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ZanzabumX [31]
3 years ago
7

Need quick answer for this 9/8-3s/8=3/8

Mathematics
2 answers:
marshall27 [118]3 years ago
7 0

Answer:

s = 2

Step-by-step explanation:

GuDViN [60]3 years ago
6 0

ANSWER:

\longrightarrow\sf\frac{9}{8}  -  \frac{3s}{8}  =  \frac{3}{8}

\longrightarrow\sf \frac{3s}{8} =  \frac{9}{8}  -  \frac{3}{8}

\longrightarrow\sf \frac{3s}{\cancel{8}}  =  \frac{9 - 3}{\cancel{8}}

\longrightarrow\sf\:3s = 6

\longrightarrow\sf\:s =  \frac{\cancel{6}}{\cancel{3}}

\longrightarrow\large\boxed{\sf\:s = 2}

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Let f be defined by the function f(x) = 1/(x^2+9)
riadik2000 [5.3K]

(a)

\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}

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\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt

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(b) The series

\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}

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(c) The series

\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}

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3 years ago
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