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uysha [10]
2 years ago
13

A petri dish of bacteria grow continuously at a rate of 200% each day. If the petri dish began

Mathematics
1 answer:
Molodets [167]2 years ago
4 0

Answer: 2430 bacteria.

Step-by-step explanation:

When we have a quantity A, and we have an increase of the X%, the new quantity can be written as:

A + (X%/100%)*A

In this case, we have A = 10 bacteria.

And X% = 200%.

Then if we start with 10 bacteria.

After one day, we will have:

10 bacteria  + (200%/100%)*10 bacteria

= 10 bacteria + 2*10 bacteria = 3*10 bacteria.

After another day, we will have:

3*10 bacteria + (200%/100%)*(3*10 bacteria)

3*10 bacteria  + 2*(3*10 bacteria)

3*(3*10 bacteria)

10 bacteria*(3)^2

We already can see the pattern here.

After t days, we will have:

10 bacteria*(3)^t.

This is the equation f(t) we wanted:

f(t) =10 bacteria*(3)^t

after 5 days, we will have:

f(5) = 10 bacteria*(3)^5 = 2430 bacteria.

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For every two laps Jane swims Betty swims 3 laps. If they swim a total of 35 laps how many did each girls swim
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Step-by-step explanation:

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Put the differential equation 9ty+ety′=yt2+81 into the form y′+p(t)y=g(t) and find p(t) and g(t). p(t)= help (formulas) g(t)= he
alexgriva [62]

Answer:

p(t) = \frac{9t^{3} + 729t  - 1}{e^{t}(t^{2} + 81) }

g(t) = 0

And

The differential equation 9ty + e^{t}y' = \frac{y}{t^{2} + 81 } is  linear and homogeneous

Step-by-step explanation:

Given that,

The differential equation is -

9ty + e^{t}y' = \frac{y}{t^{2} + 81 }

e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t  - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t  - 1}{e^{t}(t^{2} + 81) } ]y = 0

By comparing with y′+p(t)y=g(t), we get

p(t) = \frac{9t^{3} + 729t  - 1}{e^{t}(t^{2} + 81) }

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And

The differential equation 9ty + e^{t}y' = \frac{y}{t^{2} + 81 } is  linear and homogeneous.

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