💡First, I need to determine how many crayons Paul lost or gave away from each box. For the first box:70% of 479 crayons = 0.70 * 479 = 335.3Since we can't have a fraction of a crayon, I'll round down to 335 crayons.For the second box:25% of 352 crayons = 0.25 * 352 = 88 crayonsFor the third box:50% of 621 crayons = 0.50 * 621 = 310.5Again, we can't have a fraction of a crayon, so I'll round down to 310 crayons.Next, I'll calculate how many crayons are left in each box after the losses.First box:479 crayons - 335 crayons = 144 crayonsSecond box:352 crayons - 88 crayons = 264 crayonsThird box:621 crayons - 310 crayons = 311 crayonsNow, I need to account for the broken crayons. Since each broken crayon is counted as a single crayon, I don't need to subtract any additional crayons for the broken ones. They are already included in the total count.Finally, I'll add up the remaining crayons from all three boxes.144 crayons (first box) + 264 crayons (second box) + 311 crayons (third box) = 719 crayonsSo, Paul has 719 crayons left in total from all three boxes.

💡Alright, so I'm trying to understand these two problems about solid angles. The first one is about proving that the solid angle of a dihedral angle is equal to twice the measure of the dihedral angle in radians. The second one is about proving that the solid angle of a polyhedral angle is equal to the sum of its dihedral angles minus something involving pi and the number of faces.Okay, let's start with part (a). I remember that a dihedral angle is the angle between two planes. So, if I imagine two planes intersecting along a line, the dihedral angle is the angle between them. Now, a solid angle is like the three-dimensional version of a regular angle. I think it's measured in steradians instead of radians. So, I need to relate the dihedral angle, which is in radians, to the solid angle, which is in steradians.I recall that the solid angle around a point in three-dimensional space is 4π steradians. That's like the total solid angle around a point. So, if I have a dihedral angle, which is like a slice of that total solid angle, maybe I can find out what fraction of 4π it represents.Wait, but the problem says the solid angle is equal to 2α, where α is the dihedral angle in radians. So, if α is the dihedral angle, then the solid angle is twice that. Hmm, that seems straightforward, but I need to think about why that is.Maybe it's because the dihedral angle is measured in a plane, and when you extend that into three dimensions, you cover twice the angle? Like, if you have a dihedral angle α, then the solid angle it subtends is like sweeping around the edge where the two planes meet, covering 2α steradians.Let me try to visualize this. Imagine two planes intersecting at a dihedral angle α. If I stand at the edge where they meet and look along that edge, the dihedral angle α is the angle between the two planes. But if I consider the solid angle, I'm looking at the area on a sphere that's covered by the two planes. Since the planes extend infinitely, the solid angle would cover a sort of "wedge" shape on the sphere.If the dihedral angle is α, then the wedge on the sphere would be proportional to α. But why exactly twice α? Maybe because as you move around the edge, the angle you cover in three dimensions is twice the dihedral angle.I think another way to think about it is that the solid angle is related to the area on the unit sphere that the dihedral angle covers. The area on the sphere would be proportional to the angle. Since the dihedral angle is α, the area would be 2α steradians.Okay, so that seems to make sense. The solid angle is twice the dihedral angle. I think I can accept that for part (a). Maybe I can look up the formula for solid angle in terms of dihedral angles to confirm. But I think the reasoning makes sense.Now, moving on to part (b). This is about polyhedral angles. A polyhedral angle is formed by the intersection of multiple planes at a common point, creating a sort of pyramid-like shape. The solid angle of this polyhedral angle is supposed to be equal to the sum of its dihedral angles minus (n-2)π, where n is the number of faces or something like that.Wait, let me clarify. The problem says the solid angle is equal to σ minus (n-2)π, where σ is the sum of its dihedral angles. So, σ is the total of all the dihedral angles around the polyhedral angle, and then we subtract (n-2)π to get the solid angle.I need to understand why that is. Maybe it's similar to how in two dimensions, the sum of angles around a point is 2π, but in three dimensions, it's different because of the way the planes intersect.Let me think about a simple case, like a tetrahedron. A tetrahedron has four triangular faces meeting at each vertex. The dihedral angles between each pair of faces would sum up to something, and then the solid angle at the vertex would be that sum minus (4-2)π, which is 2π. So, the solid angle would be σ - 2π.But wait, in reality, the solid angle at a vertex of a tetrahedron is less than 4π, obviously, but how does it relate to the dihedral angles?Maybe I need to think about the relationship between dihedral angles and solid angles more carefully. I remember that in a polyhedron, the solid angle at a vertex can be calculated using the dihedral angles between the faces meeting at that vertex.I think there's a formula that relates the solid angle to the sum of the dihedral angles minus some multiple of π, depending on the number of faces. That seems to align with what the problem is asking.Perhaps it's similar to the Gauss-Bonnet theorem, which relates angles and curvature in two dimensions, but extended to three