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    StarBlazeApple's Avatar
    StarBlazeApple Posts: 2, Reputation: 1
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    #1

    Nov 23, 2010, 01:57 AM
    Number Problem
    A number is thrice another number. The sum of their reciprocals is 4. Find the numbers.
    StarBlazeApple's Avatar
    StarBlazeApple Posts: 2, Reputation: 1
    New Member
     
    #2

    Nov 23, 2010, 02:05 AM
    Number Problem
    One number is three times as large as another. The sum of their reciprocals is 4/3 (fraction). Find the two numbers.
    GeordieUK's Avatar
    GeordieUK Posts: 4, Reputation: 4
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    #3

    Nov 23, 2010, 07:27 AM
    A number is thrice another number.
    So:

    x = 3y

    The sum of their reciprocals is 4. Find the numbers.
    So:

    1/x + 1/y = 4

    Substitue x for 3y

    1/3y + 1/y = 4

    etc...

    Therefore the numbers are 1 and 1/3
    kakam's Avatar
    kakam Posts: 3, Reputation: 1
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    #4

    Dec 7, 2010, 04:54 PM
    a.$25 b.$400 c.$2500 d.$40,000

    5 = 3
    9 w use cross products.
    5 =3 Divide both sides by 5
    9 w
    5(w)=9(3)
    5w=27
    5 5
    w=27
    5

    on the map,the distance two sets of measurements,such as 1 in:5 mi.a scale drawing or scale model uses a scale to represent an object as smaller or larger than the actual object.a map is an example of a scale drawing...

    map - 1 in.
    actual 18 mi. write the scale as a fration.
    1 0.625 let x be the actual distance.
    18 x
    x.1=18(0.625) use cross products to slove.
    x=11.25

    similar,
    figures have exacly the same shape but not necessarity the same size.


    If we make a diagram and label it as such, perhaps we can make use of some similar triangles.

    DC=180, AB=120, BE=20, BC=2.5, DG=60, FH=x-120



    Solve for x.


    If we make a diagram and label it as such, perhaps we can make use of some similar triangles.

    DC=180, AB=120, BE=20, BC=2.5, DG=60, FH=x-120
    60 =-120
    2.5 20


    Solve for x.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Dec 8, 2010, 12:13 AM

    What do you mean by this? Is that a question? If so, create your own thread. If that's an answer, I don't see how it can help the original asker.

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