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2 years ago

6

In which quadrant does the point (-5, -21) lie? A. Quadrant IV B. Quadrant I C. Quadrant III D. Quadrant II

2 answers:

Klio2033 [76]2 years ago

7 0

Answer:

Quadrant 3

Step-by-step explanation:

Vilka [71]2 years ago

6 0

Answer:

Quadrant 3

Step-by-step explanation:

All negatives all fall under 3

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TanФ=-1.793

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Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

<u><em>Step(ii):-</em></u>

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

=0

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

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2 years ago