💡First, I need to determine Angelina's initial speed from her home to the grocery store. Let's denote this speed as ( v ) meters per second.The distance from her home to the grocery store is 150 meters, so the time it took her to get there is ( frac{150}{v} ) seconds.From the grocery store to the gym, she walked 200 meters at double her initial speed, which is ( 2v ). The time it took her to get to the gym is ( frac{200}{2v} = frac{100}{v} ) seconds.We are told she spent 10 seconds less on her way from the grocery to the gym than on her way from home to the grocery. This gives us the equation:[frac{150}{v} - 10 = frac{100}{v}]Solving for ( v ):[150 - 10v = 100 10v = 50 v = 5 text{ meters per second}]From the gym to the park, she strolled 250 meters at half her initial speed, which is ( frac{5}{2} ) meters per second. The time it took her to get to the park is:[frac{250}{frac{5}{2}} = frac{250 cdot 2}{5} = 100 text{ seconds}]From the park to the library, she rushed 300 meters at thrice the speed from the grocery to the gym, which is ( 3 cdot 2v = 6v ). Substituting ( v = 5 ):[6 cdot 5 = 30 text{ meters per second}]The time it took her to get to the library is:[frac{300}{30} = 10 text{ seconds}]We are also told she spent 20 seconds more on her way to the park than on her way to the library. This confirms our calculations:[100 = 10 + 20]Therefore, Angelina's speed from the park to the library was ( 30 ) meters per second.

💡Okay, so I have this function f(x) defined piecewise. It says that when x is greater than 0, f(x) is log base 3 of x, and when x is less than 0, f(x) is g(x). The problem is telling me that f(x) is an odd function, and I need to find the value of g(-1/9). Hmm, okay.First, I remember that an odd function has the property that f(-x) = -f(x) for all x in its domain. So, that means if I plug in a negative x into f(x), it's the same as taking the negative of f(x) when x is positive. That seems important here because g(x) is defined for x < 0, so g(-1/9) is f(-1/9), right?Let me write that down: g(-1/9) = f(-1/9). Now, since f is odd, f(-1/9) should be equal to -f(1/9). So, I can write that as g(-1/9) = -f(1/9). Now, f(1/9) is given because when x is positive, f(x) is log base 3 of x. So, f(1/9) = log₃(1/9). I need to compute that. Wait, log base 3 of 1/9. I know that 3 squared is 9, so 3^2 = 9. Therefore, 1/9 is 3^(-2). So, log base 3 of 3^(-2) is just -2, right? Because log base a of a^b is b. So, log₃(3^(-2)) = -2. So, f(1/9) is -2. Therefore, g(-1/9) = -f(1/9) = -(-2) = 2. Wait, let me double-check that. If f is odd, then f(-x) = -f(x). So, f(-1/9) = -f(1/9). Since f(1/9) is log₃(1/9) which is -2, then f(-1/9) is -(-2) = 2. So, g(-1/9) is 2. Is there anything else I need to consider? Maybe the domain of the function? Let's see, for x > 0, f(x) is log₃x, which is defined for x > 0, which is fine. For x < 0, it's g(x), which is defined for x < 0, so that's also fine. Wait, does the function f(x) have to be defined at x = 0? The problem doesn't specify, so maybe it's not necessary for this problem. Since we're only dealing with x = -1/9 and x = 1/9, which are both not zero, we don't have to worry about that. So, putting it all together, since f is odd, f(-1/9) = -f(1/9). f(1/9) is log base 3 of 1/9, which is -2. Therefore, f(-1/9) is 2, so g(-1/9) is 2. I think that makes sense. Let me just recap:1. f(x) is odd, so f(-x) = -f(x).2. Therefore, g(-1/9) = f(-1/9) = -f(1/9).3. f(1/9) = log₃(1/9) = -2.4. So, g(-1/9) = -(-2) = 2.Yep, that seems solid. I don't think I missed anything here. The key was remembering the definition of an odd function and applying it correctly. I initially thought maybe I had to do something more complicated with the logarithm, but it turned out straightforward once I broke it down step by step.Just to make sure, let me think about what log base 3 of 1/9 really is. Since 3^2 = 9, 3^(-2) = 1/9. So, log base 3 of 1/9 is indeed -2. That part checks out.And since f is odd, flipping the sign of the input flips the sign of the output. So, if f(1/9) is -2, then f(-1/9) must be 2. That all lines up.I think I'm confident with this answer. It's 2.**Final Answer**The value of ( gleft(-frac{1}{9}right) ) is boxed{2}.

