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oksian1 [2.3K]
3 years ago
10

ME in the coordinate plane has endpoints with coordinates (1,4) and (−10,−4). Graph ME and find two possible locations for point

R, so R divides ME into two parts with lengths in a ratio of 2:3
Mathematics
1 answer:
Solnce55 [7]3 years ago
8 0

Answer:

the answer is on google

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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
3 years ago
Saturn orbits the Sun at a distance of 1.43 × 1012 m. The mass of the Sun is 1.99 × 1030 kg.
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The answer is A 29.6 Earth Years

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5d^9\4(d)^2. Simplify
Vesna [10]

Answer: <u>5d^7/4</u>

Step By Step:

d^2/d^2 = d^9-2

= 5d^9-2 / 4

= 5d^7/ 4

7 0
3 years ago
7x + 2y + 5 - 5x + Oy - 1<br> Simplify expresson
frozen [14]

Answer:

To do this you have to combine like terms. For example, 7x and -5x.

3 0
3 years ago
I need help As Soon As possible!
Julli [10]
This should be simple, add the Amounts of snow for the first week
4.32 + 6.86 = 11.18
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The subtract the amounts of snow from each week
11.18 - 7.89 = 3.29
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C.) 3.29 is your answer
6 0
3 years ago
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