answer:1. Since f(x) = begin{cases} 1+ frac{1}{x} & (x>1) x^{2}+1 & (-1leq xleq 1) 2x+3 & (x<-1)end{cases}, we have f(-2) = -1, f[f(-2)] = f(-1) = 2, thus f{f[f(-2)]} = 1+ frac{1}{2} = frac{3}{2}. So, boxed{frac{3}{2}} is the answer for the first question. 2. Since f(a) = frac{3}{2}, it implies that a>1 or -1leq aleq 1. When a>1, we have 1+ frac{1}{a} = frac{3}{2}, thus a=2; When -1leq aleq 1, we have a^{2}+1 = frac{3}{2}, thus a=pm frac{sqrt{2}}{2}. Therefore, a=2 or a=pm frac{sqrt{2}}{2}. So, boxed{a=2 text{ or } a=pm frac{sqrt{2}}{2}} is the answer for the second question.

answer:First, we need to find the coordinates of vectors ( overrightarrow{AB} ) and ( overrightarrow{CA} ): overrightarrow{AB} = B - A = (-1 - 1, 0 - 1, 4 - 1) = (-2, -1, 3) overrightarrow{CA} = A - C = (1 - 2, 1 - (-2), 1 - 3) = (-1, 3, -2) The cosine of the angle between two vectors in space can be found using the dot product formula: cos(angle(overrightarrow{AB}, overrightarrow{CA})) = frac{overrightarrow{AB} cdot overrightarrow{CA}}{|overrightarrow{AB}| times |overrightarrow{CA}|} Calculating the dot product of ( overrightarrow{AB} ) and ( overrightarrow{CA} ): overrightarrow{AB} cdot overrightarrow{CA} = (-2) times (-1) + (-1) times 3 + 3 times (-2) = 2 - 3 - 6 = -7 The magnitude of the vectors are: |overrightarrow{AB}| = sqrt{(-2)^2 + (-1)^2 + 3^2} = sqrt{4 + 1 + 9} = sqrt{14} |overrightarrow{CA}| = sqrt{(-1)^2 + 3^2 + (-2)^2} = sqrt{1 + 9 + 4} = sqrt{14} Now we can find the cosine of the angle ( theta ): cos(angle(overrightarrow{AB}, overrightarrow{CA})) = frac{-7}{sqrt{14} times sqrt{14}} = frac{-7}{14} = -frac{1}{2} From here, we can conclude that ( theta = 120^circ ) since ( cos(120^circ) = -frac{1}{2} ). Therefore, the angle ( theta ) between ( overrightarrow{AB} ) and ( overrightarrow{CA} ) is (boxed{120^circ}).

answer:Let's denote the cost of an advanced ticket as ( x ) dollars. We know that 672 tickets were sold at the door, each costing 22.00, so the total revenue from door tickets is ( 672 times 22 ). The remaining tickets sold were advanced tickets. Since a total of 800 tickets were sold, the number of advanced tickets sold is ( 800 - 672 ). The total revenue from advanced tickets is then ( (800 - 672) times x ). The total revenue from both advanced and door tickets is given as 16,640. So, we can set up the equation: ( (800 - 672) times x + 672 times 22 = 16,640 ) Now, let's solve for ( x ): ( 128x + 672 times 22 = 16,640 ) ( 128x + 14,784 = 16,640 ) Subtract 14,784 from both sides: ( 128x = 16,640 - 14,784 ) ( 128x = 1,856 ) Now, divide both sides by 128 to solve for ( x ): ( x = frac{1,856}{128} ) ( x = 14.5 ) Therefore, the cost of an advanced ticket is boxed{14.50} .

answer:If the ratio of the side lengths of δADC to δABC is 1:2, then every side of δADC is half the length of the corresponding side of δABC. Since δABC is an equilateral triangle, all its sides are equal. Let's denote the side length of δABC as s. Then, the side length of δADC would be s/2. The perimeter of a triangle is the sum of its side lengths. For δADC, which is also an equilateral triangle (since it's similar to δABC), the perimeter is 3 times the length of one side. Given that the perimeter of δADC is 9 + 3√3, we can write: Perimeter of δADC = 3 * (s/2) = 9 + 3√3 Now, let's solve for s: 3 * (s/2) = 9 + 3√3 s/2 = (9 + 3√3) / 3 s/2 = 3 + √3 Now, multiply both sides by 2 to find the side length of δABC: s = 2 * (3 + √3) s = 6 + 2√3 Now that we have the side length of δABC, we can find its perimeter by multiplying this side length by 3 (since it's an equilateral triangle): Perimeter of δABC = 3 * s Perimeter of δABC = 3 * (6 + 2√3) Perimeter of δABC = 18 + 6√3 Therefore, the perimeter of equilateral triangle δABC is boxed{18} + 6√3.