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____ [38]
2 years ago
7

Help please help me i beg you

Mathematics
1 answer:
natita [175]2 years ago
4 0

I think it's C but also not completely sure good luck

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Approximate the sum of the convergent infinite series the least usingnumber of terms so that the error is less than .0001. What
____ [38]

Solution :

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$           $a_n= \frac{(-1)^n}{n ! 2^n} \ \ \ \ \ a_{n+1}= \frac{(-1)^{n+1}}{(n+1) ! 2^{n+1}}$

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$(n+1)! 2^{n+1} \geq 10000$

Now try, n ≥ 5

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\sum_{n=1 }^{5 } \frac{(-1)^n}{n ! 2^n}$         (with error 0.0001)

 

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$   = $\frac{-1}{1!2}+\frac{1}{2!2^2}-\frac{1}{3! 2^3}+\frac{1}{4! 2^4}-\frac{1}{5! 2^5}$

$\sum_{n=1 }^{\infty } \frac{(-1)^n}{n ! 2^n}$ = 0.6065104

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What is the nature of the roots of 6x² - 22=0?​
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x=√ 33 /3 , − √ 33 /3

Step-by-step explanation:

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