ANSWER???????????????????
1 answer:
The range of a function is the set of values that the function reach with its outputs.
In other words, a number belongs to the range of the function
if there exists a number
belonging to the domain of
such that
.
Looking at your graph, you can see that any value smaller than 2 is not reached by the graph, because the lines are only drawn from 2 upwards.
So, the answer is C, because it is the set of all numbers greater then or equal to 2.
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Since ABCD is a parallelogram, line AB and line DC are parallel and has the same value.
To solve this, equate line AB to be equal with line DC.
So,
Line AB = Line DC
(9x-14)in = (3x +4)in
Next group like terms to get the value of x
9x in-3x in = 4in+14in
= 
x = 3in
Since, we now have the value of x, substitute it to line DC’s equation.
DC=(3x+4)in
DC=(3(3) +4)in
DC=(9 +4) in
DC= 13 in
To check if the value is really correct, substitute X to AB
AB=(9x -14)in
AB=(9(3)-14)in
AB=(27-14)in
AB=13 in
To solve this, equate line AB to be equal with line DC.
So,
Line AB = Line DC
(9x-14)in = (3x +4)in
Next group like terms to get the value of x
9x in-3x in = 4in+14in
x = 3in
Since, we now have the value of x, substitute it to line DC’s equation.
DC=(3x+4)in
DC=(3(3) +4)in
DC=(9 +4) in
DC= 13 in
To check if the value is really correct, substitute X to AB
AB=(9x -14)in
AB=(9(3)-14)in
AB=(27-14)in
AB=13 in
<h3>1. Inscribe a circle ΔJKL. Identify the point of concurrency that is the center of the circle you drew. </h3>
<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>
To inscribe a circle in the attached triangle, we must follow the following steps. The resultant figure is the first one below.
1. Draw the angle bisector of angle JKL. The straight line in the green one.
2. Draw the angle bisector of angle KLJ. The straight line in the blue one.
3. Draw point D at these two lines.
4. Draw a line through point D perpendicular to side JK. The straight line in the red one.
5. Mark point G at the red line and side JK.
6. Draw a circle with center at D (point of concurrency) and a radius of DG.
<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>
To circumscribe a circle in the attached triangle, we must follow the following steps:
1. Draw the perpendicular bisector of the side FG. The straight line in the red one. The resultant figure is the second one below.
2. Draw the perpendicular bisector of the side GH. The straight line in the blue one.
3. The center of the circle is the intersection of these two lines
4. Draw a circle with center at D (point of concurrency).
<h3>3. Construct the two lines tangents to circle</h3>A tangent to a circle is a straight line that touches the circle at an only point. If a point A is outside a circle, we can draw two lines that pass through this point and are tangent to the circle at two different points each. So this is shown in the third figure. Lines red and blue are tangent to the circle at two different points.