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Iteru [2.4K]
1 year ago
9

Pls help answer i dont not need explaination :)

Mathematics
2 answers:
nadezda [96]1 year ago
3 0

Answer:

165 km ^2

Step-by-step explanation:

meriva1 year ago
3 0

Answer:

165 km²

Step-by-step explanation:

<u>First, let us find the area of the big rectangle.</u>

Area = length × Width

Area = 15km × 13 km

Area = 195 km²

<u>Now, let us find the area of the small rectangle.</u>

Area = length × width

Area = 6 km × 5 km

Area = 30 km²

<u>And now, let us find the area of the shaded region.</u>

<em>For that, we have to subtract the area of the small rectangle from the area of the big rectangle.</em>

Area = 195 km² - 30 km²

Area = <u>165 km²</u>

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An equilateral polygon has congruent sides, like a rhombus.

An equiangular polygon has congruent interior angles, like a rectangle.

In other words: Equilateral doesn't necessarily have congruent INTERIOR angles while equiangular does have/need congruent interior angles.

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3 years ago
Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the
dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

I = \int \int_E \int zdV where E is bounded by the cylinder y^2 + z^2 = 81 and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and y^2 + z^2 = 81

∴ z^2 = 81 - y^2

z = \sqrt{81 - y^2}

Thus, z lies between 0 to \sqrt{81 - y^2}

GIven curve x = 0 and y = 9x

x =\dfrac{y}{9}

As such,x lies between 0 to \dfrac{y}{9}

Given curve x = 0 , x =\dfrac{y}{9} and z = 0, y^2 + z^2 = 81

y = 0 and

y^2 = 81 \\ \\ y = \sqrt{81}  \\ \\  y = 9

∴ y lies between 0 and 9

Then I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix}    ^ {\sqrt {{81-y^2}}}_{0} \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{(\sqrt{81 -y^2})^2 }{2}-0  \end {bmatrix}     \ dxdy

I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix}  \dfrac{{81 -y^2} }{2} \end {bmatrix}     \ dxdy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0}    \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix}     \ dy

I = \int^9_{y=0}  \begin {bmatrix}  \dfrac{{81 \  y -y^3} }{18} \end {bmatrix}     \ dy

I = \dfrac{1}{18} \int^9_{y=0}  \begin {bmatrix}  {81 \  y -y^3}  \end {bmatrix}     \ dy

I = \dfrac{1}{18}  \begin {bmatrix}  {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}}  \end {bmatrix}^9_0

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I = \dfrac{1}{18}  \begin {bmatrix}  3280.5 - 1640.25  \end {bmatrix}

I = \dfrac{1}{18}  \begin {bmatrix}  1640.25  \end {bmatrix}

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