answer:Solve the system begin{cases} y=x^{2} y^{2}=x end{cases} to get x_1=0 and x_2=1. Therefore, the area of the figure enclosed by the curve y=x^2 and the curve y^2=x is S= int_{0}^{1} (sqrt{x}-x^{2})dx = left( frac {2}{3}x^{ frac {3}{2}}- frac {1}{3}x^{3} right)bigg|_{0}^{1} = frac {2}{3} x^{ frac {3}{2}}bigg|_{0}^{1} - frac {1}{3} x^{3}bigg|_{0}^{1} = frac {2}{3}- frac {1}{3}= boxed{frac {1}{3}}. Hence, the correct answer is A. To find the area of the figure enclosed by the curves y=x^2 and y^2=x, first, determine the x-coordinates of the intersection points, 0 and 1, then calculate the integral of sqrt{x}-x^{2} over the interval [0, 1]. For problems involving the area of plane figures, one should first sketch the rough shape of the figure. Based on the characteristics of the figure, select the appropriate integration variable and integrand, and determine the integration interval. The key to solving the problem is to find the antiderivative of the integrand.

answer:Let's solve the equation step by step. First, we simplify the absolute value on the right side of the equation: | 5 - 11 | = | -6 | = 6 Now the equation looks like this: | 5 - 8 ( 3 - some_number ) | - 6 = 71 Next, we add 6 to both sides of the equation to isolate the absolute value expression: | 5 - 8 ( 3 - some_number ) | = 71 + 6 | 5 - 8 ( 3 - some_number ) | = 77 Now we have two cases to consider because the absolute value of a number can be positive or negative: Case 1: 5 - 8 ( 3 - some_number ) = 77 Case 2: 5 - 8 ( 3 - some_number ) = -77 Let's solve for some_number in each case. Case 1: 5 - 8 ( 3 - some_number ) = 77 5 - 24 + 8 * some_number = 77 -19 + 8 * some_number = 77 8 * some_number = 77 + 19 8 * some_number = 96 some_number = 96 / 8 some_number = 12 Case 2: 5 - 8 ( 3 - some_number ) = -77 5 - 24 + 8 * some_number = -77 -19 + 8 * some_number = -77 8 * some_number = -77 + 19 8 * some_number = -58 some_number = -58 / 8 some_number = -7.25 However, since the absolute value expression must equal 77, and the result of the expression inside the absolute value must be either 77 or -77, we can disregard the second case because it does not satisfy the original equation. Therefore, the value of some_number is boxed{12} .

answer:Let's denote the speed of person X as Vx km/h. Since person Y started 3 hours after person X, by the time Y started, X had already covered a distance of 3Vx km. We are given that the difference in time when X was 30 km ahead of Y and when Y was 30 km ahead of X is 3 hours. Let's denote the time it takes for Y to catch up to the point where X is 30 km ahead as t1 hours, and the time it takes for Y to get 30 km ahead of X as t2 hours. When Y catches up to the point where X is 30 km ahead, the distances traveled by X and Y will be: Distance traveled by X = 3Vx + Vx * t1 Distance traveled by Y = 60 * t1 Since X is 30 km ahead at this point, we can write the equation: 3Vx + Vx * t1 = 60 * t1 + 30 When Y is 30 km ahead of X, the distances traveled by X and Y will be: Distance traveled by X = 3Vx + Vx * t2 Distance traveled by Y = 60 * t2 Since Y is 30 km ahead at this point, we can write the equation: 3Vx + Vx * t2 + 30 = 60 * t2 We are given that t2 - t1 = 3 hours. Now we have two equations and two unknowns (Vx and t1 or t2). We can solve these equations simultaneously to find the value of Vx. Let's simplify the first equation: 3Vx + Vx * t1 = 60 * t1 + 30 3Vx + Vx * t1 - 60 * t1 = 30 Vx * (3 + t1) - 60 * t1 = 30 Vx * t1 + 3Vx - 60 * t1 = 30 Now let's simplify the second equation: 3Vx + Vx * t2 + 30 = 60 * t2 Vx * t2 + 3Vx - 60 * t2 = -30 Now we can use the fact that t2 - t1 = 3: Vx * (t2 - t1) = 60 * (t2 - t1) Vx * 3 = 60 * 3 Vx = 60 So the speed of person X is boxed{60} km/h.

answer:To simplify and then evaluate the given expression 2x^{2}-(2xy-3y^{2})+2(x^{2}+xy-2y^{2}) with x=-2 and y=1, we proceed as follows: First, we simplify the expression by distributing and combining like terms: [ begin{align*} 2x^{2}-(2xy-3y^{2})+2(x^{2}+xy-2y^{2}) &= 2x^{2}-2xy+3y^{2}+2x^{2}+2xy-4y^{2} &= 2x^{2} + 2x^{2} + (-2xy + 2xy) + (3y^{2} - 4y^{2}) &= 4x^{2} - y^{2}. end{align*} ] Next, we substitute x=-2 and y=1 into the simplified expression: [ begin{align*} 4x^{2} - y^{2} &= 4(-2)^{2} - 1^{2} &= 4 times 4 - 1 &= 16 - 1 &= 15. end{align*} ] Therefore, after simplifying and evaluating the expression with the given values of x and y, we get boxed{15}.