answer:To solve this problem, we'll define the two original two-digit natural numbers as (x) and (y). Let's denote their rounded values to the nearest ten as (x_r) and (y_r). Using the given conditions, we can derive the following equations:1. The difference between the rounded numbers is the same as the difference between the original numbers: [ |x_r - y_r| = |x - y| ]2. The product of the rounded numbers is 184 greater than the product of the original numbers: [ x_r cdot y_r = x cdot y + 184 ]Since (x) and (y) are two-digit numbers, their possible rounded values can be 10, 20, 30, ..., 90. We will iterate through these possible values and check which pair satisfies the conditions above.We should implement these steps in Python using sympy:1. Define variables (x) and (y).2. Iterate through possible values for (x_r) and (y_r).3. Check for each pair if they satisfy the conditions.```pythonimport sympy as sp# Define variablesx, y = sp.symbols('x y')# Function to check if a pair (x_r, y_r) satisfies the conditionsdef find_original_numbers(): for x_r in range(10, 100, 10): for y_r in range(10, 100, 10): # Solve the system of equations eq1 = sp.Eq(abs(x_r - y_r), abs(x - y)) eq2 = sp.Eq(x_r * y_r, x * y + 184) solutions = sp.solve((eq1, eq2), (x, y)) # Filter solutions to ensure they are two-digit natural numbers for sol in solutions: x_val, y_val = sol if x_val.is_integer and y_val.is_integer and 10 <= x_val <= 99 and 10 <= y_val <= 99: return x_val, y_val return Noneoriginal_numbers = find_original_numbers()print(original_numbers)``````outputTraceback (most recent call last): File " original_numbers = find_original_numbers() File " solutions = sp.solve((eq1, eq2), (x, y))NotImplementedError: solving Abs(x - y) when the argume ... (output truncated)```It looks like sympy has trouble solving equations involving the absolute value directly. To resolve this, we can break down the absolute value into cases. Specifically, the absolute value condition ( |x_r - y_r| = |x - y| ) can be split into two cases:1. ( x_r - y_r = x - y )2. ( x_r - y_r = y - x )We will iterate through all possible values of (x_r) and (y_r), and for each pair, we will solve these two cases separately.Here is the adjusted Python code:```pythonimport sympy as sp# Define variablesx, y = sp.symbols('x y')# Function to check if a pair (x_r, y_r) satisfies the conditionsdef find_original_numbers(): for x_r in range(10, 100, 10): for y_r in range(10, 100, 10): # Case 1: x_r - y_r = x - y eq1_case1 = sp.Eq(x_r - y_r, x - y) eq2_case1 = sp.Eq(x_r * y_r, x * y + 184) solutions_case1 = sp.solve((eq1_case1, eq2_case1), (x, y)) # Filter solutions for case 1 for sol in solutions_case1: x_val, y_val = sol if x_val.is_integer and y_val.is_integer and 10 <= x_val <= 99 and 10 <= y_val <= 99: return x_val, y_val # Case 2: x_r - y_r = y - x eq1_case2 = sp.Eq(x_r - y_r, y - x) eq2_case2 = sp.Eq(x_r * y_r, x * y + 184) solutions_case2 = sp.solve((eq1_case2, eq2_case2), (x, y)) # Filter solutions for case 2 for sol in solutions_case2: x_val, y_val = sol if x_val.is_integer and y_val.is_integer and 10 <= x_val <= 99 and 10 <= y_val <= 99: return x_val, y_val return Noneoriginal_numbers = find_original_numbers()print(original_numbers)``````output(16, 26)```The original two-digit natural numbers are ( boxed{16 text{ and } 26} ).To verify, let's check the conditions given in the problem statement:1. Rounded difference condition: - The rounded values of (16) and (26) are (20) and (30), respectively. - The difference between the rounded numbers: ( |30 - 20| = 10 ) - The difference between the original numbers: ( |26 - 16| = 10 ) Both differences are equal, satisfying the first condition.2. Product condition: - The product of the rounded numbers: ( 20 times 30 = 600 ) - The product of the original numbers: ( 16 times 26 = 416 ) - The difference in products: ( 600 - 416 = 184 ) This satisfies the second condition.Thus, the original numbers ( boxed{16 text{ and } 26} ) correctly satisfy all the given conditions.

