What is the solution of 3+ x-2/x-3<_4

Mathematics

1 answer:

Ymorist [56]3 years ago

5 0

Answer:

x≤2−√6 or 0<x≤2+√6

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Type the correct answer in the box. Use a comma to separate the x- and y-coordinates of each point. The coordinates of the point

MAVERICK [17]

Answer:

On a unit circle, the point that corresponds to an angle of $0^{\circ}$ is at position $(1, \, 0)$ .

The point that corresponds to an angle of $90^{\circ}$ is at position $(0, \, 1)$ .

Step-by-step explanation:

On a cartesian plane, a unit circle is

a circle of radius $1$ ,
centered at the origin $(0, \, 0)$ .

The circle crosses the x- and y-axis at four points:

$(1, \, 0)$ ,
$(0, \, 1)$ ,
$(-1,\, 0)$ , and
$(0,\, -1)$ .

Join a point on the circle with the origin using a segment. The "angle" here likely refers to the counter-clockwise angle between the positive x-axis and that segment.

When the angle is equal to $0^\circ$ , the segment overlaps with the positive x-axis. The point is on both the circle and the positive x-axis. Its coordinates would be $(1, \, 0)$ .

To locate the point with a $90^{\circ}$ angle, rotate the $0^\circ$ segment counter-clockwise by $90^{\circ}$ . The segment would land on the positive y-axis. In other words, the $90^{\circ}$ -point would be at the intersection of the positive y-axis and the circle. Its coordinates would be $(0, \, 1)$ .