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3 years ago

What is the sum? 2x/x^2 + x-4/9x

Mathematics

2 answers:

Artist 52 [7]3 years ago

6 0

Answer:

$\frac{x+14}{9x}$

Step-by-step explanation:

I will show the algebraic steps associated with this problem:

$\frac{2x}{x^2}+\frac{x-4}{9x}\\=\frac{2x(9)+(x-4)(x)}{9x^2}\\=\frac{18x+x^2-4x}{9x^2}\\=\frac{x^2+14x}{9x^2}\\=\frac{x(x+14)}{9x^2}\\=\frac{x+14}{9x}$

This is the answer.

love history [14]3 years ago

3 0

<h2>Answer:</h2>

<u>The answer is </u><u>x+14 / 9x</u>

<h2>Step-by-step explanation:</h2>

See the attached image

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alex41 [277]

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

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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mariarad [96]

The correct answer is B.

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4 years ago