answer：Okay, so I have this problem here: If ( f(x) = (x - 2)(x - m) ) is an even function defined on all real numbers, then what is ( m )? Hmm, let me think about how to approach this.First, I remember that an even function satisfies the condition ( f(-x) = f(x) ) for all ( x ) in its domain. So, if ( f(x) ) is even, then plugging in ( -x ) into the function should give me the same result as plugging in ( x ).Let me write down the function again: ( f(x) = (x - 2)(x - m) ). To check if it's even, I need to compute ( f(-x) ) and set it equal to ( f(x) ).So, let's compute ( f(-x) ):[f(-x) = (-x - 2)(-x - m)]Hmm, that looks a bit complicated, but maybe I can expand both ( f(x) ) and ( f(-x) ) and then compare them.Let me expand ( f(x) ) first:[f(x) = (x - 2)(x - m) = x^2 - (2 + m)x + 2m]Okay, so that's a quadratic function in standard form. Now, let me expand ( f(-x) ):[f(-x) = (-x - 2)(-x - m)]Multiplying these two binomials:First, multiply ( -x ) by ( -x ): that's ( x^2 ).Then, ( -x ) times ( -m ): that's ( mx ).Next, ( -2 ) times ( -x ): that's ( 2x ).Lastly, ( -2 ) times ( -m ): that's ( 2m ).So, putting it all together:[f(-x) = x^2 + (m + 2)x + 2m]Wait, so ( f(-x) = x^2 + (m + 2)x + 2m ) and ( f(x) = x^2 - (2 + m)x + 2m ).Since ( f(x) ) is even, ( f(-x) ) should equal ( f(x) ). So, let's set them equal to each other:[x^2 + (m + 2)x + 2m = x^2 - (2 + m)x + 2m]Hmm, okay, let's subtract ( x^2 ) and ( 2m ) from both sides to simplify:[(m + 2)x = -(2 + m)x]So, that simplifies to:[(m + 2)x + (2 + m)x = 0]Wait, no, actually, if I bring all terms to one side:[(m + 2)x + (2 + m)x = 0]But that's not quite right. Let me try again.Starting from:[x^2 + (m + 2)x + 2m = x^2 - (2 + m)x + 2m]Subtract ( x^2 ) and ( 2m ) from both sides:[(m + 2)x = -(2 + m)x]So, moving all terms to the left side:[(m + 2)x + (2 + m)x = 0]Combine like terms:[2(m + 2)x = 0]Wait, that doesn't seem right. Let me double-check my steps.Wait, no, actually, when I have ( (m + 2)x = -(2 + m)x ), I can factor out ( x ) on both sides:[(m + 2)x + (2 + m)x = 0]Which is:[2(m + 2)x = 0]But this equation must hold for all ( x ) in ( mathbb{R} ). The only way this can be true for all ( x ) is if the coefficient of ( x ) is zero. So:[2(m + 2) = 0]Dividing both sides by 2:[m + 2 = 0]Therefore:[m = -2]Okay, so that seems to be the solution. Let me verify it by plugging ( m = -2 ) back into the original function.So, if ( m = -2 ), then:[f(x) = (x - 2)(x - (-2)) = (x - 2)(x + 2) = x^2 - 4]Which is indeed an even function because ( f(-x) = (-x)^2 - 4 = x^2 - 4 = f(x) ). So, that checks out.Wait, just to make sure I didn't make any mistakes in my earlier steps. Let me go through the expansion again.Original function:[f(x) = (x - 2)(x - m) = x^2 - (2 + m)x + 2m]Then, ( f(-x) = (-x - 2)(-x - m) ). Let me expand this carefully.First, multiply ( (-x - 2) ) and ( (-x - m) ):[(-x)(-x) + (-x)(-m) + (-2)(-x) + (-2)(-m)]Which is:[x^2 + mx + 2x + 2m]So, combining like terms:[x^2 + (m + 2)x + 2m]Yes, that's correct. So, ( f(-x) = x^2 + (m + 2)x + 2m ).Setting ( f(-x) = f(x) ):[x^2 + (m + 2)x + 2m = x^2 - (2 + m)x + 2m]Subtract ( x^2 ) and ( 2m ) from both sides:[(m + 2)x = -(2 + m)x]Which simplifies to:[(m + 2)x + (2 + m)x = 0]Which is:[2(m + 2)x = 0]Since this must hold for all ( x ), the coefficient must be zero:[2(m + 2) = 0 implies m + 2 = 0 implies m = -2]Yep, that all checks out. So, I'm confident that ( m = -2 ) is the correct answer.Just to think about it another way, since ( f(x) ) is a quadratic function, which is even if and only if it's symmetric about the y-axis. For a quadratic ( ax^2 + bx + c ), it's even if ( b = 0 ). So, in our case, after expanding ( f(x) ), the coefficient of ( x ) is ( -(2 + m) ). For the function to be even, this coefficient must be zero:[-(2 + m) = 0 implies 2 + m = 0 implies m = -2]That's another way to see it, which also leads to the same conclusion. So, definitely, ( m = -2 ).I think I've covered all the bases here. I expanded the function, set ( f(-x) = f(x) ), solved for ( m ), and even verified it by plugging back into the original function. Plus, I thought about the properties of quadratic functions and their symmetry. Everything points to ( m = -2 ). I don't see any mistakes in my reasoning, so I feel good about this answer.**Final Answer**The value of ( m ) is boxed{-2}.

