💡Okay, so I have this problem about a function ( y = f(x) ) that's even on all real numbers. Hmm, even functions are symmetric about the y-axis, right? So ( f(-x) = f(x) ) for all ( x ). Got that.Then, it says that for any ( x in mathbb{R} ), ( f(x + 4) = f(x) + f(2) ). Interesting. That seems like a functional equation. Maybe I can use that to find some properties of the function.Also, when ( x_1, x_2 in [0, 2] ) and ( x_1 neq x_2 ), the difference quotient ( frac{f(x_2) - f(x_1)}{x_2 - x_1} > 0 ). So that means the function is increasing on the interval [0, 2]. Since it's even, that should tell me something about the behavior on [-2, 0] as well. Probably decreasing there, right?Now, let's look at the statements one by one.**Statement (1):** ( f(2) = 0 ) and ( T = 4 ) is a period of the function ( f(x) ).Hmm. Let's see. If I plug ( x = -2 ) into the equation ( f(x + 4) = f(x) + f(2) ), I get ( f(-2 + 4) = f(-2) + f(2) ), which simplifies to ( f(2) = f(-2) + f(2) ). But since ( f ) is even, ( f(-2) = f(2) ). So substituting that in, we have ( f(2) = f(2) + f(2) ), which implies ( f(2) = 0 ). Okay, so that part checks out.Now, does ( T = 4 ) mean it's a period? Let's see. If ( f(x + 4) = f(x) + f(2) ), but we just found ( f(2) = 0 ), so ( f(x + 4) = f(x) ). That means 4 is indeed a period. So statement (1) is correct.**Statement (2):** The line ( x = 4 ) is an axis of symmetry for the function ( y = f(x) ).Since the function is even, it's symmetric about the y-axis, which is ( x = 0 ). But does that mean it's also symmetric about ( x = 4 )?Well, if the function has period 4, then shifting it by 4 units should give the same function. So, ( f(x + 4) = f(x) ). But symmetry about ( x = 4 ) would mean that ( f(4 + a) = f(4 - a) ) for any ( a ).Let's test this. Let ( a = x - 4 ). Then, ( f(4 + (x - 4)) = f(4 - (x - 4)) ) simplifies to ( f(x) = f(8 - x) ). But since the function is even, ( f(8 - x) = f(x - 8) ). But because the function has period 4, ( f(x - 8) = f(x - 8 + 8) = f(x) ). So, ( f(x) = f(x) ). That's always true, but does that mean it's symmetric about ( x = 4 )?Wait, actually, if ( f(4 + a) = f(4 - a) ), then yes, it's symmetric about ( x = 4 ). Because for any point ( a ) to the right of 4, there's a corresponding point ( a ) to the left of 4 with the same function value. So statement (2) is correct.**Statement (3):** The function ( y = f(x) ) is increasing on the interval ( [-6, -4] ).Okay, so let's think about the behavior on different intervals. We know that ( f ) is increasing on [0, 2]. Since it's even, it should be decreasing on [-2, 0]. Now, because the function has period 4, the behavior on [0, 2] repeats on [4, 6], and the behavior on [-2, 0] repeats on [-6, -4].Wait, so if it's increasing on [0, 2], then on [4, 6], it should also be increasing. But what about [-6, -4]? Since [-6, -4] is a shift of [-2, 0] by -4, and the function is periodic with period 4, the behavior on [-6, -4] should be the same as on [-2, 0], which is decreasing. So the function is decreasing on [-6, -4], not increasing. Therefore, statement (3) is incorrect.**Statement (4):** The function ( y = f(x) ) has four zeros on the interval ( [-6, 6] ).Let's see. We know ( f(2) = 0 ) from statement (1). Since the function is even, ( f(-2) = 0 ) as well. Also, because the function has period 4, ( f(2 + 4) = f(6) = f(2) = 0 ), and ( f(-2 - 4) = f(-6) = f(-2) = 0 ). So we have zeros at ( x = -6, -2, 2, 6 ). That's four zeros on the interval [-6, 6]. So statement (4) is correct.Wait, but are there any other zeros? Let me think. The function is increasing on [0, 2], so it goes from ( f(0) ) to ( f(2) = 0 ). If it's increasing, then ( f(0) < f(2) = 0 ). So ( f(0) ) is negative. Similarly, on [-2, 0], it's decreasing, so it goes from ( f(-2) = 0 ) to ( f(0) ), which