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Inga [223]
4 years ago
8

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin. r=-3-2 cos(theta)

Mathematics
2 answers:
Lemur [1.5K]4 years ago
4 0

now, let's keep in mind that cos(π) = -1, and that sin(π) = 0, thus cos(π)cos(θ) is just -cos(θ), and sin(π)sin(θ) is just 0sin(θ) or just 0.


Also let's recall that symmetry identity of cos(-θ) = cos(θ).


\bf r=-3-2cos(\theta )\\\\ -------------------------------\\\\ \stackrel{\textit{testing for the y-axis, }\theta =\pi -\theta }{r=-3-2cos(\pi -\theta )}\implies r=-3-2[cos(\pi )cos(\theta )+sin(\pi )sin(\theta )] \\\\\\ r=-3-2[-cos(\theta )+0]\implies r=-3+2cos(\theta )\qquad \otimes\\\\ -------------------------------


\bf \stackrel{\textit{testing for the x-axis, }\theta =-\theta }{r=-3-2cos(-\theta )}\implies r=-3-2cos(\theta )\qquad \checkmark\\\\ -------------------------------\\\\ \stackrel{\textit{testing for the origin, }\theta =\pi +\theta }{r=-3-2cos(\pi +\theta)}\implies r=-3-2[cos(\pi )cos(\theta )-sin(\pi )sin(\theta )] \\\\\\ r=-3-2[-cos(\theta )+0]\implies r=-3+2cos(\theta )\qquad \otimes

Vanyuwa [196]4 years ago
3 0

Answer:

Yes

Step-by-step explanation:

Since, it is a graph in polar coordinates, So we can start with the base equation of this given equation which should be:

r = cos(theta)

and its graph is a circle with centre at (0.5,0), if we multiply it by 2, then its radius will be doubled and it will become:

r =2 cos(theta)

Thus, giving a circle of radius 1. If we multiply it by -1, then:

r =-2 cos(theta)

It will give us a circle shifted 1 units to the left. If we add 3 in it, we will get a cardioid which is passing through y = 3 and y = -3 at x = 0,

r =3-2 cos(theta)

And this graph will be a symmetric graph about the x-axis. Hence the answer is Yes.

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4 years ago
A firm’s marketing manager believes that total sales for next year will follow the normal distribution, with a mean of $3.2 mill
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Answer:

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In this question, we have that:

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4 years ago
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2= 3x -19
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Same with #11.

x+8=3x-14
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8=2x-14
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--------------
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Check. 

x+8
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