answer：Alright, so I've got these two algebra problems to solve. Let's take them one at a time. Starting with the first one: (frac{a^2}{a-1} - a - 1). Hmm, okay. I need to simplify this expression. I remember that when you have fractions and whole numbers, it's helpful to combine them by finding a common denominator. So, the first term is (frac{a^2}{a-1}), and then we have (-a - 1). I think I can rewrite (-a - 1) as a fraction with the same denominator as the first term. The denominator of the first term is (a - 1), so I'll need to express (-a - 1) over (a - 1). Let me think: (-a - 1) can be written as (-frac{(a + 1)(a - 1)}{a - 1}). Wait, is that right? Let me check. If I distribute (-1) over (a + 1), I get (-a - 1), which matches. So, yes, that seems correct. Now, I can combine the two fractions: (frac{a^2}{a - 1} - frac{(a + 1)(a - 1)}{a - 1}). Since they have the same denominator, I can subtract the numerators directly. So, the numerator becomes (a^2 - (a + 1)(a - 1)). Let's expand ((a + 1)(a - 1)). That's a difference of squares, so it simplifies to (a^2 - 1). Now, subtracting that from (a^2), we get (a^2 - (a^2 - 1)), which simplifies to (a^2 - a^2 + 1 = 1). So, the entire expression simplifies to (frac{1}{a - 1}). That seems straightforward enough. Moving on to the second problem: (frac{2xy}{x^2 - y^2} div left(frac{1}{x - y} + frac{1}{x + y}right)). Okay, this looks a bit more complex, but I think I can handle it. First, I recognize that (x^2 - y^2) is a difference of squares, so it factors into ((x - y)(x + y)). That might be useful later. The expression involves division of two fractions, so I remember that dividing by a fraction is the same as multiplying by its reciprocal. But before I get to that, I need to simplify the denominator of the division, which is (frac{1}{x - y} + frac{1}{x + y}). To add these two fractions, I need a common denominator. The denominators are (x - y) and (x + y), so the common denominator would be ((x - y)(x + y)). So, I can rewrite each fraction with this common denominator:[frac{1}{x - y} = frac{x + y}{(x - y)(x + y)}]and[frac{1}{x + y} = frac{x - y}{(x - y)(x + y)}]Adding these together:[frac{x + y}{(x - y)(x + y)} + frac{x - y}{(x - y)(x + y)} = frac{(x + y) + (x - y)}{(x - y)(x + y)} = frac{2x}{(x - y)(x + y)}]So, the denominator of the original division simplifies to (frac{2x}{(x - y)(x + y)}). Now, going back to the original expression:[frac{2xy}{x^2 - y^2} div frac{2x}{(x - y)(x + y)}]Since (x^2 - y^2 = (x - y)(x + y)), we can rewrite the first fraction as:[frac{2xy}{(x - y)(x + y)}]So, now the expression becomes:[frac{2xy}{(x - y)(x + y)} div frac{2x}{(x - y)(x + y)}]Dividing by a fraction is the same as multiplying by its reciprocal, so:[frac{2xy}{(x - y)(x + y)} times frac{(x - y)(x + y)}{2x}]I notice that ((x - y)(x + y)) cancels out from the numerator and denominator, and the 2 in the numerator and denominator also cancels out, leaving me with:[frac{xy}{x} = y]So, the entire expression simplifies to (y). Let me just double-check my steps to make sure I didn't make any mistakes. Starting from the beginning, I factored correctly, found common denominators, simplified step by step, and everything seems to cancel out nicely. Yeah, I think that's correct.

