answer:A firm's capital structure is the mix of debt and equity financing that it uses to fund its operations. The target capital structure is the ideal mix of debt and equity that a firm should aim for in order to minimize its WACC. WACC is calculated as follows: WACC = (E/V) * Re + (D/V) * Rd * (1 - T) where: E = market value of equity V = total market value of the firm Re = cost of equity D = market value of debt Rd = cost of debt T = corporate tax rate The cost of equity is the return that investors require to compensate them for the risk of investing in a firm's stock. The cost of debt is the interest rate that a firm pays on its debt. By choosing a target capital structure that minimizes WACC, a firm can reduce its overall cost of capital and increase its value. This is because a lower WACC means that a firm can finance its operations at a lower cost, which in turn leads to higher profits and a higher stock price. A target capital structure is a firm's intended mix of debt and equity financing. The purpose of a target capital structure is to minimize the firm's weighted average cost of capital (WACC) over the long term. WACC is the average cost of a firm's various sources of capital, including debt, equity, and preferred stock. By choosing a target capital structure that minimizes WACC, a firm can reduce its overall cost of capital and increase its value.

answer:The equation of the tangent line is ( y = -322.11x + 77.79 ). Explanation: Given the polar curve ( f(theta) = 5thetasin(3theta) + 2cot(11theta) ), we need to find the tangent line at ( theta = frac{pi}{12} ). First, we evaluate ( fleft(frac{pi}{12}right) ) to find the point on the curve: [ fleft(frac{pi}{12}right) approx -6.54 ] The point where the tangent is drawn is ( left(frac{pi}{12}, -6.54right) ) or approximately ( (0.26, -6.54) ). Next, we calculate the slope of the tangent line using the derivative of ( f(theta) ): [ f'(theta) = 5sin(3theta) + 15thetacos(3theta) - 22csc^2(11theta) ] The slope of the tangent line at ( theta = frac{pi}{12} ) is: [ f'left(frac{pi}{12}right) approx -322.11 ] The equation of the tangent line at ( (0.26, -6.54) ) with slope ( -322.11 ) is given by the point-slope form: [ y - y_1 = m(x - x_1) ] [ y + 6.54 = -322.11(x - 0.26) ] [ y = -322.11x + 84.33 - 6.54 ] [ y = -322.11x + 77.79 ] Therefore, the equation of the tangent line is ( y = -322.11x + 77.79 ).

answer:The equation can be rewritten as two separate quadratic equations: 1) -frac{13 x^2}{sqrt{2}} - 6 sqrt{2} x + 7 sqrt{2} = 16 sqrt{2}, and 2) -frac{13 x^2}{sqrt{2}} - 6 sqrt{2} x + 7 sqrt{2} = -16 sqrt{2}. Solving each quadratic equation separately: For equation 1, we have: -13 x^2 - 12x + 23 = 0 x = frac{-(-12) pm sqrt{(-12)^2 - 4(-13)(23)}}{2(-13)} x = frac{12 pm sqrt{144 + 1196}}{-26} x = frac{12 pm sqrt{1340}}{-26} x = frac{12 pm 2sqrt{335}}{-26} For equation 2, we have: -13 x^2 - 12x - 9 = 0 x = frac{-(-12) pm sqrt{(-12)^2 - 4(-13)(-9)}}{2(-13)} x = frac{12 pm sqrt{144 - 468}}{-26} x = frac{12 pm sqrt{-324}}{-26} x = frac{12 pm 18i}{-26} (This solution is not real, so it's discarded) The real solutions to the original equation are from equation 1: x = frac{12 pm 2sqrt{335}}{-26} x = frac{1}{13} left(-6 pm sqrt{335}right) So, the set of all real solutions is: left{xto frac{1}{13} left(-6-sqrt{335}right)right}, left{xto frac{1}{13} left(-6+sqrt{335}right)right}

answer:Given the integral {eq}int frac{sqrt {ln x}}{x} dx {/eq}, we solve it using the substitution method. Let {eq}u = ln x {/eq}. Differentiating both sides with respect to {eq}x {/eq} gives: {eq}frac{du}{dx} = frac{1}{x} Rightarrow du = frac{1}{x} dx {/eq} Now, applying this substitution to the integral, we get: {eq}int frac{sqrt {ln x}}{x} dx = int sqrt{u} du {/eq} Using the power rule for integration, {eq}int x^n dx = frac{x^{n+1}}{n+1} {/eq}, we have: {eq}int sqrt{u} du = int u^{frac{1}{2}} du = frac{u^{frac{3}{2}}}{frac{3}{2}} + C = frac{2}{3}u^{frac{3}{2}} + C {/eq} Reverting back to the original variable {eq}x {/eq}, we find: {eq}frac{2}{3}u^{frac{3}{2}} + C = frac{2}{3}(ln x)^{frac{3}{2}} + C {/eq} Therefore, the integral is: {eq}color{blue}{boxed{int frac{sqrt {ln x}}{x} dx = frac{2}{3}(ln x)^{frac{3}{2}} + C}} {/eq}