answer:To solve this problem, we should follow a systematic approach:1. Identify all two-digit integers less than 30.2. Utilize a function to determine if a number is prime.3. Check sequences of five consecutive integers to see if all of them are not prime.4. Determine the largest integer in the first sequence found that matches the criteria.Let’s implement this in Python to ensure accuracy.```python# Step 1: Define a helper function to check for prime numbers.import sympydef is_prime(n): """Returns True if n is a prime number, otherwise False.""" return sympy.isprime(n)# Step 2: Generate all two-digit numbers less than 30numbers = list(range(10, 30))# Step 3: Check for sequences of five consecutive integers where none are primenon_prime_sequence = []for i in range(len(numbers) - 4): consecutive_five = numbers[i:i + 5] if all(not is_prime(num) for num in consecutive_five): non_prime_sequence = consecutive_five break# Step 4: Output the largest integer in the sequencelargest_non_prime = max(non_prime_sequence)print((non_prime_sequence, largest_non_prime))``````output([24, 25, 26, 27, 28], 28)```The sequence of five consecutive two-digit integers less than 30, where none of them are prime, is ([24, 25, 26, 27, 28]).Thus, the largest of these five integers is (boxed{28}).

answer:To solve for the distance between two points (A = (1, 3, -2)) and (B = (-2, 3, 2)) in a three-dimensional Cartesian coordinate system, we can use the distance formula:[d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}]Let's break it down into manageable steps and implement it in Python using the `sympy` library.1. Define the coordinates of points A and B.2. Use the distance formula to calculate the distance between the points.3. Print the final result.Here is the Python code to solve the problem:```pythonimport sympy as sp# Define the coordinates of points A and Bx1, y1, z1 = 1, 3, -2x2, y2, z2 = -2, 3, 2# Calculate the distance using the distance formuladistance = sp.sqrt((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2)# Print the final resultprint(distance)``````output5```The distance between points (A = (1, 3, -2)) and (B = (-2, 3, 2)) in the three-dimensional Cartesian coordinate system is (boxed{5}).

answer:To solve for ( |m - n| ) given ( mn = 4 ) and ( m + n = 5 ), we can use the properties of quadratic equations. Here's the step-by-step approach:1. We will start by considering the quadratic equation whose roots are ( m ) and ( n ).Given:[ mn = 4 ][ m + n = 5 ]The quadratic equation with roots ( m ) and ( n ) is:[ x^2 - (m+n)x + mn = 0 ]Substitute the given values:[ x^2 - 5x + 4 = 0 ]2. Solve the quadratic equation:[ x^2 - 5x + 4 = 0 ]The solutions to this quadratic equation will give us the values of ( m ) and ( n ).3. Compute the absolute difference ( |m - n| ).Let's carry out these steps using Python to ensure accuracy.```pythonimport math # Coefficients of the quadratic equation x^2 - 5x + 4 = 0 a = 1 b = -5 c = 4 # Using the quadratic formula to find the roots discriminant = b2 - 4*a*c root1 = (-b + math.sqrt(discriminant)) / (2 * a) root2 = (-b - math.sqrt(discriminant)) / (2 * a) # Calculate the absolute difference |m - n| absolute_difference = abs(root1 - root2) print(absolute_difference)``````output3.0```The absolute difference ( |m - n| ) is (boxed{3}).

answer:Let's break down the problem step-by-step and define the approach clearly before solving it with code.1. João starts with ( 1000 ) ml of water and Maria also starts with ( 1000 ) ml of water.2. The transferring rule alternates between João and Maria: - On odd days, João transfers water to Maria. - On even days, Maria transfers water to João.3. The amount of water transferred on day ( n ) is exactly ( n ) ml.Our goal is to find out how much water Maria has on the 200th day.We'll follow these steps:1. Initialize João and Maria’s water amounts.2. Simulate each transfer for 200 days.3. Alternate between João and Maria transferring water according to the day's number.4. Update their respective water amounts after each transfer.Let's implement and iterate this in Python.```python# Initialize the water amounts joao_water = 1000 # in milliliters maria_water = 1000 # in milliliters # Iterate through 200 days for day in range(1, 201): if day % 2 != 0: # odd days: João transfers to Maria transfer_amount = day joao_water -= transfer_amount maria_water += transfer_amount else: # even days: Maria transfers to João transfer_amount = day maria_water -= transfer_amount joao_water += transfer_amount # Print the final amount of water in Maria's jar print(maria_water)``````output900```After 200 days, the final amount of water in Maria's jar is:(boxed{900}) milliliters

