answer:To calculate the given expression (sqrt{3}(1 - sqrt{3}) - |-sqrt{3}| + sqrt[3]{27}), we can break it down into smaller parts and then use Python with the sympy library to compute each part step-by-step.Here are the steps:1. Compute (sqrt{3}).2. Compute (1 - sqrt{3}).3. Multiply (sqrt{3}) by the result from step 2.4. Compute the absolute value of (-sqrt{3}).5. Compute the cube root of 27.6. Combine all these results to obtain the final value of the expression.Let's write the Python code to perform these calculations:```pythonimport sympy as sp# Define the components of the expressionsqrt_3 = sp.sqrt(3)abs_negative_sqrt_3 = sp.Abs(-sqrt_3)cube_root_27 = sp.root(27, 3)# Calculate each part of the expressionpart1 = sqrt_3 * (1 - sqrt_3)part2 = abs_negative_sqrt_3part3 = cube_root_27# Combine all parts to get the final resultresult = part1 - part2 + part3print(result)``````output-sqrt(3) + sqrt(3)*(1 - sqrt(3)) + 3```The output from the code is `-sqrt(3) + sqrt(3)*(1 - sqrt(3)) + 3`. To get a simplified numerical result, we need to further simplify the expression:Let's proceed with the simplification step by step:1. Distribute (sqrt{3}) in the term (sqrt{3}(1 - sqrt{3})): [ sqrt{3} times 1 - sqrt{3} times sqrt{3} = sqrt{3} - 3 ]2. Combine the terms: [ -sqrt{3} + (sqrt{3} - 3) + 3 ]3. Simplify by combining like terms.Let's implement this in Python using sympy to get the final simplified result:```pythonimport sympy as sp# Define the components of the expressionsqrt_3 = sp.sqrt(3)abs_negative_sqrt_3 = sp.Abs(-sqrt_3)cube_root_27 = sp.root(27, 3)# Calculate each part of the expressionpart1 = sqrt_3 * (1 - sqrt_3)part2 = abs_negative_sqrt_3part3 = cube_root_27# Combine all parts to get the final resultresult = part1 - part2 + part3# Simplify the resultsimplified_result = sp.simplify(result)print(simplified_result)``````output0```The simplified result of the expression (sqrt{3}(1 - sqrt{3}) - |-sqrt{3}| + sqrt[3]{27}) is (0).So, the final answer is:[boxed{0}]

answer:To solve the problem, we need to find the values of (a) and (b) such that (f(f(x)) = 4x - 3). Given that (f(x) = ax - b), we can write the nested function (f(f(x))) and solve for (a) and (b).Steps:1. Express (f(f(x))) in terms of (a) and (b).2. Set this expression equal to (4x - 3).3. Solve for (a) and (b).4. Use the values of (a) and (b) to find (f(2)).Let's break it down and solve it using Python and SymPy.```pythonimport sympy as sp# Define symbolsx, a, b = sp.symbols('x a b')# Given function f(x) = ax - bf_x = a*x - b# Express f(f(x))f_f_x = a*(a*x - b) - b# Simplify the expressionf_f_x = sp.simplify(f_f_x)# We know that f(f(x)) = 4x - 3eq = sp.Eq(f_f_x, 4*x - 3)# Solve for a and bsolutions = sp.solve(eq, (a, b))a_val, b_val = solutions[0]# Define the function f(x) with the found values of a and bf_x_with_values = a_val*x - b_val# Find f(2)f_2 = f_x_with_values.subs(x, 2)print(f_2)``````output2*a - (2*a2 - 5)/(a + 1)```It seems that there has been a small misunderstanding. To clarify, let's go step by step to solve for (a) and (b) such that ( f(f(x)) = 4x - 3 ).1. Express (f(f(x))): Since (f(x) = ax - b), we have: [ f(f(x)) = a(ax - b) - b = a^2 x - ab - b ]2. Set this equal to (4x - 3): [ a^2 x - ab - b = 4x - 3 ]3. Equate the coefficients: - For the (x) term: (a^2 = 4) - For the constant term: (-ab - b = -3)4. Solve for (a) and (b):Let's proceed to solve these equations using SymPy in Python:```pythonimport sympy as sp# Define symbolsx, a, b = sp.symbols('x a b')# Given equationseq1 = sp.Eq(a2, 4)eq2 = sp.Eq(-a*b - b, -3)# Solve for a and bsolutions_a = sp.solve(eq1, a)solutions = []for a_val in solutions_a: b_val = sp.solve(eq2.subs(a, a_val), b)[0] solutions.append((a_val, b_val))# Use the first solution (a > 0)a_val, b_val = [sol for sol in solutions if sol[0] > 0][0]# Define the function f(x) with the found values of a and bf_x_with_values = a_val * x - b_val# Find f(2)f_2 = f_x_with_values.subs(x, 2)print(f_2)``````output3```The calculation shows that the value of ( f(2) ) is ( boxed{3} ). This confirms that with the given function ( f(x) = ax - b ) and the condition ( f(f(x)) = 4x - 3 ), the specific function parameters ( a ) and ( b ) yield ( f(2) = 3 ).

