answer:To solve the problem, we need to calculate the absolute value of (-2). The absolute value function, denoted by (|x|), returns the non-negative value of (x).In Python, we can use the `Abs` function from the `sympy` library to compute the absolute value.Let's implement this in Python:```pythonimport sympy as sp# Define the numbernumber = -2# Calculate the absolute valueabsolute_value = sp.Abs(number)# Print the resultprint(absolute_value)``````output2```The absolute value of (-2) is (boxed{2}).

answer:To solve this problem, we need to model the circle of villagers and their statements to determine the proportion of truthful villagers. Let's break the problem down:1. Understanding the Statements: - Each villager makes a statement about their right-hand neighbor. - A truthful villager tells the truth about whether their right-hand neighbor is truthful. - A deceitful villager lies about whether their right-hand neighbor is truthful.2. Modeling the Problem: - Let's represent the villagers as a list where each element is either 'T' (truthful) or 'L' (liar). - We need to find a configuration of villagers that satisfies the statements and allows us to determine the proportion of truthful villagers.3. Iterative Approach: - We will use an iterative approach to test different configurations of villagers. - For each configuration, we will check if the statements made by the villagers are consistent. - If a configuration is found where the statements are consistent, we calculate the proportion of truthful villagers.Let's implement this in Python using sympy to assist with the logical checks.```pythonimport sympy as spdef check_consistency(villagers): n = len(villagers) for i in range(n): if villagers[i] == 'T': if villagers[(i+1) % n] == 'T' and statements[i] != 'T': return False if villagers[(i+1) % n] == 'L' and statements[i] != 'L': return False else: # villagers[i] == 'L' if villagers[(i+1) % n] == 'T' and statements[i] != 'L': return False if villagers[(i+1) % n] == 'L' and statements[i] != 'T': return False return Truedef find_proportion_of_truthful(n): from itertools import product # Generate all possible statements global statements statements = ['T', 'L'] * (n // 2) # Iterate over all possible configurations of villagers for villagers in product('TL', repeat=n): if check_consistency(villagers): truthful_count = villagers.count('T') return truthful_count / nn = 8 # let's assume there are 8 villagers; you can adjust the numberproportion = find_proportion_of_truthful(n)print(proportion)``````output0.5```The proportion of truthful villagers, based on the given statements and logical consistency, is ( boxed{0.5} ). This means that half of the villagers are truthful and the other half are liars.

answer:To solve the problem, we need to compute the probability (p) that (a_1 b_1 + a_2 b_2) is a multiple of 6 after Ana's rolls but before Banana's rolls. Then we need to check the conditions under which this probability (p) equals (frac{1}{6}).Here are the steps to approach this problem:1. Model the Rolls: - Each roll of a die can result in any integer from 1 to 6. - We need to determine the values (a_1) and (a_2) can take and how they impact the condition (a_1 b_1 + a_2 b_2 equiv 0 pmod{6}).2. Calculate the Probability: - For each possible pair ((a_1, a_2)), determine the number of ((b_1, b_2)) pairs that satisfy (a_1 b_1 + a_2 b_2 equiv 0 pmod{6}). - Compute the probability for each ((a_1, a_2)) and check when this probability is (frac{1}{6}).3. Count the Favorable Outcomes: - Count how many ((a_1, a_2)) pairs lead to the desired probability (p = frac{1}{6}).Let's implement this in Python using sympy to handle modular arithmetic and simplify our calculations.```pythonimport sympy as spfrom itertools import product# Possible dice outcomesdice_outcomes = range(1, 7)# Function to compute the probability that a1*b1 + a2*b2 is a multiple of 6def compute_probability(a1, a2): count = 0 total_outcomes = 36 for b1, b2 in product(dice_outcomes, repeat=2): if (a1 * b1 + a2 * b2) % 6 == 0: count += 1 return count / total_outcomes# Count the number of (a1, a2) pairs leading to probability 1/6favorable_count = 0for a1, a2 in product(dice_outcomes, repeat=2): probability = compute_probability(a1, a2) if probability == 1/6: favorable_count += 1# Total number of (a1, a2) pairstotal_pairs = 36# Probability that p = 1/6final_probability = favorable_count / total_pairsprint(final_probability)``````output0.6666666666666666```The probability that (p = frac{1}{6}) is (boxed{frac{2}{3}}).

answer:Let's solve the problem step-by-step using Python and SymPy.1. Understand the problem: We need to determine the age of Xiaoming's grandfather. Let's denote his age by ( x ).2. Formulate the equation: According to the problem, after performing the following operations: - Subtract 15 from ( x ) - Divide the result by 4 - Subtract 6 from the quotient - Multiply the result by 10 We get exactly 100 years old. So we can write the equation as: [ 10 left( frac{x - 15}{4} - 6 right) = 100 ]3. Solve the equation using SymPy: We will solve the above equation to find the value of ( x ).Here is the Python code to solve it:```pythonimport sympy as sp# Define the variablex = sp.symbols('x')# Define the equationequation = 10 * ((x - 15) / 4 - 6) - 100# Solve the equation for xsolution = sp.solve(equation, x)# Print the solutionprint(solution[0])``````output79```The solution to the equation indicates that Xiaoming's grandfather is ( boxed{79} ) years old this year.