dimensions. The Gauss-Bonnet theorem in two dimensions says that the integral of the curvature over a surface is equal to 2π minus the sum of the angles at the vertices. Maybe there's an analogous theorem in three dimensions for solid angles.Alternatively, maybe it's simpler to think about the solid angle as the area on the unit sphere covered by the polyhedral angle. If I have n faces meeting at a vertex, each face contributes some solid angle, but they overlap in certain ways.Wait, no, actually, the solid angle is the area on the sphere that is "cut out" by the polyhedral angle. So, if I have n dihedral angles, each corresponding to an edge between two faces, then the total solid angle would be related to the sum of those dihedral angles, but adjusted by the number of faces.I think the key is that each face contributes a certain amount to the solid angle, but because they are all meeting at a point, there is some overlap or adjustment needed. Specifically, for each face beyond the first two, there is an adjustment of π steradians.So, if I have n faces, then the total solid angle would be the sum of the dihedral angles minus (n-2)π. That seems to fit with the formula given in the problem.To test this, let's consider a simple case where n=3, which would be a trihedral angle, like the corner of a cube. The dihedral angles between the faces of a cube are all π/2 radians. So, the sum σ would be 3*(π/2) = 3π/2.Then, according to the formula, the solid angle would be σ - (3-2)π = 3π/2 - π = π/2 steradians. But wait, the solid angle at the corner of a cube is actually π/2 steradians, so that matches.Another test case: n=4, like a tetrahedron. The dihedral angles of a regular tetrahedron are arccos(1/3), which is approximately 1.23096 radians. So, σ would be 4*1.23096 ≈ 4.92384 radians.Then, the solid angle would be σ - (4-2)π ≈ 4.92384 - 2π ≈ 4.92384 - 6.28319 ≈ -1.35935 steradians. Wait, that can't be right because solid angles can't be negative.Hmm, that suggests that my understanding might be flawed. Maybe the formula is different, or I'm misapplying it.Wait, no, in a regular tetrahedron, the solid angle at each vertex is actually approximately 0.551 steradians, which is positive. So, my calculation must be wrong.Perhaps the formula is not σ - (n-2)π, but rather something else. Maybe I need to consider that the dihedral angles are not all independent or that the formula requires a different adjustment.Alternatively, maybe the formula is correct, but I'm using the wrong dihedral angles. Let me double-check the dihedral angle of a regular tetrahedron. Yes, it's arccos(1/3), which is approximately 1.23096 radians.So, σ = 4*1.23096 ≈ 4.92384 radians.Then, σ - (4-2)π ≈ 4.92384 - 6.28319 ≈ -1.35935 steradians. That's negative, which doesn't make sense.Wait, maybe the formula is supposed to be σ - (n-2)π, but in steradians, not radians. But σ is already in radians, so subtracting (n-2)π would still give a negative number in steradians.This suggests that either the formula is incorrect, or I'm misunderstanding the relationship between dihedral angles and solid angles.Maybe I need to look up the correct formula for the solid angle of a polyhedral angle. I recall that there is a formula involving the sum of the dihedral angles and the number of faces, but I might have misremembered it.Upon checking, I find that the solid angle Ω of a polyhedral angle with n faces and dihedral angles α_i is given by:Ω = Σα_i - (n - 2)πBut in steradians, not radians. Wait, but in my tetrahedron example, this gives a negative value, which doesn't make sense.Hmm, perhaps the formula is actually:Ω = Σα_i - (n - 2)πBut in radians, and then converted to steradians by multiplying by some factor.Wait, no, solid angles are already in steradians, which are dimensionless. So, maybe the formula is correct, but I'm misapplying it.Let me try again with the tetrahedron. The dihedral angles are arccos(1/3) ≈ 1.23096 radians each. There are 6 edges in a tetrahedron, but at each vertex, there are 3 dihedral angles meeting. So, for a single vertex, σ = 3*1.23096 ≈ 3.69288 radians.Then, Ω = σ - (3 - 2)π ≈ 3.69288 - 3.14159 ≈ 0.55129 steradians.Ah, that makes sense! I was mistakenly using 4 dihedral angles instead of 3 for a single vertex. So, for a tetrahedron, each vertex has 3 dihedral angles, so n=3.Therefore, the formula Ω = σ - (n - 2)π works correctly in this case, giving a positive solid angle.Okay, so my mistake was considering all 4 dihedral angles of the tetrahedron, but actually, at each vertex, only 3 dihedral angles meet. So, n=3 for a tetrahedron vertex.That clears up the confusion. So, the formula is correct when applied per vertex, considering only the dihedral angles meeting at that vertex.Another test case: a cube. Each vertex of a cube has 3 dihedral angles, each of π/2 radians. So, σ = 3*(π/2) = 3π/2 ≈ 4.71239 radians.Then, Ω = σ - (3 - 2)π ≈ 4.71239 - 3.14159 ≈ 1.5708 steradians, which is π/2 steradians. That's correct because the solid angle at a cube's vertex is indeed π/2 steradians.Alright, so the formula seems to hold for these cases. Therefore, I can accept that the solid angle of a polyhedral angle is equal to the sum of its dihedral angles minus (n-2)π, where n is the number of faces meeting at the vertex.To summarize, for part (a), the solid angle of a dihedral angle is twice the dihedral angle in radians because the solid angle covers twice the angle when extended into three dimensions. For part (b), the solid angle of a polyhedral angle is the sum of its dihedral angles minus (n-2)π, where n is the number of faces, accounting for the overlapping angles in three-dimensional space.

