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Rufina [12.5K]
3 years ago
9

Solve for N: -24 - N = 12 Explain how you solved it.

Mathematics
2 answers:
kondor19780726 [428]3 years ago
5 0

Answer: n=-36

Step-by-step explanation: Hope this help :D

s344n2d4d5 [400]3 years ago
4 0

Answer:

The answer is - 36

Step-by-step explanation:

-24-N=12

or, - N = 12+24

N = - 36

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Janet’s savings account has $12.10 more than the $242 it had at the beginning of the year. What percent more was in the savings
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5% is the answer!! :)
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Use the Alternating Series Approximation Theorem to find the sum of the series sigma^infinity_n = 1 (-1)^n - 1/n! with less than
DanielleElmas [232]

Answer:

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198 -0.0000248

For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that |-0.0000248|= 0.0000248 and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198-0.0000248 =0.632118

Step-by-step explanation:

Assuming the following series:

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

We want to approximate the value for the series with less than 0.0001 of error.

First we need to ensure that the series converges. If we have a series \sum a_n where a_n = (-1)^n b_n [/tex] or a_n =(-1)^{n-1} b_n where b_n \geq 0 for all n if we satisfy the two conditions given:

1) lim_{n \to \infty} b_n =0

2) {b_n} is a decreasing sequence

Then \sum a_n is convergent. For this case we have that:

lim_{n \to \infty} \frac{1}{n!} =0

And \frac{1}{n!} because \frac{1}{n!} =\frac{1}{n (n-1)!} and \frac{1}{n(n-1)!} < \frac{1}{(n-1)!}

So then we satisfy both conditions and then the series converges. Now in order to find the approximation with the error required we can write the first terms for the series like this:

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198 -0.0000248

For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that |-0.0000248|= 0.0000248 and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198-0.0000248 =0.632118

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4 years ago
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Answer:

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Step-by-step explanation:

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