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
Mrs. sing will pay 6,30$ for 3 and 1/2 pounds of green beans
15 x 100= 1500 centimeters
1. -3/2
2. -1/2
3. 1.3
4. 1 1/2
The speed of slower bus is 118 km/hr and speed of faster bus is 133 km/hr
<h3>What is speed?</h3>
Speed is a scalar quantity that refers to "how fast an object is moving".
Given that, two buses leave a station at the same time and travel in opposite directions, one bus travels 15 km/h faster than the other, the two buses are 1004 kilometres apart after 4 hours,
Speed = distance/ time
Let the speed of the slower bus be v and the faster bus be (v+15)
According to the question,
vt + (v+15)t = 1004
vt + vt + 15t = 1004
t = 4
4v + 4v + 60 = 1004
8v = 944
v = 118
Hence, The speed of slower bus is 118 km/hr and speed of faster bus is 133 km/hr
For more references on speed, click;
brainly.com/question/7359669
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