💡Okay, so I have this problem where a sequence of natural numbers {x_n} is defined by x₁ = 2, and for n ≥ 2, x_n is the floor of 1.5 times x_{n-1}. Then, there's another sequence y_n defined as (-1)^{x_n}, and I need to prove that y_n is non-periodic. First, let me make sure I understand the sequences correctly. Starting with x₁ = 2, then x₂ would be floor(1.5 * 2) = floor(3) = 3. Then x₃ would be floor(1.5 * 3) = floor(4.5) = 4. Next, x₄ would be floor(1.5 * 4) = floor(6) = 6. Then x₅ is floor(1.5 * 6) = floor(9) = 9, and so on. So the sequence x_n is 2, 3, 4, 6, 9, 13, 19, 28, 42, 63, 94, 141, 211, 316, 474, 711, 1066, 1599, 2398, 3597, and so on. Now, y_n is defined as (-1)^{x_n}, so each term of y_n alternates between 1 and -1 depending on whether x_n is even or odd. So, looking at the x_n sequence: 2 (even), 3 (odd), 4 (even), 6 (even), 9 (odd), 13 (odd), 19 (odd), 28 (even), 42 (even), 63 (odd), 94 (even), 141 (odd), 211 (odd), 316 (even), 474 (even), 711 (odd), 1066 (even), 1599 (odd), 2398 (even), 3597 (odd), etc. So, y_n would be: 1, -1, 1, 1, -1, -1, -1, 1, 1, -1, 1, -1, -1, 1, 1, -1, 1, -1, 1, -1, etc. I need to show that this sequence y_n is non-periodic. That is, there is no period T such that y_{n+T} = y_n for all n beyond some point. To approach this, I think I should assume the opposite: suppose that y_n is periodic with period T. Then, for all n ≥ some N, y_{n+T} = y_n. Since y_n = (-1)^{x_n}, this would imply that x_{n+T} and x_n have the same parity for all n ≥ N. In other words, x_{n+T} ≡ x_n mod 2. So, if I can show that this leads to a contradiction, then y_n cannot be periodic. Let me think about how x_n behaves. Each term is floor(1.5 * x_{n-1}). So, 1.5 * x_{n-1} is either an integer or has a fractional part. If it's an integer, then x_n = 1.5 * x_{n-1}; otherwise, x_n = floor(1.5 * x_{n-1}). Looking at the initial terms, x₁ = 2, x₂ = 3, x₃ = 4, x₄ = 6, x₅ = 9, x₆ = 13, x₇ = 19, x₈ = 28, x₉ = 42, x_{10} = 63, etc. It seems like x_n is increasing and not repeating any values. If y_n were periodic, then the parities of x_n would have to repeat every T terms. That is, the sequence of even and odd x_n would repeat every T terms. But looking at the x_n sequence, the parities are: even, odd, even, even, odd, odd, odd, even, even, odd, even, odd, odd, even, even, odd, even, odd, even, odd, etc. I don't see an obvious repeating pattern here. The number of consecutive evens or odds seems to vary. For example, after x₁ = 2 (even), we have x₂ = 3 (odd), then x₃ = 4 (even), x₄ = 6 (even), so two evens, then x₅ = 9 (odd), x₆ = 13 (odd), x₇ = 19 (odd), so three odds, then x₈ = 28 (even), x₉ = 42 (even), two evens, x_{10} = 63 (odd), x_{11} = 94 (even), x_{12} = 141 (odd), x_{13} = 211 (odd), two odds, x_{14} = 316 (even), x_{15} = 474 (even), two evens, x_{16} = 711 (odd), x_{17} = 1066 (even), x_{18} = 1599 (odd), x_{19} = 2398 (even), x_{20} = 3597 (odd), etc. The number of consecutive evens or odds doesn't seem to settle into a fixed pattern. Sometimes there are two evens, sometimes three odds, sometimes two odds, etc. This suggests that the parities don't repeat periodically. But to make this rigorous, I need a more formal argument. Maybe I can analyze the recurrence relation x_n = floor(1.5 x_{n-1}) and see how the parity evolves. Let me consider the parity of x_n. Suppose x_{n-1} is even. Then 1.5 x_{n-1} is 3/2 x_{n-1}, which is an integer if x_{n-1} is even because x_{n-1} is divisible by 2. So, if x_{n-1} is even, x_n = 1.5 x_{n-1}, which is an integer. If x_{n-1} is odd, then 1.5 x_{n-1} = (3/2) x_{n-1}, which is not an integer because x_{n-1} is odd, so x_n = floor(1.5 x_{n-1}) = (3x_{n-1} - 1)/2. So, we can write:If x_{n-1} is even, x_n = (3/2) x_{n-1}.If x_{n-1} is odd, x_n = (3x_{n-1} - 1)/2.Now, let's see how the parity of x_n relates to x_{n-1}.If x_{n-1} is even, x_n = (3/2) x_{n-1}. Since x_{n-1} is even, 3/2 x_{n-1} is an integer. The parity of x_n depends on whether (3/2) x_{n-1} is even or odd. Let me denote x_{n-1} = 2k. Then x_n = 3k. So, if k is even, x_n is even; if k is odd, x_n is odd. Similarly, if x_{n-1} is odd, x_n = (3x_{n-1} - 1)/2. Let x_{n-1} = 2k + 1. Then x_n = (3(2k + 1) - 1)/2 = (6k + 3 - 1)/2 = (6k + 2)/2 = 3k + 1. So, x_n = 3k + 1. The parity of x_n depends on k. If k is even, x_n is odd; if k is odd, x_n is even. So, summarizing:- If x_{n-1} is even (x_{n-1} = 2k): - x_n = 3k - If k is even, x_n is even. - If k is odd, x_n is odd.