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ExtremeBDS [4]
4 years ago
14

Evaluate each expression for y=8 12+y=

Mathematics
2 answers:
evablogger [386]4 years ago
8 0

Answer:

12 + 8 = 20 i think

Step-by-step explanation:

if y is 8 then just add 12 and 8

artcher [175]4 years ago
6 0

Answer:

12 + 8 = 20

Step-by-step explanation:

If y = 8, just plug in 8 to where y is. 12 + 8 = 20. Hope this helped!

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Sally had 1/2 of a candy bar. She gave half of what she had to Frank. Frank gave half of HIS piece to Peter. How much of a candy
KonstantinChe [14]
1/8 of the candy bar
5 0
3 years ago
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

6 0
4 years ago
HELP!!
laiz [17]
Associative property!
The property states that when three or more numbers are added (or multiplied), the sum (or the product) is the same regardless of the grouping of the addends!
Hope this helps!
8 0
3 years ago
What is the measure of the central angle of a circle with a radius of 30 cm that intercepts an 8 cm arc?
Arlecino [84]
The answer is 360 cm
8 0
3 years ago
Write 4 different mixed decimals that equal 11 wholes
Softa [21]
There are an infinite number of answers. One possible answer is 3.5+1.2+2.4+3.9=11. The explanation of how I arrived at this follows: A mixed decimal is a whole number followed by a fractional part so an example would be 3.4 or 2.5. To get 4 different mixed decimals that sum to 11 follow these steps: 1. Start with 11 and subtract any mixed decimal from 11 that is smaller than 11. So I will subtract 3.5    11-3.5=7.52. Take the 7.5 from the last step and subtract another mixed decimal from 7.5 that is smaller than 7.5. I will subtract         1.2    7.5-1.2=6.33. Take the 6.3 and subtract another mixed decimal from 6.3 that is smaller than 6.3. I will subtract 2.4    6.3-2.4=3.94. Now I take all of the numbers that I subtracted and add them together along with the 3.9 that was left over and I get       3.5+1.2+2.4+3.9=11

I hope that this helps you! =D
5 0
3 years ago
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