Given : tan 235 = 2 tan 20 + tan 215
To Find : prove that
Solution:
tan 235 = 2 tan 20 + tan 215
Tan x = Tan (180 + x)
tan 235 = tan ( 180 + 55) = tan55
tan 215 = tan (180 + 35) = tan 35
=> tan 55 = 2tan 20 + tan 35
55 = 20 + 35
=> 20 = 55 - 35
taking Tan both sides
=> Tan 20 = Tan ( 55 - 35)
=> Tan 20 = (Tan55 - Tan35) /(1 + Tan55 . Tan35)
Tan35 = Cot55 = 1/tan55 => Tan55 . Tan35 =1
=> Tan 20 = (Tan 55 - Tan 35) /(1 + 1)
=> Tan 20 = (Tan 55 - Tan 35) /2
=> 2 Tan 20 = Tan 55 - Tan 35
=> 2 Tan 20 + Tan 35 = Tan 55
=> tan 55 = 2tan 20 + tan 35
=> tan 235 = 2tan 20 + tan 215
QED
Hence Proved
ANSWER

EXPLANATION
The explicit rule for an arithmetic sequence is given by:

It was given that:

and

The common difference is

We plug in the values into the explicit formula to get,

Expand to get,

The explicit rule is

Answer:
![x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://tex.z-dn.net/?f=x_3%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%263%261%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
According to the given situation, The computation of all x in a set of a real number is shown below:
First we have to determine the
so that 
![\left[\begin{array}{cccc}1&-3&5&-5\\0&1&-3&5\\2&-4&4&-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%5C0%261%26-3%265%5C%5C2%26-4%264%26-4%5Cend%7Barray%7D%5Cright%5D)
Now the augmented matrix is
![\left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\2&-4&4&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C2%26-4%264%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
After this, we decrease this to reduce the formation of the row echelon
![R_3 = R_3 -2R_1 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&2&-6&6\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_3%20%3D%20R_3%20-2R_1%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C0%262%26-6%266%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_3 = R_3 -2R_2 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&5\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_3%20%3D%20R_3%20-2R_2%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%265%5C%20%7C%5C%200%5C%5C0%260%260%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_2 = 4R_2 +5R_3 \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&4&-12&0\ |\ 0\\0&0&0&-4\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_2%20%3D%204R_2%20%2B5R_3%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%264%26-12%260%5C%20%7C%5C%200%5C%5C0%260%260%26-4%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_2 = \frac{R_2}{4}, R_3 = \frac{R_3}{-4} \rightarrow \left[\begin{array}{cccc}1&-3&5&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_2%20%3D%20%5Cfrac%7BR_2%7D%7B4%7D%2C%20%20R_3%20%3D%20%5Cfrac%7BR_3%7D%7B-4%7D%20%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-3%265%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%261%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_1 = R_1 +3 R_2 \rightarrow \left[\begin{array}{cccc}1&0&-4&-5\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_1%20%3D%20R_1%20%2B3%20R_2%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-4%26-5%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%26-1%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)
![R_1 = R_1 +5 R_3 \rightarrow \left[\begin{array}{cccc}1&0&-4&0\ |\ 0\\0&1&-3&0\ |\ 0\\0&0&0&-1\ |\ 0\end{array}\right]](https://tex.z-dn.net/?f=R_1%20%3D%20R_1%20%2B5%20R_3%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-4%260%5C%20%7C%5C%200%5C%5C0%261%26-3%260%5C%20%7C%5C%200%5C%5C0%260%260%26-1%5C%20%7C%5C%200%5Cend%7Barray%7D%5Cright%5D)

![x = \left[\begin{array}{c}4x_3&3x_3&x_3\\0\end{array}\right] \\\\ x_3 = \left[\begin{array}{c}4&3&1\\0\end{array}\right]](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4x_3%263x_3%26x_3%5C%5C0%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%20x_3%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D4%263%261%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
By applying the above matrix, we can easily reach an answer
Answer:
Common ratio: 3
Step-by-step explanation:
well 3 to 9 is multiplying by 3
9 to 27 is by multipling by 3
27 to 81 is by multipling by 3
Answer:

Step-by-step explanation:
To find the area of an equilateral triangle, we can apply a formula.


The side length is given a.
Plug a in the formula as the side length.
