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3 years ago

Polygon A is similar to Polygon B. Find the perimeter of Polygon B.

Mathematics

1 answer:

marissa [1.9K]3 years ago

3 0

$\bf \textit{ratios relations for similar figures}
\\\\\\
\begin{array}{cccllll}
side&area&volume\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
\frac{s}{s}=\frac{s}{s}&\frac{s^2}{s^2}=\frac{area}{area}&\frac{s^3}{s^3}=\frac{volume}{volume}
\end{array}\\\\
-----------------------------\\\\
$

$\bf thus
\\\\\\
\begin{array}{llll}
\cfrac{\textit{polygon A}}{\textit{polygon B}}\qquad &\cfrac{24^2}{15^2}=\cfrac{128}{x}
\end{array}$

solve for "x"

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Change each of the following points from rectangular coordinates to spherical coordinates and to cylindrical coordinates.

FromTheMoon [43]

Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

$\rho = \sqrt{x^{2}+y^{2}+z^{2}}$

$\phi = cos^{-1}\frac{z}{\rho}$

For angle θ:

If x > 0 and y > 0: $\theta = tan^{-1}\frac{y}{x}$ ;

If x < 0: $\theta = \pi + tan^{-1}\frac{y}{x}$ ;
If x > 0 and y < 0: $\theta = 2\pi + tan^{-1}\frac{y}{x}$ ;

Calculating:

a) (4,2,-4)

$\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}}$ = 6

$\phi = cos^{-1}(\frac{-4}{6})$

$\phi = cos^{-1}(\frac{-2}{3})$

For θ, choose 1st option:

$\theta = tan^{-1}(\frac{2}{4})$

$\theta = tan^{-1}(\frac{1}{2})$

b) (0,8,15)

$\rho = \sqrt{0^{2}+8^{2}+(15)^{2}}$ = 17

$\phi = cos^{-1}(\frac{15}{17})$

$\theta = tan^{-1}\frac{y}{x}$

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = $\frac{\pi}{2}$

c) (√2,1,1)

$\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}}$ = 2

$\phi = cos^{-1}(\frac{1}{2})$

$\phi$ = $\frac{\pi}{3}$

$\theta = tan^{-1}\frac{1}{\sqrt{2} }$

d) (−2√3,−2,3)

$\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}}$ = 5

$\phi = cos^{-1}(\frac{3}{5})$

Since x < 0, use 2nd option:

$\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }$

$\theta = \pi + \frac{\pi}{6}$

$\theta = \frac{7\pi}{6}$

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

$r=\sqrt{x^{2}+y^{2}}$

Angle θ is the same as spherical coordinate;

z = z

Calculating:

a) (4,2,-4)

$r=\sqrt{4^{2}+2^{2}}$ = $\sqrt{20}$

$\theta = tan^{-1}\frac{1}{2}$

z = -4

b) (0, 8, 15)

$r=\sqrt{0^{2}+8^{2}}$ = 8

$\theta = \frac{\pi}{2}$

z = 15

c) (√2,1,1)

$r=\sqrt{(\sqrt{2} )^{2}+1^{2}}$ = $\sqrt{3}$

$\theta = \frac{\pi}{3}$

z = 1

d) (−2√3,−2,3)

$r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}}$ = 4

$\theta = \frac{7\pi}{6}$

z = 3

5 0

3 years ago

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Levart [38]

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3 years ago

Consider the following series. 1 9 1 18 1 27 1 36 1 45 Determine whether the series is convergent or divergent. Justify your ans

Vedmedyk [2.9K]

Answer:

Diverges; the limit of the terms, an, is not 0 as n goes to infinity

Step-by-step explanation:

Convergence of Infinite Series

The series is given as the terms

1 , 9 , 1 , 18 , 27 , 1 , 36 , 1 , 45 ...

For a series to be convergent, the limit of the general term an when n goes to infinity must be 0. In such cases, the sum of all terms tends to a fixed value.

Our series has two clearly different sequences, which define it as a piecewise general term:

If n is odd, then

$a_n=1$

Otherwise, an is an arithmetic series starting from 9 with a common difference of 9, that is

$a_n=9+9(n/2-1)=4.5n$

if n is even.

If we take the limit as n goes to infinity, we don't know which one should be selected, thus we must assume that any of them is a correct option, therefore both of them should converge.

The limit of the function

$a_n=1$

is 1 when n goes to infinity. Since it's not 0, this piece is not convergent and the entire series isn't either.

But it goes worse because of the limit of the second piece

$a_n=4.5n$

when n goes to infinity is infinite.

The answer is

Diverges; the limit of the terms, an, is not 0 as n goes to infinity

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yanalaym [24]

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2 years ago

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s2008m [1.1K]

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