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3 years ago

Describe the different relationship between the drama videos and the science fiction videos

Mathematics

1 answer:

Flauer [41]3 years ago

5 0

Question

Describe the different relationship between the drama videos and the science fiction videos.

Answer:

Drama. You've heard the word. In fact, you've probably used the word yourself. Maybe you've said, 'Oh, she's just a drama queen,' or, 'I'm sick of all this drama.' In that context, the word 'drama' is something that has been blown out of proportion. Usually, a 'drama queen' is someone who is over the top, an over actor.

Science fiction (often shortened to Sci-Fi or SF) is a genre of speculative fiction, typically dealing with imaginative concepts such as advanced science and technology, spaceflight, time travel, and extraterrestrial life. Science fiction often explores the potential consequences of scientific and other innovations, and has been called a "literature of ideas".

Hope this helps!

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) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

4 years ago

Convert into slope-intercept form: <img src="https://tex.z-dn.net/?f=y-1%3Dm%28x-3%29" id="TexFormula1" title="y-1=m(x-3)" alt="

aliina [53]

Answer:

y=2x-5

Step-by-step explanation:

First simplify: y-1=2x-6

y-1=2(x-3)

First simplify and distribute everything.

y-1=2x-6

So, x equals 2 because it got distributed into the numbers inside the parenthesis. Same with the 2 and -3. They multiplied to become -6.

Since it's y-1=2x-6, you can simplify it even more so the -1 goes to the other side and turns into positive 1.

y - 1 (+ 1) = 2x -6 (+ 1)

-1(+1)=0 which leaves just the variable y on the left side.

-6(+1)=-5 which leaves 2x-5 on the right side.

This results in y=2x-5. Hope this helped ;)

6 0

3 years ago

Read 2 more answers

Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24

Nat2105 [25]

Answer:

E. 16 min 12 sec

Step-by-step explanation:

Let x represent total time taken to complete the exercise.

We have been given that Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session.

When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. This means that 90% time of exercise is equal to 24 minutes and 18 seconds.

18 seconds will be equal to 0.3 minutes.

Let us find total time of exercise as:

$\frac{90}{100}x=24.3$