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Iteru [2.4K]
3 years ago
13

Can someone help me complete my mathswatch there is 73 questions and all is about shapes if you are knowledgeable at shapes I ho

pe you can help me
Mathematics
1 answer:
Studentka2010 [4]3 years ago
3 0

Answer:

Make sure you watch all the videos that are related to the topic You are doing.

Step-by-step explanation:

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Read 2 more answers
A bacteria culture grows with a constant relative growth rate. After 2 hours there are 400 bacteria and after 8 hours the count
Andru [333]

Answer:

So, the initial population of the bacteria culture is 80.003 and the expression after t hours is P(t)=80.003 e^(0.8047\times t).

Step-by-step explanation:

Given:

A bacteria culture with constant relative rate.

No. of bacteria = 400        Time (t) = 2 hrs

No.of bacteria = 50,000    Time (t) = 8 hrs

Here the growth is exponential.

Formula to be used.

  • P(t)=P_0e^k^t where P(t) = Number of bacteria after time (t).

                                        P_0 = Initial population

According to the question.

P(2)=P_0e^2^k=400                 ...equation (i)

P(8)=P_0e^8^k =50,000            ...equation (ii)

Dividing equation (ii) with (i)

⇒ \frac{P_0e^8^k}{P_0e^2^k} =\frac{50,000}{400}

⇒ e^6^k=125

⇒ (e^k)^6=125

Now using \ln function both sides.

⇒ \ln (e^k)^6=\ln(125)

⇒ 6\ln(e^k)=\ln (125)  using \ln x^a=a\ln x rule

⇒ \ln e^k=\frac{\ln 125}{6}

And from logarithmic rule \ln(e) = 1

⇒ k=\frac{\n 125}{6}

⇒ k=0.8047

Now plugging this k value in any of the equation we can find the initial population.

a)

Then,

P_0(e^0^.^8^0^4^7^\times ^2) = 400

P_0=80.003

The initial population of the bacteria culture is 80.003

b)

Expression for the population after t hours.

P(t)=80.003\times e^0^.^8^0^4^7 ^(^t^)

So, the initial population of the bacteria culture is 80.003 and the expression after t hours is P(t)=80.003 e^(0.8047\times t).

8 0
3 years ago
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