answer：Okay, so I have this math problem here: If y is greater than 0, what percent is (1y)/20 + (3y)/10 of y? Hmm, let me try to figure this out step by step. First, I need to understand what the problem is asking. It wants to know what percentage the sum of (1y)/20 and (3y)/10 is of y. So, essentially, I need to find out what percent (1y/20 + 3y/10) is of y.Alright, let's start by simplifying the expression (1y)/20 + (3y)/10. To add these two fractions, I need a common denominator. The denominators here are 20 and 10. I know that 20 is a multiple of 10, so I can convert (3y)/10 to have a denominator of 20. To do that, I multiply both the numerator and the denominator of (3y)/10 by 2. That gives me (6y)/20. Now, my expression becomes (1y)/20 + (6y)/20. Adding these together, I get (1y + 6y)/20, which simplifies to (7y)/20. So, the sum of (1y)/20 and (3y)/10 is (7y)/20.Now, I need to find what percent (7y)/20 is of y. To find a percentage, I can use the formula:Percentage = (Part / Whole) * 100In this case, the "Part" is (7y)/20 and the "Whole" is y. So, plugging these into the formula, I get:Percentage = ((7y)/20) / y * 100Simplifying this, the y in the numerator and the y in the denominator cancel out, leaving me with:Percentage = (7/20) * 100Now, I just need to calculate 7 divided by 20. 7 divided by 20 equals 0.35. To convert this to a percentage, I multiply by 100:0.35 * 100 = 35%So, (1y)/20 + (3y)/10 is 35% of y.Wait, let me double-check my steps to make sure I didn't make any mistakes. 1. I found a common denominator for the fractions, which was 20.2. I converted (3y)/10 to (6y)/20.3. I added (1y)/20 and (6y)/20 to get (7y)/20.4. I set up the percentage formula with (7y)/20 as the part and y as the whole.5. The y's canceled out, leaving 7/20.6. I converted 7/20 to a decimal, which is 0.35.7. Multiplying by 100 gave me 35%.Everything seems to check out. I think I did this correctly. So, the final answer is 35%.

answer：First, I need to determine the elements of set ( N ) by applying the transformation ( y = 2x + 1 ) to each element in set ( M ).For each ( x ) in ( M ):- When ( x = -1 ), ( y = 2(-1) + 1 = -1 ).- When ( x = 0 ), ( y = 2(0) + 1 = 1 ).- When ( x = 1 ), ( y = 2(1) + 1 = 3 ).- When ( x = 2 ), ( y = 2(2) + 1 = 5 ).Thus, ( N = {-1, 1, 3, 5} ).Next, I find the intersection ( M cap N ) by identifying the common elements in both sets:- ( -1 ) is present in both ( M ) and ( N ).- ( 1 ) is also present in both ( M ) and ( N ).- ( 3 ) and ( 5 ) are not in ( M ), so they are not included in the intersection.Therefore, the intersection ( M cap N ) is ( {-1, 1} ).

answer：First, I recognize that I need to find the length of side ( a ) in triangle ( ABC ) where angle ( A ) is ( 60^circ ), side ( c ) is ( sqrt{3} ), and side ( b ) is ( 2sqrt{3} ).I recall that the Law of Cosines is the appropriate tool for this problem. The formula for the Law of Cosines is:[a^2 = b^2 + c^2 - 2bc cos A]Next, I substitute the given values into the formula:[a^2 = (2sqrt{3})^2 + (sqrt{3})^2 - 2 times 2sqrt{3} times sqrt{3} times cos 60^circ]I calculate each term step by step:[(2sqrt{3})^2 = 4 times 3 = 12][(sqrt{3})^2 = 3][2 times 2sqrt{3} times sqrt{3} = 2 times 2 times 3 = 12][cos 60^circ = 0.5]Substituting these values back into the equation:[a^2 = 12 + 3 - 12 times 0.5][a^2 = 15 - 6 = 9]Finally, I take the square root of both sides to find ( a ):[a = sqrt{9} = 3]

