answer:We are tasked with finding the range of the function ( f(x) = 2 cos x + sin 2x ) where ( x in mathbb{R} ). 1. Begin by expressing the function in simpler terms using trigonometric identities: [ f(x) = 2 cos x + sin(2x) ] Using the identity for (sin 2x), we get: [ sin(2x) = 2 sin x cos x ] Therefore, [ f(x) = 2 cos x + 2 sin x cos x = 2 cos x (1 + sin x) ] 2. Next, let's analyze the expression ( 2 cos x (1 + sin x) ). Notice that (cos x) is bounded between (-1) and (1) and (sin x) is similarly bounded. 3. The expression (1 + sin x) ranges between (0) (when (sin x = -1)) and (2) (when (sin x = 1)). 4. Now, we consider the behavior of (f(x)) by examining the product: [ f(x) = 2 cos x (1 + sin x) ] 5. To find the maximum and minimum values, we evaluate the critical points and special values within the trigonometric constraints. We consider ( sin x = frac{1}{2} ): - When (sin x = frac{1}{2}), we use ( cos x ) values of (cos x = pm frac{sqrt{3}}{2}): [ f(x) = 2 left(frac{sqrt{3}}{2}right) left(1 + frac{1}{2}right) = frac{3 sqrt{3}}{2} quad text{(maximum value)} ] - Similarly, [ f(x) = 2 left(-frac{sqrt{3}}{2}right) left(1 + frac{1}{2}right) = -frac{3 sqrt{3}}{2} quad text{(minimum value)} ] 6. The extreme values of (f(x)) are (frac{3sqrt{3}}{2}) and (-frac{3sqrt{3}}{2}). # Conclusion: The range of the function ( f(x) = 2 cos x + sin 2x ) is: [ boxed{left[-frac{3sqrt{3}}{2}, frac{3sqrt{3}}{2}right]} ]

answer:Since BD:DC = 2:1, we have: [ overrightarrow{D} = frac{2}{3} overrightarrow{C} + frac{1}{3} overrightarrow{B}. ] For AE:EC = 3:2, we have: [ overrightarrow{E} = frac{3}{5} overrightarrow{A} + frac{2}{5} overrightarrow{C}. ] Express overrightarrow{C} from each equation: [ overrightarrow{C} = frac{3}{2} overrightarrow{D} - frac{1}{2} overrightarrow{B} = frac{5}{2} overrightarrow{E} - frac{3}{2} overrightarrow{A}. ] Equating and simplifying, we find: [ 6 overrightarrow{D} + 3 overrightarrow{B} = 10 overrightarrow{E} - 6 overrightarrow{A}. ] Simplifying further: [ 6 overrightarrow{D} + 6 overrightarrow{A} = 10 overrightarrow{E} - 3 overrightarrow{B}. ] [ frac{6}{13} overrightarrow{D} + frac{6}{13} overrightarrow{A} = frac{10}{13} overrightarrow{E} - frac{3}{13} overrightarrow{B}. ] These coefficients lie on AD and BE, so this vector represents overrightarrow{P}. Thus: [ overrightarrow{P} = frac{6}{13} left(frac{2}{3} overrightarrow{C} + frac{1}{3} overrightarrow{B} right) + frac{6}{13} overrightarrow{A} = frac{6}{13} overrightarrow{A} + frac{2}{13} overrightarrow{B} + frac{4}{13} overrightarrow{C}. ] Therefore, (x, y, z) = boxed{left(frac{6}{13}, frac{2}{13}, frac{4}{13}right)}.

answer:Given i(z-1)=1, where i is the imaginary unit, we aim to find the value of overline{z}+z. First, we solve for z: begin{align*} i(z-1) &= 1 z-1 &= frac{1}{i} &= -i quad text{(since frac{1}{i} = -i)} z &= 1-i. end{align*} Next, we find the conjugate of z, overline{z}: begin{align*} z &= 1-i overline{z} &= 1+i. end{align*} Now, we calculate overline{z}+z: begin{align*} overline{z}+z &= (1+i) + (1-i) &= 1+i+1-i &= 2. end{align*} Therefore, the correct answer is boxed{A}.

answer:If Brenda wants to create 3 groups of erasers, with each group containing 90 erasers, then the total number of erasers she has can be calculated by multiplying the number of groups by the number of erasers in each group. Total number of erasers = Number of groups × Number of erasers in each group Total number of erasers = 3 × 90 Total number of erasers = 270 Therefore, Brenda has a total of boxed{270} erasers.