answer:(1) First, we find the intersection point P by solving the system of equations for lines l_{1} and l_{2}: begin{cases} 3x+4y-2=0 2x+y+2=0 end{cases} Solving this system, we find x=-2 and y=2. So the intersection point is P(-2, 2). Now we need to find the line passing through P(-2, 2) and the origin. The slope of this line can be found using the formula m = frac{y_2 - y_1}{x_2 - x_1} where (x_1, y_1) and (x_2, y_2) are the coordinates of two points on the line. In this case, we have (x_1, y_1) = P(-2, 2) and (x_2, y_2) = (0, 0) (origin). Substituting these values into the formula, we get: m = frac{2 - 0}{-2 - 0} = -1 Using the point-slope form of a linear equation, y - y_1 = m(x - x_1), we can write the equation of the line as: y - 2 = -1(x + 2) Simplifying this equation, we get the final answer: boxed{x + y = 0} (2) First, we find the slope of line l_{3}: x - 2y - 1 = 0. Rewriting this equation in slope-intercept form (y = mx + b), we have y = frac{1}{2}x - frac{1}{2}. So the slope of line l_{3} is m_{l_3} = frac{1}{2}. Since the line we are looking for is perpendicular to line l_{3}, its slope m_{l} will be the negative reciprocal of the slope of line l_{3}. That is, m_{l} = -frac{1}{m_{l_3}} = -frac{1}{frac{1}{2}} = -2 Now we use the point-slope form of a linear equation with point P(-2, 2) and slope m_{l} = -2 to find the equation of the line: y - 2 = -2(x + 2) Simplifying this equation, we get the final answer: boxed{2x + y + 2 = 0}

answer:To calculate the total amount of money you have, you need to multiply the number of each type of coin by its value and then add up all the values. Pennies: 9 pennies × 0.01 = 0.09 Nickels: 4 nickels × 0.05 = 0.20 Dimes: 3 dimes × 0.10 = 0.30 Now, add up all the values: 0.09 (pennies) + 0.20 (nickels) + 0.30 (dimes) = 0.59 You have a total of boxed{0.59} .

answer:1. **Identify the positions and constraints**: - The train cars are arranged in a line: square_1 square_2 square_3 square_4 square_5. - Nina is in the last car: square_1 square_2 square_3 square_4 N. 2. **Place Jess and Owen**: - Jess is right in front of Owen, making consecutive placements: - square_1 square_2 square_3 J O N - square_1 square_2 J O square_4 N - square_1 J O square_3 square_4 N 3. **Position Mark ahead of Jess**: - Mark must be ahead of Jess, limiting his positions: - square_1 square_2 square_3 J O N allows Mark in square_1 or square_2. - square_1 square_2 J O square_4 N allows Mark only in square_1. - square_1 J O square_3 square_4 N disallows Mark completely. 4. **Respect the gap between Lisa and Mark**: - There must be at least one car between Lisa and Mark: - The configuration square_1 square_2 J O square_4 N with Mark in square_1 allows Lisa to be in square_4. - Other configurations are not valid as they do not satisfy the gap condition. 5. **Final arrangement**: - The valid arrangement is square_1 M square_2 J O L N. 6. **Who is in the second car?**: - Mark is in the second car from the front. The person in the second car is text{Mark}. Conclusion: The solution is consistent with all given conditions. Mark is indeed the occupant of the second car according to the verified solution. The final answer is boxed{textbf{(C) }text{Mark}}

answer:Given that overrightarrow{a}=(1, 2), overrightarrow{b}=(x, 1), overrightarrow{u}= overrightarrow{a}+ overrightarrow{b}, and overrightarrow{v}= overrightarrow{a}- overrightarrow{b}, we can compute overrightarrow{u} and overrightarrow{v} as follows: overrightarrow{u} = (1 + x, 3) overrightarrow{v} = (1 - x, 1) (Ⅰ) Since overrightarrow{u} is parallel to overrightarrow{v}, their components must be proportional. Therefore, we have the proportion: frac{1 + x}{1 - x} = frac{3}{1} Cross-multiplying gives us: (1 + x) = 3(1 - x) Expanding the equation: 1 + x = 3 - 3x Moving all terms involving x to one side and constant terms to the other side, we get: x + 3x = 3 - 1 Combining like terms: 4x = 2 Dividing both sides by 4 gives us: x = frac{1}{2} Thus, the value of x when overrightarrow{u} is parallel to overrightarrow{v} is: boxed{x = frac{1}{2}} (Ⅱ) Since overrightarrow{u} is perpendicular to overrightarrow{v}, their dot product must be zero: (1 + x)(1 - x) + 3 cdot 1 = 0 Expanding the equation: 1 - x^2 + 3 = 0 Rearranging the terms: x^2 = 4 Taking the square root of both sides gives us two solutions: x = 2 text{ or } x = -2 Therefore, the values of x when overrightarrow{u} is perpendicular to overrightarrow{v} are: boxed{x = 2 text{ or } x = -2}