answer：First, let's define the variables. Let ( d ) represent the distance from home to work, ( v_w ) be the walking speed, and ( v_t ) be the transport speed.From the first scenario, when the person walks to work and takes transport back, the total time is 1.5 hours. This gives us the equation:[frac{d}{v_w} + frac{d}{v_t} = 1.5]In the second scenario, when the person takes transport both ways, the total time is 0.5 hours. This leads to:[frac{d}{v_t} + frac{d}{v_t} = 0.5]Simplifying this, we find:[frac{2d}{v_t} = 0.5 quad Rightarrow quad frac{d}{v_t} = 0.25]Substituting ( frac{d}{v_t} = 0.25 ) into the first equation:[frac{d}{v_w} + 0.25 = 1.5 quad Rightarrow quad frac{d}{v_w} = 1.25]Finally, to find the total time for walking both ways:[frac{d}{v_w} + frac{d}{v_w} = 2 times 1.25 = 2.5 text{ hours}]

answer：Okay, so I have this geometry problem here. It says that O is the circumcenter of triangle ABC, and it satisfies the condition that the vector OA plus the vector OB equals the vector OC. I need to find the measure of angle ACB. Hmm, okay, let me try to figure this out step by step.First, let me recall what a circumcenter is. The circumcenter of a triangle is the point where the perpendicular bisectors of the sides intersect, and it's equidistant from all three vertices of the triangle. So, OA, OB, and OC are all equal in length because they are the radii of the circumcircle. That means triangle ABC is inscribed in a circle with center O and radius OA = OB = OC.Now, the condition given is that the vector OA plus the vector OB equals the vector OC. So, in vector terms, that's:overrightarrow{OA} + overrightarrow{OB} = overrightarrow{OC}I need to visualize this. Maybe I can draw a diagram in my mind. Let me consider the vectors OA, OB, and OC as vectors from the origin O to points A, B, and C respectively. So, if I add vectors OA and OB, I get vector OC.Wait, so if I place vectors OA and OB tip-to-tail, the resultant vector would be OC. That means that point C is the result of moving from O by OA and then by OB. Hmm, so in terms of position vectors, this might imply that point C is located at the position OA + OB.But since O is the circumcenter, all three points A, B, and C lie on the circumference of the circle centered at O. So, OA, OB, and OC are all radii of this circle. Therefore, the lengths of OA, OB, and OC are equal. Let me denote the length as R, so OA = OB = OC = R.Now, if I have vectors OA and OB adding up to vector OC, then in terms of vector addition, the triangle formed by vectors OA, OB, and OC is such that OA + OB = OC. So, if I think of vectors OA and OB as sides of a parallelogram, then OC would be the diagonal. But in this case, since OA + OB = OC, it's more like a triangle rather than a parallelogram.Wait, maybe it's better to think in terms of coordinates. Let me assign coordinates to points A, B, and C with O as the origin. So, let me set O at (0, 0). Then, points A, B, and C can be represented as vectors from the origin.Let me denote vector OA as vector A, vector OB as vector B, and vector OC as vector C. So, the condition becomes:mathbf{A} + mathbf{B} = mathbf{C}Since O is the circumcenter, the magnitudes of A, B, and C are equal, so |A| = |B| = |C| = R.Now, I need to find the measure of angle ACB. Angle ACB is the angle at point C between points A, C, and B. So, in terms of vectors, this angle can be found using the dot product formula. If I can find the vectors CA and CB, then the angle between them is angle ACB.First, let's express vectors CA and CB in terms of vectors A, B, and C. Vector CA is A - C, and vector CB is B - C. But since C = A + B, we can substitute that in:Vector CA = A - (A + B) = -BVector CB = B - (A + B) = -ASo, vectors CA and CB are -B and -A respectively. Therefore, the angle between CA and CB is the same as the angle between vectors -B and -A, which is the same as the angle between vectors A and B because reversing the direction of both vectors doesn't change the angle between them.So, angle ACB is equal to the angle between vectors A and B. Let me denote this angle as θ. So, θ is the angle between vectors OA and OB.Since OA and OB are both radii of the circumcircle, the angle between them