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Math Help - Algebra 2 Problems-Help!

  1. #1
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    Algebra 2 Problems-Help!

    1) The illumination of a light source is inversely proportional to the square of the distance from the light source. If a light has an illumination of 75 foot-candles at a distance of 4 feet from the light source:

    a. Find a model for the illumination of this light

    b. What is the illumination of this light at a distance of 8 feet from the light source?

    c. How far from the light source would you need to be for the illumination to be 50 foot-candles?




    2) The weight of a body varies inversely as the square of the distance from the earth. If a man weighs 200lbs on the top of Mt. McKinley (20320 feet above the surface of the earth), how much would he weigh when he is 3000 miles above the surface of the earth.

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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by ep78 View Post
    1) The illumination of a light source is inversely proportional to the square of the distance from the light source. If a light has an illumination of 75 foot-candles at a distance of 4 feet from the light source:

    a. Find a model for the illumination of this light

    I=\frac{k}{d^2}

    75=\frac{k}{16}

    k=1200 the constant of porportionality

    b. What is the illumination of this light at a distance of 8 feet from the light source?

    For a distance of 8 feet: I=\frac{1200}{64}\Longrightarrow I=18.75

    c. How far from the light source would you need to be for the illumination to be 50 foot-candles?

    50=\frac{1200}{d^2}\Longrightarrow d=4\sqrt{6}



    Here's your first one.
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  3. #3
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    w = weight
    d = distance

     w = \frac{k}{d^2}

     w_{1} = 200  lbs
     d_{1}  = 20320 feet

     w_{2} = ???  lbs
     d_{2}  = 15840000 feet (converted from 3000 miles)

    Let's plug in  w_{1} and  d_{1} in our equation. This will allow us to determine k.

      w_{1} = \frac{k}{d_{1}^2}

     200 = \frac{k}{20320^2}

     200 = \frac{k}{412902400}

     k = 412902400 * 200 = 82580480000

    Now that we know  k and are given  d_{2} , we can use our equation once again, this time solving for w_{2} , which is unknown.

     w_{2} = \frac{82580480000}{d_{2}^2}

     w_{2} = \frac{82580480000}{15840000^2} = \frac{82580480000}{250905600000000} = 0.000329129681

    So, the man weighs about 0.00033 lbs when he is 3000 miles above the surface of the earth.

    Check my arithmetic.
    Good luck!
    -Andy
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  4. #4
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    thanks so much!
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