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2 years ago

If the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week

Mathematics

1 answer:

serg [7]2 years ago

5 0

The population after 20 weeks will be 403.42 $x_{0}$ in which $x_{0}$ is the initial population.

Given that the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week.

We are required to find the number of bacteria present after 10 weeks.

let the number of bacteria present at t is x.

So,

dx/dt∝x

dx/dt=kx

1/x dx=k dt

Now integrate both sides.

$\int\limits {1/x} \, dx$ = $\int\limits{k} \, dt$

log x=kt+log c----------1

Put t=0

log $x_{0}$ =0 +log c ( $x_{0}$ shows the population in beginning)

Cancelling log from both sides.

c= $x_{0}$

So put c= $x_{0}$ in 1

log x=kt+log $x_{0}$

log x=log $e^{kt}$ +log $x_{0}$

log x=log $e^{kt}x_{0}$

x= $e^{kt}x_{0}$

We have been given that the population triples in a week so we have to put the value of x=2 $x_{0}$ and t=1 to get the value of k.

2 $x_{0}$ = $e^{k} x_{0}$

2= $e^{k}$

log 2=k

We have to now put the value of t=20 and k=log 2 ,to get the population after 20 weeks.

x= $e^{20log 2}$ $x_{0}$

x= $e^{0.30*2}$ $x_{0}$

x= $e^{6}x_{0}$

x=403.42 $x_{0}$

Hence the population after 20 weeks will be 403.42 $x_{0}$ in which $x_{0}$ is the initial population.

Learn more about growth rate at brainly.com/question/25849702

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The given question is incomplete as the question incudes the following:

Calculate the population after 20 weeks.

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2 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

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The initial activity to carry out is to determine the limits of the region

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∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

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∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

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$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

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$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

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3 years ago

Edith’s age is 1/4 of Ronald’s age. In how many years will Edith’s age be 1/3 of Ronald’s age if Edith is 20 years old?

ruslelena [56]

Answer:

Step-by-step explanation:

x = Edith's age = 20 years

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y = 20×4 = 80 years

in z years we have

(x + z) = (y + z)/3

(20 + z) = (80 + z)/3

3(20 + z) = 80 + z

60 + 3z = 80 + z

60 + 2z = 80

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z = 10

in 10 years Edith's age will be 1/3 of Ronald's age.

then Edith will be 30, and Ronald will be 90.

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correct.

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1 year ago

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Answer:

Step-by-step explanation:

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