answer:The question of whether 0.9 recurring (often denoted as 0.999...) is equal to 1 is a classic example of a counterintuitive mathematical concept. At first glance, it may seem like 0.9 recurring is less than 1 because it's missing that final step to reach 1. However, this is where the nature of infinite series and limits comes into play. Consider the series 0.9 + 0.09 + 0.009 + ... . This is an infinite geometric series with a common ratio of 1/10. The sum of an infinite geometric series can be calculated using the formula S = a / (1 - r), where a is the first term and r is the common ratio. In this case, a = 0.9 and r = 1/10. Plugging in these values, we get S = 0.9 / (1 - 1/10) = 0.9 / (9/10) = 1. Another way to approach this is to consider the following equation: 10x = 9.999..., where x = 0.999... . If we subtract x from 10x, we get 9x = 9, and dividing both sides by 9 gives x = 1. This shows that 0.9 recurring is indeed equal to 1. The key insight here is that the infinite series 0.9 + 0.09 + 0.009 + ... converges to 1, meaning that the sum gets arbitrarily close to 1 as the number of terms increases without bound. In other words, the difference between 0.9 recurring and 1 becomes infinitesimally small, effectively making them equal. This result highlights the importance of understanding infinite series and limits in mathematics, and how they can sometimes lead to surprising conclusions.

answer:Zeno's paradox of Achilles and the tortoise is a classic example of an ancient philosophical puzzle that can be resolved using mathematical concepts, particularly infinite series and limits. The paradox goes like this: Achilles, a swift warrior, is racing a slow tortoise. To make the race more interesting, the tortoise is given a head start. When Achilles reaches the tortoise's starting point, the tortoise has moved a little bit ahead. When Achilles reaches the new position of the tortoise, the tortoise has moved a little bit ahead again, and so on. The paradox claims that Achilles can never catch up to the tortoise, because every time he reaches the tortoise's previous position, the tortoise has moved a little bit further ahead. From a mathematical perspective, this paradox can be viewed as an infinite series. Suppose the tortoise starts at position 1, and Achilles starts at position 0. When Achilles reaches the tortoise's starting point (position 1), the tortoise has moved to position 1 + 1/2. When Achilles reaches the new position of the tortoise, the tortoise has moved to position 1 + 1/2 + 1/4, and so on. The series representing the tortoise's position is an infinite geometric series with a common ratio of 1/2. Using the formula for the sum of an infinite geometric series, we can calculate the total distance the tortoise travels: 1 + 1/2 + 1/4 + ... = 1 / (1 - 1/2) = 2. This means that the tortoise will eventually travel a total distance of 2 units. However, the key point is that Achilles is traveling at a much faster speed than the tortoise. Suppose Achilles travels at a speed of 10 units per second, and the tortoise travels at a speed of 1 unit per second. Then, the time it takes Achilles to cover the distance to the tortoise's starting point is 1/10 seconds. The time it takes Achilles to cover the additional distance to the tortoise's new position is 1/20 seconds, and so on. The series representing the total time it takes Achilles to catch up to the tortoise is an infinite series as well: 1/10 + 1/20 + 1/40 + ... . This series also converges, and its sum can be calculated using the formula for the sum of an infinite geometric series. The result is that Achilles will catch up to the tortoise in a finite amount of time, specifically 2/9 seconds. In essence, the paradox is resolved by recognizing that the infinite series representing the distances and times involved converge to finite values. Achilles will eventually catch up to the tortoise, even though it seems counterintuitive at first. This resolution highlights the importance of understanding infinite series and limits in resolving apparent paradoxes.

answer:To solve this problem, we need to consider the probabilities of each event occurring in sequence. The probability of picking a red ball on the first draw is 7/10, since there are 7 red balls out of a total of 10 balls. Assuming the first ball is indeed red, there are now 6 red balls and 3 blue balls left in the basket. The probability of picking a red ball on the second draw is therefore 6/9. Finally, assuming the first two balls are red, there are now 5 red balls and 3 blue balls left in the basket. The probability of picking a blue ball on the third draw is therefore 3/8. To find the overall probability of the sequence of events (red, red, blue), we multiply the individual probabilities together: (7/10) × (6/9) × (3/8) = 126/720 = 7/40. So, the odds of picking two red balls followed by a blue ball are 7/40. This can also be expressed as a decimal probability of 0.175.

answer:Alan bought x eggs at the price of 2 per egg, so the cost of the eggs is 2x dollars. He also bought 6 chickens at the price of 8 per chicken, so the cost of the chickens is 6 * 8 = 48. The total amount of money Alan spent at the market is the sum of the cost of the eggs and the cost of the chickens: 2x + 48. We are given that the total amount of money Alan spent is 88, so we can write: 2x + 48 = 88. Subtracting 48 from both sides, we get: 2x = 40. Dividing both sides by 2, we get: x = 20. The value of x is 20. 20 The answer is: 20