💡Alright, so I have this problem here about power functions, specifically y equals x to the power of a, where a is in the set {-2, -1, -1/2, 1/3, 1/2, 1, 2, 3}. The question is asking how many of these functions are odd functions, and the options are A:2, B:3, C:4, D:5.Hmm, okay. First, I need to remember what defines an odd function. From what I recall, an odd function satisfies the condition f(-x) = -f(x) for all x in its domain. So, if I plug in -x into the function, it should be equal to the negative of the original function.Given that, let's consider the function y = x^a. For this to be an odd function, when I replace x with -x, I should get (-x)^a, and this should be equal to -x^a. So, (-x)^a = -x^a.Now, let's analyze when this equality holds. I know that (-x)^a can be written as (-1)^a * x^a. So, the equation becomes (-1)^a * x^a = -x^a. Dividing both sides by x^a (assuming x is not zero), we get (-1)^a = -1.So, the key equation here is (-1)^a = -1. Now, I need to find for which values of a in the given set this equation holds true.Let's consider each value of a one by one.1. a = -2: (-1)^(-2) = 1/(-1)^2 = 1/1 = 1. But we need it to be -1. So, this doesn't satisfy the condition. Therefore, a = -2 is not an odd function.2. a = -1: (-1)^(-1) = 1/(-1)^1 = 1/(-1) = -1. This satisfies the condition. So, a = -1 is an odd function.3. a = -1/2: (-1)^(-1/2) = 1/(-1)^(1/2). Now, (-1)^(1/2) is the square root of -1, which is an imaginary number, i. So, this would be 1/i, which is -i. But since we're dealing with real functions, this isn't applicable. So, a = -1/2 doesn't result in a real function, hence it's not an odd function in the real number context.4. a = 1/3: (-1)^(1/3) is the cube root of -1, which is -1. So, (-1)^(1/3) = -1. This satisfies the condition. Therefore, a = 1/3 is an odd function.5. a = 1/2: (-1)^(1/2) is the square root of -1, which is imaginary. Again, in the real number system, this isn't defined, so a = 1/2 doesn't result in a real function, hence not an odd function.6. a = 1: (-1)^1 = -1. This satisfies the condition. So, a = 1 is an odd function.7. a = 2: (-1)^2 = 1. This doesn't equal -1, so a = 2 is not an odd function.8. a = 3: (-1)^3 = -1. This satisfies the condition. Therefore, a = 3 is an odd function.So, summarizing the results, the values of a that make y = x^a an odd function are a = -1, 1/3, 1, and 3. That's four values in total.Wait, but hold on. I remember that for a function to be odd, it's not just about the exponent; the function must also be defined for all x in its domain. For example, when a is a fractional exponent like 1/2, the function y = x^(1/2) is only defined for x ≥ 0, which means it's not symmetric about the origin, hence not odd. Similarly, for negative exponents, like a = -2, the function y = x^(-2) is 1/x², which is even because f(-x) = 1/(-x)² = 1/x² = f(x), not -f(x). So, my initial analysis seems correct.Therefore, the number of odd functions among the given set is 4, which corresponds to option C.**Final Answer**boxed{C}

💡Okay, so I have this problem about a 5x5 square grid, and there's a triangle ABC with lattice points. I need to find another triangle DEF that's similar to ABC and has the maximum area possible within the grid. The options are 2, 2.5, 3, or 3.5. Hmm, let me think about how to approach this.First, I should recall what similar triangles are. Similar triangles have the same shape but different sizes, meaning their corresponding angles are equal, and their sides are proportional. So, if I can figure out the scaling factor between triangle ABC and triangle DEF, I can determine the area of DEF.But wait, the problem doesn't specify the coordinates of triangle ABC. It just says it's a lattice point triangle in the grid. Maybe I need to consider the maximum possible area of a similar triangle within the grid. Since the grid is 5x5, the maximum distance between any two points is the diagonal, which is 5√2. But I need to think about how to scale triangle ABC to fit within this grid.Let me think about the possible sizes of triangle ABC. If ABC is a right-angled triangle with legs of length 1 and 2, its area would be 1. But if it's a larger triangle, maybe with legs of length 2 and 3, the area would be 3. But since we're talking about similarity, the area scales with the square of the scaling factor.Wait, maybe I should consider specific triangles. For example, if ABC is a right-angled triangle with legs of length 1 and 1, its area is 0.5. If I scale it up by a factor of 2, the area becomes 2. That's one of the options. But can I scale it