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Math Help - review of algebra 2

  1. #1
    Newbie
    Joined
    Sep 2005
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    9

    Smile 3 problems on review of algebra 2

    Hi, I really need some help on these 3 problems of my algebra 2 review.
    They are:

    1. Wendy, a loan officer at a bank, has $1,000,000 to lend and is required to obtain an average return of 18% per year, If she can lend at the rate of 19% or at the rate of 16%, how much can she lend at the 16% rate and still meet her requirement?

    2. How many cubic centimeters of pure hydrochloric acid should be added to 20cc of a 30% solution of hydrochloric acid to obtain a 50% solution?

    3. The line x-2y=-4 is tangent to a circle at (0,2). The line y=2x-7 is tangent to the same circle at (3,-1). Find the center of the circle.

    Thanks for your help.
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    1,631
    Four postings that have exactly the same contents--3 problems!

    Better, or preferable is one posting with only one problem.

    So, since there are 3 problems here, I will not give detailed answers as I used to. I might run out of space/window on my monitor here.

    ----------
    1. Wendy, a loan officer at a bank, has $1,000,000 to lend and is required to obtain an average return of 18% per year, If she can lend at the rate of 19% or at the rate of 16%, how much can she lend at the 16% rate and still meet her requirement?

    Let x = part of the $1,000,000 that she can lend at 16%

    0.16(x) +0.19(1,000,000 -x) = 0.18(1,000,000) -----***
    x = (180,000 -190,000) / (0.16 -0.19)
    x = $333,333.00 -------answer.

    ---------------
    2. How many cubic centimeters of pure hydrochloric acid should be added to 20cc of a 30% solution of hydrochloric acid to obtain a 50% solution?

    Pure acid is 100% concentration.
    Let x = cc of pure acid to be added.

    1.00(x) +0.30(20) = 0.50(x+20)
    x = (0.50*20 -0.30*20) / (1.00 -0.50)
    x = 8 cc ---------answer.

    ---------------
    3. The line x-2y=-4 is tangent to a circle at (0,2). The line y=2x-7 is tangent to the same circle at (3,-1). Find the center of the circle.

    The center is radius away from tangent points (0,2) and (3,-1).
    Let (x,y) = center

    distance = sqrt[(x2 -x1)^2 +(y2 -y1)^2]

    radius = radius
    sqrt[(x-0)^2 +(y-2)^2] = sqrt[(x-3)^2 +(y- (-1))^2]
    Square both sides,
    x^2 +(y-2)^2 = (x-3)^2 +(y+1)^2
    .....
    Umm, are you sure the problem as posted is complete? No lacking information? The radius of the circle is not given?
    Last edited by ticbol; September 11th 2005 at 03:09 AM.
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  3. #3
    Newbie
    Joined
    Sep 2005
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    sorry

    Sorry about that, i really needed the answer, but thanks. Next time i'll post more conservatiely.
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