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3 years ago

(20 POINTS AND BRAINLIEST FOR BEST)

Mathematics

1 answer:

Lena [83]3 years ago

7 0

Answer:

i used be so good at this now i suck because it's been awhile

Step-by-step explanation:

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-3 5/6 make it a decimal

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Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

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

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

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

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

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

converges by comparison to the convergent p-series,

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

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

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

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Lynette bought 2CDs for the same price.She paid $23.15,including sales tax of $1.19 and a $2.00 discount coupon.What was the pri

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$11.57 per CD without tax and coupon

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7^16/7^12=7^-18/? what is the missing denominator

creativ13 [48]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\$

$\bf \cfrac{7^{16}}{7^{12}}=\cfrac{7^{-18}}{x}\implies x\cdot 7^{16}=7^{12}\cdot 7^{-18}\implies x\cdot 7^{16}=7^{12-18}
\\\\\\
x\cdot 7^{16}=7^{-6}\implies x=\cfrac{7^{-6}}{7^{16}}\implies x=\cfrac{7^{-6}\cdot 7^{-16}}{1}\implies x=7^{-6-16}
\\\\\\
\boxed{x=7^{-22}}\implies x=\cfrac{1}{7^{22}}$

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3 years ago