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ludmilkaskok [199]
2 years ago
9

Find the Simple Intrest on #60,000.00 for 3 years at 5% per annum

Mathematics
1 answer:
Alekssandra [29.7K]2 years ago
4 0

Answer:

the ans is 9000

Step-by-step explanation:

simple interest=<em> </em><em>6</em><em>0</em><em>0</em><em>0</em><em>0</em><em>*</em><em>3</em><em>*</em><em>5</em><em>/</em><em>1</em><em>0</em><em>0</em>

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If the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week
serg [7]

The population after 20 weeks will be 403.42x_{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=2x_{0} and t=1 to get the value of k.

2x_{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=403.42x_{0}

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

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

#SPJ4

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

Calculate the population after 20 weeks.

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