each pair of vertices is connected by an edge. Now, (A') and (B') are the feet of the perpendiculars from (A) and (B) to the opposite faces. That means (A') is the point where the perpendicular from (A) meets the face (BCD), and (B') is where the perpendicular from (B) meets the face (ACD).I need to show that the lines (AA') and (BB') intersect if and only if (AB) is perpendicular to (CD). Okay, so this is an "if and only if" statement, meaning I need to prove both directions: 1. If (AA') and (BB') intersect, then (AB) is perpendicular to (CD).2. If (AB) is perpendicular to (CD), then (AA') and (BB') intersect.Let me start with the first direction.**1. If (AA') and (BB') intersect, then (AB) is perpendicular to (CD).**Assume that (AA') and (BB') intersect at some point (P). Since (A') is the foot of the perpendicular from (A) to face (BCD), the line (AA') is perpendicular to face (BCD). Similarly, (BB') is perpendicular to face (ACD).Now, the point (P) lies on both (AA') and (BB'), so it must satisfy the conditions of being on both lines. Since (AA') is perpendicular to face (BCD), (P) must lie in the plane perpendicular to (BCD) passing through (A'). Similarly, (P) must lie in the plane perpendicular to (ACD) passing through (B').Wait, maybe I should think about the planes containing these faces. The face (BCD) is in one plane, and the face (ACD) is in another plane. The lines (AA') and (BB') are both perpendicular to these respective planes.If (AA') and (BB') intersect, then their intersection point (P) must lie in both planes, but since (AA') is perpendicular to (BCD) and (BB') is perpendicular to (ACD), the only way for them to intersect is if these two planes are somehow related in a specific way.I think I need to use some vector geometry here. Let me assign coordinates to the tetrahedron to make this more concrete. Maybe place point (C) at the origin, (D) along the x-axis, (B) in the xy-plane, and (A) somewhere in 3D space. But before I get too deep into coordinates, maybe there's a more geometric way to see this.Since (AA') is perpendicular to face (BCD), the vector (AA') is orthogonal to the plane (BCD). Similarly, (BB') is orthogonal to the plane (ACD). If these two lines intersect, then their direction vectors must satisfy some condition.Wait, the direction vector of (AA') is along the normal vector of face (BCD), and the direction vector of (BB') is along the normal vector of face (ACD). For these two lines to intersect, their normals must be related in a way that allows the lines to meet.Alternatively, maybe I can think about the projections of (AB) onto the faces (BCD) and (ACD). Since (A') is the foot from (A) to (BCD), (AA') is perpendicular, so (A') is the projection of (A) onto (BCD). Similarly, (B') is the projection of (B) onto (ACD).If (AA') and (BB') intersect, then the projections of (A) and (B) onto these faces must align in a specific way. Maybe this implies that (AB) is orthogonal to the line (CD), which is common to both faces (BCD) and (ACD).Let me try to formalize this. Let’s denote vectors:- Let vector (AB) be (vec{AB}).- Let vector (CD) be (vec{CD}).If (AB) is perpendicular to (CD), then (vec{AB} cdot vec{CD} = 0).Now, if (AA') and (BB') intersect, then there exists a point (P) such that (P) lies on both (AA') and (BB'). Since (AA') is perpendicular to face (BCD), the vector (AP) is perpendicular to face (BCD). Similarly, (BP) is perpendicular to face (ACD).Therefore, (AP) is parallel to the normal vector of face (BCD), and (BP) is parallel to the normal vector of face (ACD).Let me denote the normal vectors:- Normal vector of face (BCD): (vec{n_1} = vec{BC} times vec{BD})- Normal vector of face (ACD): (vec{n_2} = vec{AC} times vec{AD})Since (AP) is parallel to (vec{n_1}) and (BP) is parallel to (vec{n_2}), the vectors (AP) and (BP) are scalar multiples of (vec{n_1}) and (vec{n_2}) respectively.Now, since (P) lies on both (AA') and (BB'), we can write parametric equations for these lines:- (AA'): ( vec{r} = vec{A} + t