- If x_{n-1} is odd (x_{n-1} = 2k + 1): - x_n = 3k + 1 - If k is even, x_n is odd. - If k is odd, x_n is even.This seems a bit complicated, but maybe I can model the parity transitions. Let's denote E for even and O for odd. From the above:- If x_{n-1} is E: - x_n is E if k is E (i.e., x_{n-1} = 4m) - x_n is O if k is O (i.e., x_{n-1} = 4m + 2)- If x_{n-1} is O: - x_n is O if k is E (i.e., x_{n-1} = 4m + 1) - x_n is E if k is O (i.e., x_{n-1} = 4m + 3)So, the parity of x_n depends not just on whether x_{n-1} is even or odd, but also on higher bits of x_{n-1}. This suggests that the parity sequence is influenced by more than just the immediate previous term's parity, making it potentially non-periodic.To formalize this, suppose for contradiction that y_n is periodic with period T. Then, the parities of x_n repeat every T terms. That is, for all n ≥ N, x_{n+T} ≡ x_n mod 2.But from the recurrence relation, x_{n+T} depends on x_{n+T-1}, which depends on x_{n+T-2}, and so on, back to x_n. If the parities are repeating every T terms, then the entire sequence x_n would have to repeat every T terms in terms of their parities, but the actual values of x_n are increasing, so the sequence x_n itself cannot be periodic. However, y_n only depends on the parities, so even if x_n isn't periodic, y_n could still be periodic if the parities repeat.But given the way x_n is generated, the parities are influenced by more than just the previous parity. For example, if x_{n-1} is even, whether x_n is even or odd depends on whether x_{n-1}/2 is even or odd. Similarly, if x_{n-1} is odd, whether x_n is even or odd depends on whether (x_{n-1} - 1)/2 is even or odd.This creates a kind of dependency chain where the parity at step n depends on the value of x_{n-1}, which in turn depends on x_{n-2}, and so on. This suggests that the parity sequence is not purely determined by a finite state machine with a fixed number of states, which would be required for periodicity.Alternatively, consider that if y_n were periodic with period T, then the parities of x_n would repeat every T terms. But since x_n is defined by a linear recurrence with a multiplicative factor of 1.5, the sequence x_n grows exponentially, and the parities would have to align in a way that's consistent with this growth. However, the floor function introduces non-linearity, making the parities unpredictable in a way that doesn't settle into a repeating cycle.Another angle: suppose y_n is periodic with period T. Then, for all n, x_{n+T} ≡ x_n mod 2. Let's see what this implies about x_{n+T} and x_n.From the recurrence, x_{n+T} = floor(1.5 x_{n+T-1}), and x_n = floor(1.5 x_{n-1}).If x_{n+T} ≡ x_n mod 2, then floor(1.5 x_{n+T-1}) ≡ floor(1.5 x_{n-1}) mod 2.But this would require that 1.5 x_{n+T-1} and 1.5 x_{n-1} have the same floor parity. However, since 1.5 x_{n+T-1} = x_{n+T} + f_{n+T}, where 0 ≤ f_{n+T} < 1, and similarly for 1.5 x_{n-1} = x_n + f_n, the fractional parts f_{n+T} and f_n would have to be such that their floors have the same parity. But this seems unlikely to hold consistently over all n, especially as x_n grows.Moreover, considering that x_n is increasing, the differences x_{n+T} - x_n would have to be even for all n beyond some point. But since x_n grows exponentially, the differences would also grow, and it's not clear how they could maintain evenness consistently without some underlying periodicity in the growth, which isn't present here.Perhaps a more formal approach would be to assume periodicity and derive a contradiction. Suppose y_n is periodic with period T. Then, for all n ≥ N, y_{n+T} = y_n, which implies x_{n+T} ≡ x_n mod 2. Consider the