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3 years ago

Which of the following are vertical asymptotes of the function y= 3cot(2x)-4

Mathematics

1 answer:

alexandr1967 [171]3 years ago

8 0

The vertical asymptotes for y = 3cot(2x)-4 occur at 0,π/2, and every is x = πn/2, where n is an integer.

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Please answer this correctly

olga_2 [115]

Answer:

320

Step-by-step explanation:

so finding 80% of 400

move the decimal in 80% over twice to become a whole number

multiply 400 times 80=32,000

move the decimal back twice

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4 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E5_1%20%7B%20%5Cfrac%7BlnR%7D%7B%20R%5E%7B2%7D%20%20%7D%20%5C%2C%20dR%20"

alex41 [277]

Integrate by parts, setting

$\begin{matrix}u=\ln R&&\mathrm dv=\dfrac{\mathrm dR}{R^2}\\\mathrm du=\dfrac{\mathrm dR}R&&v=-\dfrac1R\end{matrix}$

So the integral is

$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\dfrac{\ln R}R\bigg|_{R=1}^{R=5}+\int_1^5\frac{\mathrm dR}{R^2}$
$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\left(\frac{\ln5}5-\frac{\ln1}1)-\frac1R\bigg|_{R=1}^{R=5}$
$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\frac{\ln5}5-\left(\frac15-1\right)$
$\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=\frac15(4-\ln5)$

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4 years ago

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Answer:

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Step-by-step explanation:

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