vec{n_1} )- (BB'): ( vec{r} = vec{B} + s vec{n_2} )For some parameters (t) and (s). If they intersect, then there exist (t) and (s) such that:( vec{A} + t vec{n_1} = vec{B} + s vec{n_2} )This implies:( vec{B} - vec{A} = t vec{n_1} - s vec{n_2} )Which means:( vec{AB} = t vec{n_1} - s vec{n_2} )Now, if (AB) is perpendicular to (CD), then (vec{AB} cdot vec{CD} = 0). Let's compute the dot product:( vec{AB} cdot vec{CD} = (t vec{n_1} - s vec{n_2}) cdot vec{CD} )But (vec{n_1}) is the normal to face (BCD), so it's perpendicular to both (vec{BC}) and (vec{BD}). Similarly, (vec{n_2}) is perpendicular to (vec{AC}) and (vec{AD}).Wait, (vec{CD}) is an edge common to both faces (BCD) and (ACD). So, (vec{CD}) lies in both planes. Therefore, (vec{n_1}) is perpendicular to (vec{CD}), and (vec{n_2}) is also perpendicular to (vec{CD}).Therefore, both (vec{n_1}) and (vec{n_2}) are perpendicular to (vec{CD}), which means their dot product with (vec{CD}) is zero.Thus, ( vec{AB} cdot vec{CD} = t (vec{n_1} cdot vec{CD}) - s (vec{n_2} cdot vec{CD}) = 0 - 0 = 0 ).So, if (AA') and (BB') intersect, then (vec{AB} cdot vec{CD} = 0), meaning (AB) is perpendicular to (CD).Okay, that seems to work for the first direction.**2. If (AB) is perpendicular to (CD), then (AA') and (BB') intersect.**Now, let's assume that (AB) is perpendicular to (CD). I need to show that (AA') and (BB') intersect.Again, using vector geometry, let's consider the parametric equations of (AA') and (BB'):- (AA'): ( vec{r} = vec{A} + t vec{n_1} )- (BB'): ( vec{r} = vec{B} + s vec{n_2} )We need to find (t) and (s) such that:( vec{A} + t vec{n_1} = vec{B} + s vec{n_2} )Which simplifies to:( vec{B} - vec{A} = t vec{n_1} - s vec{n_2} )Given that (AB) is perpendicular to (CD), we have (vec{AB} cdot vec{CD} = 0).But as before, since (vec{n_1}) and (vec{n_2}) are both perpendicular to (vec{CD}), their linear combination (t vec{n_1} - s vec{n_2}) is also perpendicular to (vec{CD}).Therefore, (vec{AB}) is perpendicular to (vec{CD}), and the equation ( vec{AB} = t vec{n_1} - s vec{n_2} ) must hold for some (t) and (s).This suggests that the vectors (vec{n_1}) and (vec{n_2}) span a space that includes (vec{AB}), given the perpendicularity condition. Therefore, there exist scalars (t) and (s) such that the equation is satisfied, meaning the lines (AA') and (BB') intersect.So, both directions are proven. Therefore, (AA') and (BB') intersect if and only if (AB) is perpendicular to (CD).Now, the second part of the question: Do they intersect if (AC = AD = BC = BD)?Hmm, so (AC = AD) and (BC = BD). This suggests a symmetry in the tetrahedron. Let me visualize this. If (AC = AD), then point (A) is equidistant from (C) and (D). Similarly, (B) is equidistant from (C) and (D). So, both (A) and (B) lie on the perpendicular bisector of (CD).Wait, in three dimensions, the set of points equidistant from (C) and (D) is the plane perpendicular to (CD) at its midpoint. So, both (A) and (B) lie on this plane. Therefore, the line (AB) lies entirely within this plane.But if (AB) lies in the plane perpendicular to (CD), then (AB) must be perpendicular to (CD). Because the plane is perpendicular to (CD), any line lying in that plane is also perpendicular to (CD).Therefore, if (AC = AD = BC = BD), then (AB) is perpendicular to (CD). From the first part, this implies that (AA') and (BB') intersect.So, yes, they do intersect in this case.Wait, let me double-check. If (AC = AD) and (BC = BD), then (A) and (B) are both on the perpendicular bisector plane of (CD). Therefore, the line (AB) is in that plane, which is perpendicular to (CD). Hence, (AB) is perpendicular to (CD), so by the earlier result, (AA') and (BB') intersect.Yes, that makes sense.So, to summarize:- If (AA') and (BB') intersect, then (AB) is perpendicular to (CD).- If (AB) is perpendicular to (CD), then (AA') and (BB') intersect.- When (AC = AD = BC = BD), (AB) is perpendicular to (CD), so (AA') and (BB') intersect.Therefore, the answer to the second question is yes, they do intersect under the given conditions.