answer：Okay, so I have this problem where I need to use 1024 small cubes, each with an edge length of 1, to form a rectangular prism. The volume of this prism is 1024, which makes sense because 1024 small cubes each with volume 1 will add up to 1024. After forming this prism, I need to color all six faces of it. Then, I have to find the maximum number of small cubes that are not colored on any of their faces.First, I need to figure out the dimensions of the rectangular prism. Since the volume is 1024, I need to find three integers a, b, and c such that a * b * c = 1024. Also, since we're dealing with a rectangular prism, the dimensions can be any combination of factors of 1024.I know that 1024 is 2^10, so the factors are all powers of 2. Let me list some possible dimensions:- 1 x 1 x 1024- 1 x 2 x 512- 1 x 4 x 256- 1 x 8 x 128- 1 x 16 x 64- 1 x 32 x 32- 2 x 2 x 256- 2 x 4 x 128- 2 x 8 x 64- 2 x 16 x 32- 4 x 4 x 64- 4 x 8 x 32- 4 x 16 x 16- 8 x 8 x 16These are all the possible rectangular prisms with volume 1024. Now, the question is asking for the maximum number of small cubes that are not colored on any of their faces after coloring all six faces. So, these cubes must be inside the prism, not on any of the outer layers.To find the number of such cubes, I need to consider the inner dimensions of the prism. If I remove one layer from each side of the prism, the remaining inner prism will consist of cubes that are not on any face. So, if the original dimensions are a x b x c, the inner dimensions will be (a-2) x (b-2) x (c-2), assuming a, b, c are all greater than or equal to 2.Therefore, the number of uncolored cubes is (a-2)*(b-2)*(c-2). So, to maximize this number, I need to choose the dimensions a, b, c such that (a-2)*(b-2)*(c-2) is as large as possible.Looking at the list of possible dimensions, let's compute (a-2)*(b-2)*(c-2) for each:1. 1 x 1 x 1024: (1-2)*(1-2)*(1024-2) = negative numbers, which don't make sense, so discard.2. 1 x 2 x 512: Similarly, negative, discard.3. 1 x 4 x 256: Negative, discard.4. 1 x 8 x 128: Negative, discard.5. 1 x 16 x 64: Negative, discard.6. 1 x 32 x 32: Negative, discard.7. 2 x 2 x 256: (0)*(0)*(254) = 0, discard.8. 2 x 4 x 128: (0)*(2)*(126) = 0, discard.9. 2 x 8 x 64: (0)*(6)*(62) = 0, discard.10. 2 x 16 x 32: (0)*(14)*(30) = 0, discard.11. 4 x 4 x 64: (2)*(2)*(62) = 24812. 4 x 8 x 32: (2)*(6)*(30) = 36013. 4 x 16 x 16: (2)*(14)*(14) = 39214. 8 x 8 x 16: (6)*(6)*(14) = 504So, the maximum number of uncolored cubes is 504 when the dimensions are 8 x 8 x 16.Wait, let me double-check that. If the prism is 8 x 8 x 16, then removing one layer from each side gives 6 x 6 x 14, which is indeed 6*6*14=504. That seems correct.Is there a way to get a larger number? Let me see. If I choose dimensions closer to each other, maybe that would help? For example, if I have a cube, which is as close to a cube as possible, that might maximize the inner volume. But 1024 is not a perfect cube. The cube root of 1024 is approximately 10.07, which is not an integer. So, the closest we can get is 8 x 8 x 16, which is what I have.Alternatively, if I choose 16 x 16 x 4, that would give (14)*(14)*(2)=392, which is less than 504. Similarly, 32 x 32 x 1 would give negative, which is invalid.Therefore, 8 x 8 x 16 seems to be the optimal dimensions to maximize the number of uncolored cubes.So, the maximum number of small cubes that are not colored on any of their faces is 504.

