answer:To find the value of (a) for the given function (f(x) = lg(1 - frac{a}{2^x})), we need to ensure that the function stays within its domain for (x in (4, +infty)). The domain requires that the argument of the logarithm be positive. Thus, we need:[1 - frac{a}{2^x} > 0]This implies:[1 > frac{a}{2^x} implies 2^x > a]Because (x in (4, +infty)), we need:[inf_{x > 4} 2^x > a]As (2^x) grows exponentially with increasing (x), the minimum value of (2^x) for (x > 4) is when (x) is just slightly greater than 4. Therefore, we need:[2^4 = 16 > a implies a < 16]Thus, the value of (a) must be less than 16 to keep the function within its domain. Hence, the largest value (a) can take is just slightly less than 16.Let's calculate this using Python and sympy for a formal verification.```pythonimport sympy as sp# Define the variable x and the parameter ax = sp.Symbol('x')a = sp.Symbol('a')# Define the function f(x) = log(1 - a / 2^x)f = sp.log(1 - a / 2x, 10)# Define the constraint for the domain: 1 - a / 2^x > 0constraint = 1 - a / 2x > 0# Solve the constraint for a when x > 4solution = sp.solve(constraint.subs(x, 4), a)print(solution)``````output(-oo < a) & (a < 16)```The output confirms our reasoning that (a) must be less than 16 for the function (f(x) = lg(1 - frac{a}{2^x})) to be defined over the domain (x in (4, +infty)).Thus, the largest possible value of (a) that satisfies the domain constraint is (a = 16).To summarize:The value of (a) must be such that (a < 16). Therefore, the largest value (a) can take is (boxed{16}).

answer:Let's break down the problem step by step:1. Determine the current year from the information given. If in 12 days it will be the year 2016, then today it is still the year 2015.2. Hao Hao has experienced 2 leap years up to now (i.e., up to the year 2015).3. Hao Hao's birth year is a multiple of 9.4. We need to determine how old Hao Hao will be in 2016.First, let's identify all leap years between 2015 and Hao Hao's birth year:- Leap years must be divisible by 4.- Years that are multiples of 100 are not leap years unless they're also multiples of 400.Given that Hao Hao has experienced 2 leap years, let's iterate through potential birth years that are multiples of 9 and count the leap years between those years and 2015. We'll continue this process until we find the correct year that fits the criteria.Let's start by using Python code to enumerate potential years and count the leap years between them.```pythondef count_leap_years(birth_year, end_year): """ Count the number of leap years between birth_year and end_year inclusive. """ leap_years = 0 for year in range(birth_year, end_year + 1): if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0): leap_years += 1 return leap_yearscurrent_year = 2015 # Given that 2016 is in 12 days.expected_leap_years = 2 # Hao Hao has experienced 2 leap years# Iterate through candidate birth years that are multiples of 9for birth_year in range(0, current_year, 9): if count_leap_years(birth_year, current_year) == expected_leap_years: break# Calculate Hao Hao's age in 2016hao_hao_age_in_2016 = 2016 - birth_yearprint((birth_year, hao_hao_age_in_2016))``````output(2007, 9)```From the calculations, Hao Hao was born in 2007. This can be verified since:1. 2008 is a leap year and 2012 is a leap year, making a total of 2 leap years by the end of 2015.2. 2007 is a multiple of 9.3. Hao Hao was born in 2007, so in 2016, he will be:[2016 - 2007 = 9]Thus, Hao Hao will be (boxed{9}) years old in 2016.

answer:To find the maximum value of the function ( f(x) = sqrt{3} sin 2x + 2 sin x + 4 sqrt{3} cos x ), we need to follow these steps:1. Define the function in Python using sympy.2. Find the derivative of the function.3. Find the critical points by setting the derivative to zero and solving for (x).4. Evaluate the function at the critical points to determine the maximum value.Let's implement this in Python using sympy:```pythonimport sympy as sp# Define the variable and the functionx = sp.symbols('x')f = sp.sqrt(3) * sp.sin(2 * x) + 2 * sp.sin(x) + 4 * sp.sqrt(3) * sp.cos(x)# Find the derivative of the functionf_prime = sp.diff(f, x)# Solve for critical points (where derivative is zero)critical_points = sp.solve(f_prime, x)# Evaluate the function at the critical points to find the maximum valuemax_value = -float('inf')for point in critical_points: value = f.subs(x, point) if value > max_value: max_value = valueprint(max_value)``````outputTimed out after 5 seconds.```It appears that the symbolic computation might be taking too long due to the complexity of the trigonometric functions involved. We can try simplifying the approach by evaluating the function at several critical points within one period (since trigonometric functions are periodic) and then finding the maximum value among those points.Let's re-implement the solution by evaluating the function's derivative within a limited interval and finding the maximum value. To handle the periodic nature of trigonometric functions, we can evaluate the function over the interval ([0, 2pi]):```pythonimport sympy as spimport math# Define the variable and the functionx = sp.symbols('x')f = sp.sqrt(3) * sp.sin(2 * x) + 2 * sp.sin(x) + 4 * sp.sqrt(3) * sp.cos(x)# Find the derivative of the functionf_prime = sp.diff(f, x)# Solve for critical points (where derivative is zero) within the interval [0, 2*pi]critical_points = sp.solve(f_prime, x)critical_points = [point.evalf() for point in critical_points if 0 <= point.evalf() <= 2 * math.pi]# Add the endpoints of the interval to the list of points to checkcritical_points.extend([0, 2 * math.pi])# Evaluate the function at the critical points to find the maximum valuemax_value = -float('inf')for point in critical_points: value = f.subs(x, point) if value > max_value: max_value = valueprint(max_value)``````outputTimed out after 5 seconds.```Given that the symbolic computation might be too heavy or complex for direct differentiation and solving, another approach would be to numerically evaluate the function over a dense grid of points within one period ([0, 2pi]). This way, we can approximate the maximum value without directly solving the derivative.Let's use NumPy for this numerical approach:```pythonimport numpy as np# Define the functiondef f(x): return np.sqrt(3) * np.sin(2 * x) + 2 * np.sin(x) + 4 * np.sqrt(3) * np.cos(x)# Create a dense grid of points between 0 and 2*pix_values = np.linspace(0, 2 * np.pi, 10000)# Evaluate the function at each grid pointf_values = f(x_values)# Find the maximum valuemax_value = np.max(f_values)print(max_value)``````output8.499999839590206```The numerical approach has given us an approximate maximum value of the function ( f(x) = sqrt{3} sin 2x + 2 sin x + 4 sqrt{3} cos x ) within the interval ([0, 2pi]).The maximum value obtained is approximately:[ boxed{8.5} ]This numerical approximation should be sufficiently accurate for most practical purposes.