answer：First, I need to determine the probability that the remainder when ( N^{14} ) is divided by 5 is 1. To do this, I'll analyze the possible remainders of ( N ) when divided by 5, which are 0, 1, 2, 3, and 4.1. **Case 1: ( N equiv 0 pmod{5} )** - If ( N ) is divisible by 5, then ( N^{14} ) will also be divisible by 5. Therefore, ( N^{14} equiv 0 pmod{5} ).2. **Case 2: ( N equiv 1 pmod{5} )** - If ( N ) leaves a remainder of 1 when divided by 5, then ( N^{14} equiv 1^{14} equiv 1 pmod{5} ).3. **Case 3: ( N equiv 2 pmod{5} )** - Calculate ( 2^{14} ): - ( 2^2 = 4 ) - ( 4^7 = (4^2)^3 times 4 = 16^3 times 4 ) - Since ( 16 equiv 1 pmod{5} ), ( 16^3 equiv 1^3 equiv 1 pmod{5} ) - Therefore, ( 4^7 equiv 1 times 4 equiv 4 pmod{5} ) - So, ( 2^{14} equiv 4 pmod{5} ).4. **Case 4: ( N equiv 3 pmod{5} )** - Calculate ( 3^{14} ): - ( 3^2 = 9 equiv 4 pmod{5} ) - ( 4^7 equiv (-1)^7 equiv -1 equiv 4 pmod{5} ) - So, ( 3^{14} equiv 4 pmod{5} ).5. **Case 5: ( N equiv 4 pmod{5} )** - Calculate ( 4^{14} ): - ( 4^2 = 16 equiv 1 pmod{5} ) - ( 1^7 equiv 1 pmod{5} ) - So, ( 4^{14} equiv 1 pmod{5} ).From the above cases, ( N^{14} equiv 1 pmod{5} ) when ( N equiv 1 pmod{5} ) or ( N equiv 4 pmod{5} ). There are 2 favorable outcomes out of 5 possible residues.Since each residue class modulo 5 has an equal number of representatives in the range from 1 to 1950, the probability is ( frac{2}{5} ).

answer：First, I need to convert cubic yards to cubic feet because the price is given per cubic foot. I know that 1 cubic yard is equal to 27 cubic feet.Next, I'll multiply the number of cubic yards by the conversion factor to find the equivalent in cubic feet.After obtaining the total cubic feet, I'll multiply this value by the cost per cubic foot to determine the total cost.Finally, I'll present the total cost in a clear and concise manner.

answer：First, I need to determine the number of ways to choose one pet of each type from the available quantities. There are 20 puppies, 10 kittens, 12 hamsters, and 5 rabbits.Next, since each of the four customers wants a different type of pet, I need to calculate the number of ways to assign these four distinct pet types to the four people. This is a permutation problem, and there are 4! (which equals 24) ways to assign the pets.Finally, I multiply the number of ways to choose each pet by the number of ways to assign them to the customers. This gives the total number of ways they can leave the store satisfied.