recurrence relation: x_{n+T} = floor(1.5 x_{n+T-1}), and x_n = floor(1.5 x_{n-1}).If x_{n+T} ≡ x_n mod 2, then floor(1.5 x_{n+T-1}) ≡ floor(1.5 x_{n-1}) mod 2.But 1.5 x_{n+T-1} = x_{n+T} + f_{n+T}, and 1.5 x_{n-1} = x_n + f_n, where 0 ≤ f_{n+T}, f_n < 1.So, floor(1.5 x_{n+T-1}) = x_{n+T} and floor(1.5 x_{n-1}) = x_n.Thus, x_{n+T} ≡ x_n mod 2.But x_{n+T} = floor(1.5 x_{n+T-1}) and x_n = floor(1.5 x_{n-1}).So, x_{n+T} ≡ x_n mod 2 implies that floor(1.5 x_{n+T-1}) ≡ floor(1.5 x_{n-1}) mod 2.But 1.5 x_{n+T-1} and 1.5 x_{n-1} differ by some amount, and their floors must have the same parity. This seems restrictive because the fractional parts could cause the floors to differ in parity.Alternatively, consider the difference x_{n+T} - x_n. If y_n is periodic, then x_{n+T} ≡ x_n mod 2, so x_{n+T} - x_n is even. But x_{n+T} = floor(1.5 x_{n+T-1}) and x_n = floor(1.5 x_{n-1}).So, x_{n+T} - x_n = floor(1.5 x_{n+T-1}) - floor(1.5 x_{n-1}).If x_{n+T-1} and x_{n-1} are related in some way due to periodicity, perhaps x_{n+T-1} = x_{n-1} + 2k for some k, but this would require that 1.5 x_{n+T-1} = 1.5 x_{n-1} + 3k, so floor(1.5 x_{n+T-1}) = floor(1.5 x_{n-1} + 3k). But floor(1.5 x_{n-1} + 3k) = floor(1.5 x_{n-1}) + 3k if 1.5 x_{n-1} + 3k has the same fractional part as 1.5 x_{n-1}. However, 3k is an integer, so floor(1.5 x_{n-1} + 3k) = floor(1.5 x_{n-1}) + 3k.Thus, x_{n+T} = x_n + 3k.But x_{n+T} = x_n + 3k, and since x_{n+T} - x_n is even, 3k must be even. So, k must be even, say k = 2m. Then, x_{n+T} = x_n + 6m.But then, x_{n+T+1} = floor(1.5 x_{n+T}) = floor(1.5 (x_n + 6m)) = floor(1.5 x_n + 9m).Similarly, x_{n+1} = floor(1.5 x_n).So, x_{n+T+1} = floor(1.5 x_n + 9m).But x_{n+T+1} should also be equal to x_{n+1} + 9m if the periodicity continues. However, floor(1.5 x_n + 9m) = floor(1.5 x_n) + 9m only if the fractional part of 1.5 x_n is less than 1 - (9m - floor(9m)), which is always true since 9m is integer. Wait, no, 9m is integer, so 1.5 x_n + 9m = 1.5 x_n + integer, so floor(1.5 x_n + 9m) = floor(1.5 x_n) + 9m.Thus, x_{n+T+1} = x_{n+1} + 9m.Similarly, x_{n+T+2} = floor(1.5 x_{n+T+1}) = floor(1.5 (x_{n+1} + 9m)) = floor(1.5 x_{n+1} + 13.5m).But 13.5m is not necessarily an integer unless m is even. Wait, m is an integer, so 13.5m = 27m/2, which is integer only if m is even. But m was arbitrary, so unless m is chosen such that 13.5m is integer, which would require m even, but m was defined as k = 2m, so m is integer, but 13.5m is integer only if m is even, say m = 2p. Then, m = 2p, so k = 4p.This seems to be getting into a loop where m must be chosen such that higher multiples are integers, which would require m to be a multiple of higher powers of 2, leading to m being zero, which would imply no period, a contradiction.Alternatively, perhaps this approach is too convoluted. Maybe a better way is to consider that if y_n is periodic, then the parities of x_n repeat every T terms. But since x_n is defined by a recurrence that depends on the actual value of x_{n-1}, not just its parity, the sequence x_n cannot be periodic, and thus y_n, which depends on the parities, also cannot be periodic.Wait, but y_n only depends on the parities, so even if x_n isn't periodic, y_n could still be periodic if the parities repeat. However, the way x_n is generated, the parities are influenced by more than just the previous parity, making it impossible for the parities to repeat periodically.Another approach: consider that if y_n is periodic with period T, then the sequence of parities of x_n must repeat every T terms. But since x_n is defined by x_n = floor(1.5 x_{n-1}), and 1.5 is a rational number, the sequence x_n could potentially have some periodicity in its parities. However, the floor function introduces non-linearity, disrupting any potential periodicity.Alternatively, consider that the sequence x_n grows exponentially, roughly like (1.5)^n, so the differences between consecutive terms grow as well. If y_n were periodic, the parities would have to repeat, but the increasing differences would cause the parities to change