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ANEK [815]
3 years ago
9

Can someone please help me please

Mathematics
1 answer:
Assoli18 [71]3 years ago
6 0
A
Volume of the Cylinder
Givens
H = 60 yards.
Diameter = 20 yards
pi = 3.14

Formula
V = pi * r^2 * h

Calculations
r = d/2
r = 32/2
r = 16

V =  3.14 * 16^2 * 60 
V = 48230 cubic yards [Cylinder's Volume]

Cone
<em>Formula</em> 
V = 1/3 pi r^2 H

<em>Givens</em>
pi = 3.14
r = 16 yards
h = 20 yards

<em>Sub and solve</em>
V = 1/3 3.14 * 16^2 * 20
V = 5359 cubic yards.

<em>Total Volume of the structure</em>
48230 + 5359 = 53589 Cubic Yards

<em>Water Content</em>
The answer to this part requires a proportion.

1 Cubic yard will hold 201.97 gallons.
53589 yd^3 = x

1/201.97 = 53589 /x  [ You should get a pretty big answer]
x = 201.87 * 53589
x = 10 819 092 gallons can be held by the tank. 

10 819 092 gallons <<<< answer

B 
If the height of both the cylinder and the cone remain the same. If the radius doubles in both the cylinder and the cone then the tank will hold 4 times as much. 

Total volume before doubling the radius = pi * r^2 h + 1/3 pi r^2 h
New Total Volume = pi * (2*r)^2 h + 1/3 pi * (2r)^2 h
New Total volume = pi * 4r^2 h + 1/3 pi *4 r^2 h
New Total Volume = 4 [pi r^2 h + 1/3 pi r^2 h]
but pi r^2 h + 1/3 pi r^2 h is the total volume before doubling the radius

New volume = 4 original volume. <<<<< answer to part B
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The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,
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Answer:

Step-by-step explanation:

Assuming that the differential equation is

\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right).

We need to solve it and obtain an expression for P(t) in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right).

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that k=0.04 and K=500 and the initial condition P(0)=100.

Notice that this equation is separable, then

\frac{dP}{P(1-P/K)} = kdt.

Now, intagrating in both sides of the equation

\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C.

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}.

So,

\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|.

We have obtained that:

-\ln\left| \frac{K-P}{P}\right| = kt +C

which is equivalent to

\ln\left| \frac{K-P}{P}\right|= -kt -C

Taking exponentials in both hands:

\left| \frac{K-P}{P}\right| = e^{-kt -C}

Hence,

\frac{K-P(t)}{P(t)} = Ae^{-kt}.

The next step is to substitute the given values in the statement of the problem:

\frac{500-P(t)}{P(t)} = Ae^{-0.04t}.

We calculate the value of A using the initial condition P(0)=100, substituting t=0:

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So,

\frac{500-P(t)}{P(t)} = 4e^{-0.04t}.

Finally, as we want the value of t such that P(t)=200, we substitute this last value into the above equation. Thus,

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This is equivalent to \frac{3}{8} = e^{-0.04t}. Taking logarithms we get \ln\frac{3}{8} = -0.04t. Then,

t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325.

So, the population of rats will be 200 after 25 months.

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