answer:Let's break down the problem step-by-step and use Python code to ensure the solution is accurate.1. Identify the known quantities: - Treadmill sale price = 100 - Chest of drawers sale price = frac{1}{2} of the treadmill sale price - Television sale price = 3 times the treadmill sale price2. Calculate the individual sale prices: - Chest of drawers sale price = frac{1}{2} times 100 - Television sale price = 3 times 1003. Calculate the total amount made from the three items.4. Given that these three items made up 75% of the total sales, set up an equation to find the total sum of money made at the garage sale.Let's compute the values using Python.```python# Given values treadmill_price = 100 # Calculate prices of other items chest_of_drawers_price = 0.5 * treadmill_price television_price = 3 * treadmill_price # Total amount from these three items total_from_three_items = treadmill_price + chest_of_drawers_price + television_price # These three items made up 75% of the total sales total_sales = total_from_three_items / 0.75 print(total_sales)``````output600.0```The total sum of money, in dollars, that Toby made at the garage sale is (boxed{600}).

answer:To solve this problem, we need to find the coefficient of (x^3y^3) in the expansion of ((2x-1)(x+y)^5).Let's break it down into manageable steps:1. Expand ((x + y)^5) using the binomial theorem.2. Multiply the result by ((2x - 1)).3. Identify the coefficient of the (x^3y^3) term in the expanded form.Let's implement this in Python using sympy.```pythonimport sympy as sp# Define the variablesx, y = sp.symbols('x y')# Expand (x + y)^5 using the binomial theoremexpansion = sp.expand((x + y)5)# Multiply the expansion by (2x - 1)full_expansion = sp.expand((2*x - 1) * expansion)# Extract the coefficient of x^3y^3coefficient = sp.collect(full_expansion, x).coeff(x3).coeff(y3)print(coefficient)``````output20```The coefficient of (x^3y^3) in the expansion of ((2x-1)(x+y)^5) is (boxed{20}).

answer:To find the probability that a randomly chosen element from among the first 15 rows of Pascal's Triangle is equal to 1, we need to consider the following:1. Pascal's Triangle is symmetric, and it starts with 1 at the top. Each row ( n ) of Pascal's Triangle contains ( n + 1 ) elements, where the elements are the binomial coefficients ( binom{n}{k} ) for ( k = 0, 1, ldots, n ).2. The value '1' appears at the ends of each row and also in the middle of the first row. For any row ( n ) (where ( n geq 0 )), the first and last elements ( ( binom{n}{0} ) and ( binom{n}{n} ) ) are always 1.3. To find our probability, we need to: - Count the total number of elements within the first 15 rows (rows 0 through 14). - Count how many of these elements are the value '1'. - Calculate the ratio of the number of '1's to the total number of elements.Let's go through this step-by-step using Python.First, calculate the total number of elements in the first 15 rows. Then, count how many of these elements are '1'. Lastly, find the probability.```python# Calculate the total number of elements in the first 15 rows (0 through 14) num_rows = 15 total_elements = sum(n + 1 for n in range(num_rows)) print(total_elements)``````output120```The total number of elements in the first 15 rows of Pascal's Triangle is 120.Next, we need to count how many of these elements are '1'. As mentioned, every row (n) has two '1's at the ends for all rows (n geq 1), except row 0, which has only one '1'.Let's calculate the total number of '1's in these rows.```python# Calculate the total number of '1's in the first 15 rows (0 through 14)num_ones = 1 # Row 0 has one '1'num_ones += sum(2 for n in range(1, num_rows)) # Each subsequent row from 1 to 14 has two '1'sprint(num_ones)``````output29```The total number of '1's in the first 15 rows of Pascal's Triangle is 29.Now, we can calculate the probability that a randomly chosen element from these rows is '1' by dividing the number of '1's by the total number of elements.Let's compute this probability.```python# Calculate the probability that a randomly chosen element is '1' probability_ones = num_ones / total_elements print(probability_ones)``````output0.24166666666666667```The probability that a randomly chosen element from the first 15 rows of Pascal's Triangle is equal to '1' is approximately (0.2417). Expressing this as a fraction out of 120, we get:[ boxed{frac{29}{120}} ]