Answer: The required matrix is
![T=\left[\begin{array}{ccc}-1&3\\2&4\end{array}\right] .](https://tex.z-dn.net/?f=T%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%263%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D%20.)
Step-by-step explanation: We are given to find the transition matrix from the bases B to B' as given below :
B = {(-1,2), (3, 4)) and B' = {(1, 0), (0, 1)}.
Let us consider two real numbers a, b such that
Again, let us consider reals c and d such that
Therefore, the transition matrix is given by
Thus, the required matrix is
Isosceles triangle I think
Answer:
i would go for a and c
Step-by-step explanation:
49/-7 and -21/3 are the only two equivalent to -7
1. 200 Parsecs, 652.312 Light years 2. 8 Parsecs, 26.09248 Light years ; Use this formula, p=1/P(parallax), to solve the equation. Then multiply 3.26156 (light years per parsec) by the number of parsecs.
1. p= 1/0.005 = 200*3.26156 = 652.312
200 Parsecs, 652.312 Light years
2. p=1/0.125 = 8*3.26156 = 26.09248
8 Parsecs, 26.09248 Light years
