Hello there!
To do this, we will take the square root from each side
4^2 would be 4
16 would be 4
4 = 4
I hope I helped!
~ Fire
I'm attaching the solution.. feel free to ask if you have questions.. I basically did long division. Hope this helps.
Alright I got you I’ll sub
So,first step is to write ![(fog)(-4)) =f[g(-4)] \\](https://tex.z-dn.net/?f=%20%28fog%29%28-4%29%29%20%3Df%5Bg%28-4%29%5D%20%5C%5C%20)
Now we start from inner paranthesis
,we need to first find value of
Answer:
![\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
If you have two matrices:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}a+e&b+f\\c+g&d+h\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%2Be%26b%2Bf%5C%5Cc%2Bg%26d%2Bh%5Cend%7Barray%7D%5Cright%5D)
We have:
![A=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]\\and\\B=\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D)
And we need to express as a single matrix:
![A+B=\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6+(-2)&-3+8\\10+3&-1+(-12)\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}6-2&5\\13&-1-12\end{array}\right]\\\\\\A+B=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=A%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%2B%28-2%29%26-3%2B8%5C%5C10%2B3%26-1%2B%28-12%29%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6-2%265%5C%5C13%26-1-12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA%2BB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
The answer is:
![\left[\begin{array}{cc}6&-3\\10&-1\end{array}\right]+\left[\begin{array}{cc}-2&8\\3&-12\end{array}\right]=\left[\begin{array}{cc}4&5\\13&-13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%26-3%5C%5C10%26-1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-2%268%5C%5C3%26-12%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%265%5C%5C13%26-13%5Cend%7Barray%7D%5Cright%5D)
It is expressed as a single matrix.