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3 years ago

Please help quickly!

1 answer:

7 0

Find the volume

volume=hpir^2
d/2=r=16/2=8

1.
hmm, ok, after 1/2 hour or 30 mins
30/5=6
6*1=6, 6 cubic meters
volume=6, solve for height
v=hpir^2
but r=8 all the time, solve for h
6=hpi8^2
6=hpi64
6/(64pi)=h
the depth is $\frac{3}{32 \pi}$ meters or about 0.02984155182973037545666570563235 meters
the depth is about 0.03 meters

2. the full volume is
v=6pi8^2=6pi64=384pi
1 cubic meter every 5 mins so
384pi/1 is how many 5 mininute intervals
so 384pi times 5=total minutes=1920pi minutes or about 6031.8578948924030178482752958966minutes

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4 years ago

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After driving for 20 minutes, Juan was 62 miles away from his apartment.

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The independent and dependent variables.

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If the value of a variable depends on the another, then it is called dependent variables.

If the value of a variable does not depend on the another, then it is called independent variables.

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3 years ago

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