answer:To solve this problem, we need to work with the properties of arithmetic sequences. Let's break it down step-by-step:1. Define the Arithmetic Sequence: In an arithmetic sequence, each term can be expressed as: [ a_n = a_1 + (n-1)d ] where ( a_1 ) is the first term and ( d ) is the common difference.2. Set up the Given Equation: We are provided with the sum of specific terms: [ a_3 + a_4 + a_5 + a_{13} + a_{14} + a_{15} = 8 ] Substitute the general form of the arithmetic sequence into the equation: [ (a_1 + 2d) + (a_1 + 3d) + (a_1 + 4d) + (a_1 + 12d) + (a_1 + 13d) + (a_1 + 14d) = 8 ] Combine like terms: [ 6a_1 + (2d + 3d + 4d + 12d + 13d + 14d) = 8 ] [ 6a_1 + 48d = 8 ]3. Solve for (a_1) and (d): We can express this equation as: [ 6a_1 + 48d = 8 ]4. Find (5a_7 - 2a_4): Express (a_7) and (a_4) in terms of (a_1) and (d): [ a_7 = a_1 + 6d ] [ a_4 = a_1 + 3d ] Therefore: [ 5a_7 - 2a_4 = 5(a_1 + 6d) - 2(a_1 + 3d) ] Simplify the expression: [ 5a_1 + 30d - 2a_1 - 6d = 3a_1 + 24d ]5. Solve using Python and SymPy:```pythonimport sympy as sp# Define variablesa1, d = sp.symbols('a1 d')# Given equationequation = 6*a1 + 48*d - 8# Solve for the variablessolution = sp.solve(equation, (a1, d))a1_value = solution[a1]d_value = solution[d]# Find 5a7 - 2a4a7 = a1_value + 6*d_valuea4 = a1_value + 3*d_valueresult = 5*a7 - 2*a4print(result)``````outputTraceback (most recent call last): File " a1_value = solution[a1]TypeError: list indices must be integers or slices, not Symbol```It appears that the solve function returned a list instead of a dictionary. We need to access the elements of the list directly and then substitute the values back into our expression for (5a_7 - 2a_4).Let's correct the code:```pythonimport sympy as sp# Define variablesa1, d = sp.symbols('a1 d')# Given equationequation = 6*a1 + 48*d - 8# Solve for the variablessolution = sp.solve(equation, (a1, d))a1_value, d_value = solution# Find 5a7 - 2a4a7 = a1 + 6*da4 = a1 + 3*dresult = 5*a7 - 2*a4# Substitute the solved valuesresult_value = result.subs({a1: a1_value, d: d_value})print(result_value)``````outputTraceback (most recent call last): File " a1_value, d_value = solutionValueError: not enough values to unpack (expected 2, got 1)```It seems like the equation (6a_1 + 48d = 8) potentially has a solution that isn't providing both (a_1) and (d). This might be due to the fact that (a_1) and (d) are dependent variables rather than independent. Let's solve for one variable in terms of the other.We can solve for (a_1) in terms of (d) or vice versa and then use that expression in our calculation for (5a_7 - 2a_4).Here's the adjusted approach:1. Solve for (a_1) in terms of (d): [ a_1 = frac{8 - 48d}{6} ] 2. Substitute this expression into (5a_7 - 2a_4):```pythonimport sympy as sp# Define variablesa1, d = sp.symbols('a1 d')# Given equationequation = 6*a1 + 48*d - 8# Solve for a1 in terms of da1_solution = sp.solve(equation, a1)[0]# Find a7 and a4 in terms of da7 = a1_solution + 6*da4 = a1_solution + 3*d# Expression for 5a7 - 2a4result = 5*a7 - 2*a4# Simplify the resultresult_simplified = sp.simplify(result)print(result_simplified)``````output4```The simplified expression for (5a_7 - 2a_4) yields the value 4. Thus, the value of (5a_7 - 2a_4) is (boxed{4}).

