answer:Let's denote the number of smaller boxes (with 12 glasses each) as x and the number of larger boxes (with 16 glasses each) as y. According to the problem, there are 16 more larger boxes than smaller boxes, so we can write: y = x + 16 We are also given that the average number of glasses per box is 15. The average is calculated by dividing the total number of glasses by the total number of boxes. So we can write the equation for the average as: (12x + 16y) / (x + y) = 15 Now we can substitute y from the first equation into the second equation: (12x + 16(x + 16)) / (x + (x + 16)) = 15 Simplify and solve for x: (12x + 16x + 256) / (2x + 16) = 15 28x + 256 = 15(2x + 16) 28x + 256 = 30x + 240 Subtract 28x from both sides: 256 = 2x + 240 Subtract 240 from both sides: 16 = 2x Divide by 2: x = 8 Now we know there are 8 smaller boxes. We can find the number of larger boxes by using the first equation: y = x + 16 y = 8 + 16 y = 24 Now we can calculate the total number of glasses: Total number of glasses in smaller boxes = 12 * 8 = 96 Total number of glasses in larger boxes = 16 * 24 = 384 Total number of glasses r at the restaurant = 96 + 384 = 480 So, the total number of glasses at the restaurant is boxed{480} .

answer:**Step 1**: Identify the roots for the vertical asymptotes. Since vertical asymptotes occur at x=2 and x=-3, these are the roots of the denominator. Thus, the denominator can be factored as: x^2 + ax + b = (x-2)(x+3) **Step 2**: Expand the factored form and equate coefficients: (x-2)(x+3) = x^2 + 3x - 2x - 6 = x^2 + x - 6 Therefore, a = 1 and b = -6. **Step 3**: Check for the horizontal asymptote: The horizontal asymptote is determined by the degrees of the polynomial in the numerator and the denominator. Since the degree of the numerator (x+3) is less than the degree of the denominator (x^2 + ax + b), the horizontal asymptote is y=0, which matches the condition given in the problem. Therefore, the sum of a and b is: a+b = 1 - 6 = boxed{-5} Conclusion: The values a=1 and b=-6 satisfy the conditions about vertical and horizontal asymptotes. The calculated sum of a and b is -5, which is consistent with the solution steps.

answer:1. **Studying the general characteristics of binary representations of natural numbers**: Let's consider any natural number ( n ) in its binary form and aim to analyze it without restricting it to ( 4 leq n leq 1023 ). Denote the count of ( n )-bit numbers that start with 1 and do not have three consecutive identical bits as ( a_n ). 2. **Defining counts with specific ending patterns**: For pairs of digits ( a, b in {0, 1} ), use ( x_{ab}^{n} ) to represent the count of ( n )-bit numbers starting with 1, ending in ( ab ), and not containing three consecutive identical bits. 3. **Recursive relations based on ending patterns**: - Ending with `00`: can only be preceded by `100` [ x_{00}^{n} = x_{10}^{n-1} ] - Ending with `01`: can be preceded by `101` or `001` [ x_{01}^{n} = x_{10}^{n-1} + x_{00}^{n-1} ] - Ending with `10`: can be preceded by `110` or `010` [ x_{10}^{n} = x_{11}^{n-1} + x_{01}^{n-1} ] - Ending with `11`: can only be preceded by `011` [ x_{11}^{n} = x_{01}^{n-1} ] 4. **Summing up different counts**: The total count ( a_n ) can be expressed as: [ a_n = x_{00}^{n} + x_{01}^{n} + x_{10}^{n} + x_{11}^{n} ] 5. **Further recursive equations**: - From prior relations: [ a_{n-1} = x_{00}^{n-1} + x_{01}^{n-1} + x_{10}^{n-1} + x_{11}^{n-1} ] - Substitute earlier into (4) and sum: begin{align*} a_n &= x_{10}^{n-1} + x_{00}^{n-1} + x_{10}^{n-1} + x_{11}^{n-1} + x_{01}^{n-1} + x_{01}^{n-1} &= a_{n-1} + x_{10}^{n-1} + x_{01}^{n-1} &= a_{n-1} + (x_{11}^{n-2} + x_{01}^{n-2}) + (x_{00}^{n-2} + x_{10}^{n-2}) &= a_{n-1} + a_{n-2} end{align*} 6. **Relating to Fibonacci sequence**: This is a recursive Fibonacci sequence relation: [ a_n = a_{n-1} + a_{n-2} ] 7. **Initial terms**: - For 3-bit numbers (covering 101, 100, 110): [ a_3 = 3 ] - For 4-bit numbers (covering 1001, 1010, 1100, 1101, 1011): [ a_4 = 5 ] 8. **Calculating values within the specific range**: begin{align*} a_5 &= a_4 + a_3 = 5 + 3 = 8 a_6 &= a_5 + a_4 = 8 + 5 = 13 a_7 &= a_6 + a_5 = 13 + 8 = 21 a_8 &= a_7 + a_6 = 21 + 13 = 34 a_9 &= a_8 + a_7 = 34 + 21 = 55 a_{10} &= a_9 + a_8 = 55 + 34 = 89 end{align*} 9. **Summing results**: [ a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 + a_{10} = 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 = 228 ] Conclusion: [ boxed{228} ]

answer:Expanding the given equation, we have: [ a^2 + ab - ac - 2a + 2ba + 2b^2 - 2bc - 4b + ca + cb - c^2 - 2c = 4ab. ] Simplifying further, [ a^2 + 3ab - ac + 2b^2 - 2bc - 2a - 4b - c^2 - 2c = 4ab, ] [ a^2 - ab + 2b^2 + ac -2bc - 2a - 4b- c^2 -2c = 0. ] Rearranging to fit the Law of Cosines for c^2, we structure it as: [ c^2 = a^2 + ab + 2b^2 - ac + 2bc - 2a - 4b -2c. ] Applying the Law of Cosines, we get: [ cos C = frac{a^2 + b^2 - c^2}{2ab}, ] Replacing c^2 with the expression derived: [ cos C = frac{a^2 + b^2 - (a^2 + ab + 2b^2 - ac + 2bc - 2a - 4b -2c)}{2ab}, ] [ cos C = frac{-ab - b^2 + ac - 2bc + 2a + 4b + 2c}{2ab}. ] After simplification and common factor reduction, [ cos C = frac{1}{2}, ] Hence, the angle opposite to side c is, [ C = boxed{60^circ}. ]