If log a^3=x and log a^2=y find log 22^2x+1+2^x=10 what is the value of x
1 answer:
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Answer:
45.05 seconds
Step-by-step explanation:
Use the formula: d = v t +
a
d = distance v = initial velocity (m/s)
t = time (s) a = acceleration (m/)
m is meters and s is seconds. They are units of measurement so leave them be.
Assuming the object is simply dropped, the initial velocity is 0 since the object was not moving before it was dropped.
The distance is 725 feet, which is 220.98 meters.
The acceleration is 9.81m/ since that is the acceleration of Earth's gravity, aka free fall.
Time is what we are trying to find so just leave it as the variable t.
So plug the values into the equation:
220.98m = (0)(t) + (9.81m/
)(t)
220.98m = (4.905m/)(t)
45.0519877676s = t
t = 45.05s
Remember to pay attention to units because your answer will be wrong otherwise