up more?If I scale it by √2, the area would be 0.5*(√2)^2 = 1, which is not as big. Hmm, maybe I need a different approach.Alternatively, maybe ABC is a different triangle. Suppose ABC is a triangle with sides of length 1, √2, and √5. That's a common triangle in grid problems. If I can find a similar triangle DEF that's larger, what's the maximum scaling factor?The grid is 5x5, so the maximum distance between two points is 5√2. If ABC has a side of length 1, then the scaling factor could be up to 5. But wait, scaling by 5 would make the sides 5, 5√2, and 5√5, but 5√5 is about 11.18, which is way larger than the grid. So, that's not possible.Wait, maybe the sides of ABC are 1, 2, and √5. Then, scaling by 2 would make them 2, 4, and 2√5. Is that possible within a 5x5 grid? Let's see, 2√5 is about 4.47, which is less than 5, so yes. The area would then be scaled by 2^2 = 4. If the original area was 1, the new area would be 4. But 4 isn't one of the options. Hmm.Wait, maybe the original triangle ABC has a larger area. If ABC is a triangle with base 2 and height 2, its area is 2. Then scaling it by √2 would make the area 4, but again, that's too big. Maybe scaling by 1.5? Then the area would be 4.5, which is still too big.Wait, perhaps I'm overcomplicating this. Maybe the maximum area is achieved by a specific triangle. Let me think about the grid. The largest possible triangle in a 5x5 grid would have a base of 5 and a height of 5, giving an area of 12.5, but that's not similar to ABC unless ABC is a right-angled triangle with legs 1 and 1, which would scale up by 5, giving an area of 12.5, but that's not an option.Wait, the options are 2, 2.5, 3, 3.5. So, maybe the maximum area is 2.5. How?If ABC is a triangle with area 0.5, scaling it by √5 would give an area of 0.5*(√5)^2 = 2.5. That's one of the options. Is that possible?Yes, if ABC is a right-angled triangle with legs 1 and 1, area 0.5, then scaling it by √5 would give legs √5 and √5, which is possible within the grid since √5 is about 2.236, and the grid is 5x5. So, the area would be 2.5.Alternatively, if ABC is a different triangle, maybe with area 1, scaling it by √2.5 would give area 2.5. But I think the first approach makes more sense.So, putting it all together, the maximum area of triangle DEF similar to ABC is 2.5.

💡First, I need to determine the original price per packet of coffee. Last month, 3 packets were sold for 12. To find the price per packet, I divide the total price by the number of packets:[text{Original Price per Packet} = frac{12}{3} = 4 text{ dollars}]Next, I calculate the new price per packet this month. The coffee is now sold at 4 packets for 10. Dividing the total price by the number of packets gives:[text{New Price per Packet} = frac{10}{4} = 2.5 text{ dollars}]To find the percentage decrease in price, I use the formula:[text{Percentage Decrease} = left( frac{text{Original Price} - text{New Price}}{text{Original Price}} right) times 100%]Substituting the values:[text{Percentage Decrease} = left( frac{4 - 2.5}{4} right) times 100% = left( frac{1.5}{4} right) times 100% = 0.375 times 100% = 37.5%]Therefore, the percentage decrease in the price per packet is closest to 37.5%.

💡Alright, so I need to figure out how much it costs to transport a 250-gram control module using the NASA Space Shuttle to the International Space Station. I know that the cost is 22,000 per kilogram. Okay, so first things first, I need to make sure I understand the units here. The cost is given per kilogram, but the control module's mass is given in grams. I need to convert grams to kilograms to make the units match.Let me recall, 1 kilogram is equal to 1000 grams. So, to convert grams to kilograms, I divide the number of grams by 1000. That means 250 grams divided by 1000 equals 0.25 kilograms. Okay, so the control module weighs 0.25 kilograms.Now, if it costs 22,000 to transport 1 kilogram, then to find out the cost for 0.25 kilograms, I can multiply the cost per kilogram by the number of kilograms. So, 22,000 multiplied by 0.25. Let me do that calculation.22,000 times 0.25 is the same as 22,000 divided by 4, because 0.25 is a quarter. So, 22,000 divided by 4 equals 5,500. Therefore, it should cost 5,500 to transport the 250-gram control module.Wait, let me double-check that. If 1 kilogram costs 22,000, then half a kilogram would cost 11,000, right? And a quarter kilogram, which is 250 grams, should be a quarter of 22,000. So, 22,000 divided by 4 is indeed 5,500. That makes sense.I think I've got it. The cost to transport the control module is 5,500.

