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Mathematics 72 Online
OpenStudy (anonymous):

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. (1 point) r = 4 - 4 cos θ No symmetry x-axis only y-axis only Origin only

OpenStudy (anonymous):

characterized by or exhibiting symmetry; well-proportioned, as a body or whole; regular in form or arrangement of corresponding parts. noting two points in a plane such that the line segment joining the points is bisected by an axis: Points(1, 1) and(1, −1) are symmetrical with respect to the x-axis.

OpenStudy (rational):

|dw:1432367802832:dw|

OpenStudy (anonymous):

thank you both of you ! @rational @leon549

OpenStudy (anonymous):

could you help me one more?? @rational

OpenStudy (rational):

you figured out this problem ?

OpenStudy (anonymous):

yes

OpenStudy (rational):

whats ur answer

OpenStudy (anonymous):

@rational x-axis right??

OpenStudy (rational):

Yup!

OpenStudy (anonymous):

The graph of a limacon curve is given. Without using your graphing calculator, determine which equation is correct for the graph. (1 point) [-5, 5] by [-5, 5] r = 2 + 3 cos θ r = 3 + 2 cos θ r = 2 + 2 cos θ r = 4 + cos θ

OpenStudy (rational):

look at second option r = 3 + 2 cos θ when \(\theta=\pi\), we have \(r=3+2(-1) = 1\) as desired so im inclining toward this option

OpenStudy (anonymous):

so the answer is r = 3 + 2 cos θ ???

OpenStudy (anonymous):

@rational

OpenStudy (rational):

I think so

OpenStudy (anonymous):

thank you @rational

OpenStudy (rational):

yw

Parth (parthkohli):

\[r = 4 - 4 \cos \theta\]\[r^2 = 4r - 4r\cos \theta\]\[\Longleftrightarrow x^2 + y^2 = 4 \sqrt{x^2 + y^2 } - 4x\]If you replace \(y\) with \(-y\), you get the same equation. The same isn't true for \(x\). Thus, it is symmetric about the x-axis.

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