## Me ready to start epic thread | |

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catsscratchfeverUser ID: 33747838 Canada 04/11/2013 11:09 PM Report Abusive Post Report Copyright Violation | ax^2 from A to B R(x) { bx + c from B to C B is 1000 feet horizontally from A and 100 feet higher. Since the road is smooth, R'(x) is continuous. What is the value of B? I think it involves some differentiation.. but I don't really know what to do. The options for answers are: a) .2 b) .02 c) .002 d) .0002 e) .00002 ?????????????????????????????????????????????????? |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:09 PM Report Abusive Post Report Copyright Violation | |

Anonymous Coward User ID: 28111028 Canada 04/11/2013 11:09 PM Report Abusive Post Report Copyright Violation | |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:10 PM Report Abusive Post Report Copyright Violation | The equation of a road is: Quoting: ax^2 from A to B R(x) { bx + c from B to C B is 1000 feet horizontally from A and 100 feet higher. Since the road is smooth, R'(x) is continuous. What is the value of B? I think it involves some differentiation.. but I don't really know what to do. The options for answers are: a) .2 b) .02 c) .002 d) .0002 e) .00002 ?????????????????????????????????????????????????? catsscratchfever How can I help you with your misunderstandings of the universe? |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:12 PM Report Abusive Post Report Copyright Violation | |

Anonymous Coward User ID: 28111028 Canada 04/11/2013 11:13 PM Report Abusive Post Report Copyright Violation | |

RapophisUser ID: 633354 Canada 04/11/2013 11:14 PM Report Abusive Post Report Copyright Violation | |

catsscratchfeverUser ID: 33747838 Canada 04/11/2013 11:14 PM Report Abusive Post Report Copyright Violation | The equation of a road is: Quoting: ax^2 from A to B R(x) { bx + c from B to C B is 1000 feet horizontally from A and 100 feet higher. Since the road is smooth, R'(x) is continuous. What is the value of B? I think it involves some differentiation.. but I don't really know what to do. The options for answers are: a) .2 b) .02 c) .002 d) .0002 e) .00002 ?????????????????????????????????????????????????? catsscratchfever How can I help you with your misunderstandings of the universe? sdavis642 SOOOoooooo When in doubt pick C....Come on OP..... The answer to this question depends on the relationship between the points x, A, and B. I am assuming that point A is at the origin, which is consistent with the expression for R(x), and the coordinates of point B would be (1000, 100). We have the equation 100 = R(1000) = a(1000)^2 = 1000000a Solving for a gives a = 0.0001 Because the derivative is also continuous, we also know that (ax^2)' = (bx + c)' at x = 1000 or 2ax = b at x = 1000 So b = 2(0.0001)*1000 = 0.2 OP |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:15 PM Report Abusive Post Report Copyright Violation | |

Anonymous Coward User ID: 28111028 Canada 04/11/2013 11:15 PM Report Abusive Post Report Copyright Violation | |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:16 PM Report Abusive Post Report Copyright Violation | The equation of a road is: Quoting: ax^2 from A to B R(x) { bx + c from B to C B is 1000 feet horizontally from A and 100 feet higher. Since the road is smooth, R'(x) is continuous. What is the value of B? I think it involves some differentiation.. but I don't really know what to do. The options for answers are: a) .2 b) .02 c) .002 d) .0002 e) .00002 ?????????????????????????????????????????????????? catsscratchfever How can I help you with your misunderstandings of the universe? sdavis642 SOOOoooooo When in doubt pick C....Come on OP..... The answer to this question depends on the relationship between the points x, A, and B. I am assuming that point A is at the origin, which is consistent with the expression for R(x), and the coordinates of point B would be (1000, 100). We have the equation 100 = R(1000) = a(1000)^2 = 1000000a Solving for a gives a = 0.0001 Because the derivative is also continuous, we also know that (ax^2)' = (bx + c)' at x = 1000 or 2ax = b at x = 1000 So b = 2(0.0001)*1000 = 0.2 OP :lihvlcdxuxluxggh: catsscratchfever Wow, you must have gone to college. Congratulations. Now ask me a question. |

