31-27+12-23-(-25)+(-6)= 12
The equation that models the movement of the object is:

Where,
t: time
a: acceleration due to gravity
v0: initial speed
h0: initial height
Suppose that the object falls with zero initial velocity and from a height of 38 feet.
The equation that models the problem is:
Answer:
If an object is dropped from a height of 38 feet, the function h (t) = -16t2 + 38 gives the height of the object after seconds
Answer:
The answer is "$238".
Step-by-step explanation:
Current worth
depreciates by
in 3 years.
time= 19 years
depreciates rate=?
Using formula:


calculates depreciate value in 3 year 

so,


depreciates rate= 20%


Answer:
I disagree . I don’t agree because if you have 25 bags with 20 braclets in each then you would just waste So many bags because you can fit 500 bracelets in those bags all together.
Step-by-step explanation:
HOPE THIS HELPS
PLZZ MARK BRAINLIEST
Probability that both roads from a to b are blocked is the product of the individual probabilities, i.e.
P(~ab)=0.25*0.25=0.0625
Similarly
P(~bc)=0.25*0.25=0.0625
Probability that EITHER one or both of ab and bc are blocked is the sum of the probabilities:
P(~ab ∪ ~bc)=0.0625+0.0625=0.125
(recall that one cannot travel from a to c if either ab or bc is blocked.)
Therefore the probability that there exists an open route from a to c
= P(ac) = 1-P(~ab ∪ ~bc)
= 1 - 0.125
=0.875