is negative. So between -2 and 0, it goes from 0 to negative, so it doesn't cross zero again. Similarly, on [0, 2], it goes from negative to 0, so only at 2 does it cross zero. So on each period interval, there's only one zero. So on [-6, 6], we have zeros at -6, -2, 2, 6. That's four zeros. So statement (4) is correct.Wait, but what about between -6 and -4? The function is decreasing there, right? So it goes from ( f(-6) = 0 ) to ( f(-4) ). Since it's decreasing, ( f(-4) ) would be higher than ( f(-6) = 0 ). So it doesn't cross zero again. Similarly, on [4, 6], it's increasing from ( f(4) ) to ( f(6) = 0 ). Since it's increasing, ( f(4) ) is less than 0, so it goes from negative to 0, so only at 6 does it cross zero. So yeah, only four zeros in total on [-6, 6]. So statement (4) is correct.So, summarizing:- Statement (1): Correct- Statement (2): Correct- Statement (3): Incorrect- Statement (4): CorrectTherefore, the correct statements are (1), (2), and (4).

💡First, I need to determine the area of the larger rectangle, which has dimensions of 10 units by 6 units. Calculating this area is straightforward by multiplying the length and width.Next, I need to find the dimensions of the smaller rectangle that is removed from the larger one. According to the problem, the smaller rectangle's length is reduced by 3 units from the larger rectangle's length, and its width is reduced by 2 units from the larger rectangle's width.Once I have the dimensions of the smaller rectangle, I can calculate its area by multiplying its length and width.Finally, to find the area of the resulting "L" shape, I subtract the area of the smaller rectangle from the area of the larger rectangle. This will give me the total area of the "L" shape.

💡First, I need to determine the total number of fishes Hazel and her dad caught from Lake Ontario and Erie, which is 23 fishes.Next, they caught 30 fishes from Lake Huron and Michigan.Adding these together, the total number of fishes caught from these two pairs of lakes is 23 + 30 = 53 fishes.They brought home a total of 97 fishes.To find out how many fishes they caught from Lake Superior, I subtract the total number of fishes caught from the other lakes from the overall total: 97 - 53 = 44 fishes.Therefore, they caught 44 fishes from Lake Superior.

💡Okay, so I have this problem here: Given positive integers ( x ) and ( y ) such that ( frac{1}{x} + frac{1}{3y} = frac{1}{6} ), what is the least possible value of ( xy )?Alright, let me try to figure this out step by step. I'm not too confident with these kinds of problems, but I'll give it a shot.First, the equation is ( frac{1}{x} + frac{1}{3y} = frac{1}{6} ). Hmm, I need to find positive integers ( x ) and ( y ) that satisfy this equation and then find the minimum value of their product ( xy ).Maybe I should start by simplifying the equation. Let me rewrite it:( frac{1}{x} + frac{1}{3y} = frac{1}{6} )I think the first step is to eliminate the fractions to make it easier to handle. To do that, I can find a common denominator or multiply both sides by the least common multiple (LCM) of the denominators. The denominators here are ( x ), ( 3y ), and ( 6 ). So, the LCM of these would be ( 6xy ), I think.Let me multiply both sides of the equation by ( 6xy ) to eliminate the denominators:( 6xy times left( frac{1}{x} + frac{1}{3y} right) = 6xy times frac{1}{6} )Simplifying each term:- ( 6xy times frac{1}{x} = 6y )- ( 6xy times frac{1}{3y} = 2x )- ( 6xy times frac{1}{6} = xy )So, putting it all together:( 6y + 2x = xy )Alright, so now I have the equation ( xy - 2x - 6y = 0 ). Hmm, this looks a bit complicated, but maybe I can rearrange it to make it factorable.Let me try to rearrange the terms:( xy - 2x - 6y = 0 )I can factor out an ( x ) from the first two terms:( x(y - 2) - 