answer:Sure, let's break down the problem and solve it step-by-step.The expression we need to evaluate is:[ 5(x^{3} - 3x^{2}y - 2xy^{2}) - 3(x^{3} - 5x^{2}y + 2y^{3}) + 2(-x^{3} + 5xy^{2} + 3y^{3}) ]Step 1: Simplify the expression by collecting like terms.Step 2: Substitute ( x = 2024 ) into the simplified expression and solve.Let's write the Python code to perform these steps using SymPy.```pythonimport sympy as sp# Define the variablesx, y = sp.symbols('x y')# Define the expressionexpr = 5*(x3 - 3*x2*y - 2*x*y2) - 3*(x3 - 5*x2*y + 2*y3) + 2*(-x3 + 5*x*y2 + 3*y3)# Simplify the expressionsimplified_expr = sp.simplify(expr)# Substitute x = 2024 into the simplified expressionresult = simplified_expr.subs(x, 2024)print(result)``````output0```The output of the simplified expression when ( x = 2024 ) is ( boxed{0} ).Let's break down why this happens:1. Expression Simplification: The original expression is: [ 5(x^{3} - 3x^{2}y - 2xy^{2}) - 3(x^{3} - 5x^{2}y + 2y^{3}) + 2(-x^{3} + 5xy^{2} + 3y^{3}) ]2. Collecting Like Terms: When we combine like terms, the expression simplifies to zero. This simplification happens because the terms from each group cancel each other out. Specifically: [ 5x^3 - 15x^2y - 10xy^2 - 3x^3 + 15x^2y - 6y^3 - 2x^3 + 10xy^2 + 6y^3 = 0 ]3. Simplified Expression: After combining the coefficients: [ (5x^3 - 3x^3 - 2x^3) + (-15x^2y + 15x^2y) + (-10xy^2 + 10xy^2) + (-6y^3 + 6y^3) = 0 ]Thus, the output is indeed zero when ( x = 2024 ).So, the value of the given expression when ( x = 2024 ) is ( boxed{0} ).

answer:To solve the problem, we'll break it down into smaller, manageable parts and use Python and SymPy to find the solution step-by-step.1. Understand the Problem: - Let (C1), (C2), and (C3) be the lengths of chapters 1, 2, and 3 respectively. - We're given the relationships: [ C1 = frac{1}{2}C2 quad text{and} quad C1 = 3C3 ] Therefore, we can express (C2) and (C3) in terms of (C1): [ C2 = 2C1 quad text{and} quad C3 = frac{C1}{3} ]2. Working Together: - Let (r1) and (r2) be the rates of the first and second typists respectively (in chapters per hour). - Working together, they retyped the first chapter ((C1)) in 3 hours and 36 minutes (which is (3 + frac{36}{60} = 3.6) hours): [ (r1 + r2) times 3.6 = C1 ]3. Second Chapter Typing: - For the second chapter ((C2)), the first typist worked alone for 2 hours: [ r1 times 2 ] - Then both typists worked together for the remaining 6 hours: [ r1 times 2 + (r1 + r2) times 6 = C2 ]4. Find the Typing Rates: - Use the equations derived from the first and second chapters to solve for (r1) and (r2).5. Third Chapter Typing: - We need to find the time (t) required for the second typist alone to retype the third chapter ((C3)): [ r2 times t = C3 ] [ t = frac{C3}{r2} ]Let's implement this step-by-step using Python and SymPy.```pythonimport sympy as sp# Define symbolsr1, r2, C1, t = sp.symbols('r1 r2 C1 t')# Given relationshipsC2 = 2 * C1C3 = C1 / 3# Equations from the problem statementeq1 = sp.Eq((r1 + r2) * 3.6, C1)eq2 = sp.Eq(r1 * 2 + (r1 + r2) * 6, C2)# Solve for r1 and r2solutions = sp.solve([eq1, eq2], (r1, r2))r1_value = solutions[r1]r2_value = solutions[r2]# Time required for the second typist to retype the third chapter alonetime_third_chapter = C3 / r2_value# Substitute C1 to get the numerical valuetime_third_chapter_value = time_third_chapter.subs(C1, 3.6 * (r1_value + r2_value))# Print the final resultprint(time_third_chapter_value)``````output3.00000000000000```The second typist will need ( boxed{3} ) hours to retype the third chapter alone.