in a way that doesn't align with a fixed period.Perhaps a more formal contradiction can be reached by assuming periodicity and showing that it leads to an impossible condition. Suppose y_n is periodic with period T. Then, for all n ≥ N, x_{n+T} ≡ x_n mod 2. Consider the recurrence relation: x_{n+T} = floor(1.5 x_{n+T-1}), and x_n = floor(1.5 x_{n-1}).If x_{n+T} ≡ x_n mod 2, then floor(1.5 x_{n+T-1}) ≡ floor(1.5 x_{n-1}) mod 2.But 1.5 x_{n+T-1} = x_{n+T} + f_{n+T}, and 1.5 x_{n-1} = x_n + f_n, where 0 ≤ f_{n+T}, f_n < 1.So, floor(1.5 x_{n+T-1}) = x_{n+T} and floor(1.5 x_{n-1}) = x_n.Thus, x_{n+T} ≡ x_n mod 2.But x_{n+T} = floor(1.5 x_{n+T-1}) and x_n = floor(1.5 x_{n-1}).So, x_{n+T} ≡ x_n mod 2 implies that floor(1.5 x_{n+T-1}) ≡ floor(1.5 x_{n-1}) mod 2.But 1.5 x_{n+T-1} and 1.5 x_{n-1} differ by some amount, and their floors must have the same parity. This seems restrictive because the fractional parts could cause the floors to differ in parity.Alternatively, consider the difference x_{n+T} - x_n. If y_n is periodic, then x_{n+T} ≡ x_n mod 2, so x_{n+T} - x_n is even.But x_{n+T} = floor(1.5 x_{n+T-1}) and x_n = floor(1.5 x_{n-1}).So, x_{n+T} - x_n = floor(1.5 x_{n+T-1}) - floor(1.5 x_{n-1}).If x_{n+T-1} and x_{n-1} are related in some way due to periodicity, perhaps x_{n+T-1} = x_{n-1} + 2k for some k, but this would require that 1.5 x_{n+T-1} = 1.5 x_{n-1} + 3k, so floor(1.5 x_{n+T-1}) = floor(1.5 x_{n-1} + 3k). But floor(1.5 x_{n-1} + 3k) = floor(1.5 x_{n-1}) + 3k if 1.5 x_{n-1} + 3k has the same fractional part as 1.5 x_{n-1}. However, 3k is an integer, so floor(1.5 x_{n-1} + 3k) = floor(1.5 x_{n-1}) + 3k.Thus, x_{n+T} = x_n + 3k.But x_{n+T} = x_n + 3k, and since x_{n+T} - x_n is even, 3k must be even. So, k must be even, say k = 2m. Then, x_{n+T} = x_n + 6m.But then, x_{n+T+1} = floor(1.5 x_{n+T}) = floor(1.5 (x_n + 6m)) = floor(1.5 x_n + 9m).Similarly, x_{n+1} = floor(1.5 x_n).So, x_{n+T+1} = floor(1.5 x_n + 9m).But x_{n+T+1} should also be equal to x_{n+1} + 9m if the periodicity continues. However, floor(1.5 x_n + 9m) = floor(1.5 x_n) + 9m only if the fractional part of 1.5 x_n is less than 1 - (9m - floor(9m)), which is always true since 9m is integer. Wait, no, 9m is integer, so 1.5 x_n + 9m = 1.5 x_n + integer, so floor(1.5 x_n + 9m) = floor(1.5 x_n) + 9m.Thus, x_{n+T+1} = x_{n+1} + 9m.Similarly, x_{n+T+2} = floor(1.5 x_{n+T+1}) = floor(1.5 (x_{n+1} + 9m)) = floor(1.5 x_{n+1} + 13.5m).But 13.5m is not necessarily an integer unless m is even. Wait, m is an integer, so 13.5m = 27m/2, which is integer only if m is even. But m was arbitrary, so unless m is chosen such that 13.5m is integer, which would require m even, but m was defined as k = 2m, so m is integer, but 13.5m is integer only if m is even, say m = 2p. Then, m = 2p, so k = 4p.This seems to be getting into a loop where m must be chosen such that higher multiples are integers, which would require m to be a multiple of higher powers of 2, leading to m being zero, which would imply no period, a contradiction.Therefore, assuming that y_n is periodic leads to a contradiction, implying that y_n cannot be periodic.

💡First, I need to determine between which two consecutive integers the square root of 50 falls. To do this, I'll identify the perfect squares that are just below and above 50.I know that (7^2 = 49) and (8^2 = 64). Since 50 is between 49 and 64, it means that (sqrt{50}) is between 7 and 8.Next, I need to find the product of these two consecutive integers. Multiplying 7 and 8 together gives:[7 times 8 = 56]Therefore, the product of the two integers between which (sqrt{50}) falls is 56.

