answer:1. **Identify the dimensions of the original rectangle:** Let a be the shorter side and b be the longer side. 2. **Calculate the diagonal of the original rectangle:** The diagonal d is given by Pythagoras' theorem: [ d = sqrt{a^2 + b^2} ] 3. **Define the dimensions of the new rectangle:** The base of the new rectangle is twice the sum of the diagonal and the longer side: [ text{Base} = 2(sqrt{a^2 + b^2} + b) ] The height of the new rectangle is half the difference between the diagonal and the longer side: [ text{Height} = frac{sqrt{a^2 + b^2} - b}{2} ] 4. **Calculate the area of the new rectangle:** The area A_{text{new}} is: [ A_{text{new}} = left(2(sqrt{a^2 + b^2} + b)right) cdot left(frac{sqrt{a^2 + b^2} - b}{2}right) ] Simplifying, we get: [ A_{text{new}} = (sqrt{a^2 + b^2} + b)(sqrt{a^2 + b^2} - b) ] Applying the difference of squares: [ A_{text{new}} = (sqrt{a^2 + b^2})^2 - b^2 ] [ A_{text{new}} = a^2 + b^2 - b^2 ] [ A_{text{new}} = a^2 ] Conclusion: The new rectangle has an area equal to the square of the shorter side of the original rectangle, a^2. The final answer is boxed{A}.

answer:1. Since 2S_n=3^n+3, we have 2a_1=3+3, hence a_1=3. For n > 1, we have 2S_{n-1}=3^{n-1}+3. Consequently, 2a_n=2S_n-2S_{n-1}=3^n-3^{n-1}, which implies a_n=3^{n-1}. Thus, the general term formula for the sequence is: a_n = begin{cases} 3 & text{if n=1} 3^{n-1} & text{if n > 1} end{cases} 2. Given that a_n b_n = log_3 a_n, we have b_1=frac{1}{3}. For n > 1, we have b_n=3^{1-n}log_3 3^{n-1}=(n-1)cdot 3^{1-n}. Therefore, T_1=b_1=frac{1}{3}, and for n > 1: begin{align} T_n &= b_1+b_2+b_3+ldots+b_n &= frac{1}{3} + (1cdot3^{-1} + 2cdot3^{-2} + ldots + (n-1)3^{1-n}). end{align} Multiplying both sides by 3, we get: begin{align} 3T_n &= 1 + (1cdot3^0 + 2cdot3^{-1} + ldots + (n-1)3^{2-n}), 2T_n &= frac{2}{3} + (3^0 + 3^{-1} + 3^{2-n}) - (n-1)cdot 3^{1-n} &= frac{2}{3} + frac{1-3^{1-n}}{1-3^{-1}} - (n-1)cdot 3^{1-n} &= frac{13}{6} - frac{6n+3}{2cdot 3^n}. end{align} Hence, T_n = boxed{frac{13}{12} + frac{6n+3}{4cdot 3^n}}. This formula is also valid for n=1.

answer:1. **Calculate scale factors**: - From t=0 to t=1, the population decreases, so the scale factor is 1 - frac{k}{100}. - From t=1 to t=2, the population increases, so the scale factor is 1 + frac{m}{100}. 2. **Find the overall scale factor**: [ (1 - frac{k}{100})(1 + frac{m}{100}) = 1 + frac{m}{100} - frac{k}{100} - frac{k cdot m}{10000}. ] 3. **Convert the overall scale factor to a percentage change**: - Subtracting 1: [ 1 + frac{m}{100} - frac{k}{100} - frac{km}{10000} - 1 = frac{m}{100} - frac{k}{100} - frac{km}{10000} ] - Converting to percentage: [ left(frac{m - k}{100} - frac{km}{10000}right) times 100 = m - k - frac{km}{100}. ] 4. **Conclusion**: - Therefore, the net change in the population from t=0 to t=2 is m - k - frac{km{100}%}. The final answer is A. boxed{m - k - frac{km}{100}%}

answer:**Solution**: A: For x^2=-3, since the square of a real number is always geq 0, this option is incorrect; B: x^3=-3 implies x=-sqrt[3]{3}, which has a solution, so this option is correct; C: The denominator of the fraction cannot be 0, so there is no solution for this option, making it incorrect; D: begin{vmatrix} x end{vmatrix}=-3, but the arithmetic square root of a real number is greater than 0, so this option is incorrect; Therefore, the correct choice is boxed{B}.