answer:The problem of an inscribed sphere within a cone offers a fascinating exploration of geometric relationships and spatial reasoning. To begin, let's visualize the scenario: a right cone with a radius of 6 units and a height of 8 units contains a sphere that touches the cone's base and its lateral surface. We can establish a relationship between the sphere's radius, the cone's radius, and the cone's height by exploiting the similarity of triangles. Consider a cross-section of the cone, passing through its apex and the center of the circular base. This cross-section contains the center of the inscribed sphere. From the center of the sphere, draw a radius perpendicular to the cone's lateral surface, and another radius to the base of the cone. These two radii, along with the slant height of the cone and its height, form two similar triangles. Using the similarity of these triangles, we can express the relationship between the radius of the sphere, r, the cone's radius, 6, and its height, 8. By proportionality, we find that r / (6 - r) = 8 / sqrt(6^2 + 8^2). We then solve for r. Calculating sqrt(6^2 + 8^2) as sqrt(100), which is 10, we rewrite the proportion as r / (6 - r) = 8 / 10 = 4 / 5. Cross-multiplying and solving for r yields r = 6 * 4 / 9, so the radius of the sphere is 8/3 units. Now, to find the volume of the sphere, we apply the formula V = 4/3 * π * r^3. Substituting r = 8/3, we find the volume of the sphere to be V = 4/3 * π * (8/3)^3 = 256/27 * π. Thus, the inscribed sphere has a radius of 8/3 units and a volume of 256/27 * π cubic units.

answer:As the opening angle of the cone changes, the inscribed sphere will not necessarily remain in contact with the slant height, the base, and maintain its original position. In our initial problem, the specific dimensions of the cone (radius 6 and height 8) allowed us to find a unique radius for the inscribed sphere. However, as the opening angle of the cone increases or decreases, the position and size of the inscribed sphere will change. When the opening angle increases, the cone becomes shallower and the sphere will move upward, away from the base. Conversely, when the opening angle decreases, the cone becomes steeper and the sphere will move downward, toward the base. There exists a critical angle, below which the sphere will no longer be able to touch the slant height and the base simultaneously. Beyond this point, the sphere will either be tangent only to the base or only to the slant height, but not both. To explore this further, we can express the radius of the inscribed sphere in terms of the cone's opening angle. Using the same similarity of triangles as before, we can relate the sphere's radius, r, to the cone's radius, R, and its height, H, as r / (R - r) = H / sqrt(R^2 + H^2). From this, we can derive a general expression for r in terms of R and H, which can be further expressed in terms of the opening angle. Analyzing this expression will reveal the relationship between the opening angle and the radius of the inscribed sphere. As the opening angle changes, the radius of the inscribed sphere will also change, illustrating the intricate connection between the cone's geometry and the sphere's position. Thus, the inscribed sphere's position and size are intimately tied to the cone's geometry, and changes to the opening angle will result in corresponding changes to the sphere's radius and its points of contact with the cone.

answer:This problem delves into the realm of number theory, specifically exploring the concept of coprime numbers. To begin, we note that two numbers are coprime (or relatively prime) if their greatest common divisor (GCD) is 1. In other words, they have no common positive integer divisors except for 1. To calculate the probability, we can employ a counting approach. We'll first determine the total number of ways to choose two distinct positive integers less than 1000, and then find the number of pairs that are coprime. The total number of ways to choose two distinct positive integers less than 1000 is given by the combination formula: C(999, 2) = 999 * 998 / 2. To count the number of coprime pairs, we can utilize the principle of inclusion-exclusion (PIE). The idea is to count the pairs that are not coprime and subtract them from the total. For a pair of numbers to not be coprime, they must share a common prime factor. We can iterate through each prime number less than 1000, counting the pairs that are divisible by that prime. Using the prime factorization of numbers, we can find the number of multiples of each prime p less than 1000, which is floor(999/p). Then, the number of pairs that share the prime factor p is given by floor(999/p) * (floor(999/p) - 1) / 2. However, this approach double-counts pairs that share multiple prime factors. To correct for this, we apply the principle of inclusion-exclusion, iterating over all possible combinations of prime factors and adjusting the count accordingly. The calculation becomes intricate, but fortunately, there's a more elegant solution based on the properties of coprime numbers. We can exploit the fact that the probability of two numbers being coprime is related to the Riemann zeta function. For two randomly chosen integers, the probability of being coprime can be expressed as 1 / ζ(2), where ζ(s) is the Riemann zeta function. The value of ζ(2) is known to be π^2 / 6. Thus, the probability of two distinct positive integers less than 1000 being coprime is approximately 1 / (π^2 / 6) = 6 / π^2 ≈ 0.6079. This result might seem surprising, as it implies that more than 60% of all possible pairs of distinct positive integers less than 1000 are coprime. While the exact probability is independent of the upper limit (1000 in this case), the limiting value of the probability holds as the upper bound increases. This reveals a fascinating property of coprime numbers, showcasing the intricate connections between number theory, probability, and the Riemann zeta function.

answer:Dennis had 10 rocks. His fish ate half of them, so 10/2 = 5 rocks were eaten. He was able to make the fish spit out 2 rocks, so 2 rocks were added back. Therefore, Dennis has 10 - 5 + 2 = 7 rocks left. 7 The answer is: 7