The graph would be a line along the x axis, between -3 and 3, excluding 3 and -3.
2006-11-13 09:08:36
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answer #1
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answered by jerzey79 2
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Draw thhe line x+3. This is a vertical line passing through the point (3,0) and parallel to the y-axis. The line extends forever in the both the +y and -y direction. Every point on this line x=3. So every point to the left of the line x is < 3.
Now do the same thing at the point (-3,0). You will now have a second line which passes through (-3,0) parallel to the y-axis and extending forever in both the +y and -y direction. Every point on this line is = x = -3 , But the absolute value is +3 Therefore, every point to the right of this line is less than 3.
Thus the graph of |x| =3 has a domain of -3
2006-11-13 17:32:16
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answer #2
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answered by ironduke8159 7
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Draw vertical lines at x = 3 and x = -3. Shade the area in between.
2006-11-13 17:10:27
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answer #3
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answered by Helmut 7
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All the x points that make this true is the open interval
(-3,+3) or -3
2006-11-13 17:09:09
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answer #4
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answered by modulo_function 7
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circle 3 and -3 and draw a line connecting the circles
2006-11-13 17:08:55
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answer #5
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answered by a_blue_grey_mist 7
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first make a point on 3 of the x axis
then make a straight vertical line (long ways)
shade or th answer will be to the LEFT of line you make (because it is LESS THAN)
2006-11-13 17:10:55
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answer #6
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answered by !!!!!! 3
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your graph should look like a "v" . where the vertex is 3 and it openning down. (but i dont think it can go negitive or else it wouldn't be absolute. (thats all i remeber)
2006-11-13 17:11:57
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answer #7
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answered by Anonymous
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I don't think you can...'y' is not defined here. Unless your equation is actually " | x | < 3 = y ".
2006-11-13 17:11:26
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answer #8
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answered by oskeewow13 3
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