answer:We approach this problem by converting each digit of the number 725_9 separately from base 9 to base 3. Since 9 = 3^2, each base 9 digit maps into exactly two base 3 digits. - For the digit 7_9, it is equivalent to 21_3 because 7 = 2 times 3^1 + 1 times 3^0. - For the digit 2_9, it directly maps to 02_3 because 2 = 0 times 3^1 + 2 times 3^0. - Finally, the digit 5_9 maps to 12_3, as 5 = 1 times 3^1 + 2 times 3^0. Thus, combining these converted digits, we output the base 3 representation of the number 725_9 as 725_9 = boxed{210212_3}.

answer:To prove that ( s(x) = sin x ) and ( c(x) = cos x ) given the conditions: 1. [ s'(x) = c(x), ] 2. [ c'(x) = -s(x), ] 3. [ s(0) = 1 - c(0) = 0, ] the solution involves verifying these properties for the sine and cosine functions and then using a uniqueness argument based on a given theorem. 1. Verify that ( s(x) = sin x ) and ( c(x) = cos x ) satisfy the differential equations. First we show that the derivative of (sin x) is (cos x). The derivative of (sin x) via the limit definition is: [ frac{d}{dx} sin x = lim_{h to 0} frac{sin(x+h) - sin x}{h}. ] Using the sum-to-product identities: [ sin(x+h) - sin(x) = sin x cos h + cos x sin h - sin x = cos x sin h + sin x (cos h - 1), ] we get: [ frac{sin(x+h) - sin(x)}{h} = cos x frac{sin h}{h} + sin x frac{cos h - 1}{h}. ] As ( h to 0 ): [ lim_{h to 0} frac{sin h}{h} = 1 quad text{and} quad lim_{h to 0} frac{cos h - 1}{h} = -lim_{h to 0} frac{2 sin^2(h/2)}{h} = 0. ] Thus, [ frac{d}{dx} sin x = cos x. ] 2. Verify that the derivative of ( cos x ) is ( -sin x ). Using the chain rule, with ( cos x = sinleft(x + frac{pi}{2}right) ): [ frac{d}{dx} cos x = frac{d}{dx} sinleft(x + frac{pi}{2}right) = cosleft(x + frac{pi}{2}right) = -sin x. ] 3. Check the initial conditions: [ s(0) = sin 0 = 0 quad text{and} quad c(0) = cos 0 = 1. ] Therefore, ( s(x) = sin x ) and ( c(x) = cos x ) satisfy all given conditions. Next, to prove these functions are unique given the conditions, consider any pair of functions ( s(x), c(x) ) that satisfy these conditions. Define: [ f(x) = (s(x) - sin x)^2 + (c(x) - cos x)^2. ] Observe that: [ f(0) = (s(0) - sin 0)^2 + (c(0) - cos 0)^2 = 0. ] To show ( f(x) = 0 ) for all ( x ), compute the derivative of ( f(x) ): begin{align*} f'(x) &= 2 (s(x) - sin x) (s'(x) - cos x) + 2 (c(x) - cos x) (c'(x) + sin x) &= 2 (s(x) - sin x) (c(x) - cos x) + 2 (c(x) - cos x) (-s(x) - sin x) &= 2 (s(x) - sin x) (c(x) - cos x) - 2 (c(x) - cos x) (s(x) + sin x) &= 2 left[ (s(x) - sin x)(c(x) - cos x) - (c(x) - cos x)(s(x) + sin x) right] &= 2 left[ (s(x) c(x) - s(x) cos x - sin x c(x) + sin x cos x) - (c(x) s(x) + c(x) sin x - cos x s(x) - cos x sin x)right] &= 2 left[ s(x) c(x) - s(x) cos x - sin x c(x) + sin x cos x - s(x) c(x) - c(x) sin x + cos x s(x) - cos x sin x right] &= 2 left[ s(x) c(x) - s(x) cos x - sin x c(x) + sin x cos x - s(x) c(x) - c(x) sin x + cos x s(x) - cos x sin x right] &= 2 left[ 0 right] &= 0. end{align*} Since ( f'(x) = 0 ), by the theorem that states if ( f'(x) = 0 ) on an interval, then ( f(x) ) is constant, we have: [ f(x) = 0 quad text{for all} quad x. ] Hence, [ (s(x) - sin x)^2 + (c(x) - cos x)^2 = 0 implies s(x) = sin x quad text{and} quad c(x) = cos x quad text{for all} quad x. ] (blacksquare)