Wikid SmithUser ID: 36652440 United States 04/11/2013 11:19 PM Report Abusive Post Report Copyright Violation | |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:20 PM Report Abusive Post Report Copyright Violation | |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:21 PM Report Abusive Post Report Copyright Violation | |

catsscratchfeverUser ID: 33747838 Canada 04/11/2013 11:21 PM Report Abusive Post Report Copyright Violation | ax^2 from A to B R(x) { bx + c from B to C B is 1000 feet horizontally from A and 100 feet higher. Since the road is smooth, R'(x) is continuous. What is the value of B? I think it involves some differentiation.. but I don't really know what to do. The options for answers are: a) .2 b) .02 c) .002 d) .0002 e) .00002 ?????????????????????????????????????????????????? catsscratchfever How can I help you with your misunderstandings of the universe? sdavis642 SOOOoooooo When in doubt pick C....Come on OP..... The answer to this question depends on the relationship between the points x, A, and B. I am assuming that point A is at the origin, which is consistent with the expression for R(x), and the coordinates of point B would be (1000, 100). We have the equation 100 = R(1000) = a(1000)^2 = 1000000a Solving for a gives a = 0.0001 Because the derivative is also continuous, we also know that (ax^2)' = (bx + c)' at x = 1000 or 2ax = b at x = 1000 So b = 2(0.0001)*1000 = 0.2 OP catsscratchfever Wow, you must have gone to college. Congratulations. Now ask me a question. sdavis642 I deserve none of your intellect....I have my own.... |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:22 PM Report Abusive Post Report Copyright Violation | |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:26 PM Report Abusive Post Report Copyright Violation | SOOOoooooo When in doubt pick C....Come on OP..... The answer to this question depends on the relationship between the points x, A, and B. I am assuming that point A is at the origin, which is consistent with the expression for R(x), and the coordinates of point B would be (1000, 100). We have the equation 100 = R(1000) = a(1000)^2 = 1000000a Solving for a gives a = 0.0001 Because the derivative is also continuous, we also know that (ax^2)' = (bx + c)' at x = 1000 or 2ax = b at x = 1000 So b = 2(0.0001)*1000 = 0.2 OP :lihvlcdxuxluxggh: catsscratchfever Wow, you must have gone to college. Congratulations. Now ask me a question. sdavis642 I deserve none of your intellect....I have my own.... :ljvlc;c;t;ctu;ut: catsscratchfever It must have cost a small fortune to get that mind. I underestimate your dedication to the ignorant. |

Anonymous Coward (OP)User ID: 25546614 United States 04/11/2013 11:30 PM Report Abusive Post Report Copyright Violation | |

whatever User ID: 20694328 United States 04/11/2013 11:32 PM Report Abusive Post Report Copyright Violation | |

catsscratchfeverUser ID: 33747838 Canada 04/11/2013 11:33 PM Report Abusive Post Report Copyright Violation | ...SOOOoooooo When in doubt pick C....Come on OP..... The answer to this question depends on the relationship between the points x, A, and B. I am assuming that point A is at the origin, which is consistent with the expression for R(x), and the coordinates of point B would be (1000, 100). We have the equation 100 = R(1000) = a(1000)^2 = 1000000a Solving for a gives a = 0.0001 Because the derivative is also continuous, we also know that (ax^2)' = (bx + c)' at x = 1000 or 2ax = b at x = 1000 So b = 2(0.0001)*1000 = 0.2 OP catsscratchfever Wow, you must have gone to college. Congratulations. Now ask me a question. sdavis642 I deserve none of your intellect....I have my own.... catsscratchfever It must have cost a small fortune to get that mind. I underestimate your dedication to the ignorant. sdavis642 Thank you for being........You... |

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