Here's the general formula for bacteria growth/decay problems
Af = Ai (e^kt)
where:
Af = Final amount
Ai = Initial amount
k = growth rate constant
<span>t = time
But there's another formula for a doubling problem.
</span>kt = ln(2)
So, Colby (1)
k1A = ln(2) / t
k1A = ln(2) / 2 = 0.34657 per hour.
So, Jaquan (2)
k2A = ln(2) / t
<span>k2A = ln(2) /3 = 0.23105 per hour.
</span>
We need to use the rate of Colby and Jaquan in order to get the final amount in 1 day or 24 hours.
Af1 = 50(e^0.34657(24))
Af1 = 204,800
Af2 = 204,800 = Ai2(e^0.23105(24))
<span>Af2 = 800</span>
4(3)-6(-2)
12- -12
12+12
24
if you traveled at 40 mph you would go 240 miles in 6 hours. your answer is 40 mph
Hi!
I think that there is a mistake in your question: only the first student says 5, not each student.
From the second student on, the student will say the previous number +4-1, that is the previous number +3.
So the second student will say 8, third 11, etc...
this will be repeated 9 times when Victor says his number (the first student doesn't have to change anything), so the number will increase 9 times: so it will increase 9*3=27
So Victor will say 32: 27 plus the original 5.
And all the numbers said will be:
5
8
11
14
17
20
23
26
29
32 - Victor