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Reptile [31]
3 years ago
5

A cylinder has a height of 16 cm and a radius of 5 cm. A cone has a height of 12 cm and a radius of 4 cm. If the cone is placed

inside the cylinder as shown, what is the volume of the air space surrounding the cone inside the cylinder? (Use 3.14 as an approximation of π.)
452.16 cm3

840.54 cm3

1,055.04 cm3

1,456.96 cm3
Mathematics
2 answers:
Vlad [161]3 years ago
8 0
Given:
Cylinder: height = 16 cm ; radius = 5 cm
cone: height = 12 cm ; radius = 4 cm

Volume of cylinder = 3.14 * (5cm)² * 16cm = 1,256 cm³
Volume of cone = 3.14 * (4cm)² * 12cm/3 = 200.96 cm³

Volume of air space = 1256 cm³ - 200.96 cm³ = 1,055.04 cm³ 
shepuryov [24]3 years ago
5 0
Cylinder: height = 16 cm ; radius = 5 cm
cone: height = 12 cm ; radius = 4 cm

Volume of cylinder = 3.14 * (5cm)² * 16cm = 1,256 cm³
Volume of cone = 3.14 * (4cm)² * 12cm/3 = 200.96 cm³

<span>Volume of air space = 1256 cm³ - 200.96 cm³ = </span>1,055.04 cm³ 
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Answer:

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Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

Therefore, the solution of the given differential equation is

                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

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Answer:

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