6y = 0 )Hmm, not sure if that helps. Maybe I can use a technique called Simon's Favorite Factoring Trick (SFFT). I remember hearing about this in algebra; it's a method to factor equations by adding and subtracting constants.To apply SFFT, I think I need to add a constant to both sides so that the left side becomes a product of binomials. Let me see.Looking at the equation ( xy - 2x - 6y = 0 ), I can rewrite it as:( xy - 2x - 6y + c = c )Where ( c ) is a constant I need to determine. The goal is to choose ( c ) such that the left side factors neatly.Looking at the coefficients of ( x ) and ( y ), which are -2 and -6 respectively, I think ( c ) should be the product of these coefficients, which is ( (-2) times (-6) = 12 ). So, let's add 12 to both sides:( xy - 2x - 6y + 12 = 12 )Now, let's factor the left side:( (x - 6)(y - 2) = 12 )Oh, nice! That worked out. So now, the equation is factored into ( (x - 6)(y - 2) = 12 ).Since ( x ) and ( y ) are positive integers, ( x - 6 ) and ( y - 2 ) must also be positive integers. Therefore, ( x - 6 ) and ( y - 2 ) are positive integer factors of 12.Let me list all the positive integer factor pairs of 12:1. ( 1 times 12 )2. ( 2 times 6 )3. ( 3 times 4 )4. ( 4 times 3 )5. ( 6 times 2 )6. ( 12 times 1 )These are all the possible pairs since 12 is not a very large number. Now, for each factor pair, I can find the corresponding values of ( x ) and ( y ) by adding back the constants we subtracted earlier.Let me go through each pair one by one:1. **First pair: ( 1 times 12 )** - ( x - 6 = 1 ) ⇒ ( x = 7 ) - ( y - 2 = 12 ) ⇒ ( y = 14 ) - So, ( x = 7 ) and ( y = 14 ) - Product ( xy = 7 times 14 = 98 )2. **Second pair: ( 2 times 6 )** - ( x - 6 = 2 ) ⇒ ( x = 8 ) - ( y - 2 = 6 ) ⇒ ( y = 8 ) - So, ( x = 8 ) and ( y = 8 ) - Product ( xy = 8 times 8 = 64 )3. **Third pair: ( 3 times 4 )** - ( x - 6 = 3 ) ⇒ ( x = 9 ) - ( y - 2 = 4 ) ⇒ ( y = 6 ) - So, ( x = 9 ) and ( y = 6 ) - Product ( xy = 9 times 6 = 54 )4. **Fourth pair: ( 4 times 3 )** - ( x - 6 = 4 ) ⇒ ( x = 10 ) - ( y - 2 = 3 ) ⇒ ( y = 5 ) - So, ( x = 10 ) and ( y = 5 ) - Product ( xy = 10 times 5 = 50 )5. **Fifth pair: ( 6 times 2 )** - ( x - 6 = 6 ) ⇒ ( x = 12 ) - ( y - 2 = 2 ) ⇒ ( y = 4 ) - So, ( x = 12 ) and ( y = 4 ) - Product ( xy = 12 times 4 = 48 )6. **Sixth pair: ( 12 times 1 )** - ( x - 6 = 12 ) ⇒ ( x = 18 ) - ( y - 2 = 1 ) ⇒ ( y = 3 ) - So, ( x = 18 ) and ( y = 3 ) - Product ( xy = 18 times 3 = 54 )Alright, so now I have all the possible pairs of ( x ) and ( y ) and their corresponding products ( xy ). Let me list them again for clarity:1. ( (7, 14) ) ⇒ ( xy = 98 )2. ( (8, 8) ) ⇒ ( xy = 64 )3. ( (9, 6) ) ⇒ ( xy = 54 )4. ( (10, 5) ) ⇒ ( xy = 50 )5. ( (12, 4) ) ⇒ ( xy = 48 )6. ( (18, 3) ) ⇒ ( xy = 54 )Looking at these products, the smallest one is 48, which comes from the pair ( (12, 4) ). So, ( x = 12 ) and ( y = 4 ) give the least possible value of ( xy ), which is 48.Wait, just to make sure I didn't make any mistakes, let me double-check one of the calculations. Let's take the pair ( (12, 4) ):- ( x = 12 ), ( y = 4 )- Plugging back into the original equation: ( frac{1}{12} + frac{1}{3 times 4} = frac{1}{12} + frac{1}{12} = frac{2}{12} = frac{1}{6} )- Yep, that works.And another one, say ( (10, 5) ):- ( x = 10 ), ( y = 5 )- ( frac{1}{10} + frac{1}{15} = frac{3}{30} + frac{2}{30} = frac{5}{30} = frac{1}{6} )- That also works.So, it seems like all the pairs satisfy the original equation, and the products are correctly calculated. Therefore, the minimum value of ( xy ) is indeed 48.I think I've covered all the steps and double-checked my work. It seems solid. I didn't skip any crucial steps, and all the factor pairs have been considered. So, I'm confident that 48 is the correct answer.**Final Answer**The least possible value of ( xy ) is boxed{48}.

