Question #8a89a

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Question #8a89a

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Answer 1 · 2016-12-21T22:27:22

Use the equations, $r = sqrt(x^2 + y^2)$ and $theta = pi + tan^-1(y/x)$
$(-3sqrt(3), -3) to (6, (7pi)/6)$

Explanation:

Given the point:

Use the equation:

$r = sqrt(x^2 + y^2$

$r = sqrt((-3sqrt(3))^2 + (-3)^2)$

$r = sqrt(27 + 9)$

$r = 6$

There are many equations for the angle, $theta$ ; which one you use depends on x and y:

$"[1] "theta = tan^-1(y/x); x > 0 and y ge 0$
$"[2] "theta = pi/2; x = 0 and y > 0$
$"[3] " theta = pi + tan^-1(y/x); x < 0$
$"[4] "theta = (3pi)/2; x = 0 and y < 0$
$"[5] "theta = 2pi + tan^-1(y/x); x > 0 and y < 0$

Because $x = -3sqrt(3)$ , then we do not care about the value of y and we use equation [3}:

$theta = pi + tan^-1((-3)/(-3sqrt(3))); x < 0$

$theta = (7pi)/6$

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Question #8a89a

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