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sasho [114]
3 years ago
9

(c). Under a set of controlled laboratory conditions, the size of the population P of a certain bacteria culture at time t (in s

econds) is given by the function P(t)= 3t² +3e' +10, t≥0.
(1) What is the size of the population after 1 minute?

(ii) Find the average rate of change of P at t = 2 and t = 6?

(iii) How fast is the size of the population changing after 1 minute?

[Verify your answer by MATHEMATICA and attach the printout of the commands and output

Mathematics
1 answer:
Bezzdna [24]3 years ago
5 0

(i) Since P(t) gives the population of the culture after t seconds, the population after 1 second is

P(1) = 3•1² + 3e¹ + 10 = 13 + 3e ≈ 21.155

In Mathematica, it's convenient to define a function:

P[t_] := 3t^2 + 3E^t + 10

(E is case-sensitive and must be capitalized. Alternatively, you could use Exp[t]. You can also specify that the argument t must be non-negative by entering a condition via P[t_ ;/ t >= 0], but that's not necessary.)

Then just evaluate P[1], or N[P[1]] or N <at symbol> P[1] or P[1] // N to get a numerical result.

(ii) The average rate of change of P(t) over an interval [a, b} is

(P(b) - P(a))/(b - a)

Then the ARoC between t = 2 and t = 6 is

(P(6) - P(2))/(6 - 2) ≈ 321.030

In M,

(P[6] - P[2])/(6 - 2)

and you can also include N just as before.

(iii) You want the instantaneous rate of change of P when t = 60 (since 1 minute = 60 seconds). Differentiate P :

P'(t) = 6t + 3e^t

Evaluate the derivative at t = 60 :

P'(60) = 6•60 + 3e⁶⁰ = 360 + 3e⁶⁰

The approximate value is quite large, so I'll just leave its exact value.

In M, the quickest way would be P'[60], or you can differentiate and replace (via ReplaceAll or /.) t with 60 as in D[P[t], t] /. t -> 60.

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Walter is 43 years old his daughter Paulette is 7. In how many years will Paulette be one-third of Walters age
Ksju [112]
The answer is 14 when rounded to the nearest whole number, to get this you simply must divide 43 by 3 which equals to 14.333. 

8 0
3 years ago
If varies inversely as . It is known that = 10 when = 3, find the value of when = 5.
Leto [7]

Answer:

y = 6

Step-by-step explanation:

If x varies inversely proportional as y.

x=\dfrac{k}{y}, k is constant of proportionality

or

k = xy

When x = 3 and y = 10

k = 3×10

k = 30

Put x = 5,

y=\dfrac{k}{x}\\\\y=\dfrac{30}{5}\\\\y=6

So, the value of y is 6 when x is 5.

8 0
3 years ago
Yall someone help I still have a bunch if other test and I'm so tired ;-;
g100num [7]

Me das mas información?

6 0
3 years ago
Rewrite the expression g+1/6d-3/4g+1/9g-g+1/4d in standard from by combining like terms I need this ASAP AND I NEED ALL STEPS TO
GarryVolchara [31]

Answer:

(5/12)d - (23/36)g

Step-by-step explanation:

First you can eliminate g and -g to get (1/6)d - (3/4)g + (1/9)g + (1/4)d. Then you need to get common denominators to add like terms together.

1/6 = 4/24 and 1/4 = 6/24. Add them together to get (10/24)d or (5/12)d.

-3/4 = -27/36 and 1/9 = 4/36. Add them together to get (-23/36)g.

So in standard form, (5/12)d - (23/36)g

5 0
3 years ago
Maricella solves for x in the equation 4x-2(3x-4)+4=-x+3(x+1)+1. She begins by adding –4 + 4 on the left side of the equation an
Elodia [21]

Maricella’s strategy is incorrect as the values added on the left and right side of the equation changes the original equation 4x-2(3x-4)+4=-x+3(x+1)+1  to  4x-2(3x-4)+4=-x+3(x+1)+3.

Further explanation:

Maricella begin by adding -4+4 on the left side of the equation and 1+1 on the right side of the equation 4x-2(3x-4)+4=-x+3(x+1)+1.

The value added on the left is equal to zero so it does not change the equation but the value added on the right is 2, a non-zero digit so it changes the equation as shown below.

4x-2(3x-4)+4-4+4=-x+3(x+1)+1+1+1\\4x-2(3x-4)+4+0=-x+3(x+1)+1+2\\4x-2(3x-4)+4=-x+3(x+1)+3

Therefore, the strategy used by Maricella will change the equation and so the solution of the equation will be incorrect.

One way of solving such a linear equation is by adding or subtracting the required value on each side of the equation. and then simplifying the further equation.

Similarly, the other way to solve any linear equation is by separating the variable terms on one side of the equation and constant terms on the other side of the equation.

After this simplify the linear equation to obtain the solution of the equation for x.

Whereas Maricella’s strategy fails to solve the equation 4x-2(3x-4)+4=-x+3(x+1)+1 since the values added and subtracted in the equation changes the original equation completely.

Learn more:

1. Linear equation application brainly.com/question/2479097

2. To solve one variable linear equation brainly.com/question/1682776

3. Binomial and trinomial expression brainly.com/question/1394854

Answer details

Grade: Middle school

Subject: Mathematics

Chapter: Linear equation

Keywords: equation, linear equation, Maricella, right side, left side, strategy, solve for x, adding, subtracting, variable, constant, solution of the equation, value, original equation, variable terms, constant terms.

4 0
3 years ago
Read 2 more answers
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