answer:Let gamma : [a, b] to mathbb{C} be a rectifiable curve, and consider the reversed curve Gamma : [-b, -a] to mathbb{C} defined by Gamma(t) = gamma(-t). We aim to show that int_{gamma}f dz = -int_{Gamma}f dz. Since f is continuous, it is uniformly continuous on [a, b]. Given epsilon > 0, there exists a partition P = (t_0 = a, t_1, ..., t_n = b) of [a, b] with small enough mesh size such that |f(t_k) - f(t_{k-1})| < epsilon for all k. Now, let's define a partition of [-b, -a] as Q = (-t_n, ..., -t_1, -t_0 = -a). Note that |gamma(t_k) - gamma(t_{k-1})| = |Gamma(-t_k) - Gamma(-t_{k-1})| for all k. We have: begin{align*} int_{gamma}f dz &= lim_{|P| to 0} sum_{k} f(gamma(t_k))(gamma(t_k) - gamma(t_{k-1})) int_{Gamma}f dz &= lim_{|Q| to 0} sum_{k} f(Gamma(-t_k))(Gamma(-t_k) - Gamma(-t_{k-1})) &= lim_{|Q| to 0} -sum_{k} f(Gamma(-t_{k-1}))(Gamma(-t_{k}) - Gamma(-t_{k-1})) &= lim_{|Q| to 0} -sum_{k} f(gamma(-t_{k-1}))(-gamma(-t_{k}))(-gamma'(-t_{k})) &= -lim_{|P| to 0} sum_{k} f(gamma(t_{k-1}))(gamma(t_{k}) - gamma(t_{k-1})) &= -int_{gamma}f dz end{align*} Thus, int_{gamma}f dz = -int_{Gamma}f dz, as required. This result holds for path integrals and can be extended to complex integrals by considering the real and imaginary parts of the integrand separately.

answer:According to Boyle's law, for a fixed amount of an ideal gas at constant temperature, the pressure and volume are inversely proportional. When the volume of the gas is halved, the pressure exerted by the gas becomes twice the original pressure. Mathematically, this can be expressed as: ( P_1 times V_1 = P_2 times V_2 ) Where: - ( P_1 ) is the initial pressure, - ( V_1 ) is the initial volume, - ( P_2 ) is the pressure after compression, and - ( V_2 ) is the volume after compression (which is half of ( V_1 )). Since ( V_2 = frac{1}{2} V_1 ), we have ( P_2 = 2P_1 ). Therefore, the pressure exerted by the gas doubles.

answer:When the minimum wage is increased, businesses may face higher labor costs compared to their competitors in other countries with lower minimum wages. This can make it more difficult for businesses to compete globally, as they may have to raise prices or reduce profits to cover the increased labor costs. Additionally, businesses may choose to outsource jobs to countries with lower labor costs, leading to job losses in the country with the higher minimum wage.

answer:No, the standard deviation and range do not measure dispersion in the same way. The standard deviation measures the average distance of data points from the mean, while the range measures the difference between the highest and lowest values in a dataset. Therefore, the standard deviation provides a more comprehensive measure of dispersion than the range.