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damaskus [11]
3 years ago
12

What is the solution to this system of equations?

Mathematics
2 answers:
kvasek [131]3 years ago
8 0

Answer:

(-2,1)

Step-by-step explanation:

AlexFokin [52]3 years ago
4 0

the blank is a aprt of the streen

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C I’m pretty sure, if wrong I’m sorry
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What value of Z+3÷1/2 = 1 true
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Answer - A) -5

Step 1 - 3 Divided by 1/2 = 6

Step 2 - 6 + -5 = 1


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A basketball player attempts 18 baskets in a game. He makes 8 of the attempted baskets. Which ratio describes the number of bask
EleoNora [17]

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4:9

Step-by-step explanation:

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What operation is used to make a prediction using a probability
Dmitry_Shevchenko [17]

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3 years ago
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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 <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :

\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 <em>p</em>-series,

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

(c) The series

\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 (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

5 0
3 years ago
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