💡First, I need to determine the additional area the calf can graze when the rope length is increased from 10 meters to 35 meters.The area that a calf can graze is a circular area with the rope as the radius. The formula for the area of a circle is ( A = pi r^2 ), where ( A ) is the area and ( r ) is the radius (length of the rope).First, I'll calculate the area with the original 10-meter rope:[A_1 = pi times (10, text{m})^2 = pi times 100, text{m}^2 = 100pi, text{m}^2]Next, I'll calculate the area with the increased 35-meter rope:[A_2 = pi times (35, text{m})^2 = pi times 1225, text{m}^2 = 1225pi, text{m}^2]To find the additional grassy ground the calf can graze, I'll subtract the original area from the increased area:[text{Additional area} = A_2 - A_1 = 1225pi, text{m}^2 - 100pi, text{m}^2 = (1225 - 100)pi, text{m}^2 = 1125pi, text{m}^2]Therefore, the calf can graze an additional ( 1125pi ) square meters of grassy ground with the increased rope length.

💡Okay, so I've got this problem about four motorcyclists starting from two cities, A and B, heading towards each other. The names are Petrov, Ivanov, Ivanovskiy, and Petrovsky. Petrov and Ivanov start from A, while Ivanovskiy and Petrovsky start from B. The problem says that Ivanov is twice as fast as Petrov, and Ivanovskiy is three times as fast as Petrovsky. Also, it mentions that Ivanov meets Petrovsky at the same time Petrov meets Ivanovskiy. The question is asking which meeting happens closer to city A: Ivanov meeting Ivanovskiy or Petrov meeting Petrovsky.Alright, let's break this down. First, I need to visualize the scenario. There are two cities, A and B, with four people starting at the same time from each city towards the other. So, Petrov and Ivanov are going from A to B, while Ivanovskiy and Petrovsky are going from B to A. Given that Ivanov is twice as fast as Petrov, I can denote Petrov's speed as v, so Ivanov's speed would be 2v. Similarly, Ivanovskiy is three times as fast as Petrovsky. Let's denote Petrovsky's speed as u, so Ivanovskiy's speed would be 3u.Now, the key piece of information is that Ivanov meets Petrovsky at the same moment Petrov meets Ivanovskiy. That means both meetings happen at the same time, t. So, I need to find the distances each pair has traveled by time t and set up equations accordingly.Let's denote the distance between cities A and B as D. Since all four start at the same time, the sum of the distances traveled by each pair when they meet should equal D.First, let's consider Ivanov and Petrovsky. Ivanov is going from A to B at speed 2v, and Petrovsky is going from B to A at speed u. The distance between them initially is D, so the time it takes for them to meet is t. The distance Ivanov covers is 2v*t, and the distance Petrovsky covers is u*t. Since they meet, the sum of these distances should be D:2v*t + u*t = DSimilarly, for Petrov and Ivanovskiy. Petrov is going from A to B at speed v, and Ivanovskiy is going from B to A at speed 3u. The distance Petrov covers is v*t, and the distance Ivanovskiy covers is 3u*t. Again, the sum should be D:v*t + 3u*t = DSo now I have two equations:1. 