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aliya0001 [1]
2 years ago
7

PLEASE HELP ASAP 28 POINTS!!!! PLS PLS PLS PLS PLS PLS HELP

Mathematics
2 answers:
pav-90 [236]2 years ago
4 0

Hello :)

We know that an answer in square feet would require us to calculate the area. In order to calculate the area of a rectangle, we multiply the length times the width. So, we have 6 x 2 which is 12 Therefore, the area is 12 square feet.

6[]

2

6+6+2+2=16

Have a nice day :)

frez [133]2 years ago
3 0
6 feet and 2 feet

For the area, it is length times width. 6X2=12, so that works out.
For the perimeter, it is 2l+2w. 2(6)+2(2)=12+4=16, so that checks out as well.

It doesn’t matter which one is length and which one is width, unless you left out info. But u should be good putting them in any order. Hope that helps!
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FromTheMoon [43]

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\rho = \sqrt{x^{2}+y^{2}+z^{2}}

\phi = cos^{-1}\frac{z}{\rho}

For angle θ:

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Calculating:

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\phi = cos^{-1}(\frac{-2}{3})

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\theta = tan^{-1}(\frac{2}{4})

\theta = tan^{-1}(\frac{1}{2})

b) (0,8,15)

\rho = \sqrt{0^{2}+8^{2}+(15)^{2}} = 17

\phi = cos^{-1}(\frac{15}{17})

\theta = tan^{-1}\frac{y}{x}

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = \frac{\pi}{2}

c) (√2,1,1)

\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}} = 2

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\phi = \frac{\pi}{3}

\theta = tan^{-1}\frac{1}{\sqrt{2} }

d) (−2√3,−2,3)

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

\phi = cos^{-1}(\frac{3}{5})

Since x < 0, use 2nd option:

\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }

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Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

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Angle θ is the same as spherical coordinate;

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\theta = \frac{\pi}{3}

z = 1

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r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}} = 4

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