answer:First, we factor the given polynomial. This equation features all powers of z from z^0 to z^6, allowing us to factor it as follows: [begin{align*} z^7 + z^6 + z^5 + z^4 + z^3 + z^2 + 1 &= z(z^6 + z^5 + z^4 + z^3 + z^2 + z + 1) - z + 1 &= z(z^6 + z^5 + z^4 + z^3 + z^2 + z + 1) + (1 - z). end{align*}] We notice that z^6 + z^5 + z^4 + z^3 + z^2 + z + 1 can be evaluated as frac{z^7 - 1}{z - 1}, which indicates that the roots are the seventh roots of unity except for z = 1. Thus, the roots are operatorname{cis} frac{360^circ}{7} k for k = 1, 2, 3, 4, 5, 6. The roots with a positive imaginary part are those with angles frac{360^circ}{7}, frac{720^circ}{7}, frac{1080^circ}{7}. Adding these angles, we get: [phi = frac{360^circ}{7} + frac{720^circ}{7} + frac{1080^circ}{7} = frac{2160^circ}{7} approx 308.57^circ.] Thus, phi approx boxed{309^circ} when rounded to the nearest degree.

answer:Let's call the three consecutive integers x, x+1, and x+2. Their sum is given as 21. So we can write the equation: x + (x + 1) + (x + 2) = 21 Combining like terms, we get: 3x + 3 = 21 Subtract 3 from both sides: 3x = 18 Divide both sides by 3: x = 6 So the first integer is 6, the second integer is 6 + 1 = 7, and the third integer is 6 + 2 = 8. The greatest of these three consecutive integers is boxed{8} .

answer:Let's denote the student's average grade the year before as x points. From the information given, we can set up the following equations: For the first year (the year before last year), the total points the student received is: 5 courses * x points = 5x For the second year (last year), the total points the student received is: 6 courses * 100 points = 600 points The average grade for the entire two-year period is given as 86 points. The total number of courses over the two years is 5 + 6 = 11 courses. So, the total points over the two years is: 11 courses * 86 points = 946 points Now, we can set up the equation for the total points over the two years: 5x (points from the first year) + 600 (points from the second year) = 946 (total points over two years) Solving for x: 5x + 600 = 946 5x = 946 - 600 5x = 346 x = 346 / 5 x = 69.2 Therefore, the student's average grade the year before was boxed{69.2} points.

answer:**Analysis** This problem mainly examines the general term of a sequence, involving basic knowledge of inequalities. According to the problem, we know frac{a_n}{n}= frac{1}{2}n+ frac{33}{n}- frac{1}{2}. By applying the basic inequality, we can obtain the conclusion. This is a medium-level problem. **Solution** Solution: frac{a_n}{n} =frac{1}{2} n+frac{33}{n} -frac{1}{2}, where frac{1}{2} n+frac{33}{n} geqslant 2sqrt{ frac{1}{2}ncdot frac{33}{n}} =sqrt{66}, Equality holds if and only if frac{1}{2} n=frac{33}{n}, that is, n=sqrt{66}. It is easy to know that 8 < sqrt{66} < 9, and frac{a_8}{8} < frac{a_9}{9}, Therefore, frac{a_n}{n} is minimized when n=8. Hence, the answer is boxed{8}.