answer：First, I need to determine how much Tom spent on the Batman and Superman games. The Batman game costs 13.60 and the Superman game costs 5.06. Adding these together gives a total expenditure of 18.66.Next, I compare this total expenditure to the amount Tom spent on video games, which is also 18.66. Since the amount spent on the two games matches the total expenditure, it implies that Tom did not own any games before purchasing these two.Therefore, Tom owned 0 games before purchasing the Batman and Superman games.

answer：First, I need to determine the depth of the river. I know the width of the river is 45 meters and the flow rate is 6 kilometers per hour. The volume of water flowing into the sea per minute is 9000 cubic meters.To make the units consistent, I'll convert the flow rate from kilometers per hour to meters per minute. 6 kilometers per hour is equivalent to:6 km/h = 6 * 1000 meters / 60 minutes = 100 meters per minute.Now, I can use the formula for the volume of water flowing through the river per minute:Volume = Width * Depth * Flow Rate.Plugging in the known values:9000 m³ = 45 m * Depth * 100 m/min.To find the depth, I'll rearrange the formula:Depth = 9000 m³ / (45 m * 100 m/min).Calculating the depth:Depth = 9000 / 4500 = 2 meters.Therefore, the depth of the river is 2 meters.

answer：Okay, so I've got this problem here about Doug, Dave, and Ralph painting a room. Let me try to figure out how to approach it step by step. First, the problem says Doug can paint a room in 6 hours. That means Doug's painting rate is 1 room per 6 hours, or in other words, he can paint 1/6 of a room in one hour. Similarly, Dave can paint the same room in 8 hours, so his rate is 1/8 of a room per hour. Now, Ralph joins them, and he can paint the room in 12 hours, so his rate is 1/12 of a room per hour.Okay, so if they're all painting together, their combined rate should be the sum of their individual rates. Let me write that down:Doug's rate: 1/6 per hour Dave's rate: 1/8 per hour Ralph's rate: 1/12 per hour Combined rate = 1/6 + 1/8 + 1/12Hmm, to add these fractions, I need a common denominator. The denominators are 6, 8, and 12. The least common multiple of these numbers is 24. So, I'll convert each fraction to have 24 as the denominator.1/6 = 4/24 1/8 = 3/24 1/12 = 2/24 So, adding them up: 4/24 + 3/24 + 2/24 = 9/24Simplifying 9/24, I can divide numerator and denominator by 3, which gives me 3/8. So, their combined painting rate is 3/8 of a room per hour.Now, the problem mentions that they take a break of 90 minutes for lunch. Since the total time t includes this break, I need to account for the time they're actually painting. 90 minutes is equal to 1.5 hours. So, if the total time is t hours, the actual time they spend painting is t - 1.5 hours.The amount of work done is equal to the rate multiplied by the time. Since painting one whole room is considered as 1 unit of work, the equation should be:Combined rate * Actual painting time = 1Plugging in the numbers:(3/8) * (t - 1.5) = 1But wait, looking back at the original rates, I had 1/6 + 1/8 + 1/12. Maybe I should express the combined rate as that sum instead of simplifying it to 3/8. Let me check the answer choices.Looking at the options:(A) (1/6 + 1/8 + 1/12)(t - 1.5) = 1 (B) (1/6 + 1/8)t = 1 (C) (1/6 + 1/8 + 1/12)(t + 1.5) = 1 (D) (1/6 + 1/8)(t - 1.5) = 1So, option A is exactly what I derived, using the combined rate as the sum of their individual rates and multiplying by the actual painting time (t - 1.5). Option B doesn't include Ralph, which isn't correct because Ralph is helping. Option C adds the break time, which would mean they're painting longer, which doesn't make sense. Option D also excludes Ralph, so that's not right either.Therefore, the correct equation should be option A: (1/6 + 1/8 + 1/12)(t - 1.5) = 1.I think that makes sense because they're taking a break, so the total time includes the break, but the painting only happens during t - 1.5 hours. So, the work done during that time should equal 1 complete room.