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Alona [7]
3 years ago
10

Ray MO bisects LMN LMO = 6x-20 and NMO =2x+36. solve for x and find LMN

Mathematics
2 answers:
mestny [16]3 years ago
5 0
6x - 20 = 2x + 36
6x - 2x = 36 + 20
4x = 56
x = 56/4
x = 14

m∠LMN = 6x - 20 + 2x + 36 = 8x + 16 = 8 * 14 + 16 = 128°

Snowcat [4.5K]3 years ago
5 0

Answer:  The required  value of x is 14 and the maesure of angle LMN is 128 degrees.

Step-by-step explanation:  As given in the question and shown in the attached figure below, ray MO bisects angle LMN.

Also,

m\angle LMO=6x-20,\\\\m\angle NMO=2x+36.

We are to find the value of x and the measure of angle LMN.

Since ray MO bisects angle LMN, so we must have

m\angle LMO=m\angle NMO\\\\\Rightarrow 6x-20=2x+36\\\\\Rightarrow 6x-2x=36+20\\\\\Rightarrow 4x=56\\\\\Rightarrow x=\dfrac{56}{4}\\\\\Rightarrow x=14.

And, we get

m\angle LMN=6x-20+2x+36=8x+16=8\times14+16=112+16=128.

Thus, the required  value of x is 14 and the maesure of angle LMN is 128 degrees.

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7. \sqrt{2}

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12. (2,4).

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14. (5.5,-1.5).

<h3>What is the distance between two points?</h3>

Suppose that we have two points, (x_1,y_1) and (x_2,y_2). The distance between them is given by:

D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For item 7, the points are (-3,-2) and (-4,-3), hence the distance is:

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More can be learned about the distance between two points at brainly.com/question/18345417

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8 0
2 years ago
What is the equation of the circle with center (4, 4) that passes through the point (10, 14)? HELP MEEEEE PLEASE
Gemiola [76]

Answer:

The equation of the circle can be written as:

  • \left(x-4\right)^2+\left(y-4\right)^2=136

Step-by-step explanation:

The general equation of a circle with center (h,k) and radius r is:

\left(x-h\right)^2+\left(y-k\right)^2=r^2

In our example, we know \left(h,k\right)=\left(4,4\right), as we just have to make sure we need determine \:r^2.

\left(x-4\right)^2+\left(y-4\right)^2=r^2\:\:

As the circle passes through (10, 14), that pair of values for x and y must satisfy the equation. So we have:

\left(10-4\right)^2+\left(14-4\right)^2=r^2

\mathrm{Switch\:sides}

r^2=\left(10-4\right)^2+\left(14-4\right)^2

r^2=6^2+10^2

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r^2=136

Thus the equation of the circle can be written as:

\left(x-4\right)^2+\left(y-4\right)^2=136

5 0
3 years ago
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