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Ivahew [28]
3 years ago
9

The bacteria has an initial population of 2000 cells and increases at a rate of 40% per hour. Use the exponential function to mo

del the growth/decay of bacteria.
Mathematics
2 answers:
Fynjy0 [20]3 years ago
5 0
2000% x .4 will give you your answer! 
BaLLatris [955]3 years ago
4 0

Here given, initial population of bacteria = 2000 cells

Also given, the population increases at a rate of 40% per hour.

As the population is increasing so we will use the formula for exponential growth. The formula is A = Pe^{(rt)}

Where, P = initial amount, r = growth rate, t = time, A = final amount.

So here, P = 2000, r = 40% = \frac{40}{100} = 0.4, t = time in hours

We will plug in the values in the formula. we will get,

A = 2000e^{(0.4t)}

We have got the required function to model the growth of the bacteria.

The required function is A = 2000e^{(0.4t)}

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If g(x) = |2x-9|, find g(2)+g(3)
Semenov [28]
X=5 , x=3 - just plug in 2 and 3 into x and the lines will make it positive instead of negative
5 0
2 years ago
Help me please and thankyou
klemol [59]

Answer:

\boxed{Option B}

Step-by-step explanation:

=> (-5*4)^3

Using the distributive property

=> (-5)^3*(4)^3

5 0
3 years ago
Read 2 more answers
Write an expression equivalent to h^6/h^3 in the form of h^m
Pavel [41]

Answer:

h^3 = h^m

Step-by-step explanation:

\frac{h^6}{h^3} = h^m

6 - 3 = 3

keep the base same, therefore

h^3 = h^m

Hope this helps!

3 0
3 years ago
Can someone help me please
navik [9.2K]

Answer:

the \: answer \: is \: choice \: b \:  \frac{ - 2 \sin(x) }{1 +  \cos(x) }

will give you an explanation if you want tag me on comment.

4 0
1 year ago
The half-life of caffeine in a healthy adult is 4.8 hours. Jeremiah drinks 18 ounces of caffeinated
statuscvo [17]

We want to see how long will take a healthy adult to reduce the caffeine in his body to a 60%. We will find that the answer is 3.55 hours.

We know that the half-life of caffeine is 4.8 hours, this means that for a given initial quantity of coffee A, after 4.8 hours that quantity reduces to A/2.

So we can define the proportion of coffee that Jeremiah has in his body as:

P(t) = 1*e^{k*t}

Such that:

P(4.8 h) = 0.5 = 1*e^{k*4.8}

Then, if we apply the natural logarithm we get:

Ln(0.5) = Ln(e^{k*4.8})

Ln(0.5) = k*4.8

Ln(0.5)/4.8 = k = -0.144

Then the equation is:

P(t) = 1*e^{-0.144*t}

Now we want to find the time such that the caffeine in his body is the 60% of what he drank that morning, then we must solve:

P(t) = 0.6 =  1*e^{-0.144*t}

Again, we use the natural logarithm:

Ln(0.6) = Ln(e^{-0.144*t})

Ln(0.6) = -0.144*t

Ln(0.6)/-0.144 = t = 3.55

So after 3.55 hours only the 60% of the coffee that he drank that morning will still be in his body.

If you want to learn more, you can read:

brainly.com/question/19599469

7 0
2 years ago
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