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4 years ago

What is 40 ÷299 I will mark u as brainlest

Mathematics

2 answers:

laila [671]4 years ago

4 0

Answer:

0.1338 to the nearest ten thousandth.

Step-by-step explanation:

40 / 299

= 0.1337792642

Alexandra [31]4 years ago

4 0

Answer:

0.133779264214047

Step-by-step explanation:

This is really something that needs to be done by a calculator, but if you want to do it the long way:

$.........00.13\\299\sqrt{40.000000} \\........299\\........101.0$

and continue on from there.

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An artist is creating a large butterfly sculpture outside a museum. There is a circular dot on each wing made out of a metal rin

IgorLugansk [536]

$\bf \textit{"The distance around each dot is 24}\pi \textit{ inches."}\\\\
\textit{circumference of a circle}\\\\
C=2\pi r\quad 
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r=radius\\
-----\\
C=24\pi 
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\\\\\\
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\textit{area of a circle}\\\\
A=\pi r^2\quad 
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4 years ago

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A scientist begins an experiment with a culture of 5 bacteria cells. Every hour, the number of bacteria cells triples. Which equ

olga2289 [7]

Answer:

y = 5 * (3^x)

Step-by-step explanation:A scientist begins an experiment with a culture of 5 bacteria cells.

Every hour, the number of bacteria cells triples.

Which equation can be used to represent the number of bacteria cells, y, as it relates to time in hours, x?

I think it would be y = 5 * (3^x)

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3 years ago

Which of the following set can be sets of a right triangle

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I need to solve for y

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3 years ago

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One model for the spread of a virusis that the rate of spread is proportional to the product of the fraction of the population P

Darya [45]

Answer:

The differential equation for the model is

$\frac{dP}{dt}=kP(1-P)$

The model for P is

$P(t)=\frac{1}{1-0.99e^{t/447}}$

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

Step-by-step explanation:

We can write the rate of spread of the virus as:

$\frac{dP}{dt}=kP(1-P)$

We know that P(0)=100 and P(3)=100+200=300.

We have to calculate t so that P(t)=0.9*100,000=90,000.

Solving the diferential equation

$\frac{dP}{dt}=kP(1-P)\\\\ \int \frac{dP}{P-P^2} =k\int dt\\\\-ln(1-\frac{1}{P})+C_1=kt\\\\1-\frac{1}{P}=Ce^{-kt}\\\\\frac{1}{P}=1-Ce^{-kt}\\\\P=\frac{1}{1-Ce^{-kt}}$

$P(0)= \frac{1}{1-Ce^{-kt}}=\frac{1}{1-C}=100\\\\1-C=0.01\\\\C=0.99\\\\\\P(3)= \frac{1}{1-0.99e^{-3k}}=300\\\\1-0.99e^{-3k}=\frac{1}{300}=0.99e^{-3k}=1-1/300=0.997\\\\e^{-3k}=0.997/0.99=1.007\\\\-3k=ln(1.007)=0.007\\\\k=-0.007/3=-0.00224=-1/447$

Then the model for the population infected at time t is:

$P(t)=\frac{1}{1-0.99e^{t/447}}$

Now, we can calculate t for P(t)=90,000

$P(t)=\frac{1}{1-0.99e^{t/447}}=90,000\\\\1-0.99e^{t/447}=1/90,000 \\\\0.99e^{t/447}=1-1/90,000=0.999988889\\\\e^{t/447}=1.010089787\\\\ t/447=ln(1.010089787)\\\\t=447ln(1.010089787)=447*0.010039225=4.487533$

At half day of the 4th day (t=4.488), the population infected reaches 90,000.

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4 years ago