answer：Okay, so I need to find the center of this hyperbola given by the equation (frac{(4x-8)^2}{9^2} - frac{(5y+5)^2}{7^2} = 1). Hmm, I remember that the standard form of a hyperbola is something like (frac{(x-h)^2}{a^2} - frac{(y-k)^2}{b^2} = 1) for a horizontally oriented hyperbola, where ((h, k)) is the center. So, I guess I need to rewrite the given equation to match that standard form.First, looking at the given equation, I see that both the (x) and (y) terms are inside squared terms, which is good. The (x) term is (4x - 8), and the (y) term is (5y + 5). I think I need to factor out the coefficients of (x) and (y) to make it look like the standard form.Starting with the (x) term: (4x - 8). I can factor out a 4, which gives me (4(x - 2)). So, ((4x - 8)^2) becomes ((4(x - 2))^2). Similarly, for the (y) term: (5y + 5). I can factor out a 5, which gives me (5(y + 1)). So, ((5y + 5)^2) becomes ((5(y + 1))^2).Now, substituting these back into the equation, it becomes (frac{(4(x - 2))^2}{9^2} - frac{(5(y + 1))^2}{7^2} = 1). Let me simplify the squares:(frac{16(x - 2)^2}{81} - frac{25(y + 1)^2}{49} = 1).Hmm, this still doesn't look exactly like the standard form because of the coefficients 16 and 25 in the numerators. I think I need to adjust the denominators to account for these coefficients so that the equation matches the standard form perfectly.To do that, I can divide both terms by their respective coefficients. For the (x) term, I have (frac{16(x - 2)^2}{81}), so if I divide both the numerator and the denominator by 16, it becomes (frac{(x - 2)^2}{frac{81}{16}}). Similarly, for the (y) term, (frac{25(y + 1)^2}{49}) becomes (frac{(y + 1)^2}{frac{49}{25}}) after dividing numerator and denominator by 25.So, the equation now looks like:(frac{(x - 2)^2}{left(frac{81}{16}right)} - frac{(y + 1)^2}{left(frac{49}{25}right)} = 1).Now, comparing this with the standard form (frac{(x - h)^2}{a^2} - frac{(y - k)^2}{b^2} = 1), I can see that (h = 2) and (k = -1). Therefore, the center of the hyperbola is at the point ((2, -1)).Wait, let me double-check that. The equation after simplifying is indeed in the standard form, and the terms inside the squares are ((x - 2)) and ((y + 1)), which correspond to ((x - h)) and ((y - k)). So, (h) is 2 and (k) is -1. Yeah, that seems right.I think I got it. The center is at (2, -1). To make sure, maybe I can plug in these values into the original equation and see if it makes sense. If I substitute (x = 2) and (y = -1) into the original equation, the terms inside the squares become zero, which should satisfy the equation since (0 - 0 = 0), but the equation equals 1. Hmm, that doesn't seem right. Wait, no, actually, plugging in the center into the equation should not necessarily satisfy the equation because the center is not a point on the hyperbola. The hyperbola is centered there, but it doesn't pass through that point. So, maybe that's why plugging in (2, -1) doesn't satisfy the equation. I think I was confused for a second there, but I remember that the center isn't on the hyperbola itself.So, yeah, I think I'm confident that the center is at (2, -1). I don't see any mistakes in my steps. I factored out the coefficients correctly, adjusted the denominators, and compared it to the standard form. Everything checks out.

answer：First, determine Marika's father's age when she was 10 years old in 2006. Since his age was five times Marika's age, he was (5 times 10 = 50) years old.Let (x) represent the number of years after 2006 when the father's age will be twice Marika's age. In (x) years, Marika will be (10 + x) years old, and her father will be (50 + x) years old.Set up the equation to find when the father's age is twice Marika's age:[50 + x = 2(10 + x)]Solve the equation:[50 + x = 20 + 2x 50 = 20 + x x = 30]Therefore, the year when Marika’s father will be twice her age is (2006 + 30 = 2036).