answer:First, let's simplify the expression by applying the cosine difference identity, which states cos(A - B) = cos A cos B + sin A sin B. We have cos x + cos(frac{pi}{3} - x) = cos x + cos frac{pi}{3} cos x + sin frac{pi}{3} sin x. Now, substitute the known values for cos frac{pi}{3} and sin frac{pi}{3}, which are frac{1}{2} and frac{sqrt{3}}{2}, respectively. cos x + frac{1}{2} cos x + frac{sqrt{3}}{2} sin x = frac{3}{2} cos x + frac{sqrt{3}}{2} sin x. Next, we can apply the sine sum identity, which states sin(A + B) = sin A cos B + cos A sin B. Observe that frac{3}{2} cos x + frac{sqrt{3}}{2} sin x = sqrt{3}(frac{sqrt{3}}{2} cos x + frac{1}{2} sin x) = sqrt{3} sin(x + frac{pi}{3}). Now, we can use the given equation sin(x + frac{pi}{3}) = frac{1}{3} to simplify our expression further. sqrt{3} sin(x + frac{pi}{3}) = sqrt{3} (frac{1}{3}) = boxed{frac{sqrt{3}}{3}}.

answer:# Problem Given a sequence ( {a_n} ) satisfying ( a_1 = 2 ), ( a_n > 0 ) for any ( n in mathbf{N} ), and ((n+1) a_n^2 + a_n a_{n+1} - n a_{n+1}^2 = 0). Also, given a sequence ( {b_n} ) such that ( b_n = 2^{n-1} + 1 ). 1. Find the general term (a_n) and the sum of the first (n) terms (S_n) of the sequence ( {a_n} ). 2. Find the sum of the first (n) terms (T_n) of the sequence ( {b_n} ). 3. Conjecture the relationship between (S_n) and (T_n) and explain the reasoning. (1) Finding the general term (a_n) and the sum (S_n) Given the recurrence relation: [ (n+1)a_n^2 + a_n a_{n+1} - n a_{n+1}^2 = 0 ] First, divide the entire equation by (a_{n+1}^2): [ (n+1) left(frac{a_n}{a_{n+1}}right)^2 + left(frac{a_n}{a_{n+1}}right) - n = 0 ] Let ( x = frac{a_n}{a_{n+1}} ). Then the equation simplifies to: [ (n+1)x^2 + x - n = 0 ] Solve for (x) using the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ): [ x = frac{-(1) pm sqrt{1^2 - 4(n+1)(-n)}}{2(n+1)} ] [ x = frac{-1 pm sqrt{1 + 4n(n+1)}}{2(n+1)} ] [ x = frac{-1 pm sqrt{1 + 4n^2 + 4n}}{2(n+1)} ] [ x = frac{-1 pm (2n + 1)}{2(n+1)} ] Considering ( a_n > 0 ) for ( n in mathbf{N} ), we discard the negative root: [ x = frac{-1 + 2n + 1}{2(n+1)} = frac{2n}{2(n+1)} = frac{n}{n+1} ] Thus: [ frac{a_n}{a_{n+1}} = frac{n}{n+1} implies a_{n+1} = frac{n+1}{n}a_n ] To find the general term (a_n), note that: [ a_n = a_1 cdot frac{n}{1} = 2n ] So, the general term is: [ a_n = 2n ] To find the sum of the first (n) terms ( S_n ): [ S_n = sum_{k=1}^n 2k = 2 left( 1 + 2 + cdots + n right) ] Using the formula for the sum of the first (n) natural numbers, ( sum_{k=1}^n k = frac{n(n+1)}{2} ): [ S_n = 2 left( frac{n(n+1)}{2} right) = n(n+1) = n^2 + n ] Thus: [ S_n = n^2 + n ] (2) Finding the sum (T_n) of the first (n) terms of ( {b_n} ) Given ( b_n = 2^{n-1} + 1 ): [ T_n = sum_{k=1}^n b_k = sum_{k=1}^n (2^{k-1} + 1) ] [ = left( sum_{k=1}^n 2^{k-1} right) + sum_{k=1}^n 1 ] The sum of the first (n) terms of a geometric series with first term 1 and common ratio 2 is: [ sum_{k=1}^n 2^{k-1} = 2^0 + 2^1 + cdots + 2^{n-1} = 2^n - 1 ] The sum of (n) ones is simply (n): [ T_n = (2^n - 1) + n = 2^n + n - 1 ] Thus: [ T_n = 2^n + n - 1 ] (3) Relationship between ( S_n ) and ( T_n ) We need to compare ( S_n = n^2 + n ) and ( T_n = 2^n + n - 1 ). Consider the difference: [ T_n - S_n = (2^n + n - 1) - (n^2 + n) = 2^n - n^2 - 1 ] Examine the sign of ( 2^n - n^2 - 1 ): - For ( n = 1 ): ( T_1 = 2^1 + 1 - 1 = 2 ), ( S_1 = 1^2 + 1 = 2 ), ( 2 - 2 = 0 ). - For ( n = 2 ): ( T_2 = 4 + 2 - 1 = 5 ), ( S_2 = 4 + 2 = 6 ), ( 5 - 6 = -1 ). - For ( n = 3 ): ( T_3 = 8 + 3 - 1 = 10 ), ( S_3 = 9 + 3 = 12 ), ( 10 - 12 = -2 ). - For ( n = 4 ): ( T_4 = 16 + 4 - 1 = 19 ), ( S_4 = 16 + 4 = 20 ), ( 19 - 20 = -1 ). - For ( n geq 5 ): ( 2^n ) grows faster than ( n^2 ), ensuring ( 2^n - n^2 - 1 > 0 ). Hence, for ( n geq 5 ), ( 2^n > n^2 + 1 ). Thus, the sequences satisfy: - ( T_1 = S_1 ) - ( T_n < S_n ) for ( 2 leq n < 5 ) - ( T_n > S_n ) for ( n geq 5 ) In conclusion: - For ( n = 1 ): ( T_n = S_n ) - For ( 2 leq n < 5 ): ( T_n < S_n ) - For ( n geq 5 ): ( T_n > S_n ) We conjecture that: [ boxed{T_n > S_n text{ for } n geq 5} ]

