answer:First, rewrite the equation as: [ z^2 + 2z - (16 + 8i) = 0. ] Using the quadratic formula, we find: [ z = frac{-2 pm sqrt{4 + 4(16 + 8i)}}{2} = frac{-2 pm sqrt{68 + 32i}}{2}. ] Let us denote 68 + 32i = (a + bi)^2 where a and b are real numbers: [ a^2 - b^2 + 2abi = 68 + 32i. ] Equating real and imaginary parts, we obtain: [ a^2 - b^2 = 68, ] [ ab = 16. ] We can express (b) in terms of (a:) (b = frac{16}{a}). Substituting, we get: [ a^2 - frac{256}{a^2} = 68. ] Multiplying through by (a^2) to clear the fraction: [ a^4 - 68a^2 - 256 = 0. ] This factors as ((a^2 - 80)(a^2 + 4) = 0.) We discard (a^2 + 4 = 0) as it gives non-real (a). Therefore, (a = pm sqrt{80} = pm 4sqrt{5}.) If (a = 4sqrt{5},) then (b = frac{16}{4sqrt{5}} = frac{4}{sqrt{5}} = frac{4sqrt{5}}{5}.) [ z = frac{-2 pm (4sqrt{5} + frac{4sqrt{5}}{5}i)}{2} = -1 pm 2sqrt{5} + frac{2sqrt{5}}{5}i. ] Similarly, if (a = -4sqrt{5},) then [ z = -1 pm 2sqrt{5} - frac{2sqrt{5}}{5}i. ] Thus, the roots are: [ boxed{(-1 + 2sqrt{5} + frac{2sqrt{5}}{5}i, -1 - 2sqrt{5} - frac{2sqrt{5}}{5}i)} ]

answer:Since (1+x)^{10} = a_0 + a_1(1-x) + a_2(1-x)^2 + ldots + a_{10}(1-x)^{10}, we have (2-(1-x))^{10} = a_0 + a_1(1-x) + a_2(1-x)^2 + ldots + a_{10}(1-x)^{10}. Therefore, a_8 = binom{10}{8}2^2(-1)^8 = 180. The correct answer is boxed{B: 180}.

answer:Let's denote the marks obtained by the student in physics as P, in chemistry as C, and in mathematics as M. According to the information given, the total marks obtained in physics, chemistry, and mathematics is P + C + M. It is also given that this total is 180 more than the marks obtained in physics alone. Therefore, we can write the following equation: P + C + M = P + 180 To find the average marks obtained in chemistry and mathematics, we need to find the sum of the marks in these two subjects and then divide by 2 (since there are two subjects). From the equation above, we can subtract P from both sides to isolate C + M: C + M = 180 Now, to find the average, we divide this sum by 2: Average marks in chemistry and mathematics = (C + M) / 2 Average marks in chemistry and mathematics = 180 / 2 Average marks in chemistry and mathematics = 90 Therefore, the average mark obtained by the student in chemistry and mathematics is boxed{90} .

answer:The given function is y=4sin(2x+frac{pi}{3})+1. The smallest positive period (T) of a sine function in the form y = Asin(Bx + C) + D is given by the formula T=frac{2pi}{|B|}. Here, B=2. So, the smallest positive period is: T=frac{2pi}{|2|}=pi. Therefore, the correct answer is option B: boxed{pi}. This problem primarily tests the understanding of how to determine the smallest positive period of a trigonometric function, making it a fundamental question.