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3 years ago

a rug in the shape of a regular quadrilateral has side length of 20ft. what is the perimeter of the rug?

Mathematics

2 answers:

Gre4nikov [31]3 years ago

7 0

Answer:

A rug in the shape of a regular quadrilateral has a side length of 20 . What is the perimeter of the rug?

A.4

B.5

C.40

D.80

Step-by-step explanation:

Helen [10]3 years ago

4 0

A rug in the shape of a regular quadrilateral has a side length of 20ft what is the perimeter of the rug? 4ft 5ft•• 40ft 80ft Correct me if I'm wrong!!

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iogann1982 [59]

$f(x,y)=9e^y(y^2-x^2)$
$\nabla f=\left\langle-18xe^y,9ye^y(2+y)\right\rangle$

Critical points occur where the gradient is zero. This is guaranteed whenever $x=0$ and either $y=0$ or $y=-2$ .

The Hessian matrix for this function looks like

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-18e^y&-18xe^y\\-18xe^y&9e^y(2-x^2+4y+y^2)\end{bmatrix}$

and has determinant

$|H(x,y)|=-162e^{2y}(2+x^2+4y+y^2)$

Maxima occur whenever the determinant is positive and $f_{xx}$ . Minima occur whenever both the determinant and $f_{xx}$ are positive. Saddle points occur whenever the determinant is negative.

At $(0,0)$ , you have a saddle point since the determinant reduces to -324, so $(0,0,0)$ is the saddle point.

At $(0,-2)$ , the determinant is $\dfrac{324}{e^4}>0$ and $f_{xx}(0,-2)=-\dfrac{18}{e^2}$ , so $\left(0,-2,\dfrac{36}{e^2}\right)$ is a local maximum.

No other critical points remain, so you're done.

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3 years ago

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

Distributive property can be used to expand the expression -2(3/4x+7).

When that property is used, we need to distrubute the monomial to the binomial which is the two terms.

$- 2( \frac{3}{4} x + 7) \\ \Longrightarrow \boxed{- 2( \frac{3}{4} x) + - 2(7)}$

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