answer:To determine how many Black Copper Marans (BCM) hens are there on the farm, we can break down the problem step by step and then validate our steps using Python code.1. Calculate the number of BCM chickens: - There are 100 chickens in total. - 20 percent of the chickens are BCM. 2. Calculate the number of BCM hens: - 80 percent of the BCM chickens are hens.Let's perform these calculations using Python:```python# Total number of chickenstotal_chickens = 100# Percentage of BCM chickenspercentage_bcm = 20 / 100# Number of BCM chickensnum_bcm_chickens = total_chickens * percentage_bcm# Percentage of BCM henspercentage_bcm_hens = 80 / 100# Number of BCM hensnum_bcm_hens = num_bcm_chickens * percentage_bcm_hensprint(num_bcm_hens)``````output16.0```The number of Black Copper Marans (BCM) hens on the farm is (boxed{16}).

answer:To solve the problem, we need to break down the information given and use it to set up equations that will help us determine the count of mugs for each color. Here are the steps we can take:1. Let ( R ) represent the number of red mugs.2. Hannah has three times more blue mugs than red mugs, so let ( B = 3R ).3. Hannah has 12 yellow mugs, so ( Y = 12 ).4. Hannah has half as many red mugs as yellow mugs, so ( R = 0.5 cdot Y ).Now, we can substitute the values and expressions we have to find the exact number of mugs in each color:- ( Y = 12 )- ( R = 0.5 cdot 12 )- ( B = 3 cdot R )Finally, we can calculate the total number of mugs Hannah has and subtract the known quantities to determine the number of mugs of another (not mentioned) color.Let's perform these calculations step by step using Python code to ensure accuracy.```python# Given informationyellow_mugs = 12red_mugs = 0.5 * yellow_mugsblue_mugs = 3 * red_mugs# Total number of mugs she hastotal_mugs = 40# Calculate the number of mugs of another colorknown_mugs = yellow_mugs + red_mugs + blue_mugsmugs_of_other_color = total_mugs - known_mugsprint(mugs_of_other_color)``````output4.0```The number of mugs in another color not mentioned is (boxed{4}).

answer:To determine if it is possible to transport fifty stones using seven three-ton trucks, we need to check if we can partition the set of stones such that the total weight assigned to each truck does not exceed 3000 kg.We will break the problem down into the following steps:1. Generate the list of stone weights: The stones have weights ranging from 370 kg to 468 kg.2. Check for partitions: Using dynamic programming or any combinatorial approach, check if it is possible to partition the weights into seven groups where each group has a total weight less than or equal to 3000 kg.Let's implement this in Python:```pythonimport itertools# Generate the list of stone weightsstone_weights = list(range(370, 469))# Define the maximum load capacity of one truckmax_load = 3000# Define the number of trucksnum_trucks = 7# Check if it's possible to partition the stone weights into 7 groupsdef can_partition(weights, num_groups, max_load): # If there are more groups than weights, it's not possible if num_groups > len(weights): return False # Try all combinations to partition the weights for comb in itertools.combinations(weights, len(weights) // num_groups): remaining_weights = list(weights) for weight in comb: remaining_weights.remove(weight) if sum(comb) <= max_load: if num_groups == 1: return True if can_partition(remaining_weights, num_groups - 1, max_load): return True return False# Check if we can partition the stone weightspossible = can_partition(stone_weights, num_trucks, max_load)print(possible)``````outputTimed out after 5 seconds.```The combinatorial approach in the previous code is not efficient for the given problem size, leading to a timeout. Instead, let's use a more efficient approach, specifically a variation of the knapsack problem that uses dynamic programming to check if the weights can be partitioned appropriately.Here's the updated approach:1. Generate the list of stone weights.2. Use dynamic programming to check for partitioning.```pythonimport numpy as np# Generate the list of stone weightsstone_weights = list(range(370, 469))# Check if a partition is possibledef can_partition(weights, num_trucks, max_load): total_weight = sum(weights) if total_weight > num_trucks * max_load: return False # Initialize DP array dp = np.zeros((num_trucks + 1, max_load + 1), dtype=bool) dp[0][0] = True for weight in weights: for trucks in range(num_trucks, 0, -1): for curr_load in range(max_load, weight-1, -1): dp[trucks][curr_load] = dp[trucks][curr_load] or dp[trucks-1][curr_load-weight] return dp[num_trucks][max_load]num_trucks = 7max_load = 3000possible = can_partition(stone_weights, num_trucks, max_load)print(possible)``````outputFalse```The output `False` indicates that it is not possible to transport fifty stones with weights ranging from 370 kg to 468 kg using seven three-ton trucks, where each truck has a maximum load capacity of 3000 kg.Therefore, we can conclude that:[boxed{text{No, it is not possible to transport the stones using the given trucks.}}]

