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4 years ago

Please help With the following question

Mathematics

1 answer:

Vikki [24]4 years ago

4 0

The equation youre using is y=mx+b

b is the y int which 3 and the slope between the points is 5/2 so your equation is y=5/2x+3

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Answer:

Step-by-step explanation:

Convert rectangle (x , y) to polar coordinates ( r , θ)

$x=r \cos \theta, y= r \sin \theta$

$r=\sqrt{x^2+y^2} , \theta =tan^-^1 (\frac{y}{x} )$

a) converts (9, 0) to polar coordinates ( r , θ)

$r=\sqrt{x^2+y^2} \\\\=\sqrt{9^2+0} \\\\=9$

$\theta= \tan^-^1 (\frac{0}{9} )\\\\=0$

b) Convert $(18,\frac{18}{\sqrt{3} } )$ to polar coordinates ( r, θ)

$r = \sqrt{18^2+(\frac{18}{\sqrt{3} })^2 } \\\\=\sqrt{324+108} \\\\=\sqrt{432}$

$\frac{x}{y} \theta = \tan^-^1(\frac{\frac{18}{\sqrt{3} } }{18} )\\\\= \tan ^-^1(\frac{1}{\sqrt{3} } )\\\\= \frac{\pi}{6}$

c) converts (-5, 5) to polar coordinates ( r , θ)

$r =\sqrt{(-5)^2+(5)^2} \\\\=\sqrt{50} \\\\=5\sqrt{2}$

$\theta=\tan^-^1(\frac{5}{-5} )\\\\= \tan^-^1(-1)\\\\=\frac{3\pi}{4}$

d) converts (-1, √3) to polar coordinates ( r , θ)

$r=\sqrt{(-1)^2+(\sqrt{3})^2 } \\\\= \sqrt{4} \\\\=2$

$\theta=\tan^-^1(\frac{\sqrt{3} }{-1} )\\\=\tan^-^1(-\sqrt{3} )\\\\=\frac{2\pi}{3}$

$= \frac{2\pi}{\sqrt{3} }$

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