dP/dt = 1000 * 9^t is the bacteria population's growth rate.where:P(t) is the population at time t, andP(0) = 6000.
The population after one hour would be:
P(1) = P(0) + <Integral of dP/dt from 0 to 1>
P(1) = P(0) + [ 1000 * 9^1 / ln(9) - 1000 * 9^0 / ln(9) ]
P(1) = 6000 + 3641 = 9461 would be the answer
The answer is the option C, which is: C. √29
The explanation of this exercise is:
To solve this problem you must apply the proccedure shown below:
1- You have that:
W (-2,8,-3) and X(1,4,-1)
2- Therefore, you can calculate the magnitude as below:
|WX|=√[(1-(-2)]²+(4-8)²+[(-1-(-3)]²)
|WX|=√29
8 - (6-4) : 2 = 7
that's it
Answer:
The population size is 55.9597 (≅ 56)
Step-by-step explanation:
Since the population of protozoa develops at a constant growth rate of 0.4416 per member per day.
Then;
Day 0 = 3
Day 1 = 3 + (3 x 0.4416) = 4.3248
Day 2 = 4.3248 + (4.3248 x 0.4416) = 6.2346
Day 3 = 6.2346 + (6.2346 x 0.4416) = 8.9878
Day 4 = 8.9878 + (8.9878 x 0.4416) = 12.9568
Day 5 = 12.9568 + (12.9568 x 0.4416) = 18.6785
Day 6 = 18.6785 + (18.6785 x 0.4416) = 26.9269
Day 7 = 26.9269 + (26.9269 x 0.4416) = 38.8178
Day 8 = 38.8178 + (38.8178 x 0.4416) = 55.9597
After eight days, the population size would be 55.9597 (≅ 56).
Answer:
2/4
Step-by-step explanation:
It’s very easy