💡Okay, so I have this problem about planes and lines, and I need to figure out which of the four propositions are correct. Let me try to break it down step by step.First, let me restate the problem to make sure I understand it. We have two distinct planes, α and β, and two different lines, m and n. There are four propositions given, numbered ① to ④, and I need to determine which ones are correct.Let me go through each proposition one by one.**Proposition ①:** If m is perpendicular to n and m is perpendicular to α, then n is parallel to α.Hmm, okay. So, if line m is perpendicular to both line n and plane α, does that mean line n is parallel to plane α? I'm not entirely sure. Let me visualize this.If m is perpendicular to α, that means m is a line that forms a right angle with every line in α that it intersects. Now, if m is also perpendicular to n, what does that say about n? Well, if m is perpendicular to both α and n, then n could either be parallel to α or lie within α. Wait, if n is in α, then it's not necessarily parallel; it could intersect α. But the proposition says n is parallel to α. So, is that always true?I think not necessarily. Because n could be in α, right? So, n might not be parallel to α. It could lie within α. So, proposition ① might not always hold true. Maybe it's incorrect.**Proposition ②:** If n is a subset of α, m is a subset of β, and α intersects β but is not perpendicular, then n and m are not perpendicular.Alright, so n is in plane α, m is in plane β, and α and β intersect but aren't perpendicular. Does that mean n and m can't be perpendicular?Hmm, let's think. If α and β intersect but aren't perpendicular, their dihedral angle isn't 90 degrees. But does that prevent any lines within them from being perpendicular?I think it's possible for lines in two intersecting but non-perpendicular planes to still be perpendicular. For example, imagine two planes intersecting at some angle θ, not 90 degrees. If you have a line in each plane that's at a certain angle relative to the line of intersection, they could still be perpendicular. So, proposition ② might not be correct either.**Proposition ③:** If α is perpendicular to β, their intersection is m, n is a subset of α, and n is perpendicular to m, then n is perpendicular to β.Okay, so α and β are perpendicular, intersecting along line m. Line n is in α and is perpendicular to m. Does that mean n is perpendicular to β?Let me recall some theorems. If two planes are perpendicular, then any line in one plane that is perpendicular to their line of intersection is also perpendicular to the other plane. Yes, that sounds familiar. So, if n is in α, perpendicular to m, and α is perpendicular to β, then n must be perpendicular to β. So, proposition ③ seems correct.**Proposition ④:** If m is parallel to n, n is perpendicular to α, and α is parallel to β, then m is perpendicular to β.Alright, so m is parallel to n, n is perpendicular to α, and α is parallel to β. Does that mean m is perpendicular to β?Let me think. If n is perpendicular to α, and α is parallel to β, then n must also be perpendicular to β. Since m is parallel to n, and n is perpendicular to β, then m should also be perpendicular to β. Because if two lines are parallel, and one is perpendicular to a plane, the other must be too. So, proposition ④ seems correct.So, summarizing my thoughts:- Proposition ①: Not necessarily correct because n could lie within α.- Proposition ②: Not necessarily correct because lines in intersecting planes can still be perpendicular.- Proposition ③: Correct, based on the theorem about perpendicular planes.- Proposition ④: Correct, because parallel lines and parallel planes preserve perpendicularity.Therefore, the correct propositions are ③ and ④.**Final Answer**The correct propositions are boxed{③④}.

💡Okay, so I have this geometry problem here with four statements about lines and planes. I need to figure out how many of these statements are correct. Let me go through each one step by step.First, let's restate the problem to make sure I understand it correctly. We have two distinct lines, a and b, and two distinct planes, α and β. There are four statements given, and I need to determine how many of them are true.Let me list out the statements again:① If a is perpendicular to b, and a is parallel to α, then b is parallel to α.② If a is parallel to α, and α is perpendicular to β, then a is perpendicular to β.③ If a is perpendicular to β, and α is perpendicular to β, then a is parallel to α.