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lidiya [134]
3 years ago
14

Central Park is a rectangular park in New York City. A map of Central Park in New York City. The measured length is 12.5 centime

ters. The measured width is 2.5 centimeters. Find the perimeter and the area of the scale drawing of Central Park. Round your measurements for the length and the width to the nearest half centimeter to calculate your answers.
Mathematics
1 answer:
Softa [21]3 years ago
5 0

The perimeter of scale drawing of Central Park is 30 centimeter

The area of scale drawing of Central Park is 31.25 square centimeters

<h3><u>Solution:</u></h3>

Given that Central Park is a rectangular park in New York City

The measured length is 12.5 centimeters

The measured width is 2.5 centimeters

<em><u>The perimeter of scale drawing of Central Park is given as:</u></em>

Given is a rectangular park

perimeter of rectangular = 2(length + width)

perimeter of rectangular = 2(12.5 + 2.5)

perimeter of rectangular = 2(15) = 30

Thus perimeter is 30 centimeter

<em><u>The area of scale drawing of Central Park is given as:</u></em>

area = length x width

area = 12.5 \times 2.5 = 31.25

Thus area of rectangular park is 31.25 square centimeters

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Find the mass of the solid paraboloid Dequals=​{(r,thetaθ​,z): 0less than or equals≤zless than or equals≤8181minus−r2​, 0less th
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Answer:

M = 5742π  

Step-by-step explanation:

Given:-

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                             ρ ( r, θ , z ) = 1 + z / 81

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Find the mass of the solid paraboloid

Solution:-

- The mass (M) of any solid body is given by the following triple integral formulation:

                           M = \int \int \int {p ( r ,theta, z)} \, dV\\\\

- We can write the above expression in cylindrical coordinates:

                           M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\

- Perform integration:

                           M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\

                           M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] }  \, dtheta\\\\M = 5742\pi  kg              

- The mass evaluated is M = 5742π                      

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2 years ago
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Answer:

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Step-by-step explanation:

From the question we are told that:

Dimension of ABC

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Dimension of FED

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Generally the equation for the similar triangles is mathematically given by

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