💡Okay, so I need to find the smallest positive integer ( n ) such that ( n^2 ) is divisible by 24 and ( n^3 ) is divisible by 720. Hmm, let me break this down step by step.First, I remember that if a number is divisible by another, all the prime factors of the divisor must be present in the dividend with at least the same exponents. So, maybe I should start by factoring 24 and 720 into their prime factors.Let me factorize 24. 24 divided by 2 is 12, divided by 2 again is 6, and again by 2 is 3, which is a prime. So, 24 is ( 2^3 times 3^1 ).Now, 720. Let me divide 720 by 2: 720 ÷ 2 = 360, ÷2 again = 180, ÷2 = 90, ÷2 = 45. Now, 45 is divisible by 3: 45 ÷3 = 15, ÷3 = 5, which is prime. So, 720 is ( 2^4 times 3^2 times 5^1 ).Alright, so now I know that ( n^2 ) must have at least ( 2^3 times 3^1 ) in its prime factors, and ( n^3 ) must have at least ( 2^4 times 3^2 times 5^1 ).Since ( n^2 ) has ( 2^3 times 3^1 ), that means ( n ) itself must have at least ( 2^{3/2} times 3^{1/2} ). But wait, exponents in prime factors have to be integers, right? So, ( 2^{3/2} ) is the same as ( sqrt{2^3} = sqrt{8} ), which isn't an integer. Hmm, so I guess I need to round up the exponents to the next integer to make sure ( n ) is an integer.So, for ( 2^{3/2} ), the smallest integer exponent is 2 because ( 2^2 = 4 ) which is greater than ( 2^{3/2} approx 2.828 ). Similarly, for ( 3^{1/2} ), the smallest integer exponent is 1 because ( 3^1 = 3 ) is greater than ( sqrt{3} approx 1.732 ).Therefore, from ( n^2 ) being divisible by 24, ( n ) must be divisible by ( 2^2 times 3^1 = 4 times 3 = 12 ).Now, moving on to ( n^3 ) being divisible by 720, which is ( 2^4 times 3^2 times 5^1 ). So, ( n^3 ) must have at least these exponents. That means ( n ) must have at least ( 2^{4/3} times 3^{2/3} times 5^{1/3} ).Again, these exponents aren't integers, so I need to round them up. Let's see:- For ( 2^{4/3} ), which is approximately ( 2^{1.333} ), the smallest integer exponent is 2 because ( 2^2 = 4 ) is greater than ( 2^{1.333} approx 2.5198 ).- For ( 3^{2/3} ), which is approximately ( 3^{0.666} ), the smallest integer exponent is 1 because ( 3^1 = 3 ) is greater than ( 3^{0.666} approx 2.08 ).- For ( 5^{1/3} ), which is approximately ( 5^{0.333} ), the smallest integer exponent is 1 because ( 5^1 = 5 ) is greater than ( 5^{0.333} approx 1.709 ).So, from ( n^3 ) being divisible by 720, ( n ) must be divisible by ( 2^2 times 3^1 times 5^1 = 4 times 3 times 5 = 60 ).Wait a second, so from ( n^2 ) we got that ( n ) must be divisible by 12, and from ( n^3 ) we got that ( n ) must be divisible by 60. So, which one is more restrictive? 60 is a multiple of 12, so 60 is the larger requirement. Therefore, ( n ) must be divisible by 60.But let me check if 60 actually satisfies both conditions. Let's compute ( n^2 ) and ( n^3 ) when ( n = 60 ).First, ( 60^2 = 3600 ). Let's factorize 3600. 3600 divided by 2 is 1800, ÷2 is 900, ÷2 is 450, ÷2 is 225. So, that's ( 2^4 ). Then, 225 is ( 15^2 = (3 times 5)^2 = 3^2 times 5^2 ). So, 3600 is ( 2^4 times 3^2 times 5^2 ). Now, 24 is ( 2^3 times 3^1 ). So, 3600 has ( 2^4 ) which is more than enough for ( 2^3 ), and ( 3^2 ) which is more than enough for ( 3^1 ). So, yes, 3600 is divisible by 24.Next, ( 60^3 = 216000 ). Let's factorize 216000. 