is equal to the central angle subtended by arc AB. In a circle, the central angle is twice the inscribed angle subtended by the same arc. But in this case, angle ACB is an inscribed angle subtended by arc AB. Therefore, angle ACB should be half of the central angle θ.Wait, but earlier I concluded that angle ACB is equal to θ, which is the central angle. That seems contradictory. Let me check my reasoning.I said that vectors CA and CB are -B and -A, so the angle between them is the same as the angle between A and B. But angle ACB is the angle between CA and CB, which is the same as the angle between -B and -A. Since the angle between two vectors is the same regardless of their direction, the angle between -B and -A is equal to the angle between A and B. Therefore, angle ACB is equal to θ, which is the angle between OA and OB.But in a circle, the central angle is twice the inscribed angle subtended by the same arc. So, if angle ACB is an inscribed angle subtended by arc AB, then it should be half of the central angle θ. Therefore, I must have made a mistake in my earlier reasoning.Wait, perhaps I confused the angle between vectors with the angle in the triangle. Let me clarify.Vectors OA and OB have an angle θ between them. The inscribed angle ACB subtended by arc AB is equal to half of θ. So, angle ACB = θ / 2.But earlier, I thought that angle ACB is equal to θ because vectors CA and CB are -B and -A, so the angle between them is θ. But that seems conflicting.Wait, perhaps I need to think more carefully. Let me consider the triangle ACB. Points A, B, and C are all on the circumference of the circle with center O. So, angle ACB is an inscribed angle, and it's subtended by arc AB. The central angle subtended by arc AB is angle AOB, which is θ.Therefore, angle ACB = (1/2) * angle AOB = θ / 2.But according to the vector analysis, angle between vectors CA and CB is θ, which is angle ACB. So, that suggests that θ = angle ACB, but according to the circle theorems, angle ACB should be θ / 2.This is a contradiction. Therefore, my earlier conclusion that angle ACB is equal to θ must be wrong.Wait, let me see. If vectors CA and CB are -B and -A, then the angle between them is the same as the angle between A and B, which is θ. But angle ACB is the angle between CA and CB, which is θ. But according to circle theorems, angle ACB should be half of θ. Therefore, my mistake must be in the vector analysis.Wait, perhaps the angle between vectors CA and CB is not the same as the angle between vectors A and B. Let me think again.Vectors CA = A - C = A - (A + B) = -BVectors CB = B - C = B - (A + B) = -ASo, vectors CA and CB are -B and -A. So, the angle between vectors CA and CB is the angle between -B and -A, which is the same as the angle between A and B, which is θ. So, angle ACB is θ.But according to the circle theorem, angle ACB should be half of θ. Therefore, there must be a mistake in my reasoning.Wait, perhaps the angle between vectors CA and CB is not the same as the angle between vectors A and B. Because vectors CA and CB are from point C, not from the origin. So, the angle between vectors CA and CB is not the same as the angle between vectors OA and OB.Wait, that's a crucial point. The angle between vectors CA and CB is the angle at point C, which is different from the angle between vectors OA and OB at the origin.So, my mistake was assuming that the angle between vectors CA and CB is the same as the angle between vectors OA and OB. That's incorrect because vectors CA and CB are positioned at point C, whereas vectors OA and OB are positioned at the origin.Therefore, I need another approach to find angle ACB.Let me consider the triangle ABC and its circumcircle with center O. Since O is the circumcenter, OA = OB = OC = R.Given that OA + OB = OC, which is a vector equation. So, in vector terms, OA + OB = OC.Let me square both sides of this vector equation to find the relationship between the magnitudes.So, |OA + OB|² = |OC|²Expanding the left side:|OA|² + |OB|² + 2 OA · OB = |OC|²Since |OA| = |OB| = |OC| = R, we can substitute:R² + R² + 2 OA · OB = R²Simplify:2R² + 2 OA · OB = R²Subtract R² from both sides:R² + 2 OA · OB = 0Therefore:OA · OB = -R² / 