answer:To find the range of the real number a, we proceed with the following steps: 1. We first simplify the expression for the complex number Z. To eliminate the complex denominator, we multiply the numerator and denominator by the conjugate of the denominator: Z = frac{a-1+2ai}{1-i} = frac{(a-1+2ai)(1+i)}{(1-i)(1+i)} = frac{(-a-1)+(3a-1)i}{2} 2. So the point corresponding to Z on the complex plane has coordinates left( frac{-a-1}{2}, frac{3a-1}{2} right). 3. For the point to be in the second quadrant, the real part must be negative, and the imaginary part must be positive: begin{cases} frac{-a-1}{2} < 0 frac{3a-1}{2} > 0 end{cases} 4. Solve the inequalities: - From frac{-a-1}{2} < 0, we find a > -1. - From frac{3a-1}{2} > 0, we find a > frac{1}{3}. 5. Since both conditions must be met for the point to be in the second quadrant, and the second condition a > frac{1}{3} encompasses a > -1, the range of a is left( frac{1}{3}, +infty right). Therefore, the range of the real number a is boxed{a > frac{1}{3}}.

answer:To find the coefficient of the x^{3} term in the expansion of (1-frac{1}{x})(1+x)^5, we first expand (1+x)^5 using the binomial theorem: (1-frac{1}{x})(1+x)^5 = (1-frac{1}{x})left({C}_{5}^{0} + {C}_{5}^{1}cdot x + {C}_{5}^{2}cdot x^{2} + {C}_{5}^{3}cdot x^{3} + {C}_{5}^{4}cdot x^{4} + {C}_{5}^{5}cdot x^{5}right) Expanding this, we focus on terms that will contribute to the coefficient of x^{3}: - The term {C}_{5}^{3}cdot x^{3} from (1+x)^5 will contribute directly to the coefficient of x^{3} when multiplied by 1. - The term {C}_{5}^{4}cdot x^{4} from (1+x)^5 will contribute to the coefficient of x^{3} when multiplied by -frac{1}{x}. Calculating these contributions: - The direct contribution from {C}_{5}^{3}cdot x^{3} is {C}_{5}^{3} = frac{5!}{3!(5-3)!} = 10. - The contribution from {C}_{5}^{4}cdot x^{4} when multiplied by -frac{1}{x} is -{C}_{5}^{4} = -frac{5!}{4!(5-4)!} = -5. Adding these contributions together to find the coefficient of the x^{3} term: text{Coefficient of } x^{3} = {C}_{5}^{3} - {C}_{5}^{4} = 10 - 5 = 5 Therefore, the coefficient of the x^{3} term in the expansion is boxed{5}.

answer:Let's assume the cost price (CP) of the articles is 100. When the shop owner buys the articles using false weights, he cheats by 12%. This means he gets 12% more of the articles for the same price. So, for 100, he actually gets articles worth 100 + 12% of 100 = 100 + 12 = 112. Now, he wants to make a 40% profit on the cost price. So, the selling price (SP) should be 100 + 40% of 100 = 100 + 40 = 140. Since he already has articles worth 112 for the cost price of 100, he needs to sell these articles for 140 to make a 40% profit. Let's denote the percentage by which he cheats while selling as x%. This means that when he sells, he gives x% less than the actual amount the customer pays for. To find out the value of x, we need to equate the value of the articles he gives to the customer to the selling price he wants to achieve (140). So, if he sells articles worth 112 at a price of 140, he is giving the customer articles worth 112 for 140. This means he is giving the customer (112/140) * 100% = 80% of what they paid for. Therefore, he is cheating the customer by 100% - 80% = 20%. The shop owner cheats by boxed{20%} while selling.

answer:1. **Completing the square** for (x): [ 2x^2 + 8x = 2(x^2 + 4x) = 2((x+2)^2 - 4) = 2(x+2)^2 - 8 ] 2. **Completing the square** for (y): [ 3y^2 - 9y = 3(y^2 - 3y) = 3((y-frac{3}{2})^2 - frac{9}{4}) = 3(y-frac{3}{2})^2 - frac{27}{4} ] 3. **Reformulate the equation**: [ 2(x+2)^2 - 8 + 3(y-frac{3}{2})^2 - frac{27}{4} + 12 = 0 ] Simplify and combine constants: [ 2(x+2)^2 + 3(y-frac{3}{2})^2 - 8 - frac{27}{4} + frac{48}{4} = 0 implies 2(x+2)^2 + 3(y-frac{3}{2})^2 = 1 ] 4. **Standard elliptical form**: [ frac{(x+2)^2}{frac{1}{2}} + frac{(y-frac{3}{2})^2}{frac{1}{3}} = 1 ] This gives semi-major axis (a = sqrt{frac{1}{2}}) and semi-minor axis (b = sqrt{frac{1}{3}}). 5. **Calculate area**: [ text{Area} = pi a b = pi sqrt{frac{1}{2}} sqrt{frac{1}{3}} = pi sqrt{frac{1}{6}} = frac{pi}{sqrt{6}} = frac{pi sqrt{6}}{6} ] Therefore, the area of the ellipse is (boxed{frac{pi sqrt{6}}{6}}).