answer:To solve the problem, we need to find the value of (3x^{2} - 9x - 7) given that (x^{2} - 3x = 2).Let's break it down into steps:1. Solve for (x) from the equation (x^{2} - 3x = 2).2. Substitute the values of (x) into the expression (3x^{2} - 9x - 7) and calculate the result.We'll use Python and SymPy to perform these steps.```pythonimport sympy as sp# Define the variablex = sp.symbols('x')# Given equationequation = sp.Eq(x2 - 3*x, 2)# Solve for xsolutions = sp.solve(equation, x)# Define the expression to evaluateexpression = 3*x2 - 9*x - 7# Calculate the value of the expression for each solutionresults = [expression.subs(x, sol) for sol in solutions]print(results)``````output[-41/2 + 3*(3/2 - sqrt(17)/2)2 + 9*sqrt(17)/2, -41/2 - 9*sqrt(17)/2 + 3*(3/2 + sqrt(17)/2)2]```It seems the output is still in its symbolic form and needs to be simplified to obtain the numerical values. Let's simplify the expressions and evaluate the numerical results of each solution.We can simplify the expression step-by-step, ensuring we are substituting and calculating it correctly.```pythonimport sympy as sp# Define the variablex = sp.symbols('x')# Given equationequation = sp.Eq(x2 - 3*x, 2)# Solve for xsolutions = sp.solve(equation, x)# Define the expression to evaluateexpression = 3*x2 - 9*x - 7# Calculate the value of the expression for each solutionresults = [sp.simplify(expression.subs(x, sol)) for sol in solutions]print(results)``````output[-1, -1]```The simplified results for the expression (3x^2 - 9x - 7) given the equation (x^2 - 3x = 2) are (-1) for both solutions.Therefore, the value of the algebraic expression (3x^2 - 9x - 7) is (boxed{-1}), which corresponds to option C.

answer:To solve the given problem, we can break it down into manageable steps and use sympy for symbolic computation.1. We are given two equations involving (a) and (b): [ (a+b)^2 = 17 ] [ (a-b)^2 = 11 ]2. We need to find (a^2 + b^2).We can expand and simplify the given equations:[(a+b)^2 = a^2 + 2ab + b^2 = 17][(a-b)^2 = a^2 - 2ab + b^2 = 11]Let's add these two equations to eliminate the (2ab) and (-2ab) terms:[(a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) = 17 + 11]Simplifying, we get:[2a^2 + 2b^2 = 28]Dividing both sides by 2, we obtain:[a^2 + b^2 = 14]Let's verify this using sympy to be sure.```pythonimport sympy as sp# Define variablesa, b = sp.symbols('a b')# Define the equationseq1 = (a + b)2 - 17eq2 = (a - b)2 - 11# Solve the system of equationssolutions = sp.solve([eq1, eq2], (a, b))# Substitute the solutions back into a^2 + b^2result = []for sol in solutions: a_val, b_val = sol result_value = (a_val2 + b_val2).evalf() result.append(result_value)# Ensure to take the unique result since a^2 + b^2 should be same for all valid (a, b) pairsunique_result = set(result)final_answer = unique_result.pop() # Since it should contain only one unique valueprint(final_answer)``````output14.0000000000000```The verification confirms our manual calculation. Therefore, the value of (a^2 + b^2) is (boxed{14}).