answer:Let's denote the number of pens Isaac bought as P and the number of pencils as L. We are given that Isaac bought 16 pens, so P = 16. We are also told that Isaac bought a total of 108 writing utensils, so the sum of pens and pencils is: P + L = 108 Substituting the value of P, we get: 16 + L = 108 Now, let's solve for L (the number of pencils): L = 108 - 16 L = 92 Isaac bought 92 pencils. The problem states that Isaac bought 12 more pencils than a certain multiple of the number of pens. Let's denote this multiple as M. So we can write the relationship between the number of pencils and the number of pens as: L = M * P + 12 Substituting the values of L and P, we get: 92 = M * 16 + 12 Now, let's solve for M: 92 - 12 = M * 16 80 = M * 16 Dividing both sides by 16 to solve for M: M = 80 / 16 M = 5 Therefore, the multiple of the number of pens that Isaac used to calculate the number of pencils he bought is boxed{5} .

answer:First, let's calculate the amount of tax deducted from Alex's weekly income: 10% of 500 = 0.10 * 500 = 50 Now, let's calculate the amount given as a tithe: 10% of 500 = 0.10 * 500 = 50 Next, let's calculate the total deductions (tax, water bill, and tithe): Total deductions = Tax + Water bill + Tithe Total deductions = 50 (tax) + 55 (water bill) + 50 (tithe) Total deductions = 155 Finally, let's calculate the amount of money Alex has left after all deductions: Money left = Weekly income - Total deductions Money left = 500 - 155 Money left = 345 Alex has boxed{345} left after all deductions.