answer:Let's assume the original price of the TV is P. After the first decrease of 20%, the new price becomes: P - 0.20P = 0.80P Let's call the percentage increase after the initial decrease x%. This means the price is increased by x/100 of the decreased price (0.80P). So the increased price is: 0.80P + (x/100 * 0.80P) We know that the net change in the price is a 20% increase, so the final price is: P + 0.20P = 1.20P Setting the increased price equal to the final price, we get: 0.80P + (x/100 * 0.80P) = 1.20P Now, we can solve for x: 0.80P + 0.008Px = 1.20P 0.008Px = 1.20P - 0.80P 0.008Px = 0.40P Divide both sides by 0.008P to solve for x: x = (0.40P) / (0.008P) x = 0.40 / 0.008 x = 50 So the percentage increase in the price after the initial decrease is boxed{50%} .

answer:Solve: dfrac{2}{1+i}=1-i Then, the point corresponding to 1+i is (1,1), and the distance to the origin is sqrt{2}. Therefore, the correct choice is boxed{B}. Simplifying dfrac{2}{1+i} yields the result. This question examines the basic concepts.

answer:In right triangle ABC with angle C=90^{circ}, we have sides BC=3 and AB=4. To find the missing side AC, we can use the Pythagorean theorem: [AC = sqrt{AB^2 - BC^2} = sqrt{4^2 - 3^2} = sqrt{16 - 9} = sqrt{7}] Now, we can calculate the trigonometric ratios for angles A and B: For angle A: - sin A = frac{text{opposite}}{text{hypotenuse}} = frac{BC}{AB} = frac{3}{4} - cos A = frac{text{adjacent}}{text{hypotenuse}} = frac{AC}{AB} = frac{sqrt{7}}{4} - tan A = frac{text{opposite}}{text{adjacent}} = frac{BC}{AC} = frac{3}{sqrt{7}} = frac{3sqrt{7}}{7} For angle B: - tan B = frac{text{opposite}}{text{adjacent}} = frac{AC}{BC} = frac{sqrt{7}}{3} Comparing these calculations with the given options, we find that: A: cos A=frac{4}{5} is incorrect because cos A=frac{sqrt{7}}{4}. B: sin A=frac{3}{4} is correct as shown in our calculation. C: tan B=frac{4}{3} is incorrect because tan B=frac{sqrt{7}}{3}. D: tan A=frac{3}{4} is incorrect because tan A=frac{3sqrt{7}}{7}. Therefore, the correct answer is boxed{B}.

answer:(1) From cos(B-C) - 2sin B sin C = -frac{1}{2}, we get cos(B+C) = cos(180^circ - A) = -cos A = -frac{1}{2}. Since cos A = frac{1}{2}, and knowing that 0 < A < 180^circ, we conclude that A = 60^circ text{ or } A = frac{pi}{3}. (2) Using the law of cosines, a^2 = b^2 + c^2 - 2bccos A, we substitute a = 5, b = 4, and cos A = cos frac{pi}{3} = frac{1}{2}, yielding 5^2 = 4^2 + c^2 - 2 cdot 4 cdot c cdot frac{1}{2}. Solving this equation for c gives c^2 + 4^2 - 4c = 25, which simplifies to c^2 - 4c + 16 = 25, or c^2 - 4c - 9 = 0. Solving for c using the quadratic formula yields c = 2 + sqrt{13} (since c > 0 by the triangle inequality). Now, to find the area S_{triangle ABC}, we use the formula S_{triangle ABC} = frac{1}{2}absin C, and since we found c = 2 + sqrt{13} and C = 60^circ, we have S_{triangle ABC} = frac{1}{2} cdot 4 cdot (2 + sqrt{13}) cdot sin 60^circ. Substitute sin 60^circ = sqrt{3}/2, S_{triangle ABC} = 2 cdot (2 + sqrt{13}) cdot frac{sqrt{3}}{2} = (2 sqrt{3} + sqrt{39}). Therefore, the area of triangle ABC is boxed{2 sqrt{3} + sqrt{39}}.