2Now, the dot product OA · OB can also be expressed in terms of the angle θ between vectors OA and OB:OA · OB = |OA||OB| cos θ = R² cos θSo, we have:R² cos θ = -R² / 2Divide both sides by R²:cos θ = -1/2Therefore, θ = arccos(-1/2)Which gives θ = 120 degrees or 240 degrees. But since θ is the angle between two vectors in a triangle, it must be less than 180 degrees. So, θ = 120 degrees.So, the central angle AOB is 120 degrees.Now, as per the circle theorems, the inscribed angle ACB subtended by arc AB is half of the central angle AOB. Therefore:angle ACB = (1/2) * angle AOB = (1/2) * 120 degrees = 60 degrees.Therefore, angle ACB is 60 degrees.Wait, let me verify this again because earlier I was confused about the angle between vectors CA and CB.So, angle ACB is an inscribed angle subtended by arc AB, which is half the central angle AOB. Since angle AOB is 120 degrees, angle ACB is 60 degrees.Yes, that makes sense. So, despite the vector analysis suggesting that the angle between vectors CA and CB is 120 degrees, that's actually the angle at the origin, not at point C. The angle at point C is half of that.Therefore, the measure of angle ACB is 60 degrees.I think that's the correct answer. Let me just recap to make sure I didn't miss anything.1. O is the circumcenter, so OA = OB = OC = R.2. Given OA + OB = OC in vectors.3. Squared both sides to get |OA + OB|² = |OC|².4. Expanded to find that OA · OB = -R² / 2.5. Used dot product formula to find the angle between OA and OB is 120 degrees.6. Applied circle theorem that inscribed angle is half the central angle, so angle ACB is 60 degrees.Yes, that seems solid. I don't see any mistakes in the reasoning now.**Final Answer**The measure of ( angle ACB ) is boxed{60^circ}.

answer：Okay, so I have this problem here: If |x + 1| and |y + 2x| are opposites of each other, then what is (x + y)^2004? Hmm, let me try to figure this out step by step.First, I know that opposites of each other mean that one is the negative of the other. So, if |x + 1| and |y + 2x| are opposites, then |x + 1| = -|y + 2x|. But wait, absolute values are always non-negative, right? So, the absolute value of something can't be negative. That means the only way for |x + 1| to equal -|y + 2x| is if both sides are zero. Because if |x + 1| is zero, then -|y + 2x| must also be zero, which implies |y + 2x| is zero as well.So, that gives me two equations:1. |x + 1| = 02. |y + 2x| = 0Let me solve the first equation: |x + 1| = 0. The absolute value of something is zero only when the inside is zero. So, x + 1 = 0. Solving for x, I subtract 1 from both sides: x = -1.Now, let's plug x = -1 into the second equation: |y + 2x| = 0. Substituting x, we get |y + 2*(-1)| = 0, which simplifies to |y - 2| = 0. Again, the absolute value is zero only when the inside is zero, so y - 2 = 0. Solving for y, I add 2 to both sides: y = 2.Alright, so now I have x = -1 and y = 2. The problem asks for (x + y)^2004. Let me compute x + y first: (-1) + 2 = 1. So, (1)^2004. Any number raised to any power is itself if the base is 1. So, 1^2004 is just 1.Wait, let me double-check my steps to make sure I didn't make a mistake. 1. I started by recognizing that opposites in this context mean one is the negative of the other, but since absolute values can't be negative, both must be zero. That makes sense because the only way |a| = -|b| is if both |a| and |b| are zero.2. Solving |x + 1| = 0 gives x = -1, which seems straightforward.3. Plugging x = -1 into |y + 2x| = 0 gives |y - 2| = 0, leading to y = 2. That also seems correct.4. Adding x and y: -1 + 2 = 1, and 1 raised to any power is 1. Yep, that checks out.I think that's solid. I don't see any errors in my reasoning. So, the answer should be 1.**Final Answer**boxed{1}

answer：First, there are a total of 36 delegates at the convention.Out of these, 16 delegates arrived with pre-printed name badges. This leaves 36 - 16 = 20 delegates without pre-printed badges.Half of these remaining delegates made their own hand-written name badges. So, 20 / 2 = 10 delegates made hand-written badges.The other half of the remaining delegates did not make any name badges. Therefore, 20 / 2 = 10 delegates were not wearing name badges.

