answer：First, I need to identify the coordinates of the vertices of trapezoid EFGH: E(0,0), F(0,3), G(5,0), and H(5,7).Next, I'll plot these points on a coordinate plane to visualize the shape. Points E and F are on the y-axis, while points G and H are on the line x=5. This indicates that sides EF and GH are vertical, and sides EG and FH are horizontal.To calculate the area of the trapezoid, I'll use the formula for the area of a trapezoid: [ text{Area} = frac{1}{2} times (b_1 + b_2) times h ]where ( b_1 ) and ( b_2 ) are the lengths of the two parallel sides, and ( h ) is the height (the distance between these two bases).First, I'll determine the lengths of the parallel sides:- The length of side EF is the difference in the y-coordinates of points E and F: ( 3 - 0 = 3 ) units.- The length of side GH is the difference in the y-coordinates of points G and H: ( 7 - 0 = 7 ) units.Next, I'll find the height of the trapezoid, which is the horizontal distance between the two vertical sides. This is the difference in the x-coordinates of points E and G: ( 5 - 0 = 5 ) units.Now, I'll plug these values into the area formula:[ text{Area} = frac{1}{2} times (3 + 7) times 5 = frac{1}{2} times 10 times 5 = 25 ]Therefore, the area of trapezoid EFGH is 25 square units.

answer：Okay, so I'm trying to understand this problem about the Cantor set and its unique primitive representation. The Cantor set ( C ) is defined as the set of non-negative integers whose ternary (base-3) representation contains no 1s. That means every number in ( C ) can only have 0s and 2s in their ternary digits. For example, numbers like 0, 2, 3, 6, 8, 9, etc., are in ( C ) because their ternary representations don't include the digit 1.The problem mentions a similarity transformation defined for a set ( A subseteq mathbf{Z} ) by ( aA + b ). I think this means scaling the set ( A ) by a factor of ( a ) and then shifting it by ( b ). So, if ( A ) is a set of integers, then ( aA + b ) would be all numbers of the form ( a cdot x + b ) where ( x ) is in ( A ).The task is to show four things about the unique primitive representation of ( C ):1. ( a_i > 1 ) for all ( i );2. All ( a_i ) are powers of 3;3. ( a_i > b_i ) for all ( i );4. The unique primitive representation of ( C ) is ( C = (3C) cup (3C + 2) ).Let me try to tackle each part one by one.**1. ( a_i > 1 ) for all ( i ):**Hmm, so I need to show that in the primitive representation of ( C ), each scaling factor ( a_i ) is greater than 1. First, what is a primitive representation? I think it means that the representation cannot be broken down further into smaller similarity transformations. So, ( C ) is expressed as a union of scaled and shifted copies of itself, and these scaling factors ( a_i ) are as small as possible.Given that ( C ) is a fractal, it's self-similar, meaning it can be expressed in terms of smaller copies of itself. In the case of the Cantor set, it's typically constructed by removing the middle third each time, so intuitively, the scaling factor should be 3, which is greater than 1. But why can't ( a_i ) be 1? If ( a_i = 1 ), then the transformation would just be a shift ( b_i ). But if we have ( C = bigcup (1 cdot C + b_i) ), that would imply that ( C ) is just a union of shifted copies of itself. However, since ( C ) is already defined by numbers without 1s in their ternary expansions, shifting it by some ( b_i ) wouldn't necessarily cover all of ( C ) unless the shifts are appropriately chosen. But more importantly, if ( a_i = 1 ), then the transformation doesn't actually scale the set; it just shifts it. Since the Cantor set is not just a union of shifted copies of itself but actually requires scaling down by a factor of 3 each time, the scaling factor ( a_i ) must be greater than 1. So, ( a_i > 1 ) makes sense.**2. All ( a_i ) are powers of 3:**This seems related to the self-similar structure of the Cantor set. The Cantor set is constructed by repeatedly removing the middle third, so each step involves scaling by 1/3. But in terms of the similarity transformations, the scaling factor is 3 because we're looking at how the set can be expressed as a union of scaled copies.If we think about the ternary representation, each digit corresponds to a scaling by a power of 3. So, to build the Cantor set, we need to consider scaling by 3, 9, 27, etc., which are all powers of 3. Suppose ( a_i ) wasn't a power of 3. Then, the scaling factor would introduce numbers that don't align with the ternary structure of ( C ). For example, if ( a_i = 2 ), then scaling ( C ) by 2 would produce numbers that might have 1s in their ternary expansions, which aren't in ( C ). Therefore, only scaling by powers of 3 preserves the structure of ( C ).