💡First, I recognize that the cost price (CP) is 80% of the selling price (SP). This means that CP = 0.80 × SP.Next, I calculate the profit by subtracting the cost price from the selling price: Profit = SP - CP.Substituting the value of CP from the first step into the profit formula gives: Profit = SP - 0.80 × SP = 0.20 × SP.To find the profit percentage, I divide the profit by the cost price and then multiply by 100%. This results in: Profit Percentage = (0.20 × SP) / (0.80 × SP) × 100% = 25%.Therefore, the profit percentage is 25%.

💡First, I need to determine the distance the train travels in 25 seconds. The train is moving at a speed of 35 km/hr, which I'll convert to meters per second (m/s) for consistency with the other units.To convert 35 km/hr to m/s:[35 , text{km/hr} = 35 times frac{1000}{3600} , text{m/s} = frac{35000}{3600} , text{m/s} approx 9.722 , text{m/s}]Next, I'll calculate the total distance the train travels in 25 seconds:[text{Distance} = text{Speed} times text{Time} = 9.722 , text{m/s} times 25 , text{s} = 243.05 , text{meters}]This distance includes both the length of the train and the length of the bridge. Since the train itself is 150 meters long, I'll subtract this to find the length of the bridge:[text{Length of the bridge} = text{Total distance} - text{Length of the train} = 243.05 , text{meters} - 150 , text{meters} = 93.05 , text{meters}]Therefore, the length of the bridge is approximately 93.05 meters.

💡First, I need to find the complex number ( z ) that satisfies the equation ( z + 3i = 3 - i ).To isolate ( z ), I subtract ( 3i ) from both sides of the equation:[z = 3 - i - 3i = 3 - 4i]Now that I have ( z = 3 - 4i ), I can calculate its modulus. The modulus of a complex number ( a + bi ) is given by:[|a + bi| = sqrt{a^2 + b^2}]Applying this formula to ( z = 3 - 4i ):[|z| = sqrt{3^2 + (-4)^2} = sqrt{9 + 16} = sqrt{25} = 5]Therefore, the modulus ( |z| ) is 5.

💡First, I need to determine how many games Kelly needs to give away so that she has 35 games left.She currently has 50 games. To find out how many games she should give away, I subtract the number of games she wants to have left from the total number of games she currently has.So, 50 minus 35 equals 15.Therefore, Kelly needs to give away 15 games.