answer：Okay, so I have this combination lock problem, and I need to figure out how many possible combinations there are. The lock has three numbers, n1, n2, and n3, each between 0 and 39 inclusive. There are some conditions given about their remainders when divided by 4. Let me try to break this down step by step.First, let me restate the problem in my own words. We have three numbers, n1, n2, n3. Each can be any integer from 0 to 39. The conditions are:1. n1 and n3 leave the same remainder when divided by 4. So, n1 ≡ n3 mod 4.2. n2 and n1 + 2 leave the same remainder when divided by 4. So, n2 ≡ (n1 + 2) mod 4.I need to find how many such combinations are possible. Hmm, okay. So, each of these numbers has some constraints based on their remainders modulo 4. Maybe I can model this by considering the possible remainders for each number and then counting the combinations.Let me recall that when dealing with modular arithmetic, especially modulo 4, each number can have a remainder of 0, 1, 2, or 3. So, for each number, n1, n2, n3, their possible remainders mod 4 are 0, 1, 2, 3.Given that n1 and n3 have the same remainder mod 4, let's denote r = n1 mod 4. Then, n3 mod 4 must also be r. Similarly, n2 must be congruent to (n1 + 2) mod 4, so n2 mod 4 = (r + 2) mod 4.So, if I can figure out how many choices there are for n1, n2, and n3 based on these remainders, I can compute the total number of combinations.Let me think about how many numbers between 0 and 39 inclusive give each remainder when divided by 4. Since 40 is divisible by 4, each remainder from 0 to 3 will occur exactly 10 times. For example, numbers congruent to 0 mod 4 are 0, 4, 8, ..., 36, which is 10 numbers. Similarly, for 1 mod 4: 1, 5, 9, ..., 37; 2 mod 4: 2, 6, 10, ..., 38; and 3 mod 4: 3, 7, 11, ..., 39. So, each remainder class has 10 numbers.Therefore, for each possible remainder r (0,1,2,3), there are 10 choices for n1, 10 choices for n3 (since n3 must have the same remainder as n1), and for n2, since it must be congruent to (r + 2) mod 4, there are also 10 choices for n2.So, for each r, the number of combinations is 10 (for n1) * 10 (for n2) * 10 (for n3) = 1000. But wait, hold on, that can't be right because 10*10*10 is 1000, but since r can be 0,1,2,3, that would make 4*1000=4000, which is way more than the total possible combinations. The total number of possible combinations without any restrictions is 40*40*40=64,000. Wait, 4000 is way less than 64,000, so maybe that's correct? Wait, no, 4000 is actually 1/16 of 64,000, which seems plausible because we're dealing with mod 4 conditions.Wait, let me think again. Each condition reduces the number of possibilities by a factor. Since we have two conditions, each reducing the degrees of freedom by a factor of 4, so total combinations would be 40*40*40 / (4*4) = 64,000 / 16 = 4,000. So, that seems consistent.But let me verify this step by step.First, let's consider n1. It can be any number from 0 to 39, so 40 possibilities. Now, n3 must be congruent to n1 mod 4, so once n1 is chosen, n3 is restricted to numbers with the same remainder. Since there are 10 numbers for each remainder, n3 has 10 choices.Similarly, n2 must be congruent to (n1 + 2) mod 4. So, once n1 is chosen, n2 is restricted to numbers with a specific remainder, which also has 10 choices.Therefore, for each choice of n1, we have 10 choices for n2 and 10 choices for n3. So, the total number of combinations is 40 (for n1) * 10 (for n2) * 10 (for n3) = 40*100 = 4,000.Wait, that's the same as before. So, that seems correct.Alternatively, another way to think about it is to fix the remainder r of n1. There are 4 possible remainders (0,1,2,3). For each r, n1 has 10 choices, n3 has 10 choices, and n2 must be congruent to (r + 2) mod 4, so n2 has 10 choices. Therefore, for each r, the number of combinations is 10*10*10=1000. Since there are 4 possible r's, the total is 4*1000=4000.Yes, that makes sense.Wait, but let me think again. Is there any overlap or overcounting? For example, does fixing r and then considering n1, n2, n3 for each r lead to overcounting? I don't think so because each combination is uniquely determined by the remainders of n1, n2, n3, and each r corresponds to distinct sets of numbers.Alternatively, perhaps I can model this using probability or combinatorics.The number of possible n1 is 40. For each n1, n3 must be in the same residue class mod 