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DerKrebs [107]
3 years ago
12

Geometry help. Inscribing circle help and constructing lines.

Mathematics
1 answer:
prohojiy [21]3 years ago
3 0
<h3>1. Inscribe a circle ΔJKL. 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.

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Find the equation of the line passing through the points (1,5) and (3,9)​
Arada [10]

Answer:

  • \boxed{\sf{2}}

Step-by-step explanation:

Use the slope formula.

\underline{\text{SLOPE:}}

\Longrightarrow: \sf{\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{Rise}{Run} }

\sf{y_2=9}\\\\\\\sf{y_1=5}\\\\\\\sf{x_2=3}\\\\\\\sf{x_1=1}

\sf{\dfrac{9-5}{3-1} }

Solve.

\sf{\dfrac{9-5}{3-1}=\dfrac{4}{2}=\boxed{\sf{2}}

  • <u>Therefore, the slope is 2, which is our answer.</u>

I hope this helps. Let me know if you have any questions.

8 0
2 years ago
Plz help i serouisly need help I will reward with one hundred points
Mekhanik [1.2K]

Percent increase is

100(new/old - 1)

Here that's

100(26/22 - 1) = 200/11 ≈ 18.2

Answer: third choice

6 0
3 years ago
Read 2 more answers
2x+5=12 pleas and thank you :D
Alik [6]
2x+5=12
12-5=2x
2x=7
X=3.5
3 0
3 years ago
Read 2 more answers
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