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3 years ago

Given: MO−→− bisects ∠LMN m∠LMO = 6x−20 m∠NMO = 2x+36 Solve for x and find m∠LMN.

Mathematics

1 answer:

Lapatulllka [165]3 years ago

5 0

Angles LMO and NMO are congruent, of equal measures,

x=15

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Can you tell me how to calculate R for below Matrix ?

scoray [572]

Answer:

The rank of the matrix is 3.

Step-by-step explanation:

Consider the prided information.

$x=\begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix}=\begin{bmatrix}9&2&6&5&8\\ 12&8&6&4&10\\ 3&4&0&2&1\end{bmatrix}$

Reduce the matrix in row echelon form as shown:

$R_1\:\leftrightarrow \:R_2\begin{bmatrix}12&8&6&4&10\\ 9&2&6&5&8\\ 3&4&0&2&1\end{bmatrix}\\$

$R_2\:\leftarrow \:R_2-\frac{3}{4}\cdot \:R_1\ and\ R_3\:\leftarrow \:R_3-\frac{1}{4}\cdot \:R_1\\\\\begin{bmatrix}12&8&6&4&10\\ 0&-4&\frac{3}{2}&2&\frac{1}{2}\\ 0&2&-\frac{3}{2}&1&-\frac{3}{2}\end{bmatrix}$

$R_3\:\leftarrow \:R_3+\frac{1}{2}\cdot \:R_2\\\begin{bmatrix}12&8&6&4&10\\ 0&-4&\frac{3}{2}&2&\frac{1}{2}\\ 0&0&-\frac{3}{4}&2&-\frac{5}{4}\end{bmatrix}$

$R_2\:\leftarrow \:R_2-\frac{3}{2}\cdot \:R_3\ and\ R_1\:\leftarrow \:R_1-6\cdot \:R_3\\\begin{bmatrix}12&8&0&20&0\\ 0&-4&0&6&-2\\ 0&0&1&-\frac{8}{3}&\frac{5}{3}\end{bmatrix}\\R_2\:\leftarrow \:-\frac{1}{4}\cdot \:R_2\ , \ R_1\:\leftarrow \:R_1-8\cdot \:R_2\ and\ \:R_1\:\leftarrow \frac{1}{12}\cdot \:R_1\\\begin{bmatrix}1&0&0&\frac{8}{3}&-\frac{1}{3}\\ 0&1&0&-\frac{3}{2}&\frac{1}{2}\\ 0&0&1&-\frac{8}{3}&\frac{5}{3}\end{bmatrix}$

The rank of a matrix is the number of non zeros rows.

Thus, the rank of the matrix is 3.

6 0

3 years ago

A city averages 14 hours of daylight in June (the longest days) and 10 in December (the shortest days). Assume that the number o

OLga [1]

We have that the graph will have a solution that is mathematically given as

y=12sin(6t-\pi)+2

<h3>Daylight time June trough to December</h3>

Question Parameters:

A city averages 14 hours of daylight in June (the longest days) and 10 in December (the shortest days).

Generally the equation for the Amplitude and vertical shift is mathematically given as

$A=\frac{Max+min}{2}\\\\A=\frac{10+14}{2}$

A=2

vertical shift = $\frac{Max-min}{2}$

vertical shift = $\frac{14-10}{2}$

vertical shift =1

Therefore, the time is given as

$t=\frac{2\pi}{b}$

Where

$b=\pi/6\\\\t=\frac{2\pi}{\pi/6}$

Hence,The graph will have a solution

y=12sin(6t-\pi)+2

For more information on Graph

brainly.com/question/14375099

8 0

2 years ago

The center of mass is defined as a point on the straight line between two objects with masses m1 and m2 such that . What are the

STatiana [176]

Let the point be P, and the masses be located at P1 and P2.

The centre of mass is such that the moments of masses m1 and m2 exert an equal moment at the point P,
and the distance P1P2=d
namely,

m1(mPP1)=m2(mPP2)
The distance of the centre of mass from m1 is therefore
d1=d*m2/(m1+m2)
Similarly, the distance of the centre of mass from m2 is
d2=d*m1/(m1+m2)

5 0

3 years ago

Read 2 more answers

I will give brainliest

Tju [1.3M]

Answer:

if you see the graph for x=-1

the y is equal -4

you can draw a straight line parallel y axes from point x=-1 it will cut the graph on y=-4

7 0

2 years ago

Samantha buys a car for $19,000. Looking on Kelley Blue Book, she realized that

Yuri [45]

Answer:

$16,150

Step-by-step explanation:

hope this helps with the work

4 0

3 years ago