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alekssr [168]
3 years ago
14

Name the pair of opposite rays with endpoint N.

Mathematics
1 answer:
Elza [17]3 years ago
3 0

Answer:

<h2>Possible Answers: NA and NX or NM and NC.</h2>

Step-by-step explanation:

PLS MARK ME BRAINLEIEST AND FLW ME

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C

Step-by-step explan:

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Use the scenario below to answer questions 5 &amp; 6:
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Hey there!

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Which of the following is equivalent to (5 /k) + ((k + 3) /(k + 5)) ?
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\frac{5}{k}+\frac{k+3}{k+5}=\frac{5(k+5)}{k(k+5)}+\frac{k(k+3)}{k(k+5)}=\frac{5k+25+k^2+3k}{k(k+5)}=\frac{k^2+8k+25}{k(k+5)}\\\\Answer:E
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aivan3 [116]
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5 0
2 years ago
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>
3 0
3 years ago
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