💡Alright, so I've got this system of equations to solve:begin{cases}3x + y = 2 2x - y = 8end{cases}Hmm, okay. I remember from class that there are a couple of ways to solve systems like this. One is substitution, and the other is elimination. I think elimination might be easier here because the coefficients of y in the two equations are opposites. That should make things cancel out nicely.Let me write down the equations again to visualize them better:1. (3x + y = 2)2. (2x - y = 8)If I add these two equations together, the y terms should cancel out. Let's try that:Adding equation 1 and equation 2:(3x + y + 2x - y = 2 + 8)Simplifying the left side:(3x + 2x + y - y = 5x + 0 = 5x)And the right side:(2 + 8 = 10)So now I have:(5x = 10)To solve for x, I just divide both sides by 5:(x = 10 / 5 = 2)Okay, so x is 2. Now I need to find y. I can plug this value of x back into either of the original equations. Let me choose the first one because it looks simpler:(3x + y = 2)Substituting x = 2:(3(2) + y = 2)Calculating 3 times 2:(6 + y = 2)Now, to solve for y, I subtract 6 from both sides:(y = 2 - 6 = -4)So y is -4. Let me double-check my work to make sure I didn't make any mistakes. I'll plug x = 2 and y = -4 into the second equation to verify:(2x - y = 8)Substituting the values:(2(2) - (-4) = 8)Calculating:(4 + 4 = 8)Which simplifies to:(8 = 8)That checks out! So both equations are satisfied with x = 2 and y = -4. Therefore, the solution set is (2, -4).Wait a second, let me think again. Did I do everything correctly? I added the two equations and got 5x = 10, so x = 2. Then substituted back into the first equation and got y = -4. Plugging into the second equation worked too. Yeah, that seems right.I guess another way to check is to graph both equations and see if they intersect at (2, -4). The first equation, 3x + y = 2, can be rewritten as y = -3x + 2. The second equation, 2x - y = 8, can be rewritten as y = 2x - 8. If I plot these two lines, they should intersect at (2, -4).Let me pick a few points for each line to see:For y = -3x + 2:- When x = 0, y = 2.- When x = 1, y = -1.- When x = 2, y = -4.For y = 2x - 8:- When x = 0, y = -8.- When x = 1, y = -6.- When x = 2, y = -4.So both lines pass through the point (2, -4). That confirms that my solution is correct.I think I'm confident now that the solution set is indeed (2, -4). There's no other point where these two lines intersect, so that must be the only solution.Just to recap, the steps I took were:1. Recognize that the coefficients of y in both equations are opposites, making elimination a good method.2. Added the two equations to eliminate y and solve for x.3. Substituted the value of x back into one of the original equations to find y.4. Verified the solution by plugging the values into the other equation.5. Double-checked by considering the graphical representation of the equations.This systematic approach helped me avoid mistakes and ensured that my solution is correct. I think I've got a good handle on solving systems of equations using elimination now!

💡Okay, so I have this problem here about functions, and I need to figure out which one is the even function with the smallest positive period of π and is monotonically increasing in the interval (0, π/4). The options are A: y = sin x, B: y = sin 2|x|, C: y = -cos 2x, and D: y = cos 2x.First, let me recall what an even function is. An even function satisfies the condition f(-x) = f(x) for all x in its domain. So, I need to check which of these functions are even.Starting with option A: y = sin x. I remember that sine is an odd function because sin(-x) = -sin x. So, A is not even. That means A is incorrect.Moving on to option B: y = sin 2|x|. Hmm, let's see. Since |x| is always positive, sin 2|x| will behave like sin 2x for x ≥ 0 and sin(-2x) for x < 0, but since |x| is symmetric, sin 2|x| is actually an even function because sin 2|x| = sin 2|x|. So, B is even. But wait, is it periodic? The period of sin x is 2π, but here we have sin 2|x|. The presence of |x| complicates things. Let me think. If I consider x positive, sin 2x has a period of π, but because of the absolute value, does it repeat? Actually, since |x| makes the function symmetric about the y-axis, but does it make it periodic? I don't think so. For example, sin 2|x| for x = 0 is 0, for x = π/4, it's sin(π/2) = 1, for x = π/2, it's sin π = 0, and so on. But does it repeat every π? Let me check x = π/4 and x = 5π/4. At x = 5π/4, sin 2|5π/4| = sin(5π/2) = 1. So, it seems like it might have a period of π, but actually, the function sin 2|x| isn't periodic because for negative x, it's the same as positive, but the periodicity requires that the function repeats in both directions. Wait, maybe it is periodic with period π? Let me test f(x + π) = sin 2|x + π|. Is that equal to sin 2|x|? Not necessarily. For example, take x = 0: f(0 + π) = sin 2π = 0, which is equal to f(0). But take x = π/4: f(π/4 + π) = sin 2|5π/4| = sin(5π/2) = 1, which is equal to f(π/4) = sin(π/2) = 1. Hmm, maybe it is periodic with period π? But wait, actually, the function sin 2|x| isn't defined for negative x in the same way as positive x, so I'm not sure if it's truly periodic. Maybe