**3. ( a_i > b_i ) for all ( i ):**This condition ensures that the shift ( b_i ) doesn't interfere with the scaling in a way that would cause overlaps or misalignments in the ternary digits. If ( a_i ) were less than or equal to ( b_i ), then the shifted copies might overlap incorrectly or not fit properly within the structure of ( C ). Specifically, since ( C ) is built by scaling and shifting without overlapping, the shift must be smaller than the scaling factor to maintain the integrity of the set.For example, if ( a_i = 3 ), then ( b_i ) can be 0 or 2, both of which are less than 3. If ( b_i ) were 3 or more, it would start overlapping with the next scaled copy, which isn't allowed in the construction of ( C ).**4. The unique primitive representation of ( C ) is ( C = (3C) cup (3C + 2) ):**This is the classic self-similar decomposition of the Cantor set. The set ( C ) can be expressed as two scaled and shifted copies of itself: one scaled by 3 and shifted by 0, and the other scaled by 3 and shifted by 2.In terms of ternary expansions, scaling by 3 shifts all digits one place to the left (multiplying by 3), and adding 2 adds a 2 in the least significant digit. This corresponds exactly to the construction of the Cantor set, where each step involves keeping the numbers that start with 0 or 2 in each ternary digit.So, putting it all together, the unique primitive representation must involve scaling by 3 and shifting by 0 or 2, which gives us ( C = (3C) cup (3C + 2) ).**Summary of Thoughts:**- The Cantor set's self-similar structure inherently requires scaling by powers of 3 because of its ternary construction.- The scaling factors must be greater than 1 to ensure proper scaling without overlap.- The shifts must be smaller than the scaling factors to maintain the correct alignment in the ternary digits.- The unique primitive representation is the simplest form that captures the entire set without redundancy, which is achieved by scaling by 3 and shifting by 0 and 2.I think I've got a reasonable grasp of why each of these points holds. The key is understanding the self-similar nature of the Cantor set and how scaling and shifting relate to its ternary representation. Each step in the construction corresponds to a similarity transformation, and the conditions on ( a_i ) and ( b_i ) ensure that these transformations correctly build up the set without introducing numbers that shouldn't be there.One thing I'm still a bit fuzzy on is the exact definition of a "primitive representation." I think it refers to the representation that can't be broken down further into smaller similarity transformations, meaning the scaling factors are as small as possible. In this case, since the smallest scaling factor is 3, and you can't go smaller without losing the self-similar structure, this must be the primitive representation.Also, I wonder if there's a more formal way to prove each of these points, maybe using induction or properties of ternary expansions. For example, to show that ( a_i > 1 ), perhaps assume ( a_i = 1 ) and reach a contradiction by showing that the set can't be covered properly. Similarly, for the powers of 3, maybe use properties of the ternary digits and how scaling affects them.Overall, I feel like I understand the intuition behind each part, but to make it rigorous, I'd need to formalize these arguments, possibly using the properties of the Cantor set and its ternary representation.

answer：First, I need to determine how many miles Tripp and Charlotte hiked on the first day. The total hike is 36 miles, and they have 27 miles remaining.To find the miles hiked on the first day, I subtract the remaining miles from the total hike:36 miles (total hike) - 27 miles (remaining) = 9 miles.Therefore, they hiked 9 miles on the first day.

answer：First, I need to determine Leontina's age based on the given condition. Let ( d ) represent the tens digit and ( u ) represent the units digit of her age. According to the problem, half of her age equals twice the sum of its digits, which translates to the equation:[frac{10d + u}{2} = 2(d + u)]Simplifying this equation:[10d + u = 4(d + u)][10d + u - 4d - 4u = 0][6d - 3u = 0][2d = u]This means the units digit ( u ) is twice the tens digit ( d ). Considering the possible values for ( d ) and ( u ) (both digits must be between 0 and 9), the valid ages are 12, 24, 36, and 48 years old.Since Leontina has aged beyond 40 years, her age must be 48 years. If her 48th birthday was on December 28, 1970, then her 40th anniversary would have been on December 28, 1962. Therefore, the missing digits in the year are 6 and 2.

