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3 years ago

Which table represents a function?

Mathematics

1 answer:

velikii [3]3 years ago

4 0

Answer:

Step-by-step explanation:

sorry if i am wrong

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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

My sister is crying because this is the only thing that can get her grade up I tried to help but it isn’t making any sense so pl

Liula [17]

Answer:

it make sense

Step-by-step explanation:

(X is another 2)

x-5=3

2x+1=5

we're gonna barrow 5 from the first question and get the answer in the second question

so..

5-5=0

hope it helps

7 0

3 years ago

The ratio of a pair of corresponding sides in two similar triangles is 5:3. The area of the smaller triangle is 108 in2. What is

aleksandrvk [35]

Answer:

The hypotenuse of a right triangle is always the side opposite the right angle. It is the ... The other two sides are called the opposite and adjacent sides. ... functions—sine, cosine, and tangent—which are defined using the words hypotenuse, opposite, and adjacent. ... Side ratios in right triangles as a function of the angles.

Step-by-step explanation:

4 0

3 years ago

Points A(8,4)&B(-2,4) lie on a line .AB is parallel to which axis

IRINA_888 [86]

If the line is parallel they will have the same slope

8 0

3 years ago

What is the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34 Round yo

Galina-37 [17]

We are asked to find the probability that a data value in a normal distribution is between a z-score of -1.32 and a z-score of -0.34.

The probability of a data score between two z-scores is given by formula $P(a$ .