answer:To solve this problem, we'll follow these steps:1. Calculate the cost of the second pair of shoes.2. Add the costs of both pairs to get the total amount Nina paid.Let's start by calculating the cost of the second pair of shoes, which is 50% more expensive than the first pair. Then, we'll find the total cost. We'll run this calculation in Python to ensure accuracy.```python# Cost of the first pair of shoescost_first_pair = 22# Cost of the second pair, which is 50% more expensivecost_second_pair = cost_first_pair * 1.50# Total cost for both pairs of shoestotal_cost = cost_first_pair + cost_second_pairprint(total_cost)``````output55.0```Nina paid a total of (boxed{55}) dollars for both pairs of shoes.

answer:To solve the problem, we need to follow these steps:1. Understand the Set ( A ): The set ( A = {x in mathbb{R} ,|, |x - 1| < 2} ) represents all real numbers ( x ) such that the absolute difference between ( x ) and 1 is less than 2.2. Determine the Interval: The inequality ( |x - 1| < 2 ) implies ( -2 < x - 1 < 2 ), which can be rewritten as: [ -2 + 1 < x < 2 + 1 implies -1 < x < 3 ] So, ( A = (-1, 3) ).3. Find Integer Elements in ( A ): The set ( A cap Z ) consists of the integers within the interval ((-1, 3)). These integers are ( 0, 1, 2 ).4. Sum the Elements: We sum the integers ( 0, 1, 2 ) to get the final result.Let's implement this in Python using sympy to ensure accuracy.```pythonimport sympy as sp# Define the interval for set Ainterval_A = sp.Interval(-1, 3, left_open=True, right_open=True)# Define the set of integers ZZ = sp.S.Integers# Intersection of set A and Zintersection_set = interval_A.intersect(Z)# Sum all elements in the intersection setsum_of_elements = sum(intersection_set)print(sum_of_elements)``````output3```The sum of all elements in the set ( A cap Z ) is (boxed{3}).

answer:```python# Let's calculate the expression 4^4 * 5^4result = (4 4) * (5 4)print(result)``````output160000```The value of ( 4^4 cdot 5^4 ) is (boxed{160000}).

answer:To generalize the problem to (x^{2} + y^{2} = a z^{2}), where (a) is a prime number of the form (4n + 3), we need to explore the solutions to this diophantine equation.Key Insight:The equation (x^2 + y^2 = az^2) has interesting properties when (a) is a prime number of the form (4n + 3). Specifically, such an equation has no non-trivial solutions in integers (i.e., (x, y, z in mathbb{Z})) unless (x = y = z = 0).Let's break this problem into steps and use Python and sympy to demonstrate this property. Steps:1. Identify the Prime Number: We start by identifying that (a) is a prime number of the form (4n + 3). For instance, the prime number 7 (since (7 = 4(1) + 3)).2. Set Up the Equation: The equation to solve is (x^2 + y^2 = az^2).3. Check for Integer Solutions: We will use sympy to automate checking for integer solutions. Implementation:We'll write a general Python function to check for solutions to the equation (x^2 + y^2 = az^2) given (a) is a prime of the form (4n + 3).```pythonimport sympy as sp# Define variablesx, y, z = sp.symbols('x y z')# Define the prime number 'a' of the form 4n + 3a = 7 # Example prime number of the form 4n + 3# Define the equation x^2 + y^2 = a * z^2equation = x2 + y2 - a * z2# Function to check if there are any integer solutions to the equationdef has_non_trivial_solutions(equation): # Solve the equation solutions = sp.solve(equation, (x, y, z), domain=sp.S.Integers) # Check if there are any non-trivial solutions for sol in solutions: if sol != (0, 0, 0): return True, sol return False, Nonehas_solution, solution = has_non_trivial_solutions(equation)print(f'Has non-trivial solutions: {has_solution}, Solution: {solution}')``````outputHas non-trivial solutions: True, Solution: (-sqrt(-y2 + 7*z2), y, z)```The output indicates that there are solutions of the form ((- sqrt{-y^2 + 7z^2}, y, z)). However, for integer values of (y) and (z), the term (sqrt{-y^2 + 7z^2}) must be an integer.To verify whether there are any non-trivial integer solutions, we need to ensure that (-y^2 + 7z^2) is a perfect square. Let's further explore this by substituting some integer values for (z) and checking if (y) results in an integer solution for (x). Implementation:We will create a Python function to iterate through some integer values of (z) and check the corresponding values of (y) and (x).```pythonimport sympy as spfrom math import isqrt# Define variablesx, y, z = sp.symbols('x y z')# Define the prime number 'a' of the form 4n + 3a = 7 # Example prime number of the form 4n + 3# Define the equation x^2 + y^2 = a * z^2equation = x2 + y2 - a * z2# Function to check for non-trivial integer solutionsdef find_non_trivial_solutions(a, max_z): solutions = [] for z_val in range(1, max_z + 1): for y_val in range(-max_z, max_z + 1): x_sq_val = a * z_val2 - y_val2 if x_sq_val >= 0: x_val = isqrt(x_sq_val) if x_val2 == x_sq_val and x_val != 0: solutions.append((x_val, y_val, z_val)) solutions.append((-x_val, y_val, z_val)) return solutions# Check for solutions with a reasonable upper bound for zmax_z = 100solutions = find_non_trivial_solutions(a, max_z)print(f'Non-trivial solutions: {solutions}')``````outputNon-trivial solutions: []```The result indicates there are no non-trivial integer solutions for the equation (x^2 + y^2 = 7z^2) for (z) values up to 100, which aligns with the theoretical expectation for (a) being a prime of the form (4n + 3). Summary:When (a) is a prime number of the form (4n + 3), the equation (x^2 + y^2 = az^2) generally has no non-trivial integer solutions. This is a known result in number theory, and the test with (a = 7) confirms this for the range of (z) values we checked.Thus, the generalized conclusion for the problem (x^2 + y^2 = az^2) where (a) is a prime of the form (4n + 3) is that there are no non-trivial integer solutions, i.e., (x = y = z = 0) is the only solution.The final answer is:[boxed{text{No non-trivial integer solutions}}]

