Awww man the other person answered before me :(
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.
Answer:
32/99
Step-by-step explanation:
Here's a formula you could use,
(DN x F) - NRP over D
DN = Decimal Number
F = 10 if one repeating number, 100 if two repeating numbers, 1000 if three repeating numbers, etc.
NRP = Non-repeating part of decimal number.
D = 9 if one repeating number, 99 if two repeating numbers, 999 if three repeating numbers, etc.
0.32 repeating as a fraction
(0.32 x 100) - 0/99 = 32/99
Simplest form:
32/99
Its (12x8)/2=48
the base doesn't matter in an right triangle. just use length x width and because its 3 instead of 4 angles its half