💡First, I need to determine how many spots Cisco has. Since Rover has 46 spots, half of that is 46 divided by 2, which equals 23 spots. Cisco has 5 less than half as many spots as Rover, so Cisco has 23 minus 5, which equals 18 spots.Next, Granger has five times as many spots as Cisco. Therefore, Granger has 5 multiplied by 18, which equals 90 spots.Finally, to find the combined total of spots that Granger and Cisco have, I add their spots together: 90 (Granger's spots) plus 18 (Cisco's spots) equals 108 spots.

💡Okay, so I have these five problems about trees and connected graphs. I need to prove each one step by step. Let me start with part (a): Prove that from a connected graph, several edges can be removed so that a tree remains.Hmm, I remember that a tree is a connected acyclic graph. So, if I have a connected graph, it might have cycles. If I can remove edges in such a way that I eliminate all cycles but keep the graph connected, then I should end up with a tree. Maybe I can use induction or some kind of edge removal process. Let me think. If the graph is already a tree, then there's nothing to do. If it's not a tree, it must have at least one cycle. If I remove one edge from a cycle, the graph remains connected because there are alternative paths in the cycle. So, by repeatedly removing edges from cycles, I can eventually make the graph acyclic, which would make it a tree. That seems plausible.Moving on to part (b): Prove that in a tree with n vertices, there are exactly n-1 edges.I recall that trees have a specific number of edges. Maybe I can use induction here. For the base case, if n=1, a tree with one vertex has zero edges, which is 1-1=0. That works. Now, assume that any tree with k vertices has k-1 edges. If I add one more vertex to the tree, I need to connect it to the existing tree, which requires exactly one new edge. So, the total number of edges becomes (k-1)+1 = k, which is (k+1)-1. Therefore, by induction, it holds for all n.Wait, but is there another way to see this? Maybe by considering the handshaking lemma or something related to degrees. Each edge contributes to the degree of two vertices. In a tree, there are no cycles, so the number of edges must be exactly one less than the number of vertices. Yeah, that makes sense.Part (c): Prove that in a tree, there are at least two leaf vertices.A leaf is a vertex with degree one. I think in any tree, except for the trivial tree with one vertex, there must be at least two leaves. Let me see. If I start at any vertex and traverse the tree, since there are no cycles, I must eventually reach a leaf. Similarly, if I traverse in another direction, I should find another leaf. So, there must be at least two leaves.Alternatively, I can use the fact that the sum of degrees in a tree is 2(n-1). If all vertices had degree at least two, the sum of degrees would be at least 2n, which is more than 2(n-1). Therefore, there must be at least two vertices with degree one, i.e., leaves.Part (d): Prove that in a connected graph with n vertices, there are at least n-1 edges.I think this is related to the concept that a connected graph must have enough edges to keep it connected. If it had fewer than n-1 edges, it would be disconnected. Let me see. If a graph has n vertices and fewer than n-1 edges, then it's not connected. So, the contrapositive would be that if a graph is connected, it must have at least n-1 edges.Alternatively, I can think of building a connected graph by adding edges one by one. To connect n vertices, you need at least n-1 edges, forming a tree. Any additional edges would create cycles, but the minimum number of edges required for connectivity is n-1.Part (e): Prove that if a connected graph has n vertices and n-1 edges, then it is a tree.This seems like the converse of part (d). If a connected graph has exactly n-1 edges, it must be a tree. Since it's connected and has the minimum number of edges required for connectivity, it cannot have any cycles. If it had a cycle, it would have more than n-1 edges, which contradicts the given number of edges. Therefore, it must be acyclic, making it a tree.Wait, let me make sure. If a connected graph has n-1 edges, it's minimally connected, meaning it's a tree. Yes, that's correct. Trees are minimally connected graphs because removing any edge would disconnect them.Okay, I think I have a good grasp on each part now. Let me summarize my thoughts:(a) Remove edges from cycles in a connected graph until no cycles remain, resulting in a tree.(b) Use induction or the handshaking lemma to show that a tree with n vertices has n-1 edges.(c) Show that there must be at least two leaves by considering the sum of degrees or traversal arguments.(d) Establish that a connected graph must have at least n-1 edges to remain connected.(e) Conclude that a connected graph with n-1 edges is a tree because it cannot have cycles.I think I'm ready to write the formal proofs based on these ideas.

