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murzikaleks [220]
3 years ago
6

Urgent!!!!?

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
4 0

Answer:

C) 6 feet

Step-by-step explanation:

The diagram is shown in the attachment.

Using the Pythagoras Theorem,

x^2=4^2+4^2

x^2=16+16

x^2=32

We take positive square root to obtain

x=\sqrt{32}

x=4\sqrt{2}

x=5.66ft

Rounding to the nearest feet we have x=6ft

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Answer: i got 6 83/1000

Step-by-step explanation:

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What is the longest flagpole ( in whole feet) that could be shipped in a box that measures 2ft by 2 ft by 12 ft
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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
What are the solutions
dalvyx [7]
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3 years ago
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