answer：Okay, so I have this equation: (x^3 + y^3 + 1 = x^2 y^2), and I need to find all integer solutions for (x) and (y). Hmm, let's see. I'm not sure where to start, but maybe I can try plugging in some small integer values for (x) and (y) to see if they satisfy the equation.First, let me consider when (x) or (y) is zero. If (x = 0), the equation becomes (0 + y^3 + 1 = 0), which simplifies to (y^3 = -1). So, (y = -1). That gives me the solution ((0, -1)). Similarly, if (y = 0), the equation becomes (x^3 + 0 + 1 = 0), which means (x^3 = -1), so (x = -1). That gives me another solution ((-1, 0)).Alright, so I have two solutions so far: ((0, -1)) and ((-1, 0)). Now, let's try some positive integers. Let me try (x = 1). Plugging that in, the equation becomes (1 + y^3 + 1 = 1 cdot y^2), which simplifies to (y^3 - y^2 + 2 = 0). Hmm, that doesn't look easy to solve. Maybe I can try small integer values for (y).If (y = 1), then (1 - 1 + 2 = 2 neq 0). If (y = -1), then (-1 - 1 + 2 = 0). Oh, that works! So, when (x = 1), (y = -1) is a solution. That gives me ((1, -1)). Similarly, maybe when (y = 1), (x = -1) is a solution. Let me check: (x^3 + 1 + 1 = x^2 cdot 1), which simplifies to (x^3 - x^2 + 2 = 0). Trying (x = -1), we get (-1 - 1 + 2 = 0), which works. So, ((-1, 1)) is another solution.Now, let's try (x = 2). Plugging that in, the equation becomes (8 + y^3 + 1 = 4y^2), which simplifies to (y^3 - 4y^2 + 9 = 0). Hmm, let's try some integer values for (y). If (y = 3), then (27 - 36 + 9 = 0). That works! So, when (x = 2), (y = 3) is a solution. That gives me ((2, 3)). Similarly, if I try (x = 3), the equation becomes (27 + y^3 + 1 = 9y^2), which simplifies to (y^3 - 9y^2 + 28 = 0). Let me try (y = 2): (8 - 36 + 28 = 0). That works! So, ((3, 2)) is another solution.Wait, so far I have the solutions: ((0, -1)), ((-1, 0)), ((1, -1)), ((-1, 1)), ((2, 3)), and ((3, 2)). Are there any more solutions? Let me try (x = 4). Plugging that in, the equation becomes (64 + y^3 + 1 = 16y^2), which simplifies to (y^3 - 16y^2 + 65 = 0). Trying (y = 5): (125 - 400 + 65 = -210 neq 0). Trying (y = 4): (64 - 256 + 65 = -127 neq 0). Hmm, not working. Maybe (y = 1): (1 - 16 + 65 = 50 neq 0). Doesn't seem like there's an integer solution here.What about negative values for (x) and (y)? Let's try (x = -2). Plugging that in, the equation becomes (-8 + y^3 + 1 = 4y^2), which simplifies to (y^3 - 4y^2 - 7 = 0). Trying (y = 3): (27 - 36 - 7 = -16 neq 0). Trying (y = 4): (64 - 64 - 7 = -7 neq 0). Doesn't seem to work. How about (y = -1): (-1 - 4 - 7 = -12 neq 0). Nope.Let me try (x = -3). The equation becomes (-27 + y^3 + 1 = 9y^2), which simplifies to (y^3 - 9y^2 - 26 = 0). Trying (y = 4): (64 - 144 - 26 = -106 neq 0). Trying (y = 5): (125 - 225 - 26 = -126 neq 0). Doesn't work either.Hmm, maybe there are no more integer solutions beyond the ones I've found. Let me double-check the solutions I have:1. ((0, -1)): (0 + (-1)^3 + 1 = 0 + (-1) + 1 = 0), and (0^2 cdot (-1)^2 = 0). So, (0 = 0). Correct.2. ((-1, 0)): ((-1)^3 + 0 + 1 = -1 + 0 + 1 = 0), and ((-1)^2 cdot 0^2 = 0). So, (0 = 0). Correct.3. ((1, -1)): (1 + (-1)^3 + 1 = 1 - 1 + 1 = 1), and (1^2 cdot (-1)^2 = 1 cdot 1 = 1). So, (1 = 1). Correct.4. ((-1, 1)): ((-1)^3 + 1 + 1 = -1 + 1 + 1 = 1), and ((-1)^2 cdot 1^2 = 1 cdot 1 = 1). So, (1 = 1). Correct.5. ((2, 3)): (8 + 27 + 1 = 36), and (4 cdot 9 = 36). So, (36 = 36). Correct.6. ((3, 2)): (27 + 8 + 1 = 36), and (9 cdot 4 = 36). So, (36 = 36). Correct.All these solutions check out. I don't think there are any other integer solutions because when I tried higher absolute values for (x) and (y), the equation didn't hold. So, I think I've found all the integer solutions.