2v*t + u*t = D2. v*t + 3u*t = DSince both equal D, I can set them equal to each other:2v*t + u*t = v*t + 3u*tLet's simplify this equation. Subtract v*t and u*t from both sides:2v*t + u*t - v*t - u*t = v*t + 3u*t - v*t - u*tThis simplifies to:v*t = 2u*tAssuming t ≠ 0 (since they do meet at some point), we can divide both sides by t:v = 2uSo, Petrov's speed is twice Petrovsky's speed. That's interesting. So, v = 2u.Now, let's substitute this back into one of the original equations to find the relationship between v and u. Let's use the first equation:2v*t + u*t = DSince v = 2u, substitute:2*(2u)*t + u*t = DWhich simplifies to:4u*t + u*t = DSo, 5u*t = DTherefore, u*t = D/5Similarly, v*t = 2u*t = 2*(D/5) = 2D/5So, the distance Petrov has traveled when they meet is 2D/5, and the distance Ivanovskiy has traveled is 3u*t = 3*(D/5) = 3D/5.Similarly, for the other meeting, Ivanov has traveled 2v*t = 2*(2u)*t = 4u*t = 4*(D/5) = 4D/5, and Petrovsky has traveled u*t = D/5.Wait, so when Ivanov meets Petrovsky, Ivanov has traveled 4D/5, and Petrovsky has traveled D/5. So, the meeting point is 4D/5 from A, which is closer to B.On the other hand, when Petrov meets Ivanovskiy, Petrov has traveled 2D/5, and Ivanovskiy has traveled 3D/5. So, the meeting point is 2D/5 from A, which is closer to A.Therefore, the meeting between Petrov and Ivanovskiy happens closer to city A, while the meeting between Ivanov and Petrovsky happens closer to city B.Wait, but the question is asking which meeting takes place closer to city A: Ivanov's with Ivanovskiy or Petrov's with Petrovsky.From the calculations, Petrov's meeting with Ivanovskiy is at 2D/5 from A, and Ivanov's meeting with Petrovsky is at 4D/5 from A. So, 2D/5 is closer to A than 4D/5.Therefore, the meeting closer to city A is Petrov meeting Ivanovskiy.But wait, let me double-check my calculations to make sure I didn't make a mistake.Starting from the two equations:1. 2v*t + u*t = D2. v*t + 3u*t = DSetting them equal:2v*t + u*t = v*t + 3u*tSubtracting v*t and u*t:v*t = 2u*tSo, v = 2uSubstituting back into equation 1:2*(2u)*t + u*t = D4u*t + u*t = D5u*t = D => u*t = D/5Therefore, v*t = 2u*t = 2D/5So, for the first meeting (Ivanov and Petrovsky):Distance from A: 2v*t = 4D/5Distance from B: u*t = D/5For the second meeting (Petrov and Ivanovskiy):Distance from A: v*t = 2D/5Distance from B: 3u*t = 3D/5Yes, so the meeting points are at 4D/5 and 2D/5 from A, respectively. Therefore, the meeting closer to A is Petrov meeting Ivanovskiy.Wait, but the problem says "Ivanov meets Petrovsky at the same moment Petrov meets Ivanovskiy." So, both meetings happen at the same time t, but at different locations. The one closer to A is Petrov meeting Ivanovskiy, and the one closer to B is Ivanov meeting Petrovsky.Therefore, the answer is that Petrov's meeting with Ivanovskiy takes place closer to city A.But let me think again. Is there another way to interpret the problem? Maybe I misread something.The problem states: "Ivanov meets Petrovsky at the same moment Petrov meets Ivanovskiy." So, both meetings happen simultaneously but at different points along the route. We need to determine which meeting is closer to A.From the calculations, the meeting points are at 2D/5 and 4D/5 from A. So, 2D/5 is closer to A, meaning Petrov's meeting with Ivanovskiy is closer to A.Alternatively, if we think about the distances from B, the meeting points are at 3D/5 and D/5 from B. So, D/5 is closer to B, meaning Ivanov's meeting with Petrovsky is closer to B.Therefore, the meeting closer to A is Petrov's meeting with Ivanovskiy.I think that's correct. So, the answer is that Petrov's meeting with Ivanovskiy takes place closer to city A.