answer:To solve this problem, we will follow these steps:1. Understand the Problem: The function (f(x) = x^3 + ax^2 + bx + 1) has a local extremum at (x=1), and the value of this extremum is (-2). We need to determine the value of (f(2)).2. Use the Given Information: - Since (f(x)) has a local extremum at (x=1), the first derivative (f'(x)) must be zero at (x=1). - The value of the function at (x=1) is (-2), i.e., (f(1) = -2).3. Formulate the Equations: - Calculate (f'(x)). - Set (f'(1) = 0) to find a relationship between (a) and (b). - Use (f(1) = -2) to form another equation.4. Solve the Equations: - Solve the system of equations to find (a) and (b). - Substitute (a) and (b) back into (f(x)) and compute (f(2)).Let's implement this in Python with sympy to find the value of (f(2)).```pythonimport sympy as sp# Define symbolsx, a, b = sp.symbols('x a b')# Define the functionf = x3 + a*x2 + b*x + 1# Calculate the first derivativef_prime = sp.diff(f, x)# Set up the equations based on the given conditionseq1 = f_prime.subs(x, 1) # f'(1) = 0eq2 = f.subs(x, 1) + 2 # f(1) = -2# Solve the system of equations to find a and bsolution = sp.solve([eq1, eq2], (a, b))# Substitute the solution back into the function and calculate f(2)f_2_value = f.subs({a: solution[a], b: solution[b], x: 2})print(f_2_value)``````output3```The value of ( f(2) ) given that ( f(x) = x^3 + ax^2 + bx + 1 ) has a local extremum of (-2) at ( x = 1 ) is (boxed{3}).

answer:To solve this problem, we need to find the terms of both sequences such that their values do not exceed 300. To begin with, let's define the sequences and their properties:- Sequence ( A ) is a geometric sequence: [ A_n = 2^n ] where ( A_n ) is the ( n )-th term of the sequence ( A ).- Sequence ( B ) is an arithmetic sequence: [ B_n = 20 + (n-1) times 20 = 20n ] where ( B_n ) is the ( n )-th term of the sequence ( B ).We need to generate these sequences until their values do not exceed 300. Then, we will find the least positive difference between a number from sequence ( A ) and a number from sequence ( B ).Let's use Python to determine the terms and the least positive difference between the two sequences.```python# Generate terms for sequence A (geometric sequence) A = [] n = 1 while True: term = 2 n if term > 300: break A.append(term) n += 1 # Generate terms for sequence B (arithmetic sequence) B = [] n = 1 while True: term = 20 * n if term > 300: break B.append(term) n += 1 print((A, B))``````output([2, 4, 8, 16, 32, 64, 128, 256], [20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300])```We have generated the terms for both sequences:- Sequence ( A = [2, 4, 8, 16, 32, 64, 128, 256] )- Sequence ( B = [20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300] )Next, we need to find the least positive difference between any number in sequence ( A ) and any number in sequence ( B ).Let's calculate the minimum positive difference between any pair of terms from ( A ) and ( B ).```pythonimport numpy as npmin_difference = np.inf# Find the minimum positive differencefor a in A: for b in B: difference = abs(a - b) if difference < min_difference: min_difference = differenceprint(min_difference)``````output4```The least positive difference between a number selected from sequence ( A ) (geometric sequence) and a number selected from sequence ( B ) (arithmetic sequence) is (boxed{4}).

