How do you solve #-2r<= -6# and graph the solutions?

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Answer 1 · 2016-12-20T01:59:15

The shaded region in the graph

Use # r = sqrt (x^2+y^2) >= 0# and change of sign, in an inequality,

reverses the the inequality operator.

# If -a <= -b#, then #a > b#

Converting to cartesian form,

#-sqrt(x^2+y^2)<= -3 to sqrt(x^2+y^2)>3 to x^2+y^2>9 #

This represents the region exterior to the circle #x^2+y^2=0#.

graph{x^2+y^2>9 [-10, 10, -5, 5]}