All categories

3 years ago

Identify the similar triangles in the diagram then sketch them so the corresponding sides and angles have the same orientation

Mathematics

1 answer:

Lana71 [14]3 years ago

6 0

There’s no diagram, no picture.

You might be interested in

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

a ballon is released from the of a platform that is 50 meters tall. The ballon rises at the rate of 4 meters per second. Write a

Elanso [62]

Y = 4x + 50

y is height, 4 meters per second (x), 50m initial height

7 0

3 years ago

Use long division to find the quotient below.

Svetradugi [14.3K]

X^3+3x^2-18x-3x^2-9x+54
————-
Final answer=

X^3-27x+54

I hope you understand my explanation and best of wishes!!

5 0

3 years ago

Which is a point on the circle whose center is (0,0) and whose radius is 5? A. (0,0)

Temka [501]

Answer:

(3,4)

Step-by-step explanation:

x^2+y^2=(5)^2

if you graph it you will see that (3,4) is on the circle

4 0

3 years ago

What is the final step for 2y=x-10? Subtract 2 from each side, divide each term by 2, divide each term by -2, or add 2 to each s

Papessa [141]

Answer:

y= x/2 - 5

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers