answer:Let's solve the problem step-by-step using Python and SymPy. (I) Finding the General Formula for the nth TermGiven an arithmetic sequence ({a_n}) with a common difference of (-2), and (a_3 = a_2 + a_5).1. Arithmetic Sequence Basic Properties: The nth term of an arithmetic sequence can be written as: [ a_n = a_1 + (n-1)d ] where (a_1) is the first term and (d) is the common difference (which is (-2) in this case).2. Given Condition: (a_3 = a_2 + a_5) Substituting the formula for (a_n) into this condition: [ a_1 + 2d = (a_1 + d) + (a_1 + 4d) ] Simplifying, we get: [ a_1 + 2(-2) = a_1 + (-2) + a_1 + 4(-2) ] [ a_1 - 4 = a_1 - 2 + a_1 - 8 ] Combining like terms: [ a_1 - 4 = 2a_1 - 10 ] Solving for (a_1): [ a_1 - 4 = 2a_1 - 10 implies 6 = a_1 ]Thus, the nth term of the sequence is:[a_n = 6 + (n-1)(-2)]Simplifying, we get:[a_n = 6 - 2(n-1) = 8 - 2n] (II) Finding the Maximum Value of the Sum of the First n Terms, (S_n)The sum of the first (n) terms of an arithmetic sequence is given by:[S_n = frac{n}{2} left( 2a_1 + (n-1)d right)]Substituting (a_1 = 6) and (d = -2):[S_n = frac{n}{2} left( 2(6) + (n-1)(-2) right)]Simplifying:[S_n = frac{n}{2} left( 12 - 2n + 2 right) = frac{n}{2} (14 - 2n) = n(7 - n)]To find the maximum value of (S_n), we need to find the vertex of the quadratic function (S_n = 7n - n^2).The vertex of a quadratic function (ax^2 + bx + c) is at (x = -frac{b}{2a}). Here, (a = -1) and (b = 7), so:[n = -frac{7}{2(-1)} = frac{7}{2} = 3.5]Since (n) must be an integer (number of terms), we check (S_n) at (n = 3) and (n = 4):[S_3 = 3(7 - 3) = 3 cdot 4 = 12][S_4 = 4(7 - 4) = 4 cdot 3 = 12]Thus, the maximum value of (S_n) is 12.```pythonimport sympy as sp# Given valuesd = -2a1 = 6# Expression for nth termn = sp.symbols('n')a_n = a1 + (n - 1) * da_n = sp.simplify(a_n)# Sum of first n termsS_n = sp.simplify(n / 2 * (2 * a1 + (n - 1) * d))# Find the maximum value of S_nS_n_expr = n * (7 - n)max_n = sp.solve(sp.diff(S_n_expr, n), n)S_n_max = S_n_expr.subs(n, max_n[0])print(f"General formula for a_n: {a_n}")print(f"Maximum value of S_n: {S_n_max}")``````outputGeneral formula for a_n: 8 - 2*nMaximum value of S_n: 49/4```From the output, we have the following results: (I) General Formula for (a_n):The nth term of the arithmetic sequence is given by:[a_n = 8 - 2n] (II) Maximum Value of the Sum of the First (n) Terms, (S_n):We found that the maximum value of (S_n) is (frac{49}{4}). However, this value corresponds to a non-integer (n = 3.5).Since (n) must be an integer, the maximum (S_n) occurs at the nearest integers, (n = 3) and (n = 4). For both (n = 3) and (n = 4), we calculated:[S_3 = 3(7 - 3) = 12][S_4 = 4(7 - 4) = 12]Thus, the maximum value of the sum of the first (n) terms when (n) is an integer is:[boxed{12}]The final results are:- The general formula for the nth term of the sequence is (boxed{a_n = 8 - 2n}).- The maximum value of the sum of the first (n) terms, (S_n), is (boxed{12}).

answer:To determine the degree of the monomial (-frac{3}{5}ab^2), we need to sum the exponents of all variables in the monomial.Let's break this down into manageable steps and write Python code to solve it:1. Identify the exponents of each variable: - The exponent of (a) is 1. - The exponent of (b) is 2.2. Sum the exponents to get the degree of the monomial.We can achieve this using SymPy in Python. Although this problem is simple enough to be solved manually, we'll use SymPy to ensure accuracy and demonstrate the process.```pythonimport sympy as sp# Define the variablesa, b = sp.symbols('a b')# Define the monomialmonomial = -3/5 * a * b2# The degree of the monomial is the sum of the exponents of all variablesdegree = sp.degree(monomial, gen=a) + sp.degree(monomial, gen=b)print(degree)``````output3```The degree of the monomial (-frac{3}{5}ab^2) is (boxed{3}).Here's the step-by-step reasoning:1. The exponent of (a) is 1.2. The exponent of (b) is 2.3. The sum of the exponents is (1 + 2 = 3).Thus, the degree of the monomial is 3.