④ If a is perpendicular to b, a is perpendicular to α, and b is perpendicular to β, then α is perpendicular to β.I need to analyze each of these statements one by one.Starting with statement ①: If a is perpendicular to b, and a is parallel to α, then b is parallel to α.Hmm, okay. So, a and b are two lines that are perpendicular to each other. Also, line a is parallel to plane α. Does this necessarily mean that line b is parallel to plane α?Let me visualize this. If line a is parallel to plane α, that means line a doesn't intersect plane α and lies in a direction that's parallel to the plane. Now, line b is perpendicular to line a. But does this imply that line b is parallel to plane α?Wait, not necessarily. Because if line a is parallel to plane α, line b, being perpendicular to a, could lie in a direction that's either parallel to the plane or it could intersect the plane. It depends on the orientation of b relative to α.For example, imagine plane α is the floor, and line a is a horizontal line on the wall, which is parallel to the floor. Then line b, which is perpendicular to a, could be a vertical line on the wall. But a vertical line on the wall is not parallel to the floor; it intersects the floor. So in this case, b is not parallel to α.Therefore, statement ① is not necessarily true. So I think statement ① is incorrect.Moving on to statement ②: If a is parallel to α, and α is perpendicular to β, then a is perpendicular to β.Alright, so line a is parallel to plane α, and plane α is perpendicular to plane β. Does this mean that line a is perpendicular to plane β?Let me think. If two planes are perpendicular, their dihedral angle is 90 degrees. So plane α is perpendicular to plane β. Now, line a is parallel to plane α. Does this make line a perpendicular to plane β?Hmm, not necessarily. Because line a could be lying in a direction that's parallel to the line of intersection of α and β. In that case, line a wouldn't be perpendicular to β.For example, imagine plane α is the floor and plane β is a wall. The line of intersection is the baseboard. If line a is parallel to the floor, it could be running along the baseboard, which is in both planes. But in this case, line a is not perpendicular to the wall (β); it's actually lying on the wall.Therefore, statement ② is also incorrect.Now, statement ③: If a is perpendicular to β, and α is perpendicular to β, then a is parallel to α.So, line a is perpendicular to plane β, and plane α is also perpendicular to plane β. Does this mean that line a is parallel to plane α?Let me visualize. If plane α and plane β are both perpendicular to each other, their intersection is a line. Now, line a is perpendicular to β. So line a is perpendicular to every line in β, which includes the line of intersection of α and β.But does that make line a parallel to plane α? Not necessarily. Because line a could be intersecting plane α at some point.Wait, actually, if line a is perpendicular to plane β, and plane α is also perpendicular to β, then line a could either be parallel to α or lie within α. But since a is a line and α is a plane, if a is not lying within α, it could be parallel or intersecting.But wait, if a is perpendicular to β, and α is perpendicular to β, then the direction of a is along the normal vector of β. Since α is also perpendicular to β, the normal vector of α is parallel to the normal vector of β. Therefore, line a, which is along the normal vector of β, would be parallel to plane α.Wait, no. If line a is along the normal vector of β, and plane α is perpendicular to β, then the normal vector of α is parallel to the normal vector of β. Therefore, line a, being along the normal vector of β, is actually perpendicular to plane α.Wait, that contradicts my earlier thought. Let me clarify.If plane α is perpendicular to plane β, their normal vectors are parallel. So, if line a is perpendicular to β, it's along the normal vector of β, which is parallel to the normal vector of α. Therefore, line a is parallel to the normal vector of α, which means line