216000 divided by 2 is 108000, ÷2 is 54000, ÷2 is 27000, ÷2 is 13500, ÷2 is 6750, ÷2 is 3375. So, that's ( 2^6 ). Then, 3375 is ( 15^3 = (3 times 5)^3 = 3^3 times 5^3 ). So, 216000 is ( 2^6 times 3^3 times 5^3 ).720 is ( 2^4 times 3^2 times 5^1 ). So, 216000 has ( 2^6 ) which is more than enough for ( 2^4 ), ( 3^3 ) which is more than enough for ( 3^2 ), and ( 5^3 ) which is more than enough for ( 5^1 ). So, yes, 216000 is divisible by 720.Therefore, ( n = 60 ) satisfies both conditions. But is there a smaller number that also satisfies both conditions? Let me think.If I try ( n = 60 ) divided by 2, which is 30. Let's check ( 30^2 = 900 ). Factorizing 900: ( 2^2 times 3^2 times 5^2 ). 24 is ( 2^3 times 3^1 ). So, 900 has only ( 2^2 ), which is less than ( 2^3 ). Therefore, 900 is not divisible by 24. So, 30 doesn't work.What about ( n = 40 )? ( 40^2 = 1600 ). Factorizing 1600: ( 2^6 times 5^2 ). It has ( 2^6 ), which is more than ( 2^3 ), but it doesn't have any factor of 3. So, 1600 isn't divisible by 24 because it lacks the factor of 3. So, 40 doesn't work.How about ( n = 30 times 2 = 60 )? We already saw that 60 works. What about ( n = 24 )? ( 24^2 = 576 ). Factorizing 576: ( 2^6 times 3^2 ). It has ( 2^6 ) and ( 3^2 ), so it's divisible by 24. Now, ( 24^3 = 13824 ). Factorizing 13824: ( 2^9 times 3^3 ). But 720 requires ( 5^1 ), which 13824 doesn't have. So, 24 doesn't work.What about ( n = 30 )? As I checked earlier, ( 30^2 = 900 ) isn't divisible by 24 because it only has ( 2^2 ). So, 30 doesn't work.How about ( n = 12 )? ( 12^2 = 144 ). Factorizing 144: ( 2^4 times 3^2 ). It's divisible by 24 because ( 2^4 ) covers ( 2^3 ) and ( 3^2 ) covers ( 3^1 ). Now, ( 12^3 = 1728 ). Factorizing 1728: ( 2^6 times 3^3 ). Again, it lacks the factor of 5, so it's not divisible by 720. So, 12 doesn't work.Hmm, so it seems like 60 is indeed the smallest number that satisfies both conditions because any smaller multiple either doesn't have enough factors of 2 or 3 or lacks the factor of 5.Wait, let me think again. Maybe I can find a number smaller than 60 that still includes all necessary prime factors. Let's see, 60 is ( 2^2 times 3 times 5 ). If I try to reduce the exponent of 2, say to ( 2^1 ), then ( n ) would be ( 2 times 3 times 5 = 30 ). But as I saw earlier, 30 squared is 900, which doesn't have enough factors of 2. So, that doesn't work.What if I reduce the exponent of 3? If I take ( 2^2 times 3^0 times 5 = 4 times 1 times 5 = 20 ). Then, ( 20^2 = 400 ), which is ( 2^4 times 5^2 ). It doesn't have any factor of 3, so it's not divisible by 24. So, that doesn't work either.What if I reduce the exponent of 5? If I take ( 2^2 times 3 times 5^0 = 4 times 3 times 1 = 12 ). Then, ( 12^3 = 1728 ), which doesn't have a factor of 5, so it's not divisible by 720. So, that doesn't work.Therefore, I can't reduce any of the exponents without breaking one of the conditions. So, 60 is indeed the smallest number that satisfies both ( n^2 ) divisible by 24 and ( n^3 ) divisible by 720.Just to recap, the key steps were:1. Factorize 24 and 720 into their prime factors.2. Determine the minimum exponents required for each prime in ( n ) based on the divisibility conditions for ( n^2 ) and ( n^3 ).3. Round up the exponents to the nearest integer to ensure ( n ) is an integer.4. Combine these primes to find the least common multiple, which gives the smallest ( n ).5. Verify by checking if ( n^2 ) and ( n^3 ) meet the divisibility requirements.6. Ensure no smaller number than 60 satisfies both conditions by testing potential candidates.So, after going through all that, I'm confident that 60 is the smallest positive integer that meets both conditions.