answer:To determine the value of (lambda) for which the vectors (overrightarrow{a}), (overrightarrow{b}), and (overrightarrow{c}) are coplanar, we need to check if the scalar triple product of these vectors is zero. The scalar triple product of vectors (overrightarrow{a}), (overrightarrow{b}), and (overrightarrow{c}) is given by the determinant of the matrix formed by placing these vectors as rows or columns.Given:[overrightarrow{a} = (2, -1, 2)][overrightarrow{b} = (-1, 3, -3)][overrightarrow{c} = (13, 6, lambda)]Vectors (overrightarrow{a}), (overrightarrow{b}), and (overrightarrow{c}) are coplanar if and only if:[begin{vmatrix}2 & -1 & 2 -1 & 3 & -3 13 & 6 & lambdaend{vmatrix} = 0]Let's calculate this determinant using Python and SymPy.```pythonimport sympy as sp# Define the variable lambdalambda_var = sp.symbols('lambda')# Define the matrixmatrix = sp.Matrix([ [2, -1, 2], [-1, 3, -3], [13, 6, lambda_var]])# Calculate the determinantdet = matrix.det()# Solve for lambda when the determinant is zerolambda_solution = sp.solve(det, lambda_var)print(lambda_solution)``````output[3]```The solution indicates that the value of (lambda) for which the vectors (overrightarrow{a} = (2, -1, 2)), (overrightarrow{b} = (-1, 3, -3)), and (overrightarrow{c} = (13, 6, lambda)) are coplanar is (lambda = 3).Thus, the value of (lambda) is (boxed{3}).

answer:To solve this problem, we need to maximize the profit for the newsstand owner based on the number of copies ordered each day. Let's break down the problem:1. Define Variables: - Let ( x ) be the number of copies ordered each day.2. Profit Calculation: - For the 20 days when 400 copies are sold: - If ( x leq 400 ), all copies are sold. - If ( x > 400 ), 400 copies are sold and ( x - 400 ) copies are returned. - For the 10 days when only 250 copies are sold: - If ( x leq 250 ), all copies are sold. - If ( x > 250 ), 250 copies are sold and ( x - 250 ) copies are returned.3. Revenue and Cost Calculation: - Selling price: 0.30 yuan per copy. - Buying price: 0.20 yuan per copy. - Return price: 0.05 yuan per copy.4. Formulate the Monthly Profit Function: - For the 20 days when 400 copies are sold: - Revenue: ( 0.30 times min(x, 400) ) - Cost: ( 0.20 times x ) - Return income: ( 0.05 times max(0, x - 400) ) - For the 10 days when 250 copies are sold: - Revenue: ( 0.30 times min(x, 250) ) - Cost: ( 0.20 times x ) - Return income: ( 0.05 times max(0, x - 250) )5. Combine the Profit Calculation: - Monthly profit can be calculated as the sum of daily profits over the 30 days.Let's implement the logic in Python to find the optimal number of copies to order each day and the maximum profit.```pythonimport sympy as sp# Define the variablex = sp.symbols('x', integer=True)# Constantssell_price = 0.30buy_price = 0.20return_price = 0.05# Profit calculation for the two scenariosprofit_400 = (sell_price * sp.Min(x, 400) + return_price * sp.Max(0, x - 400) - buy_price * x) * 20profit_250 = (sell_price * sp.Min(x, 250) + return_price * sp.Max(0, x - 250) - buy_price * x) * 10# Total monthly profittotal_profit = profit_400 + profit_250# Simplify the profit expressiontotal_profit_simplified = sp.simplify(total_profit)# Find the maximum profit by evaluating for all possible values of x# Assuming realistic range from 250 to 400 (inclusive) since these are the max sales on respective daysx_values = range(250, 401)profits = [total_profit_simplified.subs(x, val) for val in x_values]# Find the value of x that maximizes profitmax_profit = max(profits)optimal_x = x_values[profits.index(max_profit)]print(optimal_x, max_profit)``````output400 825.000000000000```Based on the output from the code execution:The optimal number of copies the newsstand owner should buy from the newspaper office each day is ( boxed{400} ) copies.The maximum amount of money the owner can earn in a month is ( boxed{825} ) yuan.