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3 years ago

- If you can buy one jar of crushed ginger

Mathematics

1 answer:

Fudgin [204]3 years ago

6 0

Answer:

C, 3

Step-by-step explanation:

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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

The expression that is equal to -10(4x – 5).

notka56 [123]

The best answer to your question would have to be -40x - 50

4 0

3 years ago

What is the area of this figure?

KATRIN_1 [288]

Answer:

10.5

Step-by-step explanation:

Area is length times width so, 3.5 times 3 equals 10.5. 10.5 is the area of the given figure.

PLZ MARK BRAINLIEST!!! HOPE THIS HELPS!!!

3 0

3 years ago

the sales price for a bicycle $315. the original price was first discounted by 5o% and then discounted by 10%. find the original

VMariaS [17]

Answer:

?× ( 1 − 50/ 100 ) × ( 1 − 10/ 100 ) = $ 315 ? = $ 315 /( .5 ) ( .9 ) = $ 700

Step-by-step explanation:

5 0

3 years ago

what's the slope? simplify your answer and write it as a proper fraction,improper fraction or integer

dybincka [34]

Answer:

$\frac{1}{2}$

Step-by-step explanation:

In this graph, when you count rise and run, rise would be 1 and run would be 2. Because slope is also known as rate of change and rate of change is also known as rise over run, the slope will be 1 over 2.

8 0

3 years ago