answer:Given that n divided by 6 leaves a remainder of 1, we can express n as n = 6k + 1 for some integer k. When we add 2010 to n, we are interested in finding the remainder when n + 2010 is divided by 6. We can express this operation as follows: [ n + 2010 = 6k + 1 + 2010 ] We know that 2010 can be expressed as a multiple of 6 plus a remainder. Since 2010 is divisible by 6 (as 2010 = 6 cdot 335), we can rewrite the equation as: [ n + 2010 = 6k + 1 + 6 cdot 335 ] Simplifying the equation gives us: [ n + 2010 = 6k + 6 cdot 335 + 1 ] [ n + 2010 = 6(k + 335) + 1 ] This equation shows that when n + 2010 is divided by 6, it leaves a remainder of 1, just as when n is divided by 6. Therefore, the remainder when n + 2010 is divided by 6 is boxed{1}.

answer:1. **Area of the rectangle**: [ text{Area} = 10 times 15 = 150. ] 2. **Area of the square**: Since the two hexagons form a square without overlapping, the area of the square is also 150. Let s be the side of the square. The equation for the area of the square is: [ s^2 = 150. ] Solving for s, we find: [ s = sqrt{150} = 5sqrt{6}. ] 3. **Calculating y**: Since y is defined as half the length of the side of the square: [ y = frac{s}{2} = frac{5sqrt{6}}{2}. ] Therefore, the dimension y is frac{5sqrt{6}{2}}. The final answer is boxed{textbf{(E)} frac{5sqrt{6}}{2}}

answer:To find det (mathbf{A} mathbf{B}), we use the property that the determinant of the product of two matrices is equal to the product of their determinants. This means we can calculate det (mathbf{A} mathbf{B}) as follows: [ det (mathbf{A} mathbf{B}) = (det mathbf{A})(det mathbf{B}) ] Given that det mathbf{A} = 2 and det mathbf{B} = 12, we substitute these values into our equation: [ det (mathbf{A} mathbf{B}) = (2)(12) = 24 ] Therefore, the determinant of the matrix product mathbf{A} mathbf{B} is boxed{24}.

answer:Using the relationships given, we first find tan x tan y. From cot x + cot y = 50, we can rewrite it as frac{1}{tan x} + frac{1}{tan y} = 50. This is equivalent to: [ frac{tan x + tan y}{tan x tan y} = 50. ] Given tan x + tan y = 40, substituting this in gives: [ frac{40}{tan x tan y} = 50 ] From which we obtain: [ tan x tan y = frac{40}{50} = frac{4}{5}. ] Now we apply the tangent of sum formula: [ tan(x+y) = frac{tan x + tan y}{1 - tan x tan y} = frac{40}{1 - frac{4}{5}} = frac{40}{frac{1}{5}} = 200. ] Thus, boxed{tan(x+y) = 200}.