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question:Rich likes to take long walks through town. First he walks 20 feet from his house to the sidewalk. Then he walks 200 feet down the sidewalk to the end of the road. Then he makes a left and walks double his total distance so far until he reaches the next intersection. Then he walks half the total distance up to this point again to the end of his route, before turning around and walking the same path all the way back home. How many feet did Rich walk?

answer:Let's calculate the total distance Rich walks step by step. 1. From his house to the sidewalk: 20 feet 2. Down the sidewalk to the end of the road: 200 feet So far, Rich has walked 20 + 200 = 220 feet. 3. He then makes a left and walks double his total distance so far, which is 2 * 220 = 440 feet. Now, the total distance Rich has walked is 220 + 440 = 660 feet. 4. Then he walks half the total distance up to this point again to the end of his route, which is 0.5 * 660 = 330 feet. The total distance Rich has walked to the end of his route is 660 + 330 = 990 feet. Since Rich turns around and walks the same path all the way back home, he will walk another 990 feet. Therefore, the total distance Rich walks is 990 + 990 = boxed{1980} feet.

question:Given a moving point M on a plane, the distance from M to the line y = 2 is 1 more than its distance to the point F(0, 1). (Ⅰ) Find the equation of the trajectory of point M; (Ⅱ) Suppose the curve formed by the moving point M is E, and a line l passing through point P(0, 1) intersects curve E at points A and B. If the sum of the slopes of lines OA and OB is 2 (where O is the origin), find the equation of line l.

answer:Solution: (Ⅰ) Since the distance from the moving point M to the line y = -2 is 1 more than its distance to the point F(0, 1), it can be inferred that the distance from the moving point to point F is equal to its distance to the line y = -1, According to the definition of a parabola, the trajectory of the moving point is a parabola with F as the focus and y = -1 as the directrix. Therefore, the equation is x^2 = 4y. ... (4 points) (Ⅱ) Obviously, it is not meaningful if line l is perpendicular to the x-axis, so we can assume the equation of the sought line is y = kx - 1, Substituting into the parabola equation and simplifying, we get: x^2 - 4kx + 4 = 0, ... (6 points) where Delta = 4k^2 + 8 > 0, x_1 + x_2 = -4k, x_1x_2 = 4 ... (8 points) Let point A be (x_1, y_1) and point B be (x_2, y_2), then we have frac{y_1}{x_1} + frac{y_2}{x_2} = 2, ① Since y_1 = kx_1 - 1 and y_2 = kx_2 - 1, substituting into ① and rearranging, we get k = 2, ... (11 points) Therefore, the equation of line l is y = 2x - 1. ... (12 points) Thus, the answers are: - (Ⅰ) The equation of the trajectory of point M is boxed{x^2 = 4y}. - (Ⅱ) The equation of line l is boxed{y = 2x - 1}.

question:Kathleen saved 21 in June, 46 in July, 45 in August, 32 in September, and half of what she saved in August in October. In November, she saved twice the amount she saved in September minus 20. She spent 12 on school supplies, 54 on new clothes, 37 on a birthday gift for her friend, and 25 on a book. Additionally, she wants to make a 10 donation to charity. Kathleen's aunt will give her a bonus of 25 if she saves more than 200 and donates 10 to charity. How much money will Kathleen have left if she fulfills her aunt's conditions and makes the donation?

answer:First, let's calculate how much Kathleen saved each month: June: 21 July: 46 August: 45 September: 32 October: Half of what she saved in August, which is 45 / 2 = 22.50 November: Twice the amount she saved in September minus 20, which is (2 * 32) - 20 = 64 - 20 = 44 Now, let's add up all the savings: Total savings = 21 + 46 + 45 + 32 + 22.50 + 44 = 210.50 Next, let's calculate her total expenses: Expenses = 12 (school supplies) + 54 (new clothes) + 37 (birthday gift) + 25 (book) + 10 (donation) = 138 Now, let's subtract her expenses from her total savings: Remaining money = 210.50 - 138 = 72.50 Since Kathleen has saved more than 200, her aunt will give her a bonus of 25. Let's add this to her remaining money: Remaining money with bonus = 72.50 + 25 = 97.50 Therefore, after fulfilling her aunt's conditions and making the donation, Kathleen will have boxed{97.50} left.

question:Mr. Jensen needs to determine the combined weight in pounds of three containers he intends to ship. The only working scale inaccurately measures weights less than 200 pounds or more than 300 pounds. The containers are weighed in pairs in every possible combination, yielding the results 234, 241, and 255 pounds. What is the combined weight in pounds of the three containers? textbf{(A)} 355qquad textbf{(B)} 365qquad textbf{(C)} 375qquad textbf{(D)} 380qquad textbf{(E)} 390

answer:Let the weights of the three containers be x, y, and z. From the problem description, the pair weights are: - (x, y) = 234 pounds - (y, z) = 241 pounds - (z, x) = 255 pounds We need to find the total weight x + y + z. Start by summing the pair weights: [ (x+y) + (y+z) + (z+x) = 234 + 241 + 255. ] Calculating the sum: [ 234 + 241 + 255 = 730. ] Each container's weight is counted twice in the sum of all pairings. Therefore, divide the total by 2 to find the combined weight: [ x + y + z = frac{730}{2} = 365. ] Thus, the combined weight in pounds of the three containers is 365. Conclusion: The calculation accurately provides the combined weight, and the problem statement is consistent and solvable. The final answer is boxed{textbf{(B)} 365}

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