Answer:
![\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}3&1\\4&-2\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%261%5C%5C4%26-2%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3x%2B4y%26x-2y%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The general matrix representation for this transformation would be:
![\left[\begin{array}{cc}x&y\end{array}\right] * A = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%2A%20A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3x%2B4y%26x-2y%5Cend%7Barray%7D%5Cright%5D)
As the matrix A should have the same amount of rows as columns in the firs matrix and the same amount of columns as the result matrix it should be a 2x2 matrix.
![\left[\begin{array}{cc}x&y\end{array}\right] * \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}3x+4y&x-2y\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26y%5Cend%7Barray%7D%5Cright%5D%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3x%2B4y%26x-2y%5Cend%7Barray%7D%5Cright%5D)
Solving the matrix product you have that the members of the result matrix are:
3x+4y = a*x + c*y
x - 2y = b*x + d*y
So the matrix A should be:
![\left[\begin{array}{cc}3&1\\4&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%261%5C%5C4%26-2%5Cend%7Barray%7D%5Cright%5D)
Step One
======
Find the length of FO (see below)
All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)
Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ
Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.
FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)] \
OJ = ??
[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2 Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2 Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2 Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )
Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.
I believe the answer is 77.
Explanation
539 divided by 7 is 77.
Hopefully this helps.
Answer:
i need ponits :)
Step-by-step explanation:
Answer:
I believe it is A pls tell if I am wrong