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Anon25 [30]
3 years ago
10

Find the perimeter of the pentagon with vertices M(-2,5), N(2,5), P (5, 3),

Mathematics
1 answer:
wlad13 [49]3 years ago
8 0

The perimeter of the pentagon is approximately 19.6.

<h3>Procedure - Determination of the perimeter of a pentagon</h3><h3 />

In this question we must plot the locations of each vertex on a Cartesian plane to determine the line segments that form the perimeter, whose lengths are determined by Pythagorean theorem and sum the resulting values to find the perimeter.

According to the image attached below, the line segments of the pentagon are MN, NP, PQ, QR and RM. By Pythagorean theorem we have the following lengths:

<h3>Line segment MN</h3><h3 /><h3>l_{MN} = \sqrt{[2-(-2)]^{2}+(5-5)^{2}} </h3><h3>l_{MN} = 4</h3><h3 /><h3>Line segment NP</h3><h3 /><h3>l_{NP} = \sqrt{(5-2)^{2}+(3-5)^{2}}</h3><h3>l_{NP} = \sqrt{13}</h3><h3 /><h3>Line segment PQ</h3><h3 /><h3>l_{PQ} = \sqrt{(5-5)^{2}+(0-3)^{2}}</h3><h3>l_{PQ} = 3</h3><h3 /><h3>Line segment QR</h3><h3 /><h3>l_{QR} = \sqrt{(3-5)^{2}+(3-0)^{2}}</h3><h3>l_{QR} = \sqrt{13}</h3><h3 /><h3>Line segment RM</h3><h3 />

l_{RM} = \sqrt{(-2-3)^{2}+(5-3)^{2}}

l_{RM} = \sqrt{29}

And the perimeter of the pentagon is:

p = l_{MN} + l_{NP} + l_{PQ} + l_{QR} + l_{RM} (1)

p = 4 + \sqrt{13} + 3 + \sqrt{13} + \sqrt{29}

p = 7 + 2\sqrt{13} + \sqrt{29}

p \approx 19.6

The perimeter of the pentagon is approximately 19.6. \blacksquare

To learn more on pentagons, we kindly invite to check this verified question: brainly.com/question/27476

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marissa [1.9K]

Answer: 0.02

Step-by-step explanation:

Given: Sample size : n= 525

The population proportion P=0.3

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The formula to calculate the standard error is given by :-

S.E.\sqrt{\dfrac{PQ}{n}}

\Rightarrow\ S.E.=\sqrt{\dfrac{0.3\times0.7}{525}}=0.02

Hence, the standard deviation of the sample proportions (i.e., the standard error of the proportion) is <u>0.02</u>.

7 0
3 years ago
Barney has 16 1/5 yards of fabric. To make an elf costume, he needs 5 2/5 yards of fabric. How many costumes can Barney make?
9966 [12]
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3 years ago
What is the distance between "-8" and 14 on a number line
vesna_86 [32]

Answer:

22

Step-by-step explanation:

First let's find the distance from -8 to 0= "8".

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5 0
3 years ago
Find the absolute maximum and minimum values of f on the set D. f(x,y)=2x^3+y^4, D={(x,y) | x^2+y^2&lt;=1}.
castortr0y [4]

Answer:

absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.

Step-by-step explanation:

We compute,

$ f_x = 6x^2, f_y=4y^3 $

Hence, $ f_x = f_y = 0 $  if and only if (x,y) = (0,0)

This is unique critical point of D. The boundary equation is given by

$ x^2+y^2=1$

Hence, the top half of the boundary is,

$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}

On T we have, $ f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1  \text{ for}\ -1 \leq x \leq 1$

We compute

$ \frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$

0 if and  only if x=0, x= 1/2 or x = -2.

We disregard  $ x = -2 \notin [-1,1]$

Hence, the critical points on T are (0,1) and $(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$

On the bottom half, B, we have

$ f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$

Therefore, the critical points on B are (0,-1) and $( 1/2, -\sqrt3/2)  

It remains to  evaluate f(x, y) at the points $ (0,0), (0 \pm1), (1/2, \pm \sqrt3/2) \text{ and}\  (\pm1, 0)$ .

We should consider  latter two points, $(\pm1,0)$, since they are the boundary points for the T and also  B. We compute $ f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2 $

We conclude that the  absolute maximum = f(1, 0) = 2

And the absolute minimum = f(−1, 0) = −2.

6 0
3 years ago
At a furniture manufacturer, worker A can assemble a shelving unit in 5 hours. Worker B can assemble the same shelving unit in 3
Contact [7]
Let's define the following variable:
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 Answer:
 
the time it takes for worker A and worker B to assemble a shelving unit together is:
 t = 1,875 hours
5 1
3 years ago
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