<h2>
Answer:</h2>
∠LMN is a right angle
<h2>
Step-by-step explanation:</h2>
If we want to prove that two right triangles are congruent by knowing that the corresponding hypotenuses and one leg are congruent, we begin as follows:
- Since two legs are congruent and we know this by the hash marks, then the triangle ΔLKN is isosceles.
- By definition LN ≅ NK
- If ∠LMN is a right angle, then MN is the altitude of triangle ΔLKN
- Also MN is the bisector of LK, so KM ≅ ML
- So we have two right triangles ΔLMN and ΔKM having the same lengths of corresponding sides
- In conclusion, ΔLMN ≅ ΔKMN
Answer:
Cole has 15 pencils.
Step-by-step explanation:
First, we have the equation c + o =18, c representing cole, and o representing connor.
From there, we get 5o=c
Using substitiution, we get o=3
5(3)=15
15+3=18
So we want to write a fraction that is less than 5/6 and has a denominator of 8. So lets start from 1/1 that is greater then 5/6 and doesn't have 8 as a denominator. Lets take one half from 1/1 and well get 1/2.
Answer:
(2.83 , 1 , 4)
Step-by-step explanation:
Rewrite these equations in matrix form
![\left[\begin{array}{ccc}2&2&-1\\4&-2&-2\\3&3&-4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%26-1%5C%5C4%26-2%26-2%5C%5C3%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
we can write it like this,
so to solve it we need to take the inverse of the 3 x 3 matrix A then multiply it by B.
We get the inverse of matrix A,
![A^{-1}=\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right] \\](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C)
now multiply the matrix with B
![X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2.83\\1\\4\end{array}\right] \\](https://tex.z-dn.net/?f=X%3DA%5E%7B-1%7DB%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%2F15%261%2F6%26-1%2F5%5C%5C1%2F3%26-1%2F6%260%5C%5C3%2F5%260%26-2%2F5%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C2%5C%5C-4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2.83%5C%5C1%5C%5C4%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
