Answer:
Therefore, the inverse of given matrix is

Step-by-step explanation:
The inverse of a square matrix
is
such that
where I is the identity matrix.
Consider, ![A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%263%5C%5C3%266%5Cend%7Barray%7D%5Cright%5D)








Therefore, the inverse of given matrix is

The formula for the perimeter of a rectangle is P = 2L + 2W, where L is the length and W is the width. Because we don't know either the length or the width we can't solve the problem...too many unknowns. BUT we do have some information that will help with this problem. We are told that the length is 2 feet longer than the width, so we can use that: L = W+2. Now we can make the substitution into the formula along with the value for the perimeter that was given to us: 36=2(W+2) + 2W, and 36 = 2W + 4 + 2W; 36 = 4W + 4; 32 = 4W and W = 8. Now go back to where you said that the length is 2 feet longer than the width. If the width is 8, then 8+2 = 10 for the length.
Well first you would find the area of the square on the bottom
Which would be:
0.7*0.7= .49
Then we would find the area of one of the triangles on the side:
8*0.7= 5.6* 1/2= 2.8
Next, you will multiply the 2.8 by 4, because there are 4 triangles:
2.8*4=11.2
Finally add the 11.2 to the area of the square, and you'll have your answer.
.49+11.2= 11.69
So, B is your answer
Answer: Length = 24; width =4
Step-by-step explanation
Since the ratio of the length to width is 6:1
Let the length be represented as 6x
And the width be = x
Such that that the Perimeter of the rectangle which is
Perimeter = 2(length + width) becomes
56 = 2(6x + x)
56/2 = 6x+x
28 = 7x
x = 4
Width = 4
Length = 6x= 6 x 4 = 24
Go to the right 0.2 then go up 0.02.