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algol13
3 years ago
9

Find the perimeter and the area of the polygon with the given vertices. W (11,2), X (11,8), Y (14,8), Z (14,2)

Mathematics
1 answer:
liberstina [14]3 years ago
5 0

Answer:

The perimeter of the polygon = 18 units

The area of the polygon will be A=18 square units

Step-by-step explanation:

<u><em>Computing the Perimeter of the Polygon: </em></u>

Considering the polygon with vertices

  • W (11,2)
  • X (11,8)
  • Y (14,8)
  • Z (14,2)

As the polygon is drawn in coordinate plane as shown in figure a.

From the attached figure a, we can observe that

  • The length of the side WZ = 3 units
  • The length of the side ZY = 6 units
  • The length of the side WX = 6 units
  • The length of the side XY = 3 units

So, the perimeter of the polygon can be calculated by taking the sum of the lengths of all sides i.e.

The perimeter of the polygon = WZ + ZY + WX + XY

                                                  = 3 + 6 + 6 + 3

                                                  = 18 units

<u><em>Computing the Area of the Polygon: </em></u>

Area can be calculated by multiplying the length and width of the polygon.

A=lw

So, the area will be:

A=lw

A=(3)(6)

A=18

Therefore, the area of the polygon will be A=18 square units

Keywords: area, perimeter, polygon

Learn more about area and perimeter from brainly.com/question/2328840

#learnwithBrainly

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Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie
melamori03 [73]

Answer:

The complete solution is

\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

y = e^{mx}

Differentiating with respect to x

y'= me^{mx}

Again differentiating with respect to x

y''= m ^2 e^{mx}

Putting the value of y, y' and y'' in left side of the differential equation

3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0

\Rightarrow 3m^2-8m-3=0

The auxiliary equation is

3m^2-8m-3=0

\Rightarrow 3m^2 -9m+m-3m=0

\Rightarrow 3m(m-3)+1(m-3)=0

\Rightarrow (3m+1)(m-3)=0

\Rightarrow m = 3, -\frac13

The complementary function is

y= Ae^{3x}+Be^{-\frac13 x}

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)    

For particular integral

y_p=\frac 1{(3D+1)(D-3)} .4

    =4.\frac 1{(3D+1)(D-3)} .e^{0.x}    [since e^{0.x}=1]

   =4\frac{1}{(3.0+1)(0-3)}      [ replace D by 0 , since L(0)≠0]

   =-\frac43

The complete solution is

y= C.F+P.I

\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43

4 0
3 years ago
Hey guys! I would be thankful if anyone could help me answer this question.
ExtremeBDS [4]
In order to not get confused, start by converting cm to m
h=20 cm = 0.2m and r=5cm = 0.05m

The cans are cylinders and we only need the rectangular shape, which is equal to the height x the circumference of the circle (2r(pi)), which you'll have to multiply by 200,000 for the number of cans necessary

your equation is therefore:
A=200,000h(2pi(r))
A=200,000(0.2)(2)(0.05)(pi)
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4 0
3 years ago
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Find the area of trapezium where area=5cm base=6cm height=8cm
Jlenok [28]

Answer:

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Step-by-step explanation:

[8×(5+6)]/2

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Sidana [21]

Answer:

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Step-by-step explanation:

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Circumference of a circle is given by the formula,

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torisob [31]

Answer:

first one with -1...21

Step-by-step explanation:

only one that follows quadratic formula format

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