answer:To solve the problem, we need to follow these steps:1. Identify the relationship between the area of the square and the length of its side.2. Calculate the side length of the square.3. Use the side length to calculate the perimeter of the square.Let's perform these steps in detail.1. For a square, the area ( A ) is given by: [ A = text{side}^2 ] Given the area ( A = 400 text{ cm}^2 ), we can find the side length by taking the square root of the area.2. Once we have the side length, the perimeter ( P ) of the square can be calculated using: [ P = 4 times text{side} ]We'll use Python to ensure our calculations are accurate.```pythonimport math# Given area of the squarearea = 400 # cm^2# Calculate the side length of the squareside_length = math.sqrt(area)# Calculate the perimeter of the squareperimeter = 4 * side_lengthprint((side_length, perimeter))``````output(20.0, 80.0)```The side length of the square is ( 20.0 , text{cm} ) and the perimeter of the square is ( 80.0 , text{cm} ).Therefore, the perimeter of the square is (boxed{80 , text{cm}}).

answer:To determine the weight of one banana, let's break down the problem into smaller steps and solve it using Python and SymPy. Here are the steps we will follow:1. Define the variables: - (a) for the weight of an apple - (b) for the weight of a banana - (g) for the weight of a grapefruit2. Translate the given relationships into equations: - The weight of 3 apples is equal to the weight of 1 grapefruit: (3a = g) - The weight of 4 bananas is equal to the weight of 2 apples: (4b = 2a)3. We know the weight of one grapefruit is 576 grams: (g = 576)4. Solve the equations step-by-step to find the weight of one banana.Let's implement this in Python:```pythonimport sympy as sp# Define the variablesa, b, g = sp.symbols('a b g')# Given equationseq1 = sp.Eq(3 * a, g)eq2 = sp.Eq(4 * b, 2 * a)# Given weight of one grapefruitgrapefruit_weight = 576# Substitute the weight of the grapefruit into eq1eq1_substituted = eq1.subs(g, grapefruit_weight)# Solve eq1 for the weight of an appleapple_weight = sp.solve(eq1_substituted, a)[0]# Substitute the weight of the apple into eq2eq2_substituted = eq2.subs(a, apple_weight)# Solve eq2 for the weight of a bananabanana_weight = sp.solve(eq2_substituted, b)[0]# Print the weight of one bananaprint(banana_weight)``````output96```Based on the calculations, the weight of one banana is ( boxed{96} ) grams.

answer:To determine the smallest integer that is greater than ((sqrt{3}+sqrt{2})^{6}), we'll break the problem down and use Python to ensure our calculations are accurate. Let's start by writing a function to compute ((sqrt{3}+sqrt{2})^{6}), and then determine the smallest integer greater than this result.Let's first calculate ((sqrt{3}+sqrt{2})^{6}):1. Calculate (sqrt{3}) and (sqrt{2}).2. Add them.3. Raise the sum to the power of 6.Then, we'll find the ceiling of this result to get the smallest integer greater than this value.```pythonimport math# Calculate the value of (sqrt(3) + sqrt(2))^6value = (math.sqrt(3) + math.sqrt(2)) 6# Determine the smallest integer greater than this valuesmallest_integer_greater = math.ceil(value)print(smallest_integer_greater)``````output970```The smallest integer that is greater than ((sqrt{3}+sqrt{2})^{6}) is (boxed{970}).

answer:To determine the largest value, we will evaluate each expression:[textbf{(A) } 2 + 0 + 1 + 7][textbf{(B) } 2 times 0 + 1 + 7][textbf{(C) } 2 + 0 times 1 + 7][textbf{(D) } 2 + 0 + 1 times 7][textbf{(E) } 2 times 0 times 1 times 7]Now let's solve these expressions step by step using Python to ensure accuracy.```python# Evaluating each expressionA = 2 + 0 + 1 + 7B = 2 * 0 + 1 + 7C = 2 + 0 * 1 + 7D = 2 + 0 + 1 * 7E = 2 * 0 * 1 * 7print((A, B, C, D, E))``````output(10, 8, 9, 9, 0)```The evaluated values for each expression are:[textbf{(A) } 2 + 0 + 1 + 7 = 10][textbf{(B) } 2 times 0 + 1 + 7 = 8][textbf{(C) } 2 + 0 times 1 + 7 = 9][textbf{(D) } 2 + 0 + 1 times 7 = 9][textbf{(E) } 2 times 0 times 1 times 7 = 0]From these results, the largest value is (boxed{10}).