Then I'll definitely will ask you questions about maths :)
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
![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.)
Thus, 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.)
Answer:
a) distance covered by hare d1 = 8t
b) distance covered by tortoise d2 = 5t + 550
c) ∆d = 550 - 3t
Step-by-step explanation:
Given;
Speed of hare u = 8m/s
Speed of tortoise v = 5 m/s
Initial distance of tortoise d0 = 550 m
a) using the equation of motion;
distance covered = speed × time + initial distance
d = vt + d0
For hare;
d0 = 0
Substituting the values;
d1 = 8t + 0
d1 = 8t
b)using the equation of motion;
distance covered = speed × time + initial distance
d2 = vt + d0
For tortoise;
d0 = 550m
Substituting the values;
d2 = 5t + 550
d2 = 5t + 550 m
c) the number of meters the tortoise is ahead of the hare.
∆d = distance covered by tortoise - distance covered by hare
∆d = d2 - d1
Substituting the values;
∆d = (5t + 550) - 8t
∆d = 550 - 3t
we need more info in order to solve this problem!