answer:To solve the problem, we need to evaluate the function ( f(x) ) at ( x = 7 ) and ( x = 0 ). The function is defined piecewise, so we will evaluate each piece accordingly:1. For ( x = 7 ), since ( 7 geq 1 ), we use the second piece of the function ( log_{3}(x + 2) ).2. For ( x = 0 ), since ( 0 < 1 ), we use the first piece of the function ( 3^{x} + 2 ).We will then add the two results to get ( f(7) + f(0) ).Let's implement this in Python using SymPy.```pythonimport sympy as sp# Define the piecewise function f(x)x = sp.symbols('x')f = sp.Piecewise((3x + 2, x < 1), (sp.log(x + 2, 3), x >= 1))# Evaluate f(7)f_7 = f.subs(x, 7)# Evaluate f(0)f_0 = f.subs(x, 0)# Sum the two resultsresult = f_7 + f_0print(result)``````outputlog(9)/log(3) + 3```The output (log(9)/log(3) + 3) can be simplified further.First, recall that (log_b(a)) is the logarithm of (a) to the base (b). Therefore, (log(9)/log(3)) can be interpreted in the context of the change of base formula:[log_{b}(a) = frac{log(a)}{log(b)}]Given the output, (log(9)/log(3)) is essentially the logarithm of 9 to the base 3. We know that:[log_3(9) = 2]So, the expression simplifies to:[2 + 3 = 5]Thus, (f(7) + f(0) = 5).We can confirm this by simplifying the expression using SymPy.Let's modify the Python code to simplify the expression before printing the final result:```pythonimport sympy as sp# Define the piecewise function f(x)x = sp.symbols('x')f = sp.Piecewise((3x + 2, x < 1), (sp.log(x + 2, 3), x >= 1))# Evaluate f(7)f_7 = f.subs(x, 7)# Evaluate f(0)f_0 = f.subs(x, 0)# Sum the two resultsresult = f_7 + f_0# Simplify the resultsimplified_result = sp.simplify(result)print(simplified_result)``````output5```The simplified output confirms our calculations. Therefore, the value of ( f(7) + f(0) ) is:[boxed{5}]The simplified result confirms that ( f(7) + f(0) = boxed{5} ). This completes our solution: Given the function [ f(x) = begin{cases} 3^{x}+2 & (x < 1) log_{3}(x+2) & (x geq 1)end{cases} ]the value of ( f(7) + f(0) ) is indeed (boxed{5}).

answer:To solve this problem, we need to find all natural numbers ( n ) such that ( 3^{2n} - 2^n ) is a prime number. Here are the steps to solve this problem:1. Understand the problem: We need to solve for ( n ) such that ( 3^{2n} - 2^n ) is a prime number.2. Test small values of ( n ): We will test small natural numbers (usually up to a reasonable limit like 20) to see if ( 3^{2n} - 2^n ) is prime.3. Check for primality: For each ( n ), compute ( 3^{2n} - 2^n ) and check if it is a prime number.Let's write the Python code to implement these steps. ```pythonimport sympy as sp# Define a function to check if a number is primedef is_prime(num): return sp.isprime(num)# Find values of n for which 3^(2n) - 2^n is primedef find_natural_numbers(limit): result = [] for n in range(1, limit + 1): value = 3(2*n) - 2n if is_prime(value): result.append(n) return result# Set a reasonable limit for testinglimit = 20natural_numbers = find_natural_numbers(limit)print(natural_numbers)``````output[1]```Based on the output, we found that the only natural number ( n ) within the tested range ( 1 leq n leq 20 ) for which ( 3^{2n} - 2^n ) is prime is ( n = 1 ).To confirm this result, let's manually check the value for ( n = 1 ):[3^{2 times 1} - 2^1 = 3^2 - 2 = 9 - 2 = 7]Since 7 is a prime number, ( n = 1 ) is indeed a valid solution.To ensure completeness, it may be beneficial to test higher values of ( n ) or provide a mathematical justification for why higher values might not yield prime results. However, based on the empirical evidence gathered from the initial range, we have:The only value of ( n in mathbb{N} ) such that ( 3^{2n} - 2^n ) is prime is (boxed{1}).