answer:To find the minimum value of the expression (x + frac{4}{x + 1}) when (x > 0), we can follow these steps:1. Define the function (f(x) = x + frac{4}{x + 1}).2. Find the derivative (f'(x)) and set it to zero to find the critical points.3. Check the second derivative at the critical points to determine if they are minima or maxima.4. Evaluate the function (f(x)) at the critical points to find the minimum value.Let's implement this in Python using sympy:```pythonimport sympy as sp# Define the variablex = sp.symbols('x', positive=True)# Define the functionfunc = x + 4 / (x + 1)# Compute the first derivativefirst_derivative = sp.diff(func, x)# Solve for critical points (where the first derivative is zero)critical_points = sp.solve(first_derivative, x)# Compute the second derivative to check the nature of critical pointssecond_derivative = sp.diff(first_derivative, x)# Evaluate the function at the critical pointsmin_value = Nonefor cp in critical_points: if second_derivative.subs(x, cp) > 0: # Check if it is a minimum min_value = func.subs(x, cp) breakprint(min_value)``````output3```The minimum value of the expression ( x + frac{4}{x + 1} ) when ( x > 0 ) is (boxed{3}).

answer:To determine the number of points on a "P" shape formed from a square, let's break down the shape and analyze it. 1. For a square with side length ( l ), the perimeter is ( 4l ).2. When we erase one side of the square to form the "P" shape, we are left with a perimeter that is ( 4l - l = 3l ).3. Points are placed every 1 cm along this shape.So, the number of points is equal to the length of this "P" shape plus one (since we start at the first point, then keep adding points each 1 cm).To generalize:- For a side length of ( l ), the number of points ( N ) on the "P" shape is ( 3l + 1 ).Let's verify this formula using the given example (side length of 3 cm) and then use it for the side length of 10 cm. Verification for ( l = 3 ) cm:[N = 3 times 3 + 1 = 9 + 1 = 10]This matches the given number of points Zhenya counted. Applying for ( l = 10 ) cm:[N = 3 times 10 + 1 = 30 + 1 = 31]Let's use Python to confirm this calculation.```python# function to calculate the number of points based on the side length of the squaredef calculate_points(side_length): return 3 * side_length + 1# Calculate number of points for side length of 10 cmside_length = 10points_count = calculate_points(side_length)print(points_count)``````output31```The calculation confirms that the number of points on a "P" shape formed from a square with a side length of 10 cm is:[boxed{31}]

answer:To determine the coefficient of (x^7 y^2) in the expansion of ((x-y)^8 (x+y)), we can use the binomial theorem and polynomial multiplication. Here are the steps to solve the problem:1. Expand ((x-y)^8) using the binomial theorem.2. Multiply the resulting expansion by ((x+y)).3. Identify the term with (x^7 y^2) and extract its coefficient.We'll accomplish this using Python and SymPy. Let's write the code to find the required coefficient.```pythonimport sympy as sp# Define the variablesx, y = sp.symbols('x y')# Expand (x-y)^8 using binomial expansionexpansion_1 = sp.expand((x - y)8)# Multiply the expansion by (x + y)full_expansion = sp.expand(expansion_1 * (x + y))# Extract the coefficient of x^7 * y^2coefficient = sp.Poly(full_expansion, x, y).coeff_monomial(x7 * y2)print(coefficient)``````output20```The coefficient of (x^7 y^2) in the expansion of ((x-y)^8 (x+y)) is (boxed{20}).

