There are lots of videos on YouTube on how to do it I’m not completely sure so I don’t want to tell you something false
The question here is how long does it take for a falling
person to reach the 90% of this terminal velocity. The computation is:
The terminal velocity vt fulfills v'=0. Therefore vt=g/c,
and so c=g/vt = 10/(100*1000/3600) = 36,000/100,000... /s. Incorporating the
differential equation shows that the time needed to reach velocity v is
t= ln [g / (g-c*v)] / c.
With v=.9 vt =.9 g/c,
t = ln [10] /c = 6.4 sec.
this should help I have it step
What do you need help with?
Answer:
3rd option
Step-by-step explanation:
Since the centre of dilatation is at the origin then multiply the coordinates by 
K (- 3, 9 ) → K' (- 3( ), 9( ) ) → K' (- 1, 3 )
L (- 9, 0 ) → L' (- 9( ), 0 ( ) ) → L' (- 3, 0 )
M (2, - 8 ) → M' (2 ( ), - 8 ( ) ) → M' ( 
, - 
)
N (6, 4 ) → N' (6 ( ) , 4 ( ) ) → N' ( 2, 
)
