Cual es el complementario de un angulo
1 answer:
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3x -y ⩾ 6
3x - 6 ⩾ y
now, with inequalities, what we do is, we graph the line of 3x - 6 = y, and then we shade the "true region".
if we pick a point on say hmmm (4, 0), namely x = 4 and y = 0, we can plug that in the inequality and see what we get,
3(4) - 0 ⩾ 6
12 - 0 ⩾ 6
12 ⩾ 6
is 12 really greater or equals to 6? well yes, therefore, the point (4, 0) lies on the "true region", since it's true, 12 is indeed ⩾ 6, so, where that point is, we shade.
now, the ⩾ means equals to or greater, and therefore, since the values could also equal the boundary points, the line is a solid line, because it includes the line itself, as well as the shading.
check the picture below.
3x - 6 ⩾ y
now, with inequalities, what we do is, we graph the line of 3x - 6 = y, and then we shade the "true region".
if we pick a point on say hmmm (4, 0), namely x = 4 and y = 0, we can plug that in the inequality and see what we get,
3(4) - 0 ⩾ 6
12 - 0 ⩾ 6
12 ⩾ 6
is 12 really greater or equals to 6? well yes, therefore, the point (4, 0) lies on the "true region", since it's true, 12 is indeed ⩾ 6, so, where that point is, we shade.
now, the ⩾ means equals to or greater, and therefore, since the values could also equal the boundary points, the line is a solid line, because it includes the line itself, as well as the shading.
check the picture below.