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Lubov Fominskaja [6]
3 years ago
7

The same architect wants to draw a room that has an area of 175 ft². What will be the area of the room in his drawing in square

inches?
Mathematics
1 answer:
Dominik [7]3 years ago
3 0

Answer:

25,200in²

Step-by-step explanation:

The architect wants to draw a room that has an area of 175 ft². We need to find this measurement in square inches.

<u>Square feet and square inches are different from feet and inches. </u>

We should know that a square foot is 144 square inches.

Multiply 175 * 144.

175 * 144=25200

So the same architect wants to draw a room that should have an area of 25,200in².

<em>Brainilest Appreciated. </em>

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Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

\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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\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}} = 6

\phi = cos^{-1}(\frac{-4}{6})

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

For θ, choose 1st option:

\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

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

\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} }

\theta = \pi + \frac{\pi}{6}

\theta = \frac{7\pi}{6}

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

r=\sqrt{x^{2}+y^{2}}

Angle θ is the same as spherical coordinate;

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z = 15

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r=\sqrt{(\sqrt{2} )^{2}+1^{2}} = \sqrt{3}

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

z = 1

d) (−2√3,−2,3)

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

\theta = \frac{7\pi}{6}

z = 3

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