a is perpendicular to plane α.But statement ③ says that a is parallel to α. So if line a is actually perpendicular to α, then statement ③ is incorrect.Wait, so my initial thought was wrong. Let me think again.If line a is perpendicular to β, then it's parallel to the normal vector of β. Since α is also perpendicular to β, the normal vector of α is parallel to the normal vector of β. Therefore, line a is parallel to the normal vector of α, which means line a is perpendicular to plane α.So, statement ③ claims that a is parallel to α, but actually, a is perpendicular to α. Therefore, statement ③ is incorrect.Wait, but hold on. Maybe I made a mistake here. Let me think again.If line a is perpendicular to β, then it's perpendicular to every line in β. Plane α is perpendicular to β, so their intersection is a line, say l. Now, line a is perpendicular to β, so it's perpendicular to l as well. But does that make line a parallel to α?No, because line a could be intersecting α at some point not on l. So, line a is not necessarily parallel to α. It could be intersecting α.Wait, but if line a is perpendicular to β, and α is perpendicular to β, then line a could be either parallel to α or lie within α. But since a is a line and α is a plane, if a is not lying within α, it could be parallel or intersecting.But in 3D space, if a line is perpendicular to a plane, and another plane is perpendicular to the same plane, the line can be either parallel to the second plane or lie within it.But since a is a line and α is a plane, and a is not necessarily lying within α, it could be parallel or intersecting. So, it's not necessarily parallel.Therefore, statement ③ is incorrect.Wait, but earlier I thought that line a is perpendicular to α, but that's not necessarily the case. If line a is perpendicular to β, and α is perpendicular to β, then line a could be parallel to α or lie within α, but it's not necessarily parallel.Therefore, statement ③ is incorrect.Okay, moving on to statement ④: If a is perpendicular to b, a is perpendicular to α, and b is perpendicular to β, then α is perpendicular to β.Hmm, this seems a bit more complex. Let me break it down.So, line a is perpendicular to line b. Also, line a is perpendicular to plane α, and line b is perpendicular to plane β. Does this imply that plane α is perpendicular to plane β?Let me visualize. If line a is perpendicular to plane α, then a is the normal vector of α. Similarly, line b is perpendicular to plane β, so b is the normal vector of β.Given that a is perpendicular to b, which are the normal vectors of α and β respectively, does that mean that the planes α and β are perpendicular?Yes, because if the normal vectors of two planes are perpendicular, then the planes themselves are perpendicular.So, if a is perpendicular to b, and a is the normal of α, b is the normal of β, then α is perpendicular to β.Therefore, statement ④ is correct.So, summarizing:①: Incorrect②: Incorrect③: Incorrect④: CorrectTherefore, only statement ④ is correct. So the number of correct statements is 1.Wait, but let me double-check statement ③ again because I might have made a mistake earlier.Statement ③: If a is perpendicular to β, and α is perpendicular to β, then a is parallel to α.Earlier, I thought that a could be either parallel to α or lie within α, but since a is a line and α is a plane, if a is not lying within α, it could be parallel or intersecting. But if a is perpendicular to β, and α is perpendicular to β, then a is parallel to α.Wait, no. If a is perpendicular to β, it's along the normal vector of β. Since α is perpendicular to β, the normal vector of α is parallel to the normal vector of β. Therefore, line a, being along the normal vector of β, is parallel to the normal vector of α, which means line a is perpendicular to plane α.Wait, so if line a is perpendicular to plane α, then it's not parallel to α. So statement ③ is incorrect.Yes, that's correct. So statement ③ is indeed incorrect.Therefore, only statement ④ is correct. So the answer is 1, which corresponds to option B.