answer:To solve this problem, let's follow these steps:1. Understand the problem and set up the equations: - Triangle (ABC) has area 2007. - Triangle (ADE) has area 7002. - Coordinates of points (B), (C), (D), and (E) are given.2. Formula for the area of a triangle using vertices: The area of a triangle with vertices ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)) is given by: [ text{Area} = frac{1}{2} left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) right| ]3. Set up equations for the areas: - For triangle (ABC): [ 2007 = frac{1}{2} left| A_x (0 - 0) + 0(0 - y_A) + 223(y_A - 0) right| = frac{1}{2} left| 223y_A right| = frac{223 |y_A|}{2} ] [ 2007 = frac{223 |y_A|}{2} implies |y_A| = frac{2007 times 2}{223} = 18 ] Hence, (y_A = 18) or (y_A = -18). - For triangle (ADE): [ 7002 = frac{1}{2} left| A_x(380 - 389) + 680(389 - y_A) + 689(y_A - 380) right| ] Since we know (y_A = 18) or (-18), we can plug these values in and solve for (x_A).Let's implement this in Python using sympy:```pythonimport sympy as sp# Define the variablesxA = sp.symbols('xA')yA1, yA2 = 18, -18# Define the equation for the area of triangle ADEarea_ADE = 7002equation1 = sp.Eq( (1/2) * sp.Abs(xA * (380 - 389) + 680 * (389 - yA1) + 689 * (yA1 - 380)), area_ADE)equation2 = sp.Eq( (1/2) * sp.Abs(xA * (380 - 389) + 680 * (389 - yA2) + 689 * (yA2 - 380)), area_ADE)# Solve for xAsolutions1 = sp.solve(equation1, xA)solutions2 = sp.solve(equation2, xA)# Sum of all possible x-coordinates of Atotal_solutions = solutions1 + solutions2sum_xA = sum(total_solutions)print(sum_xA)``````outputTraceback (most recent call last): File " solutions1 = sp.solve(equation1, xA)NotImplementedError: solving Abs(9*xA - 2862) when the argument is not real or imaginary.```The error occurs because SymPy is unable to directly solve equations involving the absolute value function ( text{Abs} ) in the context provided. To handle this, we can split the absolute value equation into two cases: one for the positive part and one for the negative part of the expression inside the absolute value. Let's rewrite our equations without the absolute value and solve each case separately:1. Positive part: [ 7002 = frac{1}{2} (xA (380 - 389) + 680 (389 - yA) + 689 (yA - 380)) ]2. Negative part: [ 7002 = frac{1}{2} (-xA (380 - 389) - 680 (389 - yA) - 689 (yA - 380)) ]Let's break down our Python code to handle these cases:```pythonimport sympy as sp# Define the variablesxA = sp.symbols('xA')yA1, yA2 = 18, -18# Define the expressions for the area calculations without absolute valueexpr1_positive = xA * (380 - 389) + 680 * (389 - yA1) + 689 * (yA1 - 380)expr1_negative = -xA * (380 - 389) - 680 * (389 - yA1) - 689 * (yA1 - 380)expr2_positive = xA * (380 - 389) + 680 * (389 - yA2) + 689 * (yA2 - 380)expr2_negative = -xA * (380 - 389) - 680 * (389 - yA2) - 689 * (yA2 - 380)# Set up equations for both positive and negative casesequation1_pos = sp.Eq((1/2) * expr1_positive, 7002)equation1_neg = sp.Eq((1/2) * expr1_negative, 7002)equation2_pos = sp.Eq((1/2) * expr2_positive, 7002)equation2_neg = sp.Eq((1/2) * expr2_negative, 7002)# Solve for xA in both casessolutions1_pos = sp.solve(equation1_pos, xA)solutions1_neg = sp.solve(equation1_neg, xA)solutions2_pos = sp.solve(equation2_pos, xA)solutions2_neg = sp.solve(equation2_neg, xA)# All solutions combinedtotal_solutions = solutions1_pos + solutions1_neg + solutions2_pos + solutions2_neg# Sum of all possible x-coordinates of Asum_xA = sum(total_solutions)print(sum_xA)``````output1200.00000000000```The sum of all possible ( x )-coordinates of ( A ) is (boxed{1200}).

answer:To find how many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits, we proceed as follows:1. Let the three-digit number be represented as (ABC), where (A), (B), and (C) are the digits of the number.2. The condition given is that (B), the middle digit, is the average of (A) and (C).Mathematically, this means:[B = frac{A + C}{2}] or equivalently (2B = A + C).Since (A), (B), and (C) must all be digits (0-9), we need to check all possible three-digit numbers (where (A) is from 1 to 9, and (B) and (C) from 0 to 9) and count how many of them satisfy (2B = A + C).Let's implement this in Python to find the count.```python# We will iterate over all possible three-digit numbers and count# those that satisfy the condition 2B = A + Ccount = 0# A: 1 to 9 (because it's a three-digit number)# B: 0 to 9 (any possible middle digit)# C: 0 to 9 (any possible last digit)for A in range(1, 10): for B in range(0, 10): for C in range(0, 10): if 2 * B == A + C: count += 1print(count)``````output45```The number of three-digit numbers that satisfy the property that the middle digit is the average of the first and the last digits is (boxed{45}).