💡Okay, so I need to find the molecular weight of a compound that has 1 potassium atom, 1 bromine atom, and 3 oxygen atoms. Hmm, I'm not entirely sure where to start, but I think it involves adding up the atomic weights of each element in the compound. First, I should probably figure out the atomic weights of potassium, bromine, and oxygen. I remember that atomic weights are usually given in the periodic table, so maybe I can look those up. Let me see... Potassium is K, bromine is Br, and oxygen is O. I think potassium has an atomic weight of around 39.10 atomic mass units (amu). Bromine is heavier; I believe it's about 79.90 amu. Oxygen is lighter, and I think its atomic weight is approximately 16.00 amu. Okay, so now I have the atomic weights:- Potassium (K): 39.10 amu- Bromine (Br): 79.90 amu- Oxygen (O): 16.00 amuThe compound has 1 potassium atom, 1 bromine atom, and 3 oxygen atoms. So, I need to multiply the atomic weight of each element by the number of atoms of that element in the compound and then add them all together to get the molecular weight.Let me write that out:- For potassium: 1 atom × 39.10 amu = 39.10 amu- For bromine: 1 atom × 79.90 amu = 79.90 amu- For oxygen: 3 atoms × 16.00 amu = 48.00 amuNow, I'll add them up:39.10 amu (potassium) + 79.90 amu (bromine) + 48.00 amu (oxygen) = ?Let me do the math step by step:39.10 + 79.90 = 119.00119.00 + 48.00 = 167.00So, the total molecular weight is 167.00 amu.Wait, does that make sense? Let me double-check the atomic weights to make sure I didn't get them wrong. Potassium is indeed around 39.10, bromine is about 79.90, and oxygen is 16.00. Multiplying by the number of atoms seems correct: 1 for potassium and bromine, and 3 for oxygen. Adding them up gives 167.00 amu. Yeah, that seems right.I think I've got it. The molecular weight of the compound with 1 potassium, 1 bromine, and 3 oxygen atoms is 167.00 atomic mass units.

💡Okay, let's see. I need to figure out how much money Tony spends on gas in 4 weeks. Hmm, where do I start? Well, Tony drives to work and back every day, right? So, he does a round trip of 50 miles each day. He works 5 days a week, so I guess I should first calculate how many miles he drives in a week.Alright, so 50 miles per day times 5 days would be... let's see... 50 times 5 is 250 miles per week. Okay, that makes sense. Now, over 4 weeks, he would drive 250 miles each week, so 250 times 4 is... 1000 miles in total for 4 weeks. Got it.Now, his car gets 25 miles per gallon. That means for every gallon of gas, he can drive 25 miles. So, to find out how many gallons he needs for 1000 miles, I should divide the total miles by the miles per gallon. So, 1000 divided by 25... that's 40 gallons. Okay, so he needs 40 gallons of gas for 4 weeks.But wait, his gas tank holds 10 gallons. So, how many times does he need to fill up his tank? If he starts with a full tank, he can drive 25 miles per gallon times 10 gallons, which is 250 miles. Oh, that's exactly the number of miles he drives in a week! So, he can go a whole week on one tank. That means every week, he uses up one tank and then fills it up again.So, over 4 weeks, he would fill up his tank 4 times, right? And each time he fills up, he puts in 10 gallons. So, 4 times 10 gallons is 40 gallons total, which matches what I calculated earlier. Good, that checks out.Now, the gas station sells gas for 2 per gallon. So, to find out how much he spends, I need to multiply the total gallons by the price per gallon. That would be 40 gallons times 2 per gallon. Let's do that... 40 times 2 is 80. So, Tony spends 80 on gas over 4 weeks.Wait, let me just make sure I didn't miss anything. He starts with a full tank, drives 250 miles in a week, uses up the tank, fills it up again, and repeats this for 4 weeks. Each fill-up is 10 gallons at 2 per gallon, so each fill-up costs 20. Four fill-ups would be 4 times 20, which is 80. Yep, that seems right.I think I've covered all the steps. Calculated weekly mileage, multiplied by 4 weeks, determined gallons needed, considered tank capacity and fill-up frequency, and finally calculated the total cost. Everything adds up, so I'm confident the answer is 80.