💡Okay, so I'm looking at this sequence: 12, 13, 15, 17, 111, 113, 117, 119, 123, 129, 131, 135, 139, 1415, 1417, 1421, 1425, 1431, 1437, 1445. I need to figure out what comes next. Hmm, let's break it down step by step.First, I notice that the numbers start with two digits: 12, 13, 15, 17. These look like they could be prime numbers, but they all start with a '1'. Let me check: 12 isn't prime, 13 is prime, 15 isn't, 17 is. So maybe it's not about primes. Maybe it's about numbers that start with '1' followed by something.Looking at the next set: 111, 113, 117, 119. These are all three-digit numbers starting with '1'. 111 is 111, 113 is prime, 117 is divisible by 13, 119 is 7 times 17. Not sure about the pattern here.Then we have 123, 129, 131, 135, 139. These are also three-digit numbers starting with '1'. 123 is 123, 129 is 129, 131 is prime, 135 is divisible by 5, 139 is prime. Still not clear.Next, we jump to four-digit numbers: 1415, 1417, 1421, 1425, 1431, 1437, 1445. These are all four-digit numbers starting with '14'. Let's see: 1415 is 1415, 1417 is prime, 1421 is 1421, 1425 is divisible by 5, 1431 is divisible by 3, 1437 is divisible by 3, 1445 is divisible by 5.Wait a minute, maybe the numbers are increasing by 2, then 4, then 6, etc. Let's check the differences:From 12 to 13: +113 to 15: +215 to 17: +217 to 111: +94 (That's a big jump)111 to 113: +2113 to 117: +4117 to 119: +2119 to 123: +4123 to 129: +6129 to 131: +2131 to 135: +4135 to 139: +4139 to 1415: +1276 (Another big jump)1415 to 1417: +21417 to 1421: +41421 to 1425: +41425 to 1431: +61431 to 1437: +61437 to 1445: +8So the increments are inconsistent, with some small jumps and some very large jumps. Maybe the large jumps are when the number of digits increases? From two-digit to three-digit, and then to four-digit.But even within the three-digit numbers, the increments vary: +2, +4, +2, +4, +6, +2, +4, +4, +6, +6, +8.It's not a clear arithmetic progression. Maybe it's something else.Looking back at the numbers, they all start with '1' or '14'. Maybe the numbers are being constructed by appending digits. For example, 12, 13, 15, 17 could be '1' followed by 2, 3, 5, 7, which are primes. Then 111, 113, 117, 119 could be '11' followed by 1, 3, 7, 9, which are odd numbers. Similarly, 123, 129, 131, 135, 139 could be '12' followed by 3, 9, 1, 5, 9. Not sure.Then in the four-digit numbers, 1415, 1417, 1421, 1425, 1431, 1437, 1445, it's '14' followed by 15, 17, 21, 25, 31, 37, 45. These seem like numbers that are multiples of 5 or primes, but not exactly.Maybe the pattern is that after a certain point, the numbers are constructed by appending two digits to '14', and those two digits are increasing by 2, 4, 6, etc. Let's see:1415: 151417: 17 (+2)1421: 21 (+4)1425: 25 (+4)1431: 31 (+6)1437: 37 (+6)1445: 45 (+8)So the appended numbers are increasing by 2, then 4, then 4, then 6, then 6, then 8. It seems like the increments are increasing by 2 each time, but sometimes repeating.If that's the case, the next increment would be +8 again, making the next appended number 45 + 8 = 53. So the next number would be 1453.But wait, let's check the previous increments:From 15 to 17: +217 to 21: +421 to 25: +425 to 31: +631 to 37: +637 to 45: +8So the pattern is +2, +4, +4, +6, +6, +8. It seems like the increment increases by 2 every two steps. So after +8, the next increment would be +10, making the next appended number 45 + 10 = 55. So the next number would be 1455.But I'm not entirely sure. Maybe the pattern is different. Alternatively, perhaps the appended numbers are following their own sequence, like primes or something else.Alternatively, maybe the entire number is being incremented by a certain pattern. Let's look at the four-digit numbers:1415, 1417, 1421, 1425, 1431, 1437, 1445.Differences between them:1417 - 1415 = 21421 - 1417 = 41425 - 1421 = 41431 - 1425 = 61437 - 1431 = 61445 - 1437 = 8So the differences are +2, +4, +4, +6, +6, +8.If this pattern continues, the next difference would be +8 again, making the next number 1445 + 8 = 1453.But wait, the pattern of differences seems to be increasing by 2 every two steps: +2, then +4 twice, then +6 twice, then +8 twice, etc. So after +8, the next difference would be +10, making the next number 1445 + 10 = 1455.Alternatively, if the pattern is to increase the increment by 2 each time, regardless of repetition, then after +8, the next increment would be +10, leading to 1445 + 10 = 1455.But I'm not entirely confident. Maybe I should look for another pattern.Another approach: Maybe the numbers are being formed by concatenating two numbers. For example, 12 could be 1 and 2, 13 is 1 and 3, 15 is 1 and 5, 17 is 1 and 7. Then 111 is 11 and 1, 113 is 11 and 3, 117 is 11 and 7, 119 is 11 and 9. Then 123 is 12 and 3, 129 is 12 and 9, 131 is 13 and 1, 135 is 13 and 5, 139 is 13 and 9. Then 1415 is 14 and 15, 1417 is 14 and 17, 1421 is 14 and 21, 1425 is 14 and 25, 1431 is 14 and 31, 1437 is 14 and 37, 1445 is 14 and 45.If that's the case, then the first part is increasing: 1, 1, 1, 1, 11, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14.The second part is: 2, 3, 5, 7, 1, 3, 7, 9, 3, 9, 1, 5, 9, 15, 17, 21, 25, 31, 37, 45.Looking at the second part: 2, 3, 5, 7 (primes), then 1, 3, 7, 9 (odd numbers), then 3, 9, 1, 5, 9 (not sure), then 15, 17, 21, 25, 31, 37, 45.Maybe the second part is increasing by 2, then 4, etc., similar to the differences we saw earlier.If that's the case, the next second part number after 45 would be 45 + 10 = 55, making the next number 14 and 55, which is 1455.Alternatively, if the second part is following its own pattern, maybe it's adding 2, 4, 6, etc., so after 45, adding 10 would give 55.But I'm still not entirely sure. Maybe I should look for another pattern.Another thought: Maybe the numbers are being formed by concatenating the previous number's last digit with the next number. For example, 12 ends with 2, then 13 starts with 1, but that doesn't seem to fit.Alternatively, maybe it's related to the number of letters in the English word for the number. For example, 12 is "twelve" (6 letters), 13 is "thirteen" (8 letters), but that doesn't seem to match the sequence.Alternatively, maybe it's related to binary representations or something else entirely.Wait, let's go back to the initial observation about the differences in the four-digit numbers: +2, +4, +4, +6, +6, +8. If this pattern continues, the next difference would be +8, making the next number 1445 + 8 = 1453. Alternatively, if the pattern is to increase the increment by 2 every two steps, then after +8, the next increment would be +10, making the next number 1445 + 10 = 1455.Given that the differences have been increasing by 2 every two steps, I think the next increment would be +10, leading to 1455.But I'm still a bit unsure. Maybe I should check if 1455 fits any other pattern.Looking back at the sequence, the numbers seem to be increasing, and the four-digit numbers are following a pattern where the last two digits are increasing by 2, 4, 4, 6, 6, 8. So if we continue that pattern, the next increase would be +10, making the last two digits 45 + 10 = 55, so the next number would be 1455.Alternatively, if the pattern is to keep increasing by 8, then the next number would be 1445 + 8 = 1453.But considering the pattern of increments (+2, +4, +4, +6, +6, +8), it seems like the increment increases by 2 every two steps. So after +8, the next increment would be +10, making the next number 1445 + 10 = 1455.Therefore, I think the next number in the sequence is 1455.

💡First, I need to identify all sets ( P ) that satisfy the condition ( {a} subsetneq P subseteq {a, b, c} ).This means that ( P ) must contain the element ( a ) and at least one more element from ( {b, c} ), but it cannot include all three elements ( a ), ( b ), and ( c ).By examining the possible combinations, the valid sets ( P ) are:- ( {a, b} )- ( {a, c} )There are 2 such sets.

💡First, I recognize that the problem provides two equations involving ( a ) and ( b ):[a + b = 10 quad text{and} quad a - b = 4]I need to find the value of ( a^2 - b^2 ).I recall the algebraic identity:[a^2 - b^2 = (a + b)(a - b)]This identity allows me to express ( a^2 - b^2 ) in terms of ( a + b ) and ( a - b ), which are already given in the problem.Substituting the given values into the identity:[a^2 - b^2 = (10)(4) = 40]Therefore, the value of ( a^2 - b^2 ) is 40.

💡First, I recognize that 150 is 250% of an unknown number, which I'll call ( x ).To set up the equation, I express 250% as a decimal by dividing by 100, resulting in 2.5.This gives me the equation:[ 150 = 2.5 times x ]To solve for ( x ), I divide both sides of the equation by 2.5:[ x = frac{150}{2.5} ]Performing the division:[ x = 60 ]Therefore, the number is 60.