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Morgarella [4.7K]
3 years ago
6

2 +2 =4 -2+3-1+2-4= whats the anwser

Mathematics
2 answers:
Sphinxa [80]3 years ago
7 0
This equation does not exist because 4 does not equal 2
vova2212 [387]3 years ago
5 0

Answer:

2

Step-by-step explanation:

4-2=2

2+3=5

5-1=4

4+2=6

6-4=2

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Use the Integral Test to determine whether the series is convergent or divergent
Inga [223]

Answer:

A. \sum_{n=1}^{\infty}\frac{n}{e^{15n}} converges by integral test

Step-by-step explanation:

A. At first we need to verify that the function which the series is related (\frac{n}{e^{15n}}) fills the necessary conditions to ensure that the test is effective.

*f(x) must be continuous or differentiable

*f(x) must be positive and decreasing

Let´s verify that f(x)=\frac{n}{e^{15n}} fills these conditions:

*Considering that eˣ≠0 for all x, the function f(x)=\frac{n}{e^{15n}} does not have any discontinuities, so it´s continuous

*Because eˣ is increasing:

      if a<b ,then eᵃ<eᵇ

      if 0<eᵃ<eᵇ ,then 1/eᵃ > 1/eᵇ

      if 1/eᵃ > 1/eᵇ and a<b, then a/eᵃ<b/eᵇ

  We conclude that f(x)=\frac{n}{e^{15n}} is decreasing

*Because eˣ is always positive and the sum is going from 1 to ∞, this show that f(x)=\frac{n}{e^{15n}} is positive in [1,∞).

Now we are able to use the integral test in f(x)=\frac{n}{e^{15n}} as follows:

\sum_{n=1}^{\infty}\frac{n}{e^{15n}}\ converges\ \leftrightarrow\ \int_{1}^{\infty}\frac{x}{e^{15x}}\ dx\ converges

Let´s proceed to integrate f(x) using integration by parts

\int_{1}^{\infty}\frac{x}{e^{15x}}\ dx=\int_{1}^{\infty}xe^{-15x}\ dx

Choose your U and dV like this:

U=x\ \rightarrow dU=1\\ dV=e^{-15x}\ \rightarrow V=\frac{-e^{-15x}}{15}

And continue using the formula for integration by parts:

\int_{1}^{\infty}Udv = UV|_{1}^{\infty} - \int_{1}^{\infty}Vdu

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15} \int_{1}^{\infty}e^{-15x}\ dx

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15}(\frac{-1}{15e^{15x}})|_{1}^{\infty}

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{1}{225e^{15x}}|_{1}^{\infty}

Because we are dealing with ∞, we´d rewrite it as a limit that will help us at the end of the integral:

\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}}(\frac{-x}{15e^{15x}}|_{1}^{b}-\frac{1}{225e^{15x}}|_{1}^{b})

\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}}-(\frac{-1}{15e^{15}}-\frac{1}{225e^{15}})

\int_{1}^{\infty}xe^{-15x}\ dx= ( \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}})+\frac{1}{15e^{15}}(1-\frac{1}{15})

We only have left to solve the limits, but because b goes to  ∞ and it is in an exponential function on the denominator everything goes to 0

\lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}} = 0

\int_{1}^{\infty}xe^{-15x}\ dx= \frac{1}{15e^{15}}(1-\frac{1}{15})

Showing that the integral converges, it´s the same as showing that the series converges.

By the integral test \sum_{n=1}^{\infty}\frac{n}{e^{15n}} converges

7 0
2 years ago
Which Which is the simplified form of the expression ?
Stells [14]

Answer:

I need more information to awnser this quetion but would have been happy to help:)

3 0
3 years ago
Read 2 more answers
2,250 watts to horsepower (Round answer to the nearest thousandth.
alexgriva [62]

Answer: 3.017hp

Step-by-step explanation: 745.7watts = 1hp

2250watts will be (2250/745.7)hp

i.e 3.017watts

6 0
3 years ago
Daisy has a pitcher that can hold 1000
mash [69]

Answer:

4.2

Step-by-step explanation:

Divide 1000 by 236 and you will get that result.

7 0
2 years ago
Point A is on a number line halfway between -17 and 5. Point B is halfway between point A and 0. What integer does point B reprs
Korolek [52]

Answer:

-3

Step-by-step explanation:

Point A is half way between -17 and 5. This is span has a total distance of 17 + 5 = 22. Half way is 11 units from -17 or 5. So -17 + 11 = -6.

Point B is halfway between A at -6 and 0. This is a distance of 6. So halfway is 3 units away at -3.

7 0
3 years ago
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