💡First, I recognize that the function ( y = -x + 3 ) is a linear function with a slope of (-1). Since the slope is negative, the function is decreasing, meaning that as ( x ) increases, ( y ) decreases.Given the domain ( 0 leq x leq 3 ), I need to determine the minimum value of ( y ) within this interval. Because the function is decreasing, the minimum value of ( y ) will occur at the maximum value of ( x ) within the domain.Substituting ( x = 3 ) into the function:[y = -3 + 3 = 0]Therefore, the minimum value of the function ( y = -x + 3 ) within the interval ( 0 leq x leq 3 ) is ( 0 ).

💡Alright, so I've got this problem here about complex numbers. Let me try to understand it step by step. The problem says:"Given that (i) is the imaginary unit, and the complex number (z = (a - i)^2), where (a in mathbb{R}), if the complex number (z) is a pure imaginary number, then (|z| = ) ? The options are A: 1, B: (sqrt{2}), C: 2, D: 4."Okay, so I need to find the magnitude of (z) given that (z) is purely imaginary. Let me recall what a pure imaginary number is. A pure imaginary number is a complex number that has no real part; it's only in the form (bi) where (b) is a real number and (i) is the imaginary unit.So, (z = (a - i)^2). I need to expand this expression and then set the real part to zero because (z) is purely imaginary. Once I find the value of (a), I can compute (|z|).Let me start by expanding (z = (a - i)^2). Using the formula ((x - y)^2 = x^2 - 2xy + y^2), I can expand this:(z = a^2 - 2a(i) + (i)^2).Wait, (i^2) is (-1), right? So, substituting that in:(z = a^2 - 2ai - 1).So, (z = (a^2 - 1) - 2ai).Now, since (z) is a pure imaginary number, the real part must be zero. The real part here is (a^2 - 1), and the imaginary part is (-2a). Therefore, for (z) to be purely imaginary, the real part must equal zero:(a^2 - 1 = 0).Let me solve this equation for (a):(a^2 = 1).Taking the square root of both sides:(a = pm 1).So, (a) can be either 1 or -1. Now, I need to find (|z|), the magnitude of (z). The magnitude of a complex number (z = x + yi) is given by (|z| = sqrt{x^2 + y^2}).But in this case, since (z) is purely imaginary, the real part (x) is zero, so (|z| = |y|), where (y) is the coefficient of (i).Wait, let me make sure. If (z = 0 + yi), then (|z| = |y|). So, in our case, (z = -2ai), so (y = -2a). Therefore, (|z| = | -2a | = 2|a|).But we found that (a = pm 1), so (|a| = 1). Therefore, (|z| = 2 * 1 = 2).So, the magnitude of (z) is 2, which corresponds to option C.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expanded (z = (a - i)^2) correctly to (a^2 - 2ai - 1).2. Identified the real part as (a^2 - 1) and the imaginary part as (-2a).3. Set the real part equal to zero: (a^2 - 1 = 0), leading to (a = pm 1).4. Calculated the magnitude as (|z| = | -2a | = 2|a| = 2*1 = 2).Everything seems to check out. So, the answer should be 2, which is option C.I think that's solid. I don't see any errors in my reasoning. The key was recognizing that for (z) to be purely imaginary, the real part must be zero, which allowed me to solve for (a). Then, using the definition of the magnitude of a complex number, I found (|z|).Just to be thorough, let me plug (a = 1) and (a = -1) back into the original expression for (z) and compute (|z|).First, for (a = 1):(z = (1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i).So, (|z| = | -2i | = 2).For (a = -1):(z = (-1 - i)^2 = (-1)^2 - 2*(-1)*i + i^2 = 1 + 2i - 1 = 2i).So, (|z| = |2i| = 2).In both cases, the magnitude is 2. That confirms my earlier result.I think I'm confident now that the correct answer is C: 2.**Final Answer**boxed{C}

💡Okay, so I have this problem here: I need to find how many prime numbers ( p ) satisfy the condition that ( 2p^4 - 7p^2 + 1 ) is a perfect square. The options are from A to E, with E being "None of the preceding." Hmm, let me try to figure this out step by step.First, I know that primes are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, the primes start from 2, 3, 5, 7, 11, and so on. The problem is asking about primes ( p ) such that when I plug ( p ) into the expression ( 2p^4 - 7p^2 + 1 ), the result is a perfect square. That means there exists some integer ( k ) where ( 2p^4 - 7p^2 + 1 = k^2 ).Alright, so maybe I should start by testing small prime numbers to see if they satisfy this condition. Let's try ( p = 2 ) first.For ( p = 2 ):[2(2)^4 - 7(2)^2 + 1 = 2(16) - 7(4) + 1 = 32 - 28 + 1 = 5]5 is not a perfect square because the squares around 5 are 4 ((2^2)) and 9 ((3^2)). So, ( p = 2 ) doesn't work.Next, let's try ( p = 3 ):[2(3)^4 - 7(3)^2 + 1 = 2(81) - 7(9) + 1 = 162 - 63 + 1 = 100]100 is a perfect square since (10^2 = 100). So, ( p = 3 ) is a solution.Now, let's check ( p = 5 ):[2(5)^4 - 7(5)^2 + 1 = 2(625) - 7(25) + 1 = 1250 - 175 + 1 = 1076]Is 1076 a perfect square? Let me see. The square of 32 is 1024, and 33 squared is 1089. 