answer:Calculation of net income with increased sales: {eq}begin{align*} {rmtext{Sales}} &= 13,200,000 {rmtext{Less: Cost of goods sold (40% of sales)}} &= (5,280,000) {rmtext{Less: Operating cost}} &= (1,200,000) {rmtext{Less: Depreciation}} &= (80,000) {rmtext{Earning before interest and tax}} &= 6,640,000 {rmtext{Less: Interest expenses}} &= (80,000) {rmtext{Income before tax}} &= 6,560,000 {rmtext{Less: Tax @ 40%}} &= 2,624,000 {rmtext{Net Income}} &= 3,936,000 end{align*} {/eq} Calculation of dividend payout: {eq}begin{align*} {rmtext{Earnings Per Share}} &= frac{{ 3,936,000}}{{1,000,000}} &= 3.936 end{align*} {/eq} {eq}begin{align*} {rmtext{Dividend Payout}} &= {rmtext{Earnings Per Share}} times {rmtext{Dividend Payout Percent}} &= 3.936 times 20% &= 0.7872 end{align*} {/eq} Therefore, the new dividend payout would be 0.7872.

answer:The excerpt that most effectively showcases the narrator's madness is: B. "By long suffering my nerves had been unstrung, until I trembled at the sound of my own voice . . . ." This sentence demonstrates the narrator's fragile mental state due to their prolonged suffering, which has led to an intense fear and nervousness, even at the sound of their own voice. The story takes place during the Spanish Inquisition, and the narrator's vivid descriptions of the terrifying surroundings and potential tortures contribute to the Gothic atmosphere of the tale. The pendulum, as an instrument of death, further accentuates the narrator's escalating fear and descent into madness.

answer:To show that f is a homeomorphism, we need to show that f is bijective and both f and f^{-1} are continuous. First, we show that f is bijective. Suppose f(x_1) = f(x_2) for some x_1, x_2 in [0,1]. Then, (1-x_1)a+x_1b = (1-x_2)a+x_2b (x_2-x_1)(b-a) = 0 Since b-a neq 0, we have x_1 = x_2. Therefore, f is injective. Next, we show that f is surjective. Let y in [a,b]. Then, y = (1-x)a+xb x = frac{y-a}{b-a} Since 0 leq x leq 1, we have y in f([0,1]). Therefore, f is surjective. Now, we show that f is continuous. Let epsilon > 0 be given. Choose delta = frac{epsilon}{b-a}. Then, for any x_1, x_2 in [0,1] such that |x_1 - x_2| < delta, we have |f(x_1) - f(x_2)| = |(1-x_1)a+x_1b - (1-x_2)a-x_2b| =|x_2-x_1||b-a| <delta|b-a| =epsilon Therefore, f is continuous. Finally, we show that f^{-1} is continuous. Let epsilon > 0 be given. Choose delta = frac{epsilon}{b-a}. Then, for any y_1, y_2 in [a,b] such that |y_1 - y_2| < delta, we have |f^{-1}(y_1) - f^{-1}(y_2)| = left|frac{y_1-a}{b-a} - frac{y_2-a}{b-a}right| =frac{|y_1 - y_2|}{|b-a|} <frac{delta}{|b-a|} =epsilon Therefore, f^{-1} is continuous. Since f is bijective and both f and f^{-1} are continuous, f is a homeomorphism.

answer:The mode is the value that appears most frequently in a dataset. In this case, both -3 sqrt{5} and -frac{8}{sqrt{5}} appear three times, while all other values appear only once. Therefore, the mode of the given set of numbers is -3 sqrt{5} and -frac{8}{sqrt{5}}. The mode is -3 sqrt{5} and -frac{8}{sqrt{5}}.