answer:To determine the relative positions of the given parabolas ( (1) y = x^2 - frac{1}{2} x + 2 ) and ( (2) y = x^2 + frac{1}{2} x + 2 ), we can analyze their vertices and compare their positions.For a parabola in the form ( y = ax^2 + bx + c ), the vertex can be found using the formula ( x = -frac{b}{2a} ).Let's compute the vertices of both parabolas and determine their relative positions.We'll use Python and SymPy to find these vertices and conclude the relationship between the graphs.```pythonimport sympy as sp# Define the symbolx = sp.symbols('x')# Define the equations of the parabolasy1 = x2 - (1/2)*x + 2y2 = x2 + (1/2)*x + 2# Calculate the vertices of the parabolasvertex1_x = -sp.Rational(-1/2, 2*1) # -b/(2a) for the first parabolavertex1_y = y1.subs(x, vertex1_x)vertex2_x = -sp.Rational(1/2, 2*1) # -b/(2a) for the second parabolavertex2_y = y2.subs(x, vertex2_x)# Print the verticesprint("Vertex of the first parabola (y = x^2 - 1/2 x + 2) :", (vertex1_x, vertex1_y))print("Vertex of the second parabola (y = x^2 + 1/2 x + 2):", (vertex2_x, vertex2_y))``````outputVertex of the first parabola (y = x^2 - 1/2 x + 2) : (1/4, 1.93750000000000)Vertex of the second parabola (y = x^2 + 1/2 x + 2): (-1/4, 1.93750000000000)```The output indicates the vertices of the two parabolas:1. The vertex of the first parabola ( y = x^2 - frac{1}{2} x + 2 ) is at ( left(frac{1}{4}, 1.9375right) ).2. The vertex of the second parabola ( y = x^2 + frac{1}{2} x + 2 ) is at ( left(-frac{1}{4}, 1.9375right) ).Both vertices have the same ( y )-coordinate, which means they lie on the same horizontal line. The ( x )-coordinate of the vertex of the first parabola is ( frac{1}{4} ), while the ( x )-coordinate of the vertex of the second parabola is ( -frac{1}{4} ).Since ( frac{1}{4} > -frac{1}{4} ), we can conclude that the vertex of the first parabola is to the right of the vertex of the second parabola.Thus, the graph of ( y = x^2 - frac{1}{2} x + 2 ) is to the right of the graph of ( y = x^2 + frac{1}{2} x + 2 ).Therefore, the correct answer is:[boxed{textbf{(D)} text{the graph of (1) is to the right of the graph of (2).}}]

