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Aleks [24]
3 years ago
10

I need to find how many solutions does the system have. ​

Mathematics
1 answer:
REY [17]3 years ago
4 0
From what I got there is one real solution :)

Rearrange the second equation to give y=2-2x, substitute into the first equation to give you a value for x and then using that you can work out y for the exact coordinates of intersection :)

Also they are both straight line graphs so they will either have one or no solutions

Hope this helped :)
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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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\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

\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;

z = z

Calculating:

a) (4,2,-4)

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

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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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