Determine if the graph of r = 4cos 5θ is symmetric about the x-axis, the y-axis, or the origin.

x-axis

Step-by-step explanation:

if r%28theta%29+=+r%28-theta%29.

Since 4cos%285%2Atheta%29+=+4cos%28-5%2Atheta%29, the graph is symmetric wrt the x-axis.

The graph is symmetric wrt the y-axis if r%28theta%29+=+r%28pi-theta%29.

Since 4cos%285%2Atheta%29+%3C%3E+4cos%285%2A%28pi+-+theta%29%29+=+-4cos%285%2Atheta%29+, the graph is not symmetric wrt the y-axis.

The graph is symmetric wrt the y-axis if r%28theta%29+=+r%28pi%2Btheta%29.

Since 4cos%285%2Atheta%29+%3C%3E+4cos%285%2A%28pi+%2B+theta%29%29+=+-4cos%285%2Atheta%29+, the graph is not symmetric wrt the origin.

Step-by-step explanation:

The function graph is symmetric about

- the y-axis when it is an even function

- the origin when it is an odd function.

A symmetrical graph about the x-axis is not a function graph.

f(x) is an even if and only if f(-x) = f(x).

f(x) is an odd if and only if f(-x) = -f(x).

We have the function r(θ) = 4cos(5θ)

(olny symmetry about the y-axis or about the origin)

Check r(-θ):

r(-θ) = 4cos(5(-θ)) = 4cos(-5θ) = 4cos(5θ)

used cos(-x) = cos x

We have r(-θ) = r(θ). Therefore the graph of r(θ) is symmetric about the y-axis.

answer: plz put a pic in so i can answer it

step-by-step explanation: