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question:Given that O is the coordinate origin, the focus of the parabola C:y^{2}=2pxleft(p gt 0right) is F, P is a point on C, PF is perpendicular to the x-axis, Q is a point on the x-axis, and PQbot OP. If |FQ|=6, then the directrix equation of C is ____.
answer:To solve the problem step by step, let's follow the given information and the solution closely: 1. **Identifying the Point P on the Parabola C:** Given that C:y^{2}=2px is a parabola with its focus at F and P is a point on C, we assume P is in the first quadrant for simplicity. The coordinates of P can be derived from the general form of a parabola. Since P is on the parabola and PF is perpendicular to the x-axis, the coordinates of P are (frac{p}{2}, p). This is because the vertex form of the parabola is (frac{p}{2}, 0), and P being on the parabola with a vertical distance p from the x-axis gives us P(frac{p}{2}, p). 2. **Determining the Slope of OP and PQ:** Given that PQ is perpendicular to OP, we first find the slope of OP, denoted as k_{OP}. Since O is the origin (0,0) and P is (frac{p}{2}, p), the slope k_{OP} = frac{p - 0}{frac{p}{2} - 0} = 2. Since PQ is perpendicular to OP, the slope of PQ, denoted as k_{PQ}, is the negative reciprocal of k_{OP}. Therefore, k_{PQ} = -frac{1}{2}. 3. **Equation of Line PQ:** With k_{PQ} = -frac{1}{2} and knowing that P(frac{p}{2}, p) lies on PQ, we can write the equation of line PQ as y - p = -frac{1}{2}(x - frac{p}{2}). 4. **Finding x When y=0 (Intersection with x-axis):** Setting y=0 in the equation of PQ, we solve for x: 0 - p = -frac{1}{2}(x - frac{p}{2}) Solving this gives x = frac{5p}{2}. 5. **Given |FQ| = 6 and Solving for p:** Since F is on the x-axis and |FQ| = 6, we have the distance from F to Q as 6. Given F is at (frac{p}{2}, 0), the distance |FQ| = frac{5p}{2} - frac{p}{2} = 6. Solving for p, we get p = 3. 6. **Finding the Directrix Equation:** The directrix of a parabola y^{2} = 2px is at x = -frac{p}{2}. Substituting p = 3 into this formula, we find the directrix equation to be x = -frac{3}{2}. Therefore, the directrix equation of the parabola C is boxed{x=-frac{3}{2}}.
question:If Rodney, Roger, and Ron can lift a total weight of 239 pounds, and Rodney can lift twice as much as Roger, while Roger can lift 7 pounds less than 4 times the amount Ron can lift, how much weight can Rodney lift?
answer:Let's assume the weight Ron can lift as R. Roger can lift 4R - 7 pounds. And Rodney can lift 2 times the amount Roger can lift, which is 2(4R - 7) pounds. The total weight they can lift is R + 4R - 7 + 2(4R - 7) = 239. Combining like terms, we get R + 4R - 7 + 8R - 14 = 239. Simplifying further, we get 13R - 21 = 239. Adding 21 to both sides gives us 13R = 260. Dividing both sides by 13 gives us R = 20. Therefore, Rodney can lift 2(4R - 7) = 2(4(20) - 7) = 2(80 - 7) = 2(73) = 146 pounds. So, Rodney can lift 146 pounds. 146 The answer is: 146
question:Thirteen balls numbered 1 to 13 are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and randomly removes a different ball. The probability that the sum of the two numbers on the balls removed is even is? A. frac{42}{156} B. frac{72}{156} C. frac{6}{13} D. frac{1}{2} E. frac{5}{13}
answer:1. **Identify the Total Number of Outcomes**: With 13 choices available when Jack picks one ball, 12 remain for Jill. Therefore, the total number of ways for the picks is 13 times 12 = 156 pairs. 2. **Identify Favorable Outcomes for an Even Sum**: - **Both Balls are Even**: With 7 even-numbered balls (2, 4, 6, 8, 10, 12, 14), if Jack picks an even ball, 6 even balls remain for Jill. So, there are 7 times 6 = 42 ways. - **Both Balls are Odd**: Similarly, 6 odd-numbered balls (1, 3, 5, 7, 9, 11, 13). If Jack picks an odd ball, 5 odd balls remain for Jill. So, there are 6 times 5 = 30 ways. 3. **Calculate Total Favorable Outcomes**: The total number of favorable outcomes (both pick even or both pick odd) is 42 + 30 = 72. 4. **Calculate the Probability**: The probability that the sum of the numbers on the balls is even is given by this ratio: [ text{Probability} = frac{72}{156} = frac{6}{13} ] 5. **Conclusion**: The probability that the sum of the two numbers on the balls removed is even is frac{6{13}}. The final answer is C. boxed{frac{6}{13}}
question:Leo had 400 marbles in a jar. He packed the marbles with ten marbles in each pack, and he gave some of them to his two friends, Manny and Neil. He gave Manny 1/4 of the number of packs of marbles, Neil received 1/8 of the number of packs of marbles, and he kept the rest. How many packs of marbles did Leo keep? Let's answer step by step:
answer:Step 1: Determine the total number of packs of marbles. Leo had 400 marbles and packed them in groups of 10 marbles each. To find the total number of packs, we need to divide the total number of marbles by the number of marbles in each pack: Total packs = 400 marbles / 10 marbles per pack = 40 packs Step 2: Calculate the number of packs given to Manny. Leo gave Manny 1/4 of the total packs of marbles: Packs given to Manny = (1/4) * 40 packs = 10 packs Step 3: Calculate the number of packs given to Neil. Leo gave Neil 1/8 of the total packs of marbles: Packs given to Neil = (1/8) * 40 packs = 5 packs Step 4: Calculate the number of packs Leo kept. To find the number of packs Leo kept, subtract the packs given to Manny and Neil from the total number of packs: Packs Leo kept = Total packs - (Packs given to Manny + Packs given to Neil) Packs Leo kept = 40 packs - (10 packs + 5 packs) = 40 packs - 15 packs = 25 packs Leo kept 25 packs of marbles.