💡First, I need to determine the speed of the faster train. I know that both trains start at the same time and move towards each other, covering a total distance of 444 km. When they meet, the faster train has traveled 60 km more than the slower one.Let’s denote the speed of the faster train as ( v ) km/h and the slower train as 16 km/h. Let’s assume they meet after ( t ) hours.In ( t ) hours, the slower train travels ( 16t ) km, and the faster train travels ( vt ) km. According to the problem, the faster train has traveled 60 km more than the slower train:[vt = 16t + 60]Since the total distance covered by both trains when they meet is 444 km, we can write:[16t + vt = 444]Substituting the first equation into the second:[16t + (16t + 60) = 444][32t + 60 = 444][32t = 384][t = 12 text{ hours}]Now, substitute ( t = 12 ) back into the first equation to find ( v ):[v cdot 12 = 16 cdot 12 + 60][12v = 192 + 60][12v = 252][v = 21 text{ km/h}]Therefore, the speed of the faster train is 21 km/h.

💡Okay, so I have this problem here: I need to figure out how many ways there are to put 7 balls into 3 boxes. The balls are not distinguishable, but the boxes are. And there's a constraint that no box can be empty. Hmm, okay, let's break this down.First, since the balls are not distinguishable, it means that the order in which they are placed in the boxes doesn't matter. So, it's more about how many balls go into each box rather than which specific balls go where. On the other hand, the boxes are distinguishable, so putting different numbers of balls into different boxes will count as different arrangements.Now, the constraint is that no box can be empty. That means each box must have at least one ball. If I didn't have this constraint, I could use the stars and bars method to solve it. But since each box must have at least one ball, I need to adjust the problem to account for that.Let me recall the stars and bars method. It's a combinatorial method used to solve problems related to distributing indistinguishable objects into distinguishable bins. The formula for the number of ways to distribute n indistinguishable objects into k distinguishable boxes is C(n + k - 1, k - 1), where C is the combination function.But in this case, since each box must have at least one ball, I need to ensure that each box gets at least one ball before distributing the remaining balls. So, I can start by placing one ball in each box to satisfy the constraint. That uses up 3 balls (since there are 3 boxes), leaving me with 7 - 3 = 4 balls to distribute freely among the 3 boxes.Now, with the remaining 4 balls, I can use the stars and bars method without any restrictions. So, the number of ways to distribute these 4 balls into 3 boxes is C(4 + 3 - 1, 3 - 1) = C(6, 2).Calculating C(6, 2): that's 6! / (2! * (6 - 2)!) = (6 * 5) / (2 * 1) = 15.Wait, so does that mean there are 15 ways? Let me double-check. I started by placing one ball in each box, ensuring none are empty, then distributed the remaining 4 balls. Using stars and bars for the remaining distribution, I got 15. That seems right.Alternatively, I could think of it as finding the number of integer solutions to the equation x1 + x2 + x3 = 7, where each xi >= 1. By substituting yi = xi - 1, the equation becomes y1 + y2 + y3 = 4, where yi >= 0. The number of non-negative integer solutions is C(4 + 3 - 1, 3 - 1) = C(6, 2) = 15. Yep, that matches.So, I think I'm confident that the number of ways is 15.