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Vitek1552 [10]
3 years ago
5

Determine whether the series is absolutely convergent 1-(1*3/3! (1*3*5/5!-(3*5*7)

Mathematics
1 answer:
DENIUS [597]3 years ago
4 0
It's not clear what your series is, so I'm going to take a wild guess on what it is you mean:

1-\dfrac{1\times3}{3!}+\dfrac{1\times3\times5}{5!}-\dfrac{1\times3\times5\times7}{7!}+\cdots
=\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)!}\prod_{k=1}^n(2k-1)

For the sum to be absolute convergent, the sum of the absolute value of the summand must converge, so you are really examining the convergence of

\displaystyle\sum_{n=1}^\infty\frac1{(2n-1)!}\prod_{k=1}^n(2k-1)

This is easily checked with the ratio test:

\displaystyle\lim_{n\to\infty}\left|\frac{\displaystyle\dfrac1{(2(n+1)-1)!}\prod_{k=1}^{n+1}(2k-1)}{\displaystyle\dfrac1{(2n-1)!}\prod_{k=1}^n(2k-1)}\right|=\lim_{n\to\infty}\left|\frac{\dfrac{1\times3\times5\times\cdots\times(2n-1)\times(2n+1)}{(2n+1)!}}{\dfrac{1\times3\times5\times\cdots\times(2n-1)}{(2n-1)!}}\right|
\displaystyle=\lim_{n\to\infty}\left|\frac{\dfrac{2n+1}{(2n+1)(2n)}}{\dfrac11}\right|=\lim_{n\to\infty}\dfrac1{2n}=0

Since \sum|a_n| converges by the ratio test, the series \sum a_n converges absolutely.
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lubasha [3.4K]

Answer:

<em>37.3° </em>

Step-by-step explanation:

sin β = \frac{20}{33} ⇒ β = arcsin \frac{20}{33} = <em>37.3°</em>

6 0
3 years ago
A 68-year old retiree has a pension savings of $43,000, a 401k of $75,000, and an IRA account of $62,000. Before 2008, he had 2%
Natalija [7]

Answer:

The answer is the IRA

Step-by-step explanation:

If he takes out 2% from each account,  the highest penalty will be for the IRA account which is, 6% of 62,000 is $3,720. He will also take 2% of that 62,000 which will be $1,240. The total will be $4,960.

For the 401k account the withdrawal is $1,500 and the penalty is $3,000. That is $460 less than the IRA account.

I hope this answer helps.

6 0
3 years ago
Read 2 more answers
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Paha777 [63]

5^4 and 4^3

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8 0
3 years ago
use the line graph to answer the question 9 and 10 how much did the temperature change from Sunday High to Mondays High ? what w
Marizza181 [45]

Answers:

<em>How much did the temperature change from Sunday High to Mondays High?</em>

Change = 4 °C

<em>What was the difference between the high temperatures on Friday and Wednesday?​</em>

Difference = 10 °C

Explanation:

Taking into account the graph, we get that the high temperature each day is:

Sunday: -10°C

Monday: -6 °C

Tuesday: - 4 °C

Wednesday: -6 °C

Thursday: 0 °C

Friday: 4 °C

Saturday: -2 °C

So, the change from Sunday High to Mondays High can be calculated as:

Change = Monday - Sunday

Change = -6 °C - (- 10 °C)

Change = -6 °C + 10 °C

Change = 4 °C

In the same way, the difference between the high temperatures on Friday and Wednesday can be calculated as:

Difference = Friday - Wednesday

Difference = 4 °C - (-6 °C)

Difference = 4 °C + 6 °C

Difference = 10 °C

Therefore, the answers are:

<em>How much did the temperature change from Sunday High to Mondays High?</em>

Change = 4 °C

<em>What was the difference between the high temperatures on Friday and Wednesday?​</em>

Difference = 10 °C

6 0
2 years ago
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

1 + tan(x) + tan^2(x) = 2 tan(x)

Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

(tan(x) - 1/2)^2 = -3/4

Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

4 0
3 years ago
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