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bonufazy [111]
3 years ago
11

1 quarter 2 dimes 3 pennies how munch I have

Mathematics
1 answer:
vfiekz [6]3 years ago
8 0
In total, you have 48 cents.
25+10+10+3= 48

Give 5 stars and brainliest answer if this helped please!<span>
</span>
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35

Step-by-step explanation:

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Read 2 more answers
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

                       

                     

 

3 0
2 years ago
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