answer:To solve this problem, we need to find the smallest possible sum of the digits in the decimal representation of the number ( n + 1 ) given that the sum of the digits of ( n ) is 2017.Let's break down the problem:1. ( n ) has a sum of digits equal to 2017.2. We need to find the smallest possible sum of the digits of ( n + 1 ).When incrementing a number, the sum of its digits can change if there is a carryover. For example, if ( n ) ends in a trailing sequence of 9's (e.g., 29, 299), incrementing ( n ) will result in a carryover which can decrease the overall digit sum.To solve this, we can use a brute-force approach:1. Start from the smallest number ( n ) whose digit sum is 2017.2. Increment ( n ) to find ( n + 1 ).3. Calculate the digit sum of ( n + 1 ) and check if it's smaller than 2017.We can write a Python program to do this efficiently.Since the sum of digits is given as 2017, constructing such a number involves using the largest digits first (9's), to minimize the total length of the number. Let's start by creating the number ( n ) with the sum of digits being 2017.Let's write the code to achieve this.```pythondef digit_sum(n): return sum(int(digit) for digit in str(n))# Start by finding a number whose sum of digits is exactly 2017n = 0current_sum = 0while current_sum < 2017: if current_sum + 9 <= 2017: current_sum += 9 n = n * 10 + 9 else: n = n * 10 + (2017 - current_sum) current_sum = 2017# Verify the constructed numbersum_digits_n = digit_sum(n)assert sum_digits_n == 2017, "Constructed number's digits do not sum to 2017"# Calculate n + 1n_plus_1 = n + 1# Calculate the sum of digits of n + 1sum_digits_n_plus_1 = digit_sum(n_plus_1)print((sum_digits_n, sum_digits_n_plus_1))``````output(2017, 2018)```The sum of the digits of ( n ) where ( n ) has a sum of digits equal to 2017 is correctly 2017. When incrementing this ( n ) by 1, the resulting sum of the digits of ( n+1 ) is 2018, which is not an improvement.To get the smallest possible sum of the digits, we need to re-examine ( n ) and find potential reductions by considering the carry while keeping sum of digits of ( n ) as 2017. For better results, let's find the digit sequences that would end up reducing the digit sums drastically when incremented by 1 (e.g., ending in a lot of '9's might result in a carryover reduction):1. Let's break down the number ( n ) more strategically.2. Focus on sequences where trailing 9’s can create a drop in digit sum when incremented.We need to find numbers like 299999 to trigger cascades in reduction:Let's write an algorithm to find such values effectively.```pythondef find_smaller_sum_digit_after_increment(target_sum): # Start from the largest digits possible to construct n digits = [] remained_sum = target_sum # Fill digits such that sum reaches the target while remained_sum > 0: if remained_sum >= 9: digits.append(9) remained_sum -= 9 else: digits.append(remained_sum) remained_sum = 0 # Construct n from the digits n = int(''.join(map(str, reversed(digits)))) # Calculate n + 1 and sum of digits n_plus_1 = n + 1 sum_digits_n = digit_sum(n) sum_digits_n_plus_1 = digit_sum(n_plus_1) return n, sum_digits_n, n_plus_1, sum_digits_n_plus_1# Test the function and verify resultsn, sum_digits_n, n_plus_1, sum_digits_n_plus_1 = find_smaller_sum_digit_after_increment(2017)print(((sum_digits_n, sum_digits_n_plus_1)))``````output(2017, 2)```The smallest possible sum of the digits in the decimal representation of ( n+1 ), given that the sum of the digits of ( n ) is 2017, is (boxed{2}).This drastic reduction is achieved because the constructed number consists of many trailing 9's (e.g., (3999999999999999ldots)), which produces significant carryover when incremented by 1, leading to a much smaller digit sum.

answer:To solve the problem of coloring (n) sectors of a circle using 5 colors such that no two adjacent sectors have the same color, we can use a combinatorial method known as the chromatic polynomial for cyclic graphs.Given a circle divided into (n) sectors (forming a cycle graph (C_n)), we need to determine the number of valid colorings using 5 colors. The chromatic polynomial for a cycle graph (C_n) can be defined recursively and is given by:[P(C_n, k) = (k-1)^n + (-1)^n(k-1)]where (k) is the number of colors available. Steps to solve the problem:1. Define the chromatic polynomial for a cycle graph.2. Substitute (k = 5) into the polynomial to determine the number of valid colorings.Let's implement this in Python using sympy.```pythonimport sympy as sp# Define the number of sectors and colorsn = sp.symbols('n')k = 5# Define the chromatic polynomial for a cycle graphchromatic_polynomial = (k - 1)n + (-1)n * (k - 1)# Simplify the expressionsimplified_polynomial = sp.simplify(chromatic_polynomial)# Define a function to compute the number of coloring schemes for given ndef number_of_coloring_schemes(n_value): return simplified_polynomial.subs(n, n_value)# Test the function for a given n, for example, n = 4result = number_of_coloring_schemes(4)print(result)``````output260```The number of valid coloring schemes for a circle divided into (n = 4) sectors using 5 colors such that no two adjacent sectors have the same color is (260).Now, to generalize the answer for any (n geq 4), we can use the same chromatic polynomial expression:[P(C_n, k) = (k-1)^n + (-1)^n(k-1)]where (k = 5).To provide the general formula for any (n geq 4), we can substitute (k = 5) into the chromatic polynomial and simplify it further:[P(C_n, 5) = (5-1)^n + (-1)^n(5-1) = 4^n + (-1)^n cdot 4]Hence, the number of valid coloring schemes for any (n geq 4) is:[boxed{4^n + (-1)^n cdot 4}]This formula can be used to find the number of valid coloring schemes for any specified (n).