it's not a standard periodic function because of the absolute value. So, perhaps B is not periodic, making it incorrect.Now, option C: y = -cos 2x. Let's check if it's even. cos(-2x) = cos 2x, so -cos(-2x) = -cos 2x, which is equal to f(-x) = -cos 2x = f(x). So, yes, it's even. Good. Now, what's its period? The period of cos 2x is π because the period of cos kx is 2π/k, so here k=2, so period is π. So, the smallest positive period is π, which matches the requirement. Now, is it monotonically increasing in (0, π/4)? Let's see. The derivative of y = -cos 2x is y' = 2 sin 2x. In the interval (0, π/4), 2x is in (0, π/2). Sin is positive in (0, π/2), so y' is positive, meaning the function is increasing. So, C satisfies all the conditions.Option D: y = cos 2x. Let's check. It's even because cos(-2x) = cos 2x. Its period is also π, same as C. Now, is it monotonically increasing in (0, π/4)? Let's compute the derivative: y' = -2 sin 2x. In (0, π/4), sin 2x is positive, so y' is negative, meaning the function is decreasing. So, D is incorrect.So, summarizing: A is odd, B is even but not periodic, C is even, period π, increasing in (0, π/4), D is even, period π, but decreasing in (0, π/4). Therefore, the correct answer is C.Wait, but in the initial thought process, I thought B might be periodic, but now I'm more confident it's not because of the absolute value. Also, confirming the derivative for C, it's positive in the interval, so it's increasing. Yeah, I think C is the right choice.

💡Okay, so I have this problem where Professor Smith buys a lottery ticket. He has to pick five different integers from 1 to 50, and the product of these numbers needs to be a perfect cube. Interestingly, the winning ticket also has this property. I need to find the probability that Professor Smith holds the winning ticket. The options are from 0 to 1, so it's a multiple-choice question with options A to E.First, I need to understand what a perfect cube is. A perfect cube is a number that can be written as some integer raised to the power of three. For example, 8 is a perfect cube because it's 2³, and 27 is 3³. So, the product of the five numbers Professor Smith picks must be such a number.Now, the key here is that both Professor Smith's ticket and the winning ticket have this property. So, I need to figure out how many possible combinations of five different numbers from 1 to 50 result in a product that's a perfect cube. Once I know that, the probability would be 1 divided by the number of such combinations because each combination is equally likely in a lottery.But wait, how do I find the number of such combinations? This seems tricky. Let me think about the prime factorization of numbers. For a product to be a perfect cube, all the exponents in its prime factorization must be multiples of 3. So, each prime number in the factorization of the five numbers must appear a number of times that's a multiple of 3.This means that if I pick numbers with prime factors, their combined exponents for each prime must add up to a multiple of 3. That sounds complicated because there are so many primes between 1 and 50. Maybe I can simplify this by looking at numbers that are already perfect cubes themselves because multiplying perfect cubes will give another perfect cube.Let me list the perfect cubes between 1 and 50. They are:- 1³ = 1- 2³ = 8- 3³ = 27- 4³ = 64 (which is over 50, so we can't include that)So, the perfect cubes in the range are 1, 8, and 27. Hmm, only three numbers. But Professor Smith needs to pick five different numbers. If he only picks these three, he still needs two more numbers. But those two numbers would have to be such that when multiplied with 1, 8, and 27, the product is still a perfect cube.Wait, that might not be possible because adding any other number would introduce new prime factors or change the exponents of existing ones, which might not be multiples of 3. For example, if he picks 2, which is a prime number, then the exponent of 2 in the product would be 1 (from 2) plus 3 (from 8), totaling 4, which isn't a multiple of 3. That would ruin the perfect cube property.So, maybe he can't just pick any other numbers. Maybe he needs to pick numbers that, when combined with the perfect cubes, balance out the exponents to make them multiples of 3. But this seems complicated because each number can have multiple prime factors, and ensuring all exponents are multiples of 3 is non-trivial.Alternatively, maybe the only way to have the product be a perfect cube is if all five numbers are perfect cubes themselves. But as we saw, there are only three perfect cubes between 1 and 50. So, he can't pick five different perfect cubes because there aren't enough. Therefore, it's impossible for him to pick five different numbers where the product is a perfect cube.If that's the case, then the number of possible combinations is zero. Therefore, the probability that Professor Smith holds the winning ticket is zero because there are no valid combinations that satisfy the condition.Wait, but the problem says that both Professor Smith and the winning ticket have this property. If there are no valid combinations, then how can both have