answer：Okay, so I have this equation to solve: ( y^2 + 7y + 10 = -(y + 2)(y + 8) ). Hmm, let me see how to approach this. I remember that when solving quadratic equations, it's usually helpful to get everything on one side so that the equation equals zero. That way, I can try to factor it or use the quadratic formula if needed.First, I should probably expand the right side of the equation to make things clearer. The right side is ( -(y + 2)(y + 8) ). Let me multiply those two binomials first. Multiplying ( (y + 2)(y + 8) ) using the distributive property (FOIL method):- First: ( y times y = y^2 )- Outer: ( y times 8 = 8y )- Inner: ( 2 times y = 2y )- Last: ( 2 times 8 = 16 )Adding those together: ( y^2 + 8y + 2y + 16 = y^2 + 10y + 16 ).But there's a negative sign in front of that, so the right side becomes ( -y^2 - 10y - 16 ).Now, let me rewrite the original equation with this expansion:( y^2 + 7y + 10 = -y^2 - 10y - 16 ).To get everything on one side, I'll add ( y^2 + 10y + 16 ) to both sides:Left side: ( y^2 + 7y + 10 + y^2 + 10y + 16 )Right side: ( -y^2 - 10y - 16 + y^2 + 10y + 16 = 0 )Combining like terms on the left side:( y^2 + y^2 = 2y^2 )( 7y + 10y = 17y )( 10 + 16 = 26 )So, the equation simplifies to:( 2y^2 + 17y + 26 = 0 )Now, I have a quadratic equation in standard form ( ax^2 + bx + c = 0 ), where ( a = 2 ), ( b = 17 ), and ( c = 26 ). I need to solve for ( y ). Let me see if this quadratic can be factored.Factoring a quadratic with a leading coefficient other than 1 can be a bit tricky. I'll try to factor it by looking for two numbers that multiply to ( a times c = 2 times 26 = 52 ) and add up to ( b = 17 ).Let me list the factors of 52:- 1 and 52 (sum 53)- 2 and 26 (sum 28)- 4 and 13 (sum 17)Oh, there we go! 4 and 13 multiply to 52 and add up to 17. So, I can use these numbers to break up the middle term.Rewriting the quadratic:( 2y^2 + 4y + 13y + 26 = 0 )Now, I'll group the terms:( (2y^2 + 4y) + (13y + 26) = 0 )Factor out the greatest common factor from each group:- From the first group: ( 2y(y + 2) )- From the second group: ( 13(y + 2) )So, the equation becomes:( 2y(y + 2) + 13(y + 2) = 0 )Now, I can factor out the common binomial factor ( (y + 2) ):( (y + 2)(2y + 13) = 0 )Setting each factor equal to zero gives the potential solutions:1. ( y + 2 = 0 ) → ( y = -2 )2. ( 2y + 13 = 0 ) → ( 2y = -13 ) → ( y = -frac{13}{2} )So, the solutions are ( y = -2 ) and ( y = -frac{13}{2} ).Wait, let me double-check by plugging these values back into the original equation to ensure they work.First, let's test ( y = -2 ):Left side: ( (-2)^2 + 7(-2) + 10 = 4 - 14 + 10 = 0 )Right side: ( -((-2) + 2)((-2) + 8) = -(0)(6) = 0 )Both sides equal 0, so ( y = -2 ) is a valid solution.Now, testing ( y = -frac{13}{2} ):Left side: ( left(-frac{13}{2}right)^2 + 7left(-frac{13}{2}right) + 10 )Calculating each term:- ( left(-frac{13}{2}right)^2 = frac{169}{4} )- ( 7 times -frac{13}{2} = -frac{91}{2} )- 10 remains as is.So, combining:( frac{169}{4} - frac{91}{2} + 10 )Convert all terms to quarters:( frac{169}{4} - frac{182}{4} + frac{40}{4} = frac{169 - 182 + 40}{4} = frac{27}{4} )Right side: ( -left(-frac{13}{2} + 2right)left(-frac{13}{2} + 8right) )Simplify inside the parentheses:- ( -frac{13}{2} + 2 = -frac{13}{2} + frac{4}{2} = -frac{9}{2} )- ( -frac{13}{2} + 8 = -frac{13}{2} + frac{16}{2} = frac{3}{2} )Multiply them:( -left(-frac{9}{2} times frac{3}{2}right) = -left(-frac{27}{4}right) = frac{27}{4} )Both sides equal ( frac{27}{4} ), so ( y = -frac{13}{2} ) is also a valid solution.Therefore, both solutions satisfy the original equation, and I can confidently say that the solutions are ( y = -2 ) and ( y = -frac{13}{2} ).