💡Alright, so I've got this geometry problem here. Let me try to understand it step by step. It says that on a plane, there are two fixed points, A and B. Then, for any k points P₁, P₂, ..., Pₖ on this plane, there must be two points P_i and P_j (where i and j are different) such that the absolute difference between the sines of the angles AP_iB and AP_jB is less than or equal to 1/1992. The question is asking for the minimum value of k.Hmm, okay. So, first, I need to visualize this. Points A and B are fixed, and we're considering points P₁ through Pₖ somewhere on the plane. For each of these points P_i, we form the angle AP_iB, take the sine of that angle, and then look at the differences between these sines for different points. The problem is saying that no matter how we choose these k points, there will always be at least two points whose sine of the angle AP_iB differs by at most 1/1992.I think this is related to the pigeonhole principle. The pigeonhole principle states that if you have more pigeons than pigeonholes, at least two pigeons must share a hole. Translating that to this problem, the "pigeonholes" would be intervals of size 1/1992 in the range of possible sine values, and the "pigeons" are the points P₁ through Pₖ. So, if we have enough points, at least two of them must fall into the same interval, meaning their sine values are close enough.Let me think about the range of possible values for sin(angle AP_iB). The angle at P_i between points A and B can range from 0 to π radians because it's a planar angle. The sine function of an angle in this range goes from 0 to 1 and back to 0. So, sin(0) = 0, sin(π/2) = 1, and sin(π) = 0. Therefore, the sine of angle AP_iB can take any value between 0 and 1.So, the possible values of sin(angle AP_iB) lie in the interval [0, 1]. If we divide this interval into smaller intervals, each of length 1/1992, how many such intervals would we have? Well, the total length is 1, and each interval is 1/1992, so we would have 1992 intervals. But wait, actually, if we divide [0,1] into 1992 equal parts, each part would have a length of 1/1992, right? So, we'd have 1992 subintervals.Now, according to the pigeonhole principle, if we have more points than the number of intervals, at least two points must fall into the same interval. So, if we have k points, and we have 1992 intervals, then if k is greater than 1992, meaning k is at least 1993, then at least two points must lie within the same interval. Therefore, the difference between their sine values would be at most the length of the interval, which is 1/1992.Wait, but the problem says "for any k points," so we need to ensure that no matter how we choose these k points, there will always be two points whose sine values differ by at most 1/1992. So, if k is 1993, then regardless of how we place these points, two of them must lie within the same interval of size 1/1992, hence their sine values differ by at most 1/1992.But let me make sure I'm not missing something. Is there a possibility that the sine function could have some non-uniform distribution that might affect this? Hmm, I don't think so because the pigeonhole principle is purely about the number of points and the number of intervals. As long as the sine values are distributed over the interval [0,1], regardless of how they're distributed, if you have more points than intervals, two must fall into the same interval.Another thing to consider is whether the angles can actually take on all values in [0,1] for sin(angle AP_iB). Well, as I thought earlier, the angle can range from 0 to π, and sin(angle) can take any value between 0 and 1. So, yes, the sine values can indeed cover the entire interval [0,1]. Therefore, dividing [0,1] into 1992 intervals each of length 1/1992 is valid.So, if we have k = 1993 points, then by the pigeonhole principle, at least two of them must lie within the same interval, meaning their sine values differ by at most 1/1992. Therefore, the minimal k is 1993.Wait, but let me think again. Is there a case where even with k = 1993, we might not have two points within the same interval? No, because the pigeonhole principle is a certainty. If you have more pigeons than holes, at least one hole must contain more than one pigeon. So, in this case, with 1993 points and 1992 intervals, at least one interval must contain at least two points. Therefore, their sine values must differ by at most 1/1992.Therefore, the minimal k is indeed 1993.**Final Answer**The minimum value of ( k ) is boxed{1993}.

💡Okay, so I need to find the greatest common divisor (gcd) of 187 and 119 and then express that result in base-5. Hmm, let's see. I remember that the gcd is the largest number that divides both numbers without leaving a remainder. There are different methods to find the gcd, like listing out all the factors or using the Euclidean algorithm. I think the Euclidean algorithm might be faster, especially for larger numbers like these.Alright, let's try the Euclidean algorithm. From what I recall, this method involves dividing the larger number by the smaller one and then replacing the larger number with the smaller number and the smaller number with the remainder from the division. I repeat this process until the remainder is zero, and the last non-zero remainder is the gcd.So, starting with 187 and 119:First, divide 187 by 119. How many times does 119 go into 187? Well, 119 times 1 is 119, and 119 times 2 is 238, which is too big. So, it goes in once with a remainder. Let's calculate the remainder:187 - 119 = 68Okay, so the remainder is 68. Now, I replace the numbers: the new pair is 119 and 68.Next, divide 119 by 68. How many times does 68 go into 119? 68 times 1 is 68, and 68 times 2 is 136, which is too big. So, it goes in once again. The remainder is:119 - 68 = 51Now, the new pair is 68 and 51.Divide 68 by 51. 51 times 1 is 51, and 51 times 2 is 102, which is too big. So, it goes in once with a remainder of:68 - 51 = 17Now, the new pair is 51 and 17.Divide 51 by 17. 17 times 3 is 51, so it goes in exactly 3 times with no remainder.Since the remainder is zero, the last non-zero remainder is 17. So, the gcd of 187 and 119 is 17.Now, I need to convert 17 from base-10 to base-5. To do that, I'll divide 17 by 5 and keep track of the remainders.First division: 17 divided by 5 is 3 with a remainder of 2.So, the least significant digit (rightmost) is 2.Now, take the quotient, which is 3, and divide that by 5.3 divided by 5 is 0 with a remainder of 3.So, the next digit to the left is 3.Since the quotient is now 0, we stop the process.Reading the remainders from bottom to top, we get 32.Therefore, 17 in base-5 is 32.Wait, let me double-check that. If I convert 32 base-5 back to base-10, it should be:3 * 5^1 + 2 * 5^0 = 15 + 2 = 17.Yes, that checks out.So, the gcd of 187 and 119 is 17, and in base-5, that's 32.