answer:To find the equation of the line ( l ) that passes through the intersection point ( P ) of the lines ( l_1 ) and ( l_2 ) and also passes through the origin, we can follow these steps:1. Find the intersection point ( P ) of the lines ( l_1 ) and ( l_2 ).2. Determine the slope of the line ( l ) that passes through the origin and the intersection point ( P ).3. Use the slope-point form to write the equation of the line ( l ).Let's break this down with Python and SymPy.1. Find the intersection point ( P ) of the lines ( l_1 ) and ( l_2 ): The equations of the lines are: ( l_1: 2x + y = 3 ) ( l_2: x + 4y = 2 ) We need to solve these equations simultaneously to find the intersection point ( P ).2. Determine the slope of the line ( l ) that passes through the origin and the intersection point ( P ): The slope ( m ) of a line passing through two points ( (x_1, y_1) ) and ( (x_2, y_2) ) is given by: [ m = frac{y_2 - y_1}{x_2 - x_1} ] In our case, one of the points is the origin ((0, 0)) and the other point is ( P ).3. Use the slope-point form to write the equation of the line ( l ): The slope-point form of the equation of a line is: [ y - y_1 = m(x - x_1) ] Since the line passes through the origin ((0, 0)), the equation simplifies to: [ y = mx ]Let's implement this in Python using SymPy:```pythonimport sympy as sp# Define the variablesx, y = sp.symbols('x y')# Define the equations of the lineseq1 = 2*x + y - 3eq2 = x + 4*y - 2# Solve the system of equations to find the intersection pointintersection = sp.solve((eq1, eq2), (x, y))P_x, P_y = intersection[x], intersection[y]# Calculate the slope of the line passing through the origin and the intersection pointslope = P_y / P_x# The equation of the line passing through the origin with the calculated slopeline_eq = sp.Eq(y, slope * x)print(line_eq)``````outputEq(y, x/10)```The equation of the line ( l ) that passes through the intersection point ( P ) of the lines ( l_1: 2x + y = 3 ) and ( l_2: x + 4y = 2 ), as well as the origin, is (boxed{y = frac{1}{10}x}).

answer:```python# Define the dimensions and capacity of the first boxheight1 = 2 # cmwidth1 = 3 # cmlength1 = 5 # cmcapacity1 = 40 # grams# Calculate the volume of the first boxvolume1 = height1 * width1 * length1# Define the dimensions of the second boxheight2 = 2 * height1 # twice the height of the first boxwidth2 = 3 * width1 # three times the width of the first boxlength2 = length1 # same length as the first box# Calculate the volume of the second boxvolume2 = height2 * width2 * length2# Since the amount of clay the box can hold is proportional to its volume,# we use the ratio of the volumes to calculate the capacity of the second boxcapacity2 = (volume2 / volume1) * capacity1print(capacity2)``````output240.0```The second box can hold ( n ) grams of clay, where ( n ) is calculated to be ( boxed{240} ).

answer:Let's solve this problem step-by-step using Python and SymPy. Step 1: Find the equation of the asymptotes of the hyperbola ( x^2 - frac{y^2}{4} = 1 )The general form of the hyperbola ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ) has asymptotes given by ( y = pm frac{b}{a} x ).For our hyperbola, ( a^2 = 1 ) and ( b^2 = 4 ), so ( a = 1 ) and ( b = 2 ).Thus, the asymptotes are:[ y = 2x ][ y = -2x ] Step 2: When the coordinates of ( P ) are ( (x_0, 2) ), find the equation of line ( l )Given point ( P(x_0, 2) ) is on the hyperbola:[ x_0^2 - frac{2^2}{4} = 1 ][ x_0^2 - 1 = 1 ][ x_0^2 = 2 ][ x_0 = pm sqrt{2} ]Since ( P ) is on the right branch, we take ( x_0 = sqrt{2} ).Next, the line ( l ) passes through ( P ) and has the form ( y = mx + c ). Since it intersects the asymptotes at ( A ) and ( B ) and ( P ) is the midpoint of ( AB ), we have:[ A = (a, 2a) quad text{and} quad B = (b, -2b) quad text{for some points } a text{ and } b. ]The midpoint condition gives:[ left(frac{a+b}{2}, frac{2a-2b}{2}right) = (sqrt{2}, 2) ]This gives us two equations:[ frac{a+b}{2} = sqrt{2} ][ a - b = 2 ]Solving these two equations for ( a ) and ( b ):[ a + b = 2sqrt{2} ][ a - b = 2 ]Adding these two equations:[ 2a = 2sqrt{2} + 2 implies a = sqrt{2} + 1 ]Subtracting the second equation from the first:[ 2b = 2sqrt{2} - 2 implies b = sqrt{2} - 1 ]Thus, the points ( A ) and ( B ) are:[ A = (sqrt{2} + 1, 2(sqrt{2} + 1)) ][ B = (sqrt{2} - 1, -2(sqrt{2} - 1)) ]The line passing through ( P ) and intersecting the asymptotes at ( A ) and ( B ) is the one we need:[ y - 2 = m(x - sqrt{2}) ]To find the slope ( m ), we use points ( A ) and ( B ):[ m = frac{2(sqrt{2} + 1) - (-2(sqrt{2} - 1))}{(sqrt{2} + 1) - (sqrt{2} - 1)} ][ m = frac{2(sqrt{2} + 1) + 2(sqrt{2} - 1)}{2} ][ m = 2sqrt{2} ]So, the equation of the line is:[ y - 2 = 2sqrt{2}(x - sqrt{2}) ] Step 3: Prove that ( |OA| cdot |OB| ) is a constantThe coordinates of ( O ) (origin) are ( (0,0) ). We need to find ( |OA| cdot |OB| ) where ( A ) and ( B ) are:[ A = (sqrt{2} + 1, 2(sqrt{2} + 1)) ][ B = (sqrt{2} - 1, -2(sqrt{2} - 1)) ]The distances are:[ |OA| = sqrt{(sqrt{2} + 1)^2 + (2(sqrt{2} + 1))^2} ][ |OB| = sqrt{(sqrt{2} - 1)^2 + (-2(sqrt{2} - 1))^2} ]Let's compute these using Python and SymPy.```pythonimport sympy as sp# Define symbolsx0 = sp.sqrt(2)y0 = 2# Coordinates of A and Ba_x = x0 + 1a_y = 2 * (x0 + 1)b_x = x0 - 1b_y = -2 * (x0 - 1)# Distance from origin to A and BOA = sp.sqrt(a_x2 + a_y2)OB = sp.sqrt(b_x2 + b_y2)# Product of distancesproduct_OA_OB = OA * OB# Simplify the productproduct_OA_OB_simplified = sp.simplify(product_OA_OB)print(product_OA_OB_simplified)``````output5```The product of the distances ( |OA| cdot |OB| ) is a constant and is equal to (boxed{5}). To summarize:1. The equations of the asymptotes of the hyperbola ( x^2 - frac{y^2}{4} = 1 ) are: [ y = 2x ] [ y = -2x ]2. When the coordinates of ( P ) are ( (sqrt{2}, 2) ), the equation of the line ( l ) passing through ( P ) and intersecting the asymptotes at points ( A ) and ( B ) is: [ y - 2 = 2sqrt{2}(x - sqrt{2}) ]3. The product ( |OA| cdot |OB| ) is a constant value and is equal to ( 5 ).Thus, the final answer is (boxed{5}).

answer:To solve the problem, we need to use the properties of a hyperbola and its asymptotes.1. The equation of the given hyperbola is: [ frac{x^2}{m} - y^2 = 1 quad text{with} quad m > 0. ]2. The asymptotes of a hyperbola given by the equation (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) are: [ y = pm frac{b}{a} x. ] For the hyperbola (frac{x^2}{m} - y^2 = 1), we can rewrite it in the standard form (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) by identifying (a^2 = m) and (b^2 = 1). Thus, the asymptotes are: [ y = pm frac{b}{a} x = pm frac{1}{sqrt{m}} x. ]3. We are given that one of the asymptotes of the hyperbola is (sqrt{3}x + my = 0).4. To find the relationship between the given asymptote and the standard asymptotes, compare (sqrt{3}x + my = 0) with the form (y = pm frac{1}{sqrt{m}} x): [ y = -frac{sqrt{3}}{m} x. ] Hence, we have: [ frac{1}{sqrt{m}} = frac{sqrt{3}}{m}. ]5. Solving the equation (frac{1}{sqrt{m}} = frac{sqrt{3}}{m}) will give us the value of (m).6. Once we have (m), we can find the focal length of the hyperbola, which is given by (2c), where (c = sqrt{a^2 + b^2}). Here, (a = sqrt{m}) and (b = 1).Let's implement this step-by-step in Python using SymPy:```pythonimport sympy as sp# Define the variable mm = sp.symbols('m', positive=True)# Equation based on the asymptoteasymptote_equation = sp.Eq(1/sp.sqrt(m), sp.sqrt(3)/m)# Solve for mm_value = sp.solve(asymptote_equation, m)[0]# Calculate the parameters a and ba = sp.sqrt(m_value)b = 1# Calculate the focal length 2cc = sp.sqrt(a2 + b2)focal_length = 2 * cprint(focal_length)``````output4```The focal length of the hyperbola (frac{x^2}{m} - y^2 = 1) with the given asymptote (sqrt{3}x + my = 0) is (boxed{4}).