1076 is between these two, so it's not a perfect square. So, ( p = 5 ) doesn't work.How about ( p = 7 ):[2(7)^4 - 7(7)^2 + 1 = 2(2401) - 7(49) + 1 = 4802 - 343 + 1 = 4460]Is 4460 a perfect square? Let's see. 66 squared is 4356, and 67 squared is 4489. So, 4460 is between these two, not a perfect square. So, ( p = 7 ) doesn't work either.Hmm, maybe I should try ( p = 11 ):[2(11)^4 - 7(11)^2 + 1 = 2(14641) - 7(121) + 1 = 29282 - 847 + 1 = 28436]Is 28436 a perfect square? Let's see. 168 squared is 28224, and 169 squared is 28561. So, 28436 is in between, not a perfect square. So, ( p = 11 ) doesn't work.At this point, I notice that as ( p ) increases, the value of ( 2p^4 - 7p^2 + 1 ) grows rapidly. Maybe it's unlikely that larger primes will satisfy the condition. But just to be thorough, let me check ( p = 13 ):[2(13)^4 - 7(13)^2 + 1 = 2(28561) - 7(169) + 1 = 57122 - 1183 + 1 = 55940]Is 55940 a perfect square? 236 squared is 55696, and 237 squared is 56169. So, 55940 is in between, not a perfect square. So, ( p = 13 ) doesn't work.I'm starting to think that maybe ( p = 3 ) is the only prime that works. But just to be sure, maybe I should analyze the equation more generally instead of plugging in primes one by one.Let me denote ( k^2 = 2p^4 - 7p^2 + 1 ). So, we have:[k^2 = 2p^4 - 7p^2 + 1]This looks like a quartic equation in terms of ( p ). Quartic equations can be tricky, but maybe I can factor it or find some substitution to make it simpler.Let me try to rearrange the equation:[2p^4 - 7p^2 + 1 - k^2 = 0]Hmm, not sure if that helps. Maybe I can consider this as a quadratic in terms of ( p^2 ). Let me set ( x = p^2 ), so the equation becomes:[2x^2 - 7x + 1 = k^2]So, now we have:[2x^2 - 7x + 1 = k^2]This is a quadratic in ( x ), but ( x ) itself is ( p^2 ), which is a square of a prime. Maybe I can analyze this equation modulo some number to find constraints on ( p ).Let me consider the equation modulo 3. Primes greater than 3 are either 1 or 2 modulo 3. Let's see what happens.If ( p equiv 0 mod 3 ), then ( p = 3 ), which we've already checked works.If ( p equiv 1 mod 3 ), then ( p^2 equiv 1 mod 3 ), and ( p^4 equiv 1 mod 3 ). Plugging into the equation:[2(1) - 7(1) + 1 = 2 - 7 + 1 = -4 equiv 2 mod 3]So, ( k^2 equiv 2 mod 3 ). But squares modulo 3 are 0 and 1. So, 2 is not a quadratic residue modulo 3. Therefore, ( p equiv 1 mod 3 ) cannot be a solution.Similarly, if ( p equiv 2 mod 3 ), then ( p^2 equiv 4 equiv 1 mod 3 ), and ( p^4 equiv 1 mod 3 ). So, same as above:[2(1) - 7(1) + 1 = -4 equiv 2 mod 3]Again, ( k^2 equiv 2 mod 3 ), which is impossible. Therefore, primes ( p ) greater than 3 cannot satisfy the equation because they would lead to ( k^2 equiv 2 mod 3 ), which isn't possible.Wait, but ( p = 3 ) is a prime, and we saw that it works. So, maybe ( p = 3 ) is the only solution.Just to double-check, let me consider another modulus, say modulo 4. Squares modulo 4 are 0 and 1.For ( p ) odd prime, ( p ) is either 1 or 3 modulo 4. Let's see.If ( p equiv 1 mod 4 ), then ( p^2 equiv 1 mod 4 ), and ( p^4 equiv 1 mod 4 ). Plugging into the equation:[2(1) - 7(1) + 1 = 2 - 7 + 1 = -4 equiv 0 mod 4]So, ( k^2 equiv 0 mod 4 ). That's possible because squares can be 0 modulo 4.If ( p equiv 3 mod 4 ), then ( p^2 equiv 9 equiv 1 mod 4 ), and ( p^4 equiv 1 mod 4 ). So, same as above:[2(1) - 7(1) + 1 = -4 equiv 0 mod 4]Again, ( k^2 equiv 0 mod 4 ), which is possible.So, modulo 4 doesn't eliminate any possibilities beyond what we saw with modulo 3. So, the key constraint is from modulo 3, which tells us that only ( p = 3 ) can satisfy the equation.Therefore, after checking small primes and analyzing the equation modulo 3, it seems that ( p = 3 ) is the only prime that satisfies the condition ( 2p^4 - 7p^2 + 1 ) being a perfect square.I think that's thorough enough. I don't see any other primes that would work, and the modular arithmetic backs up the conclusion that ( p = 3 ) is the only solution.**Final Answer**boxed{B}

💡Okay, so I have this problem where there are seven points inside or on a circle with a diameter of 1 unit. I need to find the smallest number b such that no matter how these seven points are placed, there will always be at least one pair of points with a distance less than or equal to b. Hmm, interesting.First, I remember something about the pigeonhole principle. It says that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. Maybe I can apply that here. If I divide the circle into regions (pigeonholes) and have the points as pigeons, then at least one region will contain at least two points. That makes sense.But how should I divide the circle? If I divide it into six equal sectors, each like a slice of a pie, each sector would have a central angle of 60 degrees because 360 divided by 6 is 60. So, each sector is a 60-degree slice. Now, if I have seven points and six sectors, by the pigeonhole principle, at least one sector must contain at least two points. Okay, that seems right.Now, the next part is figuring out the maximum distance between two points within one of these sectors. If two points are in the same sector, what's the farthest they can be apart? I think it's the length of the chord that subtends the 60-degree angle at the center of the circle. Let me recall the formula for the chord length. It's 2r sin(theta/2), where r is the radius and theta is the central angle in radians.The diameter is 1, so the radius r is 0.5. The central angle theta is 60 degrees, which is pi/3 radians. Plugging into the formula: chord length c = 2 * 0.5 * sin(pi/6). Sin(pi/6) is 0.5, so c = 1 * 0.5 = 0.5. So the chord length is 0.5 units.Wait, does that mean the maximum distance between two points in the same sector is 0.5? So, if two points are in the same sector, they can't be more than 0.5 units apart. Therefore, no matter how I place seven points in the circle, there will always be at least two points within 0.5 units of each other. So, b must be 0.5.But let me double-check. If I divide the circle into six sectors, each 60 degrees, and place two points in one sector, their maximum distance is 0.5. If I tried to place points in such a way that all are more than 0.5 apart, would that be possible? If I put one point in each sector, that's six points, each at least 0.5 apart from each other. But the seventh point would have to go into one of the sectors, making it within 0.5 of another point. So, yes, it seems like 0.5 is the minimal such b.I wonder if there's a way to make b smaller. Suppose I tried dividing the circle into more sectors, like seven. Then each sector would have a smaller angle, but the chord length would also be smaller. However, since we have seven points, each sector could potentially have one point, and the maximum distance would be even smaller. But wait, the problem is asking for the smallest b such that no matter how the seven points are placed, there's always at least one pair within b. So, if I make b smaller than 0.5, say 0.4, can I still guarantee that two points are within 0.4?If I divide the circle into six sectors, each with a chord length of 0.5, but if I want a smaller b, I might need more sectors. Let me think. If I divide the circle into seven sectors, each with a central angle of about 51.43 degrees. Then, the chord length for each sector would be 2 * 0.5 * sin(51.43/2). Let me calculate that. 51.43 degrees is approximately 0.8976 radians. Half of that is about 0.4488 radians. Sin(0.4488) is roughly 0.4339. So, chord length is 1 * 0.4339 ≈ 0.4339. So, if I divide into seven sectors, each chord is about 0.434.But wait, if I have seven points and seven sectors, each point could be in its own sector, and the maximum distance between any two points would be the chord length of the sector, which is about 0.434. But the problem is asking for the minimal b such that in any configuration, there's at least one pair within b. So, if I divide into seven sectors, the maximum distance is 0.434, but if I have seven points, each in their own sector, then the minimal distance between any two points is at least 0.434. But that contradicts the earlier thought.Wait, no. If I have seven points and seven sectors, each point in its own sector, then the minimal distance between any two points is actually the minimal distance between any two points in adjacent sectors. But the chord length between two points in adjacent sectors would be larger than the chord length within a single sector. Hmm, this is getting confusing.Maybe I should stick with the six sectors approach. Since with six sectors, each with a chord length of 0.5, and seven points, at least two points must be in the same sector, so their distance is at most 0.5. Therefore, b is 0.5. That seems solid.I think I'm overcomplicating it by considering seven sectors. The key is that with six sectors, regardless of how you place seven points, two must be in the same sector, and their distance is bounded by the chord length of that sector, which is 0.5. So, b is 0.5.Yeah, I think that's the right approach. So, the answer should be 0.5, or 1/2.

💡To determine the volume of the cylinder not occupied by the two cones, I will first calculate the volume of the cylinder and then subtract the combined volume of the two cones.**Step 1: Calculate the volume of the cylinder.**The formula for the volume of a cylinder is:[V_{text{cylinder}} = pi r^2 h]Given that the radius ( r ) is 10 cm and the height ( h ) is 30 cm:[V_{text{cylinder}} = pi (10)^2 (30) = 3000pi text{ cm}^3]**Step 2: Calculate the volume of one cone.**The formula for the volume of a cone is:[V_{text{cone}} = frac{1}{3}pi r^2 h]Given that the radius ( r ) is 10 cm and the height ( h ) is 15 cm:[V_{text{cone}} = frac{1}{3}pi (10)^2 (15) = 500pi text{ cm}^3]**Step 3: Calculate the combined volume of both cones.**Since there are two identical cones:[V_{text{both cones}} = 2 times 500pi = 1000pi text{ cm}^3]**Step 4: Calculate the volume not occupied by the cones.**Subtract the combined volume of the cones from the volume of the cylinder:[V_{text{not occupied}} = V_{text{cylinder}} - V_{text{both cones}} = 3000pi - 1000pi = 2000pi text{ cm}^3]**Final Answer:**[boxed{2000pi text{ cubic cm}}]

💡Okay, so I have this problem here about a function f(x) that's defined on all real numbers. It's an odd function, which I remember means that f(-x) = -f(x) for all x. That's useful because it relates the values of the function at positive and negative points.The problem also mentions that the period of the function f(3x + 1) is 2. Hmm, period means that the function repeats its values every certain interval. So, if the period is 2, then f(3(x + 2) + 1) should equal f(3x + 1). Let me write that down:f(3(x + 2) + 1) = f(3x + 1)Simplifying the left side, that's f(3x + 6 + 1) = f(3x + 7). So, f(3x + 7) = f(3x + 1). That suggests that the function f has a period related to 6, because 7 - 1 = 6. So, does that mean the period of f(x) is 6? Let me check.If f(3x + 7) = f(3x + 1), then replacing 3x with y, we get f(y + 6) = f(y). So yes, f(y + 6) = f(y), which means f is periodic with period 6. So, f(x + 6) = f(x) for all x. Got it.Now, we're given that f(1) = 2010. Since f is an odd function, f(-1) should be -2010. That's straightforward.The question asks for f(2009) + f(2010). I need to compute these two values and add them together.Let me start with f(2009). Since the function has a period of 6, I can subtract multiples of 6 from 2009 until I get a number between 0 and 5. Let me divide 2009 by 6 to see how many times 6 goes into 2009.2009 divided by 6 is 334 with a remainder. Let me calculate 334 * 6 = 2004. So, 2009 - 2004 = 5. Therefore, f(2009) = f(5).Now, f(5) is the same as f(5 - 6) because of the periodicity, which is f(-1). Since f is odd, f(-1) = -f(1) = -2010. So, f(2009) = -2010.Next, f(2010). Again, using the period of 6, let's divide 2010 by 6. 6 * 335 = 2010, so the remainder is 0. Therefore, f(2010) = f(0).But wait, f is an odd function, and for odd functions, f(-0) = -f(0). But since 0 is its own negative, that implies f(0) = -f(0). The only number that is equal to its own negative is 0. So, f(0) = 0. Therefore, f(2010) = 0.Adding them together: f(2009) + f(2010) = (-2010) + 0 = -2010.Let me double-check my steps to make sure I didn't make a mistake.1. f is odd: f(-x) = -f(x). Correct.2. Period of f(3x + 1) is 2: So f(3(x + 2) + 1) = f(3x + 1). Simplifies to f(3x + 7) = f(3x + 1). So, f(y + 6) = f(y). Hence, period 6. Correct.3. f(1) = 2010, so f(-1) = -2010. Correct.4. f(2009): 2009 divided by 6 is 334 with remainder 5, so f(2009) = f(5). f(5) = f(5 - 6) = f(-1) = -2010. Correct.5. f(2010): 2010 is a multiple of 6, so f(2010) = f(0) = 0. Correct.6. Sum: -2010 + 0 = -2010. Correct.I think that's solid. So, the answer should be -2010, which is option B.**Final Answer**boxed{B}