4 years ago

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Mathematics

1 answer:

svetlana [45]4 years ago

8 0

Answer:

2,250.

Step-by-step explanation:

Add all of the bills together. It says the minimum, so in order to pay all of the bills, you'll need 2,250.

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Show that a sequence {sn} coverages to a limit L if and only if the sequence {sn-L} coverages to zero.

Andreyy89

Let ${s_n}_{n\in\Bbb N}$ be a sequence that converges to $L$ . This means for any $\varepsilon>0$ , there is some $N$ such that $|s_n-L|$ for all $n>N$ . From this inequality we see that $|(s_n-L)-0|$ , so it follows that $s_n-L\to0$ .

On the other hand, let ${s_n-L}$ be a sequence that converges to 0. This means $|(s_n-L)-0|$ for all large enough $n$ , and we get the simpler inequality for free, $|s_n-L|$ , so it follows that $s_n\to L$ .