answer:Since tan{theta}=2, we can use the identity cos{(2theta+pi)}=-cos{2theta} to rewrite the expression. Then, we use the double angle formula for cosine, which is cos{2theta}=frac{cos^2{theta}-sin^2{theta}}{cos^2{theta}+sin^2{theta}}. This can be further simplified using the identity tan{theta}=frac{sin{theta}}{cos{theta}}, which gives us cos{2theta}=frac{1-tan^2{theta}}{1+tan^2{theta}}. Substituting the given value of tan{theta}=2 into this equation, we get: cos{2theta}=frac{1-tan^2{theta}}{1+tan^2{theta}}=frac{1-2^2}{1+2^2}=frac{-3}{5} Therefore, cos{(2theta+pi)}=-cos{2theta}=boxed{frac{3}{5}}. This problem primarily tests the understanding and application of basic trigonometric identities and double angle formulas.

answer:The probability of rolling a number other than 1 on one die is frac{5}{6}. Therefore, the probability of rolling zero 1's on three dice is left(frac{5}{6}right)^3 = frac{125}{216}. The probability of rolling a 1 on one die is frac{1}{6}, so the probability of rolling three 1's on three dice is left(frac{1}{6}right)^3 = frac{1}{216}. The probability of rolling exactly one 1 on three dice can be calculated by considering one die rolls a 1 and the other two do not. This probability is binom{3}{1} left(frac{1}{6}right)left(frac{5}{6}right)^2 = 3 times frac{1}{6} times left(frac{5}{6}right)^2 = frac{3 times 25}{216} = frac{75}{216}. Similarly, the probability of rolling exactly two 1's on three dice is binom{3}{2} left(frac{1}{6}right)^2 left(frac{5}{6}right) = 3 times left(frac{1}{6}right)^2 times frac{5}{6} = frac{3 times 5}{216} = frac{15}{216}. Now, calculate the expected number of 1's: [ E = 0 times frac{125}{216} + 1 times frac{75}{216} + 2 times frac{15}{216} + 3 times frac{1}{216} = frac{0 + 75 + 30 + 3}{216} = frac{108}{216} = boxed{frac{1}{2}} ]

answer:From (I), we have frac{a_n}{n} + 1 = 2^n, which can be rearranged to give a_n = n cdot 2^n - n. Let the sum of the first n terms of the sequence ncdot 2^n be T_n. So we have: T_n = 2 + 2 cdot 2^2 + 3 cdot 2^3 + ldots + n cdot 2^n, Multiply this by 2 to get: 2T_n = 2^2 + 2 cdot 2^3 + ldots + (n-1) cdot 2^n + n cdot 2^{n+1}. Subtracting these two equations, we find: -T_n = -2 - 2^2 - ldots - 2^n + n cdot 2^{n+1} = -frac{2(2^n - 1)}{2 - 1} + n cdot 2^{n+1}, which simplifies to: T_n = (n - 1) cdot 2^{n+1} + 2. Finally, we need to subtract frac{n(n+1)}{2} from T_n to get S_n because S_n is the sum of a_n which has -n in its expression: S_n = T_n - frac{n(n+1)}{2} = (n - 1) cdot 2^{n+1} + 2 - frac{n(n+1)}{2}. Thus, the sum of the first n terms of the sequence {a_n} is given by: boxed{S_n = (n - 1) cdot 2^{n+1} + 2 - frac{n(n+1)}{2}}.

answer:1. **Identify the fractions of the trip**: Jane's trip is split into three parts: - The first part is frac{1}{4} of the total trip. - The last part is frac{1}{6} of the total trip. - The middle part is given as 30 miles. 2. **Calculate the fraction for the middle part**: To find the fraction of the trip that the middle part (city streets) represents, we subtract the first and last fractions from 1: [ 1 - frac{1}{4} - frac{1}{6} = frac{12}{12} - frac{3}{12} - frac{2}{12} = frac{7}{12} ] Thus, the middle part is frac{7}{12} of the total trip. 3. **Set up the equation**: Let x be the total length of the trip in miles. Since the middle part is frac{7}{12} of the trip and equals 30 miles, we can write: [ frac{7}{12}x = 30 ] 4. **Solve for x**: To find x, solve the equation: [ x = frac{30 times 12}{7} = frac{360}{7} ] Conclusion with boxed answer: The total length of Jane's trip is frac{360{7}} miles. The final answer is boxed{textbf{(C)} frac{360}{7}}