The height at time t is
h(t) = 144 - 16t²
When t = 0, then h = 144.
Therefore the height from the ground is 144 when the object is dropped.
When the object reaches the ground, h = 0.
Therefore
144 - 16t² = 0
t² = 144/16 = 9
t = 3 s
Answer:
The object reaches the ground in 3 seconds.
<h3>
Answer: Choice A) 3.5</h3>
Work Shown:
The point (8,28) is the furthest to the right on the graph. We have x = 8 and y = 28 pair up here. Divide the y over the x
y/x = 28/8 = 3.5
side note: the tickmarks on the y axis go up by 4 each time
First we need to find k ( rate of growth)
The formula is
A=p e^kt
A future bacteria 4800
P current bacteria 4000
E constant
K rate of growth?
T time 5 hours
Plug in the formula
4800=4000 e^5k
Solve for k
4800/4000=e^5k
Take the log for both sides
Log (4800/4000)=5k×log (e)
5k=log (4800/4000)÷log (e)
K=(log(4,800÷4,000)÷log(e))÷5
k=0.03646
Now use the formula again to find how bacteria will be present after 15 Hours
A=p e^kt
A ?
P 4000
K 0.03646
E constant
T 15 hours
Plug in the formula
A=4,000×e^(0.03646×15)
A=6,911.55 round your answer to get 6912 bacteria will be present after 15 Hours
Hope it helps!
Answer:
1 1/2 miles
Step-by-step explanation:
if running 1 lap around a track is 1/2 miles then all we have to do is add it up
so we know that 1/2 miles is one lap
so if he runs two laps we would have run 1 mile
now that we know that he would have ran 1 mile in 2 laps then all we have to do it just add 1 more 1/2 mile in which would give u the answer of 1 1/2 miles or 1.5 miles
B: x=4 hope this helped!!