it? Maybe I'm missing something here.Let me double-check. Are there other numbers besides the perfect cubes that can be part of a product that's a perfect cube? For example, numbers that are not perfect cubes themselves but when multiplied together, their exponents add up to multiples of 3.For instance, if I have two numbers that are both squares, their product would be a fourth power, which isn't a cube. But if I have numbers with exponents that complement each other to reach multiples of 3, that could work. For example, if one number has a prime factor with exponent 1, another number could have the same prime factor with exponent 2, making the total exponent 3.But this requires careful selection of numbers to ensure that for every prime factor, the exponents add up to a multiple of 3. Given that we're picking five numbers, this seems highly restrictive. It might be possible, but it's not straightforward.However, considering the range is only up to 50, the number of primes is limited, but the combinations would still be complex. I'm not sure if there are any such combinations, but given the complexity, it's likely that there are very few or none.Therefore, if there are no valid combinations, the probability is zero. But the problem states that both Professor Smith and the winning ticket have this property, implying that at least one such combination exists. Maybe I need to reconsider.Perhaps there's a way to combine numbers such that their exponents balance out. For example, using numbers that are multiples of cubes or have exponents that complement each other. Let me think about specific numbers.Take the number 2. If I include 2, I need to include other numbers that have factors of 2 such that the total exponent is a multiple of 3. Similarly for other primes. This would require a systematic approach, perhaps using modular arithmetic to ensure exponents are balanced.But this is getting too involved, and I'm not sure if it's feasible within the constraints of picking five numbers from 1 to 50. Given the time I've spent and the lack of obvious solutions, I think it's safe to assume that there are no such combinations, making the probability zero.Wait, but the answer choices include 0, so maybe that's the intended answer. However, I'm not entirely confident because the problem states that both tickets have this property, suggesting that at least one combination exists. Maybe I'm missing a key insight here.Perhaps the numbers don't need to be perfect cubes themselves, but their product can be a cube through complementary exponents. For example, if one number has a prime factor with exponent 1, another has exponent 2, and another has exponent 0 (i.e., not present), but that doesn't make sense because exponents can't be negative.Wait, actually, exponents can be zero if a prime isn't present in a number. So, if I have numbers that together cover all primes with exponents adding up to multiples of 3, that could work. But this would require a very specific set of numbers, and with five numbers, it's still challenging.Given the complexity and the limited range, I think it's still unlikely that such combinations exist. Therefore, the probability is zero.But I'm still unsure because the problem implies that both tickets have this property, so maybe there is at least one combination. Maybe I need to think differently.Perhaps the numbers can be arranged such that each prime's exponent across all five numbers is a multiple of 3. For example, if each prime appears exactly three times across the five numbers, but that seems too restrictive.Alternatively, maybe the numbers are arranged in such a way that their exponents modulo 3 are zero. But again, this is too vague without specific examples.Given the time I've spent and the lack of progress, I think the safest answer is 0, as I can't find any valid combinations, and the problem might be designed to trick us into thinking it's possible when it's not.

💡First, I need to calculate the area of both gardens to determine which one is larger. For Chris's garden, which measures 30 feet by 60 feet, the area is calculated by multiplying the length by the width. So, 30 feet multiplied by 60 feet gives an area of 1800 square feet.Next, I'll calculate the area of Jordan's garden, which measures 35 feet by 55 feet. Multiplying these dimensions, 35 feet by 55 feet, results in an area of 1925 square feet.Now, to find out by how much one garden is larger than the other, I'll subtract the area of Chris's garden from the area of Jordan's garden. This gives me 1925 square feet minus 1800 square feet, which equals 125 square feet. Therefore, Jordan's garden is larger by 125 square feet.Moving on to the perimeter calculation, I'll use the formula for the perimeter of a rectangle, which is 2 times the sum of its length and width. For Chris's garden, the perimeter is 2 times (30 feet plus 60 feet), resulting in 180 feet.For Jordan's garden, the perimeter is calculated as 2 times (35 feet plus 55 feet), which also equals 180 feet. This means both gardens have the same perimeter.In summary, Jordan's garden has a larger area by 125 square feet, and both gardens share the same perimeter of 180 feet.