answer:We are given a positive integer p and the set A_{p} defined as: [ A_{p} = left{ x mid 2^{p} < x < 2^{p+1}, x = 3m, m in mathbf{N} right} ] To find the sum of all elements in A_{p}, we should follow several steps. 1. **Identify the first element in A_{p}:** Since x = 3m and x must be greater than 2^p, the smallest possible x greater than 2^p must be in the form of 3k where k is the smallest integer such that 3k > 2^p. Mathematically, the smallest element x in A_p is: [ 3 leftlceil frac{2^p}{3} rightrceil ] However, we simplify it by noting the smallest element, 2^{p} + 2, which satisfies: [ 2^{p} + 2 equiv (-1)^{p} + 2 equiv 1 + 2 equiv 0 pmod{3} ] 2. **Identify the last element in A_{p}:** The largest possible x smaller than 2^{p+1} must be in the form of 3k where k is the largest integer such that 3k < 2^{p+1}. Mathematically, the largest element x in A_p is: [ 3 leftlfloor frac{2^{p+1}}{3} rightrfloor ] Again, we simplify it by noting the largest element, 2^{p+1} - 2, which satisfies: [ 2^{p+1} - 2 equiv 2(-1)^{p} - 2 equiv 2 - 2 equiv 0 pmod{3} ] 3. **Form the arithmetic progression:** The elements of A_p forms an arithmetic progression with: - First term, a_1 = 2^p + 2 - Last term, a_n = 2^{p+1} - 2 - Common difference, d = 3 4. **Calculate the number of terms n:** The number of terms in the sequence can be found by solving: [ 2^{p+1} - 2 = 2^p + 2 + 3(n-1) ] Simplify and solve for n: [ 2^{p+1} - 2 = 2^p + 2 + 3n - 3 ] [ 2^{p+1} - 2 - 2^p - 2 + 3 = 3n ] [ 2^{p+1} - 2^p - 1 = 3n ] [ n = frac{2^{p+1} - 2^p - 1}{3} ] Simplifying further, [ n = frac{2^p(2-1) - 1}{3} ] [ n = frac{2^p - 1}{3} ] 5. **Calculate the sum S of elements in A_p:** The sum of an arithmetic sequence is given by: [ S = frac{n}{2} (a_1 + a_n) ] Substituting n, a_1, and a_n: [ S = frac{frac{2^p - 1}{3}}{2} left( 2^p + 2 + 2^{p+1} - 2 right) ] Simplify within the parentheses: [ S = frac{2^p - 1}{6} left( 2^p + 2^{p+1} right) ] [ S = frac{2^p - 1}{6} left( 2^p + 2 cdot 2^p right) ] [ S = frac{2^p - 1}{6} left( 3 cdot 2^p right) ] [ S = frac{2^p - 1}{6} cdot 3 cdot 2^p ] [ S = frac{2^p - 1}{2} cdot 2^p ] [ S = 2^{2p-1} - 2^{p-1} ] **Conclusion:** The sum of all elements in the set A_p is: [ boxed{2^{2p-1} - 2^{p-1}} ]

answer:Given that a is a solution to the equation x^{2}+x-5=0, we start by acknowledging this relationship: 1. Since a is a solution, we have a^{2}+a-5=0. 2. Rearranging this equation to solve for a^{2}+a, we get a^{2}+a=5. Now, we are asked to find the value of a^{2}+a+1. Using the result from step 2: 3. We substitute a^{2}+a with 5 in the expression a^{2}+a+1, which gives us 5+1. 4. Therefore, a^{2}+a+1=6. Hence, the correct answer is boxed{B}.