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Math Help - Pre-calculus Problems

  1. #1
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    Pre-calculus Problems

    Can someone please help me with these problems? I have tried working them out, but I can't seem to get a solution. Please see attachment. Thanks
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  2. #2
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    Quote Originally Posted by fw_mathis View Post
    Can someone please help me with these problems? I have tried working them out, but I can't seem to get a solution. Please see attachment. Thanks
    Hello,

    For #1, plug in the x values given into "x"; so,

    y = 1/(-2 - 2);

    y = 1/(-2 - 1);... and so forth

    You'll get the following answers,

    y = -1/4, -1/3, -1/2, -1, undefined (since you get 1/0), 1 for x = -2, -1, 0, 1, 2, 3, respectively.

    2.)

    x - 2*sqrt(x) = 0

    -2*sqrt(x) = -x

    2*sqrt(x) = x

    sqrt(x) = x/2

    Square both sides;

    x = x^2/4

    x - x^2/4 = 0

    (-x^2 + 4x)/4 = 0

    -x^2 + 4x = 0

    -x*(x - 4) = 0

    x = 0 V x = 4
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  3. #3
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    Quote Originally Posted by fw_mathis View Post
    Can someone please help me with these problems? I have tried working them out, but I can't seem to get a solution. Please see attachment. Thanks
    3.)

    Let (0, 0) be your starting position.

    You're at (75, 65);

    You could either find the hypotenuse or use the distance formula (which is the same thing).

    Using the distance formula:

    d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)

    d = sqrt((75 - 0)^2 + (65 - 0)^2)

    d = sqrt(75^2 + 65^2)

    d = sqrt(9850)

    d = 5*sqrt(394)

    d ~ 99.25 kilometers

    Pythag:

    a^2 + b^2 = c^2

    a = 65

    b = 75

    Which is the distance formula. Do you see how?

    65^2 + 75^2 = c^2 ...

    4.) V = s^3

    1000 = s^3

    Solve for s;

    s = 10 cm
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