answer:To find the coefficient of the term containing ( x^2 ) in the expansion of the binomial ( (sqrt{x} + frac{2}{x})^7 ), we can follow these steps:1. Understand the Binomial Expansion: The general term in the expansion of ( (a + b)^n ) is given by: [ T_k = binom{n}{k} a^{n-k} b^k ] For the binomial ( (sqrt{x} + frac{2}{x})^7 ), we can write the general term as: [ T_k = binom{7}{k} (sqrt{x})^{7-k} left(frac{2}{x}right)^k ]2. Simplify the General Term: Simplify the exponents of ( x ) in the general term: [ T_k = binom{7}{k} (sqrt{x})^{7-k} left(frac{2}{x}right)^k = binom{7}{k} (sqrt{x})^{7-k} cdot 2^k cdot x^{-k} ] [ T_k = binom{7}{k} cdot 2^k cdot x^{frac{7-k}{2} - k} ]3. Determine the Exponent of ( x ) to Match ( x^2 ): We need the exponent of ( x ) to be 2. Hence, we solve for ( k ) such that: [ frac{7 - k}{2} - k = 2 ] Simplify and solve for ( k ): [ frac{7 - k}{2} - k = 2 ] [ 7 - k - 2k = 4 ] [ 7 - 3k = 4 ] [ 3k = 3 ] [ k = 1 ]4. Substitute ( k = 1 ) into the General Term: Substitute ( k = 1 ) into the general term to find the coefficient: [ T_1 = binom{7}{1} cdot 2^1 cdot x^{frac{7-1}{2} - 1} ] [ T_1 = 7 cdot 2 cdot x^{frac{6}{2} - 1} ] [ T_1 = 14 cdot x^{3 - 1} ] [ T_1 = 14 cdot x^2 ]Thus, the coefficient of the term containing ( x^2 ) in the expansion of ( (sqrt{x} + frac{2}{x})^7 ) is ( 14 ).Let's confirm this calculation using Python and sympy:```pythonimport sympy as sp# Define the binomial expressionx = sp.symbols('x')binomial_expr = (sp.sqrt(x) + 2/x)7# Expand the binomial expressionexpanded_expr = sp.expand(binomial_expr)# Collect the coefficient of x^2coefficient = sp.collect(expanded_expr, x).coeff(x, 2)print(coefficient)``````output14```The coefficient of the term containing ( x^2 ) in the expansion of ( (sqrt{x} + frac{2}{x})^7 ) is (boxed{14}).

answer:To solve the problem, we need to follow these steps:1. Understand the sequence definition: - (a_1 = 1) - (a_{n+1} - a_n = sinleft(frac{(n+1)pi}{2}right))2. Generate an expression for (a_n): - By summing the differences (a_{n+1} - a_n) for (n) from 1 to (k-1), we can write: [ a_{k} = a_1 + sum_{n=1}^{k-1} sinleft(frac{(n+1)pi}{2}right) ] This gives us a way to find the terms of the sequence.3. Determine the pattern of (sinleft(frac{(n+1)pi}{2}right)): - The function (sinleft(frac{(n+1)pi}{2}right)) cycles every 4 terms: [ begin{align*} sinleft(frac{pi}{2}right) &= 1, sin(pi) &= 0, sinleft(frac{3pi}{2}right) &= -1, sin(2pi) &= 0. end{align*} ]4. Sum of the sequence: - Calculate the sum of the sequence (S_n = sum_{k=1}^{n} a_k).Let's implement this in Python using sympy to find (S_{2017}).```pythonimport sympy as sp# Initialize parametersn = 2017# Define the sequence functiondef a_k(k): if k == 1: return 1 else: return a_k(k-1) + sp.sin((k*sp.pi)/2)# Calculate the sum S_nS_n = sum(a_k(k) for k in range(1, n+1))print(S_n)``````outputTimed out after 5 seconds.```It seems the recursive implementation is not efficient enough for ( n = 2017 ). Let's take a different approach by leveraging the periodic nature of the sine function and avoid redundant calculations.1. Generate the sequence terms: By utilizing the periodicity of (sinleft(frac{(n+1)pi}{2}right)), let's calculate the terms iteratively.2. Sum the sequence terms: Sum the first 2017 terms of the sequence to obtain (S_{2017}).Here’s the optimized approach in Python:```pythonimport sympy as sp# Initialize parametersn = 2017# Define the initial terma_1 = 1# List to store the terms of the sequencea = [a_1]# Calculate the terms of the sequencefor i in range(1, n): a.append(a[i-1] + sp.sin((i+1)*sp.pi/2))# Calculate the sum of the first n termsS_n = sum(a)print(S_n)``````output1009```The optimized approach worked, and the sum of the first 2017 terms of the sequence ( {a_n} ) is ( boxed{1009} ).

answer:To solve this problem, we need to find the infimum (or greatest lower bound) for the function ( g(a) = a^2 - 4a + 6 ).Here are the steps we can take:1. Express the function: We have ( g(a) = a^2 - 4a + 6 ).2. Find the critical points: To find the minimum value, we take the derivative of ( g(a) ) and set it to zero.3. Evaluate the function: Evaluate ( g(a) ) at the critical points to find the infimum, which is ( g_{text{min}} ).Let's implement this in Python using sympy.```pythonimport sympy as sp# Define the variablea = sp.symbols('a')# Define the functiong = a2 - 4*a + 6# Find the derivative of the functiong_prime = sp.diff(g, a)# Solve for critical pointscritical_points = sp.solve(g_prime, a)# Evaluate the function at the critical points to find the infimuminfimum = g.subs(a, critical_points[0])print(infimum)``````output2```The infimum of the function ( g(a) = a^2 - 4a + 6 ) is (boxed{2}).