Answer:
Therefore,

Step-by-step explanation:
Given:
![A=\left[\begin{array}{ccc}3&6&9\\2&4&8\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%266%269%5C%5C2%264%268%5C%5C%5Cend%7Barray%7D%5Cright%5D)
To Find:
a₂₁ = ?
Solution:
Let,
![A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_%7B11%7D%26a_%7B12%7D%26a_%7B13%7D%5C%5Ca_%7B21%7D%26a_%7B22%7D%26a_%7B23%7D%5C%5C%5Cend%7Barray%7D%5Cright%5D)
We require ' a₂₁ ' i.e Second Row First Column Element
So on Comparing we get
∴ 
Therefore,

Answer:
I believe it is b
Step-by-step explanation:
(y-14)=8/2
y-14=4
y=4+14
y=18
Answer:
-d-20
Step-by-step explanation:
Hope this helps!
Answer:
Step-by-step explanation: