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What is the polar form of #( 11,-99 )#?

Answer 1

Evelyn Agard

Polar:
Exact: #(11sqrt(82), tan^-1(-9))#
About: #(99.61, -1.46)

Cartesian form: #(x,y)#

Polar form: #(r, theta)#

#r = sqrt(x^2+y^2)#

#r = sqrt(11^2+99^2) = 11sqrt(82)~~99.61#

#theta = tan^-1(y/x)#

#theta = tan^-1(-99/11)#

#theta = tan^-1(-9) ~~ -1.46#

Exact

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Answer 2

Evelyn Agard

The polar form of the point (11, -99) is ( r = \sqrt{11^2 + (-99)^2} ) and ( \theta = \arctan\left(\frac{-99}{11}\right) ). This gives the polar form as ( (r, \theta) = \left(\sqrt{12100}, \arctan\left(\frac{-99}{11}\right)\right) ). Simplifying further, ( (r, \theta) = (110, -1.48 , \text{rad}) ) (rounded to two decimal places).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.