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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:
the theorum used here is sas one
Answer:
35/8 if its simple multiple then
Step-by-step explanation:
7/4*5/2
35/8
Tanα=h/x
h=xtanα, we are told that x=6.5ft and α=74° so
h=6.5tan74 ft
h≈22.67 ft (to nearest hundredth of a foot)