answer:To solve this problem, we need to calculate the number of ways to allocate 10 volunteer positions to 4 schools such that each school receives at least one position. This is a combinatorial problem that can be solved using the "stars and bars" theorem with an additional constraint that each group must have at least one star.Steps:1. Use the stars and bars method to allocate ( n ) identical items (volunteer positions) into ( k ) distinct groups (schools), where each group gets at least one item.2. We need to apply a transformation to ensure that each school gets at least one position by giving each school one position first.3. Subtract the number of positions already given from the total to distribute the remaining positions.Let's implement this in Python using `sympy`:```pythonimport sympy as sp# Total number of positionsn = 10# Total number of schoolsk = 4# First, allocate one position to each school, reducing the problem to distributing (n - k) positions among k schools.remaining_positions = n - k# The number of ways to distribute (remaining_positions) positions among (k) schools is given by stars and bars formula:# Formula: C(remaining_positions + k - 1, k - 1)ways = sp.binom(remaining_positions + k - 1, k - 1)print(ways)``````outputTraceback (most recent call last): File " ways = sp.binom(remaining_positions + k - 1, k - 1)AttributeError: module 'sympy' has no attribute 'binom'```It looks like there was an issue using the `sympy` module's `binom` function correctly. In `sympy`, the correct function to use for binomial coefficients is `sp.binomial`.Let's correct the Python code:1. Use `sp.binomial` instead of `sp.binom`.Here is the corrected code:```pythonimport sympy as sp# Total number of positionsn = 10# Total number of schoolsk = 4# First, allocate one position to each school, reducing the problem to distributing (n - k) positions among k schools.remaining_positions = n - k# The number of ways to distribute (remaining_positions) positions among (k) schools is given by stars and bars formula:# Formula: C(remaining_positions + k - 1, k - 1)ways = sp.binomial(remaining_positions + k - 1, k - 1)print(ways)``````output84```It looks like there was an error in using the `binom` function from the `sympy` library. The correct function to use is `binomial`. Let's correct that and re-run the code:```pythonimport sympy as sp# Total number of positionsn = 10# Total number of schoolsk = 4# First, allocate one position to each school, reducing the problem to distributing (n - k) positions among k schools.remaining_positions = n - k# The number of ways to distribute (remaining_positions) positions among (k) schools is given by stars and bars formula:# Formula: C(remaining_positions + k - 1, k - 1)ways = sp.binomial(remaining_positions + k - 1, k - 1)print(ways)``````output84```The corrected code successfully computes the number of ways to allocate 10 volunteer positions to 4 schools, ensuring each school receives at least one position. The total number of different allocation methods is (boxed{84}).

answer:To find the perimeter of the triangle with the given points (A(1,2)), (B(1,8)), and (C(5,5)) on a Cartesian coordinate plane, we need to calculate the lengths of each side of the triangle and then sum them up.The distance between any two points ((x_1, y_1)) and ((x_2, y_2)) on a Cartesian plane can be found using the distance formula:[d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}]We will use this formula to calculate the distances (AB), (BC), and (CA). Finally, we will find the perimeter by summing these distances.Let's do this step by step.```pythonimport math# Points A, B, and CA = (1, 2)B = (1, 8)C = (5, 5)# Distance formula functiondef distance(p1, p2): return math.sqrt((p2[0] - p1[0])2 + (p2[1] - p1[1])2)# Calculate the distancesAB = distance(A, B)BC = distance(B, C)CA = distance(C, A)# Calculate the perimeter by summing up the distancesperimeter = AB + BC + CAprint(perimeter)``````output16.0```The perimeter of the triangle with vertices ( A(1,2) ), ( B(1,8) ), and ( C(5,5) ) is (boxed{16.0}).