answer:To solve this problem, we need to calculate the number of students who scored no less than 120 points, given that the mathematics scores are normally distributed with a mean of 100 and an unknown standard deviation (a). We know that approximately 60% of the students scored between 80 and 120 points.Here's a step-by-step outline of the solution:1. Determine the standard deviation (a): - 60% of the students scored between 80 and 120 points. - This implies that the probability (P(80 leq xi leq 120)) is 0.6. - Use the properties of the normal distribution to find (a).2. Calculate the Z-scores for 80 and 120: - Convert the scores 80 and 120 to their respective Z-scores using the formula (Z = frac{X - mu}{sigma}) where (mu) is the mean and (sigma) is the standard deviation.3. Use the Z-scores to find the standard deviation (a).4. Calculate the probability of scoring no less than 120 points: - Convert the score 120 to its Z-score. - Use the cumulative distribution function (CDF) to find the probability of scoring no less than 120 points.5. Calculate the number of students scoring no less than 120 points: - Multiply the probability by the total number of students.Let's implement this in Python using SymPy and SciPy.```pythonimport sympy as spfrom scipy.stats import norm# Given valuesmu = 100 # meanp = 0.60 # probability between 80 and 120total_students = 1000# Calculate the Z-scores for 80 and 120Z1 = sp.symbols('Z1')Z2 = sp.symbols('Z2')Z1_eq = sp.Eq(Z1, (80 - mu) / Z2)Z2_eq = sp.Eq(Z2, (120 - mu) / Z2)# Use the properties of the normal distribution to find Z1 and Z2 valuesZ1_val = norm.ppf(0.20) # P(Z < Z1) = 0.20Z2_val = norm.ppf(0.80) # P(Z < Z2) = 0.80# Set up the system of equationsequations = [Z1_eq.subs(Z1, Z1_val), Z2_eq.subs(Z2, Z2_val)]# Solve for the standard deviation asolution = sp.solve(equations, Z2)std_dev = solution[0]# Calculate the probability of scoring no less than 120 pointsZ_120 = (120 - mu) / std_devP_120 = 1 - norm.cdf(Z_120)# Calculate the number of students scoring no less than 120 pointsstudents_120 = total_students * P_120print(students_120)``````outputTraceback (most recent call last): File " from scipy.stats import normModuleNotFoundError: No module named 'scipy'```It seems that the environment does not have the `scipy` module installed. To work around this, we'll use the `sympy` module exclusively to calculate the essential probabilities using the error function `erf`, which is related to the normal distribution's cumulative distribution function (CDF).Here is the revised approach:1. Determine the standard deviation (a): - Use the property (P(80 leq xi leq 120) = 0.6) to find the standard deviation.2. Calculate the Z-scores for 80 and 120: - Convert the scores 80 and 120 to their respective Z-scores using the formula (Z = frac{X - mu}{sigma}).3. Use the Z-scores to find the standard deviation (a).4. Calculate the probability of scoring no less than 120 points: - Convert the score 120 to its Z-score. - Use the error function `erf` to find the probability of scoring no less than 120 points.5. Calculate the number of students scoring no less than 120 points: - Multiply the probability by the total number of students.Let's implement this approach in Python using SymPy.```pythonimport sympy as sp# Given valuesmu = 100 # meanp = 0.60 # probability between 80 and 120total_students = 1000# Define the variablesa = sp.Symbol('a', positive=True)# Calculate the Z-scores for 80 and 120Z1 = (80 - mu) / aZ2 = (120 - mu) / a# Use the erf function to solve for the standard deviation aequation = sp.Eq((sp.erf(Z2 / sp.sqrt(2)) - sp.erf(Z1 / sp.sqrt(2))) / 2, p)solution = sp.solve(equation, a)std_dev = solution[0]# Calculate the Z-score for 120Z_120 = (120 - mu) / std_dev# Probability of scoring no less than 120 pointsP_120 = 0.5 * (1 - sp.erf(Z_120 / sp.sqrt(2)))# Calculate the number of students scoring no less than 120 pointsstudents_120 = total_students * P_120print(sp.N(students_120))``````output200.000000000000```The output indicates that approximately 200 students scored no less than 120 points in the November monthly exam.Thus, the number of students who scored no less than 120 points is (boxed{200}).

answer:To solve this problem, we need to find the value of ( k ) such that ( k overrightarrow{a} + overrightarrow{b} ) is perpendicular to ( overrightarrow{a} - 3 overrightarrow{b} ).Two vectors are perpendicular if and only if their dot product is zero.Let's break it down step-by-step:1. Define the vectors (overrightarrow{a}) and (overrightarrow{b}): [ overrightarrow{a} = begin{pmatrix} 1 2 end{pmatrix}, quad overrightarrow{b} = begin{pmatrix} -3 2 end{pmatrix} ]2. Form the vectors ( k overrightarrow{a} + overrightarrow{b} ) and (overrightarrow{a} - 3 overrightarrow{b} ): [ k overrightarrow{a} + overrightarrow{b} = begin{pmatrix} k 2k end{pmatrix} + begin{pmatrix} -3 2 end{pmatrix} = begin{pmatrix} k - 3 2k + 2 end{pmatrix} ] [ overrightarrow{a} - 3 overrightarrow{b} = begin{pmatrix} 1 2 end{pmatrix} - 3 begin{pmatrix} -3 2 end{pmatrix} = begin{pmatrix} 1 + 9 2 - 6 end{pmatrix} = begin{pmatrix} 10 -4 end{pmatrix} ]3. Set up the dot product of these two vectors and set it to zero: [ begin{pmatrix} k - 3 2k + 2 end{pmatrix} cdot begin{pmatrix} 10 -4 end{pmatrix} = 0 ] [ (k - 3) cdot 10 + (2k + 2) cdot (-4) = 0 ]4. Solve for ( k ).Let's implement this in Python to find the value of ( k ):```pythonimport sympy as sp# Define the variablek = sp.symbols('k')# Define the vectorsa = sp.Matrix([1, 2])b = sp.Matrix([-3, 2])# Define the expressions for the vectors involvedvector1 = k * a + bvector2 = a - 3 * b# Calculate the dot product and solve for kdot_product = vector1.dot(vector2)solution = sp.solve(dot_product, k)print(solution)``````output[19]```The value of ( k ) such that ( k overrightarrow{a} + overrightarrow{b} ) is perpendicular to ( overrightarrow{a} - 3 overrightarrow{b} ) is (boxed{19}).