💡Okay, so I'm trying to figure out this problem about two concentric circles forming a circular track. The width of the track is 12 feet, and the diameter of the inner circle is given as d feet. I need to find the difference in the circumferences of the two circles. The options are A) 12π feet, B) 24π feet, C) 36π feet, and D) 48π feet.Alright, let's start by understanding what concentric circles are. They are circles that share the same center, right? So, one circle is inside the other, and the track is the area between them. The width of the track is 12 feet, which means the distance between the inner circle and the outer circle is 12 feet.Now, the diameter of the inner circle is d feet. Since the track is 12 feet wide, I think that means the radius of the outer circle is the radius of the inner circle plus 12 feet. Wait, is that right? Let me think. If the width of the track is 12 feet, that would be the difference between the radii of the outer and inner circles. So, if the inner circle has a diameter of d, its radius would be d/2. Then, the radius of the outer circle would be d/2 + 12.But wait, the problem says the width of the track is 12 feet. Is that the total width around the circle, or is it the width from the inner edge to the outer edge? I think it's the latter, meaning the distance from the inner circle to the outer circle is 12 feet. So, the radius of the outer circle is the radius of the inner circle plus 12 feet.Let me write that down:- Diameter of inner circle = d feet- Radius of inner circle = d/2 feet- Radius of outer circle = (d/2) + 12 feetNow, I need to find the difference in the circumferences of the two circles. The circumference of a circle is given by the formula C = 2πr, where r is the radius. Alternatively, since the diameter is given, I can also use C = πd.Let me calculate the circumference of the inner circle first:C_inner = π * dNow, for the outer circle, the diameter would be twice the radius, so:Diameter of outer circle = 2 * [(d/2) + 12] = d + 24 feetSo, the circumference of the outer circle is:C_outer = π * (d + 24)Now, to find the difference in circumferences:Difference = C_outer - C_inner = π*(d + 24) - π*d = π*d + 24π - π*d = 24πSo, the difference in the circumferences is 24π feet.Wait, let me double-check that. The key here is that the difference in diameters is 24 feet because the radius increases by 12 feet, so the diameter increases by 24 feet. Since circumference is proportional to diameter, the difference in circumference is π times the difference in diameters, which is 24π.Yes, that makes sense. So, the answer should be B) 24π feet.

💡First, I need to understand the problem by visualizing the two configurations of the books on the podium. In the first configuration, the combined visible length is 40 inches, and in the second configuration, it is 34 inches. The goal is to find the height of the podium.I'll start by defining the variables:- ( h ) represents the height of the podium.- ( l ) represents the thickness of each book.- ( w ) represents the width of each book.For the first configuration, the total visible length can be expressed as:[l + h - w = 40]This equation accounts for the thickness of one book, the height of the podium, and subtracts the overlapping width of the books.For the second configuration, the total visible length is:[w + h - l = 34]Here, the width of one book, the height of the podium, and the thickness of the other book are considered, with the overlapping thickness subtracted.To find the height of the podium, I'll add the two equations together:[(l + h - w) + (w + h - l) = 40 + 34]Simplifying this, the ( l ) and ( w ) terms cancel out:[2h = 74]Solving for ( h ):[h = 37]Therefore, the height of the podium is 37 inches.