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3 years ago

Find the distance between (–2, –3) and (–6, –12). Round the answer to the nearest hundredth, if necessary.

2 answers:

8 0

Correct answer is A.

$A(-2,-3),B(-6,-12)
\\d(A,B)= \sqrt{(-6-(-2))^2+(-12-(-3))^2} 
\\d(A,B)= \sqrt{(-6+2)^2+(-12+3)^2} 
\\d(A,B)= \sqrt{(-4)^2+(-9)^2} 
\\d(A,B)= \sqrt{16+81} 
\\d(A,B)= \sqrt{97} 
\\d(A,B) \approx 9.85$

laiz [17]3 years ago

6 0

The answer is A 9.85

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riadik2000 [5.3K]

(a)

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(b) The series

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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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liubo4ka [24]

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