answer：Okay, so I have these two algebraic expressions to factorize. Let me take them one by one and see how I can approach each.Starting with the first one: (-3x^{3}y + 6x^{2}y^{2} - 3xy^{3}). Hmm, factorizing this. I remember that the first step in factorizing is usually to look for the greatest common factor (GCF) among all the terms. Let me check each term:- The first term is (-3x^{3}y). The coefficients here are -3, and the variables are (x^3) and (y).- The second term is (6x^{2}y^{2}). The coefficient is 6, and the variables are (x^2) and (y^2).- The third term is (-3xy^{3}). The coefficient is -3, and the variables are (x) and (y^3).Looking at the coefficients: -3, 6, -3. The GCF of these numbers is 3. But since two of them are negative, I need to be careful with the sign. The GCF would actually be -3 because two of the coefficients are negative.Now, for the variables. For (x), the exponents are 3, 2, and 1. The lowest exponent is 1, so the GCF for (x) is (x). For (y), the exponents are 1, 2, and 3. The lowest exponent is 1, so the GCF for (y) is (y). Putting it together, the GCF of all the terms is (-3xy). I can factor this out from each term. Let me do that:[-3x^{3}y + 6x^{2}y^{2} - 3xy^{3} = -3xy(x^{2}) + (-3xy)(-2xy) + (-3xy)(y^{2})]Wait, let me make sure I'm distributing correctly. - Factoring (-3xy) from (-3x^{3}y) gives (x^{2}) because (-3xy times x^{2} = -3x^{3}y).- Factoring (-3xy) from (6x^{2}y^{2}) gives (-2xy) because (-3xy times -2xy = 6x^{2}y^{2}).- Factoring (-3xy) from (-3xy^{3}) gives (y^{2}) because (-3xy times y^{2} = -3xy^{3}).So, putting it all together, we have:[-3xy(x^{2} - 2xy + y^{2})]Now, looking at the quadratic expression inside the parentheses: (x^{2} - 2xy + y^{2}). This looks familiar. It seems like a perfect square trinomial. Let me check:A perfect square trinomial has the form (a^{2} - 2ab + b^{2} = (a - b)^{2}). Comparing this with (x^{2} - 2xy + y^{2}), we can see that (a = x) and (b = y). So, it factors into ((x - y)^{2}).Therefore, substituting back in, the expression becomes:[-3xy(x - y)^{2}]Alright, that seems to be the factorized form for the first expression.Moving on to the second expression: ((a^{2} + 9)^{2} - 36a^{2}). Hmm, this looks like a difference of squares. I remember that the difference of squares formula is (A^{2} - B^{2} = (A - B)(A + B)).Let me identify (A) and (B) here. The first term is ((a^{2} + 9)^{2}), so (A = a^{2} + 9). The second term is (36a^{2}), which is ((6a)^{2}), so (B = 6a).Applying the difference of squares formula:[(a^{2} + 9)^{2} - (6a)^{2} = (a^{2} + 9 - 6a)(a^{2} + 9 + 6a)]So, that gives me two factors: (a^{2} + 9 - 6a) and (a^{2} + 9 + 6a). Let me see if these can be factored further.Looking at (a^{2} + 9 - 6a), I can rearrange it as (a^{2} - 6a + 9). This is another perfect square trinomial. It fits the form (a^{2} - 2ab + b^{2} = (a - b)^{2}), where (a = a) and (b = 3). So, it factors into ((a - 3)^{2}).Similarly, looking at (a^{2} + 9 + 6a), rearranged as (a^{2} + 6a + 9), which is also a perfect square trinomial. It fits the form (a^{2} + 2ab + b^{2} = (a + b)^{2}), where (a = a) and (b = 3). So, it factors into ((a + 3)^{2}).Putting it all together, the original expression becomes:[(a - 3)^{2}(a + 3)^{2}]Wait, that seems a bit off. Let me double-check. The original expression was ((a^{2} + 9)^{2} - 36a^{2}), which we factored into ((a^{2} + 9 - 6a)(a^{2} + 9 + 6a)). Then each of those factors was a perfect square, resulting in ((a - 3)^{2}(a + 3)^{2}). That seems correct.Alternatively, I could think of it as ((a^{2} + 9)^{2} - (6a)^{2}), which is a difference of squares, so it factors into ((a^{2} + 9 - 6a)(a^{2} + 9 + 6a)), and each of those is a perfect square. So, yes, the factorization is correct.So, summarizing:1. The first expression factors into (-3xy(x - y)^{2}).2. The second expression factors into ((a - 3)^{2}(a + 3)^{2}).I think that's it. Let me just quickly verify by expanding them to see if I get back the original expressions.For the first one: (-3xy(x - y)^{2}).First, expand ((x - y)^{2}):[(x - y)^{2} = x^{2} - 2xy + y^{2}]Then multiply by (-3xy):[-3xy(x^{2} - 2xy + y^{2}) = -3x^{3}y + 6x^{2}y^{2} - 3xy^{3}]Which matches the original expression. Good.For the second one: ((a - 3)^{2}(a + 3)^{2}).First, note that ((a - 3)^{2}(a + 3)^{2} = [(a - 3)(a + 3)]^{2}).Calculating ((a - 3)(a + 3)):[(a - 3)(a + 3) = a^{2} - 9]Then squaring that:[(a^{2} - 9)^{2} = a^{4} - 18a^{2} + 81]Wait, but the original expression was ((a^{2} + 9)^{2} - 36a^{2}). Let me compute that:First, expand ((a^{2} + 9)^{2}):[(a^{2} + 9)^{2} = a^{4} + 18a^{2} + 81]Then subtract (36a^{2}):[a^{4} + 18a^{2} + 81 - 36a^{2} = a^{4} - 18a^{2} + 81]Which matches the expansion of ((a - 3)^{2}(a + 3)^{2}). So, that checks out as well.Therefore, I'm confident that the factorizations are correct.