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Yanka [14]
2 years ago
15

F(-3,-2), I(-4, 3), A(-2,3), M(1, -1) to I’(-4,-3), A’ (-2, -3), M’(1,1), F(-3,2)

Mathematics
1 answer:
icang [17]2 years ago
8 0
I am not sure what exactly the question is, but all points are a reflection across the x-axis. I hope that helps.
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tickets to a concert cost $15 online and $25 at the door. one hundred more people bought their tickets at the door than online.
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3 0
3 years ago
Read 2 more answers
For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?
Bond [772]
Given that r=\theta, then r'=1

The slope of a tangent line in the polar coordinate is given by:

m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}

Thus, we have:

m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}



Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

\sin\theta+\theta\cos\theta=0 \\  \\ \theta\cos\theta=-\sin\theta \\  \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta

Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

</span><span>θ = 0

</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488



Part B:

For vertical tangent lines, \frac{1}{m} =0

Thus, we have:

\cos\theta-\theta\sin\theta=0 \\  \\ \Rightarrow\theta\sin\theta=\cos\theta \\  \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta

</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

</span>θ = <span>4.91718592528713</span>
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tino4ka555 [31]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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