2.6 KiB
Let's use python to solve math problems. Display the final result in LaTeX.
Question: Find the coefficient of x^3
when 3(x^2 - x^3+x) +3(x +2x^3- 3x^2 + 3x^5+x^3) -5(1+x-4x^3 - x^2)
is simplifie.
from sympy import symbols, simplify
def solution():
x = symbols('x')
expr = 3*(x**2 - x**3 + x) + 3*(x + 2*x**3 - 3*x**2 + 3*x**5 + x**3) - 5*(1 + x - 4*x**3 - x**2)
simplified_expr = simplify(expr)
x3_coefficient = simplified_expr.as_coefficients_dict()[x**3]
result = x3_coefficient
return result
Question: The surface area of a sphere with radius r
is 4\pi r^2
. Including the area of its circular base, what is the total surface area of a hemisphere with radius 6 cm? Express your answer in terms of \pi
.
import math
def solution():
radius = 6
# Surface area of the hemisphere
hemisphere_area = 2 * math.pi * radius**2
# Area of the circular base
base_area = math.pi * radius**2
# Total surface area
total_surface_area = hemisphere_area + base_area
# Formatting the result in LaTeX
result = r'{}\\pi'.format(total_surface_area / math.pi)
return result
Question: Monica tosses a fair 6-sided die. If the roll is a prime number, then she wins that amount of dollars (so that, for example, if she rolls 3, then she wins 3 dollars). If the roll is composite, she wins nothing. Otherwise, she loses 3 dollars. What is the expected value of her winnings on one die toss? Express your answer as a dollar value to the nearest cent.
def solution():
# Probabilities of each outcome
prime_prob = 1 / 6
composite_prob = 1 / 3
otherwise_prob = 1 / 6
# Expected value of each outcome
prime_expected_value = (2 * prime_prob) + (3 * prime_prob) + (5 * prime_prob)
composite_expected_value = 0 * composite_prob
otherwise_expected_value = -3 * otherwise_prob
# Total expected value
total_expected_value = prime_expected_value + composite_expected_value + otherwise_expected_value
# Dollar value to the nearest cent
result = "{:.2f}".format(total_expected_value)
return result
Question: Given \mathbf{a} = \begin{pmatrix} -7 \\ 0 \\ 1 \end{pmatrix}
and \mathbf{b} = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix},
find \mathbf{a} - 3 \mathbf{b}.
Solution:
import numpy as np
def solution()
a = np.array([-7, 0, 1])
b = np.array([4, 2, -1])
result = a - 3 * b
result = r'\begin{{pmatrix}} {} \\ {} \\ {} \end{{pmatrix}}'.format(result[0], result[1], result[2])
return result