Answer:
![A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%26-4%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D)
message SHOW_ME_THE_MONEY_
Step-by-step explanation:
The matrix
![A=\left[\begin{array}{cc}1&4\\-1&-3\end{array}\right]\rightarrow |A|=(1 \times -3)-(-1\times 4)=1\\\rightarrow A^{-1}=\left[\begin{array}{cc}-3&-4\\1&1\end{array}\right] \\](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%264%5C%5C-1%26-3%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%20%7CA%7C%3D%281%20%5Ctimes%20-3%29-%28-1%5Ctimes%204%29%3D1%5C%5C%5Crightarrow%20A%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-3%26-4%5C%5C1%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C)
We can check that in fact A*A^⁻1=I_2 the identity matrix of size 2 x 2.
Now the message was divided in 1 x 2 matrices, then we have that the sequence given is the result of multiplying m by A, so to get m again we multiply now by A^⁻1. and we get the next table
Encoded message Decoded message message in letters by association
11 52 19 8 S H
-8 -9 15 23 O W
-13 -39 0 13 _ M
5 20 5 0 E _
12 56 20 8 T H
5 20 5 0 E _
-2 7 13 15 M O
9 41 14 5 N E
25 100 25 0 Y _
Then the message decoded is SHOW_ME_THE_MONEY_
Answer:
your answer would come out to be 16 square inches.
Answer:
a. f(0) = 4
b. f(6) = 8
Step-by-step explanation:
This is a piecewise function, so the actual equation is on the left, and the inequalities are on the right.
If x = 0, this would fall under the first part of the function because x < 5 is the requirement for the first part. Then, evaluating it: f(0) = 0 + 4 = 4.
If x = 6, this would fall under the second part of the function because 5 
x < 7 is the requirement. 6 falls under this category, so that means the second part of the function is used. Because it is a constant, 8, this evaluates to f(6) = 8.
Answer:
f−1 (x)= x/3 +1 /3
Step-by-step explanation:
The present of 45 Is 45.0 present
