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Math Help - my final project

  1. #16
    Newbie tracyneedshelp's Avatar
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    oh sorry i didn't know that meant square root : /

    ok i get the second one now...but can you tell me how to do the last one please?

    by the way thank you so much i was so frustrated with this i was in tears earlier. i would have failed me lass if it wasn't for you...
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  2. #17
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tracyneedshelp View Post
    ok i'm starting to think i'm just stupid...cuz you guys are giving me all this help and i still have no idea how to d this stuff...

    sorry guys for being so stupid :'[
    just calm down and go slowly, everthing will be ok. now let's walk through the last one.

    a regular hexagonal pyramid with a volume of 196 cm squared. and i need the base size and side lengths.
    so first, let's go to the volume equation, since that's what we are given, we want the volume to be 196

    \Rightarrow V = 196 = \frac {1}{3} bh

    \Rightarrow bh = 588 .....multiplied both sides by 3

    \Rightarrow b = \frac {588}{h}

    Now let's just choose a random value for the height. Let's say...12

    \Rightarrow b = \frac {588}{12} = 49

    Now let's find the dimensions that will get a hexagon (6-sided polygon) with area 49 cm^3

    We are told that:

    A = \frac {1}{2}apothem \cdot permeter

    we want A = 49

    \Rightarrow 49 = \frac {1}{2} apothem \cdot perimeter

    \Rightarrow apothem \cdot perimeter = 98

    \Rightarrow apothem = \frac {98}{perimeter}

    Now choose a random value for the perimeter, say ...i don't know...14

    \Rightarrow apothem = \frac {98}{14} = 7

    Now, can you tell me all the dimensions?
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  3. #18
    Newbie tracyneedshelp's Avatar
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    ok...i'm realy not too sure...

    but i'm thinking

    h= 12 area of the hexagon = 49?

    : / ok that's probably not right...
    Last edited by tracyneedshelp; May 26th 2007 at 06:16 PM.
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  4. #19
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tracyneedshelp View Post
    ok...i'm realy not too sure...

    but i'm thinking

    h= 12 area of the hexagon = 5.4?

    : / ok that's probably not right...
    ok, let's take this a step at a time. waht dimensions do we need first of all?

    we need the sides of the regular hexagon that will form the base (you can mention the apothem and all that stuff here, but it is not completely necessary, since to construct a regular polygon with sides of a certain length w would need a unique apothem). we also need the height and slant height of the pyramid. we can choose the slant height to be whatever we want (within reason) and the lateral area will follow.

    i already said the height was 12, so that's settled

    we said the perimeter was 14, this means the length of each side is 14/6

    these are the dimensions we need to build the object, all the other things like area and volume will fall into place by just building with these dimensions. that is, if we build a regular hexagon with each side 14/6, then the apothem will end up being 7 and the perimeter will end up being 14


    now read through all the posts that were made, and if you have any specific questions, ask away
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  5. #20
    Newbie tracyneedshelp's Avatar
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    thank you sooo much you are a life saver
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