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3 years ago

Is this a polygon? If it is not a polygon please say why.

Mathematics

2 answers:

Eva8 [605]3 years ago

8 0

Answer:

if you where trying to draw it straight then yes it is a polygon but if it is supposed to be rounded then no it is not.

Step-by-step explanation:

a plane figure with at least three straight sides and angles, and typically five or more.

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

technically yes if you make the lines straight and not overlapping and touching another line.

Step-by-step explanation:

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3 years ago

Read 2 more answers

Lloyd solves the equation 24x + 18(x – 2) = 216 for x. Some of his steps are shown.

ANTONII [103]

Answer:

24+18x-36=216

-12+18x=216

18x=216+12

18x=228

x=228/18

x=30/3

Step-by-step explanation:

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3 years ago

A skull cleaning factory cleans animal skulls of deer, buffalo and other types of animals using flesh-eating beetles. the fact

Shtirlitz [24]

The solution for this problem is:
The population is 500 times bigger since 8000/24 = 500. The population after t days is computed by:P(t) = P₀·4^(t/49)
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3 years ago

Find the distance between 2 and -2 A. -6 B -4 C. 0 D. 4

tangare [24]

Answer:

b) -4

Step-by-step explanation:

Well your at 2 and going to -2 so on a number line you be going back or left.

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2 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

8 0

3 years ago

Is this a polygon? If it is not a polygon please say why.​

Is this a polygon? If it is not a polygon please say why.