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Math Help - please help me

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    please help me

    please help me to solve that questions
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    Quote Originally Posted by m777 View Post
    please help me to solve that questions
    1. |(2/x) - 4| < 3

    This is really two inequalities:
    -3 < (2/x) - 4 < 3

    So add 4 to all "three" sides:
    -3 + 4 < (2/x) - 4 + 4 < 3 + 4

    1 < 2/x < 7

    So 1 < 2/x and 2/x < 7

    The first gives:
    0 < 2/x -1
    0 < (2-x)/x
    which is positive for x < 2.

    The second gives:
    2/x < 7
    2/x - 7 < 0
    (2-7x)/x < 0
    which is negative for x > 2/7.

    So 2/7 < x < 2.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by m777 View Post
    please help me to solve that questions
    2. Given points: (3, 4), (5, 8) and (3, 9)
    Show that they form the points of a right triangle.

    I'm not going to do all of it, just point you along.
    Slopes: If this is a right triangle then two lines connecting these points are perpendicular. Two lines are perpendicular if m1*m2 = -1, where m1 and m2 are the slopes of the lines in question. So pairing off the points, find the slopes of the lines between them (you'll have 3 lines). Now determine if any two of these are perpendicular.

    Distances: If this is a right triangle, then the distances will follow a^2 + b^2 = c^2, where c is the longest side. So find the distance between each pair of points and see if it fits the Pythagorean Theorem.

    -Dan

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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by m777 View Post
    please help me to solve that questions
    (1)Find the equation of line which has y intercept -5 and is perpendicular to the line joining the points (3, 6) and (1, 8)



    The equation of your line is
    y = mx + b.

    We know that the y-intercept is -5, so b = -5

    Now, we know this line is perpendicular to the line connecting the points (3,6) and (1,8). Call this slope M. We know then that m*M = -1. So find M, then you can find m.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by m777 View Post
    please help me to solve that questions

    Find the equation of the circle with centre at (2, -3) and whose radius is 3 times the radius of x^2 + y^2 - 2x - 4y = 11.

    The first thing we need to do is put the second circle into standard form. We do this by completing the square for both x and y:

    x^2 - 2x + 1 + y^2 - 4y + 4 = 11 + 1 + 4

    (x - 1)^2 + (y - 2)^2 = 16

    So this circle has a radius of 4. The circle we want has a radius of 3*4 = 12.

    So
    (x -2)^2 + (y+3)^2 = 144 is the desired circle.

    -Dan
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