8-9=-1 because it does not stop it goes to the negative number -1
Step-by-step explanation:
ratios of lengths = 4/3
ratios of area = 16/9
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Answer:
c=8
Step-by-step explanation:
Simplifying
3c + -15 = 17 + -1c
Reorder the terms:
-15 + 3c = 17 + -1c
Solving
-15 + 3c = 17 + -1c
Solving for variable 'c'.
Move all terms containing c to the left, all other terms to the right.
Add 'c' to each side of the equation.
-15 + 3c + c = 17 + -1c + c
Combine like terms: 3c + c = 4c
-15 + 4c = 17 + -1c + c
Combine like terms: -1c + c = 0
-15 + 4c = 17 + 0
-15 + 4c = 17
Add '15' to each side of the equation.
-15 + 15 + 4c = 17 + 15
Combine like terms: -15 + 15 = 0
0 + 4c = 17 + 15
4c = 17 + 15
Combine like terms: 17 + 15 = 32
4c = 32
Divide each side by '4'.
c = 8
Simplifying
c = 8
Answer:
Verified
Step-by-step explanation:
Let the 2x2 matrix A be in the form of:
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
Where det(A) = ad - bc # 0 so A is nonsingular:
Then the transposed version of A is
![A^T = \left[\begin{array}{cc}a&c\\b&d\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26c%5C%5Cb%26d%5Cend%7Barray%7D%5Cright%5D)
Then the inverted version of transposed A is
![(A^T)^{-1} = \frac{1}{ad - cb} \left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]](https://tex.z-dn.net/?f=%28A%5ET%29%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20cb%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-c%5C%5C-b%26d%5Cend%7Barray%7D%5Cright%5D)
The inverted version of A is:
![A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-b\\-c&d\end{array}\right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-b%5C%5C-c%26d%5Cend%7Barray%7D%5Cright%5D)
The transposed version of inverted A is:
![(A^{-1})^T = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]](https://tex.z-dn.net/?f=%28A%5E%7B-1%7D%29%5ET%20%3D%20%5Cfrac%7B1%7D%7Bad%20-%20bc%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26-c%5C%5C-b%26d%5Cend%7Barray%7D%5Cright%5D)
We can see that

So this theorem is true for 2 x 2 matrices