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Thread: length 4 times width; find perim if area is 144; find 2 cons

  1. #1
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    length 4 times width; find perim if area is 144; find 2 cons

    Ok guys I am really rusty now- Its been over 10 years I haven't done math really need the hlep.

    Question 1: The length of a rectangl is 4 times its width. If the area is 144cm2 (thats square), find its perimeter?

    Question 2: Find 2 consective odd intergers whose sum is 56.?

  2. #2
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    Re: Problems

    25 posts and you have not learned that the most effective post is the post that SHOWS WORK. You have only typed the problem statements. Please do better than that.

    For the rectangle, you need only

    1) Length * Width = Area
    2) 2*Length + 2*Width = Perimeter.

    Draw a picture and look at it. You should see it.

    For the odd integers, you need only a name for the odd integers. Write down a definition and show us what you get.

  3. #3
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    Re: Problems

    Quote Originally Posted by tkhunny
    25 posts and you have not learned that the most effective post is the post that SHOWS WORK. You have only typed the problem statements. Please do better than that.

    For the rectangle, you need only

    1) Length * Width = Area
    2) 2*Length + 2*Width = Perimeter.

    Draw a picture and look at it. You should see it.

    For the odd integers, you need only a name for the odd integers. Write down a definition and show us what you get.

    What do you mean about your statement?

    the 2 odd intergers are 53 and 3. Is this correct?

    I am still working on the perimeter

  4. #4
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    Re: Problems

    The perimeter is 72. Is that correct?

  5. #5
    Senior Member skeeter's Avatar
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    Re: Problems

    53 and 3 are not consecutive odd integers.

    as far as the perimeter, let's see how you arrived at the value 72.

  6. #6
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    Re: Problems

    Quote Originally Posted by taragbeh
    the 2 odd intergers are 53 and 3. Is this correct?
    NO. They have to be CONSECUTIVE.
    I'm just an imagination of your figment !

  7. #7
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    Re: Problems

    Quote Originally Posted by taragbeh
    The perimeter is 72. Is that correct?
    NO. Hint: w(4w) = 144
    I'm just an imagination of your figment !

  8. #8
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    Re: Problems

    Quote Originally Posted by Denis
    Quote Originally Posted by taragbeh
    The perimeter is 72. Is that correct?
    NO. Hint: w(4w) = 144
    Ok the perimetre is 36

  9. #9
    Senior Member skeeter's Avatar
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    Re: Problems

    no, the perimeter is not 36, w[sup:wme6ykz7]2[/sup:wme6ykz7] is 36.

    are you going to continue this guessing game?

  10. #10
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    Re: Problems

    Quote Originally Posted by skeeter
    no, the perimeter is not 36, w[sup:11oqw26l]2[/sup:11oqw26l] is 36.

    are you going to continue this guessing game?

    I am trying to figure it out but not sue what i am doing but i am really trying.
    ok so if

    p=2l + 36

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