4, so 10 choices. For each n1, n2 must be in a specific residue class mod 4, which is (n1 + 2) mod 4, so 10 choices. Therefore, for each n1, 10*10=100 combinations. So, total combinations are 40*100=4000.Yes, that seems consistent.Alternatively, another approach: Let's consider the total number of possible combinations without any restrictions: 40^3=64,000.Now, the conditions given are:1. n1 ≡ n3 mod 42. n2 ≡ (n1 + 2) mod 4Each of these conditions imposes a constraint that reduces the number of possible combinations.For the first condition, n1 ≡ n3 mod 4: Without any constraints, n1 and n3 can be any numbers, so 40 choices each. The number of pairs where n1 ≡ n3 mod 4 is equal to the sum over each residue class of (number of n1 in class) * (number of n3 in class). Since each residue class has 10 numbers, for each residue, it's 10*10=100. There are 4 residues, so total is 4*100=400. Therefore, the number of possible (n1, n3) pairs is 400. So, the probability that n1 ≡ n3 mod 4 is 400 / (40*40)=400/1600=1/4.Similarly, for the second condition, n2 ≡ (n1 + 2) mod 4. For a given n1, n2 must be in a specific residue class, which has 10 numbers. So, for each n1, there are 10 choices for n2. So, the number of possible (n1, n2) pairs is 40*10=400. So, the probability is 400 / (40*40)=1/4 as well.But wait, these two conditions are not independent, right? Because n2 depends on n1, and n3 also depends on n1. So, the total number of combinations is not just (1/4)*(1/4)*64,000, because the two conditions are dependent on n1.Wait, maybe it's better to think in terms of conditional probability.Given that n1 is fixed, n3 has 10 choices, and n2 has 10 choices. So, for each n1, 10*10=100 combinations. So, total is 40*100=4,000.Yes, that seems consistent with the earlier reasoning.Alternatively, thinking in terms of independent choices:- Choose n1: 40 choices.- Choose n2: must be congruent to (n1 + 2) mod 4, so 10 choices.- Choose n3: must be congruent to n1 mod 4, so 10 choices.So, total combinations: 40 * 10 * 10 = 4,000.Yes, that seems straightforward.Wait, but let me make sure I'm not missing something. Is there any case where n2 and n3 could interfere with each other? For example, if n2 and n3 had some relationship, but in this problem, n2 is only related to n1, and n3 is only related to n1. So, n2 and n3 are independent once n1 is fixed.Therefore, for each n1, n2 and n3 are independent, each with 10 choices. So, 10*10=100 for each n1, leading to 40*100=4,000.Yes, that seems correct.Alternatively, let's think about the number of possible triples (n1, n2, n3) where n1 ≡ n3 mod 4 and n2 ≡ (n1 + 2) mod 4.We can model this as:- For each residue r in {0,1,2,3}, count the number of triples where n1 ≡ r mod 4, n3 ≡ r mod 4, and n2 ≡ (r + 2) mod 4.Since each residue class has 10 numbers, for each r, the number of such triples is 10 (n1) * 10 (n2) * 10 (n3) = 1,000.There are 4 residues, so total is 4*1,000=4,000.Yes, that's the same as before.Alternatively, perhaps I can think in terms of graph theory or something else, but I think the above methods are sufficient.Wait, another way: Since n1 can be any number, and n2 and n3 are determined by n1's residue, the number of combinations is equal to the number of possible n1 multiplied by the number of possible n2 and n3 for each n1.Since n1 has 40 choices, and for each n1, n2 has 10 choices and n3 has 10 choices, the total is 40*10*10=4,000.Yes, that seems consistent.So, after considering multiple approaches—counting based on residue classes, fixing n1 and counting n2 and n3, considering the total number of combinations and the constraints—it seems that the total number of possible combinations is 4,000.Therefore, I think the answer is 4,000.**Final Answer**The number of possible combinations is boxed{4000}.

answer：First, I need to determine the length of the second movie, which is 25 minutes longer than the first movie.The first movie is 1 hour and 48 minutes long. I'll convert this time into minutes to make the addition easier.1 hour is equal to 60 minutes, so the first movie is 60 + 48 = 108 minutes long.Next, I'll add the additional 25 minutes to the first movie's duration: 108 minutes + 25 minutes = 133 minutes.Now, I'll convert 133 minutes back into hours and minutes. There are 60 minutes in an hour, so 133 minutes is equal to 2 hours and 13 minutes.Therefore, the second movie is 2 hours and 13 minutes long.