Answer:
(f^−1)′(6)=1/(f'(f^-1(6)))
(f^−1)′(6)=1/(f'(f^-6)))
I hope this helps.
The distance covered by the hare and the tortoise in t seconds are 8t and 5t respectively. (Simple Speed-Distance-Time relation)
The tortoise gets a 510m headstart so at t=0 is 1490m.
The functions representing the distance of both of them from the finish line is,
F(x)=2000-8t,. for hare
G(x)=1490-5t,. for tortoise
Answer: 60cm
Step-by-step explanation:
6 times 2 is 12
12 times 5 is 60
