The distance is 3.16 (label whatever is appropriate)
Answer: D , B, A, C
Step-by-step explanation:
Answer:
D
Step-by-step explanation:
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>
Answer:
-4x=4+4
x=8/-4
x= -2
not sure how I'm supposed to verify
