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Math Help - Lots of Pre-Calculus review problems (HELP!)

  1. #1
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    Lots of Pre-Calculus review problems (HELP!)

    Hey so school's starting against soon and obviously I've procrastinated on my summer work.

    If any of you would be so kind as to maybe choose a problem and help me with it I would be greatly appreciative. Of course, I want to actually understand the problems, so I don't want just answers otherwise I would just go copy one of my friends.

    Hopefully this can be helpful to others going into Calc BC as well.















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  2. #2
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    Pls do not post such a large document.

    Post each problem and your attempts to solve it in a separate post.

    Answers to Q1 & 2 look correct.
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  3. #3
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    For number 6, \log_b y=\frac{\log_k y}{\log_k y} for any value of k

    And the natural logarithm is the log functionn with base e, so \log_b y=\frac{\ln y}{\ln b} and now you should be able to do that problem



    For number 8, call the ? y, and now the perimeter of your fence is x+x+y=200 and the enlcosed area (A)=x*y=5500 sq ft, so you have two equations and two unknowns and you should be able to solve that



    (you're not going to get much more help on this forum until the users see some more effort on the rest of the problems)
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    okay I'll try some more problems in a bit but I don't really know how to do math on the computer y'know?

    EDIT:

    Okay for problem #7 plugged all those numbers into the x y chart that you get when you choose "stat" on your calculator than went to calculate linear regression and got

    y=ax+b
    a= -2.257
    b= 22341.75

    but that makes almost zero sense to me. Help here?
    Last edited by Jph93; August 18th 2009 at 05:26 PM.
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  5. #5
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    Quote Originally Posted by Jph93 View Post
    Okay for problem #7 plugged all those numbers into the x y chart that you get when you choose "stat" on your calculator than went to calculate linear regression and got

    y=ax+b
    a= -2.257
    b= 22341.75

    but that makes almost zero sense to me. Help here?
    This cannot be correct, the values for salary in your table are increasing over time therefore your value for a (or gradient value) should be positive.
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  6. #6
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    Quote Originally Posted by artvandalay11 View Post
    For number 6, \log_b y=\frac{\log_k y}{\log_k y} for any value of k

    And the natural logarithm is the log functionn with base e, so \log_b y=\frac{\ln y}{\ln b} and now you should be able to do that problem
    I don't really get this.

    so \log_5 (x+3)=\frac{\log_9 (x+3)}{\log_9 (x+3)}=1? What?

    For the second part you're saying \log_5(x+3)=\frac{ln(x+3)}{ln(5)} ?

    That would make the answer what?
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  7. #7
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    Quote Originally Posted by pickslides View Post
    This cannot be correct, the values for salary in your table are increasing over time therefore your value for a (or gradient value) should be positive.
    Stupid mistake. How does a=256.5085714 and b=15326.47 look?
    Last edited by Jph93; August 18th 2009 at 05:53 PM.
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  8. #8
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    A post like this abuses the whole spirit of MHF. I suggest you ask your teacher for solutions or answers for checking purposes. Thread closed.
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