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koban [17]
3 years ago
10

MATH

Mathematics
1 answer:
ale4655 [162]3 years ago
7 0

Answer:

the answer is 3

Step-by-step explanation:

in the number 66 there are "2" circles in them and so the answer we get 2

in the number 99 there are "2" circles and so the answer there is 2

in the number 888 there are "6" circles so the answer is 6

in the number 00 there are "2" circles that's the answer

in the number 7777 there are no circles so the answer is 0

we keep on going like that......

in the number 2876 there are 3 circles 2 in the number 8 and 1 in the number 6

hope this helped :D

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m = -2/7 + π/14 + (π n_1)/7 for n_1 element Z

or m = 2/3 - (π n_2)/18 for n_2 element Z

Step-by-step explanation:

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Multiply both sides by -1:

cos(7 m + 2) sin(12 - 18 m) = 0

Split into two equations:

cos(7 m + 2) = 0 or sin(12 - 18 m) = 0

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or sin(12 - 18 m) = 0

Subtract 2 from both sides:

7 m = -2 + π/2 + π n_1 for n_1 element Z

or sin(12 - 18 m) = 0

Divide both sides by 7:

m = -2/7 + π/14 + (π n_1)/7 for n_1 element Z

or sin(12 - 18 m) = 0

Take the inverse sine of both sides:

m = -2/7 + π/14 + (π n_1)/7 for n_1 element Z

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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:

  • If x > 0 and y > 0: \theta = tan^{-1}\frac{y}{x};
  • If x < 0: \theta = \pi + tan^{-1}\frac{y}{x};
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Calculating:

a) (4,2,-4)

\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:

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Angle θ is the same as spherical coordinate;

z = z

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a) (4,2,-4)

r=\sqrt{4^{2}+2^{2}} = \sqrt{20}

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

z = -4

b) (0, 8, 15)

r=\sqrt{0^{2}+8^{2}} = 8

\theta = \frac{\pi}{2}

z = 15

c) (√2,1,1)

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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3 years ago
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