diseaser | bah I still dont get it. |

Steve|Office | ... What's not to get? d/dt |3t-4| = (|3t-4|/(3t-4))*3. |

Cecen | You normally drop the function by a power of x, right? |

diseaser | I get that.. I just can't connect some dots here to tie it all together Cecen: yes |

Steve|Office | diseaser: What aren't you getting? |

Cecen | In this case, we can't just drop it by to the power of 0, so we have to divide it by the inside function Taking x^2 and dropping it's exponent by 1 is the same as dividing by x |

diseaser | ahhhh that helps |

walczyk | 4.8 A gun is located at the bottom of a hill of constant slope p. Show that the range of the measured up the slope of the hill is 2*v_0^2*cosa*sin(a-p)/(g*cos^2p) where a is the angel of elevation of hte gun, and the maximum value of the slope range is v_0^2/(g*(1+sinp)) |

Steve|Office | That sentence doesn't make sense no matter which network you paste it in. |

diseaser | Ahh so the critical value would be -4/3 in my case actually -4/3, and 4/3 |

Steve|Office | Why -4/3? |

diseaser | because it would make the numerator 0, thus f(-4/3) = 0 ? I think 3(|3t-4|/(3t-4)) |

Steve|Office | -4/3*3 - 4 is not 0. |

diseaser | no, but it would set the numerator of that equation to 0 making the whole thing 0 s/equation/expression crap no it wouldnt, would it I'm thinking |t|, not |3t-4| hargh, sorry man my mind is running on fumes here today |

jadenbane | Hmmm TRWBW, I'm flunking on the proving that if a symmetric has two distinct eigenvalues, their eigenvectors are orthogonal. I suck. |

g35coupe | what is a square of a graph G^2 |

SeveredCross | What's G/ |

FatalError | g35coupe, perhaps the graph obtained by squaring the adjacency matrix? (just a guess, mind you) I'm not familiar with that terminology |

walczyk | Steve|Office why doesnt it make sense |

Steve|Office | " Show that the range of the measured up the slope of the hill..." |

walczyk | yeah thats how the book worded it its asking the range of the projective measured from the bottom of the hill to the impact as opposed to the x-axis |

jadenbane | How do you prove that if a symmetric has two distinct eigenvalues, their eigenvectors are orthogonal |

n0dl | hello. what would be the best approach to factor y = 2 - 15x + 9x^2 - x^3 ? |

WILDSTYLE | Hi. |