💡Alright, so I've got these two math problems to solve. Let me take them one at a time and try to work through them step by step. I'm a bit nervous because exponents and negative numbers can sometimes trip me up, but I'll take it slow and make sure I understand each part.Starting with the first problem: (4 + (-2)^3 times 5 - (-0.28) div 4). Okay, let's break this down. I remember that in math problems like this, I should follow the order of operations, which is parentheses, exponents, multiplication and division (from left to right), and then addition and subtraction (also from left to right). So, PEMDAS: Parentheses, Exponents, Multiplication and Division, Addition and Subtraction.Looking at the first part, ( (-2)^3 ). Hmm, exponents. I think when a negative number is raised to an odd power, the result is negative. So, ( (-2)^3 ) should be (-8). Let me double-check that: (-2 times -2 = 4), and then (4 times -2 = -8). Yeah, that seems right.Next, I have to multiply that result by 5. So, (-8 times 5). Multiplying a negative by a positive gives a negative result, so that should be (-40). Okay, so far so good.Now, looking at the next part: (-(-0.28) div 4). Dividing a negative by a positive. Wait, actually, it's (-(-0.28)) divided by 4. The double negative should make it positive, right? So, (-(-0.28)) is (0.28). Then, (0.28 div 4). Let me calculate that: 0.28 divided by 4. Well, 0.28 divided by 2 is 0.14, and divided by 2 again is 0.07. So, that part is (0.07).Now, putting it all back into the original expression: (4 + (-40) - (-0.28) div 4) simplifies to (4 + (-40) + 0.07). Wait, why plus 0.07? Because the division gave us a positive 0.07, right? So, adding that.So, let's compute that: 4 plus (-40) is like 4 minus 40, which is (-36). Then, adding 0.07 to that: (-36 + 0.07 = -35.93). Hmm, that seems a bit messy, but I think that's correct. Let me just go through it again to make sure I didn't make any mistakes.Starting over: (4 + (-2)^3 times 5 - (-0.28) div 4). Exponents first: ((-2)^3 = -8). Then, multiplication: (-8 times 5 = -40). Then, division: (-(-0.28) = 0.28), and (0.28 div 4 = 0.07). So, the expression becomes (4 + (-40) + 0.07). Adding those together: (4 - 40 = -36), and (-36 + 0.07 = -35.93). Yeah, that checks out.Alright, moving on to the second problem: (-1^4 - frac{1}{6} times [2 - (-3)^2]). This one looks a bit more complicated, but I'll take it step by step.First, let's handle the exponents. I see (-1^4) and ((-3)^2). Wait, is it (-1^4) or ((-1)^4)? Because the way it's written, it's (-1^4), which means the exponent applies only to 1, not the negative sign. So, (-1^4 = -(1^4) = -1). If it were ((-1)^4), that would be 1, but I think in this case, it's just (-1).Next, inside the brackets: (2 - (-3)^2). Again, exponents first. ((-3)^2) is (9), because a negative times a negative is positive. So, (2 - 9 = -7). Okay, so the expression inside the brackets simplifies to (-7).Now, the entire expression becomes: (-1 - frac{1}{6} times (-7)). Let's compute the multiplication first: (frac{1}{6} times (-7) = -frac{7}{6}). So, now we have (-1 - (-frac{7}{6})). Subtracting a negative is like adding a positive, so this becomes (-1 + frac{7}{6}).Now, let's compute (-1 + frac{7}{6}). To add these, it's easier if both numbers have the same denominator. (-1) can be written as (-frac{6}{6}), so (-frac{6}{6} + frac{7}{6} = frac{1}{6}). So, the result is (frac{1}{6}).Wait, let me go through that again to make sure. Starting with (-1^4), which is (-1). Then, inside the brackets: (2 - (-3)^2). ((-3)^2 = 9), so (2 - 9 = -7). Then, (frac{1}{6} times (-7) = -frac{7}{6}). So, the expression is (-1 - (-frac{7}{6})), which is (-1 + frac{7}{6}). Converting (-1) to sixths: (-frac{6}{6} + frac{7}{6} = frac{1}{6}). Yep, that seems right.So, summarizing both problems:1. (4 + (-2)^3 times 5 - (-0.28) div 4 = -35.93)2. (-1^4 - frac{1}{6} times [2 - (-3)^2] = frac{1}{6})I think I've got both of them figured out. The first one was a bit tricky with all the negative signs and decimals, but breaking it down step by step helped. The second one had some order of operations nuances, especially with the exponents and the negative signs, but taking it one piece at a time made it manageable. I